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?


1 Answer 1


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

  • $\begingroup$ 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. $\endgroup$
    – Andrew
    Commented Apr 22, 2022 at 6:24
  • $\begingroup$ @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. $\endgroup$
    – joseph h
    Commented Apr 22, 2022 at 7:05

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