Comprehensive Study Notes

The perturbation treatments of degenerate and nondegenerate energy levels differ. This section examines the effect of a perturbation on a nondegenerate level. If some of the energy levels of the unperturbed system are degenerate while others are nondegenerate, the treatment in this section will apply to the nondegenerate levels only.

Nondegenerate Perturbation Theory

Let $\psi{n}^{(0)}$ be the wave function of some particular unperturbed nondegenerate level with energy $E{n}^{(0)}$. Let $\psi{n}$ be the perturbed wave function into which $\psi{n}^{(0)}$ is converted when the perturbation is applied. From (9.1) and (9.7), the Schrödinger equation for the perturbed state is

\(
\begin{equation}
\hat{H} \psi{n}=\left(\hat{H}^{0}+\lambda \hat{H}^{\prime}\right) \psi{n}=E{n} \psi{n} \tag{9.8}
\end{equation}
\)

Since the Hamiltonian in (9.8) depends on the parameter $\lambda$, both the eigenfunction $\psi{n}$ and the eigenvalue $E{n}$ depend on $\lambda$ :

\(
\psi{n}=\psi{n}(\lambda, q) \quad \text { and } \quad E{n}=E{n}(\lambda)
\)

where $q$ denotes the system's coordinates. We now expand $\psi{n}$ and $E{n}$ as Taylor series (Prob. 4.1) in powers of $\lambda$ :

\(
\begin{align}
& \psi{n}=\left.\psi{n}\right|{\lambda=0}+\left.\frac{\partial \psi{n}}{\partial \lambda}\right|{\lambda=0} \lambda+\left.\frac{\partial^{2} \psi{n}}{\partial \lambda^{2}}\right|{\lambda=0} \frac{\lambda^{2}}{2!}+\cdots \tag{9.9}\
& E{n}=\left.E{n}\right|{\lambda=0}+\left.\frac{d E{n}}{d \lambda}\right|{\lambda=0} \lambda+\left.\frac{d^{2} E{n}}{d \lambda^{2}}\right|{\lambda=0} \frac{\lambda^{2}}{2!}+\cdots \tag{9.10}
\end{align}
\)

By hypothesis, when $\lambda$ goes to zero, $\psi{n}$ and $E{n}$ go to $\psi{n}^{(0)}$ and $E{n}^{(0)}$ :

\(
\begin{equation}
\left.\psi{n}\right|{\lambda=0}=\psi{n}^{(0)} \quad \text { and }\left.\quad E{n}\right|{\lambda=0}=E{n}^{(0)} \tag{9.11}
\end{equation}
\)

We introduce the following abbreviations:

\(
\begin{equation}
\left.\psi{n}^{(k)} \equiv \frac{1}{k!} \frac{\partial^{k} \psi{n}}{\partial \lambda^{k}}\right|{\lambda=0},\left.\quad E{n}^{(k)} \equiv \frac{1}{k!} \frac{d^{k} E{n}}{d \lambda^{k}}\right|{\lambda=0}, \quad k=1,2, \ldots \tag{9.12}
\end{equation}
\)

Equations (9.9) and (9.10) become

\(
\begin{align}
& \psi{n}=\psi{n}^{(0)}+\lambda \psi{n}^{(1)}+\lambda^{2} \psi{n}^{(2)}+\cdots+\lambda^{k} \psi{n}^{(k)}+\cdots \tag{9.13}\
& E{n}=E{n}^{(0)}+\lambda E{n}^{(1)}+\lambda^{2} E{n}^{(2)}+\cdots+\lambda^{k} E{n}^{(k)}+\cdots \tag{9.14}
\end{align}
\)

For $k=1,2,3, \ldots$, we call $\psi{n}^{(k)}$ and $E{n}^{(k)}$ the $\boldsymbol{k}$ th-order corrections to the wave function and energy. We shall assume that the series (9.13) and (9.14) converge for $\lambda=1$, and we hope that for a small perturbation, taking just the first few terms of the series will give a good approximation to the true energy and wave function. (Quite often, perturbation-theory series do not converge, but even so, the first few terms of a nonconvergent series can often give a useful approximation.)

We shall take $\psi{n}^{(0)}$ to be normalized: $\left\langle\psi{n}^{(0)} \mid \psi{n}^{(0)}\right\rangle=1$. Instead of taking $\psi{n}$ as normalized, we shall require that $\psi_{n}$ satisfy

\(
\begin{equation}
\left\langle\psi{n}^{(0)} \mid \psi{n}\right\rangle=1 \tag{9.15}
\end{equation}
\)

If $\psi{n}$ does not satisfy this equation, then multiplication of $\psi{n}$ by the constant $1 /\left\langle\psi{n}^{(0)} \mid \psi{n}\right\rangle$ gives a perturbed wave function with the desired property. The condition (9.15), called intermediate normalization, simplifies the derivation. Note that multiplication of $\psi{n}$ by a constant does not change the energy in the Schrödinger equation $\hat{H} \psi{n}=E{n} \psi{n}$, so use of intermediate normalization does not affect the results for the energy corrections. If desired, at the end of the calculation, the intermediate-normalized $\psi_{n}$ can be multiplied by a constant to normalize it in the usual sense.

Substitution of (9.13) into $1=\left\langle\psi{n}^{(0)} \mid \psi{n}\right\rangle$ [Eq. (9.15)] gives

\(
1=\left\langle\psi{n}^{(0)} \mid \psi{n}^{(0)}\right\rangle+\lambda\left\langle\psi{n}^{(0)} \mid \psi{n}^{(1)}\right\rangle+\lambda^{2}\left\langle\psi{n}^{(0)} \mid \psi{n}^{(2)}\right\rangle+\cdots
\)

Since this equation is true for all values of $\lambda$ in the range 0 to 1 , the coefficients of like powers of $\lambda$ on each side of the equation must be equal, as proved after Eq. (4.11). Equating the $\lambda^{0}$ coefficients, we have $1=\left\langle\psi{n}^{(0)} \mid \psi{n}^{(0)}\right\rangle$, which is satisfied since $\psi_{n}^{(0)}$ is normalized. Equating the coefficients of $\lambda^{1}$, of $\lambda^{2}$, and so on, we have

\(
\begin{equation}
\left\langle\psi{n}^{(0)} \mid \psi{n}^{(1)}\right\rangle=0, \quad\left\langle\psi{n}^{(0)} \mid \psi{n}^{(2)}\right\rangle=0, \quad \text { etc. } \tag{9.16}
\end{equation}
\)

The corrections to the wave function are orthogonal to $\psi_{n}^{(0)}$ when intermediate normalization is used.

Substituting (9.13) and (9.14) into the Schrödinger equation (9.8), we have

\(
\begin{aligned}
\left(\hat{H}^{0}+\lambda \hat{H}^{\prime}\right)\left(\psi{n}^{(0)}+\right. & \left.\lambda \psi{n}^{(1)}+\lambda^{2} \psi{n}^{(2)}+\cdots\right) \
& =\left(E{n}^{(0)}+\lambda E{n}^{(1)}+\lambda^{2} E{n}^{(2)}+\cdots\right)\left(\psi{n}^{(0)}+\lambda \psi{n}^{(1)}+\lambda^{2} \psi_{n}^{(2)}+\cdots\right)
\end{aligned}
\)

Collecting like powers of $\lambda$, we have

\(
\begin{align}
& \hat{H}^{0} \psi{n}^{(0)}+\lambda\left(\hat{H}^{\prime} \psi{n}^{(0)}+\hat{H}^{0} \psi{n}^{(1)}\right)+\lambda^{2}\left(\hat{H}^{0} \psi{n}^{(2)}+\hat{H}^{\prime} \psi{n}^{(1)}\right)+\cdots \
& \quad=E{n}^{(0)} \psi{n}^{(0)}+\lambda\left(E{n}^{(1)} \psi{n}^{(0)}+E{n}^{(0)} \psi{n}^{(1)}\right)+\lambda^{2}\left(E{n}^{(2)} \psi{n}^{(0)}+E{n}^{(1)} \psi{n}^{(1)}+E{n}^{(0)} \psi_{n}^{(2)}\right)+\cdots \tag{9.17}
\end{align}
\)

Now (assuming suitable convergence) for the two series on each side of (9.17) to be equal to each other for all values of $\lambda$, the coefficients of like powers of $\lambda$ in the two series must be equal.

Equating the coefficients of the $\lambda^{0}$ terms, we have $\hat{H}^{0} \psi{n}^{(0)}=E{n}^{(0)} \psi_{n}^{(0)}$, which is the Schrödinger equation for the unperturbed problem, Eq. (9.2), and gives us no new information. Equating the coefficients of the $\lambda^{1}$ terms, we have

\(
\begin{gather}
\hat{H}^{\prime} \psi{n}^{(0)}+\hat{H}^{0} \psi{n}^{(1)}=E{n}^{(1)} \psi{n}^{(0)}+E{n}^{(0)} \psi{n}^{(1)} \
\hat{H}^{0} \psi{n}^{(1)}-E{n}^{(0)} \psi{n}^{(1)}=E{n}^{(1)} \psi{n}^{(0)}-\hat{H}^{\prime} \psi{n}^{(0)} \tag{9.18}
\end{gather}
\)

The First-Order Energy Correction

To find $E{n}^{(1)}$ we multiply (9.18) by $\psi{m}^{(0) *}$ and integrate over all space, which gives

\(
\begin{equation}
\left\langle\psi{m}^{(0)}\right| \hat{H}^{0}\left|\psi{n}^{(1)}\right\rangle-E{n}^{(0)}\left\langle\psi{m}^{(0)} \mid \psi{n}^{(1)}\right\rangle=E{n}^{(1)}\left\langle\psi{m}^{(0)} \mid \psi{n}^{(0)}\right\rangle-\left\langle\psi{m}^{(0)}\right| \hat{H}^{\prime}\left|\psi{n}^{(0)}\right\rangle \tag{9.19}
\end{equation}
\)

where bracket notation [Eqs. (7.1) and (7.3)] is used. $\hat{H}^{0}$ is Hermitian, and use of the Hermitian property (7.12) gives for the first term on the left side of (9.19)

\(
\begin{align}
\left\langle\psi{m}^{(0)}\right| \hat{H}^{0}\left|\psi{n}^{(1)}\right\rangle & =\left\langle\psi{n}^{(1)}\right| \hat{H}^{0}\left|\psi{m}^{(0)}\right\rangle =\left\langle\psi{n}^{(1)} \mid \hat{H}^{0} \psi{m}^{(0)}\right\rangle \
& =\left\langle\psi{n}^{(1)} \mid E{m}^{(0)} \psi_{m}^{(0)}\right\rangle =E{m}^{(0)} *\left\langle\psi{n}^{(1)} \mid \psi{m}^{(0)}\right\rangle *=E{m}^{(0)}\left\langle\psi{m}^{(0)} \mid \psi{n}^{(1)}\right\rangle \tag{9.20}
\end{align*}
\)

where we used the unperturbed Schrödinger equation $\hat{H}^{0} \psi{m}^{(0)}=E{m}^{(0)} \psi{m}^{(0)}$, the fact that $E{m}^{(0)}$ is real, and (7.4). Substitution of (9.20) into (9.19) and use of the orthonormality equation $\left\langle\psi{m}^{(0)} \mid \psi{n}^{(0)}\right\rangle=\delta_{m n}$ for the unperturbed eigenfunctions gives

\(
\begin{equation}
\left(E{m}^{(0)}-E{n}^{(0)}\right)\left\langle\psi{m}^{(0)} \mid \psi{n}^{(1)}\right\rangle=E{n}^{(1)} \delta{m n}-\left\langle\psi{m}^{(0)}\right| \hat{H}^{\prime}\left|\psi{n}^{(0)}\right\rangle \tag{9.21}
\end{equation}
\)

If $m=n$, the left side of (9.21) equals zero, and (9.21) becomes

\(
\begin{equation}
E{n}^{(1)}=\left\langle\psi{n}^{(0)}\right| \hat{H}^{\prime}\left|\psi{n}^{(0)}\right\rangle=\int \psi{n}^{(0) } \hat{H}^{\prime} \psi_{n}^{(0)} d \tau \tag{9.22}
\end{}
\)

The first-order correction to the energy is found by averaging the perturbation $\hat{H}^{\prime}$ over the appropriate unperturbed wave function.

Setting $\lambda=1$ in (9.14), we have

\(
\begin{equation}
E{n} \approx E{n}^{(0)}+E{n}^{(1)}=E{n}^{(0)}+\int \psi_{n}^{(0) } \hat{H}^{\prime} \psi_{n}^{(0)} d \tau \tag{9.23}
\end{}
\)

EXAMPLE

For the anharmonic oscillator with Hamiltonian (9.3), evaluate $E^{(1)}$ for the ground state if the unperturbed system is taken as the harmonic oscillator.
The perturbation is given by Eqs. (9.3) to (9.5) as

\(
\hat{H}^{\prime}=\hat{H}-\hat{H}^{0}=c x^{3}+d x^{4}
\)

and the first-order energy correction for the state with quantum number $v$ is given by (9.22) as $E{v}^{(1)}=\left\langle\psi{v}^{(0)}\right| c x^{3}+d x^{4}\left|\psi{v}^{(0)}\right\rangle$, where $\psi{v}^{(0)}$ is the harmonic-oscillator wave function for state $v$. For the $v=0$ ground state, use of $\psi_{0}^{(0)}=(\alpha / \pi)^{1 / 4} e^{-\alpha x^{2} / 2}$ [Eq. (4.53)] gives

\(
E{0}^{(1)}=\left\langle\psi{0}^{(0)}\right| c x^{3}+d x^{4}\left|\psi{0}^{(0)}\right\rangle=\left(\frac{\alpha}{\pi}\right)^{1 / 2} \int{-\infty}^{\infty} e^{-\alpha x^{2}}\left(c x^{3}+d x^{4}\right) d x
\)

The integral from $-\infty$ to $\infty$ of the odd function $c x^{3} e^{-\alpha x^{2}}$ is zero. Use of the Appendix integral (A.10) with $n=2$ and (4.31) for $\alpha$ gives

\(
E{0}^{(1)}=2 d\left(\frac{\alpha}{\pi}\right)^{1 / 2} \int{0}^{\infty} e^{-\alpha x^{2}} x^{4} d x=\frac{3 d}{4 \alpha^{2}}=\frac{3 d h^{2}}{64 \pi^{4} \nu^{2} m^{2}}
\)

The unperturbed ground-state energy is $E{0}^{(0)}=\frac{1}{2} h \nu$ and $E{0}^{(0)}+E_{0}^{(1)}=$ $\frac{1}{2} h \nu+3 d h^{2} / 64 \pi^{4} \nu^{2} m^{2}$.

EXERCISE Consider a one-particle, one-dimensional system with $V=\infty$ for $x<0$ and for $x>l$, and $V=c x$ for $0 \leq x \leq l$, where $c$ is a constant. (a) Sketch $V$ for $c>0$. (b) Treat the system as a perturbed particle in a box and find $E^{(1)}$ for the state with quantum number $n$. Then use Eq. (3.88) to state why the answer you got is to be expected. (Partial Answer: (b) $\frac{1}{2} c l$.)

The First-Order Wave-Function Correction
For $m \neq n$, Eq. (9.21) is

\(
\begin{equation}
\left(E{m}^{(0)}-E{n}^{(0)}\right)\left\langle\psi{m}^{(0)} \mid \psi{n}^{(1)}\right\rangle=-\left\langle\psi{m}^{(0)}\right| \hat{H}^{\prime}\left|\psi{n}^{(0)}\right\rangle, \quad m \neq n \tag{9.24}
\end{equation}
\)

To find $\psi{n}^{(1)}$, we expand it in terms of the complete, orthonormal set of unperturbed eigenfunctions $\psi{m}^{(0)}$ of the Hermitian operator $\hat{H}^{0}$ :

\(
\begin{equation}
\psi{n}^{(1)}=\sum{m} a{m} \psi{m}^{(0)}, \quad \text { where } \quad a{m}=\left\langle\psi{m}^{(0)} \mid \psi_{n}^{(1)}\right\rangle \tag{9.25}
\end{equation}
\)

where Eq. (7.41) was used for the expansion coefficients $a{m}$. Use of $a{m}=\left\langle\psi{m}^{(0)} \mid \psi{n}^{(1)}\right\rangle$ in (9.24) gives

\(
\left(E{m}^{(0)}-E{n}^{(0)}\right) a{m}=-\left\langle\psi{m}^{(0)}\right| \hat{H}^{\prime}\left|\psi_{n}^{(0)}\right\rangle, \quad m \neq n
\)

By hypothesis, the level $E{n}^{(0)}$ is nondegenerate. Therefore $E{m}^{(0)} \neq E{n}^{(0)}$ for $m \neq n$, and we may divide by $\left(E{m}^{(0)}-E_{n}^{(0)}\right)$ to get

\(
\begin{equation}
a{m}=\frac{\left\langle\psi{m}^{(0)}\right| \hat{H}^{\prime}\left|\psi{n}^{(0)}\right\rangle}{E{n}^{(0)}-E_{m}^{(0)}}, \quad m \neq n \tag{9.26}
\end{equation}
\)

The coefficients $a{m}$ in the expansion (9.25) of $\psi{n}^{(1)}$ are given by (9.26) except for $a{n}$, the coefficient of $\psi{n}^{(0)}$. From the second equation in (9.25), $a{n}=\left\langle\psi{n}^{(0)} \mid \psi{n}^{(1)}\right\rangle$. Recall that the choice of intermediate normalization for $\psi{n}$ makes $\left\langle\psi{n}^{(0)} \mid \psi{n}^{(1)}\right\rangle=0$ [Eq. (9.16)]. Hence $a{n}=\left\langle\psi{n}^{(0)} \mid \psi_{n}^{(1)}\right\rangle=0$, and Eqs. (9.25) and (9.26) give the first-order correction to the wave function as

\(
\begin{equation}
\psi{n}^{(1)}=\sum{m \neq n} \frac{\left\langle\psi{m}^{(0)}\right| \hat{H}^{\prime}\left|\psi{n}^{(0)}\right\rangle}{E{n}^{(0)}-E{m}^{(0)}} \psi_{m}^{(0)} \tag{9.27}
\end{equation}
\)

The symbol $\sum_{m \neq n}$ means we sum over all the unperturbed states except state $n$.
Setting $\lambda=1$ in (9.13) and using just the first-order wave-function correction, we have as the approximation to the perturbed wave function

\(
\begin{equation}
\psi{n} \approx \psi{n}^{(0)}+\sum{m \neq n} \frac{\left\langle\psi{m}^{(0)}\right| \hat{H}^{\prime}\left|\psi{n}^{(0)}\right\rangle}{E{n}^{(0)}-E{m}^{(0)}} \psi{m}^{(0)} \tag{9.28}
\end{equation}
\)

For $\psi_{n}^{(2)}$ and the normalization of $\psi$, see Kemble, Chapter XI.

The Second-Order Energy Correction

Equating the coefficients of the $\lambda^{2}$ terms in (9.17), we get

\(
\begin{equation}
\hat{H}^{0} \psi{n}^{(2)}-E{n}^{(0)} \psi{n}^{(2)}=E{n}^{(2)} \psi{n}^{(0)}+E{n}^{(1)} \psi{n}^{(1)}-\hat{H}^{\prime} \psi{n}^{(1)} \tag{9.29}
\end{equation}
\)

Multiplication by $\psi_{m}^{(0) *}$ followed by integration over all space gives

\(
\begin{align}
&\left\langle\psi{m}^{(0)}\right| \hat{H}^{0}\left|\psi{n}^{(2)}\right\rangle-E{n}^{(0)}\left\langle\psi{m}^{(0)} \mid \psi{n}^{(2)}\right\rangle \
& \quad=E{n}^{(2)}\left\langle\psi{m}^{(0)} \mid \psi{n}^{(0)}\right\rangle+E{n}^{(1)}\left\langle\psi{m}^{(0)} \mid \psi{n}^{(1)}\right\rangle-\left\langle\psi{m}^{(0)}\right| \hat{H}^{\prime}\left|\psi_{n}^{(1)}\right\rangle \tag{9.30}
\end{align}
\)

The integral $\left\langle\psi{m}^{(0)}\right| \hat{H}^{0}\left|\psi{n}^{(2)}\right\rangle$ in this equation is the same as the integral in (9.20), except that $\psi{n}^{(1)}$ is replaced by $\psi{n}^{(2)}$. Replacement of $\psi{n}^{(1)}$ by $\psi{n}^{(2)}$ in (9.20) gives

\(
\begin{equation}
\left\langle\psi{m}^{(0)}\right| \hat{H}^{0}\left|\psi{n}^{(2)}\right\rangle=E{m}^{(0)}\left\langle\psi{m}^{(0)} \mid \psi_{n}^{(2)}\right\rangle \tag{9.31}
\end{equation}
\)

Use of (9.31) and orthonormality of the unperturbed functions in (9.30) gives

\(
\begin{equation}
\left(E{m}^{(0)}-E{n}^{(0)}\right)\left\langle\psi{m}^{(0)} \mid \psi{n}^{(2)}\right\rangle=E{n}^{(2)} \delta{m n}+E{n}^{(1)}\left\langle\psi{m}^{(0)} \mid \psi{n}^{(1)}\right\rangle-\left\langle\psi{m}^{(0)}\right| \hat{H}^{\prime}\left|\psi_{n}^{(1)}\right\rangle \tag{9.32}
\end{equation}
\)

For $m=n$, the left side of (9.32) is zero and we get

\(
\begin{align}
& E{n}^{(2)}=-E{n}^{(1)}\left\langle\psi{n}^{(0)} \mid \psi{n}^{(1)}\right\rangle+\left\langle\psi{n}^{(0)}\right| \hat{H}^{\prime}\left|\psi{n}^{(1)}\right\rangle \
& E{n}^{(2)}=\left\langle\psi{n}^{(0)}\right| \hat{H}^{\prime}\left|\psi_{n}^{(1)}\right\rangle \tag{9.33}
\end{align}
\)

since $\left\langle\psi{n}^{(0)} \mid \psi{n}^{(1)}\right\rangle=0$ [Eq. (9.16)]. Note from (9.33) that to find the second-order correction to the energy, we have to know only the first-order correction to the wave function. In fact, it can be shown that knowledge of $\psi{n}^{(1)}$ suffices to determine $E{n}^{(3)}$ also.

In general, it can be shown that if we know the corrections to the wave function through order $k$, then we can compute the corrections to the energy through order $2 k+1$ (see Bates, Vol. I, p. 184).

Substitution of (9.27) for $\psi_{n}^{(1)}$ into (9.33) gives

\(
\begin{equation}
E{n}^{(2)}=\sum{m \neq n} \frac{\left\langle\psi{m}^{(0)}\right| \hat{H}^{\prime}\left|\psi{n}^{(0)}\right\rangle}{E{n}^{(0)}-E{m}^{(0)}}\left\langle\psi{n}^{(0)}\right| \hat{H}^{\prime}\left|\psi{m}^{(0)}\right\rangle \tag{9.34}
\end{equation}
\)

since the expansion coefficients $a_{m}$ [Eq. (9.26)] are constants that can be taken outside the integral. Since $\hat{H}^{\prime}$ is Hermitian, we have

\(
\begin{aligned}
\left\langle\psi{m}^{(0)}\right| \hat{H}^{\prime}\left|\psi{n}^{(0)}\right\rangle\left\langle\psi{n}^{(0)}\right| \hat{H}^{\prime}\left|\psi{m}^{(0)}\right\rangle & =\left\langle\psi{m}^{(0)}\right| \hat{H}^{\prime}\left|\psi{n}^{(0)}\right\rangle\left\langle\psi{m}^{(0)}\right| \hat{H}^{\prime}\left|\psi{n}^{(0)}\right\rangle * \
& \left.=\left|\left\langle\psi{m}^{(0)}\right| \hat{H}^{\prime}\right| \psi{n}^{(0)}\right\rangle\left.\right|^{2}
\end{aligned}
\)

and (9.34) becomes

\(
\begin{equation}
E{n}^{(2)}=\sum{m \neq n} \frac{\left.\left|\left\langle\psi{m}^{(0)}\right| \hat{H}^{\prime}\right| \psi{n}^{(0)}\right\rangle\left.\right|^{2}}{E{n}^{(0)}-E{m}^{(0)}} \tag{9.35}
\end{equation}
\)

which is the desired expression for $E_{n}^{(2)}$ in terms of the unperturbed wave functions and energies.

Inclusion of $E_{n}^{(2)}$ in (9.14) with $\lambda=1$ gives the approximate energy of the perturbed state as

\(
\begin{equation}
E{n} \approx E{n}^{(0)}+H{n n}^{\prime}+\sum{m \neq n} \frac{\left|H{m n}^{\prime}\right|^{2}}{E{n}^{(0)}-E_{m}^{(0)}} \tag{9.36}
\end{equation}
\)

where the integrals are over the unperturbed normalized wave functions.
For formulas for higher-order energy corrections, see Bates, Volume I, pages 181-185. The form of perturbation theory developed in this section is called Rayleigh-Schrödinger perturbation theory; other approaches exist.

Discussion

Equation (9.28) shows that the effect of the perturbation on the wave function $\psi{n}^{(0)}$ is to "mix in" contributions from other states $\psi{m}^{(0)}, m \neq n$. Because of the factor $1 /\left(E{n}^{(0)}-E{m}^{(0)}\right)$, the most important contributions (aside from $\psi_{n}^{(0)}$ ) to the perturbed wave function come from states nearest in energy to state $n$.

To evaluate the first-order correction to the energy, we must evaluate only the single integral $H_{n n}^{\prime}$, whereas to evaluate the second-order energy correction, we must evaluate the matrix elements of $\hat{H}^{\prime}$ between the $n$th state and all other states $m$, and then perform the infinite sum in (9.35). In many cases the second-order energy correction cannot be evaluated exactly. It is even harder to deal with third-order and higher-order energy corrections.

The sums in (9.28) and (9.36) are sums over different states rather than sums over different energy values. If some of the energy levels (other than the $n$ th) are degenerate, we must include a term in the sums for each linearly independent wave function corresponding to the degenerate levels.

We have a sum over states in (9.28) and (9.36) because we require a complete set of functions for the expansion (9.25), and therefore we must include all linearly independent wave functions in the sum. If the unperturbed problem has continuum wave functions (for example, the hydrogen atom), we must also include an integration over the continuum functions, if we are to have a complete set. If $\psi_{E}^{(0)}$ denotes an unperturbed continuum wave function of energy $E^{(0)}$, then (9.27) and (9.35) become

\(
\begin{gather}
\psi{n}^{(1)}=\sum{m \neq n} \frac{H{m n}^{\prime}}{E{n}^{(0)}-E{m}^{(0)}} \psi{m}^{(0)}+\int \frac{H{E, n}^{\prime}}{E{n}^{(0)}-E^{(0)}} \psi{E}^{(0)} d E^{(0)} \
E{n}^{(2)}=\sum{m \neq n} \frac{\left|H{m n}^{\prime}\right|^{2}}{E{n}^{(0)}-E{m}^{(0)}}+\int \frac{\left|H{E, n}^{\prime}\right|^{2}}{E{n}^{(0)}-E^{(0)}} d E^{(0)} \tag{9.37}
\end{gather}
\)

where $H{E, n}^{\prime} \equiv\left\langle\psi{E}^{(0)}\right| \hat{H}^{\prime}\left|\psi{n}^{(0)}\right\rangle$. The integrals in these equations are over the range of continuum-state energies (for example, from zero to infinity for the hydrogen atom). The existence of continuum states in the unperturbed problem makes evaluation of $E{n}^{(2)}$ even harder.

Comparison of the Variation and Perturbation Methods

The perturbation method applies to all the bound states of a system. Although the variation theorem stated in Section 8.1 applies only to the lowest state of a given symmetry, we can use the linear variation method to treat excited bound states.

Perturbation calculations are often hard to do because of the need to evaluate the infinite sums over discrete states and integrals over continuum states that occur in the second-order and higher-order energy corrections.

In the perturbation method, one can calculate the energy much more accurately (to order $2 k+1$ ) than the wave function (to order $k$ ). The same situation holds in the variation method, where one can get a rather good energy with a rather inaccurate wave function. If one calculates properties other than the energy, the results will generally not be as reliable as the calculated energy.

The Variation-Perturbation Method

The variation-perturbation method allows one to accurately estimate $E^{(2)}$ and higher-order perturbation-theory energy corrections for the ground state of a system without evaluating the infinite sum in (9.36). The method is based on the inequality

\(
\begin{equation}
\langle u| \hat{H}^{0}-E{g}^{(0)}|u\rangle+\langle u| \hat{H}^{\prime}-E{g}^{(1)}\left|\psi{g}^{(0)}\right\rangle+\left\langle\psi{g}^{(0)}\right| \hat{H}^{\prime}-E{g}^{(1)}|u\rangle \geq E{g}^{(2)} \tag{9.38}
\end{equation}
\)

where $u$ is any well-behaved function that satisfies the boundary conditions and where the subscript $g$ refers to the ground state. For the proof of (9.38), see Hameka, Section 7-9. By taking $u$ to be a trial function with parameters that we vary to minimize the left side of (9.38), we can estimate $E{g}^{(2)}$. The function $u$ turns out to be an approximation to $\psi{g}^{(1)}$, the first-order correction to the ground-state wave function, and $u$ can then be used to estimate $E_{g}^{(3)}$ also. Similar variational integrals can be used to find higher-order corrections to the ground-state energy and wave function.