- #1

Derivator

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according to the Hellamnn-Feynman-Theorem:

if

[itex]E(\lambda):=<\Psi(\lambda)|H(\lambda)|\Psi(\lambda)>[/itex]

then

[itex]\frac{d E}{d\lambda} = <\Psi(\lambda)|\frac{d H(\lambda)}{d\lambda}|\Psi(\lambda)>[/itex]

if [itex]\lambda[/itex] is an atomic coordinate, for instance the positions of a nucleus and H is the electronic Hamiltonian, the negative force on the nucleus is given by the above expression.

Classically one calculates the force on a particle by the negative gradient of the potential energy, thus the above formula is equivalent to classical physics (since in the Born-Oppenheimer approximation, the potential in which the nuclei move is given by the expectation value of the electronic Hamiltonian.)

Despite this analogy, it is not clear to me, why the quantum mechanical force is given by [itex]\frac{d E}{d\lambda}[/itex]. Can one justify this?