2.4 Variational principles

In section 2.1 we outlined the basic principles of quantum mechanics, and in particular noted the rôle of the quantity $\langle \Psi \vert {\hat O} \vert \Psi \rangle$ as the expectation value of the observable corresponding to the operator ${\hat O}$. In that section, mention was briefly made of the relationship:

$$\langle \Psi \vert \Psi \rangle = \sum_n \left\vert \langle \chi_n \vert \Psi \rangle \right\vert^2$$ (2.41)

which is simply derived from equations 2.5, 2.6 and 2.8. If we relax the restriction on orthonormalisation, the expression for the expectation value becomes

$$\langle O \rangle = \frac{\langle \Psi \vert {\hat O} \vert \Psi \rangle}{\langle \Psi \vert \Psi \rangle} .$$ (2.42)

We now consider the expectation value of the Hamiltonian operator for the electrons, defined in equation 2.19 and reproduced here:

$${\hat H} \vert \Psi \rangle = \left[ -\frac{1}{2} \sum_i \n... ...right\vert} \right] \vert \Psi \rangle = E \vert \Psi \rangle$$ (2.43)

in which the electronic energy is now labelled $E$, and the dependence on the nuclear coordinates is suppressed since the nuclei are assumed to be static following the conclusions of section 2.2. This equation is an eigenvalue equation for a linear Hermitian operator, and as such can always be recast in the form of finding the stationary points of a functional subject to a constraint.

Consider the expectation value of the Hamiltonian $\langle E \rangle = E [ \Psi ]$ which is a functional of the wave-function, and make a small variation to the state-vector: $\vert \Psi \rangle \rightarrow \vert \Psi \rangle + \vert \delta \Psi \rangle$. The change in $E [ \Psi ]$ is given by

$$\delta E [ \Psi ] = E [ \Psi + \delta \Psi] - E[ \Psi ]$$

$$= \frac{\langle \Psi + \delta \Psi \vert {\hat H} \vert \Psi + \del... ...langle \Psi \vert {\hat H} \vert \Psi \rangle}{\langle \Psi \vert \Psi \rangle}$$

$$= \frac{\langle \delta \Psi \vert {\hat H} \vert \Psi \rangle + \la... ...\langle \Psi \vert \delta \Psi \rangle \right) + O \left( \delta \Psi^2 \right)$$

$$= \frac{1}{\langle \Psi \vert \Psi \rangle} \left[ \langle \delta \... ...rt \Psi \rangle - E [ \Psi ] \vert \Psi \rangle \right) \right\}^{\ast} \right]$$ (2.44)

neglecting changes which are second-order or higher in $\delta \Psi$ in the last line. Thus the quantity $E [ \Psi ]$ is stationary ( $\delta E [ \Psi ] = 0$) when $\vert \Psi \rangle$ is an eigenstate of ${\hat H}$ and the eigenvalue is $E [ \Psi ]$,

$${\hat H} \vert \Psi \rangle = E [ \Psi ] \vert \Psi \rangle$$ (2.45)

and this equation is the time-independent Schrödinger equation. The eigenvalues of ${\hat H}$ can therefore be found by finding the stationary values of $E [ \Psi ]$ i.e. finding the stationary values of $\langle \Psi \vert {\hat H} \vert \Psi \rangle$ subject to the constraint that $\langle \Psi \vert \Psi \rangle$ is constant. In this procedure, the eigenvalue $E$ plays the rôle of a Lagrange multiplier used to impose the constraint.

In this dissertation we will only be interested in finding the electronic ground-state $\vert \Psi_0 \rangle$ which is the eigenstate of the Hamiltonian with the lowest eigenvalue $E_0$. Suppose that we have a state close to the ground-state, but with some small error. Since the eigenstates of the Hamiltonian form a complete set, the error can be expanded as a linear combination of the excited eigenstates. The whole state can thus be written as

$$\vert \Psi \rangle = \vert \Psi_0 \rangle + \sum_{n=1}^{\infty} c_n \vert \Psi_n \rangle$$ (2.46)

where

$${\hat H} \vert \Psi_n \rangle = E_n \vert \Psi_n \rangle.$$ (2.47)

We now calculate the value of $E [ \Psi ]$:

$$E [ \Psi ] = \frac{\langle \Psi \vert {\hat H} \vert \Psi \rangle}{\langle \Psi \vert \Psi \rangle}$$

$$= \frac{ \langle \Psi_0 + \sum_{n=1}^{\infty} c_n \Psi_n \vert {\ha... ...n=1}^{\infty} c_n \Psi_n \vert \Psi_0 + \sum_{n=1}^{\infty} c_n \Psi_n \rangle}$$

$$= \frac{ \langle \Psi_0 + \sum_{n=1}^{\infty} c_n \Psi_n \vert E_0 ... ...n=1}^{\infty} c_n \Psi_n \vert \Psi_0 + \sum_{n=1}^{\infty} c_n \Psi_n \rangle}$$

$$= \frac{ E_0 + \sum_{n=1}^{\infty} \left\vert c_n \right\vert^2 E_n } {1 + \sum_{n=1}^{\infty} \left\vert c_n \right\vert^2 }$$

$$= E_0 + \sum_{n=1}^{\infty} \left\vert c_n \right\vert^2 ( E_n - E_0 ) + O \left( \left\vert c_n \right\vert^4 \right) .$$ (2.48)

By definition, $E_n > E_0$ for $n \geq 1$, so that we note two points:

• $E [ \Psi ] \geq E_0$, with equality only when $\vert \Psi \rangle = \vert \Psi_0 \rangle$ (i.e. $c_n = 0$ for $n \geq 1$),
• the error in the estimate of $E_0$ is second-order in the error in the wave-function (i.e. $c_n$).

The importance of such a variational principle is now clear. To calculate the ground-state energy $E_0$, we can minimise the functional $E [ \Psi ]$ with respect to all states $\vert \Psi \rangle$ which are antisymmetric under exchange of particles. The value of this functional gives an upper bound to the value of $E_0$, and even a relatively poor estimate of the ground-state wave-function gives a relatively good estimate of $E_0$. Eigenstates corresponding to excited states of the Hamiltonian can be found by minimising the functional with respect to states which are constructed to be orthogonal to all lower-lying states (which is usually achieved by considering the symmetries of the states) but in this work we will only ever be interested in the ground-state, and so there are no restrictions on the states other than antisymmetry.