The variational principle

Intuition

Most Hamiltonians cannot be solved exactly, but their ground-state energy can still be pinned down from above. Take any state at all and compute the average energy in it: the answer is never below the true ground energy. So guess a family of states with a few adjustable numbers, compute the average energy as a function of them, and lower it as far as it will go. The best guess is an upper bound, and usually a good one: an error in the state makes only a second-order error in the energy.

Drop a ball anywhere into a valley and it comes to rest at a height at least as great as the lowest point. Searching over where to drop it, you learn the depth of the valley from above.

The average energy of the trial state $e^{-bx^{2}}$ for $\hat{H}=\frac{\hat{p}^{2}}{2}+|x|$ , against the width parameter $b$ . Every value lies above the true ground energy $0.8086$ ; the best, $0.8129$ at $b\approx0.54$ , is only half a per cent above it.

An upper bound on the ground energy

For any state $\ket{\psi}$ in the domain of $\hat{H}$ , the average energy is at least the ground energy. Equality holds only for the ground state.

E[\psi]=\frac{\bra{\psi}\hat{H}\ket{\psi}}{\braket{\psi|\psi}}\ \ge\ E_{0}

Properties

Minimising $E[\psi]$ over a family of trial states gives the best upper bound the family allows.
If the trial state differs from the ground state by $\epsilon$ , the energy differs by order $\epsilon^{2}$ : energies come out much better than the states.
A trial state orthogonal to the ground state bounds the first excited energy from above, and so on up the ladder.
The bound holds for any Hamiltonian bounded below, in any number of dimensions and for any number of particles.
It gives no lower bound: a variational estimate can be too high, never too low.

\bra{\psi}\hat{H}\ket{\psi}\ge E_{0}\braket{\psi|\psi}\quad\text{for every}\ \ket{\psi}

Expand the trial state in the eigenstates of the Hamiltonian. Its average energy is a weighted average of the eigenvalues, with weights $|c_{n}|^{2}$ . No eigenvalue is below $E_{0}$ , so neither is any weighted average of them.

Proof steps

\ket{\psi}=\sum_{n}c_{n}\ket{E_{n}}

The eigenstates of $\hat{H}$ form a basis.

\bra{\psi}\hat{H}\ket{\psi}=\sum_{n}|c_{n}|^{2}E_{n}

Each eigenstate contributes its energy, weighted by its probability.

\sum_{n}|c_{n}|^{2}E_{n}\ge\sum_{n}|c_{n}|^{2}E_{0}

Every $E_{n}$ is at least $E_{0}$ .

\bra{\psi}\hat{H}\ket{\psi}\ge E_{0}\braket{\psi|\psi}

Since $\sum_{n}|c_{n}|^{2}=\braket{\psi|\psi}$ .

=\ \text{only if}\ c_{n}=0\ \text{whenever}\ E_{n}>E_{0}

Equality leaves no weight on higher levels: the trial state is a ground state.

Applications

Practice

Never Below

The average energy in any state is at least the ground-state energy. A variational estimate can be too high, never too low.

\bra{\psi}\hat{H}\ket{\psi}\ge E_{0}

Try it

A trial state gives $E[\psi]=-2.3$ eV. What can be said of the true ground energy $E_{0}$ ?