8.4: Quantum forces - the Hellmann-Feynman Theorem

Last updated
Save as PDF

Page ID: 28791

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

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.