Variational Bound

Purpose

The variational principle replaces a ground-state eigenvalue problem by an optimization problem. For a self-adjoint Hamiltonian $H$ bounded below, every nonzero admissible trial state gives the Rayleigh quotient

$$\mathcal R[\psi] = \frac{\langle\psi\rvert H\lvert\psi\rangle} {\langle\psi\vert\psi\rangle} \ge E_0,$$

where

$$E_0=\inf\sigma(H).$$

Thus $\mathcal R[\psi]$ is an upper bound on the bottom of the spectrum: $E_0\le\mathcal R[\psi]$. If $E_0$ is a discrete ground-state eigenvalue, optimizing a trial family estimates that eigenvalue and its eigenstate.

The proof and physical interpretation are at Variational Principle. This card collects the formulas, domain conditions, finite-basis form, and diagnostics needed in an actual calculation.

At a glance

Task	Formula or test
Rayleigh quotient	$\mathcal R[\psi]=\langle\psi\rvert H\lvert\psi\rangle/\langle\psi\vert\psi\rangle$
Ground-state bound	$E_0\le\mathcal R[\psi]$
Best value in a family	$E_{\mathrm{var}}=\inf_{\alpha}\mathcal R[\psi(\alpha)]\ge E_0$
Parameter stationarity	$\operatorname{Re}\langle\partial_i\psi\rvert(H-E_{\mathrm{var}})\lvert\psi\rangle=0$
Linear trial space	$Hc=ESc$
Residual	$\lvert r\rangle=(H-E_{\mathrm{var}})\lvert\psi\rangle$
Energy variance	$\sigma_H^2=\langle H^2\rangle-\langle H\rangle^2=\langle r\vert r\rangle$
Exact lower-state orthogonality	$\langle j\vert\psi\rangle=0$ for every $j<k$ implies $\mathcal R[\psi]\ge E_k$
Nested trial spaces	$\theta_k^{(N+1)}\le\theta_k^{(N)}$ and $\theta_k^{(N)}\ge E_k$

The excited-state and Ritz-value statements require the spectral hypotheses spelled out below. They are not consequences of merely finding several local minima of a nonlinear ansatz.

Statement and domain conditions

For an operator-domain calculation, take $0\ne\lvert\psi\rangle\in D(H)$. For many Schrödinger operators, the most general statement instead uses the closed quadratic form $q_H$ on its form domain $Q(H)$:

$$\mathcal R[\psi] = \frac{q_H[\psi]}{\langle\psi\vert\psi\rangle} \ge E_0, \qquad 0\ne\psi\in Q(H).$$

This distinction matters when the kinetic-energy form is finite even though $H\psi$ is not square integrable. In routine wave-mechanics calculations, an admissible trial function must at least be normalizable, satisfy the physical boundary and matching conditions appropriate to the chosen operator, and give finite terms in the energy form.

For a normalized state, the quotient reduces to

$$\mathcal R[\psi]=\langle H\rangle_\psi.$$

For an unnormalized state, retain the denominator. Multiplying the state by any nonzero scalar leaves the quotient unchanged.

If the spectrum has a normalized eigenbasis,

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

then the useful one-line check is

$$\mathcal R[\psi]-E_0 = \sum_n \lvert c_n\rvert^2(E_n-E_0) \ge0.$$

For continuous spectrum, the spectral sum is supplemented or replaced by the corresponding integral. The inequality still concerns $\inf\sigma(H)$ even when no normalizable ground state exists.

Equality and what the bound means

Equality holds precisely when the spectral support of $\lvert\psi\rangle$ lies at $E_0$. If the ground state is a nondegenerate eigenstate, this means

$$\mathcal R[\psi]=E_0 \quad\Longleftrightarrow\quad \lvert\psi\rangle=e^{i\chi}\lvert0\rangle$$

for a normalized trial state. If the ground eigenspace is degenerate, any normalized vector in that eigenspace saturates the bound. If $E_0$ is only a continuum threshold and is not an eigenvalue, no normalizable trial state need attain the infimum.

The direction of the inequality is easy to misread:

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

The calculated number lies above the exact ground energy. The principle alone does not supply a lower bound on $E_0$ or a two-sided error bar.

Parameterized trial states

For a family $\lvert\psi(\alpha)\rangle$,

$$E_{\mathrm{var}} = \inf_{\alpha\in\mathcal A} \frac{ \langle\psi(\alpha)\rvert H\lvert\psi(\alpha)\rangle }{ \langle\psi(\alpha)\vert\psi(\alpha)\rangle }.$$

For a real parameter $\alpha_i$, differentiation of the quotient gives

$$\frac{\partial\mathcal R}{\partial\alpha_i} = \frac{ 2\operatorname{Re} \langle\partial_i\psi\rvert (H-\mathcal R) \lvert\psi\rangle }{ \langle\psi\vert\psi\rangle }.$$

At an interior stationary point,

$$\operatorname{Re} \langle\partial_i\psi\rvert (H-E_{\mathrm{var}}) \lvert\psi\rangle =0$$

for every variational direction. The residual is therefore orthogonal, in the real tangent-space sense, to all allowed first-order changes of the ansatz. For complex parameters, vary a parameter and its complex conjugate independently, or separate them into real and imaginary parts.

A stationary point may be a local maximum or saddle within the chosen family. The upper-bound statement applies at every admissible point, but the best estimate requires the global infimum over the family. See Variational Parameters for normalization constraints and optimization geometry.

Compact example: quartic oscillator

Consider

$$H = \frac{P^2}{2m} + \frac12m\omega^2X^2 + gX^4, \qquad g\ge0,$$

and the normalized Gaussian

$$\psi_a(x) = \left(\frac{a}{\pi}\right)^{1/4} \exp\left(-\frac{ax^2}{2}\right), \qquad a>0.$$

The required moments are

$$\langle P^2\rangle=\frac{\hbar^2a}{2}, \qquad \langle X^2\rangle=\frac{1}{2a}, \qquad \langle X^4\rangle=\frac{3}{4a^2}.$$

Hence

$$E(a) = \frac{\hbar^2a}{4m} + \frac{m\omega^2}{4a} + \frac{3g}{4a^2} \ge E_0.$$

The stationary width obeys

$$\frac{\hbar^2}{4m} - \frac{m\omega^2}{4a^2} - \frac{3g}{2a^3} =0,$$

or

$$\frac{\hbar^2}{m}a^3 - m\omega^2a - 6g =0.$$

For $g=0$, the optimum is $a=m\omega/\hbar$, and the family contains the exact harmonic-oscillator ground state. For $g>0$, substituting the positive minimizing root gives a rigorous upper bound if the expectation values and minimization are evaluated exactly. Numerical quadrature and optimization errors must be tracked separately.

Linear trial spaces and the Ritz equations

Choose linearly independent basis vectors $\{\lvert\phi_j\rangle\}_{j=1}^N$ and write

$$\lvert\psi\rangle = \sum_{j=1}^N c_j\lvert\phi_j\rangle.$$

Define the Hamiltonian and overlap matrices by

$$H_{ij} = \langle\phi_i\rvert H\lvert\phi_j\rangle, \qquad S_{ij} = \langle\phi_i\vert\phi_j\rangle.$$

Then

$$\mathcal R(c) = \frac{c^\dagger Hc}{c^\dagger Sc},$$

and stationarity with respect to $c^\dagger$ gives the generalized Hermitian eigenvalue problem

$$Hc=\theta Sc.$$

If the basis is orthonormal, $S=I$. For a linearly independent nonorthogonal basis, $S$ is positive definite. Exact or near linear dependence makes $S$ singular or ill-conditioned; remove redundant directions or orthogonalize before trusting the Ritz values.

The smallest generalized eigenvalue is

$$\theta_0 = \min_{0\ne\psi\in\mathcal V_N} \mathcal R[\psi],$$

where $\mathcal V_N=\operatorname{span}\{\phi_1,\ldots,\phi_N\}$. The remaining stationary values are obtained with $S$-orthogonality constraints. The implementation details are collected at Rayleigh–Ritz Method.

Ritz bounds and nested bases

Suppose the exact discrete eigenvalues are ordered

$$E_0\le E_1\le E_2\le\cdots$$

with multiplicity, and the relevant levels lie below the essential spectrum. Order the $N$ Ritz values similarly:

$$\theta_0^{(N)} \le \theta_1^{(N)} \le \cdots \le \theta_{N-1}^{(N)}.$$

The min–max principle gives

$$E_k\le\theta_k^{(N)}, \qquad 0\le k<N.$$

If the spaces are nested, $\mathcal V_N\subset\mathcal V_{N+1}$, then

$$\theta_k^{(N+1)} \le \theta_k^{(N)}$$

whenever both sides exist. In particular, the best ground-state estimate decreases monotonically toward $E_0$ as a convergent nested basis is enlarged. Changing nonlinear basis parameters between runs can destroy literal nesting, so monotonicity is then not automatic.

These are upper bounds to exact eigenvalues under the stated spectral conditions, not guarantees that a finite basis has captured every physical feature of the eigenvectors. See Upper Bounds and the Min–Max Principle for the full theorem.

Excited states and symmetry sectors

A single unconstrained minimization always seeks the bottom of the accessible spectrum. To target $E_k$, impose exact orthogonality to all lower exact eigenspaces:

$$\langle j\vert\psi\rangle=0, \qquad j=0,\ldots,k-1.$$

Then

$$\mathcal R[\psi]\ge E_k.$$

In practice, exact lower eigenstates are rarely known. A Ritz calculation handles the orthogonality conditions within one common trial subspace and obtains all $\theta_k$ together. Sequentially forcing orthogonality to approximate lower states does not, by itself, establish a rigorous upper bound to the intended excited level.

An exact symmetry offers a cleaner route. Restricting the trial family to an invariant sector gives an upper bound on the lowest energy in that sector. For example, odd trial functions bound the lowest odd-parity energy when parity commutes with $H$, whether or not that state is the first excited state of the full problem.

Residual and variance diagnostics

For a normalized trial state, set

$$E_\psi=\langle\psi\rvert H\lvert\psi\rangle, \qquad \lvert r\rangle=(H-E_\psi)\lvert\psi\rangle.$$

If the required domain condition holds, then

$$\lVert r\rVert^2 = \langle\psi\rvert(H-E_\psi)^2\lvert\psi\rangle = \langle H^2\rangle_\psi-E_\psi^2 = \sigma_H^2.$$

The residual vanishes exactly for an eigenstate. A small residual means that $E_\psi$ lies near some point of the spectrum:

$$\operatorname{dist} \bigl(E_\psi,\sigma(H)\bigr) \le \lVert r\rVert.$$

It does not identify which eigenvalue is nearby and does not replace the ground-state upper bound. An excited eigenstate has zero variance too.

For a nondegenerate ground state separated by $\Delta=E_1-E_0>0$, let

$$w_{\mathrm{ex}} = 1-\lvert\langle0\vert\psi\rangle\rvert^2.$$

Spectral decomposition gives

$$E_\psi-E_0 \ge \Delta w_{\mathrm{ex}}, \qquad w_{\mathrm{ex}} \le \frac{E_\psi-E_0}{\Delta}.$$

This becomes a usable numerical state-error bound only when enough external spectral information is known. The variational estimate alone does not reveal $E_0$ or $\Delta$.

Designing a trial family

A compact physically informed family often outperforms a large generic one. Build in:

• the operator domain and boundary conditions;
• exact conserved quantum numbers and exchange symmetry;
• regularity or cusp behavior at singular points;
• the expected asymptotic decay;
• known weak- and strong-coupling limits;
• enough scale parameters to balance kinetic and potential energy.

Enlarging a family can only improve the exact infimum if the old family is contained in the new one. More parameters can nevertheless make numerical optimization and integration less reliable. Report convergence with respect to both ansatz size and numerical tolerances.

The energy is unusually forgiving: small excited-state amplitudes contribute quadratically to $E_\psi-E_0$. Other observables may have errors linear in those amplitudes. A close variational energy therefore does not certify a uniformly accurate wavefunction.

Calculation workflow

1. Specify $H$, its domain, the target symmetry sector, and the quantity being bounded.
2. Choose a nonzero trial family and retain the Rayleigh-quotient denominator unless normalization is exact.
3. Evaluate kinetic, potential, and overlap terms independently; check units, convergence, and boundary terms.
4. Minimize globally over allowed nonlinear parameters, or solve $Hc=\theta Sc$ for a linear space.
5. Verify that $S$ is positive definite on the retained basis and inspect its conditioning.
6. Recompute the energy and, when possible, the residual or energy variance at the optimum.
7. Enlarge the trial space in a nested way and check monotonic Ritz convergence.
8. Quote the result as an upper bound only to the level justified by symmetry, orthogonality, and min–max assumptions.

Common mistakes

• Reversing the inequality: the exact ground energy lies below the trial expectation.
• Omitting $\langle\psi\vert\psi\rangle$ for an unnormalized ansatz.
• Using a trial function outside the operator or form domain.
• Dropping boundary terms in an integration by parts when they do not vanish.
• Treating a stationary parameter value as the global variational minimum.
• Claiming excited-state bounds from approximate orthogonality alone.
• Interpreting a small variance as proof that the state is the ground state.
• Comparing separately optimized, nonnested spaces as though monotone convergence were guaranteed.
• Keeping nearly dependent basis functions until the overlap matrix produces spurious Ritz values.
• Reporting a value below a known exact ground energy without auditing quadrature, matrix elements, units, and boundary conditions.

For a symptom-based troubleshooting guide, see Common Variational Pitfalls.

Exercises

1. Show directly that the Rayleigh quotient is unchanged under $\lvert\psi\rangle\mapsto z\lvert\psi\rangle$ for every nonzero complex number $z$.

Solution

The numerator and denominator acquire the same factor:

$$\frac{ \langle z\psi\rvert H\lvert z\psi\rangle }{ \langle z\psi\vert z\psi\rangle } = \frac{ \lvert z\rvert^2\langle\psi\rvert H\lvert\psi\rangle }{ \lvert z\rvert^2\langle\psi\vert\psi\rangle } = \mathcal R[\psi].$$

This homogeneity is why one may optimize an unnormalized trial state with the full quotient.

2. Let $E_1>E_0$ and

$$\lvert\psi\rangle = \cos\theta\,\lvert0\rangle + e^{i\phi}\sin\theta\,\lvert1\rangle.$$

Compute $\mathcal R[\psi]$ and its residual norm.

Solution

With $\Delta=E_1-E_0$,

$$\mathcal R[\psi] = E_0+\Delta\sin^2\theta.$$

The phase $\phi$ drops out because $H$ is diagonal in this basis. The energy variance is

$$\sigma_H^2 = \Delta^2\sin^2\theta\cos^2\theta,$$

so

$$\lVert r\rVert = \Delta\lvert\sin\theta\cos\theta\rvert.$$

The residual vanishes at both eigenstates, $\theta=0$ and $\theta=\pi/2$, illustrating why zero variance does not identify the ground state.

3. For the quartic-oscillator Gaussian above, recover the exact harmonic result when $g=0$ and explain why the result for $g>0$ remains an upper bound.

Solution

At $g=0$, stationarity gives

$$a=\frac{m\omega}{\hbar}.$$

Substitution yields

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

The Gaussian with this width is the exact harmonic-oscillator ground state, so the bound is saturated. For $g>0$, every normalized Gaussian remains in the form domain of the confining Hamiltonian and its Rayleigh quotient is at least $E_0$. Minimizing over $a>0$ preserves that inequality even though the exact quartic ground state is not Gaussian.

4. Starting from $\mathcal R(c)=c^\dagger Hc/(c^\dagger Sc)$, derive the generalized Ritz equation by varying $c^\dagger$.

Solution

At a stationary point with $\theta=\mathcal R(c)$,

$$\delta_{c^\dagger} \left( c^\dagger Hc - \theta c^\dagger Sc \right) =0.$$

Because $\delta c^\dagger$ is arbitrary,

$$(H-\theta S)c=0,$$

or

$$Hc=\theta Sc.$$

If $S$ is positive definite, the eigenvectors can be chosen $S$-orthonormal. If $S$ is singular, the coefficients contain a redundant direction and the basis must first be reduced.

Canonical links

• Variational Principle
• Variational Parameters
• Rayleigh–Ritz Method
• Upper Bounds and the Min–Max Principle
• Harmonic-Oscillator Variational Estimate

References

• M. Reed and B. Simon, Methods of Modern Mathematical Physics IV: Analysis of Operators, Academic Press, 1978, Sec. XIII.1.
• T. Kato, Perturbation Theory for Linear Operators, 2nd ed., Springer, 1976, Ch. VI.
• R. Courant and D. Hilbert, Methods of Mathematical Physics, Vol. 1, Wiley-Interscience, 1953, Ch. VI.
• A. Messiah, Quantum Mechanics, Vol. II, North-Holland, 1962, Ch. XVI.
• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Vol. II, Wiley, 1977, Complement E XI.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994, Ch. 16.