8.4: Quantum forces - the Hellmann-Feynman Theorem
- Page ID
- 28791
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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.