Hellmann–Feynman Theorem

This compact reference card records the statement and assumptions. The canonical proof, generalized-force interpretation, perturbative curvature, degeneracy, Pulay terms, and changing-domain caveats are on Hellmann–Feynman Theorem.

Statement

Let $H(\lambda)$ have a normalized exact eigenstate $\lvert n(\lambda)\rangle$ with eigenvalue $E_n(\lambda)$:

$$H(\lambda)\lvert n(\lambda)\rangle = E_n(\lambda)\lvert n(\lambda)\rangle, \qquad \langle n(\lambda)\vert n(\lambda)\rangle=1.$$

For a smooth nondegenerate eigenbranch, the Hellmann–Feynman theorem states

$$\frac{dE_n}{d\lambda} = \left\langle n(\lambda) \left\vert \frac{\partial H}{\partial\lambda} \right\vert n(\lambda) \right\rangle.$$

Assumptions

• The eigenstate is exact and normalized.
• The eigenvalue branch is differentiable.
• The Hamiltonian domain and boundary conditions do not introduce hidden parameter dependence, or that dependence is treated explicitly.
• Degenerate levels require diagonalizing $\partial H/\partial\lambda$ in the degenerate subspace or following a smooth resolved branch.

Derivation Sketch

Differentiate the eigenvalue equation:

$$\frac{\partial H}{\partial\lambda}\lvert n\rangle + H\frac{\partial}{\partial\lambda}\lvert n\rangle = \frac{dE_n}{d\lambda}\lvert n\rangle + E_n\frac{\partial}{\partial\lambda}\lvert n\rangle.$$

Left-multiply by $\langle n\vert$. Since $\langle n\vert H=E_n\langle n\vert$, the terms involving $\partial_\lambda\lvert n\rangle$ cancel, leaving the theorem.

See the canonical theorem page for the full proof and the corresponding formula at a degeneracy.

Quantum-Mechanical Meaning

The theorem turns a parameter derivative of an exact energy into an expectation value. It explains why first-order perturbation theory gives

$$E_n^{(1)} = \langle n^{(0)}\vert V\vert n^{(0)}\rangle$$

when $H(\lambda)=H_0+\lambda V$ and the derivative is evaluated at $\lambda=0$.

In molecular and variational calculations, using approximate parameter-dependent states can require additional Pulay or basis-response terms. The exact theorem is simpler than many approximate implementations.

Canonical Links

• Hellmann–Feynman Theorem
• Nondegenerate Perturbation Theory
• First-Order Perturbation Theory
• Variational Principle
• Rayleigh-Ritz Method
• Eigenvalues and Eigenstates

Common Mistakes

• Applying the formula to approximate eigenstates without checking correction terms.
• Ignoring degeneracy.
• Forgetting parameter dependence in the basis, boundary conditions, or domain.
• Assuming the theorem gives the derivative of every observable, not just the eigenvalue.
• Confusing the theorem with a variational stationarity condition.

Quick Check

Let $H(\lambda)=H_0+\lambda V$ and suppose $\lvert n^{(0)}\rangle$ is a nondegenerate eigenstate of $H_0$. What derivative does the theorem give at $\lambda=0$?

Solution

Since $\partial H/\partial\lambda=V$, the theorem gives

$$\left. \frac{dE_n}{d\lambda} \right\rvert_{\lambda=0} = \langle n^{(0)}\vert V\vert n^{(0)}\rangle,$$

which is the first-order energy shift.

References

• H. Hellmann, Einführung in die Quantenchemie, Deuticke, 1937.
• R. P. Feynman, “Forces in Molecules,” Physical Review 56, 340–343 (1939), doi:10.1103/PhysRev.56.340.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.