The secular equation (9.83) is easier to solve if some of the off-diagonal elements of the secular determinant are zero. In the most favorable case, all the off-diagonal elements are zero, and
\(
\begin{gather}
\left|\begin{array}{cccc}
H{11}^{\prime}-E{n}^{(1)} & 0 & \cdots & 0 \
0 & H{22}^{\prime}-E{n}^{(1)} & \cdots & 0 \
\vdots & \vdots & \ddots & \vdots \
0 & 0 & \cdots & H{d d}^{\prime}-E{n}^{(1)}
\end{array}\right|=0 \tag{9.88}\
\left(H{11}^{\prime}-E{n}^{(1)}\right)\left(H{22}^{\prime}-E{n}^{(1)}\right) \cdots\left(H{d d}^{\prime}-E{n}^{(1)}\right)=0 \
E{1}^{(1)}=H{11}^{\prime}, \quad E{2}^{(1)}=H{22}^{\prime}, \quad \cdots, \quad E{d}^{(1)}=H{d d}^{\prime} \tag{9.89}
\end{gather}
\)
Now we want to find the correct zeroth-order wave functions. We shall assume that the roots (9.89) are all different. For the root $E{n}^{(1)}=H{11}^{\prime}$, the system of equations (9.81) is
\(
\begin{aligned}
& 0=0 \
& \left(H{22}^{\prime}-H{11}^{\prime}\right) c{2}=0 \
& \left(H{d d}^{\prime}-H{11}^{\prime}\right) c{d}=0
\end{aligned}
\)
Since we are assuming unequal roots, the quantities $H{22}^{\prime}-H{11}^{\prime}, \ldots, H{d d}^{\prime}-H{11}^{\prime}$ are all nonzero. Therefore, $c{2}=0, c{3}=0, \ldots, c{d}=0$. The normalization condition (9.85) gives $c{1}=1$. The correct zeroth-order wave function corresponding to the first-order perturbation energy correction $H{11}^{\prime}$ is then [Eq. (9.73)] $\phi{1}^{(0)}=\psi{1}^{(0)}$. For the root $H{22}^{\prime}$, the same reasoning gives $\phi{2}^{(0)}=\psi{2}^{(0)}$. Using each of the remaining roots, we find similarly: $\phi{3}^{(0)}=\psi{3}^{(0)}, \ldots, \phi{d}^{(0)}=\psi{d}^{(0)}$.
When the secular determinant is in diagonal form, the initially assumed wave functions $\psi{1}^{(0)}, \psi{2}^{(0)}, \ldots, \psi_{n}^{(0)}$ are the correct zeroth-order wave functions for the perturbation $\hat{H}^{\prime}$.
The converse is also true. If the initially assumed functions are the correct zerothorder functions, then the secular determinant is in diagonal form. This is seen as follows. From $\phi{1}^{(0)}=\psi{1}^{(0)}$ we know that the coefficients in the expansion $\phi{1}^{(0)}=\sum{i=1}^{d} c{i} \psi{i}^{(0)}$ are $c{1}=1, c{2}=c_{3}=\cdots=0$, so for $n=1$ the set of simultaneous equations (9.81) becomes
\(
H{11}^{\prime}-E{1}^{(1)}=0, \quad H{21}^{\prime}=0, \quad \ldots, \quad H{d 1}^{\prime}=0
\)
Applying the same reasoning to the remaining functions $\phi{n}^{(0)}$, we conclude that $H{m i}^{\prime}=0$ for $i \neq m$. Hence, use of the correct zeroth-order functions makes the secular determinant diagonal. Note also that the first-order corrections to the energy can be found by averaging the perturbation over the correct zeroth-order wave functions:
\(
\begin{equation}
E{n}^{(1)}=H{n n}^{\prime}=\left\langle\phi{n}^{(0)}\right| \hat{H}^{\prime}\left|\phi{n}^{(0)}\right\rangle \tag{9.90}
\end{equation}
\)
a result mentioned in Eq. (9.86).
Often, instead of being in diagonal form, the secular determinant is in block-diagonal form. For example, we might have
\(
\left|\begin{array}{cccc}
H{11}^{\prime}-E{n}^{(1)} & H{12}^{\prime} & 0 & 0 \tag{9.91}\
H{21}^{\prime} & H{22}^{\prime}-E{n}^{(1)} & 0 & 0 \
0 & 0 & H{33}^{\prime}-E{n}^{(1)} & H{34}^{\prime} \
0 & 0 & H{43}^{\prime} & H{44}^{\prime}-E{n}^{(1)}
\end{array}\right|=0
\)
The secular determinant in (9.91) has the same form as the secular determinant in the linear-variation secular equation (8.65) with $S{i j}=\delta{i j}$. By the same reasoning used to show that two of the variation functions are linear combinations of $f{1}$ and $f{2}$ and two are linear combinations of $f{3}$ and $f{4}$ [Eq. (8.69)], it follows that two of the correct zeroth-order wave functions are linear combinations of $\psi{1}^{(0)}$ and $\psi{2}^{(0)}$ and two are linear combinations of $\psi{3}^{(0)}$ and $\psi{4}^{(0)}$ :
\(
\begin{array}{ll}
\phi{1}^{(0)}=c{1} \psi{1}^{(0)}+c{2} \psi{2}^{(0)}, & \phi{2}^{(0)}=c{1}^{\prime} \psi{1}^{(0)}+c{2}^{\prime} \psi{2}^{(0)} \
\phi{3}^{(0)}=c{3} \psi{3}^{(0)}+c{4} \psi{4}^{(0)}, & \phi{4}^{(0)}=c{3}^{\prime} \psi{3}^{(0)}+c{4}^{\prime} \psi{4}^{(0)}
\end{array}
\)
where primes were used to distinguish different coefficients.
When the secular determinant of degenerate perturbation theory is in block-diagonal form, the secular equation breaks up into two or more smaller secular equations, and the set of simultaneous equations (9.81) for the coefficients $c_{i}$ breaks up into two or more smaller sets of simultaneous equations.
Conversely, if we have, say, a fourfold-degenerate unperturbed level, and we happen to know that $\phi{1}^{(0)}$ and $\phi{2}^{(0)}$ are each linear combinations of $\psi{1}^{(0)}$ and $\psi{2}^{(0)}$ only, while $\phi{3}^{(0)}$ and $\phi{4}^{(0)}$ are each linear combinations of $\psi{3}^{(0)}$ and $\psi{4}^{(0)}$ only, we deal with two second-order secular determinants rather than a fourth-order secular determinant.
How can we choose the right zeroth-order wave functions in advance and thereby simplify the secular equation? Suppose there is an operator $\hat{A}$ that commutes with both $\hat{H}^{0}$ and $\hat{H}^{\prime}$. Then we can choose the unperturbed functions to be eigenfunctions of $\hat{A}$. Because $\hat{A}$ commutes with $\hat{H}^{\prime}$, this choice of unperturbed functions will make the integrals $H{i j}^{\prime}$ vanish if $\psi{i}^{(0)}$ and $\psi{j}^{(0)}$ belong to different eigenvalues of $\hat{A}$ [see Eq. (7.50)]. Thus, if the eigenvalues of $\hat{A}$ for $\psi{1}^{(0)}, \psi{2}^{(0)}, \ldots, \psi{d}^{(0)}$ are all different, the secular determinant will be in diagonal form, and we will have the right zeroth-order wave functions. If some of the eigenvalues of $\hat{A}$ are the same, we get block-diagonal rather than diagonal form. In general, the correct zeroth-order functions will be linear combinations of those unperturbed functions that have the same eigenvalue of $\hat{A}$. (This is to be expected since $\hat{A}$ commutes with $\hat{H}=\hat{H}^{0}+\hat{H}^{\prime}$, so the perturbed eigenfunctions of $\hat{H}$ can be chosen to be eigenfunctions of $\hat{A}$.) For an example, see Prob. 9.23.