4  Linear Maps

4.1 Definition of a Linear Map

The intuition to hold in mind is that a linear map on any finite dimensional vector space is a matrix.

For any two vector spaces \(\mathcal{V}\) and \(\mathcal{W}\) defined over the same field \(\mathcal{F}\), a function \(f:V\rightarrow W\) is a linear map iff \(\forall|a\rangle,|b\rangle\in V;\alpha\in F\):

  • [L1]: \(f\left(|a\rangle+|b\rangle\right)=f\left(|a\rangle\right)+f\left(|b\rangle\right)\) (additivity)

  • [L2]: \(f\left(\alpha|a\rangle\right)=\alpha f\left(|a\rangle\right)\) (homogeneity).

4.2 Examples of Linear Maps

  • An \(m\times n\) real matrix \(A\) acting on a vector \(|a\rangle\in\mathbb{R}^{n}\) returns a vector \(A|a\rangle\in\mathbb{R}^{m}\) (note the different dimensions of the spaces). The most familiar example is \(m=n\).

  • A function \(f\left(x\right)=\alpha x\) for \(\alpha,x\in\mathbb{R}\) is a linear map.

  • Derivatives acting on smooth functions, and integrals acting on integrable functions, are both linear maps.

4.3 Key Example: Dual Spaces

A key example of a linear operator is the bra \(\langle v|\). In physics we are used to thinking of the bra as the Hermitian conjugate of the vector \(|v\rangle\). That is:

\[|v\rangle=\left(\begin{array}{c} x_{1}\\ x_{2}\\ x_{3}\\ \vdots\\ x_{N} \end{array}\right)\implies\langle v|=\left(x^{*}_{1},x^{*}_{2},x^{*}_{3},\ldots x^{*}_{N}\right).\]

This leads us to think of bras and kets on somewhat equal footing. However, mathematically they are at first sight rather different objects. Specifically, while \(|v\rangle\) is a vector:

\[|v\rangle\in V\]

\(\langle v|\) is a linear map:

\[\langle v|:V\rightarrow\mathbb{C}.\]

I.e. the bra is defined as a map which takes in a vector and returns a complex number (by making an inner product):

\[\langle a|b\rangle\in\mathbb{C}.\]

While a ket, a.k.a. vector, \(|v\rangle\) lives in a vector space \(\mathcal{V}\), the bra \(\langle v|\) lives in a ‘dual vector space’ \(\mathcal{V}^{*}\), abbreviated to ‘dual space’. Pleasingly, the dual space itself becomes a vector space if endowed with addition and scalar multiplication, as follows.

4.4 Definition of Dual Spaces

Given a vector space \(\mathcal{V}\) defined over a field \(\mathcal{F}\), a dual space \(\mathcal{V}^{*}\) is the set \(V^{*}\) of all linear maps \(\langle v|:\mathcal{V\rightarrow\mathcal{F}}\) combined with the binary operations \(+\) (addition) and \(\cdot\) (scalar multiplication) obeying the following axioms.

\(\forall\langle a|,\langle b|\in V^{*}\); \(|v\rangle\in V\); \(\alpha\in F\):

  • [V*1]: \(\left(\langle a|+\langle b|\right)|v\rangle=\langle a|v\rangle+\langle b|v\rangle\) (distributivity)

  • [V*2]: \(\left(\alpha\langle a|\right)|v\rangle=\alpha\langle a|v\rangle\) (associativity).

Hence, dual spaces are vector spaces (given [V*1] and [V*2]) and we can think of bras and kets on equal footing, but this was not obvious a priori.

4.5 Self-Adjoint vs Hermitian Operators

For notational convenience in this section, I will denote the inner product \(\langle\cdot|\cdot\rangle\) as \(\langle\cdot,\cdot\rangle\).

4.5.1 The Hermitian adjoint

Consider an inner product space \(\mathcal{I}=\left(\mathcal{V},\langle\cdot,\cdot\rangle\right)\) and a linear map \(\hat{A}:\mathcal{I}\rightarrow\mathcal{I}\). The Hermitian adjoint \(\hat{A}^{\dagger}:\mathcal{I}\rightarrow\mathcal{I}\) is defined as follows.

\(\forall|a\rangle,|b\rangle\in\mathcal{I}\):

  • [H1]: \(\langle a,\hat{A}b\rangle\triangleq\langle\hat{A}^{\dagger}a,b\rangle\) (Hermitian adjoint).

We can equivalently write this in Dirac notation as:

  • [H1]: \(\langle a|\left(\hat{A}|b\rangle\right)\triangleq\left(\left(\langle b|\hat{A}^{\dagger}\right)|a\rangle\right)^{*}\) (Hermitian adjoint).

4.5.2 Definition of Self-adjoint Operators

When specifying an operator we must specify the domain on which it acts. In quantum mechanics we specify this as part of the problem.

A self-adjoint operator on a complex inner product space \(\mathcal{I}=\left(\mathcal{V},\langle\cdot|\cdot\rangle\right)\) is a linear map \(A:\mathcal{V}\rightarrow\mathcal{V}\) that is its own adjoint with the same domain.

If \(\mathcal{V}\) is finite-dimensional, the domain of the operator and its adjoint are always the same, and self-adjoint is equivalent to Hermitian. If the domain of the operator and its adjoint are not the same, the operator is Hermitian but not self-adjoint. See the lectures for examples.

Importantly, the nice properties of self-adjoint operators mentioned below do not necessarily hold for Hermitian operators. For example, self-adjoint operators have real eigenvalues, but Hermitian operators need not (consider the `eigenstate’ \(\exp(\kappa x),\kappa\in\mathbb{R}\) of the momentum operator!).

In physics we often use the phrase Hermitian when we really mean self-adjoint. Here’s a simple example of the difference.

Consider the momentum operator for a quantum particle in a 1D box. We know that \(\hat{p}=-i\hbar\partial/\partial x\). Is it Hermitian? Is it self-adjoint?

For the particle in a box we require wavefunctions to be smooth (infinitely differentiable) and to vanish at the boundaries. Hence the domain of \(\hat{p}\), \(D(\hat{p})\), is

\[D\left(\hat{p}\right)=\left\{\psi=C^{\infty}\left[0,1\right]|\psi(0)=\psi(1)=0\right\}\]

where \(C^\infty\) denotes functions that can be differentiated an infinite number of times. The domain of \(\hat{p}^\dagger\) is defined to be

\[\phi\in D\left(\hat{p}^\dagger\right)\implies\exists\chi|\langle\psi,\hat{p}\phi\rangle=\langle\chi,\psi\rangle\forall\psi\in D\left(\hat{p}\right).\]

We can check that \(\hat{p}\) is Hermitian:

\[\begin{align} \langle\psi,\hat{p}\phi\rangle=&\int_{0}^{1}\psi^{*}(x)\left(-i\hbar\phi'\right)\textrm{d}x\\ &=-i\hbar\left[\psi^{*}\phi\right]_{0}^{1}+\int_{0}^{1}\left(-i\hbar\psi^{*}(x)\right)'\phi\textrm{d}x\\ &=\langle\hat{p}\psi,\phi\rangle \end{align}\]

where the last line follows because \(\psi\) vanishes at the boundaries. But this is true regardless of the form of \(\phi\). So the domain of the adjoint has no restrictions, other than the requirement of smoothness. Hence,

\[D\left(\hat{p}^{\dagger}\right)=\left\{\psi=C^{\infty}\left[0,1\right]\right\}.\]

This is a much bigger set of functions than \(\psi\), as there are no boundary conditions imposed! Hence

\[D\left(\hat{p}\right)\subset D\left(\hat{p}^{\dagger}\right)\]

and so \(\hat{p}\) is not self-adjoint. Hence it does not necessarily have real eigenvalues! Perhaps this is obvious: what functions, obeying the boundary conditions, are eigenstates of \(\hat{p}\)?

Now consider a slightly different problem: the particle on a ring. In this case we take the less restrictive boundary conditions

\[D\left(\hat{p}\right)=\left\{\psi=C^{\infty}\left[0,1\right]|\psi(0)=\psi(1)\right\}\]

where we only require that the wavefunction returns to itself at the ends of the domain (so it is a quantum particle living on a ring). This time you can check that the boundary term only vanishes if \(\phi(0)=\phi(1)\). But that’s the same condition as is applied to \(\psi\). Hence, this time, \(\hat{p}\) is both Hermitian and self adjoint: \(D\left(\hat{p}^{\dagger}\right)=D\left(\hat{p}\right)\).

Now we can easily write down normalised eigenfunctions obeying the boundary conditions:

\[\hat{p}|p_n\rangle=p_n |p_n\rangle\]

with

\[\langle x|p_n\rangle=\exp\left(\pm i\hbar p_n x\right)\]

and

\[\hbar p_n\in 2\pi\mathbb{N}.\]

This makes physical sense if you think about momentum: a particle cannot have a well defined momentum, for all time, if trapped in a box; but it can if it can fly out of one end of the box and into the other.

4.5.3 Properties of self-adjoint operators

  • [\(l\)H1] (involubility): \(\left(\hat{A}^{\dagger}\right)^{\dagger}=\hat{A}\)

  • [\(l\)H2] (compatibility of inverse and Hermitian conjugate): \(\left(\hat{A}^{\dagger}\right)^{-1}=\left(\hat{A}^{-1}\right)^{\dagger}\)

  • [\(l\)H3] (conjugate linearity)

    • [\(l\)H3.1]: \(\left(\hat{A}+\hat{B}\right)^{\dagger}=\hat{A}^{\dagger}+\hat{B}^{\dagger}\)

    • [\(l\)H3.2]: \(\left(\alpha\hat{A}\right)^{\dagger}=\alpha^{*}\hat{A}^{\dagger}\)

  • [\(l\)H4] (anti-distributivity): \(\left(\hat{A}\hat{B}\right)^{\dagger}=\hat{B}^{\dagger}\hat{A}^{\dagger}\) .

We can also prove two important theorems.

Theorem [\(t\)H1] (Self-adjoint operators have real eigenvalues):

\[\hat{A}=\hat{A}^{\dagger}\quad\&\quad\hat{A}|a_{n}\rangle=a_{n}|a_{n}\rangle\implies a_{n}\in\mathbb{R}.\]

Proof:

consider

\[\langle a_{n}|\left(\hat{A}-\hat{A}^{\dagger}\right)|a_{n}\rangle=0\]

which follows from the definition of self-adjoint. Hence

\[\langle a_{n}|\hat{A}|a_{n}\rangle=\langle a_{n}|\hat{A}^{\dagger}|a_{n}\rangle.\]

Now use [H1]:

\[\begin{aligned} \langle a_{n}|\hat{A}^{\dagger}|a_{n}\rangle & =\left(\langle a_{n}|\hat{A}|a_{n}\rangle\right)^{*} \end{aligned}\]

to give

\[\langle a_{n}|\hat{A}|a_{n}\rangle=\left(\langle a_{n}|\hat{A}|a_{n}\rangle\right)^{*}.\]

The theorem requires that \(|a_{n}\rangle\) is an eigenvector of \(\hat{A}\), and so:

\[\begin{aligned} a_{n}\langle a_{n}|a_{n}\rangle & =\left(a_{n}\langle a_{n}|a_{n}\rangle\right)^{*}\\ & =a^{*}_{n}\left(\langle a_{n}|a_{n}\rangle\right)^{*}. \end{aligned}\]

From [I1]:

\[\langle a_{n}|a_{n}\rangle=\left(\langle a_{n}|a_{n}\rangle\right)^{*}\]

and so

\[\begin{aligned} a_{n}\langle a_{n}|a_{n}\rangle & =a^{*}_{n}\langle a_{n}|a_{n}\rangle \end{aligned}\]

or

\[a_{n}=a^{*}_{n}.\]

QED

Theorem [\(t\)H2] (eigenvectors of \(\hat{A}\) with non-degenerate eigenvalues are orthogonal):

\[\forall\hat{A}|a_{n}\rangle=a_{n}|a_{n}\rangle,\quad\hat{A}=\hat{A}^{\dagger}:\]

\[a_{n}\ne a_{m}\implies\langle a_{n}|a_{m}\rangle=0.\]

Proof:

Now consider

\[\langle a_{n}|\left(\hat{A}-\hat{A}^{\dagger}\right)|a_{m}\rangle=0\]

which is still zero by definition of self-adjoint. Hence

\[\begin{aligned} \langle a_{n}|\hat{A}|a_{m}\rangle & =\langle a_{n}|\hat{A}^{\dagger}|a_{m}\rangle\\ & =\left(\langle a_{m}|\hat{A}|a_{n}\rangle\right)^{*}. \end{aligned}\]

These are again assumed to be eigenvectors, so:

\[\begin{aligned} a_{m}\langle a_{n}|a_{m}\rangle & =\left(a_{n}\langle a_{m}|a_{n}\rangle\right)^{*}\\ & =a^{*}_{n}\langle a_{n}|a_{m}\rangle \end{aligned}\]

and from [\(t\)H1]:

\[a_{m}\langle a_{n}|a_{m}\rangle=a_{n}\langle a_{n}|a_{m}\rangle.\]

Hence

\[\left(a_{m}-a_{n}\right)\langle a_{n}|a_{m}\rangle=0.\]

Since \(a_{m}\ne a_{n}\), the inner product must equal zero.

QED

4.5.4 Examples of Self-Adjoint Operators

It is a postulate of quantum mechanics that all observable quantities are represented by self-adjoint operators. Therefore examples ought to include:

  • the Hamiltonian \(\hat{H}\) (whose corresponding observable is the energy \(E\))

  • the position operator \(\hat{x}\) (whose corresponding observable is the position \(x\))

  • the momentum operator \(\hat{p}\) (whose corresponding observable is the momentum \(p\)).

However, whether these are really self-adjoint depends on boundary conditions, and is a subtle point we will return to.

4.5.5 Key Example of a Self-adjoint Operator: the Identity Operator

One operator that is guaranteed to exist in any inner product space is the identity operator \(\mathbb{I}\).

Theorem [\(t\mathbb{I}\)1] (Existence of \(\mathbb{I}\)):

\[\forall\mathcal{I}\exists\mathbb{I}:\mathcal{I}\rightarrow\mathcal{I}|\forall|a\rangle\in\mathcal{I},\mathbb{I}|a\rangle=|a\rangle.\]

Proof: simply observe that \(|a\rangle\) is in \(\mathcal{I}\) by definition!

Theorem [\(t\mathbb{I}\)2] (Hermiticity of \(\mathbb{I}\)):

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

Proof:

From [H1]:

\[\langle a|\mathbb{I}|b\rangle=\left(\langle b|\mathbb{I}^{\dagger}|a\rangle\right)^{*}. \tag{4.1}\]

Since

\[\mathbb{I}|b\rangle=|b\rangle\]

it follows that

\[\begin{aligned} \langle a|\mathbb{I}|b\rangle & =\langle a|b\rangle\\ & =\left(\langle b|a\rangle\right)^{*} \end{aligned}\]

where the second line follows from [I1]. Substituting into Equation 4.1 gives

\[\left(\langle b|a\rangle\right)^{*}=\left(\langle b|\mathbb{I}^{\dagger}|a\rangle\right)^{*}\]

so

\[\langle b|a\rangle=\langle b|\mathbb{I}^{\dagger}|a\rangle.\]

Since this is true for all \(\langle b|\), it follows that

\[|a\rangle=\mathbb{I}^{\dagger}|a\rangle.\]

But we also have that

\[|a\rangle=\mathbb{I}|a\rangle.\]

Since this is true for all \(|a\rangle\), it follows that \(\mathbb{I}=\mathbb{I}^{\dagger}\). QED