The quantum-mechanical operators that represent physical quantities are linear (Section 3.1). These operators must meet an additional requirement, which we now discuss.
Definition of Hermitian Operators
Let $\hat{A}$ be the linear operator representing the physical property $A$. The average value of $A$ is [Eq. (3.88)]
\(
\langle A\rangle=\int \Psi^{*} \hat{A} \Psi d \tau
\)
where $\Psi$ is the state function of the system. Since the average value of a physical quantity must be a real number, we demand that
\(
\begin{gather}
\langle A\rangle=\langle A\rangle^{} \
\int \Psi^{} \hat{A} \Psi d \tau=\left[\int \Psi^{} \hat{A} \Psi d \tau\right]^{}=\int\left(\Psi^{}\right)^{}(\hat{A} \Psi)^{} d \tau \
\int \Psi^{} \hat{A} \Psi d \tau=\int \Psi(\hat{A} \Psi)^{} d \tau \tag{7.6}
\end{gather*}
\)
Equation (7.6) must hold for every function $\Psi$ that can represent a possible state of the system; that is, it must hold for all well-behaved functions $\Psi$. A linear operator that satisfies (7.6) for all well-behaved functions is called a Hermitian operator (after the mathematician Charles Hermite).
Many texts define a Hermitian operator as a linear operator that satisfies
\(
\begin{equation}
\int f^{} \hat{A} g d \tau=\int g(\hat{A} f)^{} d \tau \tag{7.7}
\end{equation}
\)
for all well-behaved functions $f$ and $g$. Note especially that on the left side of (7.7) $\hat{A}$ operates on $g$, but on the right side $\hat{A}$ operates on $f$. For the special case $f=g$, (7.7) reduces to (7.6). Equation (7.7) is apparently a more stringent requirement than (7.6), but we shall prove that (7.7) is a consequence of (7.6). Therefore the two definitions of a Hermitian operator are equivalent.
We begin the proof by setting $\Psi=f+c g$ in (7.6), where $c$ is an arbitrary constant. This gives
\(
\begin{aligned}
& \int(f+c g)^{} \hat{A}(f+c g) d \tau=\int(f+c g)[\hat{A}(f+c g)] d \tau \
& \int\left(f^{}+c^{} g^{}\right) \hat{A} f d \tau+\int\left(f^{}+c^{} g^{}\right) \hat{A} c g d \tau \
& =\int(f+c g)(\hat{A} f)^{} d \tau+\int(f+c g)(\hat{A} c g)^{} d \tau \
& \int f^{} \hat{A} f d \tau+c^{} \int g^{} \hat{A} f d \tau+c \int f^{} \hat{A} g d \tau+c^{} c \int g^{} \hat{A} g d \tau \
& =\int f(\hat{A} f)^{} d \tau+c \int g(\hat{A} f)^{} d \tau+c^{} \int f(\hat{A} g)^{} d \tau+c c^{} \int g(\hat{A} g)^{} d \tau
\end{aligned}
\)
By virtue of (7.6), the first terms on each side of this last equation are equal to each other; likewise, the last terms on each side are equal. Therefore
\(
\begin{equation}
c^{} \int g^{} \hat{A} f d \tau+c \int f^{} \hat{A} g d \tau=c \int g(\hat{A} f)^{} d \tau+c^{} \int f(\hat{A} g)^{} d \tau \tag{7.8}
\end{equation}
\)
Setting $c=1$ in (7.8), we have
\(
\begin{equation}
\int g^{} \hat{A} f d \tau+\int f^{} \hat{A} g d \tau=\int g(\hat{A} f)^{} d \tau+\int f(\hat{A} g)^{} d \tau \tag{7.9}
\end{equation}
\)
Setting $c=i$ in (7.8), we have, after dividing by $i$,
\(
\begin{equation}
-\int g^{} \hat{A} f d \tau+\int f^{} \hat{A} g d \tau=\int g(\hat{A} f)^{} d \tau-\int f(\hat{A} g)^{} d \tau \tag{7.10}
\end{equation}
\)
We now add (7.9) and (7.10) to get (7.7). This completes the proof.
Therefore, a Hermitian operator $\hat{A}$ is a linear operator that satisfies
\(
\begin{equation}
\int f_{m}^{} \hat{A} f{n} d \tau=\int f{n}\left(\hat{A} f_{m}\right)^{} d \tau \tag{7.11}
\end{equation}
\)
where $f{m}$ and $f{n}$ are arbitrary well-behaved functions and the integrals are definite integrals over all space. Using the bracket and matrix-element notations, we write
\(
\begin{align}
\left\langle f{m}\right| \hat{A}\left|f{n}\right\rangle & =\left\langle f{n}\right| \hat{A}\left|f{m}\right\rangle^{} \tag{7.12}\
\langle m| \hat{A}|n\rangle & =\langle n| \hat{A}|m\rangle^{} \tag{7.13}\
A{m n} & =\left(A{n m}\right)^{} \tag{7.14}
\end{align*}
\)
The two sides of (7.12) differ by having the functions interchanged and the complex conjugate taken.
Examples of Hermitian Operators
Let us show that some of the operators we have been using are indeed Hermitian. For simplicity, we shall work in one dimension. To prove that an operator is Hermitian, it suffices to show that it satisfies (7.6) for all well-behaved functions. However, we shall make things a bit harder by proving that (7.11) is satisfied.
First consider the one-particle, one-dimensional potential-energy operator. The right side of (7.11) is
\(
\begin{equation}
\int{-\infty}^{\infty} f{n}(x)\left[V(x) f_{m}(x)\right] d x \tag{7.15}
\end{}
\)
We have $V^{*}=V$, since the potential energy is a real function. Moreover, the order of the factors in (7.15) does not matter. Therefore,
\(
\int{-\infty}^{\infty} f{n}\left(V f{m}\right)^{*} d x=\int{-\infty}^{\infty} f{n} V^{*} f{m}^{} d x=\int{-\infty}^{\infty} f{m}^{} V f_{n} d x
\)
which proves that $V$ is Hermitian.
The operator for the $x$ component of linear momentum is $\hat{p}_{x}=-i \hbar d / d x$ [Eq. (3.23)]. For this operator, the left side of (7.11) is
\(
-i \hbar \int{-\infty}^{\infty} f{m}^{*}(x) \frac{d f_{n}(x)}{d x} d x
\)
Now we use the formula for integration by parts:
\(
\begin{equation}
\int{a}^{b} u(x) \frac{d v(x)}{d x} d x=\left.u(x) v(x)\right|{a} ^{b}-\int_{a}^{b} v(x) \frac{d u(x)}{d x} d x \tag{7.16}
\end{equation}
\)
Let
\(
u(x) \equiv-i \hbar f{m}^{*}(x), \quad v(x) \equiv f{n}(x)
\)
Then
\(
\begin{equation}
-i \hbar \int{-\infty}^{\infty} f{m}^{} \frac{d f{n}}{d x} d x=-\left.i \hbar f{m}^{} f{n}\right|{-\infty} ^{\infty}+i \hbar \int{-\infty}^{\infty} f{n}(x) \frac{d f_{m}^{}(x)}{d x} d x \tag{7.17}
\end{}
\)
Because $f{m}$ and $f{n}$ are well-behaved functions, they vanish at $x= \pm \infty$. (If they didn't vanish at infinity, they wouldn't be quadratically integrable.) Therefore, (7.17) becomes
\(
\int{-\infty}^{\infty} f{m}^{}\left(-i \hbar \frac{d f{n}}{d x}\right) d x=\int{-\infty}^{\infty} f{n}\left(-i \hbar \frac{d f{m}}{d x}\right)^{} d x
\)
which is the same as (7.11) and proves that $\hat{p}_{x}$ is Hermitian. The proof that the kineticenergy operator is Hermitian is left to the reader. The sum of two Hermitian operators can be shown to be Hermitian. Hence the Hamiltonian operator $\hat{H}=\hat{T}+\hat{V}$ is Hermitian.
Theorems about Hermitian Operators
We now prove some important theorems about the eigenvalues and eigenfunctions of Hermitian operators.
Since the eigenvalues of the operator $\hat{A}$ corresponding to the physical quantity $A$ are the possible results of a measurement of $A$ (Section 3.3), these eigenvalues should all be real numbers. We now prove that the eigenvalues of a Hermitian operator are real numbers.
We are given that $\hat{A}$ is Hermitian. Translating these words into an equation, we have [Eq. (7.11)]
\(
\begin{equation}
\int f_{m}^{} \hat{A} f{n} d \tau=\int f{n}\left(\hat{A} f_{m}\right)^{} d \tau \tag{7.18}
\end{equation}
\)
for all well-behaved functions $f{m}$ and $f{n}$. We want to prove that every eigenvalue of $\hat{A}$ is a real number. Translating this into equations, we want to show that $a{i}=a{i}^{*}$, where the eigenvalues $a{i}$ satisfy $\hat{A} g{i}=a{i} g{i}$; the functions $g_{i}$ are the eigenfunctions.
To introduce the eigenvalues $a{i}$ into (7.18), we write (7.18) for the special case where $f{m}=g{i}$ and $f{n}=g_{i}$ :
\(
\int g{i}^{*} \hat{A} g{i} d \tau=\int g{i}\left(\hat{A} g{i}\right)^{*} d \tau
\)
Use of $\hat{A} g{i}=a{i} g_{i}$ gives
\(
\begin{gather}
a{i} \int g{i}^{} g{i} d \tau=\int g{i}\left(a{i} g{i}\right)^{} d \tau=a_{i}^{} \int g{i} g{i}^{} d \tau \
\left(a{i}-a{i}^{}\right) \int\left|g_{i}\right|^{2} d \tau=0 \tag{7.19}
\end{gather*}
\)
Since the integrand $\left|g{i}\right|^{2}$ is never negative, the only way the integral in (7.19) could be zero would be if $g{i}$ were zero for all values of the coordinates. However, we always reject $g{i}=0$ as an eigenfunction on physical grounds. Hence the integral in (7.19) cannot be zero. Therefore, $\left(a{i}-a{i}^{*}\right)=0$, and $a{i}=a_{i}^{*}$. We have proved:
THEOREM 1. The eigenvalues of a Hermitian operator are real numbers.
To help become familiar with bracket notation, we shall repeat the proof of Theorem 1 using bracket notation. We begin by setting $m=i$ and $n=i$ in (7.13) to get $\langle i| \hat{A}|i\rangle=\langle i| \hat{A}|i\rangle^{}$. Choosing the function with index $i$ to be an eigenfunction of $\hat{A}$ and using the eigenvalue equation $\hat{A} g{i}=a{i} g{i}$, we have $\langle i| a{i}|i\rangle=\langle i| a_{i}|i\rangle^{}$. Therefore $a{i}\langle i \mid i\rangle=a{i}^{}\langle i \mid i\rangle^{}=a{i}^{*}\langle i \mid i\rangle$ and $\left(a{i}-a{i}^{*}\right)\langle i \mid i\rangle=0$. So $a{i}=a_{i}^{*}$, where (7.4) with $m=n$ was used.
We showed that two different particle-in-a-box energy eigenfunctions $\psi{i}$ and $\psi{j}$ are orthogonal, meaning that $\int{-\infty}^{\infty} \psi{i}^{*} \psi{j} d x=0$ for $i \neq j$ [Eq. (2.26)]. Two functions $f{1}$ and $f_{2}$ of the same set of coordinates are said to be orthogonal if
\(
\begin{equation}
\int f_{1}^{} f_{2} d \tau=0 \tag{7.20}
\end{}
\)
where the integral is a definite integral over the full range of the coordinates. We now prove the general theorem that the eigenfunctions of a Hermitian operator are, or can be chosen to be, mutually orthogonal. Given that
\(
\begin{equation}
\hat{B} F=s F, \quad \hat{B} G=t G \tag{7.21}
\end{equation}
\)
where $F$ and $G$ are two linearly independent eigenfunctions of the Hermitian operator $\hat{B}$, we want to prove that
\(
\int F^{*} G d \tau \equiv\langle F \mid G\rangle=0
\)
We begin with Eq. (7.12), which expresses the Hermitian nature of $\hat{B}$ :
\(
\langle F| \hat{B}|G\rangle=\langle G| \hat{B}|F\rangle^{*}
\)
Using (7.21), we have
\(
\begin{aligned}
\langle F| t|G\rangle & =\langle G| s|F\rangle^{} \
t\langle F \mid G\rangle & =s^{}\langle G \mid F\rangle^{*}
\end{aligned}
\)
Since eigenvalues of Hermitian operators are real (Theorem 1), we have $s^{}=s$. Use of $\langle G \mid F\rangle^{}=\langle F \mid G\rangle$ [Eq. (7.4)] gives
\(
\begin{gathered}
t\langle F \mid G\rangle=s\langle F \mid G\rangle \
(t-s)\langle F \mid G\rangle=0
\end{gathered}
\)
If $s \neq t$, then
\(
\begin{equation}
\langle F \mid G\rangle=0 \tag{7.22}
\end{equation}
\)
We have proved that two eigenfunctions of a Hermitian operator that correspond to different eigenvalues are orthogonal. The question now is: Can we have two independent eigenfunctions that have the same eigenvalue? The answer is yes. In the case of degeneracy, we have the same eigenvalue for more than one independent eigenfunction. Therefore, we can only be certain that two independent eigenfunctions of a Hermitian operator are orthogonal to each other if they do not correspond to a degenerate eigenvalue. We now show that in the case of degeneracy we may construct eigenfunctions that will be orthogonal to one another. We shall use the theorem proved in Section 3.6, that any linear combination of eigenfunctions corresponding to a degenerate eigenvalue is an eigenfunction with the same eigenvalue. Let us therefore suppose that $F$ and $G$ are independent eigenfunctions that have the same eigenvalue:
\(
\hat{B} F=s F, \quad \hat{B} G=s G
\)
We take linear combinations of $F$ and $G$ to form two new eigenfunctions $g{1}$ and $g{2}$ that will be orthogonal to each other. We choose
\(
g{1} \equiv F, \quad g{2} \equiv G+c F
\)
where the constant $c$ will be chosen to ensure orthogonality. We want
\(
\begin{gathered}
\int g{1}^{*} g{2} d \tau=0 \
\int F^{}(G+c F) d \tau=\int F^{} G d \tau+c \int F^{*} F d \tau=0
\end{gathered}
\)
Hence choosing
\(
\begin{equation}
c=-\int F^{} G d \tau / \int F^{} F d \tau \tag{7.23}
\end{equation}
\)
we have two orthogonal eigenfunctions $g{1}$ and $g{2}$ corresponding to the degenerate eigenvalue. This procedure (called Schmidt or Gram-Schmidt orthogonalization) can be extended to the case of $n$-fold degeneracy to give $n$ linearly independent orthogonal eigenfunctions corresponding to the degenerate eigenvalue.
Thus, although there is no guarantee that the eigenfunctions of a degenerate eigenvalue are orthogonal, we can always choose them to be orthogonal, if we desire, by using the Schmidt (or some other) orthogonalization method. In fact, unless stated otherwise, we shall always assume that we have chosen the eigenfunctions to be orthogonal:
\(
\begin{equation}
\int g_{i}^{} g_{k} d \tau=0, \quad i \neq k \tag{7.24}
\end{}
\)
where $g{i}$ and $g{k}$ are independent eigenfunctions of a Hermitian operator. We have proved:
THEOREM 2. Two eigenfunctions of a Hermitian operator $\hat{B}$ that correspond to different eigenvalues are orthogonal. Eigenfunctions of $\hat{B}$ that belong to a degenerate eigenvalue can always be chosen to be orthogonal.
An eigenfunction can usually be multiplied by a constant to normalize it, and we shall assume, unless stated otherwise, that all eigenfunctions are normalized:
\(
\begin{equation}
\int g_{i}^{} g_{i} d \tau=1 \tag{7.25}
\end{}
\)
The exception is where the eigenvalues form a continuum, rather than a discrete set of values. In this case, the eigenfunctions are not quadratically integrable. Examples are the linear-momentum eigenfunctions, the free-particle energy eigenfunctions, and the hydrogenatom continuum energy eigenfunctions.
Using the Kronecker delta, defined by $\delta{i k} \equiv 1$ if $i=k$ and $\delta{i k} \equiv 0$ if $i \neq k$ [Eq. (2.28)], we can combine (7.24) and (7.25) into one equation:
\(
\begin{equation}
\int g_{i}^{} g{k} d \tau=\langle i \mid k\rangle=\delta{i k} \tag{7.26}
\end{}
\)
where $g{i}$ and $g{k}$ are eigenfunctions of some Hermitian operator.
As an example, consider the spherical harmonics. We shall prove that
\(
\begin{equation}
\int{0}^{2 \pi} \int{0}^{\pi}\left[Y_{l}^{m}(\theta, \phi)\right] Y{l^{\prime}}^{m^{\prime}}(\theta, \phi) \sin \theta d \theta d \phi=\delta{l, l^{\prime}} \delta_{m, m^{\prime}} \tag{7.27}
\end{}
\)
where the $\sin \theta$ factor comes from the volume element in spherical coordinates, (5.78). The spherical harmonics are eigenfunctions of the Hermitian operator $\hat{L}^{2}$ [Eq. (5.104)]. Since eigenfunctions of a Hermitian operator belonging to different eigenvalues are orthogonal, we conclude that the integral in (7.27) is zero unless $l=l^{\prime}$. Similarly, since the $Y{l}^{m}$ functions are eigenfunctions of $\hat{L}{z}$ [Eq. (5.105)], we conclude that the integral in (7.27) is zero unless $m=m^{\prime}$. Also, the multiplicative constant in $Y_{l}^{m}$ [Eq. (5.147) of Prob. 5.34] has been chosen so that the spherical harmonics are normalized [Eq. (6.117)]. Therefore (7.27) is valid.
The integral $\langle f| \hat{B}|g\rangle$ can be simplified if either $f$ or $g$ is an eigenfunction of the Hermitian operator $\hat{B}$. If $\hat{B} g=c g$, where $c$ is a constant, then
\(
\langle f| \hat{B}|g\rangle=\langle f \mid \hat{B} g\rangle=\langle f \mid c g\rangle=c\langle f \mid g\rangle
\)
If $\hat{B} f=k f$, where $k$ is a constant, then use of the Hermitian property of $\hat{B}$ gives
\(
\langle f| \hat{B}|g\rangle=\langle g| \hat{B}|f\rangle^{}=\langle g \mid \hat{B} f\rangle^{}=\langle g \mid k f\rangle^{}=k^{}\langle g \mid f\rangle^{*}=k\langle f \mid g\rangle
\)
since the eigenvalue $k$ is real. The relation $\langle f| \hat{B}|g\rangle=k\langle f \mid g\rangle$ shows that the Hermitian operator $\hat{B}$ can act to the left in $\langle f| \hat{B}|g\rangle$.
A proof of the uncertainty principle is outlined in Prob. 7.60.