8.7.1: Eigenfunctions of Operators are Orthogonal

Learning Objectives

• Understand the properties of a Hermitian operator and their associated eigenstates

Eigenfunctions of a Hermitian operator are orthogonal if they have different eigenvalues.

Because of this theorem, we can identify orthogonal functions easily without having to integrate or conduct an analysis based on symmetry or other considerations.

Proof

\(ψ\) and \(φ\) are two eigenfunctions of the operator  with real eigenvalues \(a_1\) and \(a_2\), respectively. Since the eigenvalues are real, \(a_1^* = a_1\) and \(a_2^* = a_2\).

\[\hat {A} \psi = a_1 \psi \nonumber \]

\[\hat {A}^* \psi ^* = a_2 \psi ^* \nonumber \]

Multiply the first equation by \(φ^*\) and the second by \(ψ\) and integrate.

\[\int \psi ^* \hat {A} \psi \,d\tau = a_1 \int \psi ^* \psi \,d\tau \nonumber \]

\[\int \psi \hat {A}^* \psi ^* \,d\tau = a_2 \int \psi \psi ^* \,d\tau \label {4-45} \]

Subtract the two equations in Equation \ref{4-45} to obtain

\[\int \psi ^*\hat {A} \psi \,d\tau - \int \psi \hat {A} ^* \psi ^* \,d\tau = (a_1 - a_2) \int \psi ^* \psi \,d\tau \label {4-46} \]

The left-hand side of Equation \ref{4-46} is zero because \(\hat {A}\) is Hermitian yielding

\[ 0 = (a_1 - a_2 ) \int \psi ^* \psi \, d\tau \label {4-47} \]

If \(a_1\) and \(a_2\) in Equation \ref{4-47} are not equal, then the integral must be zero. This result proves that nondegenerate eigenfunctions of the same operator are orthogonal.

\(\square\)

Two wavefunctions, \(\psi_1(x)\) and \(\psi_2(x)\), are said to be orthogonal if

\[\int_{-\infty}^{\infty}\psi_1^\ast \psi_2 \,dx = 0. \label{4.5.1} \]

Consider two eigenstates of \(\hat{A}\), \(\psi_a(x)\) and \(\psi_{a'}(x)\), which correspond to the two different eigenvalues \(a\) and \(a'\), respectively. Thus,

\[A\psi_a = a \psi_a \label{4.5.2} \]

\[A\psi_a' = a' \psi_a' \label{4.5.3} \]

Multiplying the complex conjugate of the first equation by \(\psi_{a'}(x)\), and the second equation by \(\psi^*_{a'}(x)\), and then integrating over all \(x\), we obtain

\[ \int_{-\infty}^\infty (A \psi_a)^\ast \psi_{a'} dx = a \int_{-\infty}^\infty\psi_a^\ast \psi_{a'} dx, \label{ 4.5.4} \]

\[ \int_{-\infty}^\infty \psi_a^\ast (A \psi_{a'}) dx = a' \int_{-\infty}^{\infty}\psi_a^\ast \psi_{a'} dx. \label{4.5.5} \]

However, from Equation \(\ref{4-46}\), the left-hand sides of the above two equations are equal. Hence, we can write

\[(a-a') \int_{-\infty}^\infty\psi_a^\ast \psi_{a'} dx = 0. \nonumber \]

By assumption, \(a \neq a'\), yielding

\[\int_{-\infty}^\infty\psi_a^\ast \psi_{a'} dx = 0. \nonumber \]

In other words, eigenstates of an Hermitian operator corresponding to different eigenvalues are automatically orthogonal.

The eigenvalues of operators associated with experimental measurements are all real.

Example8.7.1.1

Draw graphs and use them to show that the particle-in-a-box wavefunctions for \(\psi(n = 2)\) and \(\psi(n = 3)\) are orthogonal to each other.

Solution

The two PIB wavefunctions are qualitatively similar when plotted

These wavefunctions are orthogonal when

\[\int_{-\infty}^{\infty} \psi(n=2) \psi(n=3) dx =0 \nonumber \]

and when the PIB wavefunctions are substituted this integral becomes

\[\begin{align*} \int_0^L \sqrt{\dfrac{2}{L}} \sin \left( \dfrac{2n}{L}x \right) \sqrt{\dfrac{2}{L}} \sin \left( \dfrac{2n}{L}x \right) dx &= ? \\[4pt] \dfrac{2}{L} \int_0^L \sin \left( \dfrac{2}{L}x \right) \sin \left( \dfrac{3}{L}x \right) &= ? \end{align*} \nonumber \]

We can expand the integrand using trigonometric identities to help solve the integral, but it is easier to take advantage of the symmetry of the integrand, specifically, the \(\psi(n=2)\) wavefunction is even (blue curves in above figure) and the \(\psi(n=3)\) is odd (purple curve). Their product (even times odd) is an odd function and the integral over an odd function is zero. Therefore \(\psi(n=2)\) and \(\psi(n=3)\) wavefunctions are orthogonal.

This can be repeated an infinite number of times to confirm the entire set of PIB wavefunctions are mutually orthogonal as the Orthogonality Theorem guarantees.

Orthogonality of Degenerate Eigenstates

Consider two eigenstates of \(\hat{A}\), \(\psi_a\) and \(\psi'_a\), which correspond to the same eigenvalue, \(a\). Such eigenstates are termed degenerate. The above proof of the orthogonality of different eigenstates fails for degenerate eigenstates. Note, however, that any linear combination of \(\psi_a\) and \(\psi'_a\) is also an eigenstate of \(\hat{A}\) corresponding to the eigenvalue \(a\). Thus, even if \(\psi_a\) and \(\psi'_a\) are not orthogonal, we can always choose two linear combinations of these eigenstates which are orthogonal. For instance, if \(\psi_a\) and \(\psi'_a\) are properly normalized, we can define the overlap integral

\[S= \int_{-\infty}^\infty \psi_a^\ast \psi_a' dx ,\label{ 4.5.10} \]

It is easily demonstrated (but not here) that

\[\psi_a'' = \frac{\vert S\vert}{\sqrt{1-\vert S\vert^2}}\left(\psi_a - S^{-1} \psi_a'\right) \label{4.5.11} \]

is a properly normalized eigenstate of \(\hat{A}\), corresponding to the eigenvalue \(a\), which is orthogonal to \(\psi_a\). It is straightforward to generalize the above argument to three or more degenerate eigenstates. Hence, we conclude that the eigenstates of a Hermitian operator are, or can be chosen to be, mutually orthogonal.

Theorem: Gram-Schmidt Orthogonalization

Degenerate eigenfunctions are not automatically orthogonal, but can be made so mathematically via the Gram-Schmidt Orthogonalization.

The above theorem argues that if the eigenvalues of two eigenfunctions are the same then the functions are said to be degenerate and linear combinations of the degenerate functions can be formed that will be orthogonal to each other. Since the two eigenfunctions have the same eigenvalues, the linear combination also will be an eigenfunction with the same eigenvalue. The proof of this theorem shows us one way to produce orthogonal degenerate functions.

Proof

If \(\psi_a\) and \(\psi'_a\) are degenerate, but not orthogonal, we can define a new composite wavefunction \(\psi_a'' = \psi'_a - S\psi_a\) where \(S\) is the overlap integral:

\[S= \langle \psi_a | \psi'_a \rangle \nonumber \]

then \(\psi_a\) and \(\psi_a'' \) will be orthogonal.

\[\begin{align*} \langle \psi_a | \psi_a'' \rangle &= \langle \psi_a | \psi'_a - S\psi_a \rangle \\[4pt] &= \cancelto{S}{\langle \psi_a | \psi'_a \rangle} - S \cancelto{1}{\langle \psi_a |\psi_a \rangle} \\[4pt] &= S - S =0 \end{align*} \nonumber \]

Oct 19, 2019, 9:08 AM

\[ \int \psi'_a^* \psi_a d\tau \nonumber \]

then \(\psi_a\) and \(Φ\) will be orthogonal.

\[ \begin{align*} \int \psi_A^* Φ d\tau &= \int \psi_a^* (\psi'_a - S \psi_a ) d\tau \\[4pt] &= \int \psi_a^* \psi'_a d\tau - S \int \psi_a^*\psi_a d\tau \\[4pt] &= S - S = 0 \end{align*} \nonumber \]

\(\square\)

Exercise 8.7.1.2

Find \(N\) that normalizes \(\psi\) if \(\psi = N(φ_1 − Sφ_2)\) where \(φ_1\) and \(φ_2\) are normalized wavefunctions and \(S\) is their overlap integral.

\[S= \langle φ_1 | φ_2 \rangle \nonumber \]

Answer

Remember that to normalize an arbitrary wavefunction, we find a constant \(N\) such that \(\langle \psi | \psi \rangle = 1\). This equates to the following procedure:

\[ \begin{align*} \langle\psi | \psi\rangle =\left\langle N\left(φ_{1} - Sφ_{2}\right) | N\left(φ_{1} - Sφ_{2}\right)\right\rangle &= 1 \\[4pt] N^2\left\langle \left(φ_{1} - Sφ_{2}\right) | \left(φ_{1}-Sφ_{2}\right)\right\rangle &=1 \\[4pt] N^2 \left[ \cancelto{1}{\langle φ_{1}|φ_{1}\rangle} - S \cancelto{S}{\langle φ_{2}|φ_{1}\rangle} - S \cancelto{S}{\langle φ_{1}|φ_{2}\rangle} + S^2 \cancelto{1}{\langle φ_{2}| φ_{2}\rangle} \right] &= 1 \\[4pt] N^2(1 - S^2 \cancel{-S^2} + \cancel{S^2})&=1 \\[4pt] N^2(1-S^2) &= 1 \end{align*} \nonumber \]

therefore

\[N = \dfrac{1}{\sqrt{1-S^2}} \nonumber \]

We conclude that the eigenstates of operators are, or can be chosen to be, mutually orthogonal.

Contributors and Attributions

• Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)