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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:

\begin{displaymath}
\langle \Psi \vert \Psi \rangle = \sum_n \left\vert \langle \chi_n \vert \Psi \rangle
\right\vert^2
\end{displaymath} (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
\begin{displaymath}
\langle O \rangle = \frac{\langle \Psi \vert {\hat O} \vert \Psi \rangle}{\langle \Psi \vert
\Psi \rangle} .
\end{displaymath} (2.42)

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

\begin{displaymath}
{\hat H} \vert \Psi \rangle =
\left[ -\frac{1}{2} \sum_i \n...
...right\vert} \right]
\vert \Psi \rangle = E \vert \Psi \rangle
\end{displaymath} (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

$\displaystyle \delta E [ \Psi ]$ $\textstyle =$ $\displaystyle E [ \Psi + \delta \Psi] - E[ \Psi ]$  
  $\textstyle =$ $\displaystyle \frac{\langle \Psi + \delta \Psi \vert {\hat H} \vert \Psi + \del...
...langle \Psi \vert {\hat H} \vert \Psi \rangle}{\langle \Psi \vert \Psi \rangle}$  
  $\textstyle =$ $\displaystyle \frac{\langle \delta \Psi \vert {\hat H} \vert \Psi \rangle + \la...
...\langle \Psi \vert
\delta \Psi \rangle \right) + O \left( \delta \Psi^2 \right)$  
  $\textstyle =$ $\displaystyle \frac{1}{\langle \Psi \vert \Psi \rangle} \left[ \langle \delta \...
...rt \Psi \rangle - E [ \Psi ] \vert \Psi \rangle \right)
\right\}^{\ast} \right]$ (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 $E [ \Psi ]$,
\begin{displaymath}
{\hat H} \vert \Psi \rangle = E [ \Psi ] \vert \Psi \rangle
\end{displaymath} (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

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

where
\begin{displaymath}
{\hat H} \vert \Psi_n \rangle = E_n \vert \Psi_n \rangle.
\end{displaymath} (2.47)

We now calculate the value of $E [ \Psi ]$:
$\displaystyle E [ \Psi ]$ $\textstyle =$ $\displaystyle \frac{\langle \Psi \vert {\hat H} \vert \Psi \rangle}{\langle \Psi \vert \Psi
\rangle}$  
       
  $\textstyle =$ $\displaystyle \frac{ \langle \Psi_0 + \sum_{n=1}^{\infty} c_n \Psi_n \vert
{\ha...
...n=1}^{\infty} c_n
\Psi_n \vert \Psi_0 + \sum_{n=1}^{\infty} c_n \Psi_n
\rangle}$  
       
  $\textstyle =$ $\displaystyle \frac{ \langle \Psi_0 + \sum_{n=1}^{\infty} c_n \Psi_n \vert
E_0 ...
...n=1}^{\infty} c_n
\Psi_n \vert \Psi_0 + \sum_{n=1}^{\infty} c_n \Psi_n
\rangle}$  
       
  $\textstyle =$ $\displaystyle \frac{ E_0 + \sum_{n=1}^{\infty} \left\vert c_n \right\vert^2 E_n }
{1 + \sum_{n=1}^{\infty} \left\vert c_n \right\vert^2 }$  
       
  $\textstyle =$ $\displaystyle E_0 + \sum_{n=1}^{\infty} \left\vert c_n \right\vert^2 ( E_n - E_0 ) +
O \left( \left\vert c_n \right\vert^4 \right) .$ (2.48)

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

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.


next up previous contents
Next: 3. Quantum Mechanics of Up: 2. Many-body Quantum Mechanics Previous: 2.3 Identical particles   Contents
Peter Haynes