Comprehensive Study Notes

We now consider the perturbation treatment of an energy level whose degree of degeneracy is $d$. We have $d$ linearly independent unperturbed wave functions corresponding to the degenerate level. We shall use the labels $1,2, \ldots, d$ for the states of the degenerate level, without implying that these are necessarily the lowest-lying states. The unperturbed Schrödinger equation is

\(
\begin{equation}
\hat{H}^{0} \psi{n}^{(0)}=E{n}^{(0)} \psi_{n}^{(0)} \tag{9.67}
\end{equation}
\)

with

\(
\begin{equation}
E{1}^{(0)}=E{2}^{(0)}=\cdots=E_{d}^{(0)} \tag{9.68}
\end{equation}
\)

The perturbed problem is

\(
\begin{gather}
\hat{H} \psi{n}=E{n} \psi_{n} \tag{9.69}\
\hat{H}=\hat{H}^{0}+\lambda \hat{H}^{\prime} \tag{9.70}
\end{gather}
\)

As $\lambda$ goes to zero, the eigenvalues in (9.69) go to the eigenvalues in (9.67); we have $\lim {\lambda \rightarrow 0} E{n}=E_{n}^{(0)}$. Figure 9.2 shows this for a hypothetical system with six states and a threefold-degenerate unperturbed level. Note that the perturbation splits the degenerate energy level. In some cases the perturbation may have no effect on the degeneracy or may only partly remove the degeneracy.

As $\lambda \rightarrow 0$, the eigenfunctions satisfying (9.69) approach eigenfunctions satisfying (9.67). Does this mean that $\lim {\lambda \rightarrow 0} \psi{n}=\psi{n}^{(0)}$ ? Not necessarily. If $E{n}^{(0)}$ is nondegenerate, there is a unique normalized eigenfunction $\psi{n}^{(0)}$ of $\hat{H}^{0}$ with eigenvalue $E{n}^{(0)}$, and we can be sure that $\lim {\lambda \rightarrow 0} \psi{n}=\psi{n}^{(0)}$. However, if $E{n}^{(0)}$ is the eigenvalue of the $d$-fold degenerate level, then (Section 3.6) any linear combination

\(
\begin{equation}
c{1} \psi{1}^{(0)}+c{2} \psi{2}^{(0)}+\cdots+c{d} \psi{d}^{(0)} \tag{9.71}
\end{equation}
\)

is a solution of (9.67) with eigenvalue (9.68). The set of linearly independent normalized functions $\psi{1}^{(0)}, \psi{2}^{(0)}, \ldots, \psi_{d}^{(0)}$, which we use as eigenfunctions corresponding to the states

FIGURE 9.2 Effect of a perturbation on energy levels.
of the degenerate level, is not unique. Using (9.71), we can construct an infinite number of sets of $d$ linearly independent normalized eigenfunctions for the degenerate level. As far as the unperturbed problem is concerned, one such set is as good as another. For example, for the three degenerate $2 p$ states of the hydrogen atom, we can use the $2 p{1}, 2 p{0}$, and $2 p{-1}$ functions; the $2 p{x}, 2 p{y}$, and $2 p{z}$, functions; or some other set of three linearly independent functions constructed as linear combinations of the members of one of the preceding sets. For the perturbed eigenfunctions that correspond to the $d$-fold degenerate unperturbed level, all we can say is that as $\lambda$ approaches zero they each approach a linear combination of unperturbed eigenfunctions:

\(
\begin{equation}
\lim {\lambda \rightarrow 0} \psi{n}=\sum{i=1}^{d} c{i} \psi_{i}^{(0)}, \quad 1 \leq n \leq d \tag{9.72}
\end{equation}
\)

Our first task is thus to determine the correct zeroth-order wave functions (9.72) for the perturbation $\hat{H}^{\prime}$. Calling these correct zeroth-order functions $\phi_{n}^{(0)}$, we have

\(
\begin{equation}
\phi{n}^{(0)}=\lim {\lambda \rightarrow 0} \psi{n}=\sum{i=1}^{d} c{i} \psi{i}^{(0)}, \quad 1 \leq n \leq d \tag{9.73}
\end{equation}
\)

Each different function $\phi_{n}^{(0)}$ has a different set of coefficients in (9.73). The correct set of zeroth-order functions depends on what the perturbation $\hat{H}^{\prime}$ is.

The treatment of the $d$-fold degenerate level proceeds like the nondegenerate treatment of Section 9.2, except that instead of $\psi{n}^{(0)}$ we use $\phi{n}^{(0)}$. Instead of Eqs. (9.13) and (9.14), we have

\(
\begin{array}{ll}
\psi{n}=\phi{n}^{(0)}+\lambda \psi{n}^{(1)}+\lambda^{2} \psi{n}^{(2)}+\cdots, & n=1,2, \ldots, d \
E{n}=E{d}^{(0)}+\lambda E{n}^{(1)}+\lambda^{2} E{n}^{(2)}+\cdots, & n=1,2, \ldots, d \tag{9.75}
\end{array}
\)

where (9.68) was used. Substitution into $\hat{H} \psi{n}=E{n} \psi_{n}$ gives

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

Equating the coefficients of $\lambda^{0}$ in this equation, we get $\hat{H}^{0} \phi{n}^{(0)}=E{d}^{(0)} \phi{n}^{(0)}$. By the theorem of Section 3.6, each linear combination $\phi{n}^{(0)}(n=1,2, \ldots, d)$ is an eigenfunction of $\hat{H}^{0}$ with eigenvalue $E_{d}^{(0)}$, and this equation gives no new information.

Equating the coefficients of the $\lambda^{1}$ terms, we get

\(
\begin{gather}
\hat{H}^{0} \psi{n}^{(1)}+\hat{H}^{\prime} \phi{n}^{(0)}=E{d}^{(0)} \psi{n}^{(1)}+E{n}^{(1)} \phi{n}^{(0)} \
\hat{H}^{0} \psi{n}^{(1)}-E{d}^{(0)} \psi{n}^{(1)}=E{n}^{(1)} \phi{n}^{(0)}-\hat{H}^{\prime} \phi{n}^{(0)}, \quad n=1,2, \ldots, d \tag{9.76}
\end{gather}
\)

We now multiply $(9.76)$ by $\psi_{m}^{(0) *}$ and integrate over all space, where $m$ is one of the states corresponding to the $d$-fold degenerate unperturbed level under consideration; that is, $1 \leq m \leq d$. We get

\(
\begin{array}{r}
\left\langle\psi{m}^{(0)}\right| \hat{H}^{0}\left|\psi{n}^{(1)}\right\rangle-E{d}^{(0)}\left\langle\psi{m}^{(0)} \mid \psi{n}^{(1)}\right\rangle=E{n}^{(1)}\left\langle\psi{m}^{(0)} \mid \phi{n}^{(0)}\right\rangle-\left\langle\psi{m}^{(0)}\right| \hat{H}^{\prime}\left|\phi{n}^{(0)}\right\rangle, \
1 \leq m \leq d \tag{9.77}
\end{array}
\)

From Eq. (9.20), we have $\left\langle\psi{m}^{(0)}\right| \hat{H}^{0}\left|\psi{n}^{(1)}\right\rangle=E{m}^{(0)}\left\langle\psi{m}^{(0)} \mid \psi{n}^{(1)}\right\rangle$. From (9.68), $E{m}^{(0)}=E{d}^{(0)}$ for $1 \leq m \leq d$, so $\left\langle\psi{m}^{(0)}\right| \hat{H}^{0}\left|\psi{n}^{(1)}\right\rangle=E{d}^{(0)}\left\langle\psi{m}^{(0)} \mid \psi{n}^{(1)}\right\rangle$, and the left side of (9.77) equals zero. Equation (9.77) becomes

\(
\left\langle\psi{m}^{(0)}\right| \hat{H}^{\prime}\left|\phi{n}^{(0)}\right\rangle-E{n}^{(1)}\left\langle\psi{m}^{(0)} \mid \phi_{n}^{(0)}\right\rangle=0, \quad m=1,2, \ldots, d
\)

Substitution of the linear combination (9.73) for $\phi_{n}^{(0)}$ gives

\(
\begin{equation}
\sum{i=1}^{d} c{i}\left\langle\psi{m}^{(0)}\right| \hat{H}^{\prime}\left|\psi{i}^{(0)}\right\rangle-E{n}^{(1)} \sum{i=1}^{d} c{i}\left\langle\psi{m}^{(0)} \mid \psi_{i}^{(0)}\right\rangle=0 \tag{9.78}
\end{equation}
\)

The zeroth-order wave functions $\psi_{i}^{(0)}(i=1,2, \ldots, d)$ of the degenerate level can always be chosen to be orthonormal, and we shall assume this has been done:

\(
\begin{equation}
\left\langle\psi{m}^{(0)} \mid \psi{i}^{(0)}\right\rangle=\delta_{m i} \tag{9.79}
\end{equation}
\)

for $m$ and $i$ in the range 1 to $d$. Equation (9.78) becomes

\(
\begin{equation}
\sum{i=1}^{d}\left[\left\langle\psi{m}^{(0)}\right| \hat{H}^{\prime}\left|\psi{i}^{(0)}\right\rangle-E{n}^{(1)} \delta{m i}\right] c{i}=0, \quad m=1,2, \ldots, d \tag{9.80}
\end{equation}
\)

This is a set of $d$ linear, homogeneous equations in the $d$ unknowns $c{1}, c{2}, \ldots, c{d}$, which are the coefficients in the correct zeroth-order wave function $\phi{n}^{(0)}$ in (9.73). Writing out (9.80), we have

\(
\begin{align}
& \left(H{11}^{\prime}-E{n}^{(1)}\right) c{1}+H{12}^{\prime} c{2}+\cdots+H{1 d}^{\prime} c{d}=0 \
& H{21}^{\prime} c{1}+\left(H{22}^{\prime}-E{n}^{(1)}\right) c{2}+\cdots+H{2 d}^{\prime} c{d}=0 \
& \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \tag{9.81}\
& H{d 1}^{\prime} c{1}+H{d 2}^{\prime} c{2}+\cdots+\left(H{d d}^{\prime}-E{n}^{(1)}\right) c{d}=0 \
& \quad H{m i}^{\prime} \equiv\left\langle\psi{m}^{(0)}\right| \hat{H}^{\prime}\left|\psi{i}^{(0)}\right\rangle
\end{align}
\)

For this set of linear homogeneous equations to have a nontrivial solution, the determinant of the coefficients must vanish (Section 8.4):

\(
\begin{align}
& \operatorname{det}\left[\left\langle\psi{m}^{(0)}\right| \hat{H}^{\prime}\left|\psi{i}^{(0)}\right\rangle-E{n}^{(1)} \delta{m i}\right]=0 \tag{9.82}\
& \left|\begin{array}{cccc}
H{11}^{\prime}-E{n}^{(1)} & H{12}^{\prime} & \cdots & H{1 d}^{\prime} \
H{21}^{\prime} & H{22}^{\prime}-E{n}^{(1)} & \cdots & H{2 d}^{\prime} \
\vdots & \vdots & \ddots & \vdots \
H{d 1}^{\prime} & H{d 2}^{\prime} & \cdots & H{d d}^{\prime}-E{n}^{(1)}
\end{array}\right|=0 \tag{9.83}
\end{align}
\)

The secular equation (9.83) is an algebraic equation of degree $d$ in $E{n}^{(1)}$. It has $d$ roots, $E{1}^{(1)}, E{2}^{(1)}, \ldots, E{d}^{(1)}$, which are the first-order corrections to the energy of the $d$-fold degenerate unperturbed level. If the roots are all different, then the first-order perturbation correction has split the $d$-fold degenerate unperturbed level into $d$ different perturbed levels of energies (correct through first order):

\(
E{d}^{(0)}+E{1}^{(1)}, \quad E{d}^{(0)}+E{2}^{(1)}, \quad \ldots, \quad E{d}^{(0)}+E{d}^{(1)}
\)

If two or more roots of the secular equation are equal, the degeneracy is not completely removed in first order. In the rest of this section, we shall assume that all the roots of (9.83) are different.

Having found the $d$ first-order energy corrections, we go back to the set of equations (9.81) to find the unknowns $c_{i}$, which determine the correct zeroth-order wave functions. To find the correct zeroth-order function

\(
\begin{equation}
\phi{n}^{(0)}=c{1} \psi{1}^{(0)}+c{2} \psi{2}^{(0)}+\cdots+c{d} \psi_{d}^{(0)} \tag{9.84}
\end{equation}
\)

corresponding to the root $E{n}^{(1)}$, we solve (9.81) for $c{2}, c{3}, \ldots, c{d}$ in terms of $c{1}$ and then find $c{1}$ by normalization. Use of (9.79) in $\left\langle\phi{n}^{(0)} \mid \phi{n}^{(0)}\right\rangle=1$ gives (Prob. 9.21)

\(
\begin{equation}
\sum{k=1}^{d}\left|c{k}\right|^{2}=1 \tag{9.85}
\end{equation}
\)

For each root $E{n}^{(1)}, n=1,2, \ldots, d$, we have a different set of coefficients $c{1}, c{2}, \ldots, c{d}$, giving a different correct zeroth-order wave function.

In the next section, we shall show that

\(
\begin{equation}
E{n}^{(1)}=\left\langle\phi{n}^{(0)}\right| \hat{H}^{\prime}\left|\phi_{n}^{(0)}\right\rangle, \quad n=1,2, \ldots, d \tag{9.86}
\end{equation}
\)

which is similar to the nondegenerate-case formula (9.22), except that the correct zerothorder functions have to be used.

Using procedures similar to those for the nondegenerate case, one can now find the first-order corrections to the correct zeroth-order wave functions and the secondorder energy corrections. For the results, see Bates, Volume I, pages 197-198; Hameka, pages 230-231.

As an example, consider the effect of a perturbation $\hat{H}^{\prime}$ on the lowest degenerate energy level of a particle in a cubic box. We have three states corresponding to this level: $\psi{211}^{(0)}, \psi{121}^{(0)}$, and $\psi_{112}^{(0)}$. These unperturbed wave functions are orthonormal, and the secular equation (9.83) is

\(
\left|\begin{array}{ccc}
\langle 211| \hat{H}^{\prime}|211\rangle-E{n}^{(1)} & \langle 211| \hat{H}^{\prime}|121\rangle & \langle 211| \hat{H}^{\prime}|112\rangle \tag{9.87}\
\langle 121| \hat{H}^{\prime}|211\rangle & \langle 121| \hat{H}^{\prime}|121\rangle-E{n}^{(1)} & \langle 121| \hat{H}^{\prime}|112\rangle \
\langle 112| \hat{H}^{\prime}|211\rangle & \langle 112| \hat{H}^{\prime}|121\rangle & \langle 112| \hat{H}^{\prime}|112\rangle-E_{n}^{(1)}
\end{array}\right|=0
\)

Solving this equation, we find the first-order energy corrections: $E{1}^{(1)}, E{2}^{(1)}, E{3}^{(1)}$. The triply degenerate unperturbed level is split into three levels of energies (through first order): $\left(6 h^{2} / 8 m a^{2}\right)+E{1}^{(1)},\left(6 h^{2} / 8 m a^{2}\right)+E{2}^{(1)},\left(6 h^{2} / 8 m a^{2}\right)+E{3}^{(1)}$. Using each of the roots $E{1}^{(1)}, E{2}^{(1)}, E_{3}^{(1)}$, we get a different set of simultaneous equations (9.81). Solving each set, we find three sets of coefficients, which determine the three correct zeroth-order wave functions.

If you are familiar with matrix algebra, note that solving (9.83) and (9.81) amounts to finding the eigenvalues and eigenvectors of the matrix whose elements are $\left\langle\psi{m}^{(0)}\right| \hat{H}^{\prime}\left|\psi{i}^{(0)}\right\rangle$.