Comprehensive Study Notes

In the previous section, we proved the orthogonality of the eigenfunctions of a Hermitian operator. We now discuss another important property of these functions; this property allows us to expand an arbitrary well-behaved function in terms of these eigenfunctions.

We have used the Taylor-series expansion (Prob. 4.1) of a function as a linear combination of the nonnegative integral powers of $(x-a)$. Can we expand a function as a linear combination of some other set of functions besides $1,(x-a),(x-a)^{2}, \ldots$ ? The answer is yes, as was first shown by Fourier in 1807. A Fourier series is an expansion of a function as a linear combination of an infinite number of sine and cosine functions. We shall not go into detail about Fourier series, but shall simply look at one example.

Expansion of a Function Using Particle-in-a-Box Wave Functions

Let us consider expanding a function in terms of the particle-in-a-box stationary-state wave functions, which are [Eq. (2.23)]

\(
\begin{equation}
\psi_{n}=\left(\frac{2}{l}\right)^{1 / 2} \sin \left(\frac{n \pi x}{l}\right), \quad n=1,2,3, \ldots \tag{7.28}
\end{equation}
\)

for $x$ between 0 and $l$. What are our chances for representing an arbitrary function $f(x)$ in the interval $0 \leq x \leq l$ by a series of the form

\(
\begin{equation}
f(x)=\sum{n=1}^{\infty} a{n} \psi{n}=\left(\frac{2}{l}\right)^{1 / 2} \sum{n=1}^{\infty} a_{n} \sin \left(\frac{n \pi x}{l}\right), \quad 0 \leq x \leq l \tag{7.29}
\end{equation}
\)

where the $a{n}$ 's are constants? Substitution of $x=0$ and $x=l$ in (7.29) gives the restrictions that $f(0)=0$ and $f(l)=0$. In other words, $f(x)$ must satisfy the same boundary conditions as the $\psi{n}$ functions. We shall also assume that $f(x)$ is finite, single-valued, and continuous, but not necessarily differentiable. With these assumptions it can be shown that the expansion (7.29) is valid. We shall not prove (7.29) but will simply illustrate its use to represent a function.

Before we can apply (7.29) to a specific $f(x)$, we must derive an expression for the expansion coefficients $a{n}$. We start by multiplying (7.29) by $\psi{m}^{*}$ :

\(
\begin{equation}
\psi_{m}^{} f(x)=\sum{n=1}^{\infty} a{n} \psi{m}^{*} \psi{n}=\left(\frac{2}{l}\right) \sum{n=1}^{\infty} a{n} \sin \left(\frac{n \pi x}{l}\right) \sin \left(\frac{m \pi x}{l}\right) \tag{7.30}
\end{}
\)

Now we integrate this equation from 0 to $l$. Assuming the validity of interchanging the integration and the infinite summation, we have

\(
\int{0}^{l} \psi{m}^{} f(x) d x=\sum{n=1}^{\infty} a{n} \int{0}^{l} \psi{m}^{} \psi{n} d x=\sum{n=1}^{\infty} a{n}\left(\frac{2}{l}\right) \int{0}^{l} \sin \left(\frac{n \pi x}{l}\right) \sin \left(\frac{m \pi x}{l}\right) d x
\)

We proved the orthonormality of the particle-in-a-box wave functions [Eq. (2.27)]. Therefore, the last equation becomes

\(
\begin{equation}
\int{0}^{l} \psi{m}^{} f(x) d x=\sum{n=1}^{\infty} a{n} \delta_{m n} \tag{7.31}
\end{}
\)

The type of sum in (7.31) occurs often. Writing it in detail, we have

\(
\begin{gather}
\sum{n=1}^{\infty} a{n} \delta{m n}=a{1} \delta{m, 1}+a{2} \delta{m, 2}+\cdots+a{m} \delta{m, m}+a{m+1} \delta{m, m+1}+\cdots \
=0+0+\cdots+a{m}+0+\cdots \
\sum{n=1}^{\infty} a{n} \delta{m n}=a{m} \tag{7.32}
\end{gather}
\)

Thus, since $\delta_{m n}$ is zero except when the summation index $n$ is equal to $m$, all terms but one vanish, and (7.31) becomes

\(
\begin{equation}
a{m}=\int{0}^{l} \psi_{m}^{} f(x) d x \tag{7.33}
\end{}
\)

which is the desired expression for the expansion coefficients.

FIGURE 7.1 Function to be expanded in terms of particle-in-a-box functions.

Changing $m$ to $n$ in (7.33) and substituting it into (7.29), we have

\(
\begin{equation}
f(x)=\sum{n=1}^{\infty}\left[\int{0}^{l} \psi_{n}^{} f(x) d x\right] \psi_{n}(x) \tag{7.34}
\end{}
\)

This is the desired expression for the expansion of an arbitrary well-behaved function $f(x)(0 \leq x \leq l)$ as a linear combination of the particle-in-a-box wave functions $\psi{n}$. Note that the definite integral $\int{0}^{l} \psi_{n}^{*} f(x) d x$ is a number and not a function of $x$.

We now use (7.29) to represent a specific function, the function of Fig. 7.1, which is defined by

\(
\begin{array}{lrl}
f(x) & =x \quad \text { for } \quad 0 \leq x \leq \frac{1}{2} l \
f(x)=l-x \quad \text { for } \quad \frac{1}{2} l \leq x \leq l \tag{7.35}
\end{array}
\)

To find the expansion coefficients $a_{n}$, we substitute (7.28) and (7.35) into (7.33):

\(
\begin{aligned}
& a{n}=\int{0}^{l} \psi{n}^{*} f(x) d x=\left(\frac{2}{l}\right)^{1 / 2} \int{0}^{l} \sin \left(\frac{n \pi x}{l}\right) f(x) d x \
& a{n}=\left(\frac{2}{l}\right)^{1 / 2} \int{0}^{l / 2} x \sin \left(\frac{n \pi x}{l}\right) d x+\left(\frac{2}{l}\right)^{1 / 2} \int_{l / 2}^{l}(l-x) \sin \left(\frac{n \pi x}{l}\right) d x
\end{aligned}
\)

Using the Appendix integral (A.1), we find

\(
\begin{equation}
a_{n}=\frac{(2 l)^{3 / 2}}{n^{2} \pi^{2}} \sin \left(\frac{n \pi}{2}\right) \tag{7.36}
\end{equation}
\)

Using (7.36) in the expansion (7.29), we have [since $\sin (n \pi / 2)$ equals zero for $n$ even and equals +1 or -1 for $n$ odd]

\(
\begin{align}
& f(x)=\frac{4 l}{\pi^{2}}\left[\sin \left(\frac{\pi x}{l}\right)-\frac{1}{3^{2}} \sin \left(\frac{3 \pi x}{l}\right)+\frac{1}{5^{2}} \sin \left(\frac{5 \pi x}{l}\right)-\cdots\right] \
& f(x)=\frac{4 l}{\pi^{2}} \sum_{n=1}^{\infty}(-1)^{n+1} \frac{1}{(2 n-1)^{2}} \sin \left[(2 n-1) \frac{\pi x}{l}\right] \tag{7.37}
\end{align}
\)

where $f(x)$ is given by (7.35). Let us check (7.37) at $x=\frac{1}{2} l$. We have

\(
\begin{equation}
f\left(\frac{l}{2}\right)=\frac{4 l}{\pi^{2}}\left(1+\frac{1}{3^{2}}+\frac{1}{5^{2}}+\frac{1}{7^{2}}+\cdots\right) \tag{7.38}
\end{equation}
\)

FIGURE 7.2 Plots of (a) the error and (b) the percent error in the expansion of the function of Fig. 7.1 in terms of particle-in-a-box wave functions when 1 and 5 terms are taken in the expansion.

Tabulating the right side of (7.38) as a function of the number of terms we take in the infinite series, we get:

If we take an infinite number of terms, the series should sum to $\frac{1}{2} l$, which is the value of $f\left(\frac{1}{2} l\right)$. Assuming the validity of the series, we have the interesting result that the infinite sum in parentheses in (7.38) equals $\pi^{2} / 8$. Figure 7.2 plots $f(x)-\sum{n=1}^{k} a{n} \psi{n}$ [where $f, a{n}$, and $\psi_{n}$ are given by (7.35), (7.36), and (7.28)] for $k$ values of 1 and 5 . As $k$, the number of terms in the expansion, increases, the series comes closer to $f(x)$, and the difference between $f$ and the series goes to zero.

Expansion of a Function in Terms of Eigenfunctions

We have seen an example of the expansion of a function in terms of a set of functionsthe particle-in-a-box energy eigenfunctions. Many different sets of functions can be used to expand an arbitrary function. A set of functions $g{1}, g{2}, \ldots, g{i}, \ldots$ is said to be a complete set if every well-behaved function $f$ that obeys the same boundary conditions as the $g{i}$ functions can be expanded as a linear combination of the $g_{i}$ 's according to

\(
\begin{equation}
f=\sum{i} a{i} g_{i} \tag{7.39}
\end{equation}
\)

where the $a{i}$ 's are constants. Of course, it is understood that $f$ and the $g{i}$ 's are all functions of the same set of variables. The limits have been omitted from the sum in (7.39). It is understood that this sum goes over all members of the complete set. By virtue of theorems of Fourier analysis (which we have not proved), the particle-in-a-box energy eigenfunctions can be shown to be a complete set.

We now postulate that the set of eigenfunctions of every Hermitian operator that represents a physical quantity is a complete set. (Completeness of the eigenfunctions can be proved in many cases, but must be postulated in the general case.) Thus, every wellbehaved function that satisfies the same boundary conditions as the set of eigenfunctions can be expanded according to (7.39). Equation (7.29) is an example of (7.39).

The harmonic-oscillator wave functions are given by a Hermite polynomial $H_{v}$ times an exponential factor [Eq. (4.86) of Prob. 4.21c]. By virtue of the expansion postulate, any well-behaved function $f(x)$ can be expanded as a linear combination of harmonicoscillator energy eigenfunctions:

\(
f(x)=\sum{n=0}^{\infty} a{n}\left(2^{n} n!\right)^{-1 / 2}(\alpha / \pi)^{1 / 4} H_{n}\left(\alpha^{1 / 2} x\right) e^{-\alpha x^{2} / 2}
\)

How about using the hydrogen-atom bound-state wave functions to expand an arbitrary function $f(r, \theta, \phi)$ ? The answer is that these functions do not form a complete set, and we cannot expand $f$ using them. To have a complete set, we must use all the eigenfunctions of a particular Hermitian operator. In addition to the bound-state eigenfunctions of the hydrogen-atom Hamiltonian, we have the continuum eigenfunctions, corresponding to ionized states. If the continuum eigenfunctions are included along with the bound-state eigenfunctions, then we have a complete set. (For the particle in a box and the harmonic oscillator, there are no continuum functions.) Equation (7.39) implies an integration over the continuum eigenfunctions, if there are any. Thus, if $\psi{n l m}(r, \theta, \phi)$ is a bound-state wave function of the hydrogen atom and $\psi{E l m}(r, \theta, \phi)$ is a continuum eigenfunction, then (7.39) becomes

\(
f(r, \theta, \phi)=\sum{n=1}^{\infty} \sum{l=0}^{n-1} \sum{m=-l}^{l} a{n l m} \psi{n l m}(r, \theta, \phi)+\sum{l=0}^{\infty} \sum{m=-l}^{l} \int{0}^{\infty} a{l m}(E) \psi{E l m}(r, \theta, \phi) d E
\)

As another example, consider the eigenfunctions of $\hat{p}_{x}$ [Eq. (3.36)]:

\(
g_{k}=e^{i k x / \hbar}, \quad-\infty<k<\infty
\)

Here the eigenvalues are all continuous, and the eigenfunction expansion (7.39) of an arbitrary function $f$ becomes

\(
f(x)=\int_{-\infty}^{\infty} a(k) e^{i k x / \hbar} d k
\)

The reader with a good mathematical background may recognize this integral as very nearly the Fourier transform of $a(k)$.

Let us evaluate the expansion coefficients in $f=\sum{i} a{i} g{i}$ [Eq. (7.39)], where the $g{i}$ functions are the complete set of eigenfunctions of a Hermitian operator. The procedure is the same as that used to derive (7.33). We multiply $f=\sum{i} a{i} g{i}$ by $g{k}^{*}$ and integrate over all space:

\(
\begin{gather}
g_{k}^{} f=\sum{i} a{i} g{k}^{*} g{i} \
\int g{k}^{*} f d \tau=\sum{i} a{i} \int g{k}^{} g{i} d \tau=\sum{i} a{i} \delta{i k}=a{k} \
a{k}=\int g_{k}^{} f d \tau \tag{7.40}
\end{gather*}
\)

where we used the orthonormality of the eigenfunctions of a Hermitian operator: $\int g{k}^{*} g{i} d \tau=\delta{i k}$ [Eq. (7.26)]. The procedure that led to (7.40) will be used often and is worth remembering. Substitution of (7.40) for $a{i}$ in $f=\sum{i} a{i} g_{i}$ gives

\(
\begin{equation}
f=\sum{i}\left[\int g{i}^{} f d \tau\right] g{i}=\sum{i}\left\langle g{i} \mid f\right\rangle g{i} \tag{7.41}
\end{}
\)

EXAMPLE

Let $F(x)=x(l-x)$ for $0 \leq x \leq l$ and $F(x)=0$ elsewhere. Expand $F$ in terms of the particle-in-a-box energy eigenfunctions $\psi_{n}=(2 / l)^{1 / 2} \sin (n \pi x / l)$ for $0 \leq x \leq l$.

We begin by noting that $F(0)=0$ and $F(l)=0$, so $F$ obeys the same boundary conditions as the $\psi{n}$ 's and can be expanded using the $\psi{n}$ 's. The expansion is $F=\sum{n=1}^{\infty} a{n} \psi{n}$, where $a{n}=\int \psi_{n}^{*} F d \tau$ [Eqs. (7.39) and (7.40)]. Thus

\(
a{n}=\int \psi{n}^{*} F d \tau=\left(\frac{2}{l}\right)^{1 / 2} \int_{0}^{l}\left(\sin \frac{n \pi x}{l}\right) x(l-x) d x=\frac{2^{3 / 2} l^{5 / 2}}{n^{3} \pi^{3}}\left[1-(-1)^{n}\right]
\)

where details of the integral evaluation are left as a problem (Prob. 7.18). The expansion $F=\sum{n=1}^{\infty} a{n} \psi_{n}$ is

\(
x(l-x)=\frac{4 l^{2}}{\pi^{3}} \sum_{n=1}^{\infty} \frac{1-(-1)^{n}}{n^{3}} \sin \frac{n \pi x}{l}, \quad \text { for } 0 \leq x \leq l
\)

EXERCISE Let $G(x)=1$ for $0 \leq x \leq l$ and $G(x)=0$ elsewhere. Expand $G$ in terms of the particle-in-a-box energy eigenfunctions. Since $G$ is not zero at 0 and at $l$, the expansion will not represent $G$ at these points but will represent $G$ elsewhere. Use the first 7 nonzero terms of the expansion to calculate $G$ at $x=\frac{1}{4} l$. Repeat this using the first 70 nonzero terms (use a programmable calculator). (Answers: 0.1219, 0.9977.)

A useful theorem is the following:
THEOREM 3. Let the functions $g{1}, g{2}, \ldots$ be the complete set of eigenfunctions of the Hermitian operator $\hat{A}$, and let the function $F$ be an eigenfunction of $\hat{A}$ with eigenvalue $k$ (that is, $\hat{A} F=k F$ ). Then if $F$ is expanded as $F=\sum{i} a{i} g{i}$, the only nonzero coefficients $a{i}$ are those for which $g{i}$ has the eigenvalue $k$. (Because of degeneracy, several $g{i}$ 's may have the same eigenvalue $k$.)

Thus in the expansion of $F$, we include only those eigenfunctions that have the same eigenvalue as $F$. The proof of Theorem 3 follows at once from $a{k}=\int g{k}^{*} F d \tau$ [Eq. (7.40)]; if $F$ and $g{k}$ correspond to different eigenvalues of the Hermitian operator $\hat{A}$, they will be orthogonal [Eq. (7.22)] and $a{k}$ will vanish.

We shall occasionally use a notation (called ket notation) in which the function $f$ is denoted by the symbol $|f\rangle$. There doesn't seem to be any point to this notation, but in advanced formulations of quantum mechanics, it takes on a special significance. In ket notation, Eq. (7.41) reads

\(
\begin{equation}
|f\rangle=\sum{i}\left|g{i}\right\rangle\left\langle g{i} \mid f\right\rangle=\sum{i}|i\rangle\langle i \mid f\rangle \tag{7.42}
\end{equation}
\)

Ket notation is conveniently used to specify eigenfunctions by listing their eigenvalues. For example, the hydrogen-atom wave function with quantum numbers $n, l, m$ is denoted by $\psi_{n l m}=|n l m\rangle$.

The contents of Sections 7.2 and 7.3 can be summarized by the statement that the eigenfunctions of a Hermitian operator form a complete, orthonormal set, and the eigenvalues are real.