The Rayleigh-Ritz Variational Method

For a given Hamiltonian ${H}$ we minimise the expectation value of the energy over a sub-set of states $\vert\Psi\rangle$ that are linear combinations of $n$ given states $\vert\psi_i\rangle$,

$$E = \frac{\langle \Psi\vert H\vert \Psi\rangle }{\langle \Psi\vert\Psi\rangle },\quad \vert\Psi\rangle= \sum_{i=1}^n x_i \vert\psi_i\rangle.$$ (3.2)

The $\vert\psi_i\rangle$ are assumed to be normalised but not necessarily mutually orthogonal, i.e., one can have $\langle \psi_i\vert\psi_j\rangle\ne 0$.

The energy $E=E(x_1,...,x_n)$ is therefore minimized with respect to the $n$ coefficients $x_i$, $i=1,...,n$. It can be written as

$$E = _{x_1,...,x_n} \frac{\sum_{i,j=1}^nx_i^*H_{ij}x_j}{\sum_{i,j=1}^nx_i^*S_{ij}x_j} \equiv _{\bf x} \frac{{\bf x}^{\dagger}\underline{\underline{H}}{\bf x}}{{\bf x}^{\dagger}\underline{\underline{S}}{\bf x}},$$ (3.3)

where one has introduced the matrices $\underline{\underline{H}}$ and $\underline{\underline{S}}$ with matrix elements

$$H_{ij}=\langle \psi_i\vert H\vert\psi_j\rangle,\quad S_{ij}=\langle \psi_i\vert\psi_j\rangle.$$ (3.4)

We find the minimum of

$$f({\bf x}) \equiv \frac{{\bf x}^{\dagger}\underline{\underline{H}}{\bf x}}{{\bf x}^{\dagger}\underline{\underline{S}}{\bf x}}$$ (3.5)

by setting the gradient to zero. We treat $x$ and its complex conjugate $x^*$ as independent variables and calculate

$$\frac{\partial}{\partial x_k^*} \left( {\bf x}^{\dagger}\underline{\underline{H}}{\bf x}\right) = \frac{\partial}{\partial x_k^*} \sum_{ij} x_i^*H_{ij}x_j = \sum_j H_{kj}x_j= (\underline{\underline{H}}{\bf x} )_k$$

$$\leadsto {\bf\nabla}^* \left( {\bf x}^{\dagger}\underline{\underline{H}}{\bf x}\right) = \underline{\underline{H}}{\bf x}$$ (3.6)

Correspondingly,

$${\bf\nabla}^* \left( {\bf x}^{\dagger}\underline{\underline{S}}{\bf x}\right) = \underline{\underline{S}}{\bf x}.$$ (3.7)

Thus,

$${\bf\nabla}^* f({\bf x}) = \frac{\underline{\underline{H}}{\bf x}}{{\bf x}^{\dagger}\underli... ...ine{\underline{S}}) {\bf x}}{{\bf x}^{\dagger}\underline{\underline{S}}{\bf x}}$$

$$\leadsto (\underline{\underline{H}} -E \underline{\underline{S}}) {\bf x}=0,$$ (3.8)

since $E=f({\bf x})$ at the minimum! A necessary condition for a minimum therefore is the equation $(\underline{\underline{H}} -E \underline{\underline{S}}) {\bf x}=0$, which has a solution for ${\bf x}$ only if

$${\rm det}\left\vert \underline{\underline{H}} -E \underline{\underline{S}}\right\vert=0.$$ (3.9)

Exercise: Check which equations one obtains when taking the derivative ${\bf\nabla}$ instead of ${\bf\nabla}^*$ !

We summarise:

$$E_{\rm Rayleigh-Ritz} \equiv \frac{\langle \Psi\vert H\vert \Psi\rangle }{\langle \Psi\vert\Psi\rangle },\quad \vert\Psi\rangle= \sum_{i=1}^n x_i \vert\psi_i\rangle$$ (3.10)

$$\leadsto (\underline{\underline{H}} -E \underline{\underline{S}}) {\bf x}=0$$

$$H_{ij} \equiv \langle \psi_i\vert H\vert\psi_j\rangle,\quad S_{ij}\equiv\langle \psi_i\vert\psi_j\rangle, {\bf x}\equiv(x_1,...,x_n)^T.$$

The minimization problem thus led us to an eigenvalue problem.