Trial Wavefunctions

A trial wavefunction is a guessed state with one or more adjustable parameters. Choosing a good family is the art of the variational method; optimizing the parameters is the mechanics. This lesson walks through both for the cleanest possible example, the one-dimensional harmonic oscillator.

Choosing a family

A useful trial family should satisfy three things: it must be normalizable, it should respect the symmetries and boundary behaviour of the problem, and it should contain at least one knob to turn. For a smooth, symmetric, confining potential like the harmonic oscillator

\hat{H} = \frac{\hat{p}^2}{2m} + \frac{1}{2} m \omega^2 \hat{x}^2,

a natural guess is a single Gaussian centred at the origin,

\psi_\alpha(x) = e^{-\alpha x^2}, \qquad \alpha > 0.

It is square-integrable, even in $x$ (matching the symmetric potential), and decays at infinity. Its one parameter $\alpha$ controls the width: large $\alpha$ is a narrow spike, small $\alpha$ is a broad spread.

Computing the energy

With Gaussian integrals the kinetic and potential expectation values work out (in natural units $\hbar = m = \omega = 1$ , after normalizing) to

\langle \hat{T} \rangle = \frac{\alpha}{2}, \qquad \langle \hat{V} \rangle = \frac{1}{8\alpha},

so the variational energy is

E(\alpha) = \frac{\alpha}{2} + \frac{1}{8\alpha}.

This captures the universal tension behind every confined ground state: shrinking the wavefunction (large $\alpha$ ) lowers the potential energy but, by the uncertainty principle, raises the kinetic energy, and vice versa. The optimum balances the two.

Optimizing

Minimize over $\alpha$ by setting the derivative to zero:

\frac{dE}{d\alpha} = \frac{1}{2} - \frac{1}{8\alpha^2} = 0 \;\;\Longrightarrow\;\; \alpha^* = \frac{1}{2}.

Evaluating there,

E(\alpha^*) = \frac{1}{2}\cdot\frac{1}{2} + \frac{1}{8 \cdot \tfrac{1}{2}} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}.

In these units the exact ground-state energy is $\tfrac{1}{2}\hbar\omega = \tfrac{1}{2}$ . The Gaussian trial state nails it exactly — because the true oscillator ground state is a Gaussian, so the family contains the right answer. This is the best case: when the trial family includes the exact solution, the variational minimum is the exact energy.

Try it

Minimize $E(\alpha) = \alpha/2 + 1/(8\alpha)$ over $\alpha > 0$ and return the minimum energy. Your code should return a single number.

Run your code to see the quantum state.