The variation method as presented in the last section gives information about only the ground-state energy and wave function. We now discuss extension of the variation method to excited states. (See also Section 8.5.)
Consider how we might extend the variation method to estimate the energy of the first excited state. We number the stationary states of the system $1,2,3, \ldots$ in order of increasing energy:
\(
E{1} \leq E{2} \leq E_{3} \leq \cdots
\)
We showed that for a normalized variational function $\phi$ [Eqs. (8.4) and (8.6)]
\(
\int \phi^{} \hat{H} \phi d \tau=\sum{k=1}^{\infty}\left|a{k}\right|^{2} E_{k} \quad \text { and } \quad \int \phi^{} \phi d \tau=\sum{k=1}^{\infty}\left|a{k}\right|^{2}=1
\)
where the $a{k}$ 's are the expansion coefficients in $\phi=\sum{k} a{k} \psi{k}$ [Eq. (8.2)]. We have $a{k}=\left\langle\psi{k} \mid \phi\right\rangle$ [Eq. (7.40)]. Let us restrict ourselves to normalized functions $\phi$ that are orthogonal to the true ground-state wave function $\psi{1}$. Then we have $a{1}=\left\langle\psi_{1} \mid \phi\right\rangle=0$ and
\(
\begin{equation}
\int \phi^{} \hat{H} \phi d \tau=\sum{k=2}^{\infty}\left|a{k}\right|^{2} E{k} \quad \text { and } \quad \int \phi^{*} \phi d \tau=\sum{k=2}^{\infty}\left|a_{k}\right|^{2}=1 \tag{8.16}
\end{}
\)
For $k \geq 2$, we have $E{k} \geq E{2}$ and $\left|a{k}\right|^{2} E{k} \geq\left|a{k}\right|^{2} E{2}$. Hence
\(
\begin{equation}
\sum{k=2}^{\infty}\left|a{k}\right|^{2} E{k} \geq \sum{k=2}^{\infty}\left|a{k}\right|^{2} E{2}=E{2} \sum{k=2}^{\infty}\left|a{k}\right|^{2}=E{2} \tag{8.17}
\end{equation}
\)
Combining (8.16) and (8.17), we have the desired result:
\(
\begin{equation}
\int \phi^{} \hat{H} \phi d \tau \geq E{2} \quad \text { if } \quad \int \psi{1}^{} \phi d \tau=0 \quad \text { and } \quad \int \phi^{} \phi d \tau=1 \tag{8.18}
\end{}
\)
The inequality (8.18) allows us to get an upper bound to the energy $E{2}$ of the first excited state. However, the restriction $\left\langle\psi{1} \mid \phi\right\rangle=0$ makes this method troublesome to apply.
For certain systems, it is possible to be sure that $\left\langle\psi{1} \mid \phi\right\rangle=0$ even though we do not know the true ground-state wave function. An example is a one-dimensional problem for which $V$ is an even function of $x$. In this case the ground-state wave function is always an even function, while the first excited-state wave function is odd. (All the wave functions must be of definite parity. The ground-state wave function is nodeless, and, since an odd function vanishes at the origin, the ground-state wave function must be even. The first excited-state wave function has one node and must be odd.) Therefore, for odd trial functions, it must be true that $\left\langle\psi{1} \mid \phi\right\rangle=0$; the even function $\psi_{1}$ times the odd function $\phi$ gives an odd integrand whose integral from $-\infty$ to $\infty$ is zero.
Another example is a particle moving in a central field (Section 6.1). The form of the potential energy might be such that we could not solve for the radial factor $R(r)$ in the eigenfunction. However, the angular factor in $\psi$ is a spherical harmonic [Eq. (6.16)], and spherical harmonics with different values of $l$ are orthogonal. Thus we can get an upper bound to the energy of the lowest state with any given angular momentum $l$ by using the factor $Y_{l}^{m}$ in the trial function. This result depends on the extension of (8.18) to higher excited states:
\(
\begin{equation}
\frac{\int \phi^{} \hat{H} \phi d \tau}{\int \phi^{} \phi d \tau} \geq E{k+1} \quad \text { if } \quad \int \psi{1}^{} \phi d \tau=\int \psi{2}^{*} \phi d \tau=\cdots=\int \psi{k}^{} \phi d \tau=0 \tag{8.19}
\end{equation}
\)