The Method of Finite Perturbations

This method arises because of the widespread availability of computer programs which can evaluate the molecular energy and properties using some approximate method. It relies upon the relationship between the perturbation theory expansion of the energy, and the derivatives of the energy,

$$\begin{eqnarray*}E(\lambda)&=&E^{(0)}+\lambda E^{(1)}+\lambda^2 E^{(2)} + ...\\ ......\partial\lambda} +\lambda^2\frac{\partial^2E}{\partial\lambda^2}\end{eqnarray*}$$

It follows that one can get the same results as given by the perturbation theory formulae, by using a method which gives the derivatives of the energy. One particularly simple way to do this is to use numerical differentiation. For example, supposing you wished to calculate a force constant, which is essentially the second-order energy with the perturbation being a change in the position of an atom in a molecule, you could do this simply by calculating the energy for various positions of the atom and fitting the result to a polynomial.

However it is simpler to use the Hellmann-Feynman theorem which is as follows:
If

$$\begin{eqnarray*}E(\lambda)=\langle\Psi(\lambda)\vert H(\lambda)\vert\Psi(\lambda)\rangle\end{eqnarray*}$$

and $\Psi(\lambda)$ is exact, then

$$\begin{eqnarray*}\frac{dE}{d\lambda}&=&\langle\frac{\partial\Psi}{\partial\lambd......ambda}\vert\Psi\rangle]+\langle\Psi\vert H^{(1)}\vert\Psi\rangle\end{eqnarray*}$$

But $\langle\Psi\vert\Psi\rangle=1$ implies that $\langle\frac{\partial\Psi}{\partial\lambda}\vert\Psi\rangle=0$, and so

$$\begin{eqnarray*}\frac{dE}{d\lambda}&=&\langle\Psi\vert H^{(1)}\vert\Psi\rangle\\&=&E^{(1)}(\lambda)\\&=&E^{(1)}+2\lambda E^{(2)}+...\end{eqnarray*}$$

This is the Hellmann-Feynman theorem - the first derivative of the energy is the expectation value of the change in the hamiltonian. One consequence is that we may therefore obtain E⁽²⁾ from the numerical first derivative of E⁽¹⁾ at $\lambda=0$. Obtaining numerical first derivatives through $(E^{(1)}(+\epsilon)-E^{(1)}(-\epsilon))/2\epsilon$ is a more stable procedure than obtaining it from the second derivative of the energy. Note that although the above derivation of the Hellmann-Feynman theorem is for the exact wavefunction holds for other classes for wavefunction, namely those in which all parameters have been chosen variationally to minimise energy. This includes Self-Consistent Field (SCF) wavefucntions. The reason is that we can write the derivative of the energy as,

$$\begin{eqnarray*}\frac{dE}{d\lambda}&=&\langle\Psi\vert H^{(1)}\vert\Psi\rangle+......c{\partial E}{\partial c_k}\frac{\partial c_k}{\partial \lambda}\end{eqnarray*}$$

where the c_k are all the parameters (e.g. molecular orbital coefficients) upon which the energy depends. If these parameters are chosen to minimise the energy then $\frac{\partial E}{\partial c_k} = 0$ and the Hellmann-Feynman theorem holds. The practical importance is that if we want to calculate a dipole moment then we can simply take the expectation value of the dipole moment operator i.e.

$$\begin{eqnarray*}\langle\mu\rangle=\langle\Psi\vert\mu\vert\Psi\rangle\end{eqnarray*}$$

rather than a derivative of the energy

$$\begin{eqnarray*}\langle\mu\rangle=-\frac{\partial E}{\partial F}\end{eqnarray*}$$

which is more difficult.