Helium: the Prototype Two-Electron Atom

Helium has two electrons and $Z=2$, and its Hamiltonian is one repulsion term away from two independent hydrogenic problems. That single term makes the equation non-separable and, at the same time, makes helium the cleanest laboratory for every many-electron idea: the mean field, the variational method, exchange, and the singlet-triplet split. The central field and Hartree-Fock constructions are heavy machinery; on helium they can be checked against numbers by hand.

The helium Hamiltonian

With the nucleus fixed and both electrons measured from it, the non-relativistic Hamiltonian is¹

with $r_{12}=|{\vec{r}}_1-{\vec{r}}_2|$ and $Z=2$. Drop the last term and the equation separates into two hydrogenic problems of charge $Z$; the ground state is the product $1s(1)1s(2)$ with both electrons in the $Z=2$ ground orbital. Its energy is twice the hydrogenic ground energy at $Z=2$,

writing energies in Hartree ($1\text{Ha}=27.211$ eV). The measured value is $-79.0$ eV.² The zeroth-order estimate overbinds by nearly $30$ eV, exactly the mean repulsion the product ignored. Helium is not a small perturbation away from independent electrons, and any honest treatment must handle the repulsion.

First-order perturbation theory

Treat the repulsion as a perturbation on the product ground state. The first-order shift is its expectation value in the unperturbed state,¹

The integral is done with the multipole expansion $1/r_{12}={\sum }_{\ell }(r_{<}^{\ell }/r_{>}^{\ell +1})P_{\ell }(\cos {\theta }_{12})$; only the $\ell =0$ term survives against the spherical $1s$ densities, and the remaining radial integral gives the clean fraction $5/8$. With $Z=2$,

First order lands within $5$ eV of experiment, an error of about $5\%$. The perturbation is not really small — $E^{(1)}$ is a third of $E^{(0)}$ — so the agreement is better than the method has any right to give, and the higher orders do not converge quickly. A variational estimate does better with less work.

The variational estimate

The physical flaw in the product $1s(Z=2)1s(Z=2)$ is that each electron screens the nucleus from the other, so neither sees the full charge $Z=2$. Promote the orbital charge to a variational parameter $Z^{\prime }$ and minimize the energy over it.³ For a $1s$ orbital of charge $Z^{\prime }$ in the field of the true nucleus $Z$, the pieces are

each in Hartree. The trial energy is the sum over both electrons plus the repulsion,

Setting $\mathbf{d}E/\mathbf{d}{Z}^{\prime }=0$ gives $2Z^{\prime }=2Z-5/8$, so the optimal effective charge is

and back-substitution collapses to $E_{\min }=-Z^{\prime 2}$:

The number $Z^{\prime }=1.69$ is the physics: each electron sees not the full $+2$ but a partially screened $+1.69$, the other electron's cloud cancelling about a third of one unit of charge. The variational energy, guaranteed to lie above the true ground energy, sits at $-77.5$ eV against the measured $-79.0$ eV — an error of $2\%$ from a single parameter.³

The three estimates and the datum, side by side, show each correction pulling the binding energy toward the measured value:

The first ionization energy

Removing one electron leaves the one-electron ion ${\text{He}}^{+}$, a hydrogenic system with $Z=2$ and energy $-Z^2/2=-2\text{Ha}=-54.4$ eV. The first ionization energy is the difference,

the largest first ionization energy of any element and the reason helium is inert. The closed $1s^2$ shell, tightly bound and with no low-lying vacancy, resists both losing and sharing an electron.

Para- and ortho-helium

The excited configurations put one electron in $1s$ and the other in a higher orbital $n\ell$ (the doubly excited states lie above the first ionization limit and autoionize). Two electrons in different spatial orbitals can form a spatially symmetric or antisymmetric combination, and antisymmetry of the total state pairs each with the opposite spin symmetry:⁴

• para-helium — spatially symmetric, spin singlet ($S=0$), terms ${}^1{L}_J$;
• ortho-helium — spatially antisymmetric, spin triplet ($S=1$), terms ${}^3{L}_J$.

The energy of a $1sn\ell$ configuration, to first order, is the sum of the two orbital energies plus the direct integral, shifted by the exchange integral with a sign set by the spin:

with $+$ for the singlet (symmetric space) and $-$ for the triplet. Since the exchange integral $K>0$, the triplet lies below the singlet of the same configuration by $2K$. The physical cause is the exchange hole: the antisymmetric spatial state keeps the electrons apart, lowering their Coulomb repulsion, and the parallel-spin triplet is the one forced into that arrangement.

The ground configuration $1s^2$ has no such partner. Both electrons occupy the same $1s$ spatial orbital, which is necessarily symmetric, so the spin must be the antisymmetric singlet. A $1s^2$ triplet would need a symmetric spin state on a symmetric spatial state — a fully symmetric total wave function, forbidden for fermions. There is therefore no $1s^2$ triplet, and the helium ground term is $1{}^1{S}_0$ with no low-lying triplet beneath it.

Because the electric-dipole operator does not touch spin, transitions between singlet and triplet — intercombination lines — are strongly forbidden by the $\Delta S=0$ rule. Para- and ortho-helium behave almost like two separate gases: the nineteenth century catalogued their spectra as two elements. The lowest triplet $2{}^3{S}_1$ cannot decay to the $1{}^1{S}_0$ ground state by any allowed route and is metastable, with a lifetime of about $7900$ s, an eternity on atomic timescales.⁴

The ordering triplet below singlet and its origin in the exchange integral is the physical content of the first Hund rule, which the next lessons extend from helium's two electrons to open shells of any size, once the coupling schemes and term symbols are in place.