# 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.

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