Variational Helium (Overview)

Helium is the first atom the hydrogen toolkit cannot solve. With two electrons repelling each other, the Schrödinger equation has no closed-form solution. Yet a one-parameter variational calculation gets the ground-state energy to within about $2\%$ of experiment — a striking demonstration of the method on a genuinely hard problem.

Why helium is hard

The helium Hamiltonian (atomic units, nuclear charge $Z = 2$ ) is

\hat{H} = \underbrace{-\frac{1}{2}\nabla_1^2 - \frac{2}{r_1}}_{\text{electron 1 near nucleus}} \;\underbrace{-\frac{1}{2}\nabla_2^2 - \frac{2}{r_2}}_{\text{electron 2 near nucleus}} \;+\; \underbrace{\frac{1}{|\mathbf{r}_1 - \mathbf{r}_2|}}_{\text{electron--electron repulsion}}.

The first two groups are just two independent hydrogen-like atoms; if that were all, the ground-state energy would be $2 \times (-Z^2/2) = -4$ Ha. The trouble is the final term, the $1/|\mathbf{r}_1 - \mathbf{r}_2|$ repulsion, which couples the two coordinates and forbids separation of variables.

The screening idea

Physically, each electron does not feel the full nuclear charge $Z = 2$ . The other electron spends part of its time between it and the nucleus, partially screening the charge. So a good trial state replaces $Z$ with an adjustable effective charge $Z_{\text{eff}}$ and uses a product of hydrogen-like $1s$ orbitals:

\psi(\mathbf{r}_1, \mathbf{r}_2) = \phi_{Z_{\text{eff}}}(\mathbf{r}_1)\,\phi_{Z_{\text{eff}}}(\mathbf{r}_2), \qquad \phi_{Z_{\text{eff}}}(\mathbf{r}) \propto e^{-Z_{\text{eff}} r}.

(The two electrons occupy the same spatial orbital with opposite spins, consistent with the Pauli principle, so the spatial part is symmetric and the spin part is the antisymmetric singlet.) The single parameter $Z_{\text{eff}}$ is the variational knob.

The result

Evaluating $\langle \hat{H} \rangle$ in this trial state gives an energy of the form

E(Z_{\text{eff}}) = Z_{\text{eff}}^2 - 2 Z\, Z_{\text{eff}} + \frac{5}{8} Z_{\text{eff}} = Z_{\text{eff}}^2 - \frac{27}{8} Z_{\text{eff}} \quad (Z = 2),

where the $\tfrac{5}{8}Z_{\text{eff}}$ term is the electron–electron repulsion integral. Minimizing,

\frac{dE}{dZ_{\text{eff}}} = 2 Z_{\text{eff}} - \frac{27}{8} = 0 \;\;\Longrightarrow\;\; Z_{\text{eff}} = \frac{27}{16} \approx 1.69 .

The optimum is less than 2, exactly the screening picture: each electron feels an effective charge of about $1.69$ , not the bare $2$ . The corresponding energy is

E_{\min} = -\left(\frac{27}{16}\right)^2 \approx -2.85 \text{ Ha},

against the experimental value of about $-2.90$ Ha — an error of roughly $2\%$ from a single parameter.

What to take away

Helium shows the method scaling beyond toy problems: a physically motivated trial state with one screening parameter turns an unsolvable two-body problem into a one-line minimization that lands within a couple of percent of reality. The same product-of-orbitals-with-effective-charge logic, generalized, underlies the Hartree–Fock method used throughout quantum chemistry.