8.4: Quantum forces - the Hellmann-Feynman Theorem

For many systems one is often interested in forces as well as energies. If we can write the energy of $a$ in state $\phi$ as $E = \langle \phi |\hat{H} |\phi \rangle$ and differentiate with respect to some quantity $\alpha$ then

$$\frac{dE}{d\alpha} = \langle \frac{d\phi}{ d\alpha} |\hat{H} |\phi \rangle + \langle \phi | \frac{d\hat{H}}{d\alpha} |\phi \rangle + \langle \phi |\hat{H} | \phi \rangle  \nonumber$$

But since $\hat{H} |\phi \rangle = E|\phi \rangle$ and $\langle \phi |\phi \rangle$ is 1 for normalisation:

$$\frac{dE}{d\alpha} = \langle \phi | \frac{d\hat{H}}{ d\alpha} |\phi \rangle + E \frac{d}{d\alpha} \langle \phi | \phi \rangle + \langle \phi | \frac{d\hat{H}}{d\alpha} | \phi \rangle \nonumber$$

This result is called the Hellmann-Feynman theorem: the first differential of the expectation value of the Hamiltonian with respect to any quantity does not involve differentials of the wavefunction.

For example, if $\alpha$ represents the position of a nucleus in a solid, then the force on that nucleus is the expectation value of the force operator $\frac{d\hat{H}}{d\alpha}$. It can be applied to any quantity which is a differential of the Hamiltonian provided the basis set does not change.

Caveat: if we use an incomplete basis set which depends explicitly the positions of the atoms, then we have $|\phi \rangle = \sum_{n,i} |u_{n,i} ({\bf r}) \rangle$. This give spurious so-called “Pulay” forces if $\phi$ is not an exact eigenstate.