The Variational Principle

Most Hamiltonians cannot be solved exactly. Perturbation theory helps when a problem is close to one we already understand, but many systems have no such "nearby" solvable cousin. The variational method offers a different route: it lets us bound the lowest energy of any system using nothing more than a guessed wavefunction and an integral.

The variational bound

Let $\hat{H}$ be a Hamiltonian with an unknown ground-state energy $E_0$ , the smallest eigenvalue. For any normalizable state $|\psi\rangle$ — chosen freely, not required to solve anything — define the Rayleigh quotient

E[\psi] = \frac{\langle \psi | \hat{H} | \psi \rangle}{\langle \psi | \psi \rangle}.

The variational principle states that this expectation value can never dip below the true ground-state energy:

E[\psi] \;\geq\; E_0.

Equality holds if and only if $|\psi\rangle$ is the exact ground state (or a degenerate partner of it). So every guess gives an upper bound on $E_0$ , and a better guess gives a tighter bound.

Why it is true

Expand the trial state in the (unknown) orthonormal energy eigenbasis $\{|n\rangle\}$ with eigenvalues $E_n$ , ordered so that $E_0 \leq E_1 \leq E_2 \leq \dots$ :

|\psi\rangle = \sum_n c_n |n\rangle, \qquad \langle \psi | \psi \rangle = \sum_n |c_n|^2 .

Because $\hat{H}|n\rangle = E_n |n\rangle$ ,

\langle \psi | \hat{H} | \psi \rangle = \sum_n E_n |c_n|^2 .

Now use the fact that every eigenvalue satisfies $E_n \geq E_0$ . Replacing each $E_n$ by the smallest one can only decrease the sum:

\langle \psi | \hat{H} | \psi \rangle = \sum_n E_n |c_n|^2 \;\geq\; E_0 \sum_n |c_n|^2 = E_0 \, \langle \psi | \psi \rangle .

Dividing by the positive norm $\langle \psi | \psi \rangle$ gives $E[\psi] \geq E_0$ . The single inequality $E_n \geq E_0$ is the entire content of the proof — no special properties of the Hamiltonian beyond having a bounded-below spectrum are required.

How the method is used

The power of the bound comes from making the trial state depend on adjustable parameters. Write $|\psi(\alpha)\rangle$ for a family of trial states controlled by one or more parameters $\alpha$ . Then

E(\alpha) = \frac{\langle \psi(\alpha) | \hat{H} | \psi(\alpha) \rangle}{\langle \psi(\alpha) | \psi(\alpha) \rangle} \;\geq\; E_0

for every $\alpha$ . Since the inequality holds for all $\alpha$ , the minimum over $\alpha$ is still an upper bound — and it is the best one this family can produce:

E_0 \;\leq\; \min_\alpha E(\alpha).

The recipe is therefore: pick a flexible family, compute $E(\alpha)$ , and minimize. Setting $\partial E / \partial \alpha = 0$ locates the optimum. The next lessons turn this recipe into concrete estimates.

A note on what it estimates well

Even a mediocre trial state can give a surprisingly good energy estimate. The reason is that $E(\alpha)$ is stationary at the true ground state: a first-order error in the wavefunction produces only a second-order error in the energy. That is why the variational method is prized for energies — and why the same trial state may reproduce energies far better than it reproduces other observables such as position spread or transition rates.