We now reconsider the helium atom from the standpoint of electron spin and the antisymmetry requirement. In the perturbation treatment of helium in Section 9.3, we found the zeroth-order wave function for the ground state to be $1 s(1) 1 s(2)$. To take spin into account, we must multiply this spatial function by a spin eigenfunction. We therefore consider the possible spin eigenfunctions for two electrons. We shall use the notation $\alpha(1) \alpha(2)$ to indicate a state where electron 1 has spin up and electron 2 has spin up; $\alpha(1)$ stands for
$\alpha\left(m_{s 1}\right)$. Since each electron has two possible spin states, we have at first sight the four possible spin functions:
\(
\alpha(1) \alpha(2), \quad \beta(1) \beta(2), \quad \alpha(1) \beta(2), \quad \alpha(2) \beta(1)
\)
There is nothing wrong with the first two functions, but the third and fourth functions violate the principle of indistinguishability of identical particles. For example, the third function says that electron 1 has spin up and electron 2 has spin down, which does distinguish between electrons 1 and 2. More formally, if we apply $\hat{P}_{12}$ to these functions, we find that the first two functions are symmetric with respect to interchange of the two electrons, but the third and fourth functions are neither symmetric nor antisymmetric and so are unacceptable.
What now? Recall that we ran into essentially the same situation in treating the helium excited states (Section 9.7), where we started with the functions $1 s(1) 2 s(2)$ and $2 s(1) 1 s(2)$. We found that these two functions, which distinguish between electrons 1 and 2 , are not the correct zeroth-order functions and that the correct zeroth-order functions are $2^{-1 / 2}[1 s(1) 2 s(2) \pm 2 s(1) 1 s(2)]$. This result suggests pretty strongly that instead of $\alpha(1) \beta(2)$ and $\beta(1) \alpha(2)$, we use
\(
\begin{equation}
2^{-1 / 2}[\alpha(1) \beta(2) \pm \beta(1) \alpha(2)] \tag{10.21}
\end{equation}
\)
These functions are the normalized linear combinations of $\alpha(1) \beta(2)$ and $\beta(1) \alpha(2)$ that are eigenfunctions of $\hat{P}_{12}$, that is, are symmetric or antisymmetric. When electrons 1 and 2 are interchanged, $2^{-1 / 2}[\alpha(1) \beta(2)+\beta(1) \alpha(2)]$ becomes $2^{-1 / 2}[\alpha(2) \beta(1)+\beta(2) \alpha(1)]$, which is the same as the original function. In contrast, $2^{-1 / 2}[\alpha(1) \beta(2)-\beta(1) \alpha(2)]$ becomes $2^{-1 / 2}[\alpha(2) \beta(1)-\beta(2) \alpha(1)]$, which is -1 times the original function. To show that the functions (10.21) are normalized, we have
\(
\begin{aligned}
& \sum{m{s 1}} \sum{m{s 2}} \frac{1}{\sqrt{2}}[\alpha(1) \beta(2) \pm \beta(1) \alpha(2)] \frac{1}{\sqrt{2}}[\alpha(1) \beta(2) \pm \beta(1) \alpha(2)] \
& =\frac{1}{2} \sum{m{s 1}}|\alpha(1)|^{2} \sum{m{s 2}}|\beta(2)|^{2} \pm \frac{1}{2} \sum{m{s 1}} \alpha^{}(1) \beta(1) \sum{m{s 2}} \beta^{}(2) \alpha(2) \
& \quad \pm \frac{1}{2} \sum{m{s 1}} \beta^{}(1) \alpha(1) \sum{m{s 2}} \alpha^{*}(2) \beta(2)+\frac{1}{2} \sum{m{s 1}}|\beta(1)|^{2} \sum{m{s 2}}|\alpha(2)|^{2}=1
\end{aligned}
\)
where we used the orthonormality relations (10.11) and (10.12).
Therefore, the four normalized two-electron spin eigenfunctions with the correct exchange properties are
\(
\text { symmetric: } \quad\left{\begin{array}{l}
\alpha(1) \alpha(2) \tag{10.22}\
\beta(1) \beta(2) \
{[\alpha(1) \beta(2)+\beta(1) \alpha(2)] / \sqrt{2}}
\end{array}\right.
\)
\(
\begin{equation}
\text { antisymmetric: }[\alpha(1) \beta(2)-\beta(1) \alpha(2)] / \sqrt{2} \tag{10.24}
\end{equation}
\)
We now include spin in the He zeroth-order ground-state wave function. The function $1 s(1) 1 s(2)$ is symmetric with respect to exchange. The overall electronic wave function including spin must be antisymmetric. Hence we must multiply the symmetric space function $1 s(1) 1 s(2)$ by an antisymmetric spin function. There is only one antisymmetric twoelectron spin function, so the ground-state zeroth-order wave function for the helium atom including spin is
\(
\begin{equation}
\psi^{(0)}=1 s(1) 1 s(2) \cdot 2^{-1 / 2}[\alpha(1) \beta(2)-\beta(1) \alpha(2)] \tag{10.26}
\end{equation}
\)
$\psi^{(0)}$ is an eigenfunction of $\hat{P}_{12}$ with eigenvalue -1 .
To a very good approximation, the Hamiltonian does not contain spin terms, so the energy is unaffected by inclusion of the spin factor in the ground-state wave function. Also, the ground state of helium is still nondegenerate when spin is considered.
To further demonstrate that the spin factor does not affect the energy, we shall assume we are doing a variational calculation for the He ground state using the trial function $\phi=f\left(r{1}, r{2}, r_{12}\right) 2^{-1 / 2}[\alpha(1) \beta(2)-\beta(1) \alpha(2)]$, where $f$ is a normalized function symmetric in the coordinates of the two electrons. The variational integral is
\(
\begin{aligned}
\int \phi^{} \hat{H} \phi d \tau=\sum{m{s 1}} \sum{m{s 2}} \iint f^{}\left(r{1},\right. & \left.r{2}, r{12}\right) \frac{1}{\sqrt{2}}[\alpha(1) \beta(2)-\beta(1) \alpha(2)]^{*} \
& \times \hat{H} f\left(r{1}, r{2}, r{12}\right) \frac{1}{\sqrt{2}}[\alpha(1) \beta(2)-\beta(1) \alpha(2)] d v{1} d v{2}
\end{aligned}
\)
Since $\hat{H}$ has no effect on the spin functions, the variational integral becomes
\(
\iint f * \hat{H} f d v{1} d v{2} \sum{m{s 1}} \sum{m{s 2}} \frac{1}{2}|\alpha(1) \beta(2)-\beta(1) \alpha(2)|^{2}
\)
Since the spin function (10.25) is normalized, the variational integral reduces to $\iint f^{*} \hat{H} f d v{1} d v{2}$, which is the expression we used before we introduced spin.
Now consider the excited states of helium. We found the lowest excited state to have the zeroth-order spatial wave function $2^{-1 / 2}[1 s(1) 2 s(2)-2 s(1) 1 s(2)]$ [Eq. (9.103)]. Since this spatial function is antisymmetric, we must multiply it by a symmetric spin function. We can use any one of the three symmetric two-electron spin functions, so instead of the nondegenerate level previously found, we have a triply degenerate level with the three zeroth-order wave functions
\(
\begin{gather}
2^{-1 / 2}[1 s(1) 2 s(2)-2 s(1) 1 s(2)] \alpha(1) \alpha(2) \tag{10.27}\
2^{-1 / 2}[1 s(1) 2 s(2)-2 s(1) 1 s(2)] \beta(1) \beta(2) \tag{10.28}\
2^{-1 / 2}[1 s(1) 2 s(2)-2 s(1) 1 s(2)] 2^{-1 / 2}[\alpha(1) \beta(2)+\beta(1) \alpha(2)] \tag{10.29}
\end{gather}
\)
For the next excited state, the requirement of antisymmetry of the overall wave function leads to the zeroth-order wave function
\(
\begin{equation}
2^{-1 / 2}[1 s(1) 2 s(2)+2 s(1) 1 s(2)] 2^{-1 / 2}[\alpha(1) \beta(2)-\beta(1) \alpha(2)] \tag{10.30}
\end{equation}
\)
The same considerations apply for the $1 s 2 p$ states.