V1.0 Introduction to the course
A brief introduction to this course, Introductory Quantum Mechanics. The set of videos is aimed at second-year physics undergraduates.
Notes and problems to accompany the videos are available at:
felixflicker.com/teaching
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Hello and welcome to this introductory
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video for introductory quantum mechanics
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the set of videos is going to cover a
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second year course in quantum mechanics.
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I'll assume knowledge of a standard
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first year undergraduate physics course
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so subjects such as complex numbers
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differential equations
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vectors and matrices classical mechanics
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and so on
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I'll provide problem sets and notes to
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accompany the videos
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on my website the link to which can be
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found in the youtube channel description
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I'll try to mix things up a little bit
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sometimes I'll be walking along like
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this with my dog Geoffrey
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sometimes I will instead
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be writing at this board sometimes
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i will be recording worked examples
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in a little bit more detail using pen
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and paper
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let's switch back to the field
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with regard to youtube I don't know
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whether you're seeing adverts at the
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start of this video
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if you don't wish to see the adverts you
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can use something such as adblock plus
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which is available
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for all browsers which will stop the
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adverts on youtube
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there's a setting in the bottom right
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hand corner
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which will allow you to change the
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quality of the video the videos are all
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filmed in 1080p
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fully high definition so if at any point
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you can't see what's being written on
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the board for example
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try increasing the quality setting and
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hopefully that should make things
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clearer
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and finally also in the settings
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you'll find that you can increase the
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speed of the videos up to
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two times so if you want to watch things
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a bit faster that should be possible
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so thank you very much for your time and
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I look forward to seeing you in the
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coming videos
V1.1 History of quantum mechanics
This channel, Introductory Quantum Mechanics, is a set of videos aimed at second-year physics undergraduates.
This video:
a brief history of the experimental developments which necessitated the development of quantum mechanics.
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hello in this video i'm going to give a
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brief
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overview of the history of the subject
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of quantum mechanics
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so quantum mechanics is somewhat unique
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amongst physics courses
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in seemingly requiring a historical
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background
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and i should add as a disclaimer at the
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start that i'm not a historian i'm a
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physicist
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and so you may want to go and check
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these facts yourself
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nevertheless the purpose of the video
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and the take home message from it
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is that however weird things get later
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on in the course
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quantum mechanics is firmly rooted in
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experimental observation
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it wasn't that we thought physics needed
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to get more magical so we started
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inventing strange interpretations of
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things
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what happened is that we carried out a
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set of experiments which revealed to us
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that the world is indeed more magical
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and we invented quantum mechanics in
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order to explain those observations
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so to begin with some pre-history
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in 1756 flame tests were developed
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in which you take a pure element and
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heat is over a flame and the flame
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changes colour
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it's an experiment you may have done
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yourself
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the this is now understood to be a
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result of the discrete nature of the
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atomic levels
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electrons occupy within atoms so the
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word quantum in quantum mechanics refers
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to
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discrete or separate and it's what
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happens when we
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go from our large scale macroscopic
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world which is seemingly continuous
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and go down to the smallest scale we
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kind of knew things had to get
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discrete somehow. In 1801 Thomas Young
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carried out an experiment which is now
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called Young's slits
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or the two-slit experiment in which he
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took two
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thin closely spaced slits in a piece of
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card shone light through it
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and measured a pattern on a screen and saw
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that light is able to interfere
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he took this as good evidence of the
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wave nature of light
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we'll take a look at that interference
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pattern in another experiment
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in 1850 atomic line spectra were measured
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for the first time
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we take a gas for pure element and
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pass white light through it
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and find that a set of discrete
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frequencies will be removed from the
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white light when we take that same gas
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and heat it up we see the light that's
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emitted from it is exactly that same set
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of frequencies that were
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absorbed from the light that was passed
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through it we now know this to be
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evidence
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of the discrete nature of the atomic
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levels once again
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in 1887 the
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photoelectric effect was measured by
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Heinrich Hertz
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so some gold leaf is taken and charged
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electrically
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two bits of gold leaf
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would repel from one another
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when light is shone onto the gold leaf we
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see that the charge dissipates
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the thing that was difficult to explain
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classically was that
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the frequency of the light being shone
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onto the gold leaf
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has to be above a certain threshold
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frequency there was no classical
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explanation of this
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in 1897 J J Thompson and others
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measured the behaviour of cathode rays
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so initially you might have been
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forgiven for thinking cathode rays were
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something like
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beams of light they were rays which
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could light up
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gases such as argon but unlike light
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it was possible to bend these rays using
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magnetic
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fields so what this showed us is that
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those rays were being carried
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by massive particles hence their ability
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to be deflected by magnetic fields and
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accelerated
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and so this was the first evidence for
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subatomic particles
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in this case electrons.
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We've gone
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through a set of experiments we've taken
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this world that appears
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on our everyday scales to be continuous
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and we've observed that down on the
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smaller scales that continuity
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emerges out of small
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sets of discrete things
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in 1911 the Millikan experiment was
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carried out in which
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oil drops electrically charged could be
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suspended using
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electrostatic potentials by measuring
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the required potential to levitate an
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oil drop
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it's possible to measure the charge
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accurately and it was found
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by Millikan that the charge always came
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in
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an integer multiple of some smallest
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amount which we now understand to be the
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charge of the electron
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in 1913 the Rutherford experiment was
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carried out also known as the
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Geiger-Marsden experiment after the
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people who really did most of the work
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in which alpha particles were seen to
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deflect
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from atoms alpha particles we now
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understand to be the nuclei
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of helium four atoms so
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occasionally these alpha particles would
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deflect through more than 90 degrees
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reflecting back
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so this was evident that while the atom
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is overall charged neutral
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that charge is not evenly distributed
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there's a small positively charged
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nucleus and a negative charge around
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that core
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this led to a problem for classical
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physics because if the
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distribution is as described by the
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Rutherford experiment
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why don't the negative charges fall into
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the positive charges to minimize their
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energy
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there's then a short break in the
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experiments i'd like to mention owing
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to the world war and world pandemic
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and the next i want to mention is the
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Compton scattering experiment in 1923.
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This saw the scattering of light
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by electrons and is used as evidence
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that much like we've seen both the
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particle and wave-like nature to light
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you could see a particle and wave like
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nature to electrons
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in 1923 to 1927 the Davisson-Germer
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experiment
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saw interference patterns in the
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electrons deflecting off the surface of
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nickel
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so this was further evidence for the
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fact that particles such as the electron
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can have wave-like characteristics
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so that's a set of experiments that
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does not exhaust the set of experiments
0:05:48.639,0:05:52.000
that require quantum explanations rather
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than classical
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but it's some of the key examples
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coincident with this of course were
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theoretical developments.
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Quantum mechanics has a very nice
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history of theory and experiment working
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together
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so the pre-history i think one of the
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most important things to mention before
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the development of quantum mechanics
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was Maxwell's equations you the unified
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theory of electromagnetism
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and in particular the prediction of
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electromagnetic waves so
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Maxwell's equations were written down
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in around
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1865 something like that
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and while coming before special
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relativity and quantum mechanics
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they're completely compatible with both
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so
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what this means is that light emits both
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a classical description
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which is wholly accurate provided
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there are no interactions so
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non-interacting light can be described
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completely classically in terms of waves
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but it can also be described completely
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quantum mechanically in terms of
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particles which we now call photons
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so this is a useful trick that'll allow
0:06:48.960,0:06:53.280
us to do various experiments
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in our own home of a quantum mechanical
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nature using light
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in 1900 Lord Rayleigh
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and shortly after James Jeans in 1905
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with Rayleigh
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developed a theory of the spectral
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radiance of black bodies
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so what this means is the
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set of frequencies coming off a body
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at a given temperature
0:07:16.560,0:07:19.759
so all bodies emit black body
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radiation
0:07:19.759,0:07:23.360
it's just that good absorbers are good
0:07:21.599,0:07:25.680
emitters and so
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black things such as pure carbon for
0:07:25.680,0:07:28.639
example
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being very good absorbers give very
0:07:28.639,0:07:33.680
clear black body radiation spectra
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now the Rayleigh-Jeans law as it's
0:07:33.680,0:07:37.919
called predicted that the spectral
0:07:36.160,0:07:39.680
radiance would continue to increase with
0:07:37.919,0:07:41.039
increasing frequency
0:07:39.680,0:07:43.199
this led to what's called the
0:07:41.039,0:07:43.599
ultraviolet catastrophe a prediction
0:07:43.199,0:07:45.440
that
0:07:43.599,0:07:46.800
all bodies should effectively have an
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infinite amount of energy in them that
0:07:46.800,0:07:49.520
they're giving off
0:07:47.599,0:07:51.280
as electromagnetic radiation which of
0:07:49.520,0:07:52.479
course was not what was experimentally
0:07:51.280,0:07:54.960
observed
0:07:52.479,0:07:55.759
in 1900 Max Planck came up with what's
0:07:54.960,0:07:58.319
really the first
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truly quantum theory where he came up
0:07:58.319,0:08:03.280
with a phenomenological model
0:08:00.240,0:08:05.680
which gave a very good match to the
0:08:03.280,0:08:07.599
experimentally observed black body
0:08:05.680,0:08:10.800
radiation spectra
0:08:07.599,0:08:14.479
and he did so by saying that the
0:08:10.800,0:08:16.080
light had was being emitted only in
0:08:14.479,0:08:18.720
discrete packets that he called
0:08:16.080,0:08:21.199
quanta and he identified that
0:08:18.720,0:08:23.199
the energy of one of these packets for a
0:08:21.199,0:08:26.319
wave with angular frequency omega
0:08:23.199,0:08:27.680
was given by hbar omega where hbar is
0:08:26.319,0:08:29.759
now what we call the reduced Planck's
0:08:27.680,0:08:33.599
constant
0:08:29.759,0:08:33.599
so he identified that pre-factor
0:08:34.240,0:08:38.560
his model was phenomenological meaning
0:08:36.080,0:08:40.479
that it fit the data accurately
0:08:38.560,0:08:41.839
but there was no microscopic explanation
0:08:40.479,0:08:44.800
as to where that
0:08:41.839,0:08:46.080
expression came from Plank himself
0:08:44.800,0:08:47.120
believed it just to be a mathematical
0:08:46.080,0:08:49.920
trick
0:08:47.120,0:08:51.760
so in 1905 the major development was
0:08:49.920,0:08:54.000
made by Albert Einstein
0:08:51.760,0:08:55.440
in a year in which he wrote
0:08:54.000,0:08:58.320
four papers each of which
0:08:55.440,0:08:58.800
individually revolutionized physics so
0:08:58.320,0:09:01.680
he
0:08:58.800,0:09:03.279
outlined special relativity at the same
0:09:01.680,0:09:03.920
time he explained the photoelectric
0:09:03.279,0:09:06.399
effect
0:09:03.920,0:09:07.200
by taking Planck's hypothesis seriously
0:09:06.399,0:09:09.360
and saying
0:09:07.200,0:09:11.040
rather than a mathematical tool this is
0:09:09.360,0:09:13.120
a physical statement light is really
0:09:11.040,0:09:15.600
conveyed by individual packets
0:09:13.120,0:09:16.720
quanta which we now call photons and the
0:09:15.600,0:09:18.720
energy of one of these
0:09:16.720,0:09:21.839
photons is hbar times the angular
0:09:18.720,0:09:24.399
frequency of that photon
0:09:21.839,0:09:24.399
he also
0:09:25.120,0:09:28.240
provided convincing evidence of the
0:09:27.120,0:09:30.160
atomic theory
0:09:28.240,0:09:32.560
by explaining Brownian motion of pollen
0:09:30.160,0:09:34.640
molecules in water as
0:09:32.560,0:09:36.320
them jostling around from the impacts of
0:09:34.640,0:09:37.440
individual atoms and this led to the
0:09:36.320,0:09:40.480
widespread
0:09:37.440,0:09:40.959
adoption of the atomic theory and he
0:09:40.480,0:09:42.880
also
0:09:40.959,0:09:44.720
in a fourth paper wrote down probably
0:09:42.880,0:09:47.120
the most famous equation ever
0:09:44.720,0:09:49.279
E=mc^2
0:09:47.120,0:09:50.399
in 1911 the same year as the Millikan
0:09:49.279,0:09:53.760
experiment
0:09:50.399,0:09:55.279
Niels Bohr wrote down the quantum
0:09:53.760,0:09:56.959
theory of the atom
0:09:55.279,0:09:58.480
it was again a phenomenological theory
0:09:56.959,0:10:02.240
but he found an equation
0:09:58.480,0:10:04.399
which predicted accurately the
0:10:02.240,0:10:06.079
measurements made in the atomic line
0:10:04.399,0:10:07.600
spectra for example
0:10:06.079,0:10:09.519
in terms of the electrons occupying
0:10:07.600,0:10:11.279
discrete energy levels in the atom
0:10:09.519,0:10:12.640
well we now know the Bohr model to be
0:10:11.279,0:10:14.399
incorrect but
0:10:12.640,0:10:16.079
it made accurate predictions for the
0:10:14.399,0:10:17.839
energy levels and it was really the
0:10:16.079,0:10:19.680
motivation that led to the development
0:10:17.839,0:10:22.160
of quantum theory later on
0:10:19.680,0:10:23.440
so 1911 leads to the end of what we call
0:10:22.160,0:10:25.600
old quantum theory
0:10:23.440,0:10:27.360
a set of phenomenological ideas that is
0:10:25.600,0:10:30.079
things that make good predictions
0:10:27.360,0:10:31.760
in terms of experimental observations
0:10:30.079,0:10:33.040
but which don't have a microscopic
0:10:31.760,0:10:34.800
explanation as to where they're coming
0:10:33.040,0:10:36.399
from
0:10:34.800,0:10:39.760
the next major development was
0:10:36.399,0:10:43.839
by Louis de Broglie in 1923
0:10:39.760,0:10:45.839
who hypothesized that just as
0:10:43.839,0:10:47.920
traditionally wave-like things such as
0:10:45.839,0:10:49.120
light admit a particle description as
0:10:47.920,0:10:51.040
Einstein said
0:10:49.120,0:10:53.120
perhaps traditionally
0:10:51.040,0:10:55.279
particle-like things such as electrons
0:10:53.120,0:10:56.560
admit a wave-like description and he
0:10:55.279,0:10:58.000
proposed a formula
0:10:56.560,0:11:00.079
telling us that the momentum of the
0:10:58.000,0:11:00.959
particle is linearly related to the wave
0:11:00.079,0:11:02.720
vector
0:11:00.959,0:11:04.079
of the corresponding wave and the
0:11:02.720,0:11:06.079
constant of proportionality is the
0:11:04.079,0:11:08.399
reduced Planck's constant
0:11:06.079,0:11:10.320
this brings us to 1925 which is pretty
0:11:08.399,0:11:12.959
much where this course will get us to
0:11:10.320,0:11:14.240
so in 1925 Erwin Schroedinger wrote down
0:11:12.959,0:11:17.360
the Schroedinger equation
0:11:14.240,0:11:19.200
as part of what is called wave mechanics
0:11:17.360,0:11:20.720
in the same year Werner Heisenberg and
0:11:19.200,0:11:23.680
others including Niels Bohr
0:11:20.720,0:11:24.320
wrote down matrix mechanics and these
0:11:23.680,0:11:26.959
were two
0:11:24.320,0:11:28.320
microscopic models for how particles
0:11:26.959,0:11:30.800
behave
0:11:28.320,0:11:32.480
down on the smallest scales so this has
0:11:30.800,0:11:34.320
gone beyond a phenomenological model to
0:11:32.480,0:11:36.399
give a microscopic explanation as to why
0:11:34.320,0:11:38.880
things are happening
0:11:36.399,0:11:39.920
later on that year Schroedinger
0:11:38.880,0:11:40.720
showed the equivalence of the two
0:11:39.920,0:11:42.640
approaches
0:11:40.720,0:11:44.640
uniting wave mechanics and matrix
0:11:42.640,0:11:45.920
mechanics into what we now call quantum
0:11:44.640,0:11:48.480
mechanics
0:11:45.920,0:11:50.000
so that's as far as this course is going
0:11:48.480,0:11:52.240
to take us there are of course
0:11:50.000,0:11:54.800
many further development i think the
0:11:52.240,0:11:56.079
major one being in 1935 when Einstein
0:11:54.800,0:11:57.600
Podolsky and Rosen
0:11:56.079,0:11:59.279
developed what's now called the EPR
0:11:57.600,0:12:01.440
paradox which led
0:11:59.279,0:12:03.120
after John Stuart Bell came up with a
0:12:01.440,0:12:05.760
testable prediction for it
0:12:03.120,0:12:07.440
and in 1980 Alain Aspect carried out a
0:12:05.760,0:12:09.600
set of experiments confirming
0:12:07.440,0:12:11.680
that quantum entanglement is a
0:12:09.600,0:12:13.440
fundamental property of the universe
0:12:11.680,0:12:15.600
arguably there are only two truly
0:12:13.440,0:12:17.360
fundamentally quantum effects
0:12:15.600,0:12:18.639
one is wave particle duality which we'll
0:12:17.360,0:12:20.079
see a lot of in this course
0:12:18.639,0:12:22.480
and the other is quantum entanglement
0:12:20.079,0:12:24.720
which we'll see less of.
0:12:22.480,0:12:26.399
While we'll see many other things such as
0:12:24.720,0:12:28.560
for example the Heisenberg uncertainty
0:12:26.399,0:12:30.320
principle which are undoubtedly quantum
0:12:28.560,0:12:32.639
in all the other cases other than wave
0:12:30.320,0:12:34.639
particle duality and entanglement
0:12:32.639,0:12:35.680
there are classical precedents that we
0:12:34.639,0:12:37.360
can find
0:12:35.680,0:12:38.800
it's not that quantum mechanics is less
0:12:37.360,0:12:40.480
magical the more that you study it
0:12:38.800,0:12:42.800
it's just that classical mechanics
0:12:40.480,0:12:48.639
was more magical than we gave it
0:12:42.800,0:12:48.639
credit for okay thank you for your time
V1.2 The Schrödinger equation
This channel, Introductory Quantum Mechanics, is a set of videos aimed at second-year physics undergraduates.
This video:
the time-dependent Schrödinger equation (TDSE), and how to obtain from it the time-independent Schrödinger equation (TISE).
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hello the behaviour of particles in
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quantum mechanics is governed by what's
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called the schrodinger equation
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written here in its time-dependent form
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so we sometimes call this
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the time-dependent schrodinger equation
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usually abbreviated to TDSE for short
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it was written down in 1925
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by this man Erwin Schrodginer as you can
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see a very well-dressed gentleman as
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they
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were back in the 1920s but you know
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what, it's the
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20s again so let's see if we can't
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conjure up a bit of that 1920s chic
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in this video
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it worked. Magic! OK.
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So the schrodinger equation when
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schrodinger originally wrote it down
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was intended to be a classical
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description of classical waves
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this quantity psi which we call the wave
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function was thought of
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as something like the wave function
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which would describe say water waves
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in which case it would take some value
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which would tell us about the height of
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the water wave
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as a function of position and time
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in fact as we'll see while there is a
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good description of
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the wave function as something like a
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classical wave, in fact quantum mechanics
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goes beyond classical physics as we now
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of course know. So before we
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proceed let me just make a couple of
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notational points
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when i write d(psi)/dt like this, this
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is a partial derivative with respect to
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time
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holding all spatial coordinates constant
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so we can also denote this in the
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following way
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so we denote this
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stating explicitly that
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the positions are all constant
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sometimes i'll denote this
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in a simpler manner in the following
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form
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where partial subscript t is just
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shorthand for
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d by dt acting on psi and sometimes i'll
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use
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newton's notation psi dot
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similarly a partial derivative with
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respect to x
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implies that we're holding the other
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variables constant so time
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y and z which can again be abbreviated
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just partial subscript x acting on
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psi
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and sometimes we'll write this as psi
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prime
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where this last notation will only
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really be used in one dimension where
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it's unambiguously a derivative with
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respect to x
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so there are a couple of key
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assumptions we're going to make
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throughout this course let's just wipe
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this board off and write them down
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when i say wipe the board off of course
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i mean magically clear the board
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so let's write down some key assumptions
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the first is that the schrodinger
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equation
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is non-relativistic. It's going to
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describe the behavior of some massive
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particle
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in the non-relativistic limit. There are
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relativistic extensions to it
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but we're not going to consider those in
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this course
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the second is that we're only going to
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describe single particles.
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Again there are extensions to the
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schrodinger equation which will allow it
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to deal with multiple particles
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but that's not going to be the focus of
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this course. The wave function psi
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always governs the behavior of a single
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particle
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okay so let's take a closer look at
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what's going on in the schrodinger
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equation let's clear the board again
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so let's have the schrodinger equation
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back
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so we'll look at properties of the wave
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function psi in future videos
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but for now let's just note that it's a
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complex-valued function
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and it has out the front of it an
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arbitrary
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global complex phase.
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It's a mathematical redundancy built
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into the the wave function
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it has this global phase which can be
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changed arbitrarily.
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But the relative phase between two
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different wave functions
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is important and we will see that later
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on in the course
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but for now there is this redundancy
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built into it.
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We've got
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the imaginary unit here i
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square root minus one. Imaginary
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numbers play an important role in
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quantum mechanics.
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hbar here is the reduced planck's
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constant h over 2 pi
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with units of energy multiplied by time
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and then we have this quantity here H
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which
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is called the hamiltonian let's write
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that down
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so h is defined as follows
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it's a differential
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operator we have this nabla squared in
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here
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which okay we can expand as follows
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so it's just the partial derivative
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with respect to x
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squared plus that of y squared and that
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of z squared remembering the notation
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from the last board and if this looks
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confusing just remember this is always
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acting on psi so this term here
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(d/dx)^2 acting on psi is really just
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psi
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double prime. Okay.
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So it's called the Hamiltonian
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and it's what's called an energy
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operator for the system
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so the hamiltonian acting on psi
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is going to return the energy of the
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system as we'll see
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in a second in general
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specifying the hamiltonian specifies the
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entire quantum problem we want to solve
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so when we
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write down a quantum mechanical
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description of a system we just need to
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write down the hamiltonian
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and in fact this first term is fixed
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what we what need to specify is the
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potential of the system which is much
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like what you do in classical mechanics
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as well
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so the potential we will assume in
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this course is time independent
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it doesn't need to be but we won't
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consider
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time dependent potentials here. So
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let's take a look at attempts to
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deal with
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how to solve this equation
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so the time-dependent schrodinger
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equation
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where i've written it with psi dot this
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time is a separable
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equation what we mean by this
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is that we can substitute the
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following anzatz
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we can say that psi(x,t)
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is equal to:
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-- and let's treat it in three
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dimensions in general -- is equal to
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phi(x) only
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and T(t) only
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so these are two separate functions one
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of which is only a function of time
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and one of which is only a function of
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position when we substitute
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that in
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we see that the equation reduces to the
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following form
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where the phi has pulled through the
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time derivative
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because it's not a function of time and
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this is a partial derivative
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the partial derivative of T(t) with respect
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to time
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is actually a total derivative because
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it's only a function of time
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and over here the hamiltonian because the
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potential is time independent
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hamiltonian only acts on the spatial
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coordinates so we can pull through the t
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over here rearranging slightly we get
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two different equations
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one purely in terms of t and one purely
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in terms of x
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so we've turned what was a partial
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differential equation in terms of both x
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and t
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into two separate equations one is an
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00:07:22,479 --> 00:07:24,400
ordinary differential equation in terms
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of time
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and the other is potentially still a
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partial differential equation in terms
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of positions
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so the fact that these two are equal for
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all times
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and all positions means they must both
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be equal to the same constant
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and we're going to suggestively call
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that constant E
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suggestive because it should remind us
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of an energy
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so let's number these equations let's
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call this one (i)
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and this one (ii)
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we'll move this up to the top of the
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board and we'll deal with (i) and (ii)
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separately
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so (i) simply rearranges
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to the following form
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H acting on phi is equal to
233
00:08:08,400 --> 00:08:12,080
E multiplying phi and this is what's
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called the time
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independent schrodinger equation because
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00:08:15,520 --> 00:08:19,120
phi here is now only a function of
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00:08:17,120 --> 00:08:22,639
positions not of time.
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00:08:19,120 --> 00:08:24,639
Let's put a box around it in general
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this one is very tricky to solve in fact
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in general it's impossible to solve
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00:08:27,120 --> 00:08:30,560
remember the hamiltonian here contains
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00:08:28,720 --> 00:08:32,159
all the real information in the problem
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in the form of the potential
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so in general we need to solve this
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00:08:34,080 --> 00:08:38,080
differential equation
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00:08:36,320 --> 00:08:39,440
in terms of boundary conditions which we
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specify and
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00:08:39,440 --> 00:08:42,800
you can see that it's an eigenvalue
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00:08:42,000 --> 00:08:45,440
equation
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00:08:42,800 --> 00:08:45,839
in that we need to solve this for
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00:08:45,440 --> 00:08:48,080
both
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00:08:45,839 --> 00:08:49,519
the eigenfunctions phi(x) and the
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00:08:48,080 --> 00:08:50,560
eigenenergies E
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00:08:49,519 --> 00:08:52,800
which will be the energies of the
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00:08:50,560 --> 00:08:55,279
particle let's take a look at the second
256
00:08:52,800 --> 00:08:57,519
equation there
257
00:08:55,279 --> 00:08:58,880
where i've done our favorite trick of
258
00:08:57,519 --> 00:09:01,920
multiplying through by
259
00:08:58,880 --> 00:09:05,760
dt. We can then integrate both sides
260
00:09:01,920 --> 00:09:09,360
to give the result:
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00:09:05,760 --> 00:09:11,200
that is, the time evolution of t is simply a
262
00:09:09,360 --> 00:09:12,160
phase winding. The winding of the complex
263
00:09:11,200 --> 00:09:14,240
phase
264
00:09:12,160 --> 00:09:16,480
so sticking our anzatzes back together
265
00:09:14,240 --> 00:09:19,839
again remember psi(x,t) the wave function
266
00:09:16,480 --> 00:09:22,240
is the product of T(t) and phi(x) we find that
267
00:09:19,839 --> 00:09:24,320
if we can solve the time independent
268
00:09:22,240 --> 00:09:28,399
schrodinger equation which by the way we
269
00:09:24,320 --> 00:09:28,399
sometimes call TISE
270
00:09:29,600 --> 00:09:34,800
then for free we get the time evolution
271
00:09:32,000 --> 00:09:34,800
of the wave function
272
00:09:35,600 --> 00:09:42,959
where phi(x) is the solution to the
273
00:09:39,279 --> 00:09:44,880
TISE so while this is in general
274
00:09:42,959 --> 00:09:46,880
impossible to solve there are a set of
275
00:09:44,880 --> 00:09:47,440
very important cases which is possible
276
00:09:46,880 --> 00:09:48,800
to solve
277
00:09:47,440 --> 00:09:50,800
and those are the ones that we'll be
278
00:09:48,800 --> 00:09:52,720
looking at in this course
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00:09:50,800 --> 00:09:54,560
okay thank you for your time it must have been
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00:09:52,720 --> 00:09:55,440
very hot in the 1920s if this suit is
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00:09:54,560 --> 00:09:57,600
something to go by
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00:09:55,440 --> 00:10:00,080
I think I'll turn back into my normal
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00:09:57,600 --> 00:10:00,080
clothes
284
00:10:03,200 --> 00:10:08,880
excuse me Geoffrey sorry about that
285
00:10:06,560 --> 00:10:10,560
so i'll see you in the next video where
286
00:10:08,880 --> 00:10:12,480
we will take a look at some particular
287
00:10:10,560 --> 00:10:18,959
solutions to the schrodinger equation
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00:10:12,480 --> 00:10:18,959
thank you
V1.3 Plane waves
This channel, Introductory Quantum Mechanics, is a set of videos aimed at second-year physics undergraduates.
This video:
plane wave solutions to the Schrödinger equation in the absence of a potential; compatibility with the Einstein relation E=?? and the de Broglie relation p=?k.
0:00:00.799,0:00:04.480
hello in this video we're going to take
0:00:03.120,0:00:06.160
a look at some
0:00:04.480,0:00:08.240
specific solutions to the schroedinger
0:00:06.160,0:00:12.080
equation so remember the
0:00:08.240,0:00:16.000
time independent schroedinger equation
0:00:12.080,0:00:16.800
is written in terms of the hamiltonian
0:00:16.000,0:00:20.240
acting
0:00:16.800,0:00:21.920
on some wave function phi(x) and this
0:00:20.240,0:00:23.039
is just the time independent part
0:00:21.920,0:00:24.160
remember we can always add the time
0:00:23.039,0:00:28.080
dependent part in
0:00:24.160,0:00:30.240
later this is defined to be
0:00:28.080,0:00:34.800
-hbar^2/2m . grad^2
0:00:30.240,0:00:38.079
plus V(x)
0:00:34.800,0:00:39.360
all acting on phi(x) and this thing
0:00:38.079,0:00:45.680
equals E phi(x)
0:00:42.960,0:00:47.200
so in one dimension which is the case
0:00:45.680,0:00:48.160
we'll be interested in in most of this
0:00:47.200,0:00:50.640
course
0:00:48.160,0:00:52.320
we can write this more simply by just
0:00:50.640,0:00:56.079
taking the more complicated power here
0:00:52.320,0:00:59.600
as -hbar^2/2m phi''(x)
0:00:56.079,0:01:03.280
plus
0:00:59.600,0:01:06.560
V(x) phi(x)
0:01:03.280,0:01:09.520
equals E phi(x)
0:01:06.560,0:01:11.200
so in this case it's just a
0:01:09.520,0:01:13.520
second order ordinary differential
0:01:11.200,0:01:15.280
equation
0:01:13.520,0:01:16.960
it's an eigenvalue problem we need to
0:01:15.280,0:01:20.880
find the eigenfunctions
0:01:16.960,0:01:22.720
phi(x) which solves this equation and the
0:01:20.880,0:01:23.920
corresponding eigenvalues E which will
0:01:22.720,0:01:25.360
be the energies of the
0:01:23.920,0:01:28.560
system the energies the particles can
0:01:25.360,0:01:29.360
take and in general to specify such a
0:01:28.560,0:01:30.880
problem
0:01:29.360,0:01:32.720
if we've got some physical system we
0:01:30.880,0:01:36.000
want to model with quantum mechanics
0:01:32.720,0:01:40.400
we just write down a potential
0:01:36.000,0:01:43.200
that encodes that system and then
0:01:40.400,0:01:44.560
we have to solve the time independent
0:01:43.200,0:01:46.479
Schrodinger equation
0:01:44.560,0:01:48.000
for our potential subject to boundary
0:01:46.479,0:01:51.119
conditions
0:01:48.000,0:01:52.320
so the simplest possible potential we
0:01:51.119,0:01:54.720
can consider
0:01:52.320,0:01:56.240
is just the case where the potential
0:01:54.720,0:01:59.200
is equal to zero
0:01:56.240,0:01:59.200
so the simplest case
0:02:02.159,0:02:05.759
V=0 and when the potential
0:02:04.719,0:02:07.360
is equal to zero
0:02:05.759,0:02:09.440
our time independent schrodinger
0:02:07.360,0:02:13.360
equation just reads
0:02:09.440,0:02:17.840
-hbar^2/2m phi''(x)=E phi(x)
0:02:13.360,0:02:21.360
this can be solved with
0:02:17.840,0:02:24.959
an ansatz so we can say that phi(x)
0:02:21.360,0:02:28.400
is equal to a plus or minus e to the i
0:02:24.959,0:02:28.400
plus or minus k x
0:02:29.040,0:02:33.440
where a plus or minus are just some
0:02:31.599,0:02:35.120
arbitrary coefficients
0:02:33.440,0:02:37.120
and when we substitute this in we
0:02:35.120,0:02:41.519
find that
0:02:37.120,0:02:45.360
hbar^2 k^2/2m = E
0:02:41.519,0:02:46.879
and in other words
0:02:45.360,0:02:48.480
the energy eigenvalues E that we've
0:02:46.879,0:02:53.120
solved for equal
0:02:48.480,0:02:54.800
hbar^2 k^2 / 2m
0:02:53.120,0:02:56.239
and now when we take a look back at the
0:02:54.800,0:02:57.519
problem we're trying to solve this
0:02:56.239,0:02:58.480
here's the time independent schrodinger
0:02:57.519,0:02:59.599
equation in 1D
0:02:58.480,0:03:01.840
we're trying to solve for the energy
0:02:59.599,0:03:03.280
eigenvalues. Energies of course have
0:03:01.840,0:03:05.680
two contributions they have
0:03:03.280,0:03:07.360
the potential energy term which is here
0:03:05.680,0:03:08.879
and the kinetic energy term which must
0:03:07.360,0:03:11.040
be this thing over here
0:03:08.879,0:03:14.159
so we've set the potential equal to zero
0:03:11.040,0:03:15.680
so the energy should be purely kinetic
0:03:14.159,0:03:17.760
and the kinetic energy we'd usually
0:03:15.680,0:03:21.040
expect to be able to write
0:03:17.760,0:03:24.720
p^2/2m
0:03:21.040,0:03:27.840
so this is true provided that
0:03:24.720,0:03:27.840
p = hbar k
0:03:28.000,0:03:33.840
you'll see from here that this is nothing
0:03:30.159,0:03:33.840
other than our de Broglie relation
0:03:37.280,0:03:44.080
which tells us that
0:03:41.040,0:03:47.200
all quantum particles have a
0:03:44.080,0:03:50.560
wave-like description as well and the
0:03:47.200,0:03:53.280
momentum p of a of the particle
0:03:50.560,0:03:54.959
corresponds to a wave vector k for the
0:03:53.280,0:03:57.040
wave
0:03:54.959,0:03:58.840
we can also write this as p equals h
0:03:57.040,0:04:00.480
over lambda where lambda is the
0:03:58.840,0:04:03.200
wavelength
0:04:00.480,0:04:03.519
so I said that in general solving the
0:04:03.200,0:04:04.799
time
0:04:03.519,0:04:06.000
independent schrodinger equation is
0:04:04.799,0:04:07.360
actually the tricky bit that we've
0:04:06.000,0:04:09.680
already done here.
0:04:07.360,0:04:10.799
We then get the time dependence
0:04:09.680,0:04:14.480
for free
0:04:10.799,0:04:19.840
so let's take a look at the time
0:04:14.480,0:04:19.840
dependent schrodinger equation TDSE
0:04:20.000,0:04:26.560
so this reads i h bar
0:04:23.120,0:04:28.240
psi dot (where the dot indicates
0:04:26.560,0:04:30.479
the partial derivative of psi with
0:04:28.240,0:04:34.160
respect to time holding position
0:04:30.479,0:04:36.720
constant) equals H psi
0:04:34.160,0:04:38.000
in general and in this case this
0:04:36.720,0:04:39.040
equals E psi because we've already
0:04:38.000,0:04:43.199
solved the time
0:04:39.040,0:04:44.479
independent part so psi(x,t)
0:04:43.199,0:04:46.240
adds in the time dependence
0:04:44.479,0:04:48.880
corresponding to phi(x)
0:04:46.240,0:04:50.720
that we've already solved for so we
0:04:48.880,0:04:54.000
can again solve this with an ansatz
0:04:50.720,0:04:58.080
let's say that psi(x,t)
0:04:54.000,0:05:01.120
is equal to A plus or minus e to the
0:04:58.080,0:05:04.400
i plus or minus k x
0:05:01.120,0:05:07.680
as before and this time minus omega t
0:05:04.400,0:05:09.440
so remember that the time dependence
0:05:07.680,0:05:12.560
of an energy eigenvalue always just adds
0:05:09.440,0:05:14.400
this phase winding term
0:05:12.560,0:05:15.600
when we substitute this in we bring down
0:05:14.400,0:05:18.479
a minus i
0:05:15.600,0:05:20.000
omega there's an i here already the i
0:05:18.479,0:05:20.479
cancels with the minus sign and we find
0:05:20.000,0:05:24.000
that
0:05:20.479,0:05:27.759
hbar omega psi = E psi
0:05:24.000,0:05:30.880
or in other words E equals
0:05:27.759,0:05:35.440
hbar omega
0:05:30.880,0:05:35.440
which is nothing other than our Einstein
0:05:36.840,0:05:43.199
relation
0:05:39.520,0:05:45.039
recall Einstein said:
0:05:43.199,0:05:46.800
take light which is classically
0:05:45.039,0:05:49.280
described by a a wave
0:05:46.800,0:05:50.880
and we can say that another way to
0:05:49.280,0:05:52.000
think of that is that it's made up of
0:05:50.880,0:05:55.600
individual packets of
0:05:52.000,0:05:58.080
energy called quanta or photons and
0:05:55.600,0:05:59.199
if a photon has angular frequency omega
0:05:58.080,0:06:02.639
it has energy
0:05:59.199,0:06:05.919
E another way to write this is
0:06:02.639,0:06:06.960
E = hf where f is the frequency
0:06:05.919,0:06:08.639
of the photon
0:06:06.960,0:06:09.520
now we're not describing photons with
0:06:08.639,0:06:10.479
the Schrodinger equation we're
0:06:09.520,0:06:12.400
describing
0:06:10.479,0:06:14.240
non-relativistic massive particles such
0:06:12.400,0:06:16.240
as electrons
0:06:14.240,0:06:17.280
but these two obey something like an
0:06:16.240,0:06:19.680
Einstein relation
0:06:17.280,0:06:21.199
and that's by construction part of the
0:06:19.680,0:06:24.400
Schroedinger equation
0:06:21.199,0:06:25.919
so that's the first example of a simple
0:06:24.400,0:06:29.759
solution to the Schroedinger equation
0:06:25.919,0:06:29.759
okay thank you for your time
V1.4 Amplitudes and probabilities
This channel, Introductory Quantum Mechanics, is a set of videos aimed at second-year physics undergraduates.
This video:
probability amplitudes, probability densities, probability density currents, the continuity equation and local conservation of probability, and general boundary conditions on the wave function.
0:00:00.880,0:00:05.920
hello we've seen previously
0:00:03.600,0:00:07.600
that in quantum mechanics we describe
0:00:05.920,0:00:09.360
particles in terms of a quantity called
0:00:07.600,0:00:11.599
the wave function denoted with the greek
0:00:09.360,0:00:12.960
letter psi
0:00:11.599,0:00:14.880
which tells us something about the
0:00:12.960,0:00:16.160
probability of finding the particle
0:00:14.880,0:00:18.320
but it can't tell us exactly the
0:00:16.160,0:00:19.760
probability because psi is in general a
0:00:18.320,0:00:22.160
complex number
0:00:19.760,0:00:24.160
additionally it has what's called a
0:00:22.160,0:00:25.199
global phase so it has some complex
0:00:24.160,0:00:27.039
phase out the front
0:00:25.199,0:00:28.720
which is gauge dependent meaning that
0:00:27.039,0:00:31.920
it's a mathematical choice
0:00:28.720,0:00:33.920
as to what value that phase takes. It's
0:00:31.920,0:00:36.880
a mathematical redundancy in the system
0:00:33.920,0:00:38.640
so psi itself can't tell us about
0:00:36.880,0:00:41.920
probabilities. In fact it denotes
0:00:38.640,0:00:43.440
what's called a probability amplitude
0:00:41.920,0:00:45.200
which we usually just refer to as the
0:00:43.440,0:00:47.039
amplitude
0:00:45.200,0:00:48.480
to identify the probability itself we
0:00:47.039,0:00:50.320
need to use what's called the
0:00:48.480,0:00:53.440
Born rule
0:00:50.320,0:00:56.239
which tells us the following
0:00:53.440,0:00:58.000
the square-modulus of psi multiplied by
0:00:56.239,0:00:59.920
a small line element dx
0:00:58.000,0:01:01.760
in one dimension gives the probability
0:00:59.920,0:01:05.040
to find the particle between x
0:01:01.760,0:01:07.520
and x+dx at time t. In general
0:01:05.040,0:01:09.200
this will be a small volume element in
0:01:07.520,0:01:11.200
three dimensions
0:01:09.200,0:01:12.880
so it's this quantity that we need to be
0:01:11.200,0:01:14.479
looking at for probabilities
0:01:12.880,0:01:17.600
and this has all the properties we'd
0:01:14.479,0:01:20.479
like: it's a real number
0:01:17.600,0:01:22.400
it no longer has this arbitrary global
0:01:20.479,0:01:24.799
phase out of the front
0:01:22.400,0:01:26.560
and in fact if we integrate the modulus
0:01:24.799,0:01:28.960
square of psi across
0:01:26.560,0:01:30.159
all of space we'll get the value one
0:01:28.960,0:01:31.200
because even though we don't know where
0:01:30.159,0:01:34.079
the particle is
0:01:31.200,0:01:35.280
we know that it must exist somewhere.
0:01:34.079,0:01:37.439
We can identify
0:01:35.280,0:01:40.000
this quantity here as what's called the
0:01:37.439,0:01:42.000
probability density.
0:01:40.000,0:01:44.000
This integrated over a region
0:01:42.000,0:01:46.240
of space gives the probability to find
0:01:44.000,0:01:48.240
the particle within that region
0:01:46.240,0:01:49.600
since integrating it across all of
0:01:48.240,0:01:51.200
space gives the value one
0:01:49.600,0:01:53.360
we actually have what's called the
0:01:51.200,0:01:55.360
'global conservation of probability'
0:01:53.360,0:01:56.880
the probability of finding a particle is
0:01:55.360,0:01:58.399
always a constant
0:01:56.880,0:02:00.079
in fact there's a stronger condition on
0:01:58.399,0:02:01.759
the probability which we'll take a look
0:02:00.079,0:02:03.520
at now with a worked example
0:02:01.759,0:02:09.840
let me just move over to my worked
0:02:03.520,0:02:09.840
example area
0:02:13.920,0:02:17.760
okay so we have the global conservation
0:02:15.840,0:02:19.120
of probability but to get a stronger
0:02:17.760,0:02:22.959
constraint on it
0:02:19.120,0:02:24.239
let's take a look at our probability
0:02:22.959,0:02:27.680
density and let's look at the time
0:02:24.239,0:02:27.680
derivative of it.
0:02:28.239,0:02:31.760
Our
0:02:30.560,0:02:36.000
probability density
0:02:31.760,0:02:39.200
is defined as |psi|^2
0:02:36.000,0:02:42.720
which is equal to psi*.psi
0:02:39.200,0:02:44.239
well I've omitted the
0:02:42.720,0:02:46.000
position and time dependence but they're
0:02:44.239,0:02:48.480
there
0:02:46.000,0:02:49.680
and so what we'd like is the time
0:02:48.480,0:02:51.360
derivative of this the partial
0:02:49.680,0:02:52.560
derivative of the probability density
0:02:51.360,0:02:56.160
with respect to time
0:02:52.560,0:02:57.440
holding position constant and so this
0:02:56.160,0:03:00.959
of course will be equal to
0:02:57.440,0:03:07.360
d(psi*)/dt.psi + psi*.d(psi)/dt
0:03:04.239,0:03:08.879
from the chain rule and then to
0:03:07.360,0:03:10.400
work out what these quantities are
0:03:08.879,0:03:13.440
we can use the time dependent Schrodinger
0:03:10.400,0:03:17.280
equation which tells us that
0:03:17.280,0:03:23.440
hbar psi dot equals
0:03:21.040,0:03:24.159
(I'll write the hamiltonian out in full)
0:03:23.440,0:03:30.480
(-hbar^2/2m grad^2)psi
0:03:27.360,0:03:33.840
plus
0:03:30.480,0:03:35.840
V psi and so
0:03:33.840,0:03:37.760
dividing by hbar and
0:03:35.840,0:03:41.599
multiplying by minus i
0:03:37.760,0:03:45.200
we have d(psi)/dt equals
0:03:41.599,0:03:49.440
minus i over h bar
0:03:45.200,0:03:52.560
-hbar^2/2m grad^2 psi
0:03:49.440,0:03:56.400
+ V psi
0:03:52.560,0:03:59.680
how about d(psi*)/dt
0:03:56.400,0:04:01.360
we just take the
0:03:59.680,0:04:05.200
complex conjugate of this
0:04:01.360,0:04:08.640
and so we get d(psi*)/dt
0:04:05.200,0:04:12.159
is equal to
0:04:08.640,0:04:15.760
minus hbar^2/2m grad^2 psi
0:04:12.159,0:04:19.120
plus
0:04:15.760,0:04:22.720
V psi*
0:04:19.120,0:04:24.639
where the potential is assumed real
0:04:22.720,0:04:26.800
okay so substituting both of these back
0:04:24.639,0:04:30.560
into this expression up here
0:04:26.800,0:04:33.919
we find that we have d(rho)/dt
0:04:30.560,0:04:37.040
is equal to so we get the i
0:04:33.919,0:04:40.080
over h bar out the front
0:04:37.040,0:04:50.080
-hbar^2/2m grad^2 psi* + v psi*
0:04:46.479,0:04:54.080
and all of this
0:04:50.080,0:04:57.280
gets multiplied by psi plus
0:04:54.080,0:05:00.479
well so sorry it would be uh
0:04:57.280,0:05:03.199
plus psi* times psi
0:05:00.479,0:05:05.039
dot psi yes psi dot but remember
0:05:03.199,0:05:06.720
that there was a minus sign in front of
0:05:05.039,0:05:08.800
the psi dot compared to the
0:05:06.720,0:05:12.160
d(psi*)/dt
0:05:08.800,0:05:12.160
actually this should have been a minus
0:05:12.240,0:05:16.560
and then otherwise inside it's pretty
0:05:14.240,0:05:20.560
much the same thing
0:05:16.560,0:05:20.560
with psi instead of psi*
0:05:22.080,0:05:29.280
okay so here we have V psi* psi
0:05:25.520,0:05:32.560
here we have minus V psi* psi
0:05:29.280,0:05:35.919
and so these are the same thing
0:05:32.560,0:05:38.479
so what's left
0:05:35.919,0:05:40.560
we have -hbar^2/2m
0:05:38.479,0:05:42.639
out the front of both expressions
0:05:40.560,0:05:44.720
and we can bring that out the front to
0:05:42.639,0:05:48.400
get d(rho)/dt
0:05:44.720,0:05:55.759
= - i hbar/2m
0:05:53.360,0:05:57.360
let's bring the psi the left of
0:05:55.759,0:05:59.039
grad^2 psi* because
0:05:57.360,0:06:01.520
the grad^2 only acts on the psi*
0:05:59.039,0:06:04.840
and inside these parentheses
0:06:01.520,0:06:07.680
but psi grad^2 psi*
0:06:04.840,0:06:15.600
minus psi* grad^2 psi
0:06:12.800,0:06:16.080
okay to go a bit further we need to
0:06:15.600,0:06:19.120
use
0:06:16.080,0:06:20.639
an identity from vector calculus
0:06:19.120,0:06:22.160
I'm just going to fold this paper over
0:06:20.639,0:06:25.039
just
0:06:22.160,0:06:25.039
a bit more closely
0:06:26.319,0:06:32.319
so we know that
0:06:30.880,0:06:34.240
and you can just you can derive this
0:06:32.319,0:06:35.520
yourself or look it up but we have
0:06:34.240,0:06:37.280
I'm going to look it up off my bit of
0:06:35.520,0:06:40.720
paper so if we have
0:06:37.280,0:06:41.360
f grad^2 g we can always write
0:06:40.720,0:06:44.639
this
0:06:41.360,0:06:48.479
as the divergence of f grad g
0:06:44.639,0:06:52.000
minus
0:06:48.479,0:06:55.440
(grad f).(grad g)
0:06:52.000,0:06:59.039
so
0:06:55.440,0:07:01.360
taking a look again at our
0:06:59.039,0:07:02.400
previous expression here in the first
0:07:01.360,0:07:06.880
term
0:07:02.400,0:07:06.880
f is psi and g is psi*
0:07:07.120,0:07:13.360
and in the second term those two switch
0:07:10.560,0:07:14.160
and so we end up with the result that
0:07:13.360,0:07:17.919
d(rho)/dt
0:07:14.160,0:07:23.039
equals
0:07:17.919,0:07:23.039
-i hbar/2m
0:07:24.400,0:07:30.960
divergence of i grad
0:07:27.440,0:07:34.240
psi*
0:07:30.960,0:07:38.240
minus (grad psi).(grad psi*)
0:07:38.479,0:07:45.840
minus the divergence of (switch all
0:07:41.840,0:07:45.840
psi and psi*)
0:07:52.639,0:07:59.120
like so and we see that this term is the
0:07:56.240,0:08:01.520
same as this term so these cancel
0:07:59.120,0:08:02.479
and the rest of the terms that remain
0:08:01.520,0:08:04.960
have a
0:08:02.479,0:08:06.319
divergence term on the outside and so we
0:08:04.960,0:08:10.560
can say that
0:08:06.319,0:08:13.199
d(rho)/dt is equal to minus
0:08:10.560,0:08:15.440
the divergence of some quantity we call
0:08:13.199,0:08:15.440
j
0:08:17.360,0:08:24.160
where j, which is in general
0:08:20.720,0:08:26.160
a function of position and time, is equal
0:08:24.160,0:08:27.680
to:
0:08:26.160,0:08:29.919
we've taken the minus out the front so:
0:08:41.839,0:08:45.200
and it's a vector quantity in three
0:08:44.159,0:08:49.120
dimensions
0:08:45.200,0:08:49.120
because of the grad here
0:08:51.120,0:08:55.680
so this is our stronger constraint it
0:08:54.320,0:08:57.839
tells us that not only is
0:08:55.680,0:08:59.680
probability conserved globally but it's
0:08:57.839,0:09:00.880
actually conserved locally
0:08:59.680,0:09:03.040
and what this means is that this
0:09:00.880,0:09:04.320
expression here is an expression for a
0:09:03.040,0:09:09.839
general fluid which
0:09:04.320,0:09:09.839
is what's called a continuity equation
0:09:12.959,0:09:17.200
and it tells us that imagine we have
0:09:15.279,0:09:18.640
some little box in space
0:09:17.200,0:09:21.040
which originally had some probability
0:09:18.640,0:09:22.080
density but this goes down it changes in
0:09:21.040,0:09:24.480
time
0:09:22.080,0:09:26.240
well it would be consistent with global
0:09:24.480,0:09:29.360
conservation of probability
0:09:26.240,0:09:30.480
to have another little box separated off
0:09:29.360,0:09:31.920
in space somewhere else
0:09:30.480,0:09:34.320
have the probability go up just
0:09:31.920,0:09:35.839
instantaneously that would be
0:09:34.320,0:09:37.600
compatible with global conservation of
0:09:35.839,0:09:39.360
probability but
0:09:37.600,0:09:41.680
what the continuity equation tells us is
0:09:39.360,0:09:43.600
that actually it's locally conserved
0:09:41.680,0:09:45.440
and so there must be a flow of
0:09:43.600,0:09:46.160
probability density between these two
0:09:45.440,0:09:47.920
points
0:09:46.160,0:09:49.360
that is, if the probability densities go
0:09:47.920,0:09:50.399
down in one region there must be a
0:09:49.360,0:09:52.880
divergence
0:09:50.399,0:09:54.560
of this current out of that region there
0:09:52.880,0:09:56.399
must be some flow out of this region
0:09:54.560,0:09:58.160
into that region
0:09:56.399,0:09:59.920
and so for it to go down from this box
0:09:58.160,0:10:01.760
and up in this box there must be a flow
0:09:59.920,0:10:05.839
between the two boxes
0:10:01.760,0:10:06.720
and this flow is a flow of this
0:10:05.839,0:10:10.399
quantity j(x,t)
0:10:06.720,0:10:10.399
which is called the probability
0:10:11.600,0:10:19.839
current density
0:10:21.040,0:10:24.800
there we go probability current density
0:10:23.519,0:10:27.519
so let's
0:10:24.800,0:10:29.760
take a look at some general boundary
0:10:27.519,0:10:33.070
conditions that we can apply on psi
0:10:29.760,0:10:36.299
as a result of these ideas
0:10:38.240,0:10:41.120
let's take a look at some boundary
0:10:39.279,0:10:42.560
conditions then the first boundary
0:10:41.120,0:10:43.680
condition on the wave function is that
0:10:42.560,0:10:46.800
it has to be continuous
0:10:43.680,0:10:49.519
across space
0:10:46.800,0:10:50.800
and the second condition is that the
0:10:49.519,0:10:56.079
first derivative of
0:10:50.800,0:10:56.079
psi with respect to x is also continuous
0:10:56.880,0:11:01.120
in fact the second condition doesn't
0:10:59.600,0:11:04.079
hold in all cases
0:11:01.120,0:11:05.440
but they are fairly pathological
0:11:04.079,0:11:08.640
cases where it doesn't hold
0:11:05.440,0:11:10.160
in fact psi prime the first derivative
0:11:08.640,0:11:12.800
with respect to position
0:11:10.160,0:11:14.560
is continuous provided that there's
0:11:12.800,0:11:15.279
no infinite discontinuity in the
0:11:14.560,0:11:16.320
potential
0:11:15.279,0:11:18.399
at places where there's infinite
0:11:16.320,0:11:19.279
discontinuities this may not hold so
0:11:18.399,0:11:21.839
let's add that
0:11:19.279,0:11:21.839
caveat
0:11:22.320,0:11:26.160
so these boundary conditions really come
0:11:24.320,0:11:27.600
from the fact that
0:11:26.160,0:11:29.920
we don't consider potentials which are
0:11:27.600,0:11:32.800
too pathological in fact we can
0:11:29.920,0:11:34.720
consider jumps in our potential we
0:11:32.800,0:11:36.079
can even have delta function potentials
0:11:34.720,0:11:38.560
but we don't consider anything worse
0:11:36.079,0:11:38.560
than that
0:11:38.640,0:11:41.760
a final condition which is not really a
0:11:40.079,0:11:42.560
boundary condition but which generally
0:11:41.760,0:11:44.480
applies
0:11:42.560,0:11:46.640
is that in regions where the potential
0:11:44.480,0:11:49.440
is infinite we require the wave function
0:11:46.640,0:11:49.440
to go to zero
0:11:49.680,0:11:52.880
the reason for this is that the modulus
0:11:51.519,0:11:54.399
square of the wave function is the
0:11:52.880,0:11:56.240
probability density
0:11:54.399,0:11:57.839
and we require the probability density
0:11:56.240,0:11:59.040
to be zero in regions where the
0:11:57.839,0:12:00.959
potential is infinity
0:11:59.040,0:12:03.040
in order to keep the energy finite which
0:12:00.959,0:12:06.320
is something we'd like to do
0:12:03.040,0:12:07.440
okay so let's take a look at some of the
0:12:06.320,0:12:10.639
philosophical
0:12:07.440,0:12:13.120
ideas behind the interpretation of these
0:12:10.639,0:12:13.120
objects
0:12:18.560,0:12:22.079
Dirac in his textbook gives a very good
0:12:20.480,0:12:23.519
explanation as to why we might expect
0:12:22.079,0:12:24.560
probabilities to come up in quantum
0:12:23.519,0:12:27.360
mechanics
0:12:24.560,0:12:27.920
we'd like descriptions of things such as
0:12:27.360,0:12:29.440
light
0:12:27.920,0:12:31.600
which are familiar on the macroscopic
0:12:29.440,0:12:33.600
scale but descriptions which
0:12:31.600,0:12:36.240
work down on the microscopic scale
0:12:33.600,0:12:37.680
when we see a beam of light like the
0:12:36.240,0:12:41.279
spot from this laser
0:12:37.680,0:12:43.440
pen we can of course
0:12:41.279,0:12:45.680
describe this classically and we can use
0:12:43.440,0:12:49.200
Maxwell's equations to describe it
0:12:45.680,0:12:52.160
but we'd like a quantum description
0:12:49.200,0:12:52.880
so Dirac gives us the example of taking
0:12:52.160,0:12:55.120
a crystal
0:12:52.880,0:12:57.360
something like calcite which I have here
0:12:55.120,0:12:59.519
so calcite is birefringent
0:12:57.360,0:13:00.720
which means if I shine the laser through
0:12:59.519,0:13:03.120
the calcite
0:13:00.720,0:13:03.839
my one spot should turn into two you can
0:13:03.120,0:13:05.680
see there
0:13:03.839,0:13:07.519
and as I rotate the crystal I actually
0:13:05.680,0:13:09.120
want to rotate around the other
0:13:07.519,0:13:11.040
and these two spots have different
0:13:09.120,0:13:14.000
polarizations the key thing is just that
0:13:11.040,0:13:14.000
there's two spots now
0:13:14.959,0:13:19.440
so why is this a problem well it's not a
0:13:17.920,0:13:20.959
problem on the classical scale
0:13:19.440,0:13:22.959
because we just say well there's a beam
0:13:20.959,0:13:24.000
of light it has some amplitude and some
0:13:22.959,0:13:26.959
intensity
0:13:24.000,0:13:28.079
and the intensity just splits into those
0:13:26.959,0:13:29.600
two beams
0:13:28.079,0:13:31.440
but if we're going to come up with any
0:13:29.600,0:13:32.959
kind of description of this on the scale
0:13:31.440,0:13:34.399
of single particles
0:13:32.959,0:13:36.480
it's going to have to somehow account
0:13:34.399,0:13:37.839
for each particle either going
0:13:36.480,0:13:39.600
one way or the other so there's got to
0:13:37.839,0:13:41.680
be some kind of probabilistic element to
0:13:39.600,0:13:44.639
it
0:13:41.680,0:13:46.000
the same idea leads to problems in the
0:13:44.639,0:13:48.720
idea of the atom
0:13:46.000,0:13:49.839
so we already had a problem after the
0:13:48.720,0:13:52.320
Rutherford experiment
0:13:49.839,0:13:54.399
showed that atoms have a positive
0:13:52.320,0:13:55.760
nucleus with negative electric charge
0:13:54.399,0:13:57.199
around it why doesn't the negative
0:13:55.760,0:13:59.040
charge drop into the positive and
0:13:57.199,0:14:00.720
minimize its energy
0:13:59.040,0:14:02.240
when Schroedinger wrote down his
0:14:00.720,0:14:04.240
description of the atom according to the
0:14:02.240,0:14:06.880
Schroedinger equation
0:14:04.240,0:14:08.560
he initially had the idea that the
0:14:06.880,0:14:10.639
probability current density that we've
0:14:08.560,0:14:13.040
just identified
0:14:10.639,0:14:14.240
may describe the current density of
0:14:13.040,0:14:16.320
the electron
0:14:14.240,0:14:17.680
as it orbits the nucleus but then that
0:14:16.320,0:14:20.639
gives us a problem again
0:14:17.680,0:14:22.399
because if the electron is really
0:14:20.639,0:14:24.160
orbiting the nucleus then
0:14:22.399,0:14:25.360
an electric charge
0:14:24.160,0:14:27.920
moving in a circle
0:14:25.360,0:14:30.240
radiates energy and so it should lose
0:14:27.920,0:14:32.079
its energy and drop into nucleus again
0:14:30.240,0:14:33.360
so according to Feynman in the Feynman
0:14:32.079,0:14:34.880
lectures, Schroedinger originally
0:14:33.360,0:14:37.040
interpreted the probability current
0:14:34.880,0:14:38.160
density as a literal current density of
0:14:37.040,0:14:40.000
the electron
0:14:38.160,0:14:42.160
the idea was that when we go down to the
0:14:40.000,0:14:44.320
quantum scale and we look at
0:14:42.160,0:14:46.560
individual particles such as photons in
0:14:44.320,0:14:48.560
this case each photon will be kind of
0:14:46.560,0:14:51.279
spread out in exactly the same way that
0:14:48.560,0:14:54.399
the intensity of the light is spread
0:14:51.279,0:14:55.920
but that can't be true because when we
0:14:54.399,0:14:57.680
observe the particles we always observe
0:14:55.920,0:14:58.959
them at a single location
0:14:57.680,0:15:00.399
it also can't be true because if we
0:14:58.959,0:15:02.399
think about the electron trying to orbit
0:15:00.399,0:15:03.920
the nucleus it would be losing energy
0:15:02.399,0:15:05.360
this is where Born came along with the
0:15:03.920,0:15:06.560
Born rule and why the Born rule was so
0:15:05.360,0:15:08.959
important
0:15:06.560,0:15:10.560
he told us that the current density was
0:15:08.959,0:15:11.600
not the current density of a smeared out
0:15:10.560,0:15:14.079
electron
0:15:11.600,0:15:15.600
it's the current density of a flow of
0:15:14.079,0:15:17.040
probability but when you look for the
0:15:15.600,0:15:17.920
electron you always find it in one
0:15:17.040,0:15:20.000
specific place
0:15:17.920,0:15:21.199
the electron is not spread out quantum
0:15:20.000,0:15:22.639
particles aren't spread out their
0:15:21.199,0:15:24.079
probabilities are spread out
0:15:22.639,0:15:26.399
according to their amplitudes described
0:15:24.079,0:15:27.519
by the wave function and Born told us
0:15:26.399,0:15:29.920
that the modulus square of that
0:15:27.519,0:15:30.959
amplitude gives us the probability of
0:15:29.920,0:15:33.600
finding the particle
0:15:30.959,0:15:36.240
in a little region okay thanks for your
0:15:33.600,0:15:36.240
time
V1.5 Two slit demo
This channel, Introductory Quantum Mechanics, is a set of videos aimed at second-year physics undergraduates.
This video:
a demonstration of Young's two slit experiment.
0:00:00.719,0:00:03.840
hello in this video i'm going to give a
0:00:02.720,0:00:05.600
quick demonstration
0:00:03.840,0:00:07.359
as to how to do the two-slit experiment
0:00:05.600,0:00:10.480
in your own room
0:00:07.359,0:00:12.160
so we're going to need two narrow
0:00:10.480,0:00:14.320
closely spaced slits
0:00:12.160,0:00:15.679
through which we pass light
0:00:14.320,0:00:16.480
the light can interfere through the two
0:00:15.679,0:00:18.400
different slits
0:00:16.480,0:00:20.000
and hit some screen where we can measure
0:00:18.400,0:00:22.240
the interference pattern
0:00:20.000,0:00:23.920
and therefore deduce properties of wave
0:00:22.240,0:00:26.160
superposition and so on
0:00:23.920,0:00:28.160
so for the screen that we're going to
0:00:26.160,0:00:29.840
hit into to measure the pattern we can
0:00:28.160,0:00:32.640
just use the wall that'll be fine so
0:00:29.840,0:00:34.160
there's a webcam focused on the wall
0:00:32.640,0:00:35.440
for the divider between the two
0:00:34.160,0:00:37.600
slits that's the tricky bit because we
0:00:35.440,0:00:39.440
need something extremely thin
0:00:37.600,0:00:41.200
but we have lots of extremely thin
0:00:39.440,0:00:42.239
things coming out of our heads: our
0:00:41.200,0:00:45.600
hairs
0:00:42.239,0:00:47.920
so what I have here is a bit of
0:00:45.600,0:00:49.440
card where I've taped a hair across it
0:00:47.920,0:00:50.800
there's a slit I've cut
0:00:49.440,0:00:52.480
with some scissors but that's just to
0:00:50.800,0:00:54.399
cut the amount of light passing through
0:00:52.480,0:00:56.800
down so it's not too intense
0:00:54.399,0:00:57.600
and stuck vertically across that you
0:00:56.800,0:00:58.960
can see the bit of
0:00:57.600,0:01:00.800
duct tape where I've used it to tape
0:00:58.960,0:01:04.239
across here
0:01:00.800,0:01:04.239
vertically like this
0:01:04.320,0:01:10.400
so if I shine the laser through
0:01:07.760,0:01:10.400
through that
0:01:10.880,0:01:15.360
slit I'll get so I'm just going to use
0:01:13.520,0:01:18.400
this green laser here which
0:01:15.360,0:01:20.240
has a wavelength of 568 nanometers
0:01:18.400,0:01:22.240
it's just quite nice and bright so it's
0:01:20.240,0:01:24.400
convenient for this
0:01:22.240,0:01:25.600
so you can see the colour there okay so
0:01:24.400,0:01:28.960
if I shine it
0:01:25.600,0:01:32.159
through the slit
0:01:28.960,0:01:34.720
at a location where
0:01:32.159,0:01:36.159
there's no hair you'll just see a
0:01:34.720,0:01:38.479
spot on the wall
0:01:36.159,0:01:39.439
I've turned the exposure right down on
0:01:38.479,0:01:40.960
the camera
0:01:39.439,0:01:42.880
but there you see when it hits the hair
0:01:40.960,0:01:46.159
you see that pattern picks up
0:01:42.880,0:01:46.159
there we go just there
0:01:48.159,0:01:53.119
I'm using a pack of cards to get the height
0:01:49.759,0:01:55.840
just right to line up with the slit
0:01:53.119,0:01:55.840
there we go
0:02:00.640,0:02:05.360
there we go and it's convenient to
0:02:03.600,0:02:06.799
remove that central spot which isn't really
0:02:05.360,0:02:08.560
contributing to the pattern
0:02:06.799,0:02:10.479
to do that you can use something like
0:02:08.560,0:02:12.319
this: this is just a block of ink for
0:02:10.479,0:02:13.280
calligraphy but something dark and quite
0:02:12.319,0:02:14.959
thin
0:02:13.280,0:02:17.440
and we can place that in the way of the
0:02:14.959,0:02:18.879
central spot
0:02:17.440,0:02:23.840
and then if I just turn it back onto the
0:02:18.879,0:02:23.840
hair again there we go
0:02:26.640,0:02:30.720
so now we get the pattern without the
0:02:28.400,0:02:33.920
central
0:02:30.720,0:02:36.160
spot and actually you can
0:02:33.920,0:02:37.760
measure, if you know the distance to
0:02:36.160,0:02:39.760
the wall with the ruler
0:02:37.760,0:02:40.879
and you can measure the spacing between
0:02:39.760,0:02:42.800
these
0:02:40.879,0:02:44.800
different peaks which you can see here
0:02:42.800,0:02:47.120
which I estimate to be something like
0:02:44.800,0:02:49.519
around four millimeters a distance to the wall
0:02:47.120,0:02:51.920
of about 40 centimeters
0:02:49.519,0:02:52.879
you can use use this to deduce the
0:02:51.920,0:02:54.480
width of the hair
0:02:52.879,0:02:56.160
and I think mine's probably about
0:02:54.480,0:03:00.480
something like 100 times the
0:02:56.160,0:03:02.720
the wavelength of the light
0:03:00.480,0:03:04.480
I estimate and so that would make it
0:03:02.720,0:03:08.239
about
0:03:04.480,0:03:10.400
50 or so
0:03:08.239,0:03:11.680
micrometers with the width of this hair
0:03:10.400,0:03:12.800
and that seems to tally with what we
0:03:11.680,0:03:17.519
expect
0:03:12.800,0:03:17.519
okay thank you for your time
V2.1a Scattering from a potential step (part I)
This channel, Introductory Quantum Mechanics, is a set of videos aimed at second-year physics undergraduates.
This video:
the forms of solutions to the TISE in regions of constant potential. Continued in video V2.1b.
0:00:02.399,0:00:04.720
Hello
0:00:03.360,0:00:07.040
in this video we're going to take a look
0:00:04.720,0:00:08.320
at solving the time independent
0:00:07.040,0:00:11.519
Schroedinger equation
0:00:08.320,0:00:12.960
in cases where we are scattering from a
0:00:11.519,0:00:15.280
potential step
0:00:12.960,0:00:16.320
so this is a particular instance of
0:00:15.280,0:00:17.760
solving the time independent
0:00:16.320,0:00:20.400
Schroedinger equation
0:00:17.760,0:00:23.359
in a case with a constant potential so
0:00:20.400,0:00:23.359
let's write that down
0:00:23.600,0:00:27.199
so our potential is going to be set
0:00:25.359,0:00:29.039
equal to V_0
0:00:27.199,0:00:31.840
and the form of the equation then looks
0:00:29.039,0:00:31.840
like this in one dimension
0:00:32.079,0:00:36.480
and the solutions to the equation in
0:00:34.960,0:00:39.120
general take different forms
0:00:36.480,0:00:40.000
depending on whether E the energy of the
0:00:39.120,0:00:42.640
particle
0:00:40.000,0:00:43.200
is greater than or less than V_0
0:00:42.640,0:00:45.520
the
0:00:43.200,0:00:47.600
constant potential in the problem so
0:00:45.520,0:00:52.480
first the case that E > V_0
0:00:48.559,0:00:52.480
we have plane wave solutions
0:00:53.600,0:00:57.360
so they take this form in general and
0:00:56.320,0:00:58.000
you can check quite easily by
0:00:57.360,0:01:00.640
substituting
0:00:58.000,0:01:01.199
those into here that these will solve
0:01:00.640,0:01:03.840
the
0:01:01.199,0:01:04.479
Schroedinger equation for particular
0:01:03.840,0:01:07.760
values
0:01:04.479,0:01:10.640
of k which we'll check in a second
0:01:07.760,0:01:13.600
the second case again for E > V_0
0:01:11.520,0:01:15.119
but well these correspond to
0:01:13.600,0:01:16.159
travelling waves when we add in the time
0:01:15.119,0:01:17.600
dependence
0:01:16.159,0:01:19.119
in the way that we always can add back
0:01:17.600,0:01:19.680
in the time dependence in terms of the
0:01:19.119,0:01:23.600
constant
0:01:19.680,0:01:23.600
phase winding
0:01:24.320,0:01:27.840
as a function of time these are two
0:01:26.799,0:01:30.240
traveling waves
0:01:27.840,0:01:31.759
but in certain cases depending on the
0:01:30.240,0:01:34.560
boundary conditions it can still be
0:01:31.759,0:01:36.240
important to look at standing waves
0:01:34.560,0:01:38.159
much like we can look at solutions to
0:01:36.240,0:01:39.920
the wave equation for sound say
0:01:38.159,0:01:42.320
we can have travelling waves of sound but
0:01:39.920,0:01:45.040
we can also have standing waves of sound
0:01:42.320,0:01:45.920
when trapped between two walls for
0:01:45.040,0:01:48.079
example
0:01:45.920,0:01:49.759
so in another case for E > V_0
0:01:48.079,0:01:51.920
we can have
0:01:49.759,0:01:55.520
well let's just write this down these
0:01:51.920,0:01:59.040
are travelling waves or plane waves
0:01:55.520,0:01:59.040
and we can also have standing waves
0:01:59.200,0:02:02.960
where of course you can in fact make
0:02:01.200,0:02:04.560
this form of solution
0:02:02.960,0:02:06.079
in terms of cosines and sines
0:02:04.560,0:02:08.000
from this solution in terms of
0:02:06.079,0:02:09.360
exponentials but this is going to be a
0:02:08.000,0:02:11.360
particularly convenient form
0:02:09.360,0:02:12.879
to consider in those cases where
0:02:11.360,0:02:14.480
standing waves are
0:02:12.879,0:02:15.840
the relevant things to consider let me
0:02:14.480,0:02:18.000
just magic my globe away for a second
0:02:15.840,0:02:19.920
with a quick slap of this board
0:02:18.000,0:02:22.160
good so let's write down that these are
0:02:19.920,0:02:24.480
standing waves
0:02:22.160,0:02:26.319
good now in the case where
0:02:24.480,0:02:27.920
E < V_0
0:02:26.319,0:02:29.440
the solutions are going to be
0:02:27.920,0:02:31.680
exponentially decaying
0:02:29.440,0:02:33.280
instead of being either travelling or
0:02:31.680,0:02:36.800
standing
0:02:33.280,0:02:39.920
so they take the following form where
0:02:36.800,0:02:43.120
kappa here has taken the place of k
0:02:39.920,0:02:44.080
and this is so that both k in
0:02:43.120,0:02:45.760
this case and
0:02:44.080,0:02:47.760
sorry k in this case and kappa in this
0:02:45.760,0:02:51.519
case are real numbers
0:02:47.760,0:02:52.319
so this looks like it doesn't
0:02:51.519,0:02:54.640
have an analogue
0:02:52.319,0:02:55.599
in terms of classical waves we have
0:02:54.640,0:02:58.319
both of these
0:02:55.599,0:02:59.200
in terms of sound and light this
0:02:58.319,0:03:00.720
looks like
0:02:59.200,0:03:02.400
there shouldn't be an analogue to just
0:03:00.720,0:03:05.519
something like a wave that just
0:03:02.400,0:03:07.280
exponentially grows or or decreases
0:03:05.519,0:03:08.800
but actually they do exist
0:03:07.280,0:03:11.840
classically and they're what's called
0:03:08.800,0:03:11.840
evanescent waves
0:03:12.000,0:03:15.280
or you might prefer to call them
0:03:12.959,0:03:17.519
evanescent solutions to the
0:03:15.280,0:03:18.560
equation because they're rather
0:03:17.519,0:03:19.120
strange to try and think of them as
0:03:18.560,0:03:21.360
waves.
0:03:19.120,0:03:24.560
I'll give you some examples of those
0:03:21.360,0:03:24.560
in a classical context separately.
0:03:24.879,0:03:28.799
So these are the general types of
0:03:26.879,0:03:31.760
solution we're going to be looking for
0:03:28.799,0:03:32.159
depending on how our energy relates to
0:03:31.760,0:03:35.120
our
0:03:32.159,0:03:36.159
potential so let's set up a particular
0:03:35.120,0:03:37.840
problem of interest
0:03:36.159,0:03:40.159
which is scattering from a potential
0:03:37.840,0:03:47.120
step so we'll clear this off and
0:03:40.159,0:03:47.120
specify the problem
V2.1b Scattering from a potential step (part II)
This channel, Introductory Quantum Mechanics, is a set of videos aimed at second-year physics undergraduates.
This video:
(continuing from video V2.1a) scattering from a potential step in the case that the energy of the particle is greater than that of the step; setting up the problem, and solving for the probability amplitudes for reflection and transmission. Continued in video V2.1c.
0:00:04.480,0:00:06.399
good
0:00:05.120,0:00:08.320
okay so the potential we're going to
0:00:06.399,0:00:10.000
consider takes the following form
0:00:08.320,0:00:11.679
it's constant it has two
0:00:10.000,0:00:14.160
different constant values in different
0:00:11.679,0:00:14.160
regions
0:00:14.240,0:00:20.480
okay and let's draw it
0:00:17.359,0:00:22.720
okay so we're going to consider
0:00:20.480,0:00:23.600
a plane wave incident from the left from
0:00:22.720,0:00:26.000
over here
0:00:23.600,0:00:27.359
and we'll consider the cases E the
0:00:26.000,0:00:28.560
energy of that plane wave is greater
0:00:27.359,0:00:31.119
than V_0 first
0:00:28.560,0:00:31.760
and then E < V_0
0:00:31.119,0:00:33.840
second
0:00:31.760,0:00:37.120
and so we can think of this V_0 as
0:00:33.840,0:00:37.120
an actual potential step
0:00:38.160,0:00:40.800
so this region here as we'll see is
0:00:39.520,0:00:42.079
what's called the classically forbidden
0:00:40.800,0:00:43.360
region before we get to that
0:00:42.079,0:00:44.800
let's look at what happens to a plain
0:00:43.360,0:00:46.160
wave trying to travel over the top of
0:00:44.800,0:00:49.200
this
0:00:46.160,0:00:52.160
to do so let's head over to the worked
0:00:49.200,0:00:52.160
example area
0:00:57.600,0:01:04.479
okay so let's get the
0:01:00.879,0:01:07.200
board reproduced up there all right
0:01:04.479,0:01:09.280
so we can label the different regions
0:01:07.200,0:01:12.479
region one and region two that's called
0:01:09.280,0:01:15.520
region one this one over here
0:01:12.479,0:01:16.880
where x < 0 and
0:01:15.520,0:01:19.920
region two
0:01:16.880,0:01:23.360
this one up here where
0:01:19.920,0:01:25.520
x > 0
0:01:23.360,0:01:26.880
now we're going to send a plane wave in
0:01:25.520,0:01:29.439
from the left
0:01:26.880,0:01:30.000
so that's going to take the following
0:01:29.439,0:01:34.000
form so
0:01:30.000,0:01:34.000
let's animate that as it comes in
0:01:38.320,0:01:43.360
and it's going to be in region one
0:01:48.000,0:01:56.320
and we'll have
0:01:52.720,0:01:57.360
phi_in(x) equals...
0:01:56.320,0:01:59.119
okay so there's going to be a constant
0:01:57.360,0:01:59.680
pre-factor in general but it'll take the
0:01:59.119,0:02:02.880
form
0:01:59.680,0:02:05.200
e to the i k x
0:02:02.880,0:02:08.160
which is the form of a right going
0:02:05.200,0:02:09.840
travelling wave
0:02:08.160,0:02:11.360
remember we're always going to when
0:02:09.840,0:02:13.200
we put the time dependence back in this
0:02:11.360,0:02:15.680
will always multiply by
0:02:13.200,0:02:17.520
(and let's just put this in brackets here)
0:02:15.680,0:02:20.560
e to the minus i omega t
0:02:21.120,0:02:24.800
and kx minus omega t is the form of a
0:02:23.920,0:02:28.080
right travelling
0:02:24.800,0:02:29.760
plane wave okay so we can safely ignore
0:02:28.080,0:02:30.720
this bit for now
0:02:29.760,0:02:33.519
we'll just deal with the time
0:02:30.720,0:02:35.599
independent part so
0:02:33.519,0:02:37.280
phi_in is going to take the form
0:02:35.599,0:02:40.319
e to the ikx
0:02:37.280,0:02:42.319
in region 1 there will be some reflected
0:02:40.319,0:02:44.800
wave
0:02:42.319,0:02:45.360
so it's still in region 1 and we can
0:02:44.800,0:02:48.959
call it
0:02:45.360,0:02:50.160
phi_reflected as a function of x
0:02:48.959,0:02:52.160
it can potentially have a different
0:02:50.160,0:02:54.239
amplitude
0:02:52.160,0:02:55.200
at the front and it's going to take the
0:02:54.239,0:02:58.319
form e to the
0:02:55.200,0:03:00.319
sorry minus i k x because
0:02:58.319,0:03:01.920
that's going to be a left travelling wave
0:03:00.319,0:03:05.680
which you can see up here hopefully so
0:03:01.920,0:03:07.200
it'll reflect back
0:03:05.680,0:03:10.239
and it'll take the form of a
0:03:07.200,0:03:14.319
left-going traveling wave
0:03:10.239,0:03:14.319
in region 2 then
0:03:16.640,0:03:20.400
we will only have a right going wave
0:03:19.519,0:03:21.760
because
0:03:20.400,0:03:23.360
the wave was incident from the left
0:03:21.760,0:03:24.959
there's a broken symmetry just given
0:03:23.360,0:03:26.720
by our starting condition of sending
0:03:24.959,0:03:29.760
the particle in from the left
0:03:26.720,0:03:32.400
so in region 2 we'll have phi_transmitted
0:03:29.760,0:03:33.920
they may have a
0:03:32.400,0:03:37.280
different amplitude t
0:03:33.920,0:03:40.640
e to the i and let's call it k prime
0:03:37.280,0:03:41.840
x this doesn't mean a derivative it
0:03:40.640,0:03:43.440
is just a different label for a
0:03:41.840,0:03:45.280
different k
0:03:43.440,0:03:47.120
okay so let's just look at the
0:03:45.280,0:03:49.840
normalizations on these first
0:03:47.120,0:03:51.360
i haven't put one on here if you
0:03:49.840,0:03:53.519
recall the previous videos
0:03:51.360,0:03:54.959
these are going to be amplitudes
0:03:53.519,0:03:57.120
the modulus square of these should give
0:03:54.959,0:03:58.239
us the probability densities
0:03:57.120,0:03:59.840
and we'd like to integrate our
0:03:58.239,0:04:01.439
probability densities across all of
0:03:59.840,0:04:02.959
space to be able to get one.
0:04:01.439,0:04:04.879
Now actually the case of plane waves is
0:04:02.959,0:04:06.959
a bit unphysical and they're what
0:04:04.879,0:04:09.439
are called non-normalizable
0:04:06.959,0:04:10.319
so you can't really have a plane wave it
0:04:09.439,0:04:11.599
doesn't
0:04:10.319,0:04:13.519
it's not a valid solution to the
0:04:11.599,0:04:14.879
Schroedinger equation because those waves
0:04:13.519,0:04:16.000
are not normalizable
0:04:14.879,0:04:18.000
as you can probably imagine from the
0:04:16.000,0:04:19.759
fact that they exist across all of space
0:04:18.000,0:04:21.680
so in reality what we would form is a
0:04:19.759,0:04:22.560
wave packet which would have a finite
0:04:21.680,0:04:24.400
extent
0:04:22.560,0:04:25.759
it'd be like a plane wave for some
0:04:24.400,0:04:26.880
region of space and that thing would be
0:04:25.759,0:04:28.400
moving along
0:04:26.880,0:04:30.400
it's a more difficult case to consider
0:04:28.400,0:04:33.199
mathematically
0:04:30.400,0:04:33.600
but but it's possible to form such
0:04:33.199,0:04:34.880
things
0:04:33.600,0:04:36.479
so we're just going to consider the
0:04:34.880,0:04:37.520
easiest case of plain waves but they're
0:04:36.479,0:04:38.800
not normalizable
0:04:37.520,0:04:40.400
so our pre-factors out the front
0:04:38.800,0:04:41.840
actually don't really matter but the
0:04:40.400,0:04:45.040
relative size of them does matter so
0:04:41.840,0:04:45.040
we're going to solve for r and t
0:04:45.280,0:04:50.800
okay so we've got the the setup then
0:04:47.840,0:04:52.800
so in region one
0:04:50.800,0:04:54.000
oh and sorry we're going to
0:04:52.800,0:04:57.199
consider the case
0:04:54.000,0:04:59.440
E > V_0
0:04:57.199,0:05:00.240
hence the form of these all taking
0:04:59.440,0:05:03.280
the form of
0:05:00.240,0:05:05.440
travelling waves plane waves and in
0:05:03.280,0:05:07.520
region one
0:05:05.440,0:05:08.720
our time independent Schroedinger
0:05:07.520,0:05:12.080
equation reads
0:05:08.720,0:05:15.680
-hbar^2/2 m phi''(x)
0:05:12.080,0:05:16.639
V(x) is 0 in
0:05:15.680,0:05:20.000
that region
0:05:16.639,0:05:23.680
so this just equals E phi
0:05:20.000,0:05:27.360
if we substitute oh and sorry
0:05:23.680,0:05:30.479
and in region one phi in region one
0:05:27.360,0:05:33.600
is equal to phi_in
0:05:30.479,0:05:37.680
plus phi_reflected
0:05:33.600,0:05:42.320
and in region two phi_two is just
0:05:37.680,0:05:46.080
phi_transmitted so in region 1
0:05:42.320,0:05:49.199
we can insert phi
0:05:46.080,0:05:49.199
region 1 into here
0:05:50.080,0:05:54.000
taking 2 derivatives of this with
0:05:52.960,0:05:57.520
respect to x
0:05:54.000,0:06:00.639
brings down i k twice from both of these
0:05:57.520,0:06:01.440
and so we just find i k squared is minus
0:06:00.639,0:06:04.400
k squared
0:06:01.440,0:06:06.160
we find that hbar squared k squared
0:06:04.400,0:06:09.840
over 2m
0:06:06.160,0:06:12.639
phi 1 equals E phi
0:06:09.840,0:06:13.680
1 and we can cancel the phi 1's out and
0:06:12.639,0:06:16.800
find the energy
0:06:13.680,0:06:17.600
in region 1 is equal to
0:06:17.600,0:06:21.840
h bar squared k squared over 2m
0:06:29.120,0:06:35.600
and this is in region 1.
0:06:32.240,0:06:35.600
In region two
0:06:36.479,0:06:41.120
our Schroedinger equation reads
0:06:38.479,0:06:51.759
-hbar^2/2m phi_2'' + V_0 phi_2 = E phi_2.
0:06:52.400,0:06:58.880
Substitute our right going
0:06:55.840,0:07:02.319
travelling wave into there brings down i
0:06:58.880,0:07:06.240
k prime twice and we find that h bar
0:07:02.319,0:07:09.440
squared k prime squared over 2m
0:07:06.240,0:07:11.759
plus v_0 equals E cancelling the
0:07:09.440,0:07:14.880
phi_2s out
0:07:11.759,0:07:18.080
and so in this case we have that
0:07:14.880,0:07:19.919
k prime is equal to
0:07:18.080,0:07:22.319
multiplied by 2m take the V_0
0:07:19.919,0:07:26.080
across to m
0:07:22.319,0:07:26.080
E - V_0
0:07:26.400,0:07:32.800
square root that divided by hbar
0:07:30.639,0:07:33.919
whereas from the top equation here we
0:07:32.800,0:07:38.000
could have said that k
0:07:33.919,0:07:40.720
equals 2 m e
0:07:38.000,0:07:41.840
over h bar so k and k prime are
0:07:40.720,0:07:44.160
different because of the
0:07:41.840,0:07:45.120
the different V values the different
0:07:44.160,0:07:48.000
potential and different features oh
0:07:45.120,0:07:48.000
sorry you can't quite see that
0:07:48.160,0:07:51.680
okay so k prime and k are wave
0:07:50.639,0:07:52.800
vectors and the wave vectors are
0:07:51.680,0:07:55.840
different in the different regions
0:07:52.800,0:07:55.840
because of the different potentials
0:07:56.240,0:08:02.800
okay so
0:08:00.000,0:08:03.520
good oh yes so now what we'd like to do
0:08:02.800,0:08:05.680
is
0:08:03.520,0:08:06.960
to solve using our boundary
0:08:05.680,0:08:09.360
conditions
0:08:06.960,0:08:13.120
so let me bring this back up here so
0:08:09.360,0:08:16.080
remember our general boundary conditions
0:08:13.120,0:08:16.080
from a previous video
0:08:20.479,0:08:23.599
for the time independent Schroedinger
0:08:22.080,0:08:26.720
equation
0:08:23.599,0:08:30.800
first tells us that phi
0:08:26.720,0:08:30.800
is continuous in space
0:08:31.120,0:08:34.479
and so what this tells us in the present
0:08:33.039,0:08:37.680
case is that
0:08:34.479,0:08:40.880
phi_1 evaluated at
0:08:37.680,0:08:41.680
x=0 where the two meet equals
0:08:40.880,0:08:45.760
phi_2
0:08:41.680,0:08:45.760
at x=0
0:08:46.240,0:08:53.040
and the second condition
0:08:49.360,0:08:54.000
is that phi prime the partial derivative
0:08:53.040,0:08:58.080
of phi with respect
0:08:54.000,0:08:58.080
to position x is also continuous
0:08:59.600,0:09:03.120
and so this gives us a condition for
0:09:01.600,0:09:04.320
the derivatives we're at the point where
0:09:03.120,0:09:06.880
the wave functions meet
0:09:04.320,0:09:07.920
and we find that phi_1 prime let's
0:09:06.880,0:09:13.120
just write 0
0:09:07.920,0:09:15.600
equals phi_2 prime and 0.
0:09:13.120,0:09:17.120
okay so taking condition 1 and
0:09:15.600,0:09:21.040
substituting
0:09:17.120,0:09:23.680
the expression for our amplitudes
0:09:21.040,0:09:25.760
we have that so let me just show you
0:09:23.680,0:09:28.720
again over here
0:09:25.760,0:09:30.160
so phi one is phi_in plus phi_r so
0:09:28.720,0:09:32.000
we're going to substitute x=0
0:09:30.160,0:09:34.399
into here so we get one from this
0:09:32.000,0:09:35.360
1+r from this one and that's going to
0:09:34.399,0:09:37.839
equal t
0:09:35.360,0:09:37.839
from this one
0:09:42.560,0:09:46.160
so what we're doing is substituting the
0:09:45.200,0:09:47.920
expressions
0:09:46.160,0:09:49.200
for the wave functions into this
0:09:47.920,0:09:52.560
expression
0:09:49.200,0:09:54.959
and then this is just a typo here okay
0:09:52.560,0:09:56.080
and for two we need to take the first
0:09:54.959,0:09:59.360
derivative
0:09:56.080,0:10:00.560
of these things so we're going to bring
0:09:59.360,0:10:04.320
down an ik
0:10:00.560,0:10:07.760
from this one minus i k from this one
0:10:04.320,0:10:07.760
and then i k prime from this one
0:10:07.920,0:10:14.560
and so we have that i k
0:10:11.440,0:10:18.000
e to the i k x
0:10:14.560,0:10:21.600
minus i k r
0:10:18.000,0:10:23.120
e to the minus i k x
0:10:21.600,0:10:25.040
and that thing is going to evaluate
0:10:23.120,0:10:28.560
at x=0
0:10:25.040,0:10:32.079
is equal to i k prime
0:10:28.560,0:10:35.200
t e to the i k prime x
0:10:32.079,0:10:36.800
also evaluated at x=0 just
0:10:35.200,0:10:39.920
taking the first derivative
0:10:36.800,0:10:43.519
and so we find cancelling out the i's
0:10:39.920,0:10:46.560
we find that k times
0:10:43.519,0:10:50.079
1-r equals k
0:10:46.560,0:10:50.079
prime times t
0:10:51.120,0:10:54.640
okay or in other words we can say
0:10:53.360,0:10:59.120
that 1-r
0:10:54.640,0:11:01.200
equals k prime over k times t
0:10:59.120,0:11:02.399
so we now have enough to solve for
0:11:01.200,0:11:06.560
the amplitudes
0:11:02.399,0:11:08.480
we have two expressions
0:11:06.560,0:11:10.079
two unknowns remember k and k prime are
0:11:08.480,0:11:12.240
known because they're specified by the
0:11:10.079,0:11:15.360
energy and V_0
0:11:12.240,0:11:17.120
and so we can
0:11:15.360,0:11:21.440
take the sum of these two things to get
0:11:17.120,0:11:21.440
t and the difference to get r
0:11:24.399,0:11:29.839
okay so we're just going to take this
0:11:26.240,0:11:29.839
equation one and equation 2.
0:11:33.519,0:11:37.440
if we add them together we find that we
0:11:35.680,0:11:41.200
get
0:11:37.440,0:11:44.800
2 equals t times
0:11:41.200,0:11:47.200
1 plus k prime over k
0:11:44.800,0:11:48.640
or in other words our transmission
0:11:47.200,0:11:51.920
amplitude t
0:11:48.640,0:11:57.839
is equal to 2/(1+k'/k)
0:11:51.920,0:11:57.839
which equals
0:12:01.360,0:12:07.360
2k/(k+k')
0:12:05.920,0:12:10.399
and taking the difference of the
0:12:07.360,0:12:13.920
equations we find that
0:12:10.399,0:12:18.399
2r=t(1-k'/k)
0:12:18.959,0:12:25.839
or r equals
0:12:31.839,0:12:37.040
this and we can substitute in the value
0:12:34.480,0:12:39.839
of t that we just found
0:12:37.040,0:12:40.880
this equals the twos will cancel we
0:12:39.839,0:12:45.440
get k/(k+k')
0:12:40.880,0:12:47.279
times
0:12:45.440,0:12:49.040
let's multiply this through so we get
0:12:47.279,0:12:55.440
(k-k')/k
0:12:52.399,0:12:55.760
cancel the k's and we find that
0:12:55.440,0:12:58.880
r=(k-k')/(k+k')
0:12:58.880,0:13:05.040
so this is our
0:13:02.320,0:13:05.040
reflection
0:13:05.519,0:13:07.839
amplitude
0:13:09.839,0:13:14.639
and this this equals t
0:13:16.959,0:13:21.839
is our transmission
0:13:22.880,0:13:25.600
amplitude
0:13:28.800,0:13:31.680
so they have something to do with
0:13:30.240,0:13:33.200
probabilities but remember they're not
0:13:31.680,0:13:35.680
probabilities themselves
0:13:33.200,0:13:37.680
because amplitudes in general can be
0:13:35.680,0:13:39.519
complex numbers
0:13:37.680,0:13:41.600
they have this arbitrary global phase
0:13:39.519,0:13:44.800
associated with them
0:13:41.600,0:13:46.079
so to work out the probability of
0:13:44.800,0:13:49.440
the particle reflecting or
0:13:47.680,0:13:50.399
transmitting over the top of this
0:13:49.440,0:13:52.720
barrier
0:13:50.399,0:13:53.600
we we need to look at something
0:13:52.720,0:13:54.880
slightly different
0:13:53.600,0:13:57.600
now it turns out that the relevant
0:13:54.880,0:13:59.120
quantity to consider is our probability
0:13:57.600,0:14:00.480
current density that we saw in a
0:13:59.120,0:14:03.040
previous video
0:14:00.480,0:14:16.639
so let's head back over to the board for
0:14:03.040,0:14:16.639
a second and take a quick look at that
V2.1c Scattering from a potential step (part III)
This channel, Introductory Quantum Mechanics, is a set of videos aimed at second-year physics undergraduates.
This video:
(continuing from video V2.1b) scattering from a potential step in the case that the energy of the particle is greater than that of the step; probability current densities for reflected and transmitted waves. Continued in V2.1d.
0:00:06.720,0:00:10.719
okay
0:00:07.440,0:00:11.599
so we have our phi_in our plane wave
0:00:10.719,0:00:14.160
going in
0:00:11.599,0:00:14.960
phi_r our plane wave reflecting and
0:00:14.160,0:00:16.800
phi_t
0:00:14.960,0:00:18.480
our plane wave transmitting we've solved
0:00:16.800,0:00:19.439
for the amplitudes for reflection and
0:00:18.480,0:00:21.199
transmission
0:00:19.439,0:00:22.880
to solve for the probabilities we need
0:00:21.199,0:00:23.519
to use our probability current density
0:00:22.880,0:00:26.720
we derived
0:00:23.519,0:00:30.400
in a previous video so the general form
0:00:26.720,0:00:33.520
looks like this where psi
0:00:30.400,0:00:36.160
is the time dependent wave function
0:00:33.520,0:00:37.040
in fact for all our plane waves the
0:00:36.160,0:00:40.480
time dependent
0:00:37.040,0:00:40.480
solution takes the following form
0:00:40.559,0:00:44.800
that is it's just the time
0:00:43.040,0:00:45.600
independent solution that we've been
0:00:44.800,0:00:49.760
working with
0:00:45.600,0:00:51.199
multiplied by a winding complex phase
0:00:49.760,0:00:52.800
where all the complex phases take the
0:00:51.199,0:00:54.239
same form here they wind at the same
0:00:52.800,0:00:57.120
rate given by the energy
0:00:54.239,0:00:58.719
of the particle and so we get the
0:00:57.120,0:00:59.600
expression for our probability current
0:00:58.719,0:01:02.559
density
0:00:59.600,0:01:05.680
for our time independent wave
0:01:02.559,0:01:05.680
functions of this form
0:01:06.400,0:01:09.520
that is pretty much the same form but
0:01:07.840,0:01:11.200
with phi in place of psi
0:01:09.520,0:01:13.360
so let's use this to calculate the
0:01:11.200,0:01:15.119
probabilities rather than the amplitudes
0:01:13.360,0:01:18.799
of reflection and transmission in the
0:01:15.119,0:01:18.799
two regions in another worked example
0:01:22.640,0:01:26.320
okay so let's look at the probability
0:01:25.280,0:01:29.439
current densities
0:01:26.320,0:01:29.439
in the two different regions
0:01:30.640,0:01:37.439
so from the board we had that j(x)
0:01:34.400,0:01:40.960
is equal to i h bar over 2
0:01:37.439,0:01:44.320
m phi d by dx
0:01:40.960,0:01:48.399
phi star minus phi star
0:01:44.320,0:01:51.680
d by dx phi
0:01:48.399,0:01:53.759
in region one
0:01:51.680,0:01:55.119
or rather it's actually the
0:01:53.759,0:02:00.320
reflection
0:01:55.119,0:02:04.960
part we'd like so we'd like j_reflected
0:02:00.320,0:02:07.840
is equal to as a function of x
0:02:04.960,0:02:08.720
we're going to insert phi reflected into
0:02:07.840,0:02:11.760
here
0:02:08.720,0:02:15.599
and remember phi so we've got phi_in
0:02:11.760,0:02:18.640
is equal to
0:02:15.599,0:02:22.080
e to the i k x
0:02:18.640,0:02:24.800
phi_reflected as e to the minus
0:02:22.080,0:02:27.280
i k x and that's multiplied sorry i
0:02:24.800,0:02:30.640
missed the prefactor by r
0:02:27.280,0:02:34.000
and phi transmitted is e
0:02:30.640,0:02:37.280
to the i k prime x
0:02:34.000,0:02:40.959
multiplied by t so if we stick phi_r
0:02:37.280,0:02:44.400
into here we find
0:02:40.959,0:02:47.599
i h bar over 2m out the front
0:02:44.400,0:02:48.480
let's write that in full for the
0:02:47.599,0:02:51.920
first one so we have
0:02:48.480,0:02:57.519
r e to the minus i
0:02:51.920,0:03:01.040
k x d by d x of phi star so phi star
0:02:57.519,0:03:04.239
that's d by dx
0:03:01.040,0:03:07.280
of r star
0:03:04.239,0:03:10.720
e to the plus
0:03:07.280,0:03:10.720
i k x
0:03:12.400,0:03:19.760
minus phi star is r star
0:03:16.640,0:03:23.360
e to the i k x and
0:03:19.760,0:03:27.360
d by d x of r e
0:03:23.360,0:03:27.360
to the minus i k x
0:03:29.120,0:03:36.239
and this equals i h bar over 2
0:03:32.239,0:03:38.640
m the r is a constant
0:03:36.239,0:03:39.360
so the derivative doesn't
0:03:38.640,0:03:41.040
change it
0:03:39.360,0:03:42.959
so we have: ... and that's true for both
0:03:41.040,0:03:47.040
cases so we're going to have a modulus r
0:03:42.959,0:03:50.720
squared out the front here
0:03:47.040,0:03:52.720
d by dx acting on e to the ikx brings
0:03:50.720,0:03:55.760
down an ik
0:03:52.720,0:03:57.360
so we get i k and then and it just
0:03:55.760,0:03:59.519
leaves it e to the ikx still
0:03:57.360,0:04:01.360
the e to the ikx cancels with the minus
0:03:59.519,0:04:04.400
i kx
0:04:01.360,0:04:05.599
and so that's all we get and then we
0:04:04.400,0:04:07.840
would have a minus here
0:04:05.599,0:04:09.680
e to the i k x d by d x on e to the
0:04:07.840,0:04:12.480
minus psi k x brings down a minus i
0:04:09.680,0:04:13.040
k minus and minus is plus so i actually
0:04:12.480,0:04:16.720
get a plus
0:04:13.040,0:04:20.000
ik and so we find that j_r
0:04:16.720,0:04:21.919
is equal to that's 2 i k
0:04:20.000,0:04:24.000
here 2 cancels with a 2 here we get
0:04:21.919,0:04:27.040
minus h bar
0:04:24.000,0:04:30.479
over m modulus r
0:04:27.040,0:04:33.840
squared k
0:04:30.479,0:04:36.639
if we'd have done the same for
0:04:33.840,0:04:36.639
phi_in
0:04:37.840,0:04:42.240
we would just have found that k goes to
0:04:40.400,0:04:43.759
minus k as you can see here and we lose
0:04:42.240,0:04:46.880
the modulus r squared
0:04:43.759,0:04:50.880
so we find that j_in
0:04:46.880,0:04:55.840
is equal to plus h bar over
0:04:50.880,0:04:56.960
m and we still get the k here
0:04:55.840,0:04:58.880
let's take a quick look at what we've
0:04:56.960,0:05:00.000
got here sorry let's move that over
0:04:58.880,0:05:03.039
slightly so you can see it
0:05:00.000,0:05:06.240
good so look at j_in
0:05:03.039,0:05:07.039
we have h bar k over m but remember
0:05:06.240,0:05:09.520
hbar k
0:05:07.039,0:05:10.639
from our de Broglie relation is just p
0:05:09.520,0:05:14.560
the momentum
0:05:10.639,0:05:17.280
and so this is p over m but p over m it
0:05:14.560,0:05:19.120
should just be our velocity
0:05:17.280,0:05:20.960
right classically momentum would be mass
0:05:19.120,0:05:22.560
times velocity and so
0:05:20.960,0:05:24.479
we can kind of think of this as the
0:05:22.560,0:05:25.680
velocity
0:05:24.479,0:05:26.960
or related to the velocity of the
0:05:25.680,0:05:28.800
particle remember it's a probability
0:05:26.960,0:05:31.919
current density
0:05:28.800,0:05:34.000
so it tells us how the
0:05:31.919,0:05:35.199
probability to find the particle evolves
0:05:34.000,0:05:38.320
with time but when we look we always
0:05:35.199,0:05:41.520
find the particle in one place
0:05:38.320,0:05:43.680
so the probability of reflection
0:05:41.520,0:05:47.600
let's call it R
0:05:43.680,0:05:50.160
is let's say it's defined to be
0:05:47.600,0:05:51.919
it really is this is just a sort of
0:05:50.160,0:05:54.320
physical way to motivate it it's
0:05:51.919,0:05:55.280
the probability current density for the
0:05:54.320,0:05:58.800
reflected wave
0:05:55.280,0:06:02.240
divided by the probability
0:05:58.800,0:06:05.039
current density of the in-going wave
0:06:02.240,0:06:06.400
and so in this case we get divided by
0:06:05.039,0:06:10.000
the hbar
0:06:06.400,0:06:11.840
k over m and we just get minus
0:06:10.000,0:06:13.840
r squared and sorry it should really be
0:06:11.840,0:06:14.960
the modulus of this because it's a
0:06:13.840,0:06:18.080
probability
0:06:14.960,0:06:21.840
the so we have the modulus of this
0:06:18.080,0:06:21.840
and so we get |r|^2
0:06:21.919,0:06:26.080
the minus sign of course is coming about
0:06:23.520,0:06:29.120
because j is a vector quantity
0:06:26.080,0:06:31.680
so the in-going wave has positive
0:06:29.120,0:06:32.960
velocity if you like positive current
0:06:31.680,0:06:34.400
probability current density
0:06:32.960,0:06:35.600
so the reflected wave must have a
0:06:34.400,0:06:36.240
negative one because it's a vector
0:06:35.600,0:06:37.680
quantity
0:06:36.240,0:06:39.440
so you can think of k as being negative
0:06:37.680,0:06:41.919
in this case
0:06:39.440,0:06:43.360
if we do the same thing for
0:06:41.919,0:06:48.400
phi_transmitted
0:06:43.360,0:06:50.800
we find that j_transmitted
0:06:48.400,0:06:52.319
is equal to: ... so it's rightgoing
0:06:50.800,0:06:52.880
you can do this explicitly i'm just
0:06:52.319,0:06:56.639
going to
0:06:52.880,0:07:00.080
state the result so we're going to get
0:06:56.639,0:07:05.039
a plus h bar over
0:07:00.080,0:07:08.080
m but we don't get a k we get a k prime
0:07:05.039,0:07:09.599
and we get a modulus
0:07:08.080,0:07:11.280
t squared much like we have the modulus
0:07:09.599,0:07:14.840
r squared for the other case
0:07:11.280,0:07:18.080
and so the probability of
0:07:14.840,0:07:21.199
transmission which is
0:07:18.080,0:07:25.440
defined to be j transmitted over
0:07:21.199,0:07:29.599
j incident is equal to ...
0:07:25.440,0:07:32.880
divided by h bar k over m and we get
0:07:29.599,0:07:35.599
modulus t squared k prime over
0:07:32.880,0:07:37.120
k and again sorry it's really
0:07:35.599,0:07:38.240
technically the modulus of this just in
0:07:37.120,0:07:41.840
case there's any
0:07:38.240,0:07:46.479
sign coming in there okay
0:07:41.840,0:07:48.160
so it's not just modulus t squared
0:07:46.479,0:07:49.520
the reason for the k prime of the k you
0:07:48.160,0:07:50.800
can think of it as the particle having
0:07:49.520,0:07:51.680
different velocities in the different
0:07:50.800,0:07:54.000
regions
0:07:51.680,0:07:54.960
and so you're saying more likely to find
0:07:54.000,0:07:57.680
a particle
0:07:54.960,0:07:58.080
if it's moving slowly in a region is
0:07:57.680,0:07:59.360
one
0:07:58.080,0:08:02.560
loose way to think of what's going on
0:07:59.360,0:08:05.680
here so we have the probability
0:08:02.560,0:08:08.080
for reflection and transmission
0:08:05.680,0:08:09.440
and in fact we can substitute back in
0:08:08.080,0:08:10.960
the energy E and V_0
0:08:09.440,0:08:12.479
into this expression if we want
0:08:10.960,0:08:14.479
to get it back in terms of
0:08:12.479,0:08:15.680
of the energy of the particle and
0:08:14.479,0:08:19.680
V_0
0:08:15.680,0:08:20.479
okay so let's take a look now at the
0:08:19.680,0:08:22.720
case that
0:08:20.479,0:08:27.840
E < V_0 and we'll switch
0:08:22.720,0:08:27.840
back to the board for a second
V2.1d Scattering from a potential step (part IV)
This channel, Introductory Quantum Mechanics, is a set of videos aimed at second-year physics undergraduates.
This video:
(continuing from video V2.1c) scattering from a potential step in the case that the energy of the particle is less than that of the step; probability amplitudes for reflected and transmitted waves; probability current densities for reflection and transmission.
0:00:08.480,0:00:10.240
okay
0:00:08.960,0:00:16.320
so this time we're going to consider the
0:00:10.240,0:00:20.080
case E < V_0 send a
0:00:16.320,0:00:20.080
plane wave in as before from the left
0:00:22.480,0:00:26.160
and by the way when i'm drawing the wave
0:00:24.160,0:00:28.880
functions here on this plot
0:00:26.160,0:00:30.080
it's really a bit of a mix of of two
0:00:28.880,0:00:31.439
different diagrams
0:00:30.080,0:00:33.520
in the picture here this is showing the
0:00:31.439,0:00:35.440
potential so the y-axis here is the
0:00:33.520,0:00:38.320
potential as a function of position
0:00:35.440,0:00:40.480
i'm drawing the waves coming in like
0:00:38.320,0:00:41.920
this
0:00:40.480,0:00:44.079
and of course really i'm here i'm
0:00:41.920,0:00:46.079
plotting the amplitude of the
0:00:44.079,0:00:47.200
wave function in the region as a
0:00:46.079,0:00:49.280
function of position
0:00:47.200,0:00:50.800
it's a very mixed notation but
0:00:49.280,0:00:53.199
it's something we tend to do
0:00:50.800,0:00:54.719
hopefully the idea is clear that this is
0:00:53.199,0:00:55.920
a wave coming in and this just happens
0:00:54.719,0:00:57.199
to be the potential landscape it's
0:00:55.920,0:01:00.719
moving in
0:00:57.199,0:01:03.359
now this is phi_in
0:01:00.719,0:01:04.159
coming in this way as before we're going
0:01:03.359,0:01:07.199
to find
0:01:04.159,0:01:07.199
reflected coming back
0:01:11.680,0:01:15.439
but this time the because
0:01:14.479,0:01:17.759
E < V_0
0:01:15.439,0:01:19.680
past the region over here is
0:01:17.759,0:01:22.080
what's called classically forbidden
0:01:19.680,0:01:23.360
so if we just take a step back for a
0:01:22.080,0:01:24.400
second and think about what would happen
0:01:23.360,0:01:26.880
classically
0:01:24.400,0:01:27.920
the potential here could be something
0:01:26.880,0:01:29.759
like a height a height is a
0:01:27.920,0:01:33.280
gravitational potential
0:01:29.759,0:01:35.840
so we could have a height that's
0:01:33.280,0:01:36.400
zero over here and then V_0 over
0:01:35.840,0:01:38.320
here
0:01:36.400,0:01:40.720
and this would be like some kind of wall
0:01:38.320,0:01:42.640
and if we send in a classical particle
0:01:40.720,0:01:45.680
as in say throw a tennis ball
0:01:42.640,0:01:46.880
in the direction of a wall if the
0:01:45.680,0:01:50.399
tennis ball has
0:01:46.880,0:01:52.000
this energy a gravitational energy
0:01:50.399,0:01:53.280
that's larger than that of the wall as
0:01:52.000,0:01:56.159
in it's going over the wall
0:01:53.280,0:01:56.880
then sure it can go over here but if
0:01:56.159,0:02:00.079
we send in
0:01:56.880,0:02:01.920
a tennis ball whose gravitational
0:02:00.079,0:02:03.520
potential energy is less than that of
0:02:01.920,0:02:04.000
the top of the wall it should bounce
0:02:03.520,0:02:05.360
back
0:02:04.000,0:02:07.040
of course this is then classically
0:02:05.360,0:02:08.319
forbidden tennis ball can't go into the
0:02:07.040,0:02:10.720
wall
0:02:08.319,0:02:12.720
in fact we'll see that in the
0:02:10.720,0:02:14.480
Schroedinger equation it's allowed
0:02:12.720,0:02:17.040
the form of the solutions are now our
0:02:14.480,0:02:19.280
evanescent type solutions
0:02:17.040,0:02:20.080
exponentially growing or decreasing and
0:02:19.280,0:02:21.840
in this case
0:02:20.080,0:02:24.319
only the decreasing solution can be
0:02:21.840,0:02:24.319
relevant
0:02:24.720,0:02:27.840
again not worrying too much about the
0:02:26.480,0:02:30.560
fact that i'm mixing
0:02:27.840,0:02:31.599
my notations with the potential and the
0:02:30.560,0:02:32.959
wave function
0:02:31.599,0:02:36.640
so we'll have an exponentially
0:02:32.959,0:02:38.800
decreasing amplitude over here
0:02:36.640,0:02:40.239
rather that's right an exponentially
0:02:38.800,0:02:43.360
decreasing
0:02:40.239,0:02:44.480
wave function because otherwise
0:02:43.360,0:02:46.400
the wave function would have to blow up
0:02:44.480,0:02:48.840
at infinity if we allow the positive
0:02:46.400,0:02:50.239
solution this barrier goes up to
0:02:48.840,0:02:53.360
infinity
0:02:50.239,0:02:53.680
okay so let's take a look at solving
0:02:53.360,0:02:55.599
this
0:02:53.680,0:02:58.879
explicitly and looking at the
0:02:55.599,0:02:58.879
classically forbidden region
0:03:02.800,0:03:05.760
okay so we'll go a little bit faster
0:03:04.480,0:03:07.760
than we did in the last one but the
0:03:05.760,0:03:11.120
working is very much the same
0:03:07.760,0:03:13.840
so in region one
0:03:11.120,0:03:14.800
we'll have phi_1 which is equal to
0:03:13.840,0:03:20.640
phi_in + phi_reflected
0:03:18.239,0:03:21.360
and this can be written as e to the i k x
0:03:21.360,0:03:27.120
plus r e to the minus i k x just as
0:03:24.560,0:03:29.360
before nothing's changing in region one
0:03:27.120,0:03:30.400
substitute into the time independent
0:03:29.360,0:03:33.440
Schroedinger equation
0:03:30.400,0:03:34.159
and we'll find that e equals h bar
0:03:33.440,0:03:37.120
squared k
0:03:34.159,0:03:39.120
squared over 2m just like before in
0:03:37.120,0:03:42.159
region 2 however
0:03:39.120,0:03:42.720
we'll have phi in region 2 is equal to
0:03:42.159,0:03:45.040
phi
0:03:42.720,0:03:46.000
transmitted remember t for transmitted
0:03:45.040,0:03:48.879
not for
0:03:46.000,0:03:49.599
time or anything like that and this is
0:03:48.879,0:03:52.799
equal to
0:03:49.599,0:03:56.319
t e to the minus
0:03:52.799,0:03:58.239
kappa x; kappa used
0:03:56.319,0:03:59.599
instead of k just to indicate that it's
0:03:58.239,0:04:01.680
really a different quantity to a wave
0:03:59.599,0:04:03.360
vector
0:04:01.680,0:04:05.280
substitutes into the time independent
0:04:03.360,0:04:08.640
Schroedinger equation in region 2
0:04:05.280,0:04:09.680
and we find that e equals minus h bar
0:04:08.640,0:04:12.959
squared
0:04:09.680,0:04:16.239
kappa squared over 2m plus
0:04:12.959,0:04:18.479
V_0 so
0:04:16.239,0:04:19.840
the minus here coming from the fact
0:04:18.479,0:04:21.440
that there was a minus ready in the
0:04:19.840,0:04:23.360
time independent Schroedinger equation
0:04:21.440,0:04:24.720
and we don't cancel it with a an i
0:04:23.360,0:04:28.000
squared as we do
0:04:24.720,0:04:30.880
in the case of plane waves okay
0:04:28.000,0:04:33.280
so boundary conditions as before in the
0:04:30.880,0:04:36.240
first boundary condition is that the
0:04:33.280,0:04:37.120
wave function is continuous so phi in
0:04:36.240,0:04:39.440
region one
0:04:37.120,0:04:40.479
at x=0 is equal to phi in
0:04:39.440,0:04:43.759
region two
0:04:40.479,0:04:47.440
at x=0 which tells us that
0:04:43.759,0:04:50.560
1+r=t just like before
0:04:47.440,0:04:50.960
boundary condition 2 the first oh dear
0:04:50.560,0:04:53.440
sorry
0:04:50.960,0:04:54.840
the first derivative of the wave
0:04:53.440,0:04:57.280
function must also
0:04:54.840,0:04:59.840
match the first derivative is also
0:04:57.280,0:04:59.840
continuous
0:04:59.919,0:05:03.600
and in this case what we find this gives
0:05:02.560,0:05:07.280
us is that
0:05:03.600,0:05:07.520
i k (1 - r) that's what we got before
0:05:07.280,0:05:10.560
on
0:05:07.520,0:05:14.160
in region 1 is now equal to minus
0:05:10.560,0:05:16.080
kappa t in region 2.
0:05:14.160,0:05:18.240
so slightly different form in fact
0:05:16.080,0:05:21.440
rewriting we have
0:05:18.240,0:05:25.360
1-r equals multiply through by minus
0:05:21.440,0:05:30.560
i we get i kappa over k
0:05:25.360,0:05:33.600
t just as before we have two equations
0:05:30.560,0:05:34.960
two unknowns r and t the amplitudes
0:05:33.600,0:05:35.600
for reflection and transmission around
0:05:34.960,0:05:38.080
the kappa
0:05:35.600,0:05:39.600
and k are both already known from
0:05:38.080,0:05:42.000
kappa from this equation k from this
0:05:39.600,0:05:42.000
equation
0:05:42.080,0:05:46.400
and using these two equations
0:05:45.039,0:05:49.680
we can derive
0:05:46.400,0:05:53.280
the probability of reflection which is k
0:05:49.680,0:05:56.240
minus i kappa over k
0:05:53.280,0:05:57.199
plus i kappa and the probability of
0:05:56.240,0:06:00.800
transmission
0:05:57.199,0:06:03.759
is 2 k/(k+i kappa)
0:06:00.800,0:06:04.800
in fact if you compare to
0:06:03.759,0:06:06.400
the results we got
0:06:04.800,0:06:08.560
for the case that energy was greater
0:06:06.400,0:06:10.000
than V_0 you'll see that
0:06:08.560,0:06:12.160
because all we've really done is we've
0:06:10.000,0:06:16.160
changed what was e to the
0:06:12.160,0:06:18.720
i k prime x to e to the minus kappa x
0:06:16.160,0:06:19.440
we've just changed i k prime to minus
0:06:18.720,0:06:20.639
kappa
0:06:19.440,0:06:23.600
and if you make those substitutions
0:06:20.639,0:06:26.240
you'll find you get the same results
0:06:23.600,0:06:27.840
all right so let's flip the page around
0:06:26.240,0:06:29.039
and look at the probabilities so those
0:06:27.840,0:06:30.000
are the amplitudes
0:06:29.039,0:06:32.479
actually let's just comment on this
0:06:30.000,0:06:35.840
quickly so we have the amplitude
0:06:32.479,0:06:37.120
for the particle to exist in region two
0:06:35.840,0:06:38.720
is non-zero
0:06:37.120,0:06:41.120
so remember region two is classically
0:06:38.720,0:06:42.800
forbidden but there's an
0:06:41.120,0:06:44.240
amplitude to find the particle there
0:06:42.800,0:06:45.680
what that means is that if we perform a
0:06:44.240,0:06:47.039
measurement we could find the particle
0:06:45.680,0:06:50.000
in region two
0:06:47.039,0:06:51.440
how does this tally
0:06:50.000,0:06:53.680
with the fact that there isn't enough
0:06:51.440,0:06:54.960
energy for the particle to be there
0:06:53.680,0:06:57.120
the answer is that our measurement
0:06:54.960,0:06:58.000
device will provide the energy if
0:06:57.120,0:06:59.759
it's to observe
0:06:58.000,0:07:01.680
the particle and really what we're doing
0:06:59.759,0:07:03.520
is changing the boundary conditions on
0:07:01.680,0:07:04.720
the problem we're changing the potential
0:07:03.520,0:07:06.000
when we make the measurement we'll
0:07:04.720,0:07:07.280
see more of that in the quantum
0:07:06.000,0:07:10.639
tunnelling
0:07:07.280,0:07:10.639
problem in the next video
0:07:11.759,0:07:15.759
nevertheless we let's take a look at
0:07:14.400,0:07:18.000
the probability
0:07:15.759,0:07:19.599
for the probability current densities in
0:07:18.000,0:07:24.000
the different regions
0:07:19.599,0:07:24.000
so let me just write this down like this
0:07:24.960,0:07:31.440
there we go partial subscript
0:07:28.160,0:07:32.080
x again means d phi* in this case by
0:07:31.440,0:07:35.039
dx
0:07:32.080,0:07:35.039
holding time constant
0:07:35.199,0:07:38.800
and so what we get is that phi
0:07:37.680,0:07:42.319
for phi_in
0:07:38.800,0:07:44.639
which equals e to the i k x
0:07:42.319,0:07:45.440
substitutes into here and just as before
0:07:44.639,0:07:48.560
we find
0:07:45.440,0:07:51.840
j_in equals h bar k over
0:07:48.560,0:07:54.960
m phi
0:07:51.840,0:07:57.840
transmitted equals the
0:07:54.960,0:07:58.479
transmission amplitude e to the minus
0:07:57.840,0:08:01.280
kappa
0:07:58.479,0:08:04.400
x substitute that into here let me just
0:08:01.280,0:08:04.400
write it out like this again
0:08:05.680,0:08:10.319
so we need this we substitute phi and
0:08:09.039,0:08:13.840
phi* into here
0:08:10.319,0:08:13.840
evaluating it we find this
0:08:15.759,0:08:19.120
so we bring down a minus kappa in both
0:08:18.639,0:08:20.720
cases
0:08:19.120,0:08:22.479
but now the minus sign between the two
0:08:20.720,0:08:24.800
terms remember it's this term
0:08:22.479,0:08:26.240
minus the complex conjugate but the two
0:08:24.800,0:08:28.639
terms are the same because phi
0:08:26.240,0:08:30.479
is now real and so actually we find that
0:08:28.639,0:08:33.279
these perfectly cancel out
0:08:30.479,0:08:34.240
we find that j_transmitted=0
0:08:33.279,0:08:37.760
and so therefore
0:08:34.240,0:08:41.519
the probability of transmission
0:08:37.760,0:08:44.640
which is equal to |j_t/j_in|
0:08:41.519,0:08:46.160
is equal to zero
0:08:44.640,0:08:48.640
this is what we call the
0:08:46.160,0:08:50.240
probability flux
0:08:48.640,0:08:51.920
the probability that the probability
0:08:50.240,0:08:53.839
current is going to
0:08:53.839,0:08:56.959
a traveling wave in the classically
0:08:56.000,0:08:58.160
forbidden region
0:08:56.959,0:09:00.080
so even though there's an amplitude to
0:08:58.160,0:09:01.760
find the particle over there
0:09:00.080,0:09:03.360
the probability current
0:09:01.760,0:09:04.880
density is zero so there's no
0:09:03.360,0:09:06.800
current propagating through that
0:09:04.880,0:09:09.200
region if we look at
0:09:06.800,0:09:11.360
the probability for reflection which is
0:09:09.200,0:09:14.160
|j_reflected / j_incident|
0:09:11.360,0:09:15.279
in fact the expression is the
0:09:14.160,0:09:17.440
same as we found
0:09:15.279,0:09:19.519
in the last video still
0:09:17.440,0:09:22.399
|r|^2
0:09:19.519,0:09:23.760
and if we substitute it in we find
0:09:22.399,0:09:28.640
for r we had
0:09:23.760,0:09:32.160
(k- i kappa)/(k+ i kappa)
0:09:28.640,0:09:34.480
and if you do the modulus square of this
0:09:32.160,0:09:37.519
you'll find out that it equals one
0:09:34.480,0:09:39.120
which must have been the case because
0:09:37.519,0:09:40.560
the probability of transmission plus
0:09:39.120,0:09:41.920
the probability of reflection is always
0:09:40.560,0:09:44.240
equal to one
0:09:41.920,0:09:47.519
from conservation of probability okay
0:09:44.240,0:09:47.519
thank you for your time
V2.2 Quantum tunnelling
This channel, Introductory Quantum Mechanics, is a set of videos aimed at second-year physics undergraduates.
This video:
quantum tunnelling (aka barrier penetration), scattering over a finite-width potential barrier, and resonant transmission.
0:00:00.560,0:00:04.480
hello in this video we're going to be
0:00:02.720,0:00:07.600
looking at quantum tunnelling
0:00:04.480,0:00:09.760
here's the potential of the problem and
0:00:07.600,0:00:11.280
let's draw it
0:00:09.760,0:00:13.280
so it's very much like the potential
0:00:11.280,0:00:14.000
step we considered in the previous two
0:00:13.280,0:00:15.360
videos
0:00:14.000,0:00:17.359
except for rather than going off to
0:00:15.360,0:00:20.800
infinity at positive x
0:00:17.359,0:00:24.560
it comes back down again to zero
0:00:20.800,0:00:26.720
in potential so
0:00:24.560,0:00:28.320
let the form of the solutions again
0:00:26.720,0:00:30.080
depend on which region we're in
0:00:28.320,0:00:32.000
let's call these regions one two and
0:00:30.080,0:00:34.399
three
0:00:32.000,0:00:35.600
and the solutions in one two and
0:00:34.399,0:00:37.200
three will depend on whether
0:00:35.600,0:00:38.879
the energy of the particle is greater
0:00:37.200,0:00:39.600
than the barrier height or less than the
0:00:38.879,0:00:41.520
barrier height
0:00:39.600,0:00:43.280
just like before when it's greater we'll
0:00:41.520,0:00:45.520
have plane waves in all
0:00:43.280,0:00:47.039
three regions when it's less
0:00:45.520,0:00:49.680
than the barrier height we'll have plane
0:00:47.039,0:00:52.960
waves in regions one and three
0:00:49.680,0:00:56.320
but evanescent waves in region two
0:00:52.960,0:00:56.320
so we can draw them something like this
0:00:56.719,0:01:00.480
I won't draw the plane waves because we
0:00:58.160,0:01:02.879
know what those look like.
0:01:00.480,0:01:03.760
For the energy less than the barrier
0:01:02.879,0:01:06.960
height
0:01:03.760,0:01:10.840
we'll have a plane wave coming in
0:01:06.960,0:01:12.080
and then we'll have either exponentially
0:01:10.840,0:01:15.040
growing
0:01:12.080,0:01:16.320
or exponentially decreasing solutions in
0:01:15.040,0:01:19.680
the barrier
0:01:16.320,0:01:21.360
and then a plane wave over here
0:01:19.680,0:01:23.439
and again I'd like to reiterate that
0:01:21.360,0:01:25.600
when I draw these pictures of the
0:01:23.439,0:01:27.200
wave functions of course they're on
0:01:25.600,0:01:30.240
different axes.
0:01:27.200,0:01:32.400
They don't have the same y-axis here
0:01:30.240,0:01:33.439
it's just a convenient schematic
0:01:32.400,0:01:34.720
notation
0:01:33.439,0:01:36.400
the other thing that's a bit
0:01:34.720,0:01:37.040
misleading about it is that of course
0:01:36.400,0:01:38.560
these are the
0:01:37.040,0:01:40.799
time independent solutions but I've just
0:01:38.560,0:01:43.040
taken them at a particular time
0:01:40.799,0:01:44.640
as time evolves the phase of these
0:01:43.040,0:01:45.360
waves changes so you can think of moving
0:01:44.640,0:01:46.720
along
0:01:45.360,0:01:48.320
and so this point we're moving up and
0:01:46.720,0:01:48.720
down so the fact that I draw them matching
0:01:48.320,0:01:50.799
here
0:01:48.720,0:01:52.159
is really just a convention and in fact
0:01:50.799,0:01:54.079
you can't even draw them matched
0:01:52.159,0:01:55.759
at the other end note that we're going
0:01:54.079,0:01:57.600
to need both the exponentially
0:01:55.759,0:02:00.159
increasing and decreasing
0:01:57.600,0:02:02.719
evanescent solutions in this case in
0:02:00.159,0:02:04.880
order to match the boundary conditions
0:02:02.719,0:02:05.920
okay so let's take a look at the form of
0:02:04.880,0:02:08.800
the waves in each
0:02:05.920,0:02:11.440
region let's clear the board good in
0:02:08.800,0:02:14.000
region one we have the following form
0:02:11.440,0:02:14.800
that is just like with the potential
0:02:14.000,0:02:16.800
step problem
0:02:14.800,0:02:18.879
we're going to send in a plane wave from
0:02:16.800,0:02:21.520
the left we'll set the
0:02:18.879,0:02:23.280
amplitude to one by convention and we'll
0:02:21.520,0:02:26.640
get a reflected wave travelling
0:02:23.280,0:02:28.879
back in the left direction
0:02:26.640,0:02:30.000
reflected back from the barrier. In
0:02:28.879,0:02:31.840
region three
0:02:30.000,0:02:34.560
that is on this side of the barrier
0:02:31.840,0:02:37.200
we'll have the following form
0:02:34.560,0:02:38.160
so a transmitted wave heading over to
0:02:37.200,0:02:40.959
the right again
0:02:38.160,0:02:42.640
but this time note that k is the same
0:02:40.959,0:02:44.239
k as appears in phi_1
0:02:42.640,0:02:45.840
this is because the potential is the
0:02:44.239,0:02:47.440
same in both regions
0:02:45.840,0:02:50.239
the potential is set to zero in both
0:02:47.440,0:02:52.959
regions so in particular the
0:02:50.239,0:02:55.120
energy eigenvalues in these regions are
0:02:52.959,0:02:56.959
as follows
0:02:55.120,0:02:58.159
that is the energy is h bar squared k
0:02:56.959,0:03:00.480
squared over 2m
0:02:58.159,0:03:01.920
for both regions one and three so it
0:03:00.480,0:03:03.440
really is the same k
0:03:01.920,0:03:05.360
in region two it depends on whether the
0:03:03.440,0:03:06.800
energy is greater than or less than
0:03:05.360,0:03:08.080
V_0 for the case that energy is
0:03:06.800,0:03:10.000
greater than V_0 we have plane
0:03:08.080,0:03:12.239
waves
0:03:10.000,0:03:13.200
where k prime is different now
0:03:12.239,0:03:14.879
because it solves
0:03:13.200,0:03:16.840
the energy eigenvalues of the time
0:03:14.879,0:03:18.480
independent Schroedinger equation are as
0:03:16.840,0:03:20.480
follows
0:03:18.480,0:03:23.120
so because of this additional V_0,
0:03:20.480,0:03:24.480
k prime does not equal k
0:03:23.120,0:03:26.799
and when E < V_0 we
0:03:24.480,0:03:29.920
have evanescent waves
0:03:26.799,0:03:32.239
of this form and the energy eigenvalues
0:03:29.920,0:03:34.799
look like this
0:03:32.239,0:03:35.680
that is we have a minus sign in front of
0:03:34.799,0:03:38.959
the
0:03:35.680,0:03:42.080
h bar squared kappa squared over 2m
0:03:38.959,0:03:43.200
and this then ensures that for
0:03:42.080,0:03:46.879
real kappa
0:03:43.200,0:03:47.920
E - V_0 is negative or
0:03:46.879,0:03:49.599
V_0 - E
0:03:47.920,0:03:51.519
is positive which is true when the
0:03:49.599,0:03:54.400
energy is less than V_0
0:03:51.519,0:03:55.120
and up here we have real k prime
0:03:54.400,0:03:58.400
giving
0:03:55.120,0:04:01.680
E - V_0 is positive so
0:03:58.400,0:04:03.599
notice that if we can solve the case for...
0:04:01.680,0:04:04.799
so quantum tunnelling occurs for when E
0:04:03.599,0:04:06.000
is less than V_0
0:04:04.799,0:04:08.080
we have to get through a classically
0:04:06.000,0:04:08.720
forbidden region when E>V_0
0:04:08.720,0:04:12.640
we're simply scattering over a potential
0:04:10.560,0:04:14.560
barrier of finite width
0:04:12.640,0:04:16.639
it's a tiny bit easier to solve
0:04:14.560,0:04:18.560
they're pretty much the same
0:04:16.639,0:04:19.680
in terms of difficulty I know that we
0:04:18.560,0:04:21.919
can solve
0:04:19.680,0:04:23.360
the case of E > V_0
0:04:21.919,0:04:24.960
and then get the solutions for
0:04:23.360,0:04:28.560
E < V_0 for free
0:04:24.960,0:04:31.040
by making the following observation
0:04:28.560,0:04:32.800
we can just substitute ik'
0:04:31.040,0:04:34.400
in place of kappa
0:04:32.800,0:04:36.160
and that will switch the form of the
0:04:34.400,0:04:38.400
solution from here to here
0:04:36.160,0:04:39.759
and it will also make this change here
0:04:38.400,0:04:40.479
so we can solve one of these problems
0:04:39.759,0:04:42.240
and get the other
0:04:40.479,0:04:44.000
solutions for free so let's deal with
0:04:42.240,0:04:46.880
the plane waves predominantly
0:04:44.000,0:04:48.800
this a tiny bit easier so let's do our
0:04:46.880,0:04:51.919
usual thing we have to just match the
0:04:48.800,0:04:53.199
wave functions at the boundaries using
0:04:51.919,0:04:56.400
the boundary conditions
0:04:53.199,0:05:00.560
we now have four unknowns
0:04:56.400,0:05:02.479
r, t, a, and b, but we also have
0:05:00.560,0:05:04.639
four matching conditions so let's
0:05:02.479,0:05:06.800
write those down
0:05:04.639,0:05:10.400
so first let's just reproduce the
0:05:06.800,0:05:13.520
picture a little bit smaller down here
0:05:10.400,0:05:16.080
so first we have that the wave function
0:05:13.520,0:05:18.880
must be continuous in space
0:05:16.080,0:05:20.639
so at x = 0 we have to match
0:05:18.880,0:05:22.160
wave functions one and two
0:05:20.639,0:05:24.320
and they must be equal to each other
0:05:22.160,0:05:26.800
substituting this into
0:05:24.320,0:05:30.080
the forms of the wave
0:05:26.800,0:05:32.720
functions we find the following result
0:05:30.080,0:05:33.360
1 + r = a + b
0:05:32.720,0:05:34.639
next we have to match
0:05:33.360,0:05:37.440
make sure that the derivatives are
0:05:34.639,0:05:39.039
continuous substituting we find this
0:05:37.440,0:05:41.360
result
0:05:39.039,0:05:42.160
and we have the same two conditions
0:05:41.360,0:05:45.440
to apply
0:05:42.160,0:05:45.440
at x = L
0:05:46.080,0:05:49.440
continuous wave function continuous
0:05:47.680,0:05:52.560
derivative and these gives the
0:05:49.440,0:05:54.880
following two conditions
0:05:52.560,0:05:57.520
okay so it's not particularly intuitive
0:05:54.880,0:05:59.840
we have four equations and four unknowns
0:05:57.520,0:06:01.520
and we can solve for them we'll do
0:05:59.840,0:06:04.560
so in the problem sets
0:06:01.520,0:06:06.160
well I'll just take the solution here
0:06:04.560,0:06:08.240
in particular we're interested in say
0:06:06.160,0:06:11.199
the transmission amplitude t
0:06:08.240,0:06:13.039
to get into this region and in
0:06:11.199,0:06:15.600
particular we're really interested in
0:06:13.039,0:06:17.520
the probability of transmission
0:06:15.600,0:06:19.360
which remember is given by the
0:06:17.520,0:06:21.919
ratio of the
0:06:19.360,0:06:23.919
probability current density in the
0:06:21.919,0:06:26.080
transmitted region to
0:06:23.919,0:06:29.039
the probability current density ingoing
0:06:27.280,0:06:30.800
and if we evaluate that for the case
0:06:29.039,0:06:31.680
that E > V_0 to plane
0:06:30.800,0:06:36.080
waves everywhere
0:06:31.680,0:06:37.919
we find the following result so
0:06:36.080,0:06:39.280
the probability of transmission the
0:06:37.919,0:06:41.440
ratio of
0:06:39.280,0:06:43.039
the probability current density in
0:06:41.440,0:06:45.280
region 3 to that
0:06:43.039,0:06:46.479
ingoing in this case it is just
0:06:45.280,0:06:48.800
|t|^2
0:06:46.479,0:06:50.160
the the velocities of the particles you
0:06:48.800,0:06:51.199
can think of as the same in regions one
0:06:50.160,0:06:53.360
and three because the
0:06:51.199,0:06:54.880
potentials are the same and we get
0:06:53.360,0:06:57.360
this expression here
0:06:54.880,0:06:58.160
now probably the most interesting
0:06:57.360,0:07:00.960
thing about it
0:06:58.160,0:07:02.240
is that we have this
0:07:00.960,0:07:06.160
sin^2(k'L) term
0:07:02.240,0:07:10.400
meaning that whenever
0:07:06.160,0:07:12.160
k'L = n pi for integer n
0:07:10.400,0:07:13.759
we have what's called
0:07:12.160,0:07:16.960
'resonant transmission'
0:07:13.759,0:07:17.680
meaning when this is fulfilled this
0:07:16.960,0:07:20.560
is zero
0:07:17.680,0:07:22.720
this whole thing is zero and we have
0:07:20.560,0:07:25.919
probability of transmission equals one
0:07:22.720,0:07:27.680
so we can tune E,
0:07:25.919,0:07:29.919
V_0, or L to get this condition to
0:07:27.680,0:07:30.479
be fulfilled and it's rather
0:07:29.919,0:07:32.080
interesting
0:07:30.479,0:07:34.160
perhaps the more philosophically
0:07:32.080,0:07:38.080
profound case is when E < V_0
0:07:34.160,0:07:38.800
so a similar-looking expression
0:07:38.080,0:07:41.280
to this
0:07:38.800,0:07:42.000
except for we now have a sinh
0:07:41.280,0:07:43.680
instead of a sine
0:07:42.000,0:07:45.599
so we lose our resonant transmission
0:07:43.680,0:07:48.720
condition sorry let me just
0:07:45.599,0:07:50.000
make a note of that name so
0:07:48.720,0:07:51.199
with E > V_0 we can have
0:07:50.000,0:07:53.280
resonant transmission
0:07:51.199,0:07:54.960
when e is less than V_0 we can't
0:07:53.280,0:07:56.400
but it's somewhat miraculous that we can
0:07:54.960,0:07:57.520
even have transmission at all
0:07:56.400,0:07:59.759
given that we have to pass through a
0:07:57.520,0:08:02.639
barrier which is classically forbidden
0:07:59.759,0:08:04.319
so quantum objects can quantum
0:08:02.639,0:08:05.039
mechanically tunnel through barriers to
0:08:04.319,0:08:07.840
which they couldn't
0:08:05.039,0:08:09.919
normally pass classically so this is the
0:08:07.840,0:08:13.120
basis of an experimental technique
0:08:09.919,0:08:16.960
called scanning tunnelling microscopy
0:08:13.120,0:08:19.759
you bring down a measurement tip
0:08:16.960,0:08:21.680
close to a sample and you can by
0:08:19.759,0:08:23.199
applying a bias voltage you can tunnel
0:08:21.680,0:08:23.759
electrons either from the tip into the
0:08:23.199,0:08:26.000
sample
0:08:23.759,0:08:27.120
or from the sample into the tip and
0:08:26.000,0:08:28.560
what tends to be done there are
0:08:27.120,0:08:29.599
different methods of operating this but
0:08:28.560,0:08:32.719
they tend to run it in
0:08:29.599,0:08:33.919
the constant current mode in which you
0:08:33.919,0:08:37.120
vary the height of your tip in order to
0:08:35.599,0:08:39.039
maintain the same current
0:08:37.120,0:08:41.120
coming through as you move the tip along
0:08:39.039,0:08:42.959
so then you record the height of the tip
0:08:41.120,0:08:45.120
at different places and you can use this
0:08:42.959,0:08:45.839
to map out the effective height of the
0:08:45.120,0:08:48.000
surface
0:08:45.839,0:08:49.680
down to the atomic scale so there's a
0:08:48.000,0:08:52.800
let me get a picture up for you here
0:08:49.680,0:08:54.000
so here's a picture of an STM image as
0:08:52.800,0:08:55.360
it's called
0:08:54.000,0:08:56.880
where you're really seeing individual
0:08:55.360,0:08:59.040
atoms so it's an incredibly powerful
0:08:56.880,0:09:01.200
technique
0:08:59.040,0:09:03.200
this also resolves a bit of a paradox we
0:09:01.200,0:09:05.040
had in the previous two videos where we
0:09:03.200,0:09:07.600
looked at scattering from
0:09:05.040,0:09:08.480
an infinitely long step and we saw that
0:09:07.600,0:09:10.720
there was no
0:09:08.480,0:09:12.160
probability for transmission in the
0:09:10.720,0:09:13.040
sense of probability current getting
0:09:12.160,0:09:14.560
through
0:09:13.040,0:09:16.560
but there was nevertheless an amplitude
0:09:14.560,0:09:17.839
for transmission and I said at the time
0:09:16.560,0:09:18.320
that what happens is if you were to
0:09:17.839,0:09:19.920
measure
0:09:18.320,0:09:21.920
a particle there you could find one
0:09:19.920,0:09:23.680
there but you'd be providing the energy
0:09:21.920,0:09:25.760
using your measurement device
0:09:23.680,0:09:27.519
so now we see what happens really
0:09:25.760,0:09:28.640
you're changing the form of the solution
0:09:27.519,0:09:29.680
originally we had the step that went off
0:09:28.640,0:09:31.120
to infinity
0:09:29.680,0:09:32.640
when you bring a measurement device in
0:09:31.120,0:09:33.440
you're actually changing the potential
0:09:32.640,0:09:35.600
landscape
0:09:33.440,0:09:37.680
you're making it possible for the
0:09:35.600,0:09:38.720
particle to exist inside the detector
0:09:37.680,0:09:40.399
and so you're bringing the step back
0:09:38.720,0:09:42.240
down again
0:09:40.399,0:09:43.920
so you're really turning the infinite
0:09:42.240,0:09:46.240
step problem into
0:09:43.920,0:09:46.959
the finite length barrier problem and
0:09:46.240,0:09:49.279
that's how
0:09:46.959,0:09:50.480
you're able to measure the particle in
0:09:49.279,0:09:53.760
that region
0:09:50.480,0:09:55.279
so I'm going to show you it so
0:09:53.760,0:09:56.880
this seems like an incredibly quantum
0:09:55.279,0:09:59.200
phenomenon and it is really
0:09:56.880,0:10:00.720
but there is a precedent for it in
0:09:59.200,0:10:03.200
terms of evanescent waves
0:10:00.720,0:10:04.079
in light and I'll show you that in a
0:10:03.200,0:10:07.200
separate video
0:10:04.079,0:10:07.200
thank you
V2.3 Evanescent waves demo
This channel, Introductory Quantum Mechanics, is a set of videos aimed at second-year physics undergraduates.
This video:
a demonstration of coupling to evanescent light waves in order to transform an amplitude for transmission into a fully fledged probability current density. As Maxwell's equations are compatible with quantum mechanics this can either be described as a classical wave effect or the quantum tunnelling of photons through a classically-forbidden region. The demo is followed by a brief discussion of why quantum tunnelling is so magical.
Music: Angelo Badalamenti - Audrey's Dance (Twin Peaks OST)
0:00:00.719,0:00:04.400
Hello welcome to my kitchen where I'm
0:00:03.120,0:00:06.240
going to show a quick demonstration
0:00:04.400,0:00:06.720
about quantum tunnelling and evanescent
0:00:06.240,0:00:09.679
waves
0:00:06.720,0:00:10.080
using light so what I have here is a
0:00:09.679,0:00:13.440
nice
0:00:10.080,0:00:16.080
block of perspex
0:00:13.440,0:00:17.680
and let me just turn the light off here
0:00:16.080,0:00:19.520
wait for the camera to adjust
0:00:17.680,0:00:21.119
so I'm going to shine a laser into the
0:00:19.520,0:00:23.519
block I found that
0:00:21.119,0:00:24.560
red is probably the most effective for
0:00:23.519,0:00:27.199
showing this
0:00:24.560,0:00:29.519
so if I shine the laser at a nice steep
0:00:27.199,0:00:31.519
angle like this
0:00:29.519,0:00:33.600
I get a spot coming through on the back
0:00:31.519,0:00:34.880
you can see over here here's a nice spot
0:00:33.600,0:00:36.000
and you can even see the beam going
0:00:34.880,0:00:37.360
through and you can see some of it
0:00:36.000,0:00:38.800
reflecting so there's some
0:00:37.360,0:00:40.559
probability for reflection some
0:00:38.800,0:00:43.520
probability for transmission
0:00:40.559,0:00:44.239
if I increase the angle of incidence
0:00:43.520,0:00:46.559
like this
0:00:44.239,0:00:47.920
eventually there we go I get total
0:00:46.559,0:00:50.399
internal reflection
0:00:47.920,0:00:52.160
so there's now no beam coming through at
0:00:50.399,0:00:52.719
all the probability of transmission is
0:00:52.160,0:00:54.160
zero
0:00:52.719,0:00:56.079
there's a bit of ambient light coming
0:00:54.160,0:00:57.440
from around the sides here
0:00:56.079,0:00:59.359
but just to show you here's what happens
0:00:57.440,0:01:02.000
when a full beam goes through
0:00:59.359,0:01:03.760
can you see that it's much bigger and
0:01:02.000,0:01:05.199
here's where it's disappeared again
0:01:03.760,0:01:08.560
it's totally internal
0:01:05.199,0:01:10.560
reflected there okay
0:01:08.560,0:01:12.159
so when we're getting total internal
0:01:10.560,0:01:14.320
reflection the probability
0:01:12.159,0:01:16.400
for transmission the probability current
0:01:14.320,0:01:19.280
density on the other side
0:01:16.400,0:01:20.240
is zero but there's nevertheless an
0:01:19.280,0:01:23.600
amplitude
0:01:20.240,0:01:26.159
to detect a photon out the back there
0:01:23.600,0:01:26.799
so what we could do in order to show
0:01:26.159,0:01:29.520
that
0:01:26.799,0:01:32.079
is remember in the quantum problem if
0:01:29.520,0:01:35.520
we have the infinitely long
0:01:32.079,0:01:36.799
barrier then we won't get any
0:01:35.520,0:01:37.840
probability current density in that
0:01:36.799,0:01:39.439
region
0:01:37.840,0:01:41.360
but if we can make the barrier finite
0:01:39.439,0:01:42.240
length by coupling to some measurement
0:01:41.360,0:01:43.920
device
0:01:42.240,0:01:45.840
we can actually get a probability
0:01:43.920,0:01:48.000
current in that other region
0:01:45.840,0:01:49.759
so what I'd like to show you is I'd like
0:01:48.000,0:01:52.159
to take this prism
0:01:49.759,0:01:53.040
right here and I'd like to place it
0:01:52.159,0:01:56.320
behind
0:01:53.040,0:01:59.920
the block over here and
0:01:56.320,0:02:02.240
leave a little air gap and have the
0:01:59.920,0:02:03.280
prism coupled to the evanescent wave out
0:02:02.240,0:02:05.680
the back of the
0:02:03.280,0:02:07.520
perspex block and and take some of the
0:02:05.680,0:02:09.679
reflected power away and divert it
0:02:07.520,0:02:12.160
and make a transmitted wave now that's
0:02:09.679,0:02:14.480
not going to happen
0:02:12.160,0:02:15.200
so there's the beam going
0:02:14.480,0:02:17.520
through
0:02:15.200,0:02:19.599
here's total internal reflection and
0:02:17.520,0:02:21.440
here's me placing the prism here
0:02:19.599,0:02:23.440
and it doesn't steal any of the
0:02:21.440,0:02:25.280
reflected power
0:02:23.440,0:02:26.560
it's not because quantum mechanics is
0:02:25.280,0:02:30.239
wrong it's just because
0:02:26.560,0:02:31.440
the wavelength of this laser is 650
0:02:30.239,0:02:35.040
nanometers
0:02:31.440,0:02:37.280
so in order to
0:02:35.040,0:02:38.879
couple to that exponentially dying
0:02:37.280,0:02:42.800
evanescent wave I'd need to get
0:02:38.879,0:02:45.760
within a few wavelengths of the light
0:02:42.800,0:02:48.080
and 650 nanometers is somewhere between
0:02:45.760,0:02:51.120
a 20th and a 200th of the
0:02:48.080,0:02:52.800
width of a hair so I'm not going to
0:02:51.120,0:02:53.360
realistically get this prism close
0:02:52.800,0:02:55.920
enough
0:02:53.360,0:02:58.480
to the block in order to couple to it
0:02:55.920,0:03:00.400
to take any significant power away
0:02:58.480,0:03:02.319
I can do a bit of a cheat though which
0:03:00.400,0:03:03.680
is that I need to find a material
0:03:02.319,0:03:05.440
or some kind of surface that I can get
0:03:03.680,0:03:07.200
close enough to the back of this
0:03:05.440,0:03:08.720
perspex block that I can couple the
0:03:07.200,0:03:12.159
evanescent wave inside
0:03:08.720,0:03:12.959
off the back and the trick is that
0:03:12.159,0:03:15.519
I can just
0:03:12.959,0:03:18.400
pour some water in there because of
0:03:15.519,0:03:19.440
course water is going to be
0:03:18.400,0:03:21.920
let's set up the total internal
0:03:19.440,0:03:23.440
reflection the water is going to get
0:03:21.920,0:03:26.799
close enough to the back there right
0:03:23.440,0:03:26.799
now it's a bit of a cheat
0:03:27.680,0:03:34.879
whoopsy daisy let's get that back there
0:03:31.519,0:03:34.879
hopefully when it settles down
0:03:36.560,0:03:40.959
I think we need a little bit more water
0:03:38.080,0:03:40.959
just to bring the height up
0:03:49.840,0:03:52.400
there we go so now you can see on the
0:03:50.959,0:03:53.599
back wall we're getting that beam
0:03:52.400,0:03:55.280
through can you see that
0:03:53.599,0:03:57.120
there we go so the reason it's a little
0:03:55.280,0:03:58.799
bit of a cheat is just that
0:03:57.120,0:04:00.400
you know what I'm really doing is
0:03:58.799,0:04:02.239
just changing the refractive index of
0:04:00.400,0:04:03.360
the material out the back of the perspex
0:04:02.239,0:04:05.280
block
0:04:03.360,0:04:06.640
but that kind of explains why this
0:04:05.280,0:04:09.840
had to work, right?
0:04:06.640,0:04:11.360
Because you know that if I put a
0:04:09.840,0:04:13.040
higher refractive index material out the
0:04:11.360,0:04:14.799
back like water I will change the
0:04:13.040,0:04:15.439
critical angle and get a beam to go
0:04:14.799,0:04:18.400
through
0:04:15.439,0:04:20.079
but how if all the individual photons
0:04:18.400,0:04:21.519
were reflecting before how would they
0:04:20.079,0:04:23.600
know to go through if you change the
0:04:21.519,0:04:25.360
stuff behind the perspex block
0:04:23.600,0:04:28.000
and the answer is that there's actually
0:04:25.360,0:04:31.199
a probability sorry an amplitude
0:04:28.000,0:04:32.639
for transmission but since I'm using
0:04:31.199,0:04:35.840
water here
0:04:32.639,0:04:38.000
let me just turn this light on and so
0:04:35.840,0:04:39.120
sorry and the reason light is so good
0:04:38.000,0:04:42.240
for showing this is
0:04:39.120,0:04:43.919
as we said before that light
0:04:42.240,0:04:45.600
is it can be thought of either as
0:04:43.919,0:04:46.400
quantum or classical. Maxwell's
0:04:45.600,0:04:48.479
equations
0:04:46.400,0:04:49.919
are compatible with quantum mechanics
0:04:48.479,0:04:53.600
so you can either think of this
0:04:49.919,0:04:55.120
as a beam of light
0:04:53.600,0:04:56.560
in a classical manner or you can think
0:04:55.120,0:04:57.919
of that beam of light as being made up
0:04:56.560,0:05:01.199
of individual quanta of
0:04:57.919,0:05:02.479
energy called photons and and those
0:05:01.199,0:05:04.160
descriptions will be compatible with
0:05:02.479,0:05:06.800
each other so you can think of this
0:05:04.160,0:05:07.280
as an evanescent wave for classical
0:05:06.800,0:05:08.720
light
0:05:07.280,0:05:10.560
or you can think of it as quantum
0:05:08.720,0:05:12.880
tunnelling of the photons through
0:05:10.560,0:05:14.000
the classically forbidden region as it
0:05:12.880,0:05:17.280
were
0:05:14.000,0:05:19.199
okay so since I've got the cup of water
0:05:17.280,0:05:21.600
here let me just adjust the focus on the
0:05:19.199,0:05:21.600
camera
0:05:22.080,0:05:26.720
there we go so there's actually a
0:05:25.199,0:05:29.120
an even better demonstration you can do
0:05:26.720,0:05:30.560
with water you can see down here that we
0:05:29.120,0:05:32.479
have total internal reflection in the
0:05:30.560,0:05:32.960
cup you can't see the tip of my finger
0:05:32.479,0:05:34.720
right
0:05:32.960,0:05:36.000
here you can see it and down here you
0:05:34.720,0:05:38.080
can't because
0:05:36.000,0:05:40.960
there's total internal reflection if I
0:05:38.080,0:05:43.039
just get it a tiny bit damp
0:05:40.960,0:05:44.320
and I place my finger on the back
0:05:43.039,0:05:46.720
here
0:05:44.320,0:05:47.840
you see my fingerprint come about that's
0:05:46.720,0:05:50.000
quite clear isn't it
0:05:47.840,0:05:51.840
so what's happening is that the
0:05:50.000,0:05:52.800
ridges of my fingerprint are getting
0:05:51.840,0:05:55.919
close enough
0:05:52.800,0:05:56.960
to the water and the
0:05:55.919,0:05:58.560
plastic
0:05:56.960,0:05:59.680
that they can couple to that evanescent
0:05:58.560,0:06:00.960
wave so I should be getting total
0:05:59.680,0:06:04.479
internal reflection
0:06:00.960,0:06:06.000
a probability of reflection 1
0:06:04.479,0:06:07.360
but by placing something close enough to
0:06:06.000,0:06:08.639
the back I can actually couple to the
0:06:07.360,0:06:11.360
evanescent wave and I can
0:06:08.639,0:06:12.639
steal some of that power away so as some
0:06:11.360,0:06:14.800
of the reflective power and it turns
0:06:12.639,0:06:16.720
into transmitted and so then I get a
0:06:14.800,0:06:18.160
propagating wave out the back
0:06:16.720,0:06:20.080
but you can see how sensitive it is
0:06:18.160,0:06:22.479
because the
0:06:20.080,0:06:24.720
troughs of my fingerprint are too far
0:06:22.479,0:06:27.520
away to have any significant coupling
0:06:24.720,0:06:29.120
hence you see dark where sorry see
0:06:27.520,0:06:30.240
light where the troughs are because I'm
0:06:29.120,0:06:31.440
not coupling and you've got total
0:06:30.240,0:06:33.360
internal reflection
0:06:31.440,0:06:34.560
and you see dark where the the peaks of
0:06:33.360,0:06:36.560
my fingerprint are
0:06:34.560,0:06:38.240
because that is coupling to the
0:06:36.560,0:06:38.560
evanescent wave and directing the power
0:06:38.240,0:06:41.120
out
0:06:38.560,0:06:44.000
so I'm getting quantum tunnelling into
0:06:41.120,0:06:46.400
my finger of these individual photons
0:06:44.000,0:06:48.319
so you can think of it with light either
0:06:46.400,0:06:50.800
as a classical or a quantum effect
0:06:48.319,0:06:52.800
it doesn't make the quantum problem any
0:06:50.800,0:06:55.440
less magical it just means that
0:06:52.800,0:06:56.000
classical waves are more magical
0:06:55.440,0:06:59.120
let's
0:06:56.000,0:07:03.840
take a look at some of that
0:06:59.120,0:07:03.840
magic in a different room
0:07:06.800,0:07:10.319
In terms of a classical particle however
0:07:08.639,0:07:13.680
it's really quite weird
0:07:10.319,0:07:17.120
imagine we take a box like this
0:07:13.680,0:07:19.120
which contains two halves separated by a
0:07:17.120,0:07:21.360
finite potential barrier
0:07:19.120,0:07:22.800
we take a classical particle such as
0:07:21.360,0:07:26.000
this marble
0:07:22.800,0:07:26.000
place it into one of the halves
0:07:26.400,0:07:31.840
and no matter how much shaking up we
0:07:27.840,0:07:31.840
give it
0:07:34.960,0:07:45.840
we'll always find it in that same half
0:07:47.680,0:07:50.960
imagine now it's a quantum particle
0:07:49.120,0:07:52.879
however and when we
0:07:50.960,0:07:54.319
give it the same shaking up half the
0:07:52.879,0:07:58.650
time we might expect to find it
0:07:54.319,0:08:01.730
in the other half
0:08:04.319,0:08:08.400
depending on whether the outside of the
0:08:06.160,0:08:13.840
box counts as a finite barrier or not
0:08:08.400,0:08:13.840
I might even expect to find
0:08:18.400,0:08:28.400
it's not in the box at all.
0:08:26.319,0:08:28.400
Thank you.
V3.1 The infinite potential well
This channel, Introductory Quantum Mechanics, is a set of videos aimed at second-year physics undergraduates.
This video:
energy eigenfunctions and eigenvalues of the infinite potential well (aka particle in a 1D box).
0:00:00.399,0:00:04.799
hi in this video we're going to take a
0:00:02.480,0:00:07.040
look at the infinite potential well
0:00:04.799,0:00:08.880
also sometimes known as particle in a
0:00:07.040,0:00:12.320
box
0:00:08.880,0:00:15.360
the potential is as follows so
0:00:12.320,0:00:17.760
0 within the region [0,L] along the
0:00:15.360,0:00:19.039
1d line or infinity outside of that
0:00:17.760,0:00:22.080
region
0:00:19.039,0:00:24.080
and let's draw it so
0:00:22.080,0:00:25.840
when we say the potential is infinity
0:00:24.080,0:00:27.840
here and here this really just means
0:00:25.840,0:00:30.560
that the particle can't exist there
0:00:27.840,0:00:32.000
so our boundary conditions take a
0:00:30.560,0:00:34.480
slightly different form this time we
0:00:32.000,0:00:36.480
have that
0:00:34.480,0:00:37.760
the
0:00:36.480,0:00:39.680
wave function at
0:00:37.760,0:00:40.879
position x=0 and at
0:00:39.680,0:00:42.960
position x=L
0:00:40.879,0:00:44.559
must be equal to zero so we're no longer
0:00:42.960,0:00:46.000
looking for the wave function to be
0:00:44.559,0:00:48.800
continuous between the regions
0:00:46.000,0:00:50.640
we use the fact that the wave function
0:00:48.800,0:00:52.079
must vanish in these two regions here
0:00:50.640,0:00:54.160
and so must vanish on the boundaries of
0:00:52.079,0:00:56.160
those regions
0:00:54.160,0:00:57.760
so within the well itself where we want
0:00:56.160,0:00:59.520
to solve for the particle
0:00:57.760,0:01:01.520
we have the potential equal zero it's
0:00:59.520,0:01:03.199
still a constant so we're going to have
0:01:01.520,0:01:05.280
our
0:01:03.199,0:01:06.799
plane wave type solutions but in this
0:01:05.280,0:01:07.760
case we're going to be considering
0:01:06.799,0:01:09.680
standing waves
0:01:07.760,0:01:13.439
rather than travelling waves so the
0:01:09.680,0:01:16.320
solutions look like this
0:01:13.439,0:01:16.720
causing the sign I mean you can apply
0:01:16.320,0:01:18.080
these
0:01:16.720,0:01:19.840
boundary conditions I've labeled them
0:01:18.080,0:01:21.119
both 1 because that's they're kind of
0:01:19.840,0:01:22.640
taking the place of
0:01:21.119,0:01:24.960
what used to be condition 1 let's call
0:01:22.640,0:01:30.799
them condition 1
0:01:24.960,0:01:30.799
and condition 2.
0:01:30.880,0:01:34.000
so applying those boundary conditions to
0:01:33.280,0:01:37.840
this state
0:01:34.000,0:01:40.479
boundary condition 1 tells us
0:01:37.840,0:01:40.880
that phi of zero equals zero so if we
0:01:40.479,0:01:42.720
stick
0:01:40.880,0:01:44.159
x=0 into here this one
0:01:42.720,0:01:48.240
disappears anyway
0:01:44.159,0:01:48.240
and we find that 0=A
0:01:52.560,0:01:55.840
and using condition 2
0:01:57.360,0:02:00.560
we only have B sin(k x) left we
0:02:00.159,0:02:03.040
stick
0:02:00.560,0:02:03.600
x=L into it find it must equal
0:02:03.040,0:02:07.200
zero
0:02:03.600,0:02:10.640
and so we need sin(k L)=0
0:02:07.200,0:02:14.239
and so we find that
0:02:10.640,0:02:18.319
k L = n pi
0:02:14.239,0:02:21.599
where n is any integer
0:02:18.319,0:02:24.879
so let's label those k's by that integer
0:02:21.599,0:02:26.239
n and our solutions for the
0:02:24.879,0:02:30.080
eigenenergies
0:02:26.239,0:02:32.400
are as follows h bar squared k squared
0:02:30.080,0:02:34.640
over 2m but where k is now labeled by n
0:02:32.400,0:02:35.440
and we can substitute this expression
0:02:34.640,0:02:39.280
in here for
0:02:35.440,0:02:41.920
kn so this equals
0:02:39.280,0:02:46.400
hbar^2/2m (n pi /L)^2
0:02:43.200,0:02:48.000
so and let's label the energies by
0:02:46.400,0:02:50.160
that integer n as well
0:02:48.000,0:02:52.480
so we have an infinite tower of
0:02:50.160,0:02:54.480
different energy eigenstates which solve
0:02:52.480,0:02:57.200
the Schroedinger equation in the well
0:02:54.480,0:02:58.879
given by this expression here
0:02:57.200,0:03:00.560
for each there's a corresponding
0:02:58.879,0:03:01.760
eigenfunction so these are the energy
0:03:00.560,0:03:03.519
eigenvalues that solve the time
0:03:01.760,0:03:05.599
independent Schroedinger equation
0:03:03.519,0:03:06.640
and our energy eigenfunctions we've
0:03:05.599,0:03:09.920
solved for
0:03:06.640,0:03:13.360
here let's just rewrite this equation
0:03:09.920,0:03:13.360
taking these two things into account
0:03:13.680,0:03:17.519
so this is our time independent solution
0:03:16.400,0:03:18.000
the solution to the time independent
0:03:17.519,0:03:20.720
Schroedinger
0:03:18.000,0:03:22.560
equation as always we can add back in
0:03:20.720,0:03:23.920
our time dependence very easily for the
0:03:22.560,0:03:26.480
energy eigenvalues
0:03:23.920,0:03:27.360
and in fact if we do that we remember
0:03:26.480,0:03:29.200
that write the time
0:03:27.360,0:03:31.040
dependent wave function as psi rather
0:03:29.200,0:03:33.680
than phi and we can update this as
0:03:31.040,0:03:35.599
follows
0:03:33.680,0:03:37.440
so when we add the time dependence back
0:03:35.599,0:03:40.080
in we can also label our
0:03:37.440,0:03:41.280
energy eigenstates by n our energy
0:03:40.080,0:03:44.000
eigenfunctions
0:03:41.280,0:03:46.319
we have the time independent form
0:03:44.000,0:03:49.440
multiplied by our winding phase factor
0:03:46.319,0:03:49.920
e^(- i E_n t / hbar) where
0:03:49.440,0:03:53.120
E_n
0:03:49.920,0:03:56.400
are the energy eigenvalues okay so let's
0:03:53.120,0:03:56.799
plot those solutions on the next
0:03:56.400,0:04:01.840
board
0:03:56.799,0:04:01.840
so we'll just move this up to the corner
0:04:02.080,0:04:07.840
so for n equals one we have this form
0:04:05.840,0:04:11.040
with plus the modulus of psi one which
0:04:07.840,0:04:14.720
is time independent
0:04:11.040,0:04:15.599
similarly psi two with a node now in the
0:04:14.720,0:04:19.840
center
0:04:15.599,0:04:21.519
psi 3 and so on and each additional
0:04:19.840,0:04:23.520
as n increases we increase the number of
0:04:21.519,0:04:26.800
nodes in the well if we plot
0:04:23.520,0:04:30.400
the potential again over there
0:04:26.800,0:04:33.840
it is conventional to plot these
0:04:30.400,0:04:34.479
energy eigenfunctions on this plot now
0:04:33.840,0:04:37.919
of course
0:04:34.479,0:04:41.280
that's slightly incorrect because the
0:04:37.919,0:04:42.000
y axis here is the potential whereas the
0:04:41.280,0:04:44.639
y-axis over
0:04:42.000,0:04:45.440
here is the modulus of the wave function
0:04:44.639,0:04:48.000
even worse
0:04:45.440,0:04:49.199
we sometimes like to write what we
0:04:48.000,0:04:52.320
should write down as
0:04:49.199,0:04:53.680
psi without the modulus sign on it which
0:04:52.320,0:04:56.960
would in this case come
0:04:53.680,0:04:58.240
through down like this
0:04:56.960,0:04:59.840
but of course that's not really accurate
0:04:58.240,0:05:01.600
because psi without the modulus is a
0:04:59.840,0:05:03.680
complex number and we have this
0:05:01.600,0:05:04.880
complex winding so what you might prefer
0:05:03.680,0:05:06.560
to think
0:05:04.880,0:05:09.039
if we take the example of psi_2 for
0:05:06.560,0:05:10.400
example and rewrite it
0:05:09.039,0:05:11.680
we might write to like to write
0:05:10.400,0:05:13.280
something like this where we've drawn a
0:05:11.680,0:05:16.000
sine
0:05:13.280,0:05:16.560
but as that complex phase winds this can
0:05:16.000,0:05:19.759
of course
0:05:16.560,0:05:21.759
switch through to this solution
0:05:19.759,0:05:24.880
into negative sine but in general
0:05:21.759,0:05:28.240
actually what we really have here is
0:05:24.880,0:05:30.240
the real part of psi plotted here
0:05:28.240,0:05:32.560
and we have another axis coming out of
0:05:30.240,0:05:36.000
the board at 90 degrees which is
0:05:32.560,0:05:37.919
the imaginary part of
0:05:36.000,0:05:39.840
psi_2 in this case and really what
0:05:37.919,0:05:42.479
happens is that this
0:05:39.840,0:05:44.479
original wave function winds around
0:05:42.479,0:05:47.520
the axis in the complex plane
0:05:44.479,0:05:50.560
as it evolves in time okay all that
0:05:47.520,0:05:54.800
taken into account we can plot
0:05:50.560,0:05:59.919
the psi_1 over there
0:05:54.800,0:06:02.400
psi_2 where this is kind of a snapshot
0:05:59.919,0:06:04.800
of the complex wave function where it
0:06:02.400,0:06:07.280
happens to be purely real and
0:06:04.800,0:06:08.319
the positive value if this distance
0:06:07.280,0:06:10.560
here is one
0:06:08.319,0:06:11.919
so this the height of this would be
0:06:10.560,0:06:14.160
E_1
0:06:11.919,0:06:15.600
and height of this one would be E_2 again
0:06:14.160,0:06:17.919
this is all really conventional
0:06:15.600,0:06:19.039
the energies can be separated
0:06:17.919,0:06:20.800
according to potential they have the
0:06:19.039,0:06:23.600
same units
0:06:20.800,0:06:25.280
but the the form of the wave function
0:06:23.600,0:06:28.400
doesn't really fit on this plot
0:06:25.280,0:06:31.199
and finally psi_3
0:06:28.400,0:06:32.479
we're again drawing a snapshot of this
0:06:31.199,0:06:34.240
where happens to be real
0:06:32.479,0:06:36.319
now I'm partly showing this just because
0:06:34.240,0:06:38.479
this is so common to show these things
0:06:36.319,0:06:40.240
that you'll see it in textbooks and on
0:06:38.479,0:06:42.319
wikipedia for example
0:06:40.240,0:06:43.759
but it's slightly sloppy notation but
0:06:42.319,0:06:45.039
hopefully one that's nevertheless quite
0:06:43.759,0:06:47.039
intuitive
0:06:45.039,0:06:48.479
okay so in the next video we're going
0:06:47.039,0:06:48.960
to take a look at some properties of
0:06:48.479,0:06:51.199
these
0:06:48.960,0:06:52.080
different energy eigenfunctions that
0:06:51.199,0:06:54.400
solve the
0:06:52.080,0:06:56.240
problem in particular they're a
0:06:54.400,0:06:57.599
form of what's called a bound state
0:06:56.240,0:07:00.160
so all these states are bound into the
0:06:57.599,0:07:02.000
well there's an infinite number of them
0:07:00.160,0:07:04.479
when we call it a well even though it's
0:07:02.000,0:07:05.280
above zero in terms of potential you can
0:07:04.479,0:07:06.639
think of it as
0:07:05.280,0:07:08.319
well we can just shift down the infinite
0:07:06.639,0:07:09.680
parts down to zero and this is an
0:07:08.319,0:07:10.240
infinitely deep well with an infinite
0:07:09.680,0:07:12.319
number
0:07:10.240,0:07:14.720
of these energy eigenfunctions in it
0:07:12.319,0:07:17.840
each with their own energy eigenvalue
0:07:14.720,0:07:17.840
okay thanks your time
V3.2 Normalisation
This channel, Introductory Quantum Mechanics, is a set of videos aimed at second-year physics undergraduates.
This video:
finding the normalisations of the infinite potential well energy eigenstates.
0:00:00.480,0:00:03.199
hello in this video we're going to take
0:00:02.560,0:00:05.359
a look at
0:00:03.199,0:00:08.480
normalisation of the wave function
0:00:05.359,0:00:12.240
sometimes spelled with the z here
0:00:08.480,0:00:14.160
so we know from previous videos that the
0:00:12.240,0:00:16.400
modulus square of the wave function
0:00:14.160,0:00:17.520
gives us the probability density at a
0:00:16.400,0:00:19.199
point
0:00:17.520,0:00:20.640
and integrated across all of space we
0:00:19.199,0:00:22.240
expect this to give us one
0:00:20.640,0:00:24.160
because although we're not sure where
0:00:22.240,0:00:25.199
the particle is we know that it must
0:00:24.160,0:00:26.640
exist somewhere
0:00:25.199,0:00:30.080
mathematically we can write this down in
0:00:26.640,0:00:31.760
the following way
0:00:30.080,0:00:34.239
which we saw previously leads to the
0:00:31.760,0:00:37.280
conservation of global probability
0:00:34.239,0:00:39.920
let's take a look at a worked example
0:00:37.280,0:00:40.320
treating again the infinite potential
0:00:39.920,0:00:44.160
well
0:00:40.320,0:00:47.520
from video V3.1 where we'll use this
0:00:44.160,0:00:50.719
to establish the pre-factor
0:00:47.520,0:00:50.719
on the wave function
0:00:55.360,0:01:02.559
okay so we've previously seen
0:00:59.199,0:01:04.400
the infinite potential well and
0:01:02.559,0:01:07.040
just to resketch it here
0:01:04.400,0:01:08.720
we have a potential that looks like this
0:01:07.040,0:01:10.960
let's say this is x=0 and this
0:01:08.720,0:01:13.840
is x=L
0:01:10.960,0:01:13.840
this is the potential
0:01:14.479,0:01:17.600
and it's infinity in these two regions
0:01:16.880,0:01:21.439
and zero in
0:01:17.600,0:01:24.840
in the middle here we've seen that the
0:01:21.439,0:01:28.240
energy eigenstates take the form
0:01:24.840,0:01:40.000
phi_n(x)=B sin(n pi x/L)
0:01:36.240,0:01:42.159
which comes about
0:01:40.000,0:01:43.439
from requiring that the wave function
0:01:42.159,0:01:45.360
vanish at
0:01:43.439,0:01:47.920
the points where the potential goes
0:01:45.360,0:01:50.000
to infinity
0:01:47.920,0:01:53.280
so the question is what's this
0:01:50.000,0:01:55.040
coefficient here this pre-factor
0:01:53.280,0:01:57.360
it wasn't determined by the boundary
0:01:55.040,0:02:00.560
conditions but actually we can always
0:01:57.360,0:02:01.119
use the normalisation the fact
0:02:00.560,0:02:03.759
that
0:02:01.119,0:02:05.119
the integral of the probability
0:02:03.759,0:02:08.239
density across all of space
0:02:05.119,0:02:09.840
equals one to solve for the
0:02:08.239,0:02:12.640
prefactor out the front
0:02:09.840,0:02:13.680
so the condition we have is that one has
0:02:12.640,0:02:15.520
to equal
0:02:13.680,0:02:16.879
the integral of the probability density
0:02:15.520,0:02:20.560
which is given by
0:02:16.879,0:02:22.080
|phi_n(x)|^2
0:02:20.560,0:02:24.319
we integrate from minus infinity to
0:02:22.080,0:02:27.280
infinity but the wave function is zero
0:02:24.319,0:02:28.480
everywhere except for [0,L] so we
0:02:27.280,0:02:32.319
can just integrate
0:02:28.480,0:02:35.200
over [0,L]
0:02:32.319,0:02:37.200
and this is enough to solve for B up to
0:02:35.200,0:02:40.400
a global phase
0:02:37.200,0:02:43.200
so to do it we need to use
0:02:40.400,0:02:43.920
well so let's substitute this in first so
0:02:43.200,0:02:47.519
we have
0:02:43.920,0:02:48.959
the integral from 0 to L of |B|^2
0:02:47.519,0:02:50.720
because remember it can be
0:02:48.959,0:02:54.000
complex in general
0:02:50.720,0:02:57.760
sin^2( n pi x/L)
0:02:59.840,0:03:03.680
now to do this integral of
0:03:02.400,0:03:06.640
sin^2
0:03:03.680,0:03:07.360
and in general integrals for problems
0:03:06.640,0:03:09.680
to do with the
0:03:07.360,0:03:11.599
infinite well we're going to use a
0:03:09.680,0:03:14.720
couple of relations so we can use
0:03:11.599,0:03:19.920
cos^2(theta) + sin^2(theta) = 1
0:03:18.400,0:03:23.680
and the other one that tends to be useful is
0:03:19.920,0:03:28.480
cos^2(theta) - sin^2(theta) = cos(2theta)
0:03:29.120,0:03:32.640
so we have a sine squared so we want to
0:03:30.720,0:03:33.599
take this one minus this one and divide
0:03:32.640,0:03:35.599
by two
0:03:33.599,0:03:37.519
so we have that one equals we can bring
0:03:35.599,0:03:40.159
the B squared out of the integral
0:03:37.519,0:03:42.720
and actually we're going to have a
0:03:40.159,0:03:42.720
half as well
0:03:43.760,0:03:47.680
so i'm just going to take this equation
0:03:44.959,0:03:50.080
the top one and subtract this equation
0:03:47.680,0:03:51.280
and then divide by 2 to get sine squared
0:03:50.080,0:03:56.159
so we're going to have
0:03:51.280,0:03:56.159
1 - cos(2 theta)
0:03:59.200,0:04:02.239
so we have one
0:04:01.519,0:04:05.920
equals
0:04:02.239,0:04:08.959
|B|^2/2
0:04:05.920,0:04:08.959
this one just gives us L
0:04:09.040,0:04:14.080
and then we have minus integral
0:04:12.000,0:04:17.040
|B|^2 / 2
0:04:14.080,0:04:17.759
integrate this and we have sine sorry
0:04:17.040,0:04:22.880
i've put
0:04:17.759,0:04:22.880
theta here so sorry 2 theta
0:04:25.280,0:04:30.240
is equal to 2 n pi x / L
0:04:31.360,0:04:37.680
so this is going to integrate to
0:04:34.639,0:04:42.479
sin(2 theta) which is sin(2 n pi x/L)
0:04:37.680,0:04:46.880
and we need to divide by
0:04:42.479,0:04:49.440
2 and pi and multiply by L
0:04:46.880,0:04:51.040
and stick in the limit sorry to L but we
0:04:49.440,0:04:53.520
see that
0:04:51.040,0:04:54.160
when we put the limit 0 in
0:04:53.520,0:04:57.120
there that's
0:04:54.160,0:04:57.759
0 because sin(0)=0 we put
0:04:57.120,0:05:01.280
the L in
0:04:57.759,0:05:04.720
we get 2 n pi and the L to cancel
0:05:01.280,0:05:07.360
but 2 n pi for integer n
0:05:04.720,0:05:08.240
sin( 2 n pi ) is always zero so
0:05:07.360,0:05:10.320
actually this thing
0:05:08.240,0:05:11.919
is always equal to zero and so we've
0:05:10.320,0:05:15.199
solved and found
0:05:11.919,0:05:18.880
that |B|^2
0:05:15.199,0:05:23.759
is equal to 2/L or
0:05:18.880,0:05:26.880
|B| = sqrt(2/L)
0:05:23.759,0:05:30.880
and therefore we found our normalization
0:05:26.880,0:05:32.479
our energy eigenstates for
0:05:30.880,0:05:35.520
the infinite potential well
0:05:32.479,0:05:38.880
are given by
0:05:35.520,0:05:41.919
sqrt(2/L) sin( n pi x/L)
0:05:38.880,0:05:43.840
and now the probability to find
0:05:41.919,0:05:46.400
any particle described by this wave
0:05:43.840,0:05:50.720
function across all of space
0:05:46.400,0:05:50.720
is equal to one which is what we like
0:05:51.120,0:05:56.960
okay so thank you for your time
V3.3 Stationary states
This channel, Introductory Quantum Mechanics, is a set of videos aimed at second-year physics undergraduates.
This video:
showing that energy eigenstates have time-independent probability densities.
0:00:02.639,0:00:05.040
hello
0:00:03.600,0:00:06.960
in this rather brief video we're going
0:00:05.040,0:00:08.639
to take a look at stationary states
0:00:06.960,0:00:11.280
when we write down a solution to our
0:00:08.639,0:00:14.160
time dependent Schroedinger equation
0:00:11.280,0:00:15.200
psi(x,t) we're always
0:00:14.160,0:00:17.680
bearing in mind
0:00:15.200,0:00:19.920
that we're writing this in a separable
0:00:17.680,0:00:24.000
form
0:00:19.920,0:00:27.039
phi(x) multiplied by T(t)
0:00:24.000,0:00:29.359
we saw in the first lecture
0:00:27.039,0:00:31.279
the first set of videos that we can
0:00:29.359,0:00:32.000
always solve for T(t) and find that it
0:00:31.279,0:00:34.399
gives us
0:00:32.000,0:00:35.120
a complex phase winding since we're
0:00:34.399,0:00:36.880
looking now
0:00:35.120,0:00:38.719
at bound states we can label our
0:00:36.880,0:00:40.399
eigenstates by an integer
0:00:38.719,0:00:41.760
n labeling the states there's an
0:00:40.399,0:00:44.000
infinite number of them in the
0:00:41.760,0:00:45.280
infinite potential well and then our
0:00:44.000,0:00:47.520
general solution takes the following
0:00:45.280,0:00:47.520
form
0:00:47.760,0:00:51.680
so we can label our wave function psi by
0:00:50.480,0:00:54.640
the integer n
0:00:51.680,0:00:55.440
phi is labeled by the same n and our
0:00:54.640,0:01:00.160
phase winding
0:00:55.440,0:01:03.199
is given by the energy eigenvalue E_n
0:01:00.160,0:01:05.600
so if we look at the probability density
0:01:03.199,0:01:05.600
for this
0:01:05.920,0:01:11.360
given by |psi|^2 we
0:01:09.040,0:01:14.479
find that for these energy
0:01:11.360,0:01:16.799
states we use the complex phase and we
0:01:14.479,0:01:19.759
simply find that
0:01:16.799,0:01:20.960
we have the modulus square of the time
0:01:19.759,0:01:23.040
independent part
0:01:20.960,0:01:24.080
so this is why we say that energy
0:01:23.040,0:01:25.920
eigenstates
0:01:24.080,0:01:28.000
are what are called stationary states
0:01:25.920,0:01:31.040
the probability density for them is
0:01:28.000,0:01:32.880
constant in time
0:01:31.040,0:01:34.159
and when we say energy eigenstates this
0:01:32.880,0:01:36.320
is synonymous with
0:01:34.159,0:01:38.159
energy eigenfunctions this statement
0:01:36.320,0:01:39.200
holds completely generally it's not just
0:01:38.159,0:01:41.680
for bound states
0:01:39.200,0:01:43.119
energy eigenstates are always stationary
0:01:41.680,0:01:46.320
states
0:01:43.119,0:01:47.360
as you can derive from the substitution
0:01:46.320,0:01:49.759
of this ansatz
0:01:47.360,0:01:53.200
into your Schroedinger equation okay
0:01:49.759,0:01:53.200
thank you for your time
V3.4 Orthonormality of eigenstates
This channel, Introductory Quantum Mechanics, is a set of videos aimed at second-year physics undergraduates.
This video:
showing that the energy eigenstates of the infinite potential well are orthogonal and normalised (orthonormal).
0:00:00.399,0:00:04.480
hello in this video we're going to take
0:00:02.240,0:00:07.759
a look at the orthonormality
0:00:04.480,0:00:11.200
of different energy eigenstates so
0:00:07.759,0:00:14.920
we have from the previous video that all
0:00:11.200,0:00:16.720
physical states in quantum mechanics are
0:00:14.920,0:00:20.320
normalized
0:00:16.720,0:00:23.439
but in addition for any two different
0:00:20.320,0:00:24.480
energy eigenstates of the infinite
0:00:23.439,0:00:26.560
potential well
0:00:24.480,0:00:29.199
are also also orthogonal to each other
0:00:26.560,0:00:32.399
by which we mean the following
0:00:29.199,0:00:34.480
the integral of phi_n* . phi_m
0:00:32.399,0:00:36.160
over all of space in this case over
0:00:34.480,0:00:40.559
the range of the well
0:00:36.160,0:00:40.559
is equal to zero if n doesn't equal m
0:00:40.719,0:00:44.000
together these conditions tell us
0:00:42.559,0:00:46.640
that the states are
0:00:44.000,0:00:48.239
both normalized and orthogonal and we
0:00:46.640,0:00:51.520
abbreviate this to 'orthonormal'
0:00:48.239,0:00:51.520
so the statement is as follows
0:00:51.760,0:00:55.600
so the integral of phi_n* phi_m over
0:00:54.640,0:00:57.600
all of space
0:00:55.600,0:00:59.840
is equal to the kronecker delta which is
0:00:57.600,0:01:02.399
defined to be 1 if n=m
0:00:59.840,0:01:04.159
and 0 if n doesn't equal n. This is an
0:01:02.399,0:01:08.080
incredibly useful relation
0:01:04.159,0:01:11.119
which we'll put to much use in the coming
0:01:08.080,0:01:12.960
videos it doesn't just hold for energy
0:01:11.119,0:01:13.840
eigenstates of the infinite potential
0:01:12.960,0:01:16.320
well it holds
0:01:13.840,0:01:17.520
for a much broader class of problems and
0:01:16.320,0:01:19.759
we'll see much more of that
0:01:17.520,0:01:21.360
when we come to look at matrix mechanics
0:01:19.759,0:01:23.200
later on in the course
0:01:21.360,0:01:25.360
for now let's look at a worked example
0:01:23.200,0:01:26.960
where we show this explicitly
0:01:25.360,0:01:30.000
for the eigenstates of the
0:01:26.960,0:01:30.000
infinite potential well
0:01:32.880,0:01:35.759
who's a good boy
0:01:37.119,0:01:40.159
I've got a cameo from Geoffrey in the
0:01:38.840,0:01:42.880
background
0:01:40.159,0:01:44.720
all right so let's take a look at
0:01:42.880,0:01:46.479
the energy eigenstates of the infinite
0:01:44.720,0:01:48.399
potential well
0:01:46.479,0:01:50.320
another quick reminder as to what the
0:01:48.399,0:01:54.880
potential looks like
0:01:50.320,0:02:00.960
we have potential going from 0 to L
0:01:54.880,0:02:00.960
along x up to infinity here
0:02:04.240,0:02:07.360
and 0 within the well and we've seen
0:02:06.799,0:02:10.520
that
0:02:07.360,0:02:14.239
the normalized
0:02:10.520,0:02:14.239
eigenfunctions look like this
0:02:20.720,0:02:23.840
and this time we'd like to prove that
0:02:22.480,0:02:26.480
two
0:02:23.840,0:02:27.120
eigenfunctions are orthogonal to each
0:02:26.480,0:02:29.599
other
0:02:27.120,0:02:31.440
so what this means is that
0:02:29.599,0:02:32.000
the integral from minus infinity to infinity
0:02:31.440,0:02:42.080
of phi_n(x)*.phi_m(x) dx
0:02:38.160,0:02:45.200
equals delta_{nm}
0:02:42.080,0:02:48.560
which by definition equals
0:02:45.200,0:02:51.840
1 if n=m 0 if
0:02:48.560,0:02:51.840
n doesn't equal m
0:02:52.160,0:02:55.440
orthogonal of course implies
0:02:54.400,0:02:56.800
something
0:02:55.440,0:02:58.319
that there's some similarity to two
0:02:56.800,0:02:59.920
vectors being at 90 degrees and we'll
0:02:58.319,0:03:01.519
see when we study matrix mechanics that
0:02:59.920,0:03:05.200
there's actually a very close analogy
0:03:01.519,0:03:08.000
to that idea so let's
0:03:05.200,0:03:08.640
show this for the the eigenstates
0:03:08.000,0:03:10.720
of the
0:03:08.640,0:03:11.920
infinite well let's just substitute this
0:03:10.720,0:03:14.959
expression into here
0:03:11.920,0:03:17.040
so we get a 2/L out the front
0:03:14.959,0:03:19.760
the integral only goes from zero to L
0:03:17.040,0:03:22.959
because phi is 0 outside of that range
0:03:19.760,0:03:32.959
sin(n pi x/L)sin(m pi x/L)dx
0:03:30.959,0:03:34.720
okay we're going to need to use another
0:03:32.959,0:03:37.760
trigonometric identity
0:03:34.720,0:03:41.440
this time we need to use that
0:03:37.760,0:03:51.760
cos(A+B)=cos(A)cos(B)-sin(A)sin(B)
0:03:48.480,0:03:55.200
and therefore
0:03:51.760,0:04:04.720
cos(A-B)=cos(A)cos(B)+sin(A)sin(B)
0:04:06.080,0:04:10.080
okay so if we put A = n pi x/L
0:04:09.840,0:04:13.280
B = m pi x/L
0:04:10.080,0:04:16.160
we want to
0:04:13.280,0:04:17.680
add this one sorry subtract this one
0:04:16.160,0:04:18.880
from this one and that'll give us the
0:04:17.680,0:04:21.280
two sines
0:04:18.880,0:04:23.840
sorry you can't quite see that so we
0:04:21.280,0:04:25.360
want to subtract this expression from
0:04:23.840,0:04:28.800
this expression that will give us
0:04:25.360,0:04:32.240
2sin(A)sin(B) over here
0:04:28.800,0:04:33.520
and so overall we'll have so we'll
0:04:32.240,0:04:35.840
have a factor of two we bring out the
0:04:33.520,0:04:39.360
front so we have 1/L
0:04:35.840,0:04:42.639
integral from zero to L
0:04:39.360,0:04:48.000
cos(A-B) so it's
0:04:42.639,0:04:48.000
cos((n-m) pi x/L)
0:04:48.160,0:04:57.840
minus cos((n+m)pi x/L)
0:05:00.160,0:05:03.280
this integrates to
0:05:03.680,0:05:10.560
these will become sines
0:05:07.360,0:05:13.680
sin((n-m)pi x /L)/((n-m)pi)
0:05:13.680,0:05:18.920
and multiplied by L; minus
0:05:17.120,0:05:28.880
sin((n+m)pi x/L)/((n+m)pi)
0:05:24.720,0:05:32.160
multiply by L
0:05:28.880,0:05:33.199
between zero and L okay well the Ls
0:05:32.160,0:05:36.720
cancel so that's good
0:05:33.199,0:05:38.000
news when we stick in
0:05:36.720,0:05:40.479
sin(0)=0 so the zero
0:05:38.000,0:05:43.039
limit is always zero when we substitute
0:05:40.479,0:05:43.039
the L in
0:05:43.280,0:05:47.120
this one is always going to be zero
0:05:44.960,0:05:48.479
because n+m
0:05:47.120,0:05:50.400
for integer n and m is always an
0:05:48.479,0:05:52.400
integer as the sum of two integers is an
0:05:50.400,0:05:55.360
integer
0:05:52.400,0:05:56.479
and so this is ... substitute in
0:05:55.360,0:05:58.400
here these two cancel
0:05:56.479,0:06:00.080
so it's an integer times pi. Sine of that
0:05:58.400,0:06:03.199
is always zero so this one
0:06:00.080,0:06:04.080
disappears this one also almost
0:06:03.199,0:06:06.160
disappears
0:06:04.080,0:06:10.080
the zero limit disappears when we put L
0:06:06.160,0:06:12.560
in so we have sin((n-m)pi)
0:06:10.080,0:06:14.000
well n-m is also an integer for
0:06:12.560,0:06:16.240
integer n and m
0:06:14.000,0:06:18.319
the only problem is when n equals m this
0:06:16.240,0:06:21.280
is zero. sin(0)=0 -- good --
0:06:18.319,0:06:23.600
but n-m is also 0 on the bottom
0:06:21.280,0:06:25.199
and 0 divided by 0 is undefined
0:06:23.600,0:06:28.080
so let's substitute it in the only one
0:06:25.199,0:06:31.919
we need to worry about
0:06:28.080,0:06:38.639
sin((n-m)pi)/((n-m)pi)
0:06:36.800,0:06:40.400
so to work out what that is we use
0:06:38.639,0:06:41.280
l'Hopital's rule differentiate the top
0:06:40.400,0:06:42.960
and bottom
0:06:41.280,0:06:45.440
and we can differentiate them with
0:06:42.960,0:06:49.599
respect to, say, n-m
0:06:45.440,0:06:53.520
so this thing must equal
0:06:49.599,0:06:57.919
as we multiply this we get
0:06:53.520,0:06:57.919
cos((n-m)pi)
0:06:58.000,0:07:01.280
differentiate with respect to n-m
0:06:59.680,0:07:03.919
and we just get a pi on the bottom
0:07:01.280,0:07:04.560
cancel those this is evaluated at n
0:07:03.919,0:07:08.240
equals
0:07:04.560,0:07:10.319
m and so this thing equals one
0:07:08.240,0:07:12.319
and so we found that if we'd put
0:07:10.319,0:07:14.080
anything else anything other than n
0:07:12.319,0:07:15.520
equals m into here it would have
0:07:14.080,0:07:17.039
disappeared because if n
0:07:15.520,0:07:18.560
and m are different integers this is
0:07:17.039,0:07:21.280
non-zero this is zero
0:07:18.560,0:07:22.880
the whole thing is zero if n equals m it
0:07:21.280,0:07:23.919
evaluates to one and that's precisely
0:07:22.880,0:07:26.240
what we wanted to show
0:07:23.919,0:07:27.680
we've shown the orthogonality of
0:07:26.240,0:07:29.120
these eigenstates and in fact we've
0:07:27.680,0:07:31.440
shown that they're orthonormal because
0:07:29.120,0:07:35.520
we've already normalized the eigenstates
0:07:31.440,0:07:37.759
so the normalized eigenstates
0:07:35.520,0:07:39.840
of the states within the infinite
0:07:37.759,0:07:42.080
well
0:07:39.840,0:07:44.720
are orthonormal okay thanks for your
0:07:42.080,0:07:44.720
time
V3.5 Fourier decomposition
This channel, Introductory Quantum Mechanics, is a set of videos aimed at second-year physics undergraduates.
This video:
decomposing arbitrary functions (which match the boundary conditions) into weighted sums of energy eigenstates of the infinite potential well. Deriving the time dependence of arbitrary wave functions prepared in the well.
0:00:00.240,0:00:03.120
hello in this video we're going to take
0:00:02.560,0:00:06.160
a look
0:00:03.120,0:00:08.480
at fourier decomposition of functions
0:00:06.160,0:00:10.000
in the particular case of the infinite
0:00:08.480,0:00:13.280
potential well
0:00:10.000,0:00:16.400
so we showed in a previous video that
0:00:13.280,0:00:18.400
the eigenstates in the infinite
0:00:16.400,0:00:21.760
potential well the energy eigenstates
0:00:18.400,0:00:22.880
are all orthonormal in fact there's
0:00:21.760,0:00:24.800
a stronger condition
0:00:22.880,0:00:26.240
the energy eigenstates the infinite
0:00:24.800,0:00:30.240
potential well form what's called a
0:00:26.240,0:00:32.079
complete orthonormal basis
0:00:30.240,0:00:33.920
what we mean by this is that any
0:00:32.079,0:00:36.239
arbitrary function
0:00:33.920,0:00:36.960
of position within the well can be
0:00:36.239,0:00:39.280
written
0:00:36.960,0:00:40.399
as a sum of different energy
0:00:39.280,0:00:42.879
eigenstates
0:00:40.399,0:00:44.000
just clear the board okay so
0:00:42.879,0:00:48.320
mathematically what we're saying
0:00:44.000,0:00:50.800
is this: an arbitrary function
0:00:48.320,0:00:52.160
f(x) can be written as a sum from n
0:00:50.800,0:00:54.640
equals one to infinity
0:00:52.160,0:00:56.000
of phi_n(x) our energy eigenstates in the
0:00:54.640,0:00:57.920
infinite potential well
0:00:56.000,0:00:59.039
multiplied by potentially complex
0:00:57.920,0:01:02.239
coefficients
0:00:59.039,0:01:05.360
f_n. It's a form of Fourier analysis
0:01:02.239,0:01:07.280
in fact again it's not just
0:01:05.360,0:01:08.479
eigenstates for the infinite
0:01:07.280,0:01:10.240
potential well this
0:01:08.479,0:01:11.600
occurs in a much broader class of
0:01:10.240,0:01:13.280
cases in
0:01:11.600,0:01:15.600
quantum mechanics and we'll take a
0:01:13.280,0:01:17.439
closer look at the cases it applies to
0:01:15.600,0:01:19.600
when we come to matrix mechanics later
0:01:17.439,0:01:22.000
on
0:01:19.600,0:01:23.119
for now you can think of it in a very
0:01:22.000,0:01:26.080
close analogy
0:01:23.119,0:01:27.439
to expanding an arbitrary vector in
0:01:26.080,0:01:29.360
an n-dimensional vector space
0:01:27.439,0:01:30.640
in terms of the n basis vectors in the
0:01:29.360,0:01:33.040
space
0:01:30.640,0:01:33.840
we'll see that that analogy is indeed
0:01:33.040,0:01:36.640
very close
0:01:33.840,0:01:38.240
later on in the course for now in order
0:01:36.640,0:01:40.159
to make this useful we need a method of
0:01:38.240,0:01:42.320
solving for these complex coefficients
0:01:40.159,0:01:43.840
f_n and we can do that quite simply
0:01:42.320,0:01:46.399
using the orthonormality
0:01:43.840,0:01:47.040
of the energy eigenstates so we can
0:01:46.399,0:01:49.840
write the
0:01:47.040,0:01:50.159
following we can multiply from the left
0:01:49.840,0:01:53.200
by
0:01:50.159,0:01:54.960
phi_m*(x) and then integrate
0:01:53.200,0:01:56.640
from minus infinity to infinity in fact
0:01:54.960,0:01:59.680
this will only go over the
0:01:56.640,0:02:00.880
well because the phi_m are zero
0:01:59.680,0:02:02.159
outside as well
0:02:00.880,0:02:04.799
and when we do this on the right hand
0:02:02.159,0:02:04.799
side we have
0:02:04.880,0:02:08.239
the following so we've multiplied by
0:02:07.439,0:02:09.599
phi_m*(x)
0:02:08.239,0:02:11.360
in from the left and then we've
0:02:09.599,0:02:14.080
integrated dx
0:02:11.360,0:02:15.760
the phi_m*(x) can pass through the sum
0:02:14.080,0:02:16.480
and the integral dx can also pass
0:02:15.760,0:02:18.319
through the sum
0:02:16.480,0:02:20.400
that's because sums and integrals
0:02:18.319,0:02:22.879
commute
0:02:20.400,0:02:24.560
f_n are coefficients to be determined but
0:02:22.879,0:02:25.920
they're not a function of x so they can
0:02:24.560,0:02:29.680
come outside the integral
0:02:25.920,0:02:29.680
and so this quantity here
0:02:29.840,0:02:36.000
from our orthonormality condition is
0:02:32.480,0:02:39.040
just the kronecker delta
0:02:36.000,0:02:41.680
which is one if m equals n
0:02:39.040,0:02:42.480
and zero otherwise so when we sum over
0:02:41.680,0:02:44.319
n
0:02:42.480,0:02:45.760
the kronecker delta selects out the
0:02:44.319,0:02:48.640
case that n equals m
0:02:45.760,0:02:50.319
every other term is zero and so this
0:02:48.640,0:02:52.319
expression over here reduces to the
0:02:50.319,0:02:56.160
following
0:02:52.319,0:02:58.480
that is just f_m so
0:02:56.160,0:02:59.200
overall then if we just switch let's
0:02:58.480,0:03:01.040
swap these
0:02:59.200,0:03:02.480
the sides of these two things and switch
0:03:01.040,0:03:05.760
the label m to n
0:03:02.480,0:03:08.000
and we have the result
0:03:05.760,0:03:09.680
so we can write any arbitrary function
0:03:08.000,0:03:12.560
f(x) which
0:03:09.680,0:03:14.239
lies within the same
0:03:12.560,0:03:14.800
boundaries as the infinite potential
0:03:14.239,0:03:18.080
well
0:03:14.800,0:03:19.760
as a decomposition of energy eigenstates
0:03:18.080,0:03:21.599
in the potential well and the
0:03:19.760,0:03:23.519
coefficients we use in that expansion
0:03:21.599,0:03:25.040
we can determine from this simple
0:03:23.519,0:03:28.239
formula here
0:03:25.040,0:03:29.920
so this is already very useful it's
0:03:28.239,0:03:32.319
as useful as Fourier decomposition.
0:03:32.319,0:03:35.519
A particularly important use of this is
0:03:35.040,0:03:37.840
that
0:03:35.519,0:03:38.560
we can specify some starting wave
0:03:37.840,0:03:40.080
function
0:03:38.560,0:03:42.080
so it doesn't need to be an eigenstate
0:03:40.080,0:03:43.840
say we prepare a state
0:03:42.080,0:03:45.680
which is say a position eigenstate in
0:03:43.840,0:03:48.400
the well we identify the particle
0:03:45.680,0:03:48.879
at a particular position or we can put
0:03:48.879,0:03:52.640
in general quite different wave
0:03:51.599,0:03:54.000
functions in the well
0:03:52.640,0:03:56.799
depending on what types of measurements
0:03:54.000,0:03:59.280
we made and so on if we prepare them
0:03:56.799,0:04:00.560
or we can prepare
0:03:59.280,0:04:02.879
them by specifying
0:04:00.560,0:04:03.599
the amplitude at each point in the
0:04:02.879,0:04:05.840
well
0:04:03.599,0:04:06.720
but then we'd like to know how the wave
0:04:05.840,0:04:09.599
function
0:04:06.720,0:04:11.040
varies with time after that now the
0:04:09.599,0:04:12.879
time-dependent Schroedinger equation should
0:04:11.040,0:04:14.400
tell us the time evolution of any state
0:04:12.879,0:04:16.479
not just energy eigenstates
0:04:14.400,0:04:17.440
and this is how it does it because
0:04:16.479,0:04:20.479
remember
0:04:17.440,0:04:21.519
the time evolution of the energy
0:04:20.479,0:04:24.240
eigenstates
0:04:21.519,0:04:24.639
is trivial we know how to solve this and
0:04:24.240,0:04:26.800
so
0:04:24.639,0:04:28.080
we apply the same reasoning here we find
0:04:26.800,0:04:31.280
that starting off
0:04:28.080,0:04:33.840
in f(x) at time t=0
0:04:31.280,0:04:36.080
the result for f(x,t) at later times is
0:04:33.840,0:04:38.880
as follows
0:04:36.080,0:04:40.160
okay so specifying the wave function
0:04:38.880,0:04:41.840
at an initial instant
0:04:40.160,0:04:44.080
we know its behavior for all future
0:04:41.840,0:04:45.600
times because it must obey the time
0:04:44.080,0:04:47.919
dependent Schroedinger equation so it's
0:04:45.600,0:04:49.600
dictated
0:04:47.919,0:04:50.960
that is provided no measurement is made
0:04:49.600,0:04:54.800
to the system measurement
0:04:50.960,0:04:58.080
is very strange as we'll see shortly
0:04:54.800,0:05:00.880
so we specify our arbitrary
0:04:58.080,0:05:01.680
function of position its time evolution
0:05:00.880,0:05:03.759
we work out
0:05:01.680,0:05:05.840
by decomposing it into energy
0:05:03.759,0:05:07.600
eigenstates each of which
0:05:05.840,0:05:11.440
we know the time dependence of and it
0:05:07.600,0:05:13.280
takes this trivial form
0:05:11.440,0:05:15.840
the time dependence of the state itself
0:05:13.280,0:05:17.280
need no longer be trivial
0:05:15.840,0:05:18.880
because we're summing up different
0:05:17.280,0:05:21.120
energy eigenstates here
0:05:18.880,0:05:22.960
in fact this is a general expression of
0:05:21.120,0:05:23.520
quantum superposition which we'll take a
0:05:22.960,0:05:27.680
look at
0:05:23.520,0:05:27.680
in the next video thanks for your time
V4.1 Quantum superposition
This channel, Introductory Quantum Mechanics, is a set of videos aimed at second-year physics undergraduates.
This video:
linear combinations (superpositions) of solutions to the SchrÃ¶dinger equation are also solutions; superpositions of stationary states need not themselves be stationary states; normalisation of superposed states; finding the probability of a superposition to give a particular outcome.
0:00:01.520,0:00:05.759
hello in this video we're going to take
0:00:03.360,0:00:08.880
a look at quantum superposition
0:00:05.759,0:00:12.800
the central story of the video is going
0:00:08.880,0:00:12.800
to be that quantum mechanics is linear
0:00:12.880,0:00:16.160
what we mean by that is that if two wave
0:00:15.599,0:00:18.800
functions
0:00:16.160,0:00:19.600
psi one and psi two which are independently
0:00:18.800,0:00:22.720
solutions
0:00:19.600,0:00:25.279
to the time-dependent Schroedinger equation
0:00:22.720,0:00:27.279
then any linear combination of them
0:00:25.279,0:00:29.279
is also a solution
0:00:27.279,0:00:30.560
that is wave functions of a linear
0:00:29.279,0:00:33.040
superposition
0:00:30.560,0:00:34.000
in general as we saw in a previous video
0:00:33.040,0:00:36.800
we can
0:00:34.000,0:00:36.800
write the following
0:00:37.360,0:00:41.600
if we can write some arbitrary wave
0:00:39.600,0:00:44.719
function down
0:00:41.600,0:00:45.760
we can decompose it in terms of energy
0:00:44.719,0:00:49.280
eigenstates
0:00:45.760,0:00:51.360
psi_n and
0:00:49.280,0:00:53.120
any linear combination like this will
0:00:51.360,0:00:55.280
also be a solution to the schrodinger
0:00:53.120,0:00:57.680
equation so it's not just a sum of two
0:00:55.280,0:00:59.199
it's a sum of any number of wave
0:00:57.680,0:01:02.800
functions which individually solve
0:00:59.199,0:01:02.800
the time dependent Schroedinger equation
0:01:02.879,0:01:06.640
let's clear the board just move this up
0:01:04.239,0:01:06.640
to the top
0:01:07.760,0:01:11.840
and just to reiterate if psi_n is a
0:01:10.720,0:01:12.960
solution to the time-dependent
0:01:11.840,0:01:14.960
Schroedinger equation
0:01:12.960,0:01:17.439
then any arbitrary function which is
0:01:14.960,0:01:20.400
written as a superposition a sum
0:01:17.439,0:01:21.200
of different psi_n weighted by
0:01:20.400,0:01:24.000
coefficients
0:01:21.200,0:01:24.479
f_n which are in general complex this is
0:01:24.000,0:01:29.040
also
0:01:24.479,0:01:31.200
a solution these are energy eigenstates
0:01:29.040,0:01:32.240
that we've been dealing with so far if
0:01:31.200,0:01:37.360
we perform
0:01:32.240,0:01:37.360
a measurement on f(x,t)
0:01:37.759,0:01:42.000
a measurement of energy this is where
0:01:40.960,0:01:42.960
things get a little bit strange in
0:01:42.000,0:01:45.600
quantum mechanics
0:01:42.960,0:01:48.159
we will always find exactly one of
0:01:45.600,0:01:51.200
the energy eigenstates
0:01:48.159,0:01:52.479
so let's write that down. A measurement
0:01:51.200,0:01:55.600
of the energy of
0:01:52.479,0:01:57.840
f(x,t) will reveal one energy E_n
0:01:55.600,0:01:59.439
so even though we prepare the state in a
0:01:57.840,0:02:01.360
superposition of different energy
0:01:59.439,0:02:03.280
eigenstates when we make a measurement
0:02:01.360,0:02:04.960
of the energy we only find one of the
0:02:03.280,0:02:08.080
energies
0:02:04.960,0:02:09.759
and as a result of that the wave
0:02:08.080,0:02:10.399
function will change as a result of that
0:02:09.759,0:02:12.879
measurement
0:02:10.399,0:02:14.640
it'll no longer be f(x,t) it will
0:02:12.879,0:02:18.560
then be the energy eigenstate
0:02:14.640,0:02:18.560
corresponding to that eigenenergy
0:02:19.120,0:02:22.800
after measuring eigenenergy eigenvalue
0:02:22.560,0:02:27.120
e_n
0:02:22.800,0:02:30.239
the state is psi_n(x,t)
0:02:27.120,0:02:30.239
and that's with certainty
0:02:30.560,0:02:34.160
whether you want to say that the state
0:02:32.000,0:02:36.000
has changed depends on your
0:02:34.160,0:02:38.080
interpretation of quantum mechanics
0:02:36.000,0:02:39.680
in the standard interpretation we tend
0:02:38.080,0:02:40.959
to teach at university, the Copenhagen
0:02:39.680,0:02:42.560
interpretation,
0:02:40.959,0:02:45.040
this process is called
0:02:42.560,0:02:45.040
'wave function collapse'
0:02:45.440,0:02:49.840
but this is an interpretational question
0:02:47.840,0:02:52.480
in the many worlds theory for example
0:02:49.840,0:02:53.680
wavefunction collapse does not exist
0:02:52.480,0:02:56.319
there is a different process for
0:02:53.680,0:02:59.440
explaining how a state prepared
0:02:56.319,0:03:02.080
in f(x,t) can, when measured
0:02:59.440,0:03:04.319
according to its energy, change into
0:03:02.080,0:03:08.560
psi_n(x,t) in fact it doesn't change
0:03:04.319,0:03:11.280
it just appears to change to us so
0:03:08.560,0:03:12.800
the probability when performing an
0:03:11.280,0:03:15.920
energy measurement on f
0:03:12.800,0:03:16.800
to find energy E_n and for the state
0:03:15.920,0:03:19.280
subsequently to be
0:03:16.800,0:03:20.480
psi_n is given by the modulus square of
0:03:19.280,0:03:22.879
the coefficient
0:03:20.480,0:03:23.519
assuming this is correctly normalized
0:03:22.879,0:03:25.040
which all
0:03:23.519,0:03:27.200
physical states in quantum mechanics
0:03:25.040,0:03:30.000
are
0:03:27.200,0:03:32.560
the probability to find the result E_n
0:03:30.000,0:03:35.680
in an energy measurement of f(x,t)
0:03:32.560,0:03:36.239
|f_n|^2 before the
0:03:35.680,0:03:38.640
measurement
0:03:36.239,0:03:39.280
after the measurement it will be in
0:03:38.640,0:03:40.799
state psi_n
0:03:39.280,0:03:42.799
and it'll have energy E_n with
0:03:40.799,0:03:43.840
probability one so the state really has
0:03:42.799,0:03:45.360
changed
0:03:43.840,0:03:47.200
we'll look at some of the more
0:03:45.360,0:03:48.159
philosophical interpretations
0:03:47.200,0:03:51.360
surrounding this
0:03:48.159,0:03:54.239
in a separate video for now let's
0:03:51.360,0:03:55.680
take a look at what this means for the
0:03:54.239,0:03:58.480
time dependence of states
0:03:55.680,0:04:01.599
let's just clear the board so let's take
0:03:58.480,0:04:04.799
a look at the time dependence
0:04:01.599,0:04:06.560
in the absence of measurement the
0:04:04.799,0:04:08.159
time dependence for state is dictated
0:04:06.560,0:04:09.120
entirely by the time dependent
0:04:08.159,0:04:11.680
Schroedinger equation
0:04:09.120,0:04:13.040
so we have that the state psi evolves
0:04:11.680,0:04:14.799
unitarily
0:04:13.040,0:04:18.400
according to the time-dependent Schroedinger
0:04:14.799,0:04:18.400
equation in the absence of measurement
0:04:18.720,0:04:23.199
the word unitarily here we'll see in
0:04:21.600,0:04:24.560
more detail what this means later on in
0:04:23.199,0:04:27.120
the course
0:04:24.560,0:04:28.960
but for now all you need to know is that
0:04:27.120,0:04:30.960
a unitary evolution of the wave function
0:04:28.960,0:04:32.160
simply preserves the normalization if
0:04:30.960,0:04:33.520
you start with a normalized wave
0:04:32.160,0:04:34.160
function which you must for a physical
0:04:33.520,0:04:36.720
state
0:04:34.160,0:04:37.280
it remains normalized for all subsequent
0:04:36.720,0:04:39.440
times
0:04:37.280,0:04:41.280
and this is built into the schrodinger
0:04:39.440,0:04:42.320
equation
0:04:41.280,0:04:44.240
okay so we'd like to look at the time
0:04:42.320,0:04:45.759
dependence of states remember if we have
0:04:44.240,0:04:47.440
an energy eigenstate
0:04:45.759,0:04:50.960
the probability density of that
0:04:47.440,0:04:53.280
eigenstate is time independent
0:04:50.960,0:04:55.680
so if this is our energy eigenstate
0:04:53.280,0:04:59.680
labeled with subscript n
0:04:55.680,0:05:03.280
then the probability density is this
0:04:59.680,0:05:04.880
so rho_n is modulus square of psi n
0:05:03.280,0:05:06.800
this is a function of time but the
0:05:04.880,0:05:09.039
result is completely equal to
0:05:06.800,0:05:10.320
phi_n only a function of
0:05:09.039,0:05:12.479
position x
0:05:10.320,0:05:13.759
and that's because of this form of the
0:05:12.479,0:05:15.600
time evolution
0:05:13.759,0:05:17.120
however when we take a quantum
0:05:15.600,0:05:18.000
superposition of two different energy
0:05:17.120,0:05:20.240
eigenstates
0:05:18.000,0:05:22.960
the result the resulting probability
0:05:20.240,0:05:25.280
density need not be time independent
0:05:22.960,0:05:26.880
so consider this state alpha psi one
0:05:25.280,0:05:28.960
plus b psi two where
0:05:26.880,0:05:30.160
psi one and psi two are different energy
0:05:28.960,0:05:33.919
eigen states
0:05:30.160,0:05:33.919
the probability density is this
0:05:34.320,0:05:38.080
it's equal to modulus psi squared again
0:05:36.720,0:05:41.440
and expanding
0:05:38.080,0:05:43.199
the product we find this result
0:05:41.440,0:05:45.280
and if we stick in the forms of the
0:05:43.199,0:05:48.160
energy eigenstates again
0:05:45.280,0:05:48.160
we find the result
0:05:48.560,0:05:52.160
so it takes this form from this term
0:05:51.759,0:05:54.160
here
0:05:52.160,0:05:55.360
psi one is time dependent but when we
0:05:54.160,0:05:57.199
take the modulus square of it because
0:05:55.360,0:05:58.960
it's an energy eigenstate the result is
0:05:57.199,0:06:01.600
the same as phi one squared
0:05:58.960,0:06:02.479
modulus which is time independent same
0:06:01.600,0:06:06.319
with phi two
0:06:02.479,0:06:09.600
but the cross-terms are in general
0:06:06.319,0:06:10.000
time dependent the result is still
0:06:09.600,0:06:11.680
real
0:06:10.000,0:06:13.199
because it's a probability density so it
0:06:11.680,0:06:14.560
better be real and we know it must be
0:06:13.199,0:06:15.199
real because we're taking the modulus
0:06:14.560,0:06:18.400
square
0:06:15.199,0:06:18.800
of some number which is always real
0:06:18.400,0:06:20.639
but
0:06:18.800,0:06:22.720
it may in general be time dependent even
0:06:20.639,0:06:26.000
though the states from which it's a sum
0:06:22.720,0:06:28.479
have separate time
0:06:26.000,0:06:30.560
independent probability
0:06:28.479,0:06:32.639
densities
0:06:30.560,0:06:34.560
okay now let's take a look at the
0:06:32.639,0:06:36.960
normalization of this superposition so
0:06:34.560,0:06:39.039
let's clear the board again
0:06:36.960,0:06:41.680
so look at normalization we'll pick the
0:06:39.039,0:06:43.600
same state as before
0:06:41.680,0:06:46.080
and remember that all physical states
0:06:43.600,0:06:48.639
must be normalized
0:06:46.080,0:06:50.319
which means that the modulus square of
0:06:48.639,0:06:50.880
the wave function integrated across all
0:06:50.319,0:06:53.840
of space
0:06:50.880,0:06:56.240
must equal one we ensure that our energy
0:06:53.840,0:06:58.000
eigenstates are normalized
0:06:56.240,0:06:59.599
and so this places a condition on the
0:06:58.000,0:07:00.880
possible alphas and betas we can take in
0:06:59.599,0:07:03.919
our superposition
0:07:00.880,0:07:07.280
in this case the condition is this
0:07:03.919,0:07:09.199
which equals again just expanding the
0:07:07.280,0:07:12.000
product into these four terms
0:07:09.199,0:07:13.680
but we know that not only are our energy
0:07:12.000,0:07:15.199
eigenstates normalized they're also
0:07:13.680,0:07:18.080
orthogonal to one another
0:07:15.199,0:07:19.440
so that condition by definition means
0:07:18.080,0:07:22.800
that this term is zero
0:07:19.440,0:07:23.280
because this is psi one and psi two
0:07:22.800,0:07:25.919
are
0:07:23.280,0:07:28.720
orthogonal so by definition this is zero
0:07:25.919,0:07:28.720
same with this term
0:07:28.800,0:07:32.000
and we also have that these two states
0:07:30.880,0:07:33.759
are normalized
0:07:32.000,0:07:36.160
so putting this together we find the
0:07:33.759,0:07:36.160
following
0:07:36.240,0:07:41.039
that is one equals the modulus of alpha
0:07:38.319,0:07:42.960
squared plus the modulus of beta squared
0:07:41.039,0:07:44.080
so this gives us our normalization that
0:07:42.960,0:07:47.120
must appear on
0:07:44.080,0:07:48.720
this wave function as the following that
0:07:47.120,0:07:49.440
is we now have a properly normalized
0:07:48.720,0:07:53.919
wave function
0:07:49.440,0:07:56.000
psi for arbitrary complex alpha and beta
0:07:53.919,0:07:57.280
if we perform an energy measurement on
0:07:56.000,0:07:59.520
psi
0:07:57.280,0:08:01.039
it's not itself in energy eigenstates we
0:07:59.520,0:08:03.759
won't certainly get
0:08:01.039,0:08:05.199
any particular energy we can say for
0:08:03.759,0:08:08.319
certainty that we won't get
0:08:05.199,0:08:10.479
any result other than E_1 or E_2 because
0:08:08.319,0:08:12.240
the amplitude for any other energy
0:08:10.479,0:08:13.520
eigenstate is zero
0:08:12.240,0:08:16.319
it didn't need to be but it is in this
0:08:13.520,0:08:17.840
particular choice
0:08:16.319,0:08:20.720
and the probability that we'll find
0:08:17.840,0:08:23.680
energy E_1 is given by
0:08:20.720,0:08:25.360
the modulus of alpha squared divided by
0:08:23.680,0:08:26.000
the normalization which would be this
0:08:25.360,0:08:27.840
thing squared
0:08:26.000,0:08:29.120
so it'll be modulus alpha squared
0:08:27.840,0:08:32.320
divided by alpha modulus
0:08:29.120,0:08:34.080
squared plus modulus beta squared and
0:08:32.320,0:08:35.039
similarly the probability for finding
0:08:34.080,0:08:36.880
energy E_2
0:08:35.039,0:08:38.560
will be the modulus of beta squared
0:08:36.880,0:08:40.640
divided by the normalization
0:08:38.560,0:08:41.919
so this guarantees that the total
0:08:40.640,0:08:44.320
probability to find
0:08:41.919,0:08:46.480
the particle in some energy is equal to
0:08:44.320,0:08:48.160
one
0:08:46.480,0:08:49.760
as always when we carry out our
0:08:48.160,0:08:51.279
normalization notice that the
0:08:49.760,0:08:52.959
normalization condition
0:08:51.279,0:08:54.640
only places a constraint on the
0:08:52.959,0:08:58.000
magnitude not the phase
0:08:54.640,0:08:59.760
so we're left with a global phase (which
0:08:58.000,0:09:03.200
is again ambiguous) out the front
0:08:59.760,0:09:04.720
of this wave function but the relative
0:09:03.200,0:09:06.720
phase between these two
0:09:04.720,0:09:08.720
contributions to the superposition is
0:09:06.720,0:09:10.959
important so that would change things
0:09:08.720,0:09:12.880
but there's a global complex phase at
0:09:10.959,0:09:14.560
the front of this which is arbitrary
0:09:12.880,0:09:16.399
because the global phase in quantum
0:09:14.560,0:09:19.680
mechanics is unobservable
0:09:16.399,0:09:19.680
okay thanks for your time
V4.2 The finite potential well
This channel, Introductory Quantum Mechanics, is a set of videos aimed at second-year physics undergraduates.
This video:
establishing properties of the bound states within the finite potential well.
0:00:01.040,0:00:03.439
hello in this video we're going to take
0:00:02.960,0:00:05.520
a look
0:00:03.439,0:00:06.640
at the finite potential well we've
0:00:05.520,0:00:07.520
looked previously at the infinite
0:00:06.640,0:00:08.960
potential well
0:00:07.520,0:00:10.960
now we're going to bring that potential
0:00:08.960,0:00:14.400
down to a finite value so
0:00:10.960,0:00:17.920
the potential is as follows
0:00:14.400,0:00:19.760
so zero within a region of x as before
0:00:17.920,0:00:21.039
and V_0 otherwise where V_0 is
0:00:19.760,0:00:22.720
no longer infinity
0:00:21.039,0:00:24.560
this time we've set it up so that the
0:00:22.720,0:00:27.599
well is symmetric about zero
0:00:24.560,0:00:28.640
it just makes the maths slightly
0:00:27.599,0:00:31.679
simpler to work with
0:00:28.640,0:00:34.399
let's draw it
0:00:31.679,0:00:34.880
so it's symmetric about zero and it has
0:00:34.399,0:00:39.280
height V_0
0:00:34.880,0:00:41.680
and this is zero down here
0:00:39.280,0:00:43.760
now let's try to guess what the
0:00:41.680,0:00:47.200
solutions look like
0:00:43.760,0:00:48.879
so down at the bottom of the well
0:00:47.200,0:00:49.920
we'll have some bound state down here
0:00:48.879,0:00:50.960
just like we did in the infinite
0:00:49.920,0:00:52.960
potential well
0:00:50.960,0:00:54.960
I'm going to sketch it in this slightly
0:00:52.960,0:00:58.079
dodgy way of drawing the wave functions
0:00:54.960,0:00:59.199
onto the potential plot at a snapshot in
0:00:58.079,0:01:02.000
time
0:00:59.199,0:01:03.680
so before the lowest energy state looked
0:01:02.000,0:01:06.559
like this in fact let's draw that
0:01:03.680,0:01:06.559
in a different colour
0:01:09.280,0:01:14.320
so the solution to the infinite
0:01:10.640,0:01:14.320
potential well looked like this
0:01:14.479,0:01:17.600
we're going to have something like that
0:01:15.840,0:01:19.360
but there's no longer a requirement that
0:01:17.600,0:01:20.720
the wave function vanishes at this point
0:01:19.360,0:01:22.000
because remember the wave function only
0:01:20.720,0:01:24.000
has to vanish in regions where the
0:01:22.000,0:01:27.040
potential is infinity
0:01:24.000,0:01:28.479
so using this as a motivation
0:01:27.040,0:01:31.840
we can guess that the lowest energy
0:01:28.479,0:01:31.840
state might look something like this
0:01:35.920,0:01:40.000
that is it takes the form of standing
0:01:39.119,0:01:41.680
waves within the
0:01:40.000,0:01:43.360
well but it doesn't need to vanish at
0:01:41.680,0:01:43.840
the edge of the well and outside of the
0:01:43.360,0:01:45.600
well
0:01:43.840,0:01:48.000
it isn't zero it's exponentially
0:01:45.600,0:01:49.600
decreasing remember our solutions both
0:01:48.000,0:01:51.920
to the infinite potential well here
0:01:49.600,0:01:52.960
and also to our scattering from an
0:01:51.920,0:01:56.719
infinitely long
0:01:52.960,0:01:59.520
but finite height potential step
0:01:56.719,0:02:00.399
so that's the lowest energy state we
0:01:59.520,0:02:02.719
might guess
0:02:00.399,0:02:04.079
similarly that if we have another bound
0:02:02.719,0:02:07.280
state up here
0:02:04.079,0:02:07.280
it might look something like this
0:02:07.520,0:02:11.599
where again this is a snapshot in time
0:02:09.200,0:02:14.800
of the real part of the wavefunction psi
0:02:11.599,0:02:17.520
and this will again come off to
0:02:14.800,0:02:17.520
zero like this
0:02:19.040,0:02:22.480
and we might have higher energy bound
0:02:21.040,0:02:23.760
states in the well there's no
0:02:22.480,0:02:25.200
requirement for there to be an infinite
0:02:23.760,0:02:26.720
number of bound states which there was
0:02:25.200,0:02:29.040
in the infinite potential well
0:02:26.720,0:02:30.800
we can only fit some number in before
0:02:29.040,0:02:31.920
the energy of the states is higher than
0:02:30.800,0:02:33.680
that of the well
0:02:31.920,0:02:36.239
and when we do that we then expect the
0:02:33.680,0:02:37.599
solutions to be plane waves there will
0:02:36.239,0:02:39.519
be some kind of boundary condition going
0:02:37.599,0:02:42.160
on here because of the edges of the
0:02:39.519,0:02:43.599
well but in general we'll have a
0:02:42.160,0:02:45.680
continuum of different
0:02:43.599,0:02:47.040
plane wave states above the well so
0:02:45.680,0:02:48.640
there's an infinite number of these
0:02:47.040,0:02:51.040
there's only a finite number of states
0:02:48.640,0:02:53.519
trapped within the well
0:02:51.040,0:02:54.319
taking a look at the forms in the
0:02:53.519,0:02:55.599
different regions
0:02:54.319,0:02:57.920
we can guess that we'll have standing
0:02:55.599,0:02:59.680
waves in here remember our general
0:02:57.920,0:03:00.959
types of wave solution are either plane
0:02:59.680,0:03:02.400
waves other
0:03:00.959,0:03:04.720
types of solution for regions with
0:03:02.400,0:03:06.720
constant potential are the plane waves
0:03:04.720,0:03:08.640
which we have up here they're
0:03:06.720,0:03:09.440
standing waves a form of plane wave but
0:03:08.640,0:03:10.959
where we have
0:03:09.440,0:03:12.480
equal contributions from left- and right-
0:03:10.959,0:03:13.519
going waves which we expect within the
0:03:12.480,0:03:15.440
well
0:03:13.519,0:03:17.680
and they can be evanescent waves
0:03:15.440,0:03:20.080
exponentially increasing or decreasing
0:03:17.680,0:03:20.800
and note that over here our our physical
0:03:20.080,0:03:22.400
guess
0:03:20.800,0:03:26.000
was that we have exponentially
0:03:22.400,0:03:29.760
increasing solutions in this region
0:03:26.000,0:03:32.959
and a decreasing solution over here
0:03:29.760,0:03:34.720
okay so let's write down the forms of
0:03:32.959,0:03:36.000
those wave functions and substitute in
0:03:34.720,0:03:38.319
the boundary conditions on the next
0:03:36.000,0:03:42.159
board
0:03:38.319,0:03:42.159
sorry didn't erase it one more go
0:03:43.599,0:03:47.840
sorry the board's playing up
0:03:48.159,0:03:51.599
oh sorry I must have it set to change my
0:03:50.879,0:03:55.360
clothes
0:03:51.599,0:03:58.640
right oh hi okay got it
0:03:55.360,0:03:59.760
got it okay all right so wave functions
0:03:58.640,0:04:01.360
in the different regions
0:03:59.760,0:04:03.120
it depends on whether the energy is
0:04:01.360,0:04:05.120
greater than or less than V_0
0:04:03.120,0:04:06.480
if it's greater than V_0 we're
0:04:05.120,0:04:07.519
just back to plane wave solutions in
0:04:06.480,0:04:09.200
all three regions
0:04:07.519,0:04:10.640
and we're solving exactly the same
0:04:09.200,0:04:12.480
problem as
0:04:10.640,0:04:14.159
the scattering over the top of a
0:04:12.480,0:04:14.959
potential barrier which you've seen in a
0:04:14.159,0:04:17.280
previous video
0:04:14.959,0:04:18.000
so let's only consider the bound states
0:04:17.280,0:04:20.400
which lie
0:04:18.000,0:04:21.120
with energy less than V_0 in that
0:04:20.400,0:04:22.240
case
0:04:21.120,0:04:24.240
we have the following results in
0:04:22.240,0:04:26.880
the different regions
0:04:24.240,0:04:27.680
so let's call it region one; x is to the
0:04:26.880,0:04:30.479
left of the
0:04:27.680,0:04:31.520
well so less than -L/2 we
0:04:30.479,0:04:34.400
have phi one
0:04:31.520,0:04:34.880
equals unknown coefficient a
0:04:34.400,0:04:37.919
times e^(i kappa x)
0:04:34.880,0:04:40.080
it must be exponentially increasing
0:04:37.919,0:04:42.960
in order to die off at x equals minus
0:04:40.080,0:04:46.240
infinity rather than blow up
0:04:42.960,0:04:48.960
similarly in region 3
0:04:46.240,0:04:50.720
that is x>L/2. In
0:04:48.960,0:04:53.040
region 3 you must have the form
0:04:50.720,0:04:54.000
unknown constant d times e^(-kappa x)
0:04:53.040,0:04:56.000
where
0:04:54.000,0:04:57.680
kappa is the same kappa as appeared in
0:04:56.000,0:04:58.320
region 1 because the potentials are the
0:04:57.680,0:05:00.800
same
0:04:58.320,0:05:01.520
both equal to V_0 that is both of
0:05:00.800,0:05:02.800
these
0:05:01.520,0:05:04.560
when substituted into the time
0:05:02.800,0:05:06.960
independent Schroedinger equation gives
0:05:04.560,0:05:08.880
the form
0:05:06.960,0:05:10.800
minus h bar squared kappa squared over
0:05:08.880,0:05:13.520
2m plus V_0
0:05:10.800,0:05:14.240
and this ensures that we can have real
0:05:13.520,0:05:16.080
kappa
0:05:14.240,0:05:18.479
for E