# Proving the Feynman-Hellmann Theorem in quantum mechanics

Concerning the Feynman-Hellmann theorem can someone point me on how solve this:

If $$H E = E |E\rangle$$ and assuming $$H$$ is depending on a variable $$\lambda$$ eg., $$H = H(\lambda)$$ then $$\langle \frac{\partial H (\lambda)}{ \partial \lambda} \rangle = \frac{\partial E (\lambda)}{ \partial \lambda}$$. And states of $$H$$ are orthogonal and normalised. I'm confuse since professor did not show form of hamiltonian. How do you differentiate it if you do not know its form?

If H depends on $$\lambda$$ then so to do its its eigenvalues. That is $$E = E(λ)$$. And since $$E$$ is the eigenvalue corresponding to the state $$|E\rangle$$ then

$$E(\lambda) |E\rangle = H(\lambda)|E\rangle$$ or

$$E(\lambda)\langle E|E\rangle = \langle E|H(\lambda)|E\rangle$$ so

$$E(\lambda) = \langle E|H(\lambda)|E\rangle$$

due to the orthonormality condition (we could have just written this expression down since it is the expression for the expected value of $$H(\lambda))$$. Now differentiate both sides with respect to $$\lambda$$ to get

$$\frac{\partial E(\lambda)}{\partial \lambda} = \frac{\partial }{\partial \lambda}\langle E|H(\lambda)|E\rangle \\$$

$$\hspace{5cm} =\frac{\partial \langle E |}{\partial \lambda} H(\lambda)|E\rangle + \langle E| \frac{\partial H(\lambda)}{\partial \lambda} |E \rangle + \langle E | H(\lambda) \frac{\partial |E \rangle}{\partial \lambda} \\$$

Again use the fact that the states are orthonormal and the fact that $$H|E\rangle =E|E\rangle$$ and you should get the result you need $$\frac{\partial E(\lambda)}{\partial \lambda}= \langle \frac{\partial H(\lambda)}{\partial \lambda} \rangle$$

• I realize this answer is 1+ years old and already accepted, but just as a small clarification for the record in case anyone stumbles on this in the future, "use the fact that the states are orthonormal" here means that since $\langle E | E\rangle= 1$, then $$\frac{\partial \langle E |}{\partial \lambda} | E \rangle + \langle E | \frac{\partial | E \rangle}{\partial \lambda} = \frac{\partial}{\partial \lambda} \langle E | E \rangle = \frac{\partial}{\partial \lambda} 1 = 0.$$ It does not mean $\langle E | \frac{\partial | E \rangle}{\partial \lambda} = 0$, which is not generally true. Commented Apr 22, 2022 at 6:24
• @Andrew Yes, the term is zero, and (ortho) normality means <E|E>=1 and so you take the derivative of a constant which is 0. I did not what to include every point and hoped OP would find that obvious. But it is a good point nevertheless Andrew, and maybe I could have used better terminology. Thanks. Commented Apr 22, 2022 at 7:05