1  Canonical Field Quantization

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In this chapter we will see in full detail how to build a quantum field. We will focus on the simplest case of a real, scalar, non-relativistic field. As we will see, this will turn out to be a quantum theory of sound propagating through a crystal. This is the phonon field, whose particle excitations are called phonons.

References

• N. W. Ashcroft and N. D. Mermin, Solid State Physics (Harcourt 1976), Appendix L

• R. P. Feynman, Statistical Mechanics (Advanced Book Classics, 1972) Chapter 6

• A. Altland and B. Simons, Condensed Matter Field Theory (Cambridge University Press, 2010), Chapter 1

• T. Lancaster and S. J. Blundell, Quantum Field Theory for the Gifted Amateur (Oxford University Press, 2014), Section I

• S. M. Girvin and K. Yang, Modern Condensed Matter Physics (Cambridge University Press, 2019), Section 6.2

1.1 Recap: Quantum Harmonic Oscillators

Recall the Time Independent Schroedinger Equation (TISE) for the quantum harmonic oscillator:

\[\hat{H}|n\rangle=\left(\frac{\hat{p}^{2}}{2m}+\frac{1}{2}m\omega^{2}\hat{x}^{2}\right)|n\rangle=E_{n}|n\rangle.\]

The potential is shown in the position basis in Fig. 1.1.

Figure 1.1: The quantum harmonic oscillator potential \(V\left(x\right)=\frac{1}{2}m\omega^{2}x^{2}\) and first few eigenstates.

An elegant way to find all eigenstates and eigenvalues is to use creation and annihilation operators (aka ladder operators, aka raising and lowering operators). The intuition developed in this simple QM problem forms the entire basis of QFT.

1.1.1 Creation and Annihilation Operators

The raising operator is

\[\boxed{\hat{a}^{\dagger}\triangleq\sqrt{\frac{m\omega}{2}}\left(\hat{x}-\frac{i}{m\omega}\hat{p}\right).} \tag{1.1}\]

It is non-Hermitian. Its hermitian conjugate is the lowering operator

\[\boxed{\hat{a}=\sqrt{\frac{m\omega}{2}}\left(\hat{x}+\frac{i}{m\omega}\hat{p}\right).} \tag{1.2}\]

This gives

\[\begin{aligned} \hat{a}^{\dagger}\hat{a} & =\frac{m\omega}{2}\left(\hat{x}^{2}+\frac{\hat{p}^{2}}{m^{2}\omega^{2}}+\frac{i}{m\omega}\left(\hat{x}\hat{p}-\hat{p}\hat{x}\right)\right). \end{aligned}\]

The term in nested parentheses is just the commutator

\[\left[\hat{x},\hat{p}\right]=i\hat{\mathbb{I}}\]

where \(\hat{\mathbb{I}}\) is the identity operator. Hence,

\[\hat{a}^{\dagger}\hat{a}=\frac{m\omega}{2}\hat{x}^{2}+\frac{\hat{p}^{2}}{2m\omega}-\frac{1}{2}\hat{\mathbb{I}}\]

and so

\[\omega\left(\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\hat{\mathbb{I}}\right)=\hat{H}.\]

Therefore the TISE can be written

\[\boxed{\hat{H}|n\rangle=\omega\left(\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\hat{\mathbb{I}}\right)|n\rangle=E_{n}|n\rangle.} \tag{1.3}\]

1.1.2 Commutation relations

The commutator of the creation and annihilation operators is

\[\begin{aligned} \left[\hat{a},\hat{a}^{\dagger}\right] & =\frac{m\omega}{2}\left[\hat{x}+\frac{i}{m\omega}\hat{p},\hat{x}-\frac{i}{m\omega}\hat{p}\right]\\ & =-i\left[\hat{x},\hat{p}\right] \end{aligned}\]

and so \[\boxed{\left[\hat{a},\hat{a}^{\dagger}\right]=\hat{\mathbb{I}}}\]

The commutator of the operators with the Hamiltonian is therefore:

\[\begin{aligned} \left[\hat{H},\hat{a}^{\dagger}\right] & =\left[\omega\left(\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\hat{\mathbb{I}}\right),\hat{a}^{\dagger}\right]\\ & =\omega\left(\hat{a}^{\dagger}\hat{a}\hat{a}^{\dagger}-\hat{a}^{\dagger}\hat{a}^{\dagger}\hat{a}\right)\\ & =\omega\hat{a}^{\dagger}\left[\hat{a},\hat{a}^{\dagger}\right] \end{aligned}\]

giving

\[\boxed{\left[\hat{H},\hat{a}^{\dagger}\right]=\omega\hat{a}^{\dagger}.} \tag{1.4}\]

Similarly,

\[\left[\hat{H},\hat{a}\right]=-\omega\hat{a}.\]

1.1.3 Energy eigenstates and eigenvalues

Using Eq. 1.4 we can see the effect of \(\hat{a}^{\dagger}\) on the TISE of Eq. 1.3:

\[\begin{aligned} \hat{H}|n\rangle & =E_{n}|n\rangle\\ & \downarrow\nonumber \\ \hat{a}^{\dagger}\hat{H}|n\rangle & =E_{n}\hat{a}^{\dagger}|n\rangle\\ & \downarrow\nonumber \\ \left(\hat{H}\hat{a}^{\dagger}-\left[\hat{H},\hat{a}^{\dagger}\right]\right)|n\rangle & =E_{n}\hat{a}^{\dagger}|n\rangle\\ & \downarrow\nonumber \\ \left(\hat{H}\hat{a}^{\dagger}-\omega\hat{a}^{\dagger}\right)|n\rangle & =E_{n}\hat{a}^{\dagger}|n\rangle \end{aligned}\]

and the final result

\[\boxed{\hat{H}\left(\hat{a}^{\dagger}|n\rangle\right)=\left(E_{n}+\omega\right)\left(\hat{a}^{\dagger}|n\rangle\right).}\]

That is, if \(|n\rangle\) is an eigenstate of the harmonic oscillator with eigenvalue \(E_{n}\), then \(\hat{a}^{\dagger}|n\rangle\) is an eigenstate with eigenvalue \(E_{n}+\omega\). Repeating the process \(l\) times we find

\[\hat{H}\left(\left(\hat{a}^{\dagger}\right)^{l}|n\rangle\right)=\left(E_{n}+l\omega\right)\left(\left(\hat{a}^{\dagger}\right)^{l}|n\rangle\right).\]

This tells us that the energy levels are evenly spaced, and that

\[\left(\hat{a}^{\dagger}\right)^{l}|n\rangle\propto|n+l\rangle. \tag{1.5}\]

Similarly, we find

\[\hat{H}\left(\hat{a}|n\rangle\right)=\left(E_{n}-\omega\right)\left(\hat{a}|n\rangle\right).\]

The creation and annihilation operators move the state up or down the rungs of an energy ladder with evenly spaced rungs. While there exist an infinite number of rungs, the energies do not stretch down to negative energies. To see this, first note that

\[\left|\hat{a}|n\rangle\right|^{2}\ge0\]

because the thing on the left, whatever it is, is the square modulus of something, and that is always \(\ge0\). Expanding it out we have

\[\begin{aligned} \langle n|\hat{a}^{\dagger}\hat{a}|n\rangle & \ge0\\ \langle n|\frac{1}{\omega}\hat{H}-\frac{1}{2}|n\rangle & \ge0\\ \langle n|\frac{1}{\omega}E_{n}-\frac{1}{2}|n\rangle & \ge0\\ E_{n} & \ge\frac{\omega}{2}. \end{aligned}\]

Therefore there is a lowest-energy state, a lowest rung to the ladder. This is the ground state which we denote \(|0\rangle\). This is just a convenient label for a ket. It is not the number zero! You can think of it as shorthand for \(|\psi_{n=0}\rangle\). The ground state has the property that

\[\boxed{\hat{a}|0\rangle=0}\]

where the right hand side really is the number \(0\), so that any further action of lowering operators continues to return \(0\). We can find the energy of this state straightforwardly:

\[\begin{aligned} \hat{H}|0\rangle & =\omega\left(\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\hat{\mathbb{I}}\right)|0\rangle\\ & =\frac{\omega}{2}|0\rangle \end{aligned}\]

so \(E_{0}=\omega/2\). This is the ground state energy, also called the zero-point energy. Combining the results, we see that

\[\hat{H}|n\rangle=\omega\left(\hat{a}^{\dagger}\hat{a}+\frac{1}{2}\right)|n\rangle=\omega\left(n+\frac{1}{2}\right)|n\rangle\]

for integer \(n\ge0\). Defining the ‘number operator’

\[\hat{n}\triangleq\hat{a}^{\dagger}\hat{a}\]

we see that

\[\hat{n}|n\rangle=n|n\rangle.\]

1.1.4 Eigenstates

To find the energy eigenstates in the position basis we can use the definition of the lowering operator from Eq. 1.2:

\[\begin{aligned} \sqrt{\frac{m\omega}{2}}\left(\hat{x}+\frac{i}{m\omega}\hat{p}\right)|0\rangle & =0\\ & \downarrow\nonumber \\ \left(x+\frac{1}{m\omega}\frac{\text{d}}{\text{d}x}\right)\phi_{0}\left(x\right) & =0. \end{aligned}\]

This is a first order linear ODE which has the solution

\[\phi_{0}\left(x\right)\propto\exp\left(-\frac{m\omega}{2}x^{2}\right).\]

All excited states can then be found by acting the creation operator on this state repeatedly. For example,

\[\begin{aligned} \phi_{1}\left(x\right) & \propto\hat{a}^{\dagger}\phi_{0}\left(x\right)\\ & =\left(x-\frac{1}{m\omega}\frac{\text{d}}{\text{d}x}\right)\exp\left(-\frac{m\omega}{2}x^{2}\right)\\ & =2x\exp\left(-\frac{m\omega}{2}x^{2}\right). \end{aligned}\]

1.2 Second quantization

Note Added: within single-particle quantum mechanics, the following description of the nth excited state of the harmonic oscillator as n bosons is really only an analogy. It strictly becomes precise only when we move to QFT and allow multi-particle states.

• Eq. 1.5 tells us we can create the \(n^{\text{th}}\) excited state of the harmonic oscillator by acting \(n\) raising operators on the ground state.

• Since all rungs of the ladder are evenly spaced with spacing \(\omega\), we can also interpret the \(n^{\text{th}}\) excited state as the presence of \(n\) identical particles each of energy \(\omega\).

• This is the origin of the name ‘creation and annihilation operators’: they create and annihilate particles in a second-quantised description.

• ‘First quantization’ is the realisation that classical particles exhibit wave-like properties in QM. We call the process of rewriting a problem in terms of creation and annihilation operators ‘second quantization’. It is the realisation that classical waves gain particle-like properties in QM. The descriptions are equivalent.

• The fact that we can fit multiple particles into the same state identifies \(\hat{a}^{\dagger}\) as the creation operator of a boson.

1.3 \(N\) Isolated Quantum Harmonic Oscillators

Now consider \(N\) independent QHOs, each of mass \(M\) and angular frequency \(\omega_{0}\), evenly spaced with spacing \(a\) along a chain. (We reserve the symbol \(m\) for something else, as we will see.) Let \(\hat{u}_{n}\) be the displacement of oscillator \(n\) from its equilibrium position, and \(\hat{p}_{n}\) its momentum. Assume periodic boundary conditions, so that site \(N+1\) is equivalent to site \(1\) (i.e. the chain is really a ring). The Hamiltonian is

\[\hat{H}=\sum^{N}_{n=1}\frac{\hat{p}^{2}_{n}}{2M}+\frac{1}{2}M\omega^{2}_{0}\hat{u}^{2}_{n}. \tag{1.6}\]

We can rewrite this using creation and annihilation operators, each with a site label:

\[\hat{H}=\sum^{N}_{n=1}\omega_{0}\left(\hat{a}^{\dagger}_{n}\hat{a}_{n}+\frac{1}{2}\hat{\mathbb{I}}\right).\]

Here, \(\hat{a}^{\dagger}_{n}\) creates an excitation of energy \(\omega_{0}\) at site \(n\). We can also take advantage of periodicity (translational symmetry) to Fourier transform. The allowed wavevectors are \(k=2\pi j/Na\) for integer \(j\), and we take the \(N\) distinct values lying in the Brillouin zone \(-\pi/a<k\le\pi/a\) (see Conventions). Using

\[\begin{aligned} \hat{a}_{k} & =\frac{1}{\sqrt{N}}\sum^{N}_{n=1}\exp\left(-ikan\right)\hat{a}_{n}\\ \hat{a}_{n} & =\frac{1}{\sqrt{N}}\sum_{k}\exp\left(ikan\right)\hat{a}_{k} \end{aligned}\]

(so that \(\hat{a}^{\dagger}_{n}=N^{-1/2}\sum_{k}\exp\left(-ikan\right)\hat{a}^{\dagger}_{k}\)) gives

\[\begin{aligned} \hat{H} & =\sum_{n}\omega_{0}\hat{a}^{\dagger}_{n}\hat{a}_{n}+\frac{1}{2}N\omega_{0}\hat{\mathbb{I}}\\ & =\frac{1}{N}\sum_{n,k,q}\omega_{0}\exp\left(i\left(q-k\right)an\right)\hat{a}^{\dagger}_{k}\hat{a}_{q}+\frac{1}{2}N\omega_{0}\hat{\mathbb{I}}\nonumber \\ & =\sum_{k,q}\omega_{0}\delta_{kq}\hat{a}^{\dagger}_{k}\hat{a}_{q}+\frac{1}{2}N\omega_{0}\hat{\mathbb{I}}\\ & =\sum_{k}\omega_{0}\hat{a}^{\dagger}_{k}\hat{a}_{k}+\frac{1}{2}N\omega_{0}\hat{\mathbb{I}} \end{aligned}\]

where we used the useful expression

\[\frac{1}{N}\sum^{N}_{n=1}\exp\left(i\left(k-q\right)an\right)=\delta_{kq} \tag{1.7}\]

which follows since the sum is over equally distributed points around the unit circle. Since there are exactly \(N\) values of \(k\), we can also write \(N\omega_{0}/2=\sum_{k}\omega_{0}/2\), giving

\[\begin{aligned} \hat{H} & =\sum_{k}\omega_{0}\left(\hat{a}^{\dagger}_{k}\hat{a}_{k}+\frac{1}{2}\hat{\mathbb{I}}\right). \end{aligned}\]

In this case, \(\hat{a}^{\dagger}_{k}\) creates an excitation of energy \(\omega_{0}\) with wavevector \(k\).

The Hamiltonian in Eq. 1.6 is overly simplified: the QHOs are uncoupled, and so it is hard to imagine how one might physically create an excitation across multiple oscillators. It is a remarkable result, which we will derive shortly, that when interactions are turned on between the oscillators, provided translational invariance is preserved, the effect is merely to add a wavevector dependence to the angular frequency \(\omega_{0}\rightarrow\omega_{k}\). That is:

\[\hat{H}_{\text{interacting}}=\sum_{k}\omega_{k}\left(\hat{a}^{\dagger}_{k}\hat{a}_{k}+\frac{1}{2}\hat{\mathbb{I}}\right)\]

(although the new \(\hat{a}_{k}\) will no longer simply be Fourier transforms of the \(\hat{a}_{n}\)). Before seeing how this works, let us first build intuition by looking at the classical problem.

1.4 \(N\) Coupled Classical Harmonic Oscillators

Consider a classical chain of \(N\) balls of mass \(M\), with neighbouring balls connected by springs of stiffness \(K\) (Fig. 1.2). To retain the isolated oscillators of the previous section, suppose each ball is also tethered to its own equilibrium position, so that on its own it would oscillate at angular frequency \(\omega_{0}\) (think of a row of pendula coupled by springs). For sound in an ordinary crystal there is no such tether, and \(\omega_{0}=0\); we keep it general since it will turn out to play an important role. The balls oscillate about equilibrium positions \(an\) (where \(a\) is a lattice constant and \(n\in\left[1,N\right]\)). Let the position of ball \(n\) at time \(t\) be \(an+u_{n,t}\), so that \(u_{n,t}\) is its displacement from equilibrium. From now on we will only care about the (small) displacements \(u_{n,t}\).

1.4.1 Recap: forces on balls and springs

Wobbling one of the balls will lead to a vibration propagating through the chain. To find the dispersion relation of this propagating mode, consider the force on the \(n^{\text{th}}\) ball (using Newton’s 2nd law):

\[M\ddot{u}_{n,t}=-M\omega^{2}_{0}u_{n,t}+K\left(u_{n+1,t}-u_{n,t}\right)-K\left(u_{n,t}-u_{n-1,t}\right). \tag{1.8}\]

Figure 1.2: A classical chain of balls and springs.

We need the normal modes of oscillation. To find them, first take advantage of the translational symmetry by applying a Fourier transform (in the conventions of Conventions, with angular wavevector \(k\) in the Brillouin zone and angular frequency \(\omega\)):

\[u_{n,t}=\frac{1}{\sqrt{N}}\sum_{k}\int\frac{\text{d}\omega}{2\pi}\exp\left(i\left(kan-\omega t\right)\right)Q_{k,\omega}.\]

Inserting the Fourier transform into the equations of motion gives:

\[\begin{aligned} & M\sum_{k}\int\frac{\text{d}\omega}{2\pi}\exp\left(i\left(kan-\omega t\right)\right)\left(-\omega^{2}\right)Q_{k,\omega}\nonumber \\ & =\sum_{k}\int\frac{\text{d}\omega}{2\pi}\exp\left(i\left(kan-\omega t\right)\right)Q_{k,\omega}\left(-M\omega^{2}_{0}+K\left(\exp\left(ika\right)-1\right)-K\left(1-\exp\left(-ika\right)\right)\right). \end{aligned}\]

We can inverse Fourier transform to find a separate equation for each pair of \(k\) and \(\omega\). Using \(\exp\left(ika\right)-2+\exp\left(-ika\right)=-4\sin^{2}\left(ka/2\right)\),

\[-M\omega^{2}Q_{k,\omega}=-\left(M\omega^{2}_{0}+4K\sin^{2}\left(\frac{ka}{2}\right)\right)Q_{k,\omega}\]

so that non-zero solutions require \(\omega=\pm\omega_{k}\), where the notation \(\omega_{k}\) emphasises that this is a specific solution \(\omega\left(k\right)\):

\[\omega^{2}_{k}=\omega^{2}_{0}+\frac{4K}{M}\sin^{2}\left(\frac{ka}{2}\right). \tag{1.9}\]

(We take \(\omega_{k}\ge0\); since \(u_{n,t}\) is real, the solutions at \(\pm\omega_{k}\) combine into a single real normal mode.) Around \(ka\approx0\) (long wavelengths) we have

\[\omega^{2}_{k}\approx\omega^{2}_{0}+c^{2}k^{2},\qquad c\triangleq a\sqrt{\frac{K}{M}}.\]

For \(\omega_{0}=0\) the mode is linearly dispersing, \(\omega_{k}\approx c\left|k\right|\), and \(c\) is the speed of sound along the chain. For \(\omega_{0}\neq0\) the relation has exactly the form of the relativistic dispersion \(E^{2}=m^{2}+p^{2}\), with the speed of sound playing the role of the speed of light and \(\omega_{0}\) playing the role of a mass. Keep this in mind!

1.4.2 The classical Lagrangian

Thanks to work carried out in Bristol by Aharonov and Bohm, we know that in quantum mechanics we must deal with potentials rather than forces. To set up the quantum problem we must therefore rephrase the classical problem: construct the Lagrangian, and from there the Hamiltonian, of the classical system. We will then quantize this in Section 1.5.

The classical balls and springs model is governed by the Lagrangian

\[L=\sum^{N}_{n=1}\frac{1}{2}M\dot{u}^{2}_{n}-\frac{1}{2}M\omega^{2}_{0}u^{2}_{n}-\frac{1}{2}K\left(u_{n+1}-u_{n}\right)^{2}. \tag{1.10}\]

Note that each spring appears exactly once in the sum (the spring between balls \(n\) and \(n+1\) appears only in the \(n^{\text{th}}\) term). To check this is equivalent to Eq. 1.8, we can derive the classical equations of motion using the Euler Lagrange equations:

\[\left(\frac{\partial L}{\partial u_{n}}\right)_{\dot{u}_{n}}-\frac{\text{d}}{\text{d}t}\left(\frac{\partial L}{\partial\dot{u}_{n}}\right)_{u_{n}}=0\]

along with the identities

\[\left(\frac{\partial u_{m}}{\partial u_{n}}\right)_{\dot{u}_{m}}=\left(\frac{\partial\dot{u}_{m}}{\partial\dot{u}_{n}}\right)_{u_{m}}=\delta_{nm}\]

where we have introduced the

Kronecker delta:

\[\delta_{nm}\triangleq\begin{cases} \begin{array}{c} 0,\\ 1, \end{array} & \begin{array}{c} n\ne m\\ n=m. \end{array}\end{cases}\]

We find

\[\begin{aligned} \left(\frac{\partial L}{\partial u_{n}}\right)_{\dot{u}_{n}} & =\sum^{N}_{m=1}-M\omega^{2}_{0}u_{m}\delta_{m,n}-K\left(u_{m+1}-u_{m}\right)\left(\delta_{m+1,n}-\delta_{m,n}\right)\\ & =-M\omega^{2}_{0}u_{n}-K\left(u_{n}-u_{n-1}\right)+K\left(u_{n+1}-u_{n}\right) \end{aligned}\]

and

\[\begin{aligned} \left(\frac{\partial L}{\partial\dot{u}_{n}}\right)_{u_{n}} & =M\dot{u}_{n} \end{aligned}\]

so that the Euler Lagrange equations read

\[M\ddot{u}_{n}=-M\omega^{2}_{0}u_{n}+K\left(u_{n+1}-u_{n}\right)-K\left(u_{n}-u_{n-1}\right)\]

which is Eq. 1.8, as expected.

1.4.3 The Classical Hamiltonian

To find the Hamiltonian, we need the momentum canonically conjugate to the position. This is defined to be:

\[\begin{aligned} p_{n} & \triangleq\left(\frac{\partial L}{\partial\dot{u}_{n}}\right)_{u_{n}} \end{aligned}\]

which in this case gives

\[p_{n}=M\dot{u}_{n}.\]

The Hamiltonian is then the Legendre transform of the Lagrangian:

\[\begin{aligned} H & \triangleq\sum_{n}p_{n}\dot{u}_{n}-L. \end{aligned}\]

In this case we find

\[H=\sum_{n}\frac{p^{2}_{n}}{2M}+\frac{1}{2}M\omega^{2}_{0}u^{2}_{n}+\frac{1}{2}K\left(u_{n+1}-u_{n}\right)^{2}. \tag{1.11}\]

This is the classical Hamiltonian for the balls and springs model. It is a constant of motion corresponding to the total energy of the system.

1.5 \(N\) Coupled Quantum Harmonic Oscillators

Now let’s combine these ideas, with a chain of coupled quantum harmonic oscillators. Our aim will be to confirm the intuition that we can decompose the problem into one of \(N\) independent QHOs, one for each wavevector \(k\). To create the quantum problem we can ‘canonically quantize’ the classical Hamiltonian in Eq. 1.11. This simply means promoting observables to operators:

\[\hat{H}=\sum^{N}_{n=1}\frac{\hat{p}^{2}_{n}}{2M}+\frac{1}{2}M\omega^{2}_{0}\hat{u}^{2}_{n}+\frac{1}{2}K\left(\hat{u}_{n+1}-\hat{u}_{n}\right)^{2}\]

which obey the ‘canonical commutation relations’:

\[\begin{aligned} \left[\hat{u}_{n},\hat{p}_{m}\right] & =i\hat{\mathbb{I}}\delta_{nm}\\ \left[\hat{u}_{n},\hat{u}_{m}\right] & =0\\ \left[\hat{p}_{n},\hat{p}_{m}\right] & =0. \end{aligned}\]

Note that operators on different sites always commute. We can again take advantage of translational invariance to define

\[\begin{aligned} \hat{u}_{n} & =\frac{1}{\sqrt{N}}\sum_{k}\exp\left(ikan\right)\hat{Q}_{k}\\ \hat{p}_{n} & =\frac{1}{\sqrt{N}}\sum_{k}\exp\left(-ikan\right)\hat{\Pi}_{k} \end{aligned}\]

and

\[\begin{aligned} \hat{Q}_{k} & =\frac{1}{\sqrt{N}}\sum^{N}_{n=1}\exp\left(-ikan\right)\hat{u}_{n}\\ \hat{\Pi}_{k} & =\frac{1}{\sqrt{N}}\sum^{N}_{n=1}\exp\left(ikan\right)\hat{p}_{n} \end{aligned}\]

where all \(k\) sums run over the \(N\) wavevectors in the Brillouin zone. (The opposite sign in the exponent for \(\hat{\Pi}_{k}\) is chosen so that it is the momentum conjugate to \(\hat{Q}_{k}\).) Note that these transformed co-ordinates also obey canonical commutation relations:

\[\begin{aligned} \left[\hat{Q}_{k},\hat{\Pi}_{q}\right] & =\frac{1}{N}\sum_{n,m}\exp\left(ia\left(qm-kn\right)\right)\left[\hat{u}_{n},\hat{p}_{m}\right]\\ & =\frac{1}{N}\sum_{n,m}\exp\left(ia\left(qm-kn\right)\right)i\hat{\mathbb{I}}\delta_{nm}\\ & =\frac{1}{N}\sum_{n}\exp\left(ia\left(q-k\right)n\right)i\hat{\mathbb{I}}\\ & =i\hat{\mathbb{I}}\delta_{kq} \end{aligned}\]

using Eq. 1.7 in the last line. Similarly,

\[\left[\hat{Q}_{k},\hat{Q}_{q}\right]=\left[\hat{\Pi}_{k},\hat{\Pi}_{q}\right]=0.\]

However, \(\hat{Q}_{k}\) and \(\hat{\Pi}_{k}\) are no longer Hermitian. Rather,

\[\hat{Q}^{\dagger}_{k}=\hat{Q}_{-k},\qquad\hat{\Pi}^{\dagger}_{k}=\hat{\Pi}_{-k}.\]

Substituting into the Hamiltonian, and using Eq. 1.7 to do the sums over \(n\), the three terms become

\[\begin{aligned} \sum_{n}\hat{p}^{2}_{n} & =\sum_{k}\hat{\Pi}_{-k}\hat{\Pi}_{k}\\ \sum_{n}\hat{u}^{2}_{n} & =\sum_{k}\hat{Q}_{-k}\hat{Q}_{k}\\ \sum_{n}\left(\hat{u}_{n+1}-\hat{u}_{n}\right)^{2} & =\sum_{k}\left(\exp\left(-ika\right)-1\right)\left(\exp\left(ika\right)-1\right)\hat{Q}_{-k}\hat{Q}_{k}=\sum_{k}4\sin^{2}\left(\frac{ka}{2}\right)\hat{Q}_{-k}\hat{Q}_{k} \end{aligned}\]

so that in these new variables the Hamiltonian reads

\[\begin{aligned} \hat{H} & =\sum_{k}\frac{1}{2M}\hat{\Pi}_{-k}\hat{\Pi}_{k}+\frac{1}{2}M\omega^{2}_{k}\hat{Q}_{-k}\hat{Q}_{k}\\ & =\sum_{k}\frac{1}{2M}\hat{\Pi}^{\dagger}_{k}\hat{\Pi}_{k}+\frac{1}{2}M\omega^{2}_{k}\hat{Q}^{\dagger}_{k}\hat{Q}_{k} \end{aligned} \tag{1.12}\]

with exactly the same \(\omega_{k}\) as the classical dispersion relation, Eq. 1.9:

\[\omega^{2}_{k}=\omega^{2}_{0}+\frac{4K}{M}\sin^{2}\left(\frac{ka}{2}\right).\]

Unfortunately Eq. 1.12 makes it clear that states at \(k\) and \(-k\) are coupled, which is not quite what we want. If we can decouple the modes, we can find the eigenstates and eigenvalues using creation and annihilation operators as for the single harmonic oscillator. A clever definition of creation operator allows us to do this while simultaneously decoupling \(k\) from \(-k\) (the two steps can also be done separately, with more algebra). We define a different creation operator for each mode \(k\):

\[\boxed{\hat{a}^{\dagger}_{k}\triangleq\sqrt{\frac{M\omega_{k}}{2}}\left(\hat{Q}_{-k}-\frac{i}{M\omega_{k}}\hat{\Pi}_{k}\right)}\]

giving (using \(\omega_{-k}=\omega_{k}\))

\[\begin{aligned} \hat{a}_{k} & =\sqrt{\frac{M\omega_{k}}{2}}\left(\hat{Q}^{\dagger}_{-k}+\frac{i}{M\omega_{k}}\hat{\Pi}^{\dagger}_{k}\right)\\ & =\sqrt{\frac{M\omega_{k}}{2}}\left(\hat{Q}_{k}+\frac{i}{M\omega_{k}}\hat{\Pi}_{-k}\right). \end{aligned}\]

Note that \(\hat{a}^{\dagger}_{k}\ne\hat{a}_{-k}\). These obey the commutation relations:

\[\begin{aligned} \left[\hat{a}_{k},\hat{a}^{\dagger}_{q}\right] & =\left[\sqrt{\frac{M\omega_{k}}{2}}\left(\hat{Q}_{k}+\frac{i}{M\omega_{k}}\hat{\Pi}^{\dagger}_{k}\right),\sqrt{\frac{M\omega_{q}}{2}}\left(\hat{Q}^{\dagger}_{q}-\frac{i}{M\omega_{q}}\hat{\Pi}_{q}\right)\right]\\ & =\frac{1}{2}M\sqrt{\omega_{k}\omega_{q}}\left[\hat{Q}_{k}+\frac{i}{M\omega_{k}}\hat{\Pi}^{\dagger}_{k},\hat{Q}^{\dagger}_{q}-\frac{i}{M\omega_{q}}\hat{\Pi}_{q}\right]\\ & =\frac{i}{2}\sqrt{\omega_{k}\omega_{q}}\left(\frac{1}{\omega_{k}}\left[\hat{\Pi}^{\dagger}_{k},\hat{Q}^{\dagger}_{q}\right]-\frac{1}{\omega_{q}}\left[\hat{Q}_{k},\hat{\Pi}_{q}\right]\right)\\ & =\frac{i}{2}\sqrt{\omega_{k}\omega_{q}}\left(-\frac{1}{\omega_{k}}i\delta_{k,q}\hat{\mathbb{I}}-\frac{1}{\omega_{q}}i\delta_{k,q}\hat{\mathbb{I}}\right)\\ & =\delta_{k,q}\hat{\mathbb{I}} \end{aligned}\]

and

\[\left[\hat{a}_{k},\hat{a}_{q}\right]=\left[\hat{a}^{\dagger}_{k},\hat{a}^{\dagger}_{q}\right]=0.\]

We can now re-arrange to get

\[\begin{aligned} \hat{Q}_{k} & =\sqrt{\frac{1}{2M\omega_{k}}}\left(\hat{a}_{k}+\hat{a}^{\dagger}_{-k}\right)\\ \hat{\Pi}_{k} & =i\sqrt{\frac{M\omega_{k}}{2}}\left(\hat{a}^{\dagger}_{k}-\hat{a}_{-k}\right) \end{aligned}\]

and so

\[\begin{aligned} \hat{H} & =\sum_{k}\frac{1}{2M}\hat{\Pi}^{\dagger}_{k}\hat{\Pi}_{k}+\frac{1}{2}M\omega^{2}_{k}\hat{Q}^{\dagger}_{k}\hat{Q}_{k}\\ & =\sum_{k}\frac{\omega_{k}}{4}\left\{ \left(\hat{a}_{k}-\hat{a}^{\dagger}_{-k}\right)\left(\hat{a}^{\dagger}_{k}-\hat{a}_{-k}\right)+\left(\hat{a}^{\dagger}_{k}+\hat{a}_{-k}\right)\left(\hat{a}_{k}+\hat{a}^{\dagger}_{-k}\right)\right\} \\ & =\sum_{k}\frac{\omega_{k}}{4}\left(\hat{a}_{k}\hat{a}^{\dagger}_{k}+\hat{a}^{\dagger}_{-k}\hat{a}_{-k}+\hat{a}^{\dagger}_{k}\hat{a}_{k}+\hat{a}_{-k}\hat{a}^{\dagger}_{-k}\right)\\ & =\sum_{k}\omega_{k}\left(\frac{1}{2}\hat{a}^{\dagger}_{k}\hat{a}_{k}+\frac{1}{2}\hat{a}^{\dagger}_{-k}\hat{a}_{-k}+\frac{1}{2}\hat{\mathbb{I}}\right) \end{aligned}\]

where the cross terms cancelled in the third line, and in the last line I have used the commutators to put creation operators to the left of annihilation operators. The modes are now nicely decoupled (\(k\) and \(-k\) terms appear separately). Finally, because the Brillouin zone is symmetric, replacing \(k\rightarrow-k\) simply relabels the terms in a sum over \(k\): \(\sum_{k}f\left(-k\right)=\sum_{k}f\left(k\right)\) (at the zone boundary, \(k=\pi/a\) and \(-\pi/a\) are the same point). Since also \(\omega_{-k}=\omega_{k}\), the second term is equal to the first, and we have

\[\hat{H}=\sum_{k}\omega_{k}\left(\hat{a}^{\dagger}_{k}\hat{a}_{k}+\frac{1}{2}\hat{\mathbb{I}}\right).\]

Hence, using the Fourier transform we have written the coupled QHOs in terms of independent creation and annihilation operators. The number operator

\[\hat{n}_{k}\triangleq\hat{a}^{\dagger}_{k}\hat{a}_{k}\]

counts the number of particles in mode \(k\), i.e. with wavevector \(k\).

The \(k=0\) mode. If \(\omega_{0}=0\), as for sound in a real crystal, then \(\omega_{k=0}=0\). This mode is a uniform translation of the whole chain, and its Hamiltonian \(\hat{\Pi}^{\dagger}_{0}\hat{\Pi}_{0}/2M\) is that of a free particle (the centre of mass of the chain) rather than an oscillator: the definition of \(\hat{a}_{k=0}\) breaks down, since it involves dividing by \(\omega_{k=0}\). The simplest fix is to keep a small \(\omega_{0}>0\), which pins the chain in place and gaps out this mode; we will do so from now on (alternatively, treat the centre of mass separately). Note that \(\omega_{k=0}=\omega_{0}\).

1.5.1 Excitations of a Quantum Chain: Fock Space

Each operator \(\hat{a}^{\dagger}_{k}\) works exactly like a creation operator acts on a single QHO. Acting it on the ground state creates an excitation with a well defined wavevector. There are \(N\) independent \(k\) modes in the Brillouin zone; for notational simplicity let’s just number them \(k=1,2,\ldots,N\). The ground state, or vacuum state, \(|\Omega\rangle\) is a product of \(N\) independent modes:

\[|\Omega\rangle=|0_{k=1}\rangle\otimes|0_{k=2}\rangle\otimes|0_{k=3}\rangle\otimes|0_{k=4}\rangle\otimes\ldots\]

which can be written compactly as

\[|\Omega\rangle=|0000\ldots\rangle.\]

Acting a creation operator with wavevector \(k\) raises mode \(k\), at a cost of \(\omega_{k}\). Then

\[\hat{a}^{\dagger}_{k=3}|\Omega\rangle=|00100\ldots\rangle\]

or

\[\frac{1}{\sqrt{3!4!2!}}\left(\hat{a}^{\dagger}_{k=3}\right)^{3}\left(\hat{a}^{\dagger}_{k=2}\right)^{4}\left(\hat{a}^{\dagger}_{k=1}\right)^{2}|\Omega\rangle=|24300\ldots\rangle\]

and so on (the prefactor normalises the state, since \(\hat{a}^{\dagger}|n\rangle=\sqrt{n+1}|n+1\rangle\)). There’s no problem with exciting any given mode multiple times, just as there’s no problem with exciting a single QHO to its \(n^{\text{th}}\) energy level. Note that \(|24300\ldots\rangle\) is just shorthand for a tensor product of \(N\) separate kets:

\[|24300\ldots\rangle\triangleq|k=1,n=2\rangle\otimes|k=2,n=4\rangle\otimes|k=3,n=3\rangle\otimes|k=4,n=0\rangle\otimes\ldots\]

In single-particle QM we consider kets living in a Hilbert space (with some caveats!). The states here live in a Fock space: the direct sum

\[\mathcal{F}=\mathcal{H}_{0}\oplus\mathcal{H}_{1}\oplus\mathcal{H}_{2}\oplus\ldots\]

where \(\mathcal{H}_{n}\) is the Hilbert space of states containing \(n\) excitations in total (\(\mathcal{H}_{0}\) contains only \(|\Omega\rangle\)). Fock space is itself a Hilbert space; what is new is that particle number is not fixed, and operators such as \(\hat{a}^{\dagger}_{k}\) move states between the different \(\mathcal{H}_{n}\). The occupation number states \(|n_{1}n_{2}n_{3}\ldots\rangle\) form an orthonormal basis for it.

You can equally well create an excitation localised at site \(n\) by acting the creation operator

\[\hat{a}^{\dagger}_{n}\triangleq\frac{1}{\sqrt{N}}\sum_{k}\exp\left(-ikan\right)\hat{a}^{\dagger}_{k}\]

on the vacuum (for the uncoupled chain this is exactly the creation operator of oscillator \(n\); once the springs are on, we simply take it as the definition of ‘an excitation at \(n\)’):

\[\begin{aligned} \hat{a}^{\dagger}_{n}|\Omega\rangle & =\frac{1}{\sqrt{N}}\sum_{k}\exp\left(-ikan\right)\hat{a}^{\dagger}_{k}|\Omega\rangle\\ & =\frac{1}{\sqrt{N}}\left(\exp\left(-ik_{1}an\right)\hat{a}^{\dagger}_{k_{1}}|000\ldots\rangle+\exp\left(-ik_{2}an\right)\hat{a}^{\dagger}_{k_{2}}|000\ldots\rangle+\ldots\right)\\ & =\frac{1}{\sqrt{N}}\left(\exp\left(-ik_{1}an\right)|100\ldots\rangle+\exp\left(-ik_{2}an\right)|010\ldots\rangle+\ldots\right) \end{aligned}\]

where \(k_{1},k_{2},\ldots\) are the allowed wavevectors. You can think by analogy of plucking a guitar string at a given position along its length. The effect is to excite all the harmonics of the string with different amplitudes.

1.6 Classical Field Theory

So far we have worked with discrete positions \(n\). A field is a continuum, so to proceed we must smooth out our description. Our motivation is that we are typically only interested in lengthscales / wavelengths significantly larger than a lattice spacing, and so we can ‘coarse grain’ or smooth over the lattice details.

1.6.1 Classical Lagrangian Field Theory

In the classical balls and springs model we can make a conceptual leap by approximating our discrete displacements by a continuum field. This amounts to making the change

\[u_{n,t}\rightarrow\varphi_{\vb{x},t}. \tag{1.13}\]

I have introduced bold \(\vb{x}\) to emphasise that this works in any number \(d\) of spatial dimensions, and because we will shortly need to differentiate between \(d\)-vectors \(\vb{x}\) and spacetime vectors \(x^{\mu}=\left(t,\vb{x}\right)^{\mu}\). Eq. 1.13 is the defining expression of field theory. While conceptually simple, the consequences are profound. Rather than individual balls oscillating about discrete sites \(n\) with displacements \(u_{n,t}\), we instead have a continuous field defined at all positions \(\vb{x}\) and times \(t\). You can think of it as a flexible membrane, like the surface of a drum. In crystals the result leads to simplified mathematics. In the standard model, however, there are no individual sites: only the smooth field exists, and it does so across all of spacetime. It is worth taking the time to think about Eq. 1.13 conceptually before proceeding.

The following might help (or it might do the opposite!). In classical mechanics it is common to denote generalised co-ordinates as \(q_{n,t}\). In this alternative notation, when we coarse grain and smooth out, we simply switch \(q_{n,t}\rightarrow q_{\vb{x},t}\), interpolating the discrete variable \(n\) with continuous positions \(\vb{x}\). Then we just relabel \(q_{\vb{x},t}\rightarrow\varphi_{\vb{x},t}\) as a matter of convention.

Coarse-graining requires three replacements. The displacements become the field; differences between neighbours become derivatives; and since there is one ball per length \(a\), sums over sites become integrals:

\[u_{n}\rightarrow\varphi_{x},\qquad u_{n+1}-u_{n}\rightarrow a\partial_{x}\varphi_{x}+\mathcal{O}\left(a^{2}\right),\qquad\sum_{n}\rightarrow\int\frac{\text{d}x}{a}.\]

The Lagrangian of Eq. 1.10 then coarse-grains as follows:

\[\begin{aligned} L & =\sum_{n}\frac{1}{2}M\dot{u}^{2}_{n}-\frac{1}{2}M\omega^{2}_{0}u^{2}_{n}-\frac{1}{2}K\left(u_{n+1}-u_{n}\right)^{2}\\ & \downarrow\text{coarse-grain}\\ L & \approx\int\frac{\text{d}x}{a}\left(\frac{1}{2}M\dot{\varphi}^{2}_{x}-\frac{1}{2}M\omega^{2}_{0}\varphi^{2}_{x}-\frac{1}{2}Ka^{2}\left(\partial_{x}\varphi_{x}\right)^{2}\right)\\ & =\int\text{d}x\left(\frac{1}{2}\rho\dot{\varphi}^{2}_{x}-\frac{1}{2}\rho\omega^{2}_{0}\varphi^{2}_{x}-\frac{1}{2}Y\left(\partial_{x}\varphi_{x}\right)^{2}\right) \end{aligned}\]

where \(\rho\triangleq M/a\) is the mass per unit length and \(Y\triangleq Ka\) is the stiffness of the continuum (its Young’s modulus). We kept only the lowest order in \(a\), since we are considering coarse-grained (smooth) fields. The same procedure works in \(d\) dimensions (on a hypercubic lattice with springs along each axis), with \(\sum_{n}\rightarrow\int\text{d}^{d}\vb{x}/a^{d}\) and \(\left(\partial_{x}\varphi\right)^{2}\rightarrow\left(\nabla\varphi\right)^{2}\), giving \(\rho=M/a^{d}\) and \(Y=Ka^{2-d}\). The Lagrangian is \(L=\int\text{d}^{d}\vb{x}\mathcal{L}\) with the

Lagrangian density of the coarse-grained chain:

\[\mathcal{L}=\frac{1}{2}\rho\dot{\varphi}^{2}_{\vb{x}}-\frac{1}{2}Y\left(\nabla\varphi_{\vb{x}}\right)^{2}-\frac{1}{2}\rho\omega^{2}_{0}\varphi^{2}_{\vb{x}}. \tag{1.14}\]

Its Euler Lagrange equation (derived below) is \(\rho\ddot{\varphi}=Y\nabla^{2}\varphi-\rho\omega^{2}_{0}\varphi\): a wave equation with speed

\[c=\sqrt{\frac{Y}{\rho}}=a\sqrt{\frac{K}{M}}\]

which is precisely the speed of sound we found from the discrete chain, in any dimension. As hoped, we still have our classical sound mode propagating in the coarse-grained theory.

There are now only two parameters that matter, so let’s tidy up. First, absorb the density into the field, \(\varphi\rightarrow\varphi/\sqrt{\rho}\) (equivalently, measure the field in units where \(\rho=1\)):

\[\mathcal{L}=\frac{1}{2}\dot{\varphi}^{2}_{\vb{x}}-\frac{1}{2}c^{2}\left(\nabla\varphi_{\vb{x}}\right)^{2}-\frac{1}{2}\omega^{2}_{0}\varphi^{2}_{\vb{x}}.\]

Second, the only velocity in the problem is now \(c\), so we can choose to measure time in units in which \(c=1\): the medium’s own ‘natural units’, just as \(\hbar=c=1\) for the vacuum. Finally, rename \(\omega_{0}\triangleq m\). Then

\[\mathcal{L}=\frac{1}{2}\dot{\varphi}^{2}_{\vb{x}}-\frac{1}{2}\left(\nabla\varphi_{\vb{x}}\right)^{2}-\frac{1}{2}m^{2}\varphi^{2}_{\vb{x}}=\frac{1}{2}\partial^{\mu}\varphi\partial_{\mu}\varphi-\frac{1}{2}m^{2}\varphi^{2}. \tag{1.15}\]

This is the Klein Gordon Lagrange density, which we will meet again in Chapter 2. The on-site frequency of the individual oscillators has become the mass of the field (this is why the balls’ mass was called \(M\): the symbol \(m\) is conventionally reserved for the mass of the field). For sound in a crystal \(m=0\). To restore \(c\) in any result below, replace \(\vb{k}\rightarrow c\vb{k}\) in frequencies.

As before, neglect the \(t\) label since everything is evaluated at the same time. The Lagrangian is

\[L=\int\text{d}^{d}\vb{x}\mathcal{L}\left(\varphi_{\vb{x}},\dot{\varphi}_{\vb{x}},\nabla\varphi_{\vb{x}}\right)\]

and action

\[S=\int\text{d}tL=\int\text{d}^{D}x\mathcal{L}\]

where \(D=d+1\). Now that \(c=1\), this notation emphasises that the action is a Lorentz invariant scalar. We can find the Euler Lagrange equations using the principle of least action:

\[\begin{aligned} S\left[\varphi+\lambda\epsilon\right] & =\int\text{d}t\int\text{d}^{d}\vb{x}\left\{ \frac{1}{2}\left(\dot{\varphi}_{\vb{x}}+\lambda\dot{\epsilon}_{\vb{x}}\right)^{2}-\frac{1}{2}\left(\nabla\varphi_{\vb{x}}+\lambda\nabla\epsilon_{\vb{x}}\right)^{2}-\frac{1}{2}m^{2}\left(\varphi_{\vb{x}}+\lambda\epsilon_{\vb{x}}\right)^{2}\right\} \\ & \Downarrow\\ \left.\left(\frac{\partial S\left[\varphi+\lambda\epsilon\right]}{\partial\lambda}\right)_{\varphi,\epsilon}\right|_{\lambda=0} & =\int\text{d}t\int\text{d}^{d}\vb{x}\left\{ \dot{\varphi}_{\vb{x}}\dot{\epsilon}_{\vb{x}}-\nabla\varphi_{\vb{x}}\cdot\nabla\epsilon_{\vb{x}}-m^{2}\varphi_{\vb{x}}\epsilon_{\vb{x}}\right\} \\ & =\int\text{d}t\int\text{d}^{d}\vb{x}\,\epsilon_{\vb{x}}\left\{ -\ddot{\varphi}_{\vb{x}}+\nabla^{2}\varphi_{\vb{x}}-m^{2}\varphi_{\vb{x}}\right\} \end{aligned}\]

using integration by parts with respect to time in the first term, and space in the second. Setting this equal to zero for all \(\epsilon\) requires that

\[\ddot{\varphi}_{\vb{x}}=\nabla^{2}\varphi_{\vb{x}}-m^{2}\varphi_{\vb{x}}.\]

For \(m=0\) this is the classical wave equation (\(\ddot{\varphi}=c^{2}\nabla^{2}\varphi\) with \(c\) restored); for \(m\ne0\) it is the Klein Gordon equation.

NB you will need to understand the principle of coarse graining, but will not need to memorise the particular case detailed above.

1.6.2 Classical Hamiltonian Field Theory

As before, we need the momentum conjugate to the field. This is given by

\[\pi_{\vb{x},t}\triangleq\left(\frac{\partial\mathcal{L}}{\partial\dot{\varphi}_{\vb{x},t}}\right)_{\varphi_{\vb{x},t}}\]

where we will continue to neglect the \(t\) label. In our case we find

\[\pi_{\vb{x}}=\dot{\varphi}_{\vb{x}}.\]

The Hamiltonian is again the Legendre transform of the Lagrangian:

\[H\triangleq\left(\int\text{d}^{d}\vb{x}\pi_{\vb{x}}\dot{\varphi}_{\vb{x}}\right)-L\]

and in our case this gives

\[H=\int\text{d}^{d}\vb{x}\left(\frac{1}{2}\pi^{2}_{\vb{x}}+\frac{1}{2}\left(\nabla\varphi_{\vb{x}}\right)^{2}+\frac{1}{2}m^{2}\varphi^{2}_{\vb{x}}\right).\]

1.7 Quantum Field Theory of the Harmonic Chain

1.7.1 Canonical Field Quantization

Finally we are ready to create our first quantum field. To do so, we simply promote our classical fields to non-commuting quantum operators:

\[\begin{aligned} \varphi_{\vb{x}} & \rightarrow\hat{\varphi}_{\vb{x}}\\ \pi_{\vb{x}} & \rightarrow\hat{\pi}_{\vb{x}}. \end{aligned}\]

We are now explicitly justified in neglecting the \(t\) label: these are operators at a single time, \(t=0\), at which the Schroedinger and Heisenberg pictures coincide. We will restore the time dependence in Section 1.8. The quantum fields obey the canonical commutation relations

\[\left[\hat{\varphi}_{\vb{x}},\hat{\varphi}_{\vb{y}}\right]=\left[\hat{\pi}_{\vb{x}},\hat{\pi}_{\vb{y}}\right]=0\]

and

\[\left[\hat{\varphi}_{\vb{x}},\hat{\pi}_{\vb{y}}\right]=i\delta^{d}\left(\vb{x}-\vb{y}\right)\hat{\mathbb{I}}.\]

Note that in this description, unlike in single-particle quantum mechanics, positions are not themselves operators. Rather, positions should be thought of as like the fixed discrete lattice sites \(n\), while the field is like \(u_{n}\), the displacement of the ball from that site.

Our Hamiltonian has translational symmetry. We can again take advantage of this by working in the Fourier basis. We keep the system in a box of volume \(V\) with periodic boundary conditions, and use the unitary normalisation of Conventions:

\[\begin{aligned} \hat{\varphi}_{\vb{x}} & =\frac{1}{\sqrt{V}}\sum_{\vb{k}}\exp\left(i\vb{k}\cdot\vb{x}\right)\hat{\Phi}_{\vb{k}}\\ \hat{\pi}_{\vb{x}} & =\frac{1}{\sqrt{V}}\sum_{\vb{k}}\exp\left(-i\vb{k}\cdot\vb{x}\right)\hat{\Pi}_{\vb{k}} \end{aligned} \tag{1.16}\]

and the inverse transforms

\[\begin{aligned} \hat{\Phi}_{\vb{k}} & =\frac{1}{\sqrt{V}}\int_{V}\text{d}^{d}\vb{x}\exp\left(-i\vb{k}\cdot\vb{x}\right)\hat{\varphi}_{\vb{x}}\\ \hat{\Pi}_{\vb{k}} & =\frac{1}{\sqrt{V}}\int_{V}\text{d}^{d}\vb{x}\exp\left(i\vb{k}\cdot\vb{x}\right)\hat{\pi}_{\vb{x}}. \end{aligned} \tag{1.17}\]

Important point: in going to the field description, we have moved to describing real space with a continuum, meaning we use integrals over \(\vb{x}\) instead of a discrete sum over lattice sites. However, in a box with periodic boundary conditions, \(\vb{k}\) remains discrete, \(\vb{k}=2\pi\vb{n}/L\): an integer number of whole wavelengths must fit around the system. Moreover, since our field is a coarse-graining of the chain, it is only meaningful for wavelengths longer than the lattice spacing, so we keep the wavevectors within the Brillouin zone, \(\left|k_{i}\right|\le\pi/a\). There are then exactly as many modes as there were balls, and every sum over \(\vb{k}\) below is finite. In Chapter 2 we will take the infinite-volume limit, in which \(\vb{k}\) becomes continuous, using the box-to-continuum dictionary. In general, whether \(\vb{k}\) is discrete or continuous depends on the boundary conditions.

The new fields obey the canonical commutation relations

\[\left[\hat{\Phi}_{\vb{k}},\hat{\Phi}_{\vb{q}}\right]=\left[\hat{\Pi}_{\vb{k}},\hat{\Pi}_{\vb{q}}\right]=0\]

and

\[\begin{aligned} \left[\hat{\Phi}_{\vb{k}},\hat{\Pi}_{\vb{q}}\right] & =\frac{1}{V}\int\text{d}^{d}\vb{x}\int\text{d}^{d}\vb{y}\exp\left(i\left(\vb{q}\cdot\vb{y}-\vb{k}\cdot\vb{x}\right)\right)\left[\hat{\varphi}_{\vb{x}},\hat{\pi}_{\vb{y}}\right]\\ & =\frac{1}{V}\int\text{d}^{d}\vb{x}\int\text{d}^{d}\vb{y}\exp\left(i\left(\vb{q}\cdot\vb{y}-\vb{k}\cdot\vb{x}\right)\right)i\delta^{d}\left(\vb{x}-\vb{y}\right)\hat{\mathbb{I}}\\ & =\frac{1}{V}\int\text{d}^{d}\vb{x}\exp\left(i\left(\vb{q}-\vb{k}\right)\cdot\vb{x}\right)i\hat{\mathbb{I}}\\ & =i\delta_{\vb{k},\vb{q}}\hat{\mathbb{I}} \end{aligned}\]

using the useful relation

\[\int_{V}\text{d}^{d}\vb{x}\exp\left(i\left(\vb{q}-\vb{k}\right)\cdot\vb{x}\right)=V\delta_{\vb{k},\vb{q}}.\]

As for the chain, the Fourier components of the field are not Hermitian (although \(\hat{\varphi}_{\vb{x}}\) itself is):

\[\begin{aligned} \hat{\Phi}{}^{\dagger}_{\vb{k}} & =\hat{\Phi}_{-\vb{k}}\\ \hat{\Pi}{}^{\dagger}_{\vb{k}} & =\hat{\Pi}_{-\vb{k}}. \end{aligned}\]

1.7.2 The Hamiltonian of the Quantum Field

We canonically quantize our field Hamiltonian as

\[\begin{aligned} H & =\int\text{d}^{d}\vb{x}\left(\frac{1}{2}\pi^{2}_{\vb{x}}+\frac{1}{2}\left(\nabla\varphi_{\vb{x}}\right)^{2}+\frac{1}{2}m^{2}\varphi^{2}_{\vb{x}}\right)\\ & \downarrow\text{canonically quantize}\nonumber \\ \hat{H} & =\int\text{d}^{d}\vb{x}\left(\frac{1}{2}\hat{\pi}^{2}_{\vb{x}}+\frac{1}{2}\left(\nabla\hat{\varphi}_{\vb{x}}\right)^{2}+\frac{1}{2}m^{2}\hat{\varphi}^{2}_{\vb{x}}\right) \end{aligned}\]

and Fourier transforming using Eq. 1.16 gives, term by term,

\[\begin{aligned} \int\text{d}^{d}\vb{x}\hat{\pi}^{2}_{\vb{x}} & =\frac{1}{V}\sum_{\vb{k},\vb{q}}\int\text{d}^{d}\vb{x}\exp\left(-i\left(\vb{k}+\vb{q}\right)\cdot\vb{x}\right)\hat{\Pi}_{\vb{k}}\hat{\Pi}_{\vb{q}}=\sum_{\vb{k}}\hat{\Pi}_{-\vb{k}}\hat{\Pi}_{\vb{k}}\\ \int\text{d}^{d}\vb{x}\left(\nabla\hat{\varphi}_{\vb{x}}\right)^{2} & =\frac{1}{V}\sum_{\vb{k},\vb{q}}\left(i\vb{k}\right)\cdot\left(i\vb{q}\right)\int\text{d}^{d}\vb{x}\exp\left(i\left(\vb{k}+\vb{q}\right)\cdot\vb{x}\right)\hat{\Phi}_{\vb{k}}\hat{\Phi}_{\vb{q}}=\sum_{\vb{k}}\vb{k}^{2}\hat{\Phi}_{-\vb{k}}\hat{\Phi}_{\vb{k}}\\ \int\text{d}^{d}\vb{x}\hat{\varphi}^{2}_{\vb{x}} & =\sum_{\vb{k}}\hat{\Phi}_{-\vb{k}}\hat{\Phi}_{\vb{k}} \end{aligned}\]

(in each case the \(\vb{x}\) integral gives \(V\delta_{\vb{k},-\vb{q}}\)), so that

\[\hat{H}=\sum_{\vb{k}}\left\{ \frac{1}{2}\hat{\Pi}^{\dagger}_{\vb{k}}\hat{\Pi}_{\vb{k}}+\frac{1}{2}\omega^{2}_{\vb{k}}\hat{\Phi}^{\dagger}_{\vb{k}}\hat{\Phi}_{\vb{k}}\right\} \tag{1.18}\]

where we have defined

\[\omega_{\vb{k}}\triangleq\sqrt{\vb{k}^{2}+m^{2}}.\]

Restoring \(c\) this is \(\omega_{\vb{k}}=\sqrt{c^{2}\vb{k}^{2}+\omega^{2}_{0}}\): the long-wavelength limit of the chain’s dispersion, Eq. 1.9, with no stray factors. This is exactly the form of Eq. 1.12 with \(M\rightarrow1\).

1.7.3 Creation and Annihilation Operators

The procedure now follows Section 1.5 exactly. Defining the field creation operator

\[\hat{a}^{\dagger}_{\vb{k}}\triangleq\sqrt{\frac{\omega_{\vb{k}}}{2}}\left(\hat{\Phi}_{-\vb{k}}-\frac{i}{\omega_{\vb{k}}}\hat{\Pi}_{\vb{k}}\right)\]

we find

\[\begin{aligned} \left[\hat{a}_{\vb{k}},\hat{a}^{\dagger}_{\vb{q}}\right] & =\delta_{\vb{k},\vb{q}}\hat{\mathbb{I}} \end{aligned}\]

and

\[\left[\hat{a}_{\vb{k}},\hat{a}_{\vb{q}}\right]=\left[\hat{a}^{\dagger}_{\vb{k}},\hat{a}^{\dagger}_{\vb{q}}\right]=0.\]

We can now re-arrange to get

\[\begin{aligned} \hat{\Phi}_{\vb{k}} & =\sqrt{\frac{1}{2\omega_{\vb{k}}}}\left(\hat{a}_{\vb{k}}+\hat{a}^{\dagger}_{-\vb{k}}\right)\\ \hat{\Pi}_{\vb{k}} & =i\sqrt{\frac{\omega_{\vb{k}}}{2}}\left(\hat{a}^{\dagger}_{\vb{k}}-\hat{a}_{-\vb{k}}\right) \end{aligned}\]

and so the Hamiltonian is

\[\begin{aligned} \hat{H} & =\sum_{\vb{k}}\left\{ \frac{1}{2}\hat{\Pi}^{\dagger}_{\vb{k}}\hat{\Pi}_{\vb{k}}+\frac{1}{2}\omega^{2}_{\vb{k}}\hat{\Phi}^{\dagger}_{\vb{k}}\hat{\Phi}_{\vb{k}}\right\} \\ & =\sum_{\vb{k}}\left\{ \frac{\omega_{\vb{k}}}{4}\left(\hat{a}_{\vb{k}}-\hat{a}^{\dagger}_{-\vb{k}}\right)\left(\hat{a}^{\dagger}_{\vb{k}}-\hat{a}_{-\vb{k}}\right)+\frac{\omega_{\vb{k}}}{4}\left(\hat{a}^{\dagger}_{\vb{k}}+\hat{a}_{-\vb{k}}\right)\left(\hat{a}_{\vb{k}}+\hat{a}^{\dagger}_{-\vb{k}}\right)\right\} \\ & =\sum_{\vb{k}}\frac{\omega_{\vb{k}}}{4}\left(\hat{a}^{\dagger}_{\vb{k}}\hat{a}_{\vb{k}}+\hat{a}_{\vb{k}}\hat{a}^{\dagger}_{\vb{k}}+\hat{a}^{\dagger}_{-\vb{k}}\hat{a}_{-\vb{k}}+\hat{a}_{-\vb{k}}\hat{a}^{\dagger}_{-\vb{k}}\right)\\ & =\sum_{\vb{k}}\frac{\omega_{\vb{k}}}{4}\left(\hat{a}^{\dagger}_{\vb{k}}\hat{a}_{\vb{k}}+\left(\hat{a}^{\dagger}_{\vb{k}}\hat{a}_{\vb{k}}+\left[\hat{a}_{\vb{k}},\hat{a}^{\dagger}_{\vb{k}}\right]\right)+\hat{a}^{\dagger}_{-\vb{k}}\hat{a}_{-\vb{k}}+\left(\hat{a}^{\dagger}_{-\vb{k}}\hat{a}_{-\vb{k}}+\left[\hat{a}_{-\vb{k}},\hat{a}^{\dagger}_{-\vb{k}}\right]\right)\right)\\ & =\sum_{\vb{k}}\omega_{\vb{k}}\left(\hat{a}^{\dagger}_{\vb{k}}\hat{a}_{\vb{k}}+\frac{1}{2}\hat{\mathbb{I}}\right) \end{aligned} \tag{1.19}\]

where in the last line we used \(\left[\hat{a}_{\vb{k}},\hat{a}^{\dagger}_{\vb{k}}\right]=\hat{\mathbb{I}}\) and relabelled \(\vb{k}\rightarrow-\vb{k}\) in the \(-\vb{k}\) terms (the set of allowed \(\vb{k}\) is symmetric).

1.7.4 The zero-point energy

The ground state energy is

\[E_{0}=\frac{1}{2}\sum_{\vb{k}}\omega_{\vb{k}}.\]

In our box, with the lattice cutoff, this is a large but finite number: one zero-point energy \(\omega_{\vb{k}}/2\) per mode, and there are exactly as many modes as there were balls. Two things happen as we remove the box and the lattice.

• Infinite volume, \(V\rightarrow\infty\) at fixed lattice spacing. Using the dictionary, \(\sum_{\vb{k}}\rightarrow V\int\text{d}^{d}\vb{k}/\left(2\pi\right)^{d}\), so \(E_{0}=V\int_{\text{BZ}}\frac{\text{d}^{d}\vb{k}}{\left(2\pi\right)^{d}}\frac{\omega_{\vb{k}}}{2}\) grows in proportion to the volume. This is a harmless infrared divergence: the energy density \(E_{0}/V\) is finite. (In the continuum notation you will often see this written with a factor \(\left(2\pi\right)^{d}\delta^{d}\left(\vb{0}\right)\), which is just \(V\).)
• Continuum limit, \(a\rightarrow0\). The Brillouin zone grows without bound, and the energy density \(\sim\int^{\pi/a}\text{d}^{d}\vb{k}\left|\vb{k}\right|\sim a^{-\left(d+1\right)}\) diverges. This ultraviolet divergence is the price of pretending that the field description holds down to arbitrarily short distances.

Neither divergence has anything to do with the positions \(\vb{x}\) forming an uncountable set: what matters is the number of modes, which is the number of oscillators we started with.

1.7.5 Normal Ordering

The zero-point energy is only the first of many infinities one will encounter in QFT (once we take the limits above). This one is perfectly harmless, since the absolute energy is meaningless: it is only differences in energies that have meaning. We can simply redefine our zero of energy.

There is a more systematic way to do this, however. Note that the zero-point energy came about because we had to use the commutator \(\left[\hat{a}_{\vb{k}},\hat{a}^{\dagger}_{\vb{k}}\right]\). This reflects an ambiguity introduced in canonical quantization: classical observables commute, but quantum operators do not, so a single classical expression corresponds to many quantum operators. Define the classical mode amplitudes \(\alpha_{\vb{k}}\triangleq\sqrt{\omega_{\vb{k}}/2}\left(\Phi_{\vb{k}}+i\Pi_{-\vb{k}}/\omega_{\vb{k}}\right)\) (the classical counterparts of \(\hat{a}_{\vb{k}}\)). The classical Hamiltonian is \(H=\sum_{\vb{k}}\omega_{\vb{k}}\alpha^{*}_{\vb{k}}\alpha_{\vb{k}}\), which could equally be written \(\sum_{\vb{k}}\omega_{\vb{k}}\alpha_{\vb{k}}\alpha^{*}_{\vb{k}}\) or \(\sum_{\vb{k}}\frac{\omega_{\vb{k}}}{2}\left(\alpha^{*}_{\vb{k}}\alpha_{\vb{k}}+\alpha_{\vb{k}}\alpha^{*}_{\vb{k}}\right)\). Quantum mechanically these become \(\hat{a}^{\dagger}\hat{a}\), \(\hat{a}\hat{a}^{\dagger}=\hat{a}^{\dagger}\hat{a}+\hat{\mathbb{I}}\), or their average, which differ by constants. Writing the classical Hamiltonian in terms of \(\varphi\) and \(\pi\) and then promoting to operators silently chose the symmetric ordering, and this is where the \(\frac{1}{2}\) comes from. Every choice gives the same energy differences and the same equations of motion, so we are free to choose the most convenient.

We define the normal ordering of a chain of operators to be such that all creation operators are to the left of all annihilation operators. We denote the normal ordering of an operator \(\hat{\mathcal{O}}\) to be \(:\hat{\mathcal{O}}:\). In our case

\[\hat{H}=\sum_{\vb{k}}\frac{\omega_{\vb{k}}}{4}\left(\hat{a}^{\dagger}_{\vb{k}}\hat{a}_{\vb{k}}+\hat{a}_{\vb{k}}\hat{a}^{\dagger}_{\vb{k}}+\hat{a}^{\dagger}_{-\vb{k}}\hat{a}_{-\vb{k}}+\hat{a}_{-\vb{k}}\hat{a}^{\dagger}_{-\vb{k}}\right)\]

and so its normal ordering, placing all creation operators to the left (without worrying about the resulting commutators) is

\[:\hat{H}:=\sum_{\vb{k}}\frac{\omega_{\vb{k}}}{2}\left(\hat{a}^{\dagger}_{\vb{k}}\hat{a}_{\vb{k}}+\hat{a}^{\dagger}_{-\vb{k}}\hat{a}_{-\vb{k}}\right)=\sum_{\vb{k}}\omega_{\vb{k}}\hat{a}^{\dagger}_{\vb{k}}\hat{a}_{\vb{k}}. \tag{1.20}\]

This means that using normal ordered operators instead of the operators themselves, we subtract exactly the zero-point energy:

\[:\hat{H}:=\hat{H}-\frac{1}{2}\sum_{\vb{k}}\omega_{\vb{k}}\hat{\mathbb{I}}.\]

In general, normal ordering removes this type of term, introduced by the ordering ambiguity of quantum operators under canonical quantization, whatever the size of the box or the cutoff. It shifts the zero of energy so that the ground state has zero energy. Eq. 1.20 is a totally reasonable Hamiltonian, in which all modes are decoupled.

1.7.6 Excitations of a Quantum Field: Particles

Now let’s take a step back and inspect our handywork, as we did for the quantum chain in Section 1.5.1 (you might like to re-read that section quickly). Each operator \(\hat{a}^{\dagger}_{\vb{k}}\) works exactly like a creation operator acts on a single QHO. Acting it on the ground state creates an excitation with a well defined wavevector. The ground state, or vacuum state, \(|\Omega\rangle\) is a product of independent modes, one for each allowed wavevector. In 1D:

\[|\Omega\rangle=|0_{k_{1}}\rangle\otimes|0_{k_{2}}\rangle\otimes|0_{k_{3}}\rangle\otimes|0_{k_{4}}\rangle\otimes\ldots \tag{1.21}\]

Note that, while positions have become continuous variables (forming an uncountably infinite set), the possible wavevectors remain discrete, and with the lattice cutoff there are finitely many of them. Our boundary conditions quantize the possible wavevectors. This is quite natural: a guitar string exists at an uncountably infinite number of positions between its two ends, yet its harmonics are standing waves which fit along the string. Hence the notation in Eq. 1.21 is perfectly well defined, and the occupation number states built on it form a basis for the Fock space exactly as for the chain.¹ Let’s look at what happens when we act the field operator \(\hat{\Phi}^{\dagger}_{\vb{k}}\) on the vacuum:

\[\begin{aligned} \hat{\Phi}^{\dagger}_{\vb{k}}|\Omega\rangle & =\sqrt{\frac{1}{2\omega_{\vb{k}}}}\left(\hat{a}^{\dagger}_{\vb{k}}+\hat{a}_{-\vb{k}}\right)|\Omega\rangle\\ & =\frac{1}{\sqrt{2E_{\vb{k}}}}\hat{a}^{\dagger}_{\vb{k}}|\Omega\rangle\\ & =\frac{1}{\sqrt{2E_{\vb{k}}}}|\vb{k}\rangle\\ & =\frac{1}{2E_{\vb{k}}}|k\rangle \end{aligned} \tag{1.22}\]

where \(E_{\vb{k}}=\omega_{\vb{k}}\) and we have defined the single-particle state \(|\vb{k}\rangle\triangleq\hat{a}^{\dagger}_{\vb{k}}|\Omega\rangle\) (with \(\langle\vb{k}|\vb{q}\rangle=\delta_{\vb{k},\vb{q}}\) in the box). In the last line we have adopted the standard notation

\[|k\rangle=\sqrt{2E_{\vb{k}}}|\vb{k}\rangle \tag{1.23}\]

where \(k\) is a spacetime vector and \(\vb{k}\) a spatial one. This relativistic normalisation guarantees the Lorentz invariance of objects such as \(\langle k|q\rangle\), which is \(2E_{\vb{k}}\delta_{\vb{k},\vb{q}}\) in the box and \(2E_{\vb{k}}\left(2\pi\right)^{d}\delta^{d}\left(\vb{k}-\vb{q}\right)\) in infinite volume (see Relativistic conventions). Our theory started out non-relativistic, but since QFT is so frequently applied in relativistic settings, notation such as Eq. 1.23 has been established with this in mind.

What is \(|k\rangle\) physically? It is nothing other than a particle! We call this particle a phonon. It has crystal-momentum \(\vb{k}\), and energy \(E_{\vb{k}}=\omega_{\vb{k}}=\sqrt{\vb{k}^{2}+m^{2}}\).

Note that Eq. 1.22 means we can create a particle with momentum \(\vb{k}\) from the vacuum using either \(\hat{a}^{\dagger}_{\vb{k}}\) or \(\hat{\Phi}^{\dagger}_{\vb{k}}\) (acting on any other state, \(\hat{\Phi}^{\dagger}_{\vb{k}}\) can also annihilate a particle of momentum \(-\vb{k}\)). The latter is more natural in a relativistic setting. In textbooks you are more likely to see particles created with the field operator than with the creation operator itself, although the reason is not always discussed.

At this point you might well ask why we would expect our theory to be relativistic, given that we started from balls and springs with absolute positions \(an\) and absolute times \(t\). Essentially, the coarse-grained theory is relativistic even though the microscopic theory is not: Eq. 1.15 is Lorentz invariant, with the speed of sound playing the role of the speed of light. The lattice breaks this ‘emergent’ Lorentz invariance only at wavelengths comparable to \(a\): for example, the full dispersion Eq. 1.9 contains \(\sin^{2}\left(ka/2\right)\) rather than \(\left(ka/2\right)^{2}\). We will return to this point in the next chapter.

We can denote multi-particle states as

\[|k_{1},k_{2}\rangle=\sqrt{2E_{\vb{k}_{1}}2E_{\vb{k}_{2}}}\hat{a}^{\dagger}_{\vb{k}_{2}}\hat{a}^{\dagger}_{\vb{k}_{1}}|\Omega\rangle=2E_{\vb{k}_{1}}2E_{\vb{k}_{2}}\hat{\Phi}^{\dagger}_{\vb{k}_{2}}\hat{\Phi}^{\dagger}_{\vb{k}_{1}}|\Omega\rangle\]

(the second form holding for \(\vb{k}_{2}\ne-\vb{k}_{1}\)) and so on. You can equally well create a particle at a specific position \(\vb{x}\) by acting the real-space field operator on the vacuum. Since \(\hat{\varphi}_{\vb{x}}\) is Hermitian, Eq. 1.16 gives \(\hat{\varphi}_{\vb{x}}=\hat{\varphi}^{\dagger}_{\vb{x}}=V^{-1/2}\sum_{\vb{k}}\exp\left(-i\vb{k}\cdot\vb{x}\right)\hat{\Phi}^{\dagger}_{\vb{k}}\), so

\[\hat{\varphi}_{\vb{x}}|\Omega\rangle=\frac{1}{\sqrt{V}}\sum_{\vb{k}}\exp\left(-i\vb{k}\cdot\vb{x}\right)\frac{1}{2E_{\vb{k}}}|k\rangle.\]

This is a good place to see the infinite-volume limit in action. Using the dictionary (\(\sum_{\vb{k}}\rightarrow V\int\text{d}^{d}\vb{k}/\left(2\pi\right)^{d}\), and box states \(|k\rangle\rightarrow V^{-1/2}|k\rangle\)), this becomes

\[\begin{aligned} \hat{\varphi}_{\vb{x}}|\Omega\rangle & =\int\frac{\text{d}^{d}\vb{k}}{\left(2\pi\right)^{d}}\frac{1}{2E_{\vb{k}}}\exp\left(-i\vb{k}\cdot\vb{x}\right)|k\rangle\\ & =\int\frac{\text{d}^{D}k}{\left(2\pi\right)^{d}}\delta\left(k^{2}-m^{2}\right)\Theta\left(k^{0}\right)\exp\left(-i\vb{k}\cdot\vb{x}\right)|k\rangle \end{aligned}\]

(where \(k^{2}=k_{\mu}k^{\mu}\)) which is manifestly built from Lorentz invariant pieces. We conventionally denote

\[\hat{\varphi}_{\vb{x}}|\Omega\rangle=|x\rangle.\]

Hence, acting the field operator at position \(\vb{x}\) creates a particle at spacetime position \(x\) (here at \(t=0\)).² Again, you can think by analogy of plucking a guitar string at a given position. The effect is to excite all the harmonics of the string, with different weights. In this case, ‘plucking’ means creating a particle from the vacuum, and the guitar string is the universe itself (here, a quantum crystal).

All of this intuition will carry across when you come to creating photons rather than phonons. However, photons are particles that do not require a medium through which to travel. This should again be natural, since photons are quantized particles of light, which requires no medium through which to propagate, while phonons are quantized particles of sound (lattice vibrations in a crystal) and sound does require a medium.

1.8 Time dependence

Single-particle non-relativistic QM amounts to solving the time dependent Schroedinger equation (TDSE):

\[i\frac{\text{d}|\psi\rangle}{\text{d}t}=\hat{H}|\psi\rangle.\]

Assuming the Hamiltonian is time-independent we have the general solution

\[|\psi\left(t\right)\rangle=\exp\left(-i\hat{H}t\right)|\psi\left(0\right)\rangle\]

which you can confirm by acting \(i\text{d}/\text{d}t\) on both sides. That is, if we have the solution at one time (which we obtain by solving the time independent Schroedinger equation, TISE), we automatically have the solution at all subsequent times. It is convenient to define

the time evolution operator:

\[\hat{U}\left(t\right)\triangleq\exp\left(-i\hat{H}t\right).\]

Since \(\hat{H}\) is Hermitian, \(\hat{U}\) is unitary:

\[\hat{U}\hat{U}^{\dagger}=\hat{U}^{\dagger}\hat{U}=\hat{\mathbb{I}}.\]

For our canonically quantized field operators, such as

\[\hat{\Phi}_{\vb{k}}=\sqrt{\frac{1}{2\omega_{\vb{k}}}}\left(\hat{a}_{\vb{k}}+\hat{a}^{\dagger}_{-\vb{k}}\right)\]

we have so far not written any explicit time dependence: they are operators at \(t=0\), where the Schroedinger and Heisenberg pictures agree. In the Heisenberg picture, operators are time dependent and states are time independent. To obtain the operator \(\hat{\mathcal{O}}_{t}\) at time \(t\) from that at time \(0\) we use the time evolution operator:

\[\hat{\mathcal{O}}_{t}=\hat{U}^{\dagger}_{t}\hat{\mathcal{O}}_{0}\hat{U}_{t}\]

or

\[\hat{\mathcal{O}}_{t}=\exp\left(i\hat{H}t\right)\hat{\mathcal{O}}_{0}\exp\left(-i\hat{H}t\right). \tag{1.24}\]

To add in the explicit time dependence of our field operators \(\hat{\varphi}_{\vb{x}}\) we can therefore use:

\[\hat{\varphi}_{x}\triangleq\hat{\varphi}_{\vb{x},t}=\hat{U}^{\dagger}_{t}\hat{\varphi}_{\vb{x}}\hat{U}_{t}=\exp\left(i\hat{H}t\right)\hat{\varphi}_{\vb{x}}\exp\left(-i\hat{H}t\right).\]

Explicitly, substituting \(\hat{\Phi}_{\vb{k}}\) into Eq. 1.16 and relabelling \(\vb{k}\rightarrow-\vb{k}\) in the \(\hat{a}^{\dagger}\) term, our field operators are given by:

\[\hat{\varphi}_{\vb{x}}=\sum_{\vb{k}}\frac{1}{\sqrt{2\omega_{\vb{k}}V}}\left\{ \hat{a}^{\phantom{\dagger}}_{\vb{k}}\exp\left(i\vb{k}\cdot\vb{x}\right)+\hat{a}^{\dagger}_{\vb{k}}\exp\left(-i\vb{k}\cdot\vb{x}\right)\right\} .\]

To add the field operators’ time dependence, we need to understand the effect of time evolution on the creation and annihilation operators. But this is simple: by definition, these operators create an evenly spaced ladder of energy eigenstates. Writing \(\hat{H}=\sum_{\vb{k}}\hat{H}_{\vb{k}}\) with \(\hat{H}_{\vb{k}}=\omega_{\vb{k}}\left(\hat{a}^{\dagger}_{\vb{k}}\hat{a}_{\vb{k}}+\frac{1}{2}\hat{\mathbb{I}}\right)\), we have from Eq. 1.4:

\[\begin{aligned} \left[\hat{H}_{\vb{k}},\hat{a}^{\dagger}_{\vb{k}}\right] & =\omega_{\vb{k}}\hat{a}^{\dagger}_{\vb{k}}\\ \left[\hat{H}_{\vb{k}},\hat{a}_{\vb{k}}\right] & =-\omega_{\vb{k}}\hat{a}_{\vb{k}} \end{aligned}\]

while \(\hat{a}_{\vb{k}}\) commutes with all the other \(\hat{H}_{\vb{q}}\).

In the Schroedinger picture states evolve unitarily in time according to the time dependent Schroedinger equation. The equivalent in the Heisenberg picture is the unitary evolution of operators according to the Heisenberg equation of motion. For an operator \(\hat{A}_{H}\left(t\right)\) in the Heisenberg picture,

\[i\frac{\text{d}\hat{A}_{H}}{\text{d}t}=\left[\hat{A}_{H},\hat{H}\right].\]

For the raising operator this gives

\[i\frac{\text{d}\hat{a}^{\dagger}_{\vb{k}}}{\text{d}t}=-\omega_{\vb{k}}\hat{a}^{\dagger}_{\vb{k}}\]

which is solved by

\[\hat{a}^{\dagger}_{\vb{k}}\left(t\right)=\exp\left(i\omega_{\vb{k}}t\right)\hat{a}^{\dagger}_{\vb{k}}\left(0\right).\]

From Eq. 1.24 this gives

\[\hat{U}^{\dagger}_{t}\hat{a}^{\dagger}_{\vb{k}}\hat{U}_{t}=\exp\left(i\omega_{\vb{k}}t\right)\hat{a}^{\dagger}_{\vb{k}}.\]

The lowering operator gets the opposite sign in the exponent. Finally this gives \[\hat{\varphi}_{\vb{x},t}=\sum_{\vb{k}}\frac{1}{\sqrt{2\omega_{\vb{k}}V}}\left\{ \hat{a}^{\phantom{\dagger}}_{\vb{k}}\exp\left(-i\omega_{\vb{k}}t\right)\exp\left(i\vb{k}\cdot\vb{x}\right)+\hat{a}^{\dagger}_{\vb{k}}\exp\left(i\omega_{\vb{k}}t\right)\exp\left(-i\vb{k}\cdot\vb{x}\right)\right\}\]

or

\[\hat{\varphi}_{x}=\sum_{\vb{k}}\frac{1}{\sqrt{2\omega_{\vb{k}}V}}\left\{ \hat{a}^{\phantom{\dagger}}_{\vb{k}}\exp\left(-ik\cdot x\right)+\hat{a}^{\dagger}_{\vb{k}}\exp\left(ik\cdot x\right)\right\}\]

where \(k\cdot x=k_{\mu}x^{\mu}=\omega_{\vb{k}}t-\vb{k}\cdot\vb{x}\) (the \(+---\) signature has flipped the sign in the exponents). So overall including the time dependence explicitly just amounts to switching \(\vb{k}\cdot\vb{x}\rightarrow-k\cdot x\) in the exponents!

Finally, in infinite volume the dictionary (\(\sum_{\vb{k}}\rightarrow V\int\text{d}^{d}\vb{k}/\left(2\pi\right)^{d}\), \(\hat{a}_{\vb{k}}\rightarrow V^{-1/2}\hat{a}\left(\vb{k}\right)\)) turns this into

\[\hat{\varphi}_{x}=\int\frac{\text{d}^{d}\vb{k}}{\left(2\pi\right)^{d}}\frac{1}{\sqrt{2\omega_{\vb{k}}}}\left\{ \hat{a}\left(\vb{k}\right)\exp\left(-ik\cdot x\right)+\hat{a}^{\dagger}\left(\vb{k}\right)\exp\left(ik\cdot x\right)\right\} \tag{1.25}\]

with \(\left[\hat{a}\left(\vb{k}\right),\hat{a}^{\dagger}\left(\vb{q}\right)\right]=\left(2\pi\right)^{d}\delta^{d}\left(\vb{k}-\vb{q}\right)\). This is the form we will use from Chapter 2 onwards.

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1. Two remarks. First, the same is true in infinite volume, where \(\vb{k}\) is continuous: the states \(|\vb{k}\rangle\) are then not normalisable (just like \(|x\rangle\) and \(|p\rangle\) in single-particle QM), but the Fock space still has a countable basis. Pick any orthonormal set of normalisable wavepackets \(f_{j}\left(\vb{k}\right)\) (for example Hermite functions), and define \(\hat{a}_{j}=\int\frac{\text{d}^{d}\vb{k}}{\left(2\pi\right)^{d}}f^{*}_{j}\left(\vb{k}\right)\hat{a}\left(\vb{k}\right)\); these obey \(\left[\hat{a}_{i},\hat{a}^{\dagger}_{j}\right]=\delta_{ij}\), and the occupation number states \(|n_{1}n_{2}\ldots\rangle\) built from them form a basis. What fails in infinite volume is only the product notation of Eq. 1.21 over a continuum of labels. Second, even with countably many modes, Fock space consists of the states with finitely many excitations built on \(|\Omega\rangle\), not the whole infinite tensor product of all the single-mode spaces (which is a much larger and stranger object).

2. Strictly, ‘at’ should be read as ‘localised near’: in a relativistic theory \(\langle x|y\rangle\) for two points at equal times is not a delta function, but falls off over a Compton wavelength \(1/m\). We will return to related subtleties in Chapter 2.