A good trial state is mostly good judgement: give it the symmetry the true state must have, make it obey the right conditions at walls and at infinity, and leave a few numbers free. For a box, a simple parabola that vanishes at both walls gives the ground energy to about one per cent. For more accuracy, take the trial state to be a combination of several fixed functions with free coefficients: minimising then becomes a matrix eigenvalue problem, the Rayleigh–Ritz method, and its lowest eigenvalue is the best bound the functions allow.

A tailor measures a few numbers — chest, waist, length — and cuts a suit that fits almost perfectly. A trial state with a few well-chosen parameters fits the ground state the same way.

The ground state $\sqrt2\sin\pi x$ of a box of width 1 and the trial state $\sqrt{30}\,x(1-x)$, both normalised. They are hard to tell apart, and the trial energy, $10$ in units of $\hbar^{2}/2mL^{2}$, is only $1.3$ per cent above the exact $\pi^{2}\approx9.87$.

A trial family should respect what is known about the ground state. A linear family $\ket{\psi}=\sum_{i}c_{i}\ket{\phi_{i}}$ turns the minimisation into an eigenvalue problem for the matrix of in the functions .

• Match the symmetry: an even potential has an even ground state, with no nodes.
• Match the boundary conditions: vanish at infinite walls, decay at infinity.
• Each eigenvalue of the Rayleigh–Ritz matrix bounds the corresponding true level from above, and adding functions can only lower them.
• With an orthonormal set $\ket{\phi_{i}}$ the matrix equation is an ordinary eigenvalue problem; otherwise the overlaps $\braket{\phi_{i}|\phi_{j}}$ enter.

Write the average energy as a ratio of two quadratic forms in the coefficients. At its minimum the derivative with respect to each coefficient vanishes, and that condition is the eigenvalue equation of the matrix $H_{ij}$. The lowest eigenvalue is the minimum, and so an upper bound.

The average energy in the trial state, with orthonormal $\ket{\phi_{i}}$.

At a stationary point, changing any coefficient does not change $E$ to first order.

The derivative gives the matrix eigenvalue equation.

The smallest eigenvalue is the minimum of the ratio, and a variational upper bound.

The trial state $x(1-x)$ vanishes at both walls of a box of width 1, like the ground state. Its average energy is 10 in units of $\hbar^{2}/2m$, against the exact $\pi^{2}$.

In an even potential the ground state is even and has no node. A trial state should be even too; an odd one bounds an excited level instead.

Adding a function to a Rayleigh–Ritz basis enlarges the family of trial states, so the minimum can only go down or stay the same.

With a trial state made of orthonormal functions with free coefficients, minimising the energy means finding the lowest eigenvalue of the matrix of $\hat{H}$ in those functions.

Great work! You now know how to:

• choose trial states with the right symmetry and boundary conditions
• turn a linear trial family into a matrix eigenvalue problem
• judge a trial state by the energy it gives