In the last section, we wrote the helium-atom Hamiltonian as $\hat{H}=\hat{H}^{0}+\hat{H}^{\prime}$, where the ground-state eigenfunction $\psi{g}^{(0)}$ of $\hat{H}^{0}$ is (9.49). What happens if we use the zeroth-order perturbation-theory ground-state wave function $\psi{g}^{(0)}$ as the variation function $\phi$ in the variational integral? The variational integral $\langle\phi| \hat{H}|\phi\rangle=\langle\phi \mid \hat{H} \phi\rangle$ then becomes
\(
\begin{align}
\langle\phi| \hat{H}|\phi\rangle & =\left\langle\psi{g}^{(0)} \mid\left(\hat{H}^{0}+\hat{H}^{\prime}\right) \psi{g}^{(0)}\right\rangle=\left\langle\psi{g}^{(0)} \mid \hat{H}^{0} \psi{g}^{(0)}+\hat{H}^{\prime} \psi{g}^{(0)}\right\rangle \
& =\left\langle\psi{g}^{(0)} \mid E{g}^{(0)} \psi{g}^{(0)}\right\rangle+\left\langle\psi{g}^{(0)} \mid \hat{H}^{\prime} \psi{g}^{(0)}\right\rangle=E{g}^{(0)}+E{g}^{(1)} \tag{9.55}
\end{align}
\)
since $\hat{H} \psi{g}^{(0)}=E{g}^{(0)} \psi{g}^{(0)},\left\langle\psi{g}^{(0)} \mid \psi{g}^{(0)}\right\rangle=1$, and $E{g}^{(1)}=\left\langle\psi{g}^{(0)}\right| \hat{H}^{\prime}\left|\psi{g}^{(0)}\right\rangle$ [Eq. (9.22)]. Use of $\psi_{g}^{(0)}$ as the variation function gives the same energy result as in first-order perturbation theory.
Now consider variation functions for the helium-atom ground state. If we used $\psi_{g}^{(0)}$ [Eq. (9.49)] as the trial function, we would get the first-order perturbation result, -74.82 eV . To improve on this result, we introduce a variational parameter into (9.49). We try the normalized function
\(
\begin{equation}
\phi=\frac{1}{\pi}\left(\frac{\zeta}{a{0}}\right)^{3} e^{-\zeta r{1} / a{0}} e^{-\zeta r{2} / a_{0}} \tag{9.56}
\end{equation}
\)
which is obtained from (9.49) by replacing the true atomic number $Z$ by a variational parameter $\zeta$ (zeta). $\zeta$ has a simple physical interpretation. Since one electron tends to screen the other from the nucleus, each electron is subject to an effective nuclear charge somewhat less than the full nuclear charge $Z$. If one electron fully shielded the other from the nucleus, we would have an effective nuclear charge of $Z-1$. Since both electrons are in the same orbital, they will be only partly effective in shielding each other. We thus expect $\zeta$ to lie between $Z-1$ and $Z$.
We now evaluate the variational integral. To expedite things, we rewrite the helium Hamiltonian (9.39) as
\(
\begin{align}
\hat{H}=[ & \left.-\frac{\hbar^{2}}{2 m{e}} \nabla{1}^{2}-\frac{\zeta e^{2}}{4 \pi \varepsilon{0} r{1}}-\frac{\hbar^{2}}{2 m{e}} \nabla{2}^{2}-\frac{\zeta e^{2}}{4 \pi \varepsilon{0} r{2}}\right]+(\zeta-Z) \frac{e^{2}}{4 \pi \varepsilon{0} r{1}}+(\zeta-Z) \frac{e^{2}}{4 \pi \varepsilon{0} r{2}} \
& +\frac{e^{2}}{4 \pi \varepsilon{0} r{12}} \tag{9.57}
\end{align}
\)
where the terms involving zeta were added and subtracted. The terms in brackets in (9.57) are the sum of two hydrogenlike Hamiltonians for nuclear charge $\zeta$. Moreover, the trial function (9.56) is the product of two hydrogenlike $1 s$ functions for nuclear charge $\zeta$. Therefore, when these terms operate on $\phi$, we have an eigenvalue equation, the eigenvalue being the sum of two hydrogenlike $1 s$ energies for nuclear charge $\zeta$ :
\(
\begin{equation}
\left[-\frac{\hbar^{2}}{2 m{e}} \nabla{1}^{2}-\frac{\zeta e^{2}}{4 \pi \varepsilon{0} r{1}}-\frac{\hbar^{2}}{2 m{e}} \nabla{2}^{2}-\frac{\zeta e^{2}}{4 \pi \varepsilon{0} r{2}}\right] \phi=-\zeta^{2}(2) \frac{e^{2}}{8 \pi \varepsilon{0} a{0}} \phi \tag{9.58}
\end{equation}
\)
Using (9.57) and (9.58), we have
\(
\begin{align}
\int \phi^{} \hat{H} \phi d \tau=-\zeta^{2} \frac{e^{2}}{4 \pi \varepsilon{0} a{0}} \int & \phi^{} \phi d \tau+\frac{(\zeta-Z) e^{2}}{4 \pi \varepsilon_{0}} \int \frac{\phi^{} \phi}{r{1}} d \tau \
& +\frac{(\zeta-Z) e^{2}}{4 \pi \varepsilon{0}} \int \frac{\phi^{} \phi}{r{2}} d \tau+\frac{e^{2}}{4 \pi \varepsilon{0}} \int \frac{\phi^{} \phi}{r_{12}} d \tau \tag{9.59}
\end{align*}
\)
Let $f{1}$ be a normalized $1 s$ hydrogenlike orbital for nuclear charge $\zeta$, occupied by electron 1 . Let $f{2}$ be the same function for electron 2 :
\(
\begin{equation}
f{1}=\frac{1}{\pi^{1 / 2}}\left(\frac{\zeta}{a{0}}\right)^{3 / 2} e^{-\zeta r{1} / a{0}}, \quad f{2}=\frac{1}{\pi^{1 / 2}}\left(\frac{\zeta}{a{0}}\right)^{3 / 2} e^{-\zeta r{2} / a{0}} \tag{9.60}
\end{equation}
\)
Noting that $\phi=f{1} f{2}$, we now evaluate the integrals in (9.59):
\(
\begin{gathered}
\int \phi^{} \phi d \tau=\iint f_{1}^{} f{2}^{*} f{1} f{2} d \tau{1} d \tau{2}=\int f{1}^{} f{1} d \tau{1} \int f_{2}^{} f{2} d \tau{2}=1 \
\int \frac{\phi^{} \phi}{r{1}} d \tau=\int \frac{f{1}^{} f{1}}{r{1}} d \tau{1} \int f{2}^{} f{2} d \tau{2}=\int \frac{f_{1}^{} f{1}}{r{1}} d \tau{1} \
=\frac{1}{\pi} \frac{\zeta^{3}}{a{0}^{3}} \int{0}^{\infty} e^{-2 \zeta r{1} / a{0}} \frac{r{1}^{2}}{r{1}} d r{1} \int{0}^{\pi} \sin \theta{1} d \theta{1} \int{0}^{2 \pi} d \phi{1}=\frac{\zeta}{a{0}}
\end{gathered}
\)
where the Appendix integral (A.8) was used. Also,
\(
\int \frac{\phi^{} \phi}{r{2}} d \tau=\int \frac{f{2}^{} f{2}}{r{2}} d \tau{2}=\int \frac{f{1}^{*} f{1}}{r{1}} d \tau{1}=\frac{\zeta}{a{0}}
\)
since it doesn't matter whether the label 1 or 2 is used on the dummy variables in the definite integral. Finally, we must evaluate $\left(e^{2} / 4 \pi \varepsilon{0}\right) \int\left(\phi^{*} \phi / r{12}\right) d \tau$. This is the same as the integral (9.52) that occurred in the perturbation treatment, except that $Z$ is replaced by $\zeta$. Hence, from (9.53)
\(
\frac{e^{2}}{4 \pi \varepsilon{0}} \int \frac{\phi^{*} \phi}{r{12}} d \tau=\frac{5 \zeta e^{2}}{32 \pi \varepsilon{0} a{0}}
\)
The variational integral (9.59) thus has the value
\(
\begin{equation}
\int \phi^{} \hat{H} \phi d \tau=\left(\zeta^{2}-2 Z \zeta+\frac{5}{8} \zeta\right) \frac{e^{2}}{4 \pi \varepsilon{0} a{0}} \tag{9.61}
\end{}
\)
As a check, if we set $\zeta=Z$ in (9.61), we get the first-order perturbation-theory result, (9.50) plus (9.53).
We now vary $\zeta$ to minimize the variational integral:
\(
\begin{gather}
\frac{\partial}{\partial \zeta} \int \phi^{} \hat{H} \phi d \tau=\left(2 \zeta-2 Z+\frac{5}{8}\right) \frac{e^{2}}{4 \pi \varepsilon{0} a{0}}=0 \
\zeta=Z-\frac{5}{16} \tag{9.62}
\end{gather*}
\)
As anticipated, the effective nuclear charge lies between $Z$ and $Z-1$. Using (9.62) and (9.61), we get
\(
\begin{equation}
\int \phi^{} \hat{H} \phi d \tau=\left(-Z^{2}+\frac{5}{8} Z-\frac{25}{256}\right) \frac{e^{2}}{4 \pi \varepsilon{0} a{0}}=-\left(Z-\frac{5}{16}\right)^{2} \frac{e^{2}}{4 \pi \varepsilon{0} a{0}} \tag{9.63}
\end{}
\)
Putting $Z=2$, we get as our approximation to the helium ground-state energy $-(27 / 16)^{2}\left(e^{2} / 4 \pi \varepsilon{0} a{0}\right)=-(729 / 256) 2(13.604 \mathrm{eV})=-77.48 \mathrm{eV}$, as compared with the true value of -79.00 eV . Use of $\zeta$ instead of $Z$ has reduced the error from $5.3 \%$ to $1.9 \%$. In accord with the variation theorem, the true ground-state energy is less than the variational integral.
How can we improve our variational result? We might try a function that had the general form of (9.56), that is, a product of two functions, one for each electron:
\(
\begin{equation}
\phi=u(1) u(2) \tag{9.64}
\end{equation}
\)
However, we could try a variety of functions $u$ in (9.64), instead of the single exponential used in (9.56). A systematic procedure for finding the function $u$ that gives the lowest value of the variational integral will be discussed in Section 11.1. This procedure shows that for the best possible choice of $u$ in (9.64) the variational integral equals -77.86 eV , which is still in error by $1.5 \%$. We might ask why (9.64) does not give the true ground-state energy, no matter what form we try for $u$. The answer is that, when we write the trial function as the product of separate functions for each electron, we are making an approximation. Because of the $e^{2} / 4 \pi \varepsilon{0} r{12}$ term in the Hamiltonian, the Schrödinger equation for helium is not separable, and the true ground-state wave function cannot be written as the product of separate functions for each electron. To reach the true ground-state energy, we must go beyond a function of the form (9.64).
The Bohr model gave the correct energies for the hydrogen atom but failed when applied to helium. Hence, in the early days of quantum mechanics, it was important to show that the new theory could give an accurate treatment of helium. The pioneering work on the helium ground state was done by Hylleraas in the years 1928-1930. To allow for the effect of one electron on the motion of the other, Hylleraas used variational functions that contained the interelectronic distance $r_{12}$. One function he used is
\(
\begin{equation}
\phi=N\left[e^{-\zeta r{1} / a{0}} e^{-\zeta r{2} / a{0}}\left(1+b r_{12}\right)\right] \tag{9.65}
\end{equation}
\)
where $N$ is the normalization constant and $\zeta$ and $b$ are variational parameters. Since
\(
\begin{equation}
r{12}=\left[\left(x{2}-x{1}\right)^{2}+\left(y{2}-y{1}\right)^{2}+\left(z{2}-z_{1}\right)^{2}\right]^{1 / 2} \tag{9.66}
\end{equation}
\)
the function (9.65) goes beyond the simple product form (9.64). Minimization of the variational integral with respect to the parameters gives $\zeta=1.849, b=0.364 / a{0}$, and a ground-state energy of -78.7 eV , in error by 0.3 eV . The $1+b r{12}$ term makes the wave function larger for large values of $r{12}$. This is as it should be, because the repulsion between the electrons makes it energetically more favorable for the electrons to avoid each other. Using a more complicated six-term trial function containing $r{12}$, Hylleraas obtained an energy only 0.01 eV above the true ground-state energy.
Hylleraas's work has been extended by others. Using a 1078 -term variational function, Pekeris found a ground-state energy of $-2.903724375\left(e^{2} / 4 \pi \varepsilon{0} a{0}\right)$ [C. L. Pekeris, Phys. Rev., 115, 1216 (1959)]. With relativistic and nuclear-motion corrections added, this gave for $E{i}$, the ionization energy of helium, $E{i} / h c=198310.69 \mathrm{~cm}^{-1}$, compared with the experimental value $198310.67 \mathrm{~cm}^{-1}$. Using an improved variational function, Frankowski and Pekeris bettered Perkeris's result by obtaining the energy $-2.90372437703\left(e^{2} / 4 \pi \varepsilon{0} a{0}\right)$, a result believed to be within $10^{-11}\left(e^{2} / 4 \pi \varepsilon{0} a{0}\right)$ of the true nonrelativistic, infinite-nuclearmass ground-state energy [K. Frankowski and C. L. Pekeris, Phys. Rev., 146, 46 (1966)]. Drake and Yan used linear variational functions containing $r{12}$ to calculate the ground-state energy and many excited-state energies of He that are thought to be accurate to 1 part in $10^{14}$ or better [G. W. F. Drake and Z-C. Yan, Chem. Phys. Lett., 229, 486 (1994); Phys. Rev. A, 46, 2378 (1992)]. These workers similarly calculated Li variational energies for the ground state and two excited states with 1 part in $10^{9}$ accuracy or better [Z-C. Yan and G. W. F. Drake, Phys. Rev. A, 52, 3711 (1995)]. Adding in relativistic and nuclear motion corrections, Drake and Yan found good agreement between theoretically calculated and experimental spectroscopic transition frequencies of He and Li. By doing a series of variational calculations with increasing numbers of terms in the variation function and extrapolating to the limit of an infinite number of terms, Drake and co-workers found the following ground-state,
nonrelativistic, infinite-nuclear-mass He energy: $-2.90372437703411959831\left(e^{2} / 4 \pi \varepsilon{0} a_{0}\right)$, which is believed accurate to 21 significant figures [G. W. F. Drake et al., Phys. Rev. A, 65, 054501 (2002)].