Linear Variational Method

So far each trial state had a parameter that entered nonlinearly (a Gaussian width, an effective charge). The linear variational method instead writes the trial state as a linear combination of fixed basis functions and optimizes the coefficients. The magic is that minimizing over linear coefficients is exact and reduces to a matrix eigenvalue problem.

A basis-expansion trial state

Pick $N$ fixed, linearly independent functions $\{\,|\chi_k\rangle\,\}$ and write

|\psi\rangle = \sum_{k=1}^{N} c_k\, |\chi_k\rangle .

The coefficients $c_k$ are the variational parameters. Define the Hamiltonian matrix and overlap matrix

H_{jk} = \langle \chi_j | \hat{H} | \chi_k \rangle, \qquad S_{jk} = \langle \chi_j | \chi_k \rangle .

If the basis is orthonormal, $S_{jk} = \delta_{jk}$ and the overlap drops out.

The secular equation

Minimizing the Rayleigh quotient $E = \dfrac{\mathbf{c}^\dagger H \mathbf{c}}{\mathbf{c}^\dagger S \mathbf{c}}$ over the coefficients $\mathbf{c}$ leads to the generalized eigenvalue problem

H\,\mathbf{c} = E\, S\, \mathbf{c},

with non-trivial solutions only when the secular determinant vanishes:

\det(H - E S) = 0 .

For an orthonormal basis this is just $\det(H - E\,\mathbb{1}) = 0$ — the ordinary eigenvalue equation. The $N$ roots $E_1 \leq E_2 \leq \dots \leq E_N$ are the linear variational energies. The smallest, $E_1$ , is the best variational estimate of the ground state; remarkably, $E_2, \dots$ are upper bounds on the excited states too (the Hylleraas–Undheim / MacDonald theorem).

The two-dimensional case

With two orthonormal basis states the Hamiltonian matrix is a real symmetric $2\times 2$ matrix

H = \begin{pmatrix} a & b \\ b & d \end{pmatrix},

and the secular equation $\det(H - E\,\mathbb{1}) = 0$ gives the quadratic $(a-E)(d-E) - b^2 = 0$ , whose roots are

E_\pm = \frac{a+d}{2} \pm \sqrt{\left(\frac{a-d}{2}\right)^2 + b^2}.

The off-diagonal coupling $b$ pushes the levels apart: the lower root $E_-$ drops below the smaller diagonal entry. This level repulsion is the variational improvement over using a single basis state.

Try it

For $H = \begin{pmatrix} 1 & 1 \\ 1 & 3 \end{pmatrix}$ , compute and return the lower eigenvalue — the best two-state variational estimate of the ground-state energy.

Run your code to see the quantum state.