Variational Estimate for the Harmonic Oscillator

The harmonic oscillator is an unusually transparent test of the variational method. Its exact ground state and energy are already known, so every step of a trial-state calculation can be checked: admissibility, normalization, expectation values, optimization, the upper bound, and the quality of the optimized state.

The calculation also succeeds in the strongest possible way. A one-parameter Gaussian family contains the exact ground state, and minimizing its width recovers both the oscillator length and the zero-point energy. That success is a property of this well-matched ansatz, not a general promise that a one-parameter variational calculation will be exact.

Required background. The Variational Principle gives the upper bound and Trial Wavefunctions explains admissibility. Quantum Harmonic Oscillator supplies the exact spectrum for comparison.

Oscillator energy and natural length

Consider a particle of mass $m$ in one dimension with

$$H = \frac{P^2}{2m} + \frac12m\omega^2X^2.$$

The parameters satisfy $m\gt0$ and $\omega\gt0$.

In the position representation,

$$P = -i\hbar\frac{d}{dx},$$

so the energy quadratic form is

$$q_H[\psi] ={} \frac{\hbar^2}{2m} \int_{-\infty}^{\infty} \lvert\psi'(x)\rvert^2\,dx + \frac12m\omega^2 \int_{-\infty}^{\infty} x^2\lvert\psi(x)\rvert^2\,dx.$$

The competing terms identify a natural length,

$$\ell = \sqrt{ \frac{\hbar}{m\omega} }.$$

We will recover this scale by optimization rather than insert it into the trial state from the start.

Gaussian Trial State

Choose the normalized, real, even family

$$\psi_b(x) = \frac{1}{(\pi b^2)^{1/4}} \exp\left( -\frac{x^2}{2b^2} \right).$$

The width parameter is restricted to $b\gt0$.

The parameter $b$ controls the wavefunction width. It is not the position standard deviation. Since

$$\lvert\psi_b(x)\rvert^2 = \frac{1}{\sqrt{\pi}b} \exp\left( -\frac{x^2}{b^2} \right),$$

the variance will be $b^2/2$.

This family is admissible for every $b\gt0$: each state is square-integrable, smooth, even, and has finite kinetic and potential energy. The positivity restriction prevents a redundant sign convention and is naturally enforced by a logarithmic parameter such as $b=b_{\mathrm{ref}}e^\eta$.

The same family is often written using an inverse-square width,

$$\alpha = \frac{1}{b^2},$$

as

$$\psi_\alpha(x) = \left( \frac{\alpha}{\pi} \right)^{1/4} e^{-\alpha x^2/2}.$$

Both parameterizations describe exactly the same set of rays. Mixing their meanings is a common source of reciprocal-width errors.

Normalization and Moments

The required Gaussian integrals are

$$\int_{-\infty}^{\infty} e^{-x^2/b^2}\,dx = \sqrt{\pi}b,$$ $$\int_{-\infty}^{\infty} x^2e^{-x^2/b^2}\,dx = \frac{\sqrt{\pi}}{2}b^3,$$

and

$$\int_{-\infty}^{\infty} x^4e^{-x^2/b^2}\,dx = \frac{3\sqrt{\pi}}{4}b^5.$$

The first confirms

$$\int_{-\infty}^{\infty} \lvert\psi_b(x)\rvert^2\,dx = 1.$$

Parity gives

$$\langle X\rangle_b = 0,$$

and the second moment is

$$\langle X^2\rangle_b = \frac{b^2}{2}.$$

Thus

$$(\Delta X)_b = \frac{b}{\sqrt2}.$$

The fourth moment,

$$\langle X^4\rangle_b = \frac{3b^4}{4},$$

will later provide an energy-variance check.

Kinetic-Energy Expectation

Differentiate the trial state:

$$\psi_b'(x) = -\frac{x}{b^2}\psi_b(x),$$ $$\psi_b''(x) = \left( \frac{x^2}{b^4} - \frac{1}{b^2} \right) \psi_b(x).$$

Direct substitution gives

$$\begin{aligned} \langle T\rangle_b &= -\frac{\hbar^2}{2m} \int_{-\infty}^{\infty} \psi_b(x)\psi_b''(x)\,dx \\ &= -\frac{\hbar^2}{2m} \left( \frac{\langle X^2\rangle_b}{b^4} - \frac{1}{b^2} \right) \\ &= \frac{\hbar^2}{4mb^2}. \end{aligned}$$

Integration by parts supplies a useful check. The Gaussian and its derivative vanish at infinity, so

$$\langle T\rangle_b = \frac{\hbar^2}{2m} \int_{-\infty}^{\infty} \lvert\psi_b'(x)\rvert^2\,dx = \frac{\hbar^2}{4mb^2}.$$

This form makes positivity explicit. Narrowing the state increases its momentum gradients and makes the kinetic energy diverge as $b^{-2}$.

The corresponding momentum moments are

$$\langle P\rangle_b = 0, \qquad \langle P^2\rangle_b = \frac{\hbar^2}{2b^2},$$

and therefore

$$(\Delta P)_b = \frac{\hbar}{\sqrt2b}.$$

Every member of the family saturates

$$(\Delta X)_b(\Delta P)_b = \frac{\hbar}{2}.$$

Minimum uncertainty alone does not select the ground-state width; the Hamiltonian determines which minimum-uncertainty Gaussian has the lowest energy.

Potential-Energy Expectation

The quadratic potential requires only $\langle X^2\rangle_b$:

$$\begin{aligned} \langle V\rangle_b &= \frac12m\omega^2 \langle X^2\rangle_b \\ &= \frac{m\omega^2b^2}{4}. \end{aligned}$$

The potential term has the opposite width dependence from the kinetic term. A broad state reduces gradients but samples larger values of $x^2$.

Energy Functional

Adding the two contributions gives

$$E(b) = \langle\psi_b\rvert H\lvert\psi_b\rangle = \frac{\hbar^2}{4mb^2} + \frac{m\omega^2b^2}{4}.$$

The upper-bound theorem guarantees

$$E(b) \geq E_0$$

for every $b\gt0$, where $E_0$ is the true ground-state energy. It does not yet tell us which width gives the best bound.

Introduce the dimensionless width ratio

$$s = \frac{b}{\ell}.$$

Using $\ell^2=\hbar/(m\omega)$, the energy becomes

$$\varepsilon(s) \equiv \frac{E(b)}{\hbar\omega} = \frac{1}{4s^2} + \frac{s^2}{4}.$$

All dimensions have disappeared. The two terms are the kinetic and potential contributions in units of $\hbar\omega$.

The kinetic contribution falls as $s^{-2}$ while the potential contribution rises as $s^2$. Their sum has its unique minimum at $s=b/\ell=1$, where the two contributions are equal and the Gaussian coincides with the exact ground state.

The limiting behavior is physically sensible:

Width regime	Kinetic energy	Potential energy	Total energy
$s\to0$	diverges	tends to zero	diverges
$s=1$	$\hbar\omega/4$	$\hbar\omega/4$	$\hbar\omega/2$
$s\to\infty$	tends to zero	diverges	diverges

Optimization

Differentiating with respect to the dimensional width gives

$$\frac{dE}{db} = -\frac{\hbar^2}{2mb^3} + \frac12m\omega^2b.$$

The stationary equation is

$$m\omega^2b = \frac{\hbar^2}{mb^3},$$

or

$$b^4 = \frac{\hbar^2}{m^2\omega^2} = \ell^4.$$

Because $b\gt0$, the only stationary width is

$$b_\star = \ell.$$

The second derivative is

$$\frac{d^2E}{db^2} = \frac{3\hbar^2}{2mb^4} + \frac12m\omega^2,$$

which is positive for every $b\gt0$. In particular,

$$\left. \frac{d^2E}{db^2} \right|_{b=\ell} = 2m\omega^2 \gt0.$$

The stationary point is therefore the unique global minimum. Substitution gives

$$E_{\mathrm{var}} = E(\ell) = \frac12\hbar\omega.$$

Optimization without calculus

The dimensionless expression can also be rearranged as

$$\begin{aligned} \varepsilon(s) &= \frac12 + \frac14 \left( s- \frac1s \right)^2 \\ &\geq \frac12. \end{aligned}$$

Equality holds only at $s=1$. This is the arithmetic–geometric-mean inequality in a form that displays the nonnegative energy excess.

Logarithmic width

For numerical work, write

$$s = e^\eta, \qquad \eta\in\mathbb R.$$

Then positivity is automatic and

$$\varepsilon(\eta) = \frac12\cosh(2\eta).$$

The optimum is $\eta_\star=0$. Near it,

$$\varepsilon(\eta) = \frac12 + \eta^2 + O(\eta^4).$$

The absence of a linear term is the local stationarity expected at a variational optimum. General parameter gradients, Hessians, and reparameterizations are developed in Variational Parameters.

Why the Bound Becomes Exact

At the optimum,

$$\psi_{b_\star}(x) = \frac{1}{(\pi\ell^2)^{1/4}} \exp\left( -\frac{x^2}{2\ell^2} \right),$$

which is the exact normalized oscillator ground state. The variational family happened to contain the target eigenvector.

This can be checked without importing the known spectrum. Acting with the Hamiltonian on a general family member gives the local-energy expression

$$\frac{H\psi_b(x)}{\psi_b(x)} = \frac{\hbar^2}{2mb^2} + \frac12 \left( m\omega^2 - \frac{\hbar^2}{mb^4} \right) x^2.$$

At $b=\ell$, the coefficient of $x^2$ vanishes and

$$H\psi_\ell = \frac12\hbar\omega\,\psi_\ell.$$

Thus the optimized state satisfies the full eigenvalue equation. To establish that it is the ground state without assuming the known spectrum, use $H=\hbar\omega(a^\dagger a+1/2)\ge\hbar\omega/2$; the normalized Gaussian attains this lower bound. Zero residual by itself would identify an eigenstate, not its index.

For $b\ne\ell$, the local energy depends on $x$, so the state is not an eigenstate. Its energy variance is

$$\begin{aligned} (\Delta H)_b^2 &= \langle H^2\rangle_b - E(b)^2 \\ &= \frac{(\hbar\omega)^2}{8} \left( s^2-s^{-2} \right)^2. \end{aligned}$$

It vanishes only at $s=1$. For a normalized state in the operator domain, this variance is also the squared residual norm,

$$(\Delta H)_b^2 = \left\lVert (H-E(b))\psi_b \right\rVert^2.$$

Energy minimization and residual vanishing therefore agree at the optimum in this example. In a restricted family that does not contain an eigenstate, the minimum energy generally has a nonzero residual outside the trial tangent space.

The Variational Monte Carlo Preview recasts this same local energy as a sampled estimator and explains how its variance controls both residual diagnostics and statistical uncertainty.

Exact Comparison and Error Measures

The exact ground-state energy is

$$E_0 = \frac12\hbar\omega,$$

so the variational excess at arbitrary width is

$$E(b)-E_0 = \frac{\hbar\omega}{4} \left( s- \frac1s \right)^2.$$

In the logarithmic coordinate,

$$\frac{E(b)-E_0}{\hbar\omega} = \sinh^2\eta.$$

The squared overlap with the exact ground state is

$$\begin{aligned} F(s) &= \left\lvert \langle\psi_\ell\vert\psi_b\rangle \right\rvert^2 \\ &= \frac{2s}{1+s^2} \\ &= \frac{1}{\cosh\eta}. \end{aligned}$$

Near the optimum,

$$1-F = \frac{\eta^2}{2} + O(\eta^4),$$

while

$$\frac{E-E_0}{\hbar\omega} = \eta^2 + O(\eta^4).$$

Both are second order here, but they quantify different objects. In more complicated variational families, a small energy error need not imply that every observable or wavefunction feature is comparably accurate.

Virial and Uncertainty Interpretations

At the stationary width,

$$\langle T\rangle_\star = \langle V\rangle_\star = \frac14\hbar\omega.$$

This is the harmonic-oscillator virial relation. In the present one-parameter family, changing $b$ is a scale transformation: the kinetic term scales as $b^{-2}$ and the quadratic potential as $b^2$. Stationarity under that scaling forces the two terms to balance.

Every trial Gaussian also saturates the position–momentum uncertainty relation. The optimizer chooses the member whose position and momentum spreads balance the coefficients in the Hamiltonian. The broader physical interpretation of this balance is developed in Zero-Point Energy.

A wrong-width centered Gaussian is a squeezed vacuum relative to the oscillator’s natural quadratures. With $s=e^\eta$, its energy is

$$E = \frac12\hbar\omega\cosh(2\eta).$$

The sign convention for the squeezing parameter may reverse $\eta$, but the energy is unchanged. The state interpretation is developed in Squeezed States: First Encounter.

Adding Center and Mean Momentum

The centered real ansatz already respects the parity and time-reversal properties of the nondegenerate ground state. To see what those choices save, enlarge the family to

$$\psi_{b,x_c,p_c}(x) ={} \frac{1}{(\pi b^2)^{1/4}} \times \exp\left[ -\frac{(x-x_c)^2}{2b^2} \right] \times \exp\left[ \frac{i p_c(x-x_c)}{\hbar} \right].$$

Its moments are

$$\langle X\rangle = x_c, \qquad \langle P\rangle = p_c,$$ $$\langle X^2\rangle = x_c^2 + \frac{b^2}{2},$$

and

$$\langle P^2\rangle = p_c^2 + \frac{\hbar^2}{2b^2}.$$

Therefore

$$E(b,x_c,p_c) ={} \frac{\hbar^2}{4mb^2} + \frac{m\omega^2b^2}{4} + \frac{p_c^2}{2m} + \frac12m\omega^2x_c^2.$$

Every added term is nonnegative. The optimizer returns

$$x_{c,\star} = 0, \qquad p_{c,\star} = 0, \qquad b_\star = \ell.$$

Building exact symmetries into the ansatz removed two parameters whose optimum was known in advance. Displacements and phase-space motion become useful for different targets, such as coherent states and displaced oscillators, but they do not improve this ground-state calculation.

A Deliberately Imperfect Trial Family

Exact recovery can hide the distinction between the theorem and the ansatz. Consider instead the normalized exponential

$$\chi_a(x) = \frac{1}{\sqrt a} e^{-\lvert x\rvert/a}, \qquad a\gt0.$$

This state has the correct even parity and is in the Hamiltonian’s quadratic-form domain, but its cusp and exponential tail do not match the smooth Gaussian oscillator ground state.

Using the form expression for kinetic energy,

$$\int_{-\infty}^{\infty} \lvert\chi_a'(x)\rvert^2\,dx = \frac{1}{a^2},$$

and

$$\langle X^2\rangle_\chi = \frac{a^2}{2}.$$

Its variational energy is

$$E_\chi(a) = \frac{\hbar^2}{2ma^2} + \frac{m\omega^2a^2}{4}.$$

For $t=a/\ell$,

$$\frac{E_\chi(t)}{\hbar\omega} = \frac{1}{2t^2} + \frac{t^2}{4}.$$

Stationarity gives

$$t_\star^4 = 2, \qquad t_\star = 2^{1/4},$$

and hence

$$E_{\chi,\star} = \frac{\hbar\omega}{\sqrt2} \gt \frac12\hbar\omega.$$

This is a valid but non-exact upper bound. The optimizer has found the best state in the exponential family; it cannot remove the family’s structural mismatch.

The cusp also illustrates why the form domain matters. A naive second derivative of $e^{-\lvert x\rvert/a}$ misses a delta distribution at the origin. The positive quadratic-form expression using the weak first derivative gives the correct variational kinetic energy without pretending that $\chi_a$ lies in the strong operator domain.

Gaussian Variational Methods extends the calculation to coupled coordinates and correlated bases. Anharmonic Oscillator by Variational Methods shows what remains of the width optimization when no Gaussian is exact.

Exercises

Check the energy variance

Use the local energy and the Gaussian moments to derive

$$(\Delta H)_b^2 = \frac{(\hbar\omega)^2}{8} \left( s^2-s^{-2} \right)^2.$$

Why does its vanishing give stronger information than $dE/db=0$?

Solution

Write the local energy as

$$E_L(x) = A+Bx^2,$$

where

$$\begin{aligned} A &= \frac{\hbar^2}{2mb^2}, \\ B &= \frac12 \left( m\omega^2 - \frac{\hbar^2}{mb^4} \right). \end{aligned}$$

Only the $Bx^2$ term fluctuates. Since

$$\operatorname{Var}(X^2) = \langle X^4\rangle - \langle X^2\rangle^2 = \frac{b^4}{2},$$

we obtain

$$(\Delta H)_b^2 = B^2\frac{b^4}{2}.$$

Substituting $b=s\ell$ and $\ell^2=\hbar/(m\omega)$ yields the stated expression. It vanishes only when $s=1$.

The condition $dE/db=0$ means that the residual is orthogonal to the single tangent direction generated by changing $b$. Zero variance means the full residual vanishes, so the state is an exact eigenstate.

Extend to an isotropic oscillator in d dimensions

For

$$H = \sum_{j=1}^{d} \left( \frac{P_j^2}{2m} + \frac12m\omega^2X_j^2 \right),$$

use the product Gaussian

$$\Psi_b(\mathbf x) = \prod_{j=1}^{d} \psi_b(x_j).$$

Find the optimized width and energy.

Solution

Each Cartesian factor contributes independently,

$$E_1(b) = \frac{\hbar^2}{4mb^2} + \frac{m\omega^2b^2}{4}.$$

Therefore

$$E_d(b) = dE_1(b).$$

Multiplication by the positive constant $d$ does not change the minimizing width, so

$$b_\star = \ell.$$

The optimized energy is

$$E_{d,\star} = \frac{d}{2}\hbar\omega.$$

The product family contains the exact isotropic-oscillator ground state, so the bound is saturated.

An observable with a first-order error

Set $b=\ell$, $p_c=0$, and $x_c=\epsilon\ell$. Compare the energy error, the error in $\langle X\rangle$, and the ground-state fidelity for small $\epsilon$.

Solution

$E-E_0=\hbar\omega\epsilon^2/2$ while $\langle X\rangle=\epsilon\ell$ is first order. Gaussian integration gives $F=e^{-\epsilon^2/2}=1-\epsilon^2/2+O(\epsilon^4)$. Small energy error or large fidelity does not make every observable’s error second order.

References

• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Vol. 1, 2nd ed. (Wiley-VCH, 2020), for Gaussian wave packets, the oscillator, and variational reasoning.
• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed. (Cambridge University Press, 2018), for the oscillator and elementary variational estimates.
• M. Reed and B. Simon, Methods of Modern Mathematical Physics, Vol. IV: Analysis of Operators (Academic Press, 1978), for the min–max principle and quadratic-form formulation. Bibliographic record.
• W. Ritz, “Über eine neue Methode zur Lösung gewisser Variationsprobleme der mathematischen Physik”, Journal für die reine und angewandte Mathematik 135, 1–61 (1909), for the historical variational method.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed. (Springer, 1994), chapters on the variational principle and harmonic oscillator.