Section 9.3 applied perturbation theory to the ground state of the helium atom. We now treat the lowest excited states of He . The unperturbed energies are given by (9.48). The lowest unperturbed excited states have $n{1}=1, n{2}=2$ or $n{1}=2, n{2}=1$, and substitution in (9.48) gives
$E^{(0)}=-\frac{5 Z^{2}}{8}\left(\frac{e^{2}}{4 \pi \varepsilon{0} a{0}}\right)=-\frac{20}{8} 2\left(\frac{e^{2}}{8 \pi \varepsilon{0} a{0}}\right)=-5(13.606 \mathrm{eV})=-68.03 \mathrm{eV}$
Recall that the $n=2$ level of a hydrogenlike atom is fourfold degenerate, the $2 s$ and three $2 p$ states all having the same energy. The first excited unperturbed energy level of He is eightfold degenerate. The eight unperturbed wave functions are [Eq. (9.44)]
\(
\begin{align}
& \psi{1}^{(0)}=1 s(1) 2 s(2), \quad \psi{2}^{(0)}=2 s(1) 1 s(2), \quad \psi{3}^{(0)}=1 s(1) 2 p{x}(2), \quad \psi{4}^{(0)}=2 p{x}(1) 1 \mathrm{~s}(2) \
& \psi{5}^{(0)}=1 \mathrm{~s}(1) 2 p{y}(2), \quad \psi{6}^{(0)}=2 p{y}(1) 1 s(2), \quad \psi{7}^{(0)}=1 \mathrm{~s}(1) 2 p{z}(2), \quad \psi{8}^{(0)}=2 p{z}(1) 1 s(2) \tag{9.93}
\end{align}
\)
where $1 s(1) 2 s(2)$ signifies the product of a hydrogenlike $1 s$ function for electron 1 and a hydrogenlike $2 s$ function for electron 2 . The explicit form of $\psi_{8}^{(0)}$, for example, is (Table 6.2)
\(
\psi{8}^{(0)}=\frac{1}{4(2 \pi)^{1 / 2}}\left(\frac{Z}{a{0}}\right)^{5 / 2} r{1} e^{-Z r{1} / 2 a{0}} \cos \theta{1} \cdot \frac{1}{\pi^{1 / 2}}\left(\frac{Z}{a{0}}\right)^{3 / 2} e^{-Z r{2} / a_{0}}
\)
We chose to use the real $2 p$ hydrogenlike orbitals, rather than the complex ones.
Since the unperturbed level is degenerate, we must solve a secular equation. The secular equation (9.83) assumes that the functions $\psi{1}^{(0)}, \psi{2}^{(0)}, \ldots, \psi_{8}^{(0)}$ are orthonormal. This condition is met. For example,
\(
\begin{aligned}
\int \psi{1}^{(0)} * \psi{1}^{(0)} d \tau & =\iint 1 s(1) 2 s(2) 1 s(1) 2 s(2) d \tau{1} d \tau{2} \
& =\int|1 s(1)|^{2} d \tau{1} \int|2 s(2)|^{2} d \tau{2}=1 \cdot 1=1 \
\int \psi{3}^{(0) *} \psi{5}^{(0)} d \tau & =\int|1 s(1)|^{2} d \tau{1} \int 2 p{x}(2) * 2 p{y}(2) d \tau{2}=1 \cdot 0=0
\end{aligned}
\)
where the orthonormality of the hydrogenlike orbitals has been used.
The secular determinant contains $8^{2}=64$ elements. The operator $\hat{H}^{\prime}$ is Hermitian, and $H{i j}^{\prime}=\left(H{j i}^{\prime}\right)^{}$. Also, since $\hat{H}^{\prime}$ and $\psi{1}^{(0)}, \ldots, \psi{8}^{(0)}$ are all real, we have $\left(H_{j i}^{\prime}\right)^{}=H{j i}^{\prime}$, so $H{i j}^{\prime}=H_{j i}^{\prime}$. The secular determinant is symmetric about the principal diagonal. This cuts the labor of evaluating integrals almost in half.
By using parity considerations, we can show that most of the integrals $H{i j}^{\prime}$ are zero. First consider $H{13}^{\prime}$ :
$H{13}^{\prime}=\int{-\infty}^{\infty} \int{-\infty}^{\infty} \int{-\infty}^{\infty} \int{-\infty}^{\infty} \int{-\infty}^{\infty} \int{-\infty}^{\infty} 1 s(1) 2 s(2) \frac{e^{2}}{4 \pi \varepsilon{0} r{12}} 1 s(1) 2 p{x}(2) d x{1} d y{1} d z{1} d x{2} d y{2} d z{2}$
An $s$ hydrogenlike function depends only on $r=\left(x^{2}+y^{2}+z^{2}\right)^{1 / 2}$ and is therefore an even function. The $2 p{x}(2)$ function is an odd function of $x{2}$ [Eq. (6.119)]. $r{12}$ is given by (9.66), and if we invert all six coordinates, $r{12}$ is unchanged:
\(
r{12} \rightarrow\left[\left(-x{1}+x{2}\right)^{2}+\left(-y{1}+y{2}\right)^{2}+\left(-z{1}+z{2}\right)^{2}\right]^{1 / 2}=r{12}
\)
Hence, on inverting all six coordinates, the integrand of $H{13}^{\prime}$ goes into minus itself. Therefore [Eq. (7.64)], $H{13}^{\prime}=0$. The same reasoning gives $H{14}^{\prime}=H{15}^{\prime}=H{16}^{\prime}=H{17}^{\prime}=$ $H{18}^{\prime}=0$ and $H{23}^{\prime}=H{24}^{\prime}=H{25}^{\prime}=H{26}^{\prime}=H{27}^{\prime}=H{28}^{\prime}=0$. Now consider $H{35}^{\prime}$ :
\(
H{35}^{\prime}=\int{-\infty}^{\infty} \cdots \int{-\infty}^{\infty} 1 s(1) 2 p{x}(2) \frac{e^{2}}{4 \pi \varepsilon{0} r{12}} 1 s(1) 2 p{y}(2) d x{1} \cdots d z_{2}
\)
Suppose we invert the $x$ coordinates: $x{1} \rightarrow-x{1}$ and $x{2} \rightarrow-x{2}$. This transformation will leave $r{12}$ unchanged. The functions $1 s(1)$ and $2 p{y}(2)$ will be unaffected. However, $2 p{x}(2)$ will go over to minus itself, so the net effect will be to change the integrand of $H{35}^{\prime}$ into minus itself. Hence (Prob. 7.30), $H{35}^{\prime}=0$. Likewise, $H{36}^{\prime}=H{37}^{\prime}=H{38}^{\prime}=0$ and $H{45}^{\prime}=H{46}^{\prime}=H{47}^{\prime}=H{48}^{\prime}=0$. By considering the transformation $y{1} \rightarrow-y{1}, y{2} \rightarrow-y{2}$, we see that $H{57}^{\prime}=H{58}^{\prime}=H{67}^{\prime}=H{68}^{\prime}=0$. The secular equation is thus
\(
\left.\begin{array}{|cccccccc}
b{11} & H{12}^{\prime} & 0 & 0 & 0 & 0 & 0 & 0 \
H{12}^{\prime} & b{22} & 0 & 0 & 0 & 0 & 0 & 0 \
0 & 0 & b{33} & H{34}^{\prime} & 0 & 0 & 0 & 0 \
0 & 0 & H{34}^{\prime} & b{44} & 0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 & b{55} & H{56}^{\prime} & 0 & 0 \
0 & 0 & 0 & 0 & H{56}^{\prime} & b{66} & 0 & 0 \
0 & 0 & 0 & 0 & 0 & 0 & b{77} & H{78}^{\prime} \
0 & 0 & 0 & 0 & 0 & 0 & H{78}^{\prime} & b{88}
\end{array} \right\rvert\,=0
\)
The secular determinant is in block-diagonal form and factors into four determinants, each of second order. We conclude that the correct zeroth-order functions have the form
\(
\begin{array}{ll}
\phi{1}^{(0)}=c{1} \psi{1}^{(0)}+c{2} \psi{2}^{(0)}, & \phi{2}^{(0)}=\bar{c}{1} \psi{1}^{(0)}+\bar{c}{2} \psi{2}^{(0)} \
\phi{3}^{(0)}=c{3} \psi{3}^{(0)}+c{4} \psi{4}^{(0)}, & \phi{4}^{(0)}=\bar{c}{3} \psi{3}^{(0)}+\bar{c}{4} \psi{4}^{(0)} \
\phi{5}^{(0)}=c{5} \psi{5}^{(0)}+c{6} \psi{6}^{(0)}, & \phi{6}^{(0)}=\bar{c}{5} \psi{5}^{(0)}+\bar{c}{6} \psi{6}^{(0)} \tag{9.94}\
\phi{7}^{(0)}=c{7} \psi{7}^{(0)}+c{8} \psi{8}^{(0)}, & \phi{8}^{(0)}=\bar{c}{7} \psi{7}^{(0)}+\bar{c}{8} \psi{8}^{(0)}
\end{array}
\)
where the unbarred coefficients correspond to one root of each second-order determinant and the barred coefficients correspond to the second root.
The first determinant is
\(
\left|\begin{array}{cc}
H{11}^{\prime}-E^{(1)} & H{12}^{\prime} \tag{9.95}\
H{12}^{\prime} & H{22}^{\prime}-E^{(1)}
\end{array}\right|=0
\)
We have
\(
\begin{aligned}
H{11}^{\prime} & =\int{-\infty}^{\infty} \cdots \int{-\infty}^{\infty} 1 s(1) 2 s(2) \frac{e^{2}}{4 \pi \varepsilon{0} r{12}} 1 s(1) 2 s(2) d x{1} \cdots d z{2} \
H{11}^{\prime} & =\iint[1 s(1)]^{2}[2 s(2)]^{2} \frac{e^{2}}{4 \pi \varepsilon{0} r{12}} d \tau{1} d \tau{2} \
H{22}^{\prime} & =\iint[1 s(2)]^{2}[2 s(1)]^{2} \frac{e^{2}}{4 \pi \varepsilon{0} r{12}} d \tau{1} d \tau_{2}
\end{aligned}
\)
The integration variables are dummy variables and may be given any symbols whatever. Let us relabel the integration variables in $H{22}^{\prime}$ as follows: We interchange $x{1}$ and $x{2}$, interchange $y{1}$ and $y{2}$, and interchange $z{1}$ and $z{2}$. This relabeling leaves $r{12}$ [Eq. (9.66)] unchanged, so
\(
\begin{equation}
H{22}^{\prime}=\iint[1 s(1)]^{2}[2 s(2)]^{2} \frac{e^{2}}{4 \pi \varepsilon{0} r{12}} d \tau{2} d \tau{1}=H{11}^{\prime} \tag{9.96}
\end{equation}
\)
The same argument shows that $H{33}^{\prime}=H{44}^{\prime}, H{55}^{\prime}=H{66}^{\prime}$, and $H{77}^{\prime}=H{88}^{\prime}$.
We denote $H{11}^{\prime}$ by the symbol $J{1 s 2 s}$ :
\(
\begin{equation}
H{11}^{\prime}=J{1 s 2 s}=\iint[1 s(1)]^{2}[2 s(2)]^{2} \frac{e^{2}}{4 \pi \varepsilon{0} r{12}} d \tau{1} d \tau{2} \tag{9.97}
\end{equation}
\)
This is an example of a Coulomb integral, the name arising because $J_{1 s 2 s}$ is equal to the electrostatic energy of repulsion between an electron with probability density function $[1 s]^{2}$ and an electron with probability density function $[2 s]^{2}$.
The integral $H{12}^{\prime}$ is denoted by $K{1 s 2 s}$ :
\(
\begin{equation}
H{12}^{\prime}=K{1 s 2 s}=\iint 1 s(1) 2 s(2) \frac{e^{2}}{4 \pi \varepsilon{0} r{12}} 2 s(1) 1 s(2) d \tau{1} d \tau{2} \tag{9.98}
\end{equation}
\)
This is an exchange integral: The functions on the left and right of $e^{2} / 4 \pi \varepsilon{0} r{12}$ differ from each other by an exchange of electrons 1 and 2 .
The general definitions of the Coulomb integral $J{m n}$ and the exchange integral $K{m n}$ are
\(
\begin{align}
J{m n} & \equiv\left\langle f{m}(1) f{n}(2)\right| e^{2} / 4 \pi \varepsilon{0} r{12}\left|f{m}(1) f{n}(2)\right\rangle \tag{9.99}\
K{m n} & \equiv\left\langle f{m}(1) f{n}(2)\right| e^{2} / 4 \pi \varepsilon{0} r{12}\left|f{n}(1) f{m}(2)\right\rangle \tag{9.100}
\end{align}
\)
where the integrals go over the full range of the spatial coordinates of electrons 1 and 2 , and $f{m}$ and $f{n}$ are spatial orbitals.
Substitution of (9.96) to (9.98) into (9.95) gives
\(
\begin{align}
& \left|\begin{array}{cc}
J{1 s 2 s}-E^{(1)} & K{1 s 2 s} \
K{1 s 2 s} & J{1 s 2 s}-E^{(1)}
\end{array}\right|=0 \tag{9.101}\
& \left(J{1 s 2 s}-E^{(1)}\right)^{2}=\left(K{1 s 2 s}\right)^{2} \
& E{1}^{(1)}=J{1 s 2 s}-K{1 s 2 s}, \quad E{2}^{(1)}=J{1 s 2 s}+K{1 s 2 s} \tag{9.102}
\end{align}
\)
We now find the coefficients of the correct zeroth-order wave functions that correspond to these two roots. Use of $E_{1}^{(1)}$ in (9.81) gives
\(
\begin{aligned}
& K{1 s 2 s} c{1}+K{1 s 2 s} c{2}=0 \
& K{1 s 2 s} c{1}+K{1 s 2 s} c{2}=0
\end{aligned}
\)
Hence $c{2}=-c{1}$. Normalization gives
\(
\begin{gathered}
\left\langle\phi{1}^{(0)} \mid \phi{1}^{(0)}\right\rangle=\left\langle c{1} \psi{1}^{(0)}-c{1} \psi{2}^{(0)} \mid c{1} \psi{1}^{(0)}-c{1} \psi{2}^{(0)}\right\rangle=\left|c{1}\right|^{2}+\left|c{1}\right|^{2}=1 \
c_{1}=2^{-1 / 2}
\end{gathered}
\)
where the orthonormality of $\psi{1}^{(0)}$ and $\psi{2}^{(0)}$ was used. The zeroth-order wave function corresponding to $E_{1}^{(1)}$ is then
\(
\begin{equation}
\phi{1}^{(0)}=2^{-1 / 2}\left(\psi{1}^{(0)}-\psi_{2}^{(0)}\right)=2^{-1 / 2}[1 s(1) 2 s(2)-2 s(1) 1 s(2)] \tag{9.103}
\end{equation}
\)
Similarly, one finds the function corresponding to $E_{2}^{(1)}$ to be
\(
\begin{equation}
\phi{2}^{(0)}=2^{-1 / 2}\left(\psi{1}^{(0)}+\psi_{2}^{(0)}\right)=2^{-1 / 2}[1 s(1) 2 s(2)+2 s(1) 1 s(2)] \tag{9.104}
\end{equation}
\)
We have three more second-order determinants to deal with:
\(
\begin{align}
& \left|\begin{array}{cc}
H{33}^{\prime}-E^{(1)} & H{34}^{\prime} \
H{34}^{\prime} & H{33}^{\prime}-E^{(1)}
\end{array}\right|=0 \tag{9.105}\
& \left|\begin{array}{cc}
H{55}^{\prime}-E^{(1)} & H{56}^{\prime} \
H{56}^{\prime} & H{55}^{\prime}-E^{(1)}
\end{array}\right|=0 \tag{9.106}\
& \left|\begin{array}{cc}
H{77}^{\prime}-E^{(1)} & H{78}^{\prime} \
H{78}^{\prime} & H{77}^{\prime}-E^{(1)}
\end{array}\right|=0 \tag{9.107}
\end{align}
\)
Consider $H{33}^{\prime}$ and $H{55}^{\prime}$ :
\(
\begin{aligned}
& H{33}^{\prime}=\int{-\infty}^{\infty} \cdots \int{-\infty}^{\infty} 1 s(1) 2 p{x}(2) \frac{e^{2}}{4 \pi \varepsilon{0} r{12}} 1 s(1) 2 p{x}(2) d x{1} \cdots d z{2} \
& H{55}^{\prime}=\int{-\infty}^{\infty} \cdots \int{-\infty}^{\infty} 1 s(1) 2 p{y}(2) \frac{e^{2}}{4 \pi \varepsilon{0} r{12}} 1 s(1) 2 p{y}(2) d x{1} \cdots d z{2}
\end{aligned}
\)
These two integrals are equal-the only difference between them involves replacement of $2 p{x}(2)$ by $2 p{y}(2)$-and these two orbitals differ only in their orientation in space. More formally, if we relabel the dummy integration variables in $H{33}^{\prime}$ according to the scheme $x{2} \rightarrow y{2}, y{2} \rightarrow x{2}, x{1} \rightarrow y{1}, y{1} \rightarrow x{1}$, then $r{12}$ is unaffected and $H{33}^{\prime}$ is transformed to $H{55}^{\prime}$. Similar reasoning shows $H{77}^{\prime}=H{33}^{\prime}$. Introducing the symbol $J_{1 s 2 p}$ for these Coulomb integrals, we have
\(
H{33}^{\prime}=H{55}^{\prime}=H{77}^{\prime}=J{1 s 2 p}=\iint 1 s(1) 2 p{z}(2) \frac{e^{2}}{4 \pi \varepsilon{0} r{12}} 1 s(1) 2 p{z}(2) d \tau{1} d \tau{2}
\)
Also, the exchange integrals involving the $2 p$ orbitals are equal:
\(
H{34}^{\prime}=H{56}^{\prime}=H{78}^{\prime}=K{1 s 2 p}=\iint 1 s(1) 2 p{z}(2) \frac{e^{2}}{4 \pi \varepsilon{0} r{12}} 2 p{z}(1) 1 s(2) d \tau{1} d \tau{2}
\)
The three determinants $(9.105)$ to $(9.107)$ are thus identical and have the form
\(
\left|\begin{array}{cc}
J{1 s 2 p}-E^{(1)} & K{1 s 2 p} \
K{1 s 2 p} & J{1 s 2 p}-E^{(1)}
\end{array}\right|=0
\)
The determinant is similar to (9.101), and by analogy with (9.102)-(9.104), we get
\(
\begin{gather}
E{3}^{(1)}=E{5}^{(1)}=E{7}^{(1)}=J{1 s 2 p}-K{1 s 2 p} \tag{9.108}\
E{4}^{(1)}=E{6}^{(1)}=E{8}^{(1)}=J{1 s 2 p}+K{1 s 2 p} \tag{9.109}\
\phi{3}^{(0)}=2^{-1 / 2}\left[1 s(1) 2 p{x}(2)-1 s(2) 2 p{x}(1)\right] \
\phi{4}^{(0)}=2^{-1 / 2}\left[1 s(1) 2 p{x}(2)+1 s(2) 2 p{x}(1)\right] \
\phi{5}^{(0)}=2^{-1 / 2}\left[1 s(1) 2 p{y}(2)-1 s(2) 2 p{y}(1)\right] \tag{9.110}\
\phi{6}^{(0)}=2^{-1 / 2}\left[1 s(1) 2 p{y}(2)+1 s(2) 2 p{y}(1)\right] \
\phi{7}^{(0)}=2^{-1 / 2}\left[1 s(1) 2 p{z}(2)-1 s(2) 2 p{z}(1)\right] \
\phi{8}^{(0)}=2^{-1 / 2}\left[1 s(1) 2 p{z}(2)+1 s(2) 2 p{z}(1)\right]
\end{gather}
\)
The electrostatic repulsion $e^{2} / 4 \pi \varepsilon{0} r{12}$ between the electrons has partly removed the degeneracy. The hypothetical eightfold-degenerate unperturbed level has been split into two nondegenerate levels associated with the configuration $1 s 2 s$ and two triply degenerate levels associated with the configuration $1 s 2 p$. It might be thought that higher-order energy corrections would further resolve the degeneracy. Actually, application of an external magnetic field is required to completely remove the degeneracy. Because the $e^{2} / 4 \pi \varepsilon{0} r{12}$ perturbation has not completely removed the degeneracy, any normalized linear combinations of $\phi{3}^{(0)}$, $\phi{5}^{(0)}$, and $\phi{7}^{(0)}$ and of $\phi{4}^{(0)}, \phi{6}^{(0)}$, and $\phi{8}^{(0)}$ can serve as correct zeroth-order wave functions.
To evaluate the Coulomb and exchange integrals in $E^{(1)}$ in (9.102) and (9.108), one uses the $1 / r_{12}$ expansion given in Prob. 9.14. The results are
\(
\begin{align}
J{1 s 2 s} & =\left(\frac{17}{81}\right) \frac{Z e^{2}}{4 \pi \varepsilon{0} a_{0}}=11.42 \mathrm{eV},
\end{align} \quad J{1 s 2 p}=\left(\frac{59}{243}\right) \frac{Z e^{2}}{4 \pi \varepsilon{0} a{0}}=13.21 \mathrm{eV}, ~\left(\frac{16}{729}\right) \frac{Z e^{2}}{4 \pi \varepsilon{0} a{0}}=1.19 \mathrm{eV}, \quad K{1 s 2 p}=\left(\frac{112}{6561}\right) \frac{Z e^{2}}{4 \pi \varepsilon{0} a{0}}=0.93 \mathrm{eV}
\)
where we used $Z=2$ and $e^{2} / 8 \pi \varepsilon{0} a{0}=13.606 \mathrm{eV}$. Recalling that $E^{(0)}=-68.03 \mathrm{eV}$ [Eq. (9.92)], we get (Fig. 9.3)
FIGURE 9.3 The first excited levels of the helium atom.
\(
\begin{gathered}
E^{(0)}+E{1}^{(1)}=E^{(0)}+J{1 s 2 s}-K{1 s 2 s}=-57.8 \mathrm{eV} \
E^{(0)}+E{2}^{(1)}=E^{(0)}+J{1 s 2 s}+K{1 s 2 s}=-55.4 \mathrm{eV} \
E^{(0)}+E{3}^{(1)}=E^{(0)}+J{1 s 2 p}-K{1 s 2 p}=-55.7{5} \mathrm{eV} \
E^{(0)}+E{4}^{(1)}=E^{(0)}+J{1 s 2 p}+K_{1 s 2 p}=-53.9 \mathrm{eV}
\end{gathered}
\)
The first-order energy corrections seem to indicate that the lower of the two levels of the $1 s 2 p$ configuration lies below the higher of the two levels of the $1 s 2 s$ configuration. Study of the helium spectrum reveals that this is not so. The error is due to neglect of the higherorder perturbation-energy corrections.
Using the variation-perturbation method (Section 9.2), Knight and Scherr calculated the second- and third-order corrections $E^{(2)}$ and $E^{(3)}$ for these four excited levels. [R. E. Knight and C. W. Scherr, Rev. Mod. Phys., 35, 431 (1963); for energy corrections through 17th order, see F. C. Sanders and C. W. Scherr, Phys. Rev., 181, 84 (1969).] Figure 9.4 shows their results (which are within 0.1 eV of the experimental energies). Figure 9.4 shows that Fig. 9.3 is quite inaccurate. Since the perturbation $e^{2} / 4 \pi \varepsilon{0} r{12}$ is not really very small, a perturbation treatment that includes only the $E^{(1)}$ correction does not give accurate results.
The first-order correction to the wave function, $\psi^{(1)}$, will include contributions from other configurations (configuration interaction). When we say that a level belongs to the configuration $1 s 2 s$, we are indicating the configuration that makes the largest contribution to the true wave function.
We started with the eight degenerate zeroth-order functions (9.93). These functions have three kinds of degeneracy. There is the degeneracy between hydrogenlike functions with the same $n$, but different $l$; the $2 s$ and the $2 p$ functions have the same energy. There is the degeneracy between hydrogenlike functions with the same $n$ and $l$, but different $m$; the $2 p{1}, 2 p{0}$, and $2 p{-1}$ functions have the same energy. (For convenience we used the real functions $2 p{x, y, z}$, but we could have started with the functions $2 p{1,0,-1}$.) Finally, there is the degeneracy between functions that differ only in the interchange of the two electrons between the orbitals; the functions $\psi{1}^{(0)}=1 s(1) 2 s(2)$ and $\psi{2}^{(0)}=1 s(2) 2 s(1)$ have the same energy. This last kind of degeneracy is called exchange degeneracy. When the interelectronic repulsion $e^{2} / 4 \pi \varepsilon{0} r{12}$ was introduced as a perturbation, the exchange degeneracy and the degeneracy associated with the quantum number $l$ were removed. The degeneracy associated with $m$ remained, however; each $1 s 2 p$ helium level is triply degenerate, and we could just as well have used the $2 p{1}, 2 p{0}$, and $2 p{-1}$ orbitals instead of the real orbitals in constructing the correct zeroth-order wave functions. Let us consider the reasons for the removal of the $l$ degeneracy and the exchange degeneracy.
The interelectronic repulsion in helium makes the $2 s$ orbital energy less than the $2 p$ energy. Figures 6.9 and 6.8 show that a $2 s$ electron has a greater probability than a $2 p$ electron of being closer to the nucleus than the $1 s$ electron(s). A $2 s$ electron will not be as effectively shielded from the nucleus by the $1 s$ electrons and will therefore have a lower energy than a $2 p$ electron. [According to Eq. (6.94), the greater the nuclear charge, the lower the energy.] Mathematically, the difference between the $1 s 2 s$ and the $1 s 2 p$ energies results from the Coulomb integral $J{1 s 2 s}$ being smaller than $J{1 s 2 p}$. These Coulomb integrals
$1 s 2 p\left{\begin{array}{ll}-57.8 \mathrm{eV} & \text { FIGURE 9.4 } E^{(0)}+E^{(1)}+ \ -58.1 & \begin{array}{l}E^{(2)}+E^{(3)} \text { for the first excited } \ \text { levels of helium. Also shown }\end{array} \ 1 s 2 s \begin{cases}-58.4 & 2^{-1 / 2}[1 s(1) 2 s(2)+2 s(1) 1 s(2)] \ \text { are the correct zeroth-order } \ \text { wave functions for the } 1 s 2 s\end{cases} \ -59.2 & 2^{-1 / 2}[1 s(1) 2 s(2)-2 s(1) 1 s(2)]\end{array} \begin{array}{l}\text { levels. }\end{array}\right.$
represent the electrostatic repulsion between the appropriate charge distributions. When the $2 s$ electron penetrates the charge distribution of the $1 s$ electron, it feels a repulsion from only the unpenetrated part of the $1 s$ charge distribution. Hence the $1 s-2 s$ electrostatic repulsion is less than the $1 s-2 p$ repulsion, and the $1 s 2 s$ levels lie below the $1 s 2 p$ levels. The interelectronic repulsion in many-electron atoms lifts the $l$ degeneracy, and the orbital energies for the same value of $n$ increase with increasing $l$.
Now consider the removal of the exchange degeneracy. The functions (9.93) with which we began the perturbation treatment have each electron assigned to a definite orbital. For example, the function $\psi{1}^{(0)}=1 s(1) 2 s(2)$ has electron 1 in the $1 s$ orbital and electron 2 in the $2 s$ orbital. For $\psi{2}^{(0)}$ the opposite is true. The secular determinant was not diagonal, so the initial functions were not the correct zeroth-order wave functions. The correct zeroth-order functions do not assign each electron to a definite orbital. Thus the first two correct zeroth-order functions are
\(
\phi{1}^{(0)}=2^{-1 / 2}[1 s(1) 2 s(2)-1 s(2) 2 s(1)], \quad \phi{2}^{(0)}=2^{-1 / 2}[1 s(1) 2 s(2)+1 s(2) 2 s(1)]
\)
We cannot say which orbital electron 1 is in for either $\phi{1}^{(0)}$ or $\phi{2}^{(0)}$. This property of the wave functions of systems containing more than one electron results from the indistinguishability of identical particles in quantum mechanics and will be discussed further in Chapter 10. Since the functions $\phi{1}^{(0)}$ and $\phi{2}^{(0)}$ have different energies, the exchange degeneracy is removed when the correct zeroth-order functions are used.