The variational method

At the end of the last section, helium still owed us money: the electron repulsion is a third of the main part — hardly “small” — and first-order perturbation theory gives $-74.8\ \text{eV}$, 4 eV away from the experimental $-79.0\ \text{eV}$. To do better, grinding out second order means summing over infinitely many intermediate states — sheer misery. The root difficulty: perturbation theory needs a small parameter, and helium has none.

This section takes a completely different road. It makes no expansion whatsoever and asks nothing about the size of the perturbation. Its only weapon is a single inequality — and it is an inequality we already know how to prove.

The variational principle

Why no guess can break below the ground statebasic~5 minexpand

Let the true eigenstates of $\hat H$ be $\ket{n}$ with eigenvalues $E_0\le E_1\le E_2\le\cdots$. Take any normalised “trial state” $\ket{\psi}$ — a pure guess, not required to be any eigenstate. Since the eigenstates are complete (section 3.7), it can always be expanded:

$$\ket{\psi}=\sum_n c_n\ket{n},\qquad \sum_n|c_n|^2=1\tag{7.3.1}$$

Compute the energy expectation:

$$\braket{\psi}{\hat H\psi}=\sum_n|c_n|^2E_n \ \ge\ \sum_n|c_n|^2E_0=E_0\tag{7.3.2}$$

The middle step uses only $E_n\ge E_0$: replacing each $E_n$ by the smaller $E_0$ can only shrink the sum. Hence

$$\boxed{\ \braket{\psi}{\hat H\psi}\ \ge\ E_0\ }\tag{7.3.3}$$

valid for any normalised state, with equality exactly when $\ket{\psi}$ is the ground state itself.

In plain words: the energy expectation is the average of the eigenenergies weighted by $|c_n|^2$, and an average can never dip below the minimum. That is all there is to it.

(If the trial state is not normalised, use $\braket{\psi}{\hat H\psi}/\braket{\psi}{\psi}\ge E_0$ — same content.)

This inequality instantly becomes a machine for computing energies:

1. Guess a family of parametrised wavefunctions $\psi_\alpha$ ($\alpha$ can be one or several parameters);
2. Compute $E(\alpha)=\braket{\psi_\alpha}{\hat H\psi_\alpha}$;
3. Minimise over $\alpha$.

The minimum is still an upper bound — but it is the tightest upper bound in that family of guesses. The smarter the family, the closer the answer gets to the truth. This is the variational method: we “vary” over a family of wavefunctions to minimise the energy functional.

Warm-up: deliberately guessing the wrong shape

First, test the blade somewhere the exact answer is known, to see “what happens when the guess is wrong”. For hydrogen (true ground state: the exponential $\ee^{-r/a_0}$, $E_0=-13.6\ \text{eV}$), deliberately pick the wrong shape as a trial — a Gaussian:

$$\psi_b(r)\propto\ee^{-br^2}\tag{7.3.5}$$

with $b$ the variational parameter. Work out the kinetic and potential expectation values (both standard Gaussian integrals),

$$E(b)=\frac{3\hbar^2b}{2m}-\frac{e^2}{4\pi\varepsilon_0}\,2\sqrt{\frac{2b}{\pi}}\tag{7.3.6}$$

Set the derivative in $b$ to zero and substitute back:

$$E_{\min}=-\frac{4}{3\pi}\,\frac{me^4}{(4\pi\varepsilon_0)^2\hbar^2} \approx-11.5\ \text{eV}\tag{7.3.7}$$

The true value is $-13.6\ \text{eV}$. Two observations: the upper-bound promise held ($-11.5$ really is above $-13.6$); and although the shape was completely wrong (a Gaussian is too flat at the origin and falls off too fast at large $r$), the energy still captured 85%. That is the power of the “error is second-order small” theorem — and also its trap: an accurate energy does not mean an accurate wavefunction; $\psi_b$ is badly wrong both near the origin and far away.

The main course: the helium ground state

Time to settle the debt owed for two sections. The craft of building a trial wavefunction is writing your physical intuition into the parameters:

Each electron, besides being pulled by the nucleus ($Z=2$), is pushed away by the other electron. Seen from one electron’s viewpoint, the other’s cloud of negative charge “shades” part of the nucleus — the effective nuclear charge it feels should be less than 2, greater than 1 (the other electron is sometimes inside it and sometimes outside, so the shading is imperfect). This effect is called screening.

So guess: each electron sits in a hydrogen-like ground state of nuclear charge $Z_e$, with $Z_e$ the variational parameter,

$$\psi_{Z_e}(\vec r_1,\vec r_2)=\frac{Z_e^3}{\pi a_0^3}\, \ee^{-Z_e(r_1+r_2)/a_0}\tag{7.3.8}$$

and let energy minimisation decide “how much screening” for us.

The screened variational calculation: from E(Zₑ) to −77.5 eVadvanced~12 minexpand

Keep the books in units of $e^2/(4\pi\varepsilon_0 a_0)=27.2\ \text{eV}$ (twice the Rydberg). The Hamiltonian splits into three pieces.

Kinetic energy. The trial state is a hydrogen-like ground state at charge $Z_e$, so the one-electron kinetic energy follows from the hydrogen-like virial theorem: $\langle T\rangle=+Z_e^2/2$ (in the units above). Two electrons:

$$\langle T_1+T_2\rangle=Z_e^2\tag{7.3.9}$$

Nuclear attraction. The hydrogen-like ground state has $\langle1/r\rangle=Z_e/a_0$. The true nuclear charge is $Z=2$ (not $Z_e$!), so

$$\left\langle-\frac{Z}{r_1}-\frac{Z}{r_2}\right\rangle=-2ZZ_e\tag{7.3.10}$$

Electron repulsion. This is the only integral that genuinely has to be computed — the electrostatic interaction energy of two exponential charge clouds of charge $Z_e$ — and the result is a famous number:

$$\left\langle\frac{1}{r_{12}}\right\rangle=\frac{5}{8}Z_e\tag{7.3.11}$$

(Method: expand $1/r_{12}$ in spherical harmonics; only the spherically symmetric term survives, and what remains is elementary. The 34 eV used in the first-order calculation of section 7.1 is just this at $Z_e=2$: $\tfrac58\times2\times27.2=34.0\ \text{eV}$.)

Total:

$$E(Z_e)=\left[Z_e^2-2ZZ_e+\frac58Z_e\right]\times27.2\ \text{eV}\tag{7.3.12}$$

Sanity check: set $Z_e=Z=2$ (no screening) and get $(4-8+1.25)\times27.2=-74.8\ \text{eV}$ — exactly the first-order perturbation result. First-order perturbation theory = the variational method with its parameter left unoptimised; the two roads meet right here.

Minimise. $\dd E/\dd Z_e=2Z_e-2Z+\tfrac58=0$:

$$Z_e=Z-\frac{5}{16}=\frac{27}{16}=1.6875\tag{7.3.13}$$

The screening is $5/16\approx0.31$: on average the other electron shades about a third of one nuclear charge — squarely in the intuitively expected “between 0 and 1, on the small side” range. Substituting back (the bracket at the minimum equals exactly $-Z_e^2$):

$$E_{\min}=-\left(\frac{27}{16}\right)^2\times27.2\ \text{eV}=-77.5\ \text{eV}\tag{7.3.14}$$

The scoreboard:

Method	Result	Gap to experiment
Ignore repulsion entirely	$-108.8$ eV	29.8 eV
First-order perturbation (section 7.1)	$-74.8$ eV	4.2 eV
One-parameter variation	$-77.5$ eV	1.5 eV
Experiment	$-79.0$ eV	—

One parameter, one page of algebra, and the error drops to 2%. And this road has no end: in 1929 Hylleraas wrote $r_{12}$ explicitly into the trial function and reached 0.01 eV with a handful of parameters; modern variational calculations pin the helium ground state to a dozen-plus significant figures. The accuracy is limited only by how much imagination you are willing to pack into the trial function — a promise perturbation theory can never make.

When to use it, and where to be careful

The variational method applies startlingly widely: no small perturbation required, no nearby solvable model required — only the ability to compute $\braket{\psi}{\hat H\psi}$. But its blind spots are just as sharp.

∎Key formulas

• Variational principle

  $$\frac{\braket{\psi}{\hat H\psi}}{\braket{\psi}{\psi}}\ \ge\ E_0$$

  Holds for any trial state; equality iff ψ is the ground state

• Order of the error

  $$\ket{\psi}=\ket{0}+\epsilon\ket{\delta}\ \Rightarrow\ E(\psi)=E_0+O(\epsilon^2)$$

  Err in the wavefunction at first order, in the energy only at second

• Helium: energy functional

  $$E(Z_e)=\left[Z_e^2-2ZZ_e+\tfrac{5}{8}Z_e\right]\frac{e^2}{4\pi\varepsilon_0 a_0}$$

  At Zₑ=Z it reduces to first-order perturbation, −74.8 eV

• Helium: the minimum

  $$Z_e=Z-\tfrac{5}{16}=1.6875,\quad E_{\min}=-77.5\ \text{eV}$$

  Experiment −79.0 eV; screening ≈ 1/3 of a charge

?Self-check4 questions

1. 1.

   The fundamental reason the variational principle ⟨ψ|H|ψ⟩ ≥ E₀ holds is:

   • The energy expectation is a weighted average of the eigenvalues with weights |cₙ|², and an average cannot dip below the minimum
   • The trial wavefunction must be close to the true ground state
   • The Hamiltonian is a positive-definite operator
   • The uncertainty principle forbids energies that are too low

   Submit
2. 2.

   In the helium variational calculation, the physical meaning of Zₑ = 27/16 ≈ 1.69 is:

   • Each electron, on average, has about 1/3 of a nuclear charge shaded away by the other electron’s charge cloud
   • The true charge of the helium nucleus is not actually 2
   • The electron’s effective mass has changed
   • It is an artefact of mathematical optimisation with no physical meaning

   Submit
3. 3.

   Which of the following statements about the limitations of the variational method are correct? (Select all that apply.)

   Select all that apply

   • It gives only an upper bound, with no error range
   • Even when the energy is very accurate, the wavefunction’s details can still be poor
   • Applied directly to excited states, the upper-bound property fails if the trial state contains a ground-state component
   • It requires the Hamiltonian to contain a small parameter

   Submit
4. 4.

   Using E(Zₑ) = [Zₑ² − 4Zₑ + (5/8)Zₑ] × 27.2 eV with Zₑ = 27/16, compute the variational ground-state energy of helium, in eV (mind the sign).

   eV50% relative tolerance

   Submit

What comes next

Perturbation theory and the variational method both answer the same kind of question: where are the energy levels? But some questions are not about levels at all — how does an alpha particle escape the nucleus? How does an electron cross a barrier it could classically never climb? The protagonist of those questions is the wavefunction under the barrier, and there the wavefunction decays exponentially — no polynomial-style correction can keep up with it. The WKB approximation of the next section is built for exactly this “semiclassical” situation: when the potential varies slowly enough, the wavefunction is locally a plane wave at every point — and with that we will compute a tunnelling probability for the first time, and explain the lifetime law in nuclear physics that spans thirty orders of magnitude.