To deal with the time-independent Schrödinger equation for systems (such as atoms or molecules) that contain interacting particles, we must use approximation methods. This chapter discusses the variation method, which allows us to approximate the ground-state energy of a system without solving the Schrödinger equation. The variation method is based on the following theorem:
THE VARIATION THEOREM
Given a system whose Hamiltonian operator $\hat{H}$ is time independent and whose lowest-energy eigenvalue is $E_{1}$, if $\phi$ is any normalized, well-behaved function of the coordinates of the system's particles that satisfies the boundary conditions of the problem, then
\(
\begin{equation}
\int \phi^{} \hat{H} \phi d \tau \geq E_{1}, \quad \phi \text { normalized } \tag{8.1}
\end{}
\)
The variation theorem allows us to calculate an upper bound for the system's groundstate energy.
To prove (8.1), we expand $\phi$ in terms of the complete, orthonormal set of eigenfunctions of $\hat{H}$, the stationary-state eigenfunctions $\psi_{k}$ :
\(
\begin{equation}
\phi=\sum{k} a{k} \psi_{k} \tag{8.2}
\end{equation}
\)
where
\(
\begin{equation}
\hat{H} \psi{k}=E{k} \psi_{k} \tag{8.3}
\end{equation}
\)
Note that the expansion (8.2) requires that $\phi$ obey the same boundary conditions as the $\psi_{k}$ 's. Substitution of (8.2) into the left side of (8.1) gives
\(
\int \phi^{} \hat{H} \phi d \tau=\int \sum{k} a{k}^{} \psi{k}^{*} \hat{H} \sum{j} a{j} \psi{j} d \tau=\int \sum{k} a{k}^{} \psi_{k}^{} \sum{j} a{j} \hat{H} \psi_{j} d \tau
\)
Using the eigenvalue equation (8.3) and assuming the validity of interchanging the integration and the infinite summations, we get
\(
\begin{aligned}
\int \phi^{} \hat{H} \phi d \tau & =\int \sum{k} a{k}^{} \psi{k}^{*} \sum{j} a{j} E{j} \psi{j} d \tau=\sum{k} \sum{j} a{k}^{} a{j} E{j} \int \psi_{k}^{} \psi{j} d \tau \
& =\sum{k} \sum{j} a{k}^{*} a{j} E{j} \delta_{k j}
\end{aligned}
\)
where the orthonormality of the eigenfunctions $\psi_{k}$ was used. We perform the sum over $j$, and, as usual, the Kronecker delta makes all terms zero except the one with $j=k$, giving
\(
\begin{equation}
\int \phi^{} \hat{H} \phi d \tau=\sum{k} a{k}^{} a{k} E{k}=\sum{k}\left|a{k}\right|^{2} E_{k} \tag{8.4}
\end{equation}
\)
Since $E{1}$ is the lowest-energy eigenvalue of $\hat{H}$, we have $E{k} \geq E{1}$. Since $\left|a{k}\right|^{2}$ is never negative, we can multiply the inequality $E{k} \geq E{1}$ by $\left|a{k}\right|^{2}$ without changing the direction of the inequality sign to get $\left|a{k}\right|^{2} E{k} \geq\left|a{k}\right|^{2} E{1}$. Therefore, $\sum{k}\left|a{k}\right|^{2} E{k} \geq \sum{k}\left|a{k}\right|^{2} E_{1}$, and use of (8.4) gives
\(
\begin{equation}
\int \phi^{} \hat{H} \phi d \tau=\sum{k}\left|a{k}\right|^{2} E{k} \geq \sum{k}\left|a{k}\right|^{2} E{1}=E{1} \sum{k}\left|a_{k}\right|^{2} \tag{8.5}
\end{}
\)
Because $\phi$ is normalized, we have $\int \phi^{*} \phi d \tau=1$. Substitution of the expansion (8.2) into the normalization condition gives
\(
\begin{gather}
1=\int \phi^{} \phi d \tau=\int \sum{k} a{k}^{} \psi_{k}^{} \sum{j} a{j} \psi{j} d \tau=\sum{k} \sum{j} a{k}^{} a{j} \int \psi{k}^{} \psi{j} d \tau=\sum{k} \sum{j} a{k}^{} a{j} \delta{k j} \
1=\sum{k}\left|a{k}\right|^{2} \tag{8.6}
\end{gather}
\)
Use of (8.6) in (8.5) gives the variation theorem (8.1):
\(
\begin{equation}
\int \phi^{} \hat{H} \phi d \tau \geq E_{1}, \quad \phi \text { normalized } \tag{8.7}
\end{}
\)
Suppose we have a function $\phi$ that is not normalized. To apply the variation theorem, we multiply $\phi$ by a normalization constant $N$ so that $N \phi$ is normalized. Replacing $\phi$ by $N \phi$ in (8.7), we have
\(
\begin{equation}
|N|^{2} \int \phi^{} \hat{H} \phi d \tau \geq E_{1} \tag{8.8}
\end{}
\)
$N$ is determined by $\int(N \phi)^{} N \phi d \tau=|N|^{2} \int \phi^{} \phi d \tau=1$; so $|N|^{2}=1 / \int \phi^{*} \phi d \tau$ and (8.8) becomes
\(
\begin{equation}
\frac{\int \phi^{} \hat{H} \phi d \tau}{\int \phi^{} \phi d \tau} \geq E_{1} \tag{8.9}
\end{equation}
\)
where $\phi$ is any well-behaved function (not necessarily normalized) that satisfies the boundary conditions of the problem.
The function $\phi$ is called a trial variation function, and the integral in (8.1) [or the ratio of integrals in (8.9)] is called the variational integral. To arrive at a good approximation to the ground-state energy $E{1}$, we try many trial variation functions and look for the one that gives the lowest value of the variational integral. From (8.1), the lower the value of the variational integral, the better the approximation we have to $E{1}$. One way to disprove quantum mechanics would be to find a trial variation function that made the variational integral less than $E{1}$ for some system where $E{1}$ is known.
Let $\psi_{1}$ be the true ground-state wave function:
\(
\begin{equation}
\hat{H} \psi{1}=E{1} \psi_{1} \tag{8.10}
\end{equation}
\)
If we happened to be lucky enough to hit upon a variation function that was equal to $\psi{1}$, then, using (8.10) in (8.1), we see that the variational integral will be equal to $E{1}$. Thus the ground-state wave function gives the minimum value of the variational integral. We therefore expect that the lower the value of the variational integral, the closer the trial variational function will approach the true ground-state wave function. However, it turns out that the variational integral approaches $E{1}$ a lot faster than the trial variation function approaches $\psi{1}$, and it is possible to get a rather good approximation to $E_{1}$ using a rather poor $\phi$.
In practice, one usually puts several parameters into the trial function $\phi$ and then varies these parameters so as to minimize the variational integral. Successful use of the variation method depends on the ability to make a good choice for the trial function.
Let us look at some examples of the variation method. Although the real utility of the method is for problems to which we do not know the true solutions, we will consider problems that are exactly solvable so that we can judge the accuracy of our results.
EXAMPLE
Devise a trial variation function for the particle in a one-dimensional box of length $l$.
The wave function is zero outside the box and the boundary conditions require that $\psi=0$ at $x=0$ and at $x=l$. The variation function $\phi$ must meet these boundary conditions of being zero at the ends of the box. As noted after Eq. (4.57), the ground-state $\psi$ has no nodes interior to the boundary points, so it is desirable that $\phi$ have no interior nodes. A simple function that has these properties is the parabolic function
\(
\begin{equation}
\phi=x(l-x) \quad \text { for } 0 \leq x \leq l \tag{8.11}
\end{equation}
\)
and $\phi=0$ outside the box. Since we have not normalized $\phi$, we use Eq. (8.9). Inside the box the Hamiltonian is $-\left(\hbar^{2} / 2 m\right) d^{2} / d x^{2}$. For the numerator and denominator of (8.9), we have
\(
\begin{gather}
\int \phi^{} \hat{H} \phi d \tau=-\frac{\hbar^{2}}{2 m} \int{0}^{l}\left(l x-x^{2}\right) \frac{d^{2}}{d x^{2}}\left(l x-x^{2}\right) d x=\frac{\hbar^{2}}{m} \int{0}^{l}\left(l x-x^{2}\right) d x=\frac{\hbar^{2} l^{3}}{6 m} \tag{8.12}\
\int \phi^{} \phi d \tau=\int_{0}^{l} x^{2}(l-x)^{2} d x=\frac{l^{5}}{30} \tag{8.13}
\end{gather}
\)
Substituting in the variation theorem (8.9), we get
\(
E_{1} \leq \frac{5 h^{2}}{4 \pi^{2} m l^{2}}=0.1266515 \frac{h^{2}}{m l^{2}}
\)
From Eq. (2.20), $E_{1}=h^{2} / 8 m l^{2}=0.125 h^{2} / m l^{2}$, and the energy error is $1.3 \%$.
Since $\int|\phi|^{2} d \tau=l^{5} / 30$, the normalized form of (8.11) is $\left(30 / l^{5}\right)^{1 / 2} x(l-x)$.
Figure 7.3 shows that this function rather closely resembles the true ground-state particle-in-a-box wave function.
EXERCISE A one-particle, one-dimensional system has the potential energy function $V=V{0}$ for $0 \leq x \leq l$ and $V=\infty$ elsewhere (where $V{0}$ is a constant). (a) Use the variation function $\phi=\sin (\pi x / l)$ for $0 \leq x \leq l$ and $\phi=0$ elsewhere to estimate the ground-state energy of this system. (b) Explain why the result of (a) is the exact groundstate energy. Hint: See one of the Chapter 4 problems. (Answer: (a) $V_{0}+h^{2} / 8 m l^{2}$.)
The preceding example did not have a parameter in the trial function. The next example does.
EXAMPLE
For the one-dimensional harmonic oscillator, devise a variation function with a parameter and find the optimum value of the parameter. Estimate the ground-state energy.
The variation function $\phi$ must be quadratically integrable and so must go to zero as $x$ goes to $\pm \infty$. The function $e^{-x}$ has the proper behavior at $+\infty$ but becomes infinite at $-\infty$. The function $e^{-x^{2}}$ has the proper behavior at $\pm \infty$. However, it is dimensionally unsatisfactory, since the power to which we raise $e$ must be dimensionless. This can be seen from the Taylor series $e^{z}=1+z+z^{2} / 2!+\cdots$ [Eq. (4.44)]. Since all the terms in this series must have the same dimensions, $z$ must have the same dimensions as 1 ; that is, $z$ in $e^{z}$ must be dimensionless. Hence we modify $e^{-x^{2}}$ to $e^{-c x^{2}}$, where $c$ has units of length ${ }^{-2}$. We shall take $c$ as a variational parameter. The true ground-state $\psi$ must have no nodes. Also, since $V=\frac{1}{2} k x^{2}$ is an even function, the ground-state $\psi$ must have definite parity and must be an even function, since an odd function has a node at the origin. The trial function $e^{-c x^{2}}$ has the desired properties of having no nodes and of being an even function.
Use of (4.30) for $\hat{H}$ and Appendix integrals gives (Prob. 8.3)
\(
\begin{aligned}
\int \phi^{} \hat{H} \phi d \tau & =-\frac{\hbar^{2}}{2 m} \int{-\infty}^{\infty} e^{-c x^{2}} \frac{d^{2} e^{-c x^{2}}}{d x^{2}} d x+2 \pi^{2} v^{2} m \int{-\infty}^{\infty} x^{2} e^{-2 c x^{2}} d x \
& =\frac{\hbar^{2}}{m}\left(\frac{\pi c}{8}\right)^{1 / 2}+v^{2} m\left(\frac{\pi^{5}}{8}\right)^{1 / 2} c^{-3 / 2} \
\int \phi^{} \phi d \tau & =\int{-\infty}^{\infty} e^{-2 c x^{2}} d x=2 \int{0}^{\infty} e^{-2 c x^{2}} d x=\left(\frac{\pi}{2 c}\right)^{1 / 2}
\end{aligned}
\)
The variational integral $W$ is
\(
\begin{equation}
W \equiv \frac{\int \phi^{} \hat{H} \phi d \tau}{\int \phi^{} \phi d \tau}=\frac{\hbar^{2} c}{2 m}+\frac{\pi^{2} \nu^{2} m}{2 c} \tag{8.14}
\end{equation}
\)
We now vary $c$ to minimize the variational integral (8.14). A necessary condition that $W$ be minimized is that
\(
\begin{gather}
\frac{d W}{d c}=0=\frac{\hbar^{2}}{2 m}-\frac{\pi^{2} \nu^{2} m}{2 c^{2}} \
c= \pm \pi \nu m / \hbar \tag{8.15}
\end{gather}
\)
The negative root $c=-\pi \nu m / \hbar$ is rejected, since it would make $\phi=e^{-c x^{2}}$ not quadratically integrable. Substitution of $c=\pi \nu m / \hbar$ into (8.14) gives $W=\frac{1}{2} h \nu$. This is the exact ground-state harmonic-oscillator energy. With $c=\pi \nu m / \hbar$ the variation function $\phi$ is the same (except for being unnormalized) as the harmonic-oscillator ground-state wave function (4.53) and (4.31).
For the normalized harmonic-oscillator variation function $\phi=(2 c / \pi)^{1 / 4} e^{-c x^{2}}$, a large value of $c$ makes $\phi$ fall off very rapidly from its maximum value at $x=0$. This makes the probability density large only near $x=0$. The potential energy $V=\frac{1}{2} k x^{2}$ is low near $x=0$, so a large $c$ means a low $\langle V\rangle=\langle\phi| V|\phi\rangle$. [Note also that $\langle V\rangle$ equals the second term on the right side of (8.14).] However, because a large $c$ makes $\phi$ fall off very rapidly from its maximum, it makes $|d \phi / d x|$ large in the region near
$x=0$. From Prob. 7.7b, a large $|d \phi / d x|$ means a large value of $\langle T\rangle$ [which equals the first term on the right side of (8.14)]. The optimum value of $c$ minimizes the sum $\langle T\rangle+\langle V\rangle=W$. In atoms and molecules, the true wave function is a compromise between the tendency to minimize $\langle V\rangle$ by confining the electrons to regions of low $V$ (near the nuclei) and the tendency to minimize $\langle T\rangle$ by allowing the electron probability density to spread out over a large region.
EXERCISE Consider a one-particle, one-dimensional system with $V=0$ for $-\frac{1}{2} l \leq x \leq \frac{1}{2} l$ and $V=b \hbar^{2} / m l^{2}$ elsewhere (where $b$ is a positive constant) (Fig. 2.5 with $V_{0}=b \hbar^{2} / m l^{2}$ and the origin shifted). (a) For the variation function $\phi=(x-c)(x+c)=x^{2}-c^{2}$ for $-c \leq x \leq c$ and $\phi=0$ elsewhere, where the variational parameter $c$ satisfies $c>\frac{1}{2} l$, one finds that the variational integral $W$ is given by
\(
W=\frac{\hbar^{2}}{m l^{2}}\left[\frac{5 l^{2}}{4 c^{2}}+b\left(1-\frac{15 l}{16 c}+\frac{5 l^{3}}{32 c^{3}}-\frac{3 l^{5}}{256 c^{5}}\right)\right]
\)
Sketch $\phi$ and $V$ on the same plot. Find the equation satisfied by the value of $c$ that minimizes $W$. (b) Find the optimum $c$ and $W$ for $V_{0}=20 \hbar^{2} / \mathrm{ml}^{2}$ and compare with the true ground-state energy $2.814 \hbar^{2} / m l^{2}$ (Prob. 4.31c). (Hint: You may want to use the Solver in a spreadsheet or a programmable calculator to find $c / l$.) (Answer: (a) $48 t^{4}-24 t^{2}-128 t^{3} / b+3=0$, where $t \equiv c / l$. (b) $c=0.6715 l, W=3.454 \hbar^{2} / m l^{2}$.)