Proof of the Hellman-Feynman Theorem

Consider a system with a Hamiltonian $H(\lambda)$ that depends on some parameters $\lambda$. Let $\vert\psi(\lambda)\rangle$ be an eigenvector of $H(\lambda)$ with eigenvalue $E(\lambda)$

$$H(\lambda)\vert\psi(\lambda)\rangle = E(\lambda)\vert\psi(\lambda)\rangle$$ (16)

We further assume that $\vert\psi(\lambda)\rangle$ is normalized so that

$$\langle\psi(\lambda)\vert\psi(\lambda)\rangle = 1$$ (17)

The Hellman-Feynman theorem states that

$${dE \over d\lambda} = \left<\psi(\lambda)\left\vert{dH \over d\lambda}\right\vert\psi(\lambda)\right>$$ (18)

The proof of the Hellman-Feynman theorem is straightforward. We begin with the fact that

$$E(\lambda)= \langle \psi(\lambda)\vert H(\lambda)\vert\psi(\lambda)\rangle$$ (19)

Differentiating both sides yields

$${dE \over d\lambda} = \left<\psi(\lambda)\left\vert{dH \over... ...i\left\vert H(\lambda)\right\vert{d\psi \over d\lambda}\right>$$ (20)

Since $\vert\psi(\lambda)\rangle$ is an eigenvector of $H(\lambda)$, this can be written as

$${dE \over d\lambda} = \left<\psi(\lambda)\left\vert{dH \over d\lambda}\right\vert\psi(\... ...da)\right> + E(\lambda)\left<\psi\left\vert\right.{d\psi \over d\lambda}\right>$$

$$\left<\psi(\lambda)\left\vert{dH \over d\lambda}\right\vert\psi(\... ...ambda)\right> + \left<\psi\left\vert\right.{d\psi \over d\lambda}\right>\right]$$

However, since $\vert\psi(\lambda)\rangle$ is normalized, we have, from the normalization condition:

$$\langle\psi(\lambda)\vert\psi(\lambda)\rangle = 1$$

$$\left<{d\psi \over d\lambda}\left\vert\right.\psi(\lambda)\right> + \left<\psi\left\vert\right.{d\psi \over d\lambda}\right> = 0$$

Hence, the term in square brackets vanishes, and we have

$${dE \over d\lambda} = \left<\psi(\lambda)\left\vert{dH \over d\lambda}\right\vert\psi(\lambda)\right>$$ (21)

which is just the Hellman-Feynman theorem.