The helium atom has two electrons and a nucleus of charge $+2 e$. We shall consider the nucleus to be at rest (Section 6.6) and place the origin of the coordinate system at the nucleus. The coordinates of electrons 1 and 2 are $\left(x{1}, y{1}, z{1}\right)$ and $\left(x{2}, y{2}, z{2}\right)$; see Fig. 9.1.
If we take the nuclear charge to be $+Z e$ instead of $+2 e$, we can treat heliumlike ions such as $\mathrm{H}^{-}, \mathrm{Li}^{+}$, and $\mathrm{Be}^{2+}$. The Hamiltonian operator is
\(
\begin{equation}
\hat{H}=-\frac{\hbar^{2}}{2 m{e}} \nabla{1}^{2}-\frac{\hbar^{2}}{2 m{e}} \nabla{2}^{2}-\frac{Z e^{2}}{4 \pi \varepsilon{0} r{1}}-\frac{Z e^{2}}{4 \pi \varepsilon{0} r{2}}+\frac{e^{2}}{4 \pi \varepsilon{0} r{12}} \tag{9.39}
\end{equation}
\)
where $m{e}$ is the mass of the electron, $r{1}$ and $r{2}$ are the distances of electrons 1 and 2 from the nucleus, and $r{12}$ is the distance from electron 1 to 2 . The first two terms are the operators for the electrons' kinetic energy [Eq. (3.48)]. The third and fourth terms are the potential energies of attraction between the electrons and the nucleus. The final term is the potential energy of interelectronic repulsion [Eq. (6.58)]. Note that the potential energy of a system of interacting particles cannot be written as the sum of potential energies of the individual particles. The potential energy is a property of the system as a whole.
The Schrödinger equation involves six independent variables, three coordinates for each electron. In spherical coordinates, $\psi=\psi\left(r{1}, \theta{1}, \phi{1}, r{2}, \theta{2}, \phi{2}\right)$.
The operator $\nabla{1}^{2}$ is given by Eq. (6.6) with $r{1}, \theta{1}, \phi{1}$ replacing $r, \theta, \phi$. The variable $r{12}$ is $r{12}=\left[\left(x{1}-x{2}\right)^{2}+\left(y{1}-y{2}\right)^{2}+\left(z{1}-z{2}\right)^{2}\right]^{1 / 2}$, and by using the relations between Cartesian and spherical coordinates, we can express $r{12}$ in terms of $r{1}, \theta{1}, \phi{1}, r{2}, \theta{2}, \phi_{2}$.
Because of the $e^{2} / 4 \pi \varepsilon{0} r{12}$ term, the Schrödinger equation for helium cannot be separated in any coordinate system, and we must use approximation methods. The perturbation method separates the Hamiltonian (9.39) into two parts, $\hat{H}^{0}$ and $\hat{H}^{\prime}$, where $\hat{H}^{0}$ is the Hamiltonian of an exactly solvable problem. If we choose
\(
\begin{gather}
\hat{H}^{0}=-\frac{\hbar^{2}}{2 m{e}} \nabla{1}^{2}-\frac{Z e^{2}}{4 \pi \varepsilon{0} r{1}}-\frac{\hbar^{2}}{2 m{e}} \nabla{2}^{2}-\frac{Z e^{2}}{4 \pi \varepsilon{0} r{2}} \tag{9.40}\
\hat{H}^{\prime}=\frac{e^{2}}{4 \pi \varepsilon{0} r{12}} \tag{9.41}
\end{gather}
\)
then $\hat{H}^{0}$ is the sum of two hydrogenlike Hamiltonians, one for each electron:
\(
\begin{gather}
\hat{H}^{0}=\hat{H}{1}^{0}+\hat{H}{2}^{0} \tag{9.42}\
\hat{H}{1}^{0} \equiv-\frac{\hbar^{2}}{2 m{e}} \nabla{1}^{2}-\frac{Z e^{2}}{4 \pi \varepsilon{0} r{1}}, \quad \hat{H}{2}^{0} \equiv-\frac{\hbar^{2}}{2 m{e}} \nabla{2}^{2}-\frac{Z e^{2}}{4 \pi \varepsilon{0} r{2}} \tag{9.43}
\end{gather}
\)
The unperturbed system is a helium atom in which the two electrons exert no forces on each other. Although such a system does not exist, this does not prevent us from applying perturbation theory to this system.
Since the unperturbed Hamiltonian (9.42) is the sum of the Hamiltonians for two independent particles, we can use the separation-of-variables results of Eqs. (6.18) to (6.24) to conclude that the unperturbed wave functions have the form
\(
\begin{equation}
\psi^{(0)}\left(r{1}, \theta{1}, \phi{1}, r{2}, \theta{2}, \phi{2}\right)=F{1}\left(r{1}, \theta{1}, \phi{1}\right) F{2}\left(r{2}, \theta{2}, \phi{2}\right) \tag{9.44}
\end{equation}
\)
and the unperturbed energies are
\(
\begin{gather}
E^{(0)}=E{1}+E{2} \tag{9.45}\
\hat{H}{1}^{0} F{1}=E{1} F{1}, \quad \hat{H}{2}^{0} F{2}=E{2} F{2} \tag{9.46}
\end{gather}
\)
Since $\hat{H}{1}^{0}$ and $\hat{H}{2}^{0}$ are hydrogenlike Hamiltonians, the solutions of (9.46) are the hydrogenlike eigenfunctions and eigenvalues. From Eq. (6.94), we have
\(
\begin{gather}
E{1}=-\frac{Z^{2}}{n{1}^{2}} \frac{e^{2}}{8 \pi \varepsilon{0} a{0}}, \quad E{2}=-\frac{Z^{2}}{n{2}^{2}} \frac{e^{2}}{8 \pi \varepsilon{0} a{0}} \tag{9.47}\
E^{(0)}=-Z^{2}\left(\frac{1}{n{1}^{2}}+\frac{1}{n{2}^{2}}\right) \frac{e^{2}}{8 \pi \varepsilon{0} a{0}}, \quad \begin{aligned}
n{1} & =1,2,3, \ldots \
n{2} & =1,2,3, \ldots
\end{aligned} \tag{9.48}
\end{gather}
\)
where $a_{0}$ is the Bohr radius. Equation (9.48) gives the zeroth-order energies of states with both electrons bound to the nucleus. The He atom also has continuum states.
The lowest level has $n{1}=1, n{2}=1$, and its zeroth-order wave function is [Eq. (6.104)]
\(
\begin{equation}
\psi{1 s^{2}}^{(0)}=\frac{1}{\pi^{1 / 2}}\left(\frac{Z}{a{0}}\right)^{3 / 2} e^{-Z r{1} / a{0}} \cdot \frac{1}{\pi^{1 / 2}}\left(\frac{Z}{a{0}}\right)^{3 / 2} e^{-Z r{2} / a_{0}}=1 s(1) 1 s(2) \tag{9.49}
\end{equation}
\)
where $1 s(1) 1 s(2)$ denotes the product of hydrogenlike $1 s$ functions for electrons 1 and 2 , and where the subscript indicates that both electrons are in hydrogenlike $1 s$ orbitals. (Note that the procedure of assigning electrons to orbitals and writing the atomic wave function as the product of one-electron orbital functions is an approximation.) The energy of this unperturbed ground state is
\(
\begin{equation}
E{1 s^{2}}^{(0)}=-Z^{2}(2) \frac{e^{2}}{8 \pi \varepsilon{0} a_{0}} \tag{9.50}
\end{equation}
\)
The quantity $-e^{2} / 8 \pi \varepsilon{0} a{0}$ is the ground-state energy of the hydrogen atom (taking the nucleus to be infinitely heavy) and equals -13.606 eV [Eqs. (6.105)-(6.108)]. If the electron mass $m{e}$ in $a{0}$ is replaced by the reduced mass for ${ }^{4} \mathrm{He},-e^{2} / 8 \pi \varepsilon{0} a{0}$ is changed to -13.604 eV , and we shall use this number to (partly) correct for the nuclear motion in He. For helium, $Z=2$ and (9.50) gives $-8(13.604 \mathrm{eV})=-108.83 \mathrm{eV}$ :
\(
\begin{equation}
E_{1 s^{2}}^{(0)}=-108.83 \mathrm{eV} \tag{9.51}
\end{equation}
\)
How does this zeroth-order energy compare with the true helium ground-state energy? The experimental first ionization energy of He is 24.587 eV . The second ionization energy of He is easily calculated theoretically, since it is the ionization energy of the hydrogenlike ion $\mathrm{He}^{+}$and is equal to $2^{2}(13.604 \mathrm{eV})=54.416 \mathrm{eV}$. If we choose the zero of energy as the completely ionized atom [this choice is implicit in (9.39)], then the ground-state energy of the helium atom is $-(24.587+54.416) \mathrm{eV}=-79.00 \mathrm{eV}$. The zeroth-order energy (9.51) is in error by $38 \%$. We should have expected such a large error, since the perturbation term $e^{2} / 4 \pi \varepsilon{0} r{12}$ is not small.
The next step is to evaluate the first-order perturbation correction to the energy. The unperturbed ground state is nondegenerate, and use of (9.22) and (9.49) gives
\(
\begin{align}
E^{(1)} & =\left\langle\psi^{(0)}\right| \hat{H}^{\prime}\left|\psi^{(0)}\right\rangle \
E^{(1)}=\frac{Z^{6} e^{2}}{\left(4 \pi \varepsilon{0}\right) \pi^{2} a{0}^{6}} \int{0}^{2 \pi} \int{0}^{2 \pi} \int{0}^{\pi} \int{0}^{\pi} & \int{0}^{\infty} \int{0}^{\infty} e^{-2 Z r{1} / a{0}} e^{-2 Z r{2} / a{0}} \frac{1}{r{12}} r{1}^{2} \sin \theta{1} \
& \times r{2}^{2} \sin \theta{2} d r{1} d r{2} d \theta{1} d \theta{2} d \phi{1} d \phi_{2} \tag{9.52}
\end{align}
\)
The volume element for this two-electron problem contains the coordinates of both electrons; $d \tau=d \tau{1} d \tau{2}$. The integral in (9.52) can be evaluated by using an expansion of $1 / r_{12}$ in terms of spherical harmonics, as outlined in Prob. 9.14. One finds
\(
\begin{equation}
E^{(1)}=\frac{5 Z}{8}\left(\frac{e^{2}}{4 \pi \varepsilon{0} a{0}}\right) \tag{9.53}
\end{equation}
\)
Recalling that $\frac{1}{2} e^{2} / 4 \pi \varepsilon{0} a{0}$ equals 13.604 eV when the ${ }^{4} \mathrm{He}$ reduced mass is used, and putting $Z=2$, we find for the first-order perturbation energy correction for the helium ground state:
\(
E^{(1)}=\frac{10}{4}(13.604 \mathrm{eV})=34.01 \mathrm{eV}
\)
Our approximation to the energy is now
\(
\begin{equation}
E^{(0)}+E^{(1)}=-108.83 \mathrm{eV}+34.01 \mathrm{eV}=-74.82 \mathrm{eV} \tag{9.54}
\end{equation}
\)
which, compared with the experimental value of -79.00 eV , is in error by $5.3 \%$.
To evaluate the first-order correction to the wave function and higher-order corrections to the energy requires evaluating the matrix elements of $1 / r_{12}$ between the ground unperturbed state and all excited states (including the continuum) and performing the appropriate summations and integrations. No one has yet figured out how to evaluate directly all the contributions to $E^{(1)}$. Note that the effect of $\psi^{(1)}$ is to mix into the wave-function contributions from other configurations besides $1 s^{2}$. We call this configuration interaction. The largest contribution to the true ground-state wave function of helium comes from the $1 s^{2}$ configuration, which is the unperturbed (zeroth-order) wave function.
$E^{(2)}$ for the helium ground state has been calculated using the variation-perturbation method, Eq. (9.38). Scherr and Knight used 100-term trial functions to get extremely accurate approximations to the wave-function corrections through sixth order and thus to the energy corrections through thirteenth order [C. W. Scherr and R. E. Knight, Rev. Mod. Phys., 35, 436 (1963)]. For calculations of the energy corrections through order 401, see J. D. Baker et al., Phys. Rev. A, 41, 1247 (1990). The second-order correction $E^{(2)}$ turns out to be -4.29 eV , and $E^{(3)}$ is +0.12 eV . Through third order, we have for the groundstate energy
\(
E \approx-108.83 \mathrm{eV}+34.01 \mathrm{eV}-4.29 \mathrm{eV}+0.12 \mathrm{eV}=-78.99 \mathrm{eV}
\)
which is close to the experimental value -79.00 eV . Including corrections through thirteenth order, Scherr and Knight obtained a ground-state helium energy of $-2.90372433\left(e^{2} / 4 \pi \varepsilon{0} a{0}\right)$, which is close to the value $-2.90372438\left(e^{2} / 4 \pi \varepsilon{0} a{0}\right)$ obtained from the purely variational calculations described in the next section.
The perturbation-theory series expansion for the He-atom energy can be proved to converge [R. Ahlrichs, Phys. Rev. A, 5, 605 (1972)].
An exact wave function and energy cannot be found for the two-electron groundstate He atom, but remarkably, there exists a two-electron problem for which the exact ground-state solution of the Schrödinger equation has been found. This is a hypothetical atom (called the Hooke's-law atom or harmonium) with Hamiltonian operator
\(
\hat{H}=-\frac{\hbar^{2}}{2 m{e}} \nabla{1}^{2}-\frac{\hbar^{2}}{2 m{e}} \nabla{2}^{2}+\frac{1}{2} k\left(r{1}^{2}+r{2}^{2}\right)+\frac{e^{2}}{4 \pi \varepsilon{0} r{12}}
\)
where $r{1}$ and $r{2}$ are the distances of the electrons from the origin. For certain values of the force-constant $k$, exact ground-state wave functions and energies have been found. See en.wikipedia.org/wiki/Hooke's_atom.