# 4.5: Eigenfunctions of Operators are Orthogonal

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Consideration of the quantum mechanical description of the particle-in-a-box exposed two important properties of quantum mechanical systems. We saw that the eigenfunctions of the Hamiltonian operator are orthogonal, and we also saw that the position and momentum of the particle could not be determined exactly. We now examine the generality of these insights by stating and proving some fundamental theorems. These theorems use the Hermitian property of quantum mechanical operators that correspond to observables, which is discuss first.

## Hermitian Operators

Since the eigenvalues of a quantum mechanical operator correspond to measurable quantities, the eigenvalues must be real, and consequently a quantum mechanical operator must be Hermitian. To prove this, we start with the premises that \(ψ\) and \(φ\) are functions, \(\int d\tau\) represents integration over all coordinates, and the operator \(\hat {A}\) is *Hermitian *by definition if

\[ \int \psi ^* \hat {A} \psi \,d\tau = \int (\hat {A} ^* \psi ^* ) \psi \,d\tau \label {4-37}\]

This equation means that the complex conjugate of Â can operate on \(ψ^*\) to produce the same result after integration as Â operating on \(φ\), followed by integration. To prove that a quantum mechanical operator \(\hat {A}\) is Hermitian, consider the eigenvalue equation and its complex conjugate.

\[\hat {A} \psi = a \psi \label {4-38}\]

\[\hat {A}^* \psi ^* = a^* \psi ^* = a \psi ^* \label {4-39}\]

Note that \(a^* = a\) because the eigenvalue is real. Multiply Equation \(\ref{4-38}\) and \(\ref{4-39}\) from the left by \(ψ^*\) and \(ψ\), respectively, and integrate over the full range of all the coordinates. Note that \(ψ\) is normalized. The results are

\[ \int \psi ^* \hat {A} \psi \,d\tau = a \int \psi ^* \psi \,d\tau = a \label {4-40}\]

\[ \int \psi \hat {A}^* \psi ^* \,d \tau = a \int \psi \psi ^* \,d\tau = a \label {4-41}\]

Since both integrals equal \(a\), they must be equivalent.

\[ \int \psi ^* \hat {A} \psi \,d\tau = \int \psi \hat {A}^* \psi ^* \,d\tau \label {4-42}\]

The operator acting on the function,

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

produces a new function. Since functions commute, Equation \(\ref{4-42}\) can be rewritten as

\[ \int \psi ^* \hat {A} \psi d\tau = \int (\hat {A}^*\psi ^*) \psi d\tau \label{4-43}\]

This equality means that \(\hat {A}\) is **Hermitian**.

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

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

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

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

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

## 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, and

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

then it is easily demonstrated 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 an Hermitian operator are, or can be chosen to be, *mutually orthogonal.*

Degenerate eigenfunctions are **not **automatically orthogonal, but can be made so mathematically via the Gram-Schmidt Orthogonalization. The proof of this theorem shows us one way to produce orthogonal degenerate functions.

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

## Contributors

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