Hellmann–Feynman theorem

Remarks

The proof differentiates $E(\lambda) = \langle \psi(\lambda), H(\lambda)\psi(\lambda) \rangle$ by the product rule. The two terms with $\psi'$ add up to $E(\lambda)\big(\langle \psi', \psi \rangle + \langle \psi, \psi' \rangle\big)$, because $H(\lambda)$ is Hermitian and $\psi(\lambda)$ is an eigenvector. This sum is $E(\lambda)$ times the derivative of $\norm{\psi(\lambda)}^2 = 1$, so it vanishes, and only the term with $H'$ is left. Example: for $H(\lambda) = \begin{pmatrix} \lambda & 1 \\ 1 & -\lambda \end{pmatrix}$ the top eigenvalue is $E(\lambda) = \sqrt{1 + \lambda^2}$. The unit eigenvector is proportional to $(1, E - \lambda)$, and a short computation gives $\langle \psi, H'\psi \rangle = \lambda / \sqrt{1 + \lambda^2}$, which is indeed $E'(\lambda)$. The normalization is needed: for an eigenvector of non-constant length the right side must be divided by $\norm{\psi(\lambda)}^2$. The same formula holds for self-adjoint operators on a Hilbert space under suitable domain and differentiability conditions.

Going deeper

For a student

For $H(\lambda) = H_0 + \lambda V$ the theorem gives $E'(\lambda) = \langle \psi(\lambda), V \psi(\lambda) \rangle$. At $\lambda = 0$ this is the first-order energy correction of Rayleigh–Schrödinger perturbation theory. In quantum chemistry, taking $\lambda$ to be a nuclear coordinate turns the theorem into a formula for the force on a nucleus as an expectation value. The hypothesis that $\psi$ is differentiable can fail where two eigenvalues cross, since the eigenvectors may jump there.