8 The Spectrum of a Hermitian Operator

8.1 Hermitian Differential Operators

Recall that a linear differential operator \(A\) is Hermitian if \[\langle \psi_1 , A \cdot \psi_2 \rangle = \langle A \cdot \psi_1 , \psi_2\rangle\] for all continuous square normalisable wave functions \(\psi_1(x)\), \(\psi_2(x)\). More succinctly, a Hermitian operator obeys \(A^\dagger = A\). We demonstrated last time that position, momentum and energy are represented by Hermitian differential operators, \[\hat x = x\,, \quad\qquad \hat p = - i \hbar \displaystyle \frac{\partial}{\partial x}\,, \quad\qquad \hat H = -\displaystyle \frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + V(x) \, .\] Our working assumption is that all measurable quantities are represented by Hermitian differential operators in quantum mechanics.

A wave function \(\psi_a(x)\) is an eigenfunction of a Hermitian differential operator \(A\) with eigenvalue \(a\) if it obeys \[A \cdot \psi_a(x) = a \psi_a(x) \, .\] Such wave functions play a distinguished role in quantum mechanics due to the following observations.

• Expectation values: let us first compute the expectation value of measurements of \(A\). Assuming \(\psi_a(x)\) is normalised, we find \[\langle A \rangle = \langle \psi_a, A \cdot \psi_a\rangle = \langle \psi_a , a \psi_a \rangle = a \langle \psi_a , \psi_a \rangle = a \, .\] Similarly, \(\langle A^n \rangle = a^n\) for any positive integer \(n >0\).

• Uncertainty: the uncertainty in measurements of \(A\) is therefore \[\Delta A = \sqrt{\langle A^2\rangle - \langle A\rangle^2} = \sqrt{a^2 - a^2} = 0 \, .\]

This is therefore a wave function with a definite value \(a\) for the measurable quantity \(A\). In other words, measurements of \(A\) should yield the result \(a\) with probability \(1\).

In addition, the eigenfunctions and eigenvalues of Hermitian operators have the following important properties.

Theorem: Let \(A\) be a Hermitian operator.

1. The eigenvalues of \(A\) are real: \(a \in \mathbb{R}\).

2. Two eigenfunctions \(\psi_1(x)\), \(\psi_2(x)\) of \(A\) with distinct eigenvalues \(a_1 \neq a_2\) are orthogonal.

Proof: Suppose \(\psi_1(x)\), \(\psi_2(x)\) are eigenfunctions of \(A\) with eigenvalues \(a_1\), \(a_2\), \[A \cdot \psi_1(x) = a_1 \psi_1(x) \qquad A \cdot \psi_2(x) = a_2 \psi_2(x) \, .\] Then \[\begin{aligned} \langle \psi_1 , A\cdot \psi_2\rangle & = \langle \psi_1 , a_2\psi_2\rangle = a_2\langle \psi_1,\psi_2\rangle \\[1ex] \langle A\cdot \psi_1 , \psi_2\rangle & = \langle a_1 \psi_1 ,\psi_2\rangle = \bar a_1\langle \psi_1,\psi_2\rangle \, . \end{aligned}\] Subtracting these two equations we find \(0 = (\bar a_1 - a_2) \langle \psi_1,\psi_2\rangle\).

1. First suppose that \(a_2=a_2 = a\) and \(\psi_1 = \psi_2 = \psi\). Then we find \[(\bar a-a)\langle \psi,\psi\rangle = 0\, .\] Recall that \(\langle \psi , \psi \rangle = 0\) if and only if \(\psi(x) = 0\) identically. Therefore, provided the eigenfunction is not zero, we have \(\bar a = a\) or equivalently \(a \in \mathbb{R}\).

2. Second suppose that \(a_1 \neq a_2\). Importing the result from part (i) we now have \[(a_1-a_2)\langle \psi_1,\psi_2\rangle = 0\] and therefore \(\langle \psi_1,\psi_2\rangle = 0\).

Furthermore, an extension of the second part of this theorem is that the eigenfunctions of a Hermitian operator can be chosen to form an orthonormal basis. However, what we mean by “orthonormal basis" depends on whether the spectrum of eigenvalues is discrete or continuous.

8.2 Discrete Spectrum

Suppose \(A\) has a discrete spectrum of eigenvalues \(\{a_n\}\) labelled by an index \(n\). We will for simplicity assume the spectrum is non-degenerate: there is one linearly independent eigenfunction \(\phi_n(x)\) for each distinct eigenvalue \(a_n\).

In this case, we can choose the eigenfunctions \(\phi_n(x)\) to form a complete orthonormal basis in the standard sense. This means that as well as eigenfunctions with different eigenvalues being orthogonal, all of the eigenfunctions are normalised. In summary, \[\langle \phi_{m} , \phi_{n} \rangle = \delta_{mn} \, .\] where \(\delta_{mn}\) is the unit matrix. Furthermore, any continuous square-integrable wave function has a unique expansion \[\psi(x) = \sum_{n} c_{n} \phi_{n}(x)\] with coefficients \(c_n \in \mathbb{C}\) (yes, there are subtleties here which we ignore for now).

1. The coefficients are found using the inner product and orthonormality, \[\langle \phi_m,\psi\rangle = \sum_n c_n \langle \phi_m , \phi_n \rangle = \sum_n c_n \delta_{mn} = c_m \, .\]

2. The norm of a wave function is \[\langle \psi, \psi\rangle = \sum_{mn} \bar c_m c_n \langle \phi_m,\phi_n\rangle = \sum_n |c_n|^2 \, .\]

3. If the wave function is normalised,

   The decomposition of a wave function in an eigenbasis \(\{\phi_m\}\) of an Hermitian operator, with coefficients \(c_n\), strongly suggests \(|c_n|^2\) is the probability of finding the system in the state \(\phi_n\) for which measurement leads to the value \(a_n\).
   \[\langle \psi , \psi\rangle = \sum_n |c_n|^2 = 1 \, .\] This suggests that we interpret \(|c_n|^2\) as the probability for a measurement of \(A\) to yield the result \(a_n\): these probabilities should add up to \(1\).

Example: Energy in an Infinite Square Well

Consider the Hamiltonian operator for an infinite potential well in the region \(0<x<L\), \[\hat H = \frac{\hat p^2}{2m} = - \frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} \, .\] It is straightforward to see that \[\phi_n(x) = \sqrt{\frac{2}{L}} \sin\left( \frac{n\pi x}{L}\right) \qquad n \in \mathbb{Z}_{>0}\] are Hamiltonian eigenfunctions with eigenvalues \[E_n = \frac{\hbar^2}{2m}\left( \frac{n\pi}{L} \right)^2 \, .\] We have already shown that these eigenfunctions are orthonormal. The fact that any continuous wave function can be expressed uniquely as a linear combination of these wave functions is the content of Fourier’s theorem.

In particular, if we expand any wave function \[\psi(x) = \sum_{n > 0} c_n \phi_n(x)\] then \(|c_n|^2\) is the probability that a measurement of energy will yield the result \(E_n\). The fact that these probabilities sum to \(1\) is Parceval’s theorem.

8.3 Continuous Spectrum

A Hermitian operator \(A\) can also have a continuous spectrum of eigenvalues, say \(a \in \mathbb{R}\), or some interval in \(\mathbb{R}\). In this case, we cannot choose the eigenfunctions to form a complete orthonormal basis in the standard sense - we need a “continuous" version of the unit matrix \(\delta_{mn}\).

For this purpose, we will introduce the Dirac delta function, denoted by \(\delta(a)\). This is not a function but a ‘distribution’. This means it behaves as a function inside integrals. You can think about it roughly as a function with \[\delta(a) = \begin{cases} 0 & \quad a \neq 0 \\ \infty & \quad a = 0 \end{cases} \,\] but where the area under the function is \(1\), \[\int^\infty_{-\infty} \delta(a)\, {\rm d}a = 1 \, .\] A more precise definition is as a limit of a Gaussian function \[\delta_\epsilon(a) = \frac{1}{\epsilon \sqrt{\pi}}e^{-a^2 / \epsilon^2}\] as it becomes infinitely thin \(\epsilon \to 0^+\).

For the calculations that we need to do in this course, we will accept the following important properties of the Dirac delta function.

• For any continuous function \(f(a)\), \[\int^\infty_{-\infty} \delta(a-a') f(a') \, {\rm d}a' = f(a) \, .\] This is the continuous analogue of the discrete formula \(\sum_n \delta_{mn}f_{n} = f_m\) and so \(\delta(a-a')\) is a continuous analogue of the unit matrix \(\delta_{mn}\).

• The Dirac delta function is the Fourier transform of \(1\), \[\delta(a) = \frac{1}{2\pi} \int^\infty_{-\infty} e^{iaa'} \, {\rm d}a' \, .\]

• It follows from the above that \(\delta(a) = \delta(-a) = \overline{\delta(a)}\).

For a Hermitian operator \(A\) with a continuous spectrum, it is possible to find a basis of eigenfunctions \(\phi_a(x)\) with eigenvalues \(a \in \mathbb{R}\) such that \[\langle \phi_{a} , \phi_{a'} \rangle = \delta(a-a') \, .\] This means the eigenfunctions \(\psi_a(x)\) are not square-normalisable since \(\langle \psi_a , \psi_{a} \rangle = \infty\). Nevertheless, any square-normalisable wave function can still be uniquely expanded \[\psi(x) = \int^\infty_{-\infty} \, c(a) \, \phi_a(x)\, {\rm d}a\,,\] with complex coefficients \(c(a)\) depending continuously on \(a\).

1. The coefficients can be calculated using the inner product, \[\begin{aligned} \langle \phi_{a} , \psi \rangle & = \int^\infty_{-\infty} c(a') \, \langle \phi_{a} , \phi_{a'} \rangle\, {\rm d}a'\\ & = \int^\infty_{-\infty} c(a') \, \delta(a-a')\, {\rm d}a' \\ & = c(a) \, . \end{aligned}\] This is the continuous analogue of the discrete result \(\langle \phi_n ,\psi\rangle = c_n\).

2. The norm of a wave function can be expressed \[\begin{aligned} \langle \psi , \psi \rangle & = \int^\infty_{-\infty} \overline{c(a)} c(a') \langle \phi_{a} , \phi_{a'} \rangle\, {\rm d}a {\rm d}a' \\ & = \int^\infty_{-\infty} \overline{c(a)} c(a') \delta(a-a') \, {\rm d}a {\rm d}a'\\ & = \int^\infty_{-\infty} |c(a)|^2\,{\rm d}a \, . \end{aligned}\] This is the continuous analogue of the discrete result \(\langle \psi,\psi\rangle = \sum_n |c_n|^2\).

3. For a normalised wave function \[\langle \psi , \psi \rangle = \int^\infty_{-\infty}|c(a)|^2\, {\rm d}a = 1 \, .\] This suggests that we should treat \(|c(a)|^2\) as a probability distribution for measurements of \(A\).

Example: Momentum

An example is the momentum operator \(\hat p = -i \hbar \partial_x\) for a particle moving in one dimension. It is straightforward to see that the eigenfunctions are plane waves \(e^{ipx/\hbar}\) with eigenvalue \(p\). We choose the normalisation \[\phi_p(x) = \frac{1}{\sqrt{2\pi\hbar}} e^{ipx/\hbar} \, .\] so that \[\langle \psi_p, \psi_p'\rangle = \frac{1}{2\pi \hbar} \int^\infty_{-\infty} e^{i(p-p')x/\hbar} {\rm d}x= \delta(p-p') \, .\] as required.

Expanding a wave function as a linear combination of momentum eigenfunctions we find \[\begin{aligned} \psi(x) & = \int^{\infty}_{-\infty} c(p) \phi_p(x) \, {\rm d}p\\ & = \frac{1}{\sqrt{2\pi \hbar}}\int^{\infty}_{-\infty} c(p) \, e^{ipx/\hbar}\,{\rm d}p \, . \end{aligned}\] This is nothing other than the Fourier transform (we will see later that this is the Fourier transform between the position and momentum space wave functions).

8.4 Problems

1. More on Hermitian operators:

   A Hermitian operator obeys \(A^\dagger = A\) or equivalently \(\langle A \psi_1,\psi_2\rangle = \langle \psi_1 , A\psi_2\rangle\) for any square-normalizable \(\psi_1(x)\), \(\psi_2(x)\).

   1. Show that eigenvalues of Hermitian operators are real.

   2. Show that two eigenfunctions of a Hermitian operator with different eigenvalues are orthogonal.

   3. Show that position \(\hat x\) and momentum \(\hat p\) are Hermitian.

   4. Show that \(\hat H=\frac{\hat p^2}{2m}+V(x)\) is Hermitian.

   5. Why are measurable quantities represented by Hermitian operators?

   Solution ▶

   1. Suppose \(A \psi = a \psi\). We have, \[\begin{aligned} \langle \psi, A\psi \rangle = \langle \psi , a \psi\rangle = a \langle \psi,\psi \rangle \\ \nonumber \langle A\psi , \psi\rangle = \langle a\psi,\psi \rangle = \bar a\langle \psi,\psi\rangle \, . \end{aligned}\] Subtracting these equations we find \((a-\bar a) \langle \psi,\psi \rangle = 0\). Assuming \(\psi \neq 0\), we find \(a = \bar a\) and therefore \(a \in \mathbb{R}\).

   2. Suppose \(A \psi_1 = a_1 \psi_1\) and \(A \psi_2 = a_2 \psi_2\) with \(a_1 \neq a_2\). We have \[\begin{aligned} \langle \psi_1, A\psi_2 \rangle & = \langle \psi_1 , a_2 \psi_2\rangle = a_2 \langle \psi_1,\psi_2 \rangle \\ \nonumber \langle A\psi_1 , \psi_2\rangle & = \langle a_1\psi_1,\psi_2 \rangle = \bar a_1\langle \psi_1,\psi_2\rangle = a_1\langle \psi_1,\psi_2\rangle \, . \end{aligned}\] Subtracting these equations we find \((a_2-a_1)\langle \psi_1,\psi_2\rangle = 0\) and therefore \(\langle \psi_1,\psi_2\rangle = 0\) since \(a_1 \neq a_2\).

   3. For position, using the fact that \(x\in \mathbb{R}\), \[\begin{aligned} \langle \hat x \cdot \psi_1 , \psi_2 \rangle & = \int^\infty_{-\infty} \overline{x \psi_1(x)} \, \psi_2(x) \, dx \\ \nonumber & = \int^\infty_{-\infty} \overline{\psi_1(x)} \, x \psi_2(x) \, dx \\ & = \langle \psi_1, \hat x \cdot \psi_2\rangle \end{aligned}\] For momentum, integrating by parts, \[\begin{aligned} \langle \hat p \cdot \psi_1 , \psi_2 \rangle & = \int^\infty_{-\infty} \overline{-i\hbar\frac{\partial\psi_1(x)}{\partial x}} \psi_2(x) dx \\ \nonumber & = \int^\infty_{-\infty} i \hbar\frac{\partial\overline{ \psi_1(x)}}{\partial x} \psi_2(x) dx \\ & = \int^\infty_{-\infty} \overline{ \psi_1(x)}\left(- i \hbar\frac{\partial \psi_2(x)}{\partial x} \right) dx \\ & = \langle \psi_1, \hat p \cdot \psi_2\rangle \, . \end{aligned}\] where in passing to the third line we have integrated by parts and discarded the boundary term since \(\overline{\psi_1(x)}\psi_2(x)\) must vanish as \(x \to \pm \infty\) for square-normalizable wavefunctions.

   4. For the hamiltonian, we can use the properties of the adjoint together with the fact that \(x\) and \(p\) are Hermitian to show \[\begin{aligned} \hat H^\dagger & = \frac{(\hat p^2)^\dagger}{2m} + V(x)^\dagger \\ \nonumber & = \frac{(\hat p^\dagger)^2}{2m} + V(x^\dagger) \\ & = \frac{\hat p^2}{2m} + V(x) = \hat H \, . \end{aligned}\] where we assume the potential \(V(x)\) is a real analytic function. You can also show this directly using the differential operator \[H = - \frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x)\] inside the inner product and integrating by parts twice.

   5. In quantum mechanics, the possible outcomes of a measurement are the eigenvalues of the corresponding linear operator \(A\). Measurements must yield real numbers. If \(A\) is Hermitian, this is guaranteed from part (a).
      In quantum mechanics, if a measurement of \(A\) yields the result \(a_1\), then the probability another measurement will yield a different result \(a_2 \neq a_1\) immediately afterwards is zero. If \(A\) is Hermitian, this is guaranteed from part (b).

2. TUTORIAL 2
   Delta function Prove the following property of the Dirac delta function \(\delta(x)\) and the Heaviside step function \(\theta(x)\) by integrating with a test function \(g(x)\). \[\frac{{\rm d}\theta(x)}{{\rm d}x} = \delta(x)\,,\] Also prove \[\delta\big( f(x) \big) = \sum_{\text{$x_0$ s.t. $f(x_0)=0$}} \frac{\delta(x-x_0)}{\left|\frac{{\rm d}f(x)}{{\rm d}x}\right|} \,.\] You may want to start by doing an explicit example, e.g. by rewriting \[\int_{-\infty}^{\infty} \delta(x^2 - 4) g(x) {\rm d}x\,,\] as the sum of two integrals over \(y=x^2-4\). Be careful with signs and the order of integration limits.

   Solution ▶

   1. For a smooth test function \(g(x)\) we have \[\int_{-\infty}^{\infty} g(x) \frac{{\rm d}}{{\rm d}x }\theta(x) {\rm d x} = \int_{-\infty}^{\infty} \frac{{\rm d}}{{\rm d}x} g(x) \theta(x) {\rm d x} = -\int_{0}^{\infty} \frac{{\rm d}}{{\rm d}x} g(x) {\rm d x} = -g(x)\Big|_{0}^{\infty} = g(0)\,,\] from which the conclusion follows.

   2. Change variables to \(y=f(x)\) so that \({\rm d}y = |f'(x)| {\rm d}x\) (where the absolute sign is used so we can keep the integration limits for integration over \(y\) in natural order). This gives \[\int_{-\infty}^{\infty} g(x) \delta\big(f(x)\big) {\rm d x} = \int g\big(x=f^{-1}(y)\big) \frac{\delta(y)}{\Big|f'(x)\Big|} {\rm d y}\,.\] The integral on the right-hand side may consist of of multiple integrals if \(f(x)\) is not monotonic. The integrals get contributions at \(y=0\) only, where the integral will lead to replacement of \(x\) with the value of \(x\) at which \(y=0\). That can also be done in terms of an integral over \(x\), \[= \int_{-\infty}^{\infty} \sum_{\text{$x$ s.t. f(x)=0}} \frac{\delta(x-x_0)}{\Big|f'(x)\Big|} g(x) {\rm d}x\,,\] from which the requested result follows.

      The example would give \(f(x)=x^2-4\) so that \(f'(x)=2 x\), and \(f'(-2)=-4\) and \(f'(2)=4\). If you use \({\rm d}y = f'(x){\rm d}x\) and keep track of the integration direction, you get \[\begin{aligned} \int_{-\infty}^{\infty} \delta(x^2-4) g(x) {\rm d}x &= \int_{\infty}^{-4} \delta(y)\frac{g(x=-\sqrt{y+4})}{-4} {\rm d}y + \int_{-4}^{\infty} \delta(y)\frac{g(x=\sqrt{y+4})}{4} {\rm d}y\\[1ex] &= \int_{-4}^{\infty} \delta(y)\left[\frac{g(x=-\sqrt{y+4})}{4} + \frac{g(x=\sqrt{y+4})}{4}\right] {\rm d}y\\[1ex] &= \frac{1}{4}\Big( g(x=-2) + g(x=2)\Big)\,. \end{aligned}\]