Quantum variational principle ($\inf \sigma (H)\le ⟨\psi ,H\psi ⟩$)

The quantum variational principle estimates the lowest energy by choosing a trial wavefunction and computing its expectation value, without solving the Schrödinger equation exactly. Minimizing over a manageable family of trial wavefunctions gives an upper bound; enlarging that family can improve the estimate.

Let $H$ be a self-adjoint operator bounded below, and put $E_{\ast }=\inf \sigma (H)$. For $\Vert \psi \Vert =1$ in its operator domain $D(H)$,Trial states may also lie in the form domain $Q(H)=D((H-bI{)}^{1/2})$, where $b\le E_{\ast }$. Then use the quadratic form of a positive quantum Hamiltonian shifted by $b$:This extends the expectation value beyond $D(H)$ without requiring $H\psi$ to exist. Equality for a normalized state holds exactly in the energy eigenspace at $E_{\ast }$, if that eigenvalue exists. ^[ref].

A short proof makes the direction of the bound memorable. If $H$ has a complete orthonormal basis of energy eigenstates ${\varphi }_n$, writeThe expectation value is a weighted average of allowed energies:Equality requires every occupied energy eigenstate to have energy $E_{\ast }$. ^[ref]. With continuous spectrum, the spectral theorem replaces the sum by an integral against the scalar spectral measure; the same nonnegative-difference argument works.

For an unnormalized trial wavefunction, divide by its squared norm: $R_H[\psi ]=q_H[\psi ]/\Vert \psi {\Vert }^2$ is the form version of the Rayleigh quotient. The quantum variational principle supplies an upper bound on $E_{\ast }$; it supplies neither a lower bound on $E_{\ast }$ nor, by itself, a guarantee that a bound state exists.

To see the method at work, choose the real Gaussian wave packetThe Gaussian integral gives wavefunction normalization and the two useful position averages:Thus its position probability density is a normal distribution with standard deviation $1/\sqrt{2a}$: large $a$ means narrow confinement. In units $\hslash =m=1$, integration by parts and ${\psi }_a^{\prime }=-ax{\psi }_a$ give the kinetic energyThese smooth, rapidly decaying trial wavefunctions belong to the operator domains of both polynomial Quantum Hamiltonians below. Narrowing the Gaussian wave packet increases its kinetic energy; widening it increases the potential energy in either confining potential.

For the quantum harmonic oscillator, take $\hslash =m=\omega =1$:The trial values at $a=1/2,1,2$ are respectively $0.625,0.5,0.625$. The derivative $E_{\mathrm{ho}}^{\prime }(a)=(1-a^{-2})/4$ changes sign at $a=1$, where kinetic energy and potential energy are both $1/4$. Here the trial family contains an exact ground state:The zero-point energy remains positive even at the best width. For comparison, removing the potential leaves $E(a)=a/4\to 0$ as $a\to 0$: the free Quantum Hamiltonian on the line has spectral bottom zero but no normalizable zero-energy energy eigenstate. A sequence of ever broader trial wavefunctions need not approach an actual bound state.

For a pure quartic oscillator, keep $\hslash =m=1$ and set the quartic coefficient to one:The trial values at $a=1,2,3$ are respectively $1,0.6875,0.833333\dots$. Setting the derivative to zero givesThe sign of this derivative changes from negative to positive at the optimum, giving the smallest energy within this Gaussian wave packet family:At this width, $⟨T⟩\approx 0.454280$ and $⟨V⟩\approx 0.227140$, so $2⟨T⟩=4⟨V⟩$, as expected from the power-law quantum virial theorem for a scale-optimized trial wavefunction. The number $0.681420$ is an upper estimate of the ground-state energy, not an asserted exact energy eigenvalue. Indeed,depends on $x$ for every $a>0$, so no member of this trial family is an energy eigenstate. Nevertheless, a single width parameter already turns a difficult Schrödinger equation into a useful, explicit energy estimate.

For $\kappa >0$ and the radial exponential trial state $e^{-\alpha r}$, $\alpha >0$, put $a={\hslash }^2/(m\kappa )$. Direct radial integration gives the Rayleigh quotientThe kinetic expectation is ${\hslash }^2{\alpha }^2/(2m)$; the potential expectation follows by splitting its radial integral at $R$. The common prefactor cancels from the stationary condition, giving $a\alpha =(1+2\alpha R)e^{-2\alpha R}$. Its positive solution is unique, since the left side increases and the right side decreases. For $\epsilon =R/a\to 0$, it satisfies $a\alpha =1-2{\epsilon }^2+O({\epsilon }^3)$.

The quantum variational principle makes $E(\alpha )$ an upper bound for the ground energy. The explicit choice $\alpha =1/a$ gives $E_0(R)\le (\kappa /a)[1/2-\{1-(1+\epsilon )e^{-2\epsilon }\}/\epsilon ]$, whose first correction to $-\kappa /(2a)$ is $2\kappa R^2/(3a^3)$. The true energy is no lower than the point-Coulomb value because capping makes the potential less attractive.

Unrestricted minimization of a Rayleigh quotient seeks the ground state. To bound the $k$th excited energy from above, restrict normalized trial wavefunctions to the orthogonal complement of the exact lower-energy eigenspaces. In an exact orthonormal eigenbasis, all remaining expansion coefficients multiply energies at least $E_k$, so their weighted average is at least $E_k$. A known symmetry sector can supply some of these exclusions automatically. Orthogonality only to inaccurate approximate lower states does not automatically retain the same rigorous excited-state upper bound.

An admissible wavefunction chosen from a tractable family for the variational method. Its Hamiltonian expectation gives an upper bound on the ground-state energy; parameters can be adjusted to lower that bound.

For $V=-A{e}^{-\mu r}/r$, the normalized exponential with scale $\alpha$ gives $E(\alpha )={\hslash }^2{\alpha }^2/(2m)-4A{\alpha }^3/(\mu +2\alpha {)}^2$. Choosing $\alpha =\mu /2$ proves binding if $\mu <Am/{\hslash }^2$. Failure to obtain a negative trial expectation outside this range does not prove absence of a bound state. At equality a small positive admixture of the scale-$\mu$ exponential has a negative cross matrix element and proves binding too.

For a bound stationary state in a homogeneous potential $V(\lambda x)={\lambda }^{n}V(x)$, scaling a normalized wavefunction gives stationarity of ${\lambda }^2⟨T⟩+{\lambda }^{-n}⟨V⟩$ at $\lambda =1$. Hence $2⟨T⟩=n⟨V⟩$.