Helium is the first atom that cannot be solved exactly: a nucleus of charge $2e$ and two electrons that repel each other. Leave out the repulsion and each electron is in the 1s state of a hydrogen-like ion, for a total of $-108.8$ eV; the measured ground energy is $-79.0$ eV, so the repulsion matters a great deal. The ground state has both electrons in 1s, a symmetric spatial state, so their spins form a singlet. Its average repulsion, $(5/8)Z$ hartree, gives $-74.8$ eV at first order. Better still, let each electron see a screened nucleus of charge $\bar{Z}$ and choose $\bar{Z}$ variationally: $\bar{Z}=Z-5/16=1.69$, and $-77.5$ eV, within 2 per cent. The excited states come in singlets, parahelium, and triplets, orthohelium, the triplets lower by the exchange energy — and there is no triplet ground state, because two electrons in 1s cannot both have the same spin.

Two children on one seesaw with a parent in the middle: each is pulled toward the parent, but each also keeps the other a little farther out. Each electron of helium sees a nucleus partly hidden by the other.

The energy of helium’s trial state against its effective charge $\bar{Z}$, in hartrees: $\bar{Z}^{2}-2Z\bar{Z}+\tfrac58\bar{Z}$ with . At , the dot, it is the first-order result, ; its lowest point, at , still lies above the measured , as a trial energy must.

In atomic units, with the nucleus fixed, the two electrons of an atom of nuclear charge $Z$ have the Hamiltonian

• Without the repulsion the ground energy is $-Z^{2}$ hartree, $-108.8$ eV for $Z=2$; the measured value is $-2.904$ hartree, eV.

Take both electrons in a $1s$ orbital of adjustable charge $\bar{Z}$. The kinetic and nuclear energies follow from the hydrogen chapter. The repulsion needs an integral: averaging $1/|\mathbf{r}_{1}-\mathbf{r}_{2}|$ over directions leaves , and the radial integrals give . Minimising over gives the best screened charge and an upper bound on the energy.

Both electrons in a $1s$ orbital of charge $\bar{Z}$, the spins in the singlet.

For a $1s$ state of charge $\bar{Z}$, $\braket{-\tfrac12\nabla^{2}}=\tfrac12\bar{Z}^{2}$ and , from the hydrogen chapter’s averages.

The integral over the angle between $\mathbf{r}_{1}$ and $\mathbf{r}_{2}$ gives $(r_1+r_2-\mathrm{\mid }{r}_1-r_2\mathrm{\mid }$, and the other angle .

The remaining radial integrals of $e^{-2\bar{Z}(r_{1}+r_{2})}$, done by parts.

Add the three averages.

For $Z=2$: $\bar{Z}=\tfrac{27}{16}$ and hartree, eV, against the measured eV.

Leave out the repulsion and each electron of helium is in the 1s state of a hydrogen-like ion with $Z=2$, at $-Z^{2}\times13.6$ eV.

Each electron partly hides the nucleus from the other, so each sees an effective charge $\bar{Z}$ smaller than $Z$.

The trial energy is the average energy of an actual state, so the variational principle puts it above the true ground energy.

Both electrons of helium’s ground state are in 1s, a symmetric spatial state, so the spins form a singlet.

Great work! You now know how to:

• estimate helium’s ground energy at first order and variationally
• explain the screened charge $Z-5/16$
• tell parahelium from orthohelium