How to calculate when exact solutions run out
Removing both electrons from a helium atom takes 79.0 eV. Helium is the next simplest atom after hydrogen, but because its two electrons repel each other, its Schrödinger equation has no exact solution. How close can we get to that 79.0, and how can we tell whether an approximate answer is trustworthy?
Almost every real quantum problem is like helium: a part we can solve, plus something that spoils it. This essay describes the two most widely used ways of making progress. Perturbation theory treats the spoiling part as a small correction and asks what it changes. The variational method guesses a wavefunction and uses a guarantee built into quantum mechanics to improve the guess. Each comes with a way of judging when to trust it, and that judgement matters as much as the formulas.
Perturbation theory: correct the part you can solve
Split the energy operator as $\hat H = \hat H_0 + \hat W$, where $\hat H_0$ has known stationary states $|n\rangle$ and energies $E_n^{(0)}$, and $\hat W$ is the extra piece. The first correction to the energy of a state is the extra piece averaged over the unperturbed state:
$$E_n \approx E_n^{(0)} + \langle n|\hat W|n\rangle .$$The idea is natural: to first approximation the state does not change, so the extra energy is the average of $\hat W$ in the state we already have. The optional section shows where it comes from. The result is labelled "first order" because it is the first term of a series in powers of $\hat W$.
Worked example: helium to first order
The solvable part. Ignore the repulsion between the electrons. Each then moves independently in the field of a nucleus with charge $Z = 2$, and a hydrogen-like atom with charge $Z$ has energies $Z^2$ times hydrogen's. Both electrons go into 1s (with opposite spins), so
$$\begin{aligned} E^{(0)} &= 2\times\bigl(-Z^2\times13.6\ \text{eV}\bigr)\\[4pt] &= 2\times(-54.4\ \text{eV}) = -108.8\ \text{eV}. \end{aligned}$$That is 30 eV too low: without their repulsion the electrons would be far more tightly bound.
The first-order correction is the repulsion $e^2/4\pi\varepsilon_0 r_{12}$, where $r_{12}$ is the distance between the electrons, averaged over the product of two 1s wavefunctions. The integral is a classic, quoted here rather than derived:
$$\langle\hat W\rangle = \tfrac54\,Z\times13.6\ \text{eV} = 34.0\ \text{eV}.$$The result: $E \approx -108.8 + 34.0 = -74.8$ eV, against the measured $-79.0$ eV. Five percent off.
How much to trust it. The correction is not small: 34 eV against 109 eV, about a third. A first-order answer is only reliable when the correction is small compared with what it corrects, and here it is not, so a 5 percent error is about what one should expect. It is a good start, not a final answer.
What "small" must be small compared with
In helium the first-order correction came from the average of $\hat W$. Often that average is zero, and the effect of the perturbation comes instead from mixing states together. The simplest case shows what controls the approximation. Take two levels, at energies $0$ and $\Delta$, and a perturbation that couples them with strength $u$:
$$\hat H = \begin{pmatrix} 0 & u\\ u & \Delta\end{pmatrix}.$$This one can be solved exactly, which is what makes it a good test. The lower energy is $\tfrac12\Delta - \sqrt{\tfrac14\Delta^2 + u^2}$. The next term of perturbation theory, the second-order shift, gives
$$E_{\text{lower}} \approx -\frac{u^2}{\Delta}.$$The coupling pushes the two levels apart, the lower one down, by an amount that grows with the coupling and shrinks with the spacing. Comparing the two formulas, the approximation is just the first term of the exact result expanded in powers of $u/\Delta$. So the condition for it to work is not that $u$ is small in some absolute sense, but that it is small compared with the spacing $\Delta$ between the levels it mixes.
The same comparison shows what goes wrong when levels are nearly equal. Keep the coupling fixed and shrink the spacing: $-u^2/\Delta$ runs off to minus infinity, while the exact shift levels off at $-u$. When two states have nearly the same energy, even a weak perturbation mixes them completely, and the series is useless. That is the ammonia molecule of the formalism essay: two basis states of equal energy, where the coupling does not shift the levels slightly but replaces them with even and odd combinations. The remedy is to solve the nearly equal states exactly among themselves, as we did there, and treat the rest as the perturbation.
Optional: where the first-order formula comes from
Write the exact state and energy as the unperturbed ones plus corrections, $|\psi\rangle = |n\rangle + |\delta\rangle$ and $E = E_n^{(0)} + \delta E$, and put them into $(\hat H_0 + \hat W)|\psi\rangle = E|\psi\rangle$. Now project onto $\langle n|$, that is, take the overlap of both sides with the unperturbed state. Because $\hat H_0$ is Hermitian and $|n\rangle$ is its eigenstate, $\langle n|\hat H_0|\delta\rangle = E_n^{(0)}\langle n|\delta\rangle$, and those terms cancel against the matching ones on the right. What is left is exact:
$$\delta E\,\langle n|\psi\rangle = \langle n|\hat W|\psi\rangle .$$To lowest order, replace $|\psi\rangle$ by $|n\rangle$ on both sides. Then $\langle n|n\rangle = 1$ and $\delta E = \langle n|\hat W|n\rangle$. Keeping the first correction to $|\psi\rangle$ as well gives the second-order term, a sum over the other states $m$ of $|\langle m|\hat W|n\rangle|^2/(E_n^{(0)} - E_m^{(0)})$, which for two states is the $-u^2/\Delta$ above. The energy differences in its denominators are why small spacings are dangerous.
The variational method: a guaranteed bound
The second approach rests on one fact. For any state $|\psi\rangle$ whatsoever, the average energy $\langle\psi|\hat H|\psi\rangle$ is at least the true ground-state energy. The reason is short: expand $|\psi\rangle$ in the true energy eigenstates; the average energy is a weighted average of their energies, with positive weights adding to one, and a weighted average can never lie below the smallest thing being averaged.
So any guessed wavefunction gives an upper bound on the ground-state energy, and the lower the bound, the better the guess. Give the guess an adjustable parameter, compute the average energy as a function of that parameter, and minimise.
For helium, a natural guess is the same product of 1s wavefunctions as before, but with the nuclear charge in them replaced by an adjustable value $Z'$. The idea is that each electron partly shields the nucleus from the other, so each should see a charge somewhat less than 2. Computing the average energy for this guess and minimising gives $Z' = 2 - \tfrac{5}{16} = 1.69$, and
$$E = -2\,Z'^2\times13.6\ \text{eV} = -77.5\ \text{eV}.$$That is within 2 percent of the measured $-79.0$ eV, and, as it must be, above it. Better guesses with more parameters push the bound down. Calculations of this kind with thousands of parameters, plus small corrections for relativity, now match the measured value to many decimal places.
Two questions
1. A stronger coupling. Two levels 1.0 eV apart are coupled with $u = 0.3$ eV. Compare the second-order shift of the lower level with the exact one.
Worked answer
Second order: $-u^2/\Delta = -0.09/1.0 = -0.090$ eV. Exact: $0.5 - \sqrt{0.25 + 0.09} = 0.5 - 0.583 = -0.083$ eV. The approximation overshoots by about 8 percent, consistent with $u/\Delta = 0.3$ being small but not very small.
The easy slip is to judge the coupling as small because 0.3 eV sounds small. What matters is the ratio to the spacing. The same 0.3 eV coupling between levels 0.1 eV apart would make the series useless.
2. A result that is too good. A student's trial wavefunction for helium gives an average energy of $-80.2$ eV, lower than the measured $-79.0$ eV. They are pleased, since in the variational method lower is better. What should they conclude?
Worked answer
That something is wrong. For any state, the average energy cannot lie below the true ground-state energy, so a result below $-79.0$ eV means an error in the calculation: a wavefunction that is not normalised, an integral done wrongly, or a term left out of $\hat H$.
The tempting wrong turn is to treat "lower is better" as unlimited. It holds only above the bound. The guarantee that makes the method useful also makes it self-checking: a variational answer below the true value is not an improvement, it is a mistake.
Further reading
- Landau & Lifshitz, Quantum Mechanics (2nd ed.), §38 "Perturbations independent of time", and §20 "The variational principle".
- Born, Atomic Physics (Dover, 8th ed.), Ch. VI §4 "The Helium Atom": the atom that the old quantum theory could not handle and the new one approximates so well.
Where this leaves us
Helium's 79.0 eV can be approached in two steps. Treating the electrons' repulsion as a first-order correction gives 74.8 eV, about 5 percent off, as expected for a correction a third the size of what it corrects. Letting each electron see a screened nuclear charge and using the variational bound gives 77.5 eV, within 2 percent and guaranteed to lie above the truth. The rules for trust are as important as the formulas: a perturbation must be small compared with the spacing of the levels it mixes, and a variational energy can never go below the true one.
With these tools, quantum mechanics predicts the numbers of atoms, molecules and solids to whatever accuracy we care to pay for. What it predicts, in every case, are probabilities of measurement results. The core sequence ends by looking directly at that step: what happens, according to the equations, when a measurement is actually made and a detector clicks.