Variational Principle

The variational principle turns the ground-state problem into an optimization problem. Instead of solving the eigenvalue equation exactly, choose a family of trial states and minimize the energy expectation value.

Required background. Expectation Values and Spectral Decomposition explain how a state’s energy distribution determines its mean.

For a self-adjoint Hamiltonian $H$ bounded below, let $E_0=\inf\sigma(H)$ be the bottom of its spectrum. The energy quotient satisfies

$$E[\psi] = \frac{\langle\psi|H|\psi\rangle}{\langle\psi|\psi\rangle} \ge E_0$$

for every nonzero trial state $\lvert\psi\rangle$ in the domain of $H$. Thus every admissible trial state gives an upper bound on the spectral bottom, whether or not a normalizable ground state exists.

Finite-dimensional minimization leads to the Rayleigh–Ritz method.

The energy quotient and its domain

Let $H$ be a self-adjoint Hamiltonian with a lowest spectral value $E_0$. If a complete orthonormal eigenbasis exists, the familiar discrete statement is especially simple:

$$H\lvert n\rangle=E_n\lvert n\rangle, \qquad E_0\le E_1\le E_2\le\cdots.$$

For any normalized state

$$\lvert\psi\rangle = \sum_n c_n\lvert n\rangle, \qquad \sum_n |c_n|^2=1,$$

the energy expectation value obeys $E[\psi]\ge E_0$.

More generally, use the closed quadratic form $q_H$ associated with $H$. If $H\ge aI$, its form domain is $Q(H)=D((H-aI)^{1/2})$, and

$$q_H[\psi]=a\|\psi\|^2+\|(H-aI)^{1/2}\psi\|^2, \qquad E[\psi]=\frac{q_H[\psi]}{\|\psi\|^2}.$$

This agrees with $\langle\psi|H|\psi\rangle$ on the operator domain $D(H)$, but allows more trial functions. For a Dirichlet Schrödinger problem, continuous piecewise-linear functions with square-integrable weak derivative and zero boundary values can have a valid kinetic form $\hbar^2\int|\psi'|^2/(2m)$ even though $H\psi$ is not square-integrable. The potential form must also be well defined. Merely normalizing a function is insufficient.

Proof in an Eigenbasis

Using the spectral expansion,

$$\begin{aligned} E[\psi] &= \langle\psi|H|\psi\rangle \\ &= \sum_n |c_n|^2E_n. \end{aligned}$$

Since each $E_n\ge E_0$,

$$\sum_n |c_n|^2E_n \ge E_0\sum_n |c_n|^2 = E_0.$$

For continuous as well as discrete spectra, the spectral theorem replaces the sum by $E[\psi]=\int E\,d\mu_\psi(E)$, where the normalized spectral measure has support in $[E_0,\infty)$. The same inequality follows. For an unnormalized state divide by its squared norm.

Equality holds only when $\lvert\psi\rangle$ has support entirely inside the ground-state eigenspace. If the ground state is nondegenerate, equality means the trial state is the exact ground state up to an overall phase.

Variational Families

A practical calculation chooses a family of normalized or normalizable trial states $\lvert\psi(\alpha)\rangle$ depending on parameters $\alpha$. The variational estimate is

$$E_{\mathrm{var}} = \inf_\alpha \frac{ \langle\psi(\alpha)|H|\psi(\alpha)\rangle }{ \langle\psi(\alpha)|\psi(\alpha)\rangle }.$$

The result still satisfies

$$E_{\mathrm{var}}\ge E_0.$$

Enlarging the trial family cannot raise its infimum, provided every old state remains available. An infimum need not be attained, and a numerical local minimum need not find it. This is the basic reason variational calculations can be systematically improved, though systematic improvement of the energy does not automatically imply equally good improvement of every observable.

The derivative and tangent-space conditions used to find that optimum are derived in Variational Parameters.

Physical Interpretation

A trial state has some overlap with the true ground state and some overlap with excited states. Excited-state contamination raises the average energy because excited components carry energies larger than $E_0$:

$$E[\psi]-E_0 = \sum_n |c_n|^2(E_n-E_0).$$

The variational method therefore tries to suppress high-energy features. In wave mechanics this often means balancing localization against gradients: squeezing a wavefunction may reduce potential energy, but it increases kinetic energy.

The method is especially effective when the trial family builds in known qualitative information:

• boundary conditions,
• symmetry and parity,
• regular behavior near singular points,
• correct decay at large distance,
• cusp or short-distance structure when known,
• limiting behavior in exactly solvable regimes.

Localization and kinetic cost

For a normalized Gaussian $\psi_a(x)=(a/\pi)^{1/4}e^{-ax^2/2}$ in a harmonic oscillator,

$$E(a)=\frac{\hbar^2a}{4m}+\frac{m\omega^2}{4a}\ge\frac{\hbar\omega}{2}.$$

The first term penalizes narrow localization; the second penalizes large spatial extent. Their balance at $a=m\omega/\hbar$ saturates the bound because the family contains the exact ground state. Variational Estimate for the Harmonic Oscillator derives the integrals and compares a less suitable trial family.

Energy Accuracy Versus State Accuracy

The variational energy can be accurate even when the trial wavefunction misses some details. Suppose the exact ground state is nondegenerate and

$$\lvert\psi\rangle = c_0\lvert0\rangle + \sum_{n>0}c_n\lvert n\rangle.$$

Then

$$E[\psi]-E_0 = \sum_{n>0}|c_n|^2(E_n-E_0).$$

The energy error is second order in small excited-state amplitudes. By contrast, expectation values of other observables can have first-order errors if the observable connects the ground state to the excited components. A good variational energy is therefore not by itself proof of a uniformly good wavefunction.

Exercises

What a gap says about state accuracy

Suppose the ground state is nondegenerate and separated from all other spectral values by $\Delta>0$. For normalized $\psi$, prove $1-|\langle0|\psi\rangle|^2\le(E[\psi]-E_0)/\Delta$. Why is the gap assumption essential?

Solution

Every component orthogonal to the ground state costs at least $\Delta$, so $E[\psi]-E_0\ge\Delta\|(I-|0\rangle\langle0|)\psi\|^2$. Without a gap, spectral weight arbitrarily close to $E_0$ can have small energy cost while remaining orthogonal to the ground state. Applying the bound numerically also requires a justified estimate of $E_0$ and the gap.

An infimum without a ground state

For a free particle on the line, evaluate the Gaussian energy and send $a\to0^+$. Does the limiting normalized ground state exist?

Solution

$E(a)=\hbar^2a/(4m)\to0$, the bottom of the free spectrum. The Gaussian spreads without limit; there is no normalized constant wavefunction on the line. The infimum is zero but is not attained by a normalizable state.

References

• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.
• M. Reed and B. Simon, Methods of Modern Mathematical Physics IV: Analysis of Operators, Academic Press, 1978.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.