= Quantum variational principle
{title2=$\inf\sigma(H)\leq\langle\psi,H\psi\rangle$}
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 $E_*=\inf\sigma(H)$. For $\|\psi\|=1$ in its <operator domain> $D(H)$,
$$
\langle\psi,H\psi\rangle\geq E_*.
$$
Trial states may also lie in the form domain $Q(H)=D((H-bI)^{1/2})$, where $b\leq E_*$. Then use the <quadratic form of a positive quantum Hamiltonian> shifted by $b$:
$$
q_H[\psi]=\|(H-bI)^{1/2}\psi\|^2+b\|\psi\|^2\geq E_*\|\psi\|^2.
$$
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 $E_*$, if that <eigenvalue> exists. https://www.mat.univie.ac.at/~gerald/ftp/book-schroe/schroe.pdf[Gerald Teschl, Mathematical Methods in Quantum Mechanics, Theorem 2.19 and equations (3.50)–(3.51)]{ref}.
A short proof makes the direction of the bound memorable. If $H$ has a complete <orthonormal basis> of <energy eigenstates> $\phi_n$, write
$$
\psi=\sum_n c_n\phi_n,\qquad H\phi_n=E_n\phi_n,\qquad \sum_n|c_n|^2=1.
$$
The <expectation value> is a weighted average of allowed energies:
$$
q_H[\psi]-E_*=\sum_n|c_n|^2(E_n-E_*)\geq0.
$$
Equality requires every occupied <energy eigenstate> to have energy $E_*$. https://live.ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/61bc31b8d8bf0680c322733910a71aa0_MIT8_05F13_Chap_01.pdf[Barton Zwiebach, MIT 8.05 Wave Mechanics, section 5, Variational Principle]{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: $R_H[\psi]=q_H[\psi]/\|\psi\|^2$ is the form version of the <Rayleigh quotient>. The <quantum variational principle> supplies an upper bound on $E_*$; it supplies neither a lower bound on $E_*$ nor, by itself, a guarantee that a <bound state> exists.
To see the method at work, choose the real <Gaussian wave packet>
$$
\psi_a(x)=\left(\frac a\pi\right)^{1/4}e^{-ax^2/2},\qquad a>0,\qquad x\in\mathbb R.
$$
The <Gaussian integral> gives <wavefunction normalization> and the two useful position averages:
$$
\int_{\mathbb R}|\psi_a|^2\,dx=1,\qquad \langle x^2\rangle_a=\frac1{2a},\qquad \langle x^4\rangle_a=\frac3{4a^2}.
$$
Thus its position <probability density> is a <normal distribution> with <standard deviation> $1/\sqrt{2a}$: large $a$ means narrow confinement. In units $\hbar=m=1$, <integration by parts> and $\psi_a'=-ax\psi_a$ give the <kinetic energy>
$$
\langle T\rangle_a=\frac12\int_{\mathbb R}|\psi_a'|^2\,dx=\frac{a^2}2\langle x^2\rangle_a=\frac a4.
$$
These 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 $\hbar=m=\omega=1$:
$$
H_{\mathrm{ho}}=-\frac12\frac{d^2}{dx^2}+\frac{x^2}2,\qquad E_{\mathrm{ho}}(a)=\frac a4+\frac1{4a}.
$$
The trial values at $a=1/2,1,2$ are respectively $0.625,0.5,0.625$. The <derivative> $E_{\mathrm{ho}}'(a)=(1-a^{-2})/4$ changes sign at $a=1$, where <kinetic energy> and <potential energy> are both $1/4$. Here the trial family contains an exact <ground state>:
$$
H_{\mathrm{ho}}\psi_1=\frac12\psi_1,\qquad E_0=\frac12.
$$
The <zero-point energy> remains positive even at the best width. For comparison, removing the potential leaves $E(a)=a/4\to0$ as $a\to0$: 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>.
\Image[/quantum-variational-gaussian.png]
{title=Gaussian trial widths and the kinetic–potential energy tradeoff}
{description=Normalized <Gaussian wave packets> with $a=1/2,1,2$ have different position <probability densities>. The bars show the competing <kinetic energy> $a/4$ and harmonic <potential energy> $1/(4a)$; their sum is smallest at $a=1$.}
For a pure <quartic oscillator>, keep $\hbar=m=1$ and set the quartic coefficient to one:
$$
H_4=-\frac12\frac{d^2}{dx^2}+x^4,\qquad E_4(a)=\frac a4+\frac3{4a^2}.
$$
The trial values at $a=1,2,3$ are respectively $1,0.6875,0.833333\ldots$. Setting the <derivative> to zero gives
$$
E_4'(a)=\frac14-\frac3{2a^3}=0,\qquad a_{\mathrm{opt}}=6^{1/3}\approx1.817121.
$$
The sign of this <derivative> changes from negative to positive at the optimum, giving the smallest energy within this <Gaussian wave packet> family:
$$
E_{0,4}\leq\min_{a>0}E_4(a)=\frac38\,6^{1/3}\approx0.681420.
$$
At this width, $\langle T\rangle\approx0.454280$ and $\langle V\rangle\approx0.227140$, so $2\langle T\rangle=4\langle V\rangle$, 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,
$$
\frac{H_4\psi_a}{\psi_a}=x^4-\frac{a^2x^2}2+\frac a2
$$
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.
\Image[/quantum-variational-energy-landscape.png]
{title=Energy landscapes for harmonic and quartic Gaussian trials}
{description=The <quantum harmonic oscillator> reaches its exact <ground-state energy> $1/2$ at $a=1$. The pure <quartic oscillator> has a <Gaussian wave packet> optimum at $a=6^{1/3}$ giving only the upper bound $E_{0,4}\leq0.681420$. Dashed curves separate <kinetic energy> and <potential energy>.}
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