# Proof of Hellmann Feynman Theorem for TD wavefunctions

by Pablolopez
Tags: hellmann-feynman, physics, proof, quantum, theorem
 P: 2 Dear users, I am dealing with the proof of the Hellman Feynman-theorem for time-dependent wavefunctions given by the Wikipedia: (http://en.wikipedia.org/wiki/Hellman...heorem#Proof_2) I got stack: \begin{align} &\frac{\partial}{\partial \lambda}\langle\Phi(\textbf{r},\textbf{R},t)|\hat{H}|\Phi(\textbf{r},\t extbf{R},t)\rangle= \nonumber \\ &= i\hbar \langle \frac{\partial}{\partial \lambda} \Phi(\textbf{r},\textbf{R},t)| \frac{\partial}{\partial t} \Phi(\textbf{r},\textbf{R},t)\rangle + \langle \Phi(\textbf{r},\textbf{R},t)|\frac{\partial}{\partial \lambda}\hat{H}|\Phi(\textbf{r},\textbf{R},t)\rangle - \nonumber \\ &- i\hbar \langle \frac{\partial}{\partial t} \Phi(\textbf{r},\textbf{R},t)|\frac{\partial}{\partial \lambda} \Phi(\textbf{r},\textbf{R},t)\rangle = i\hbar\frac{d\lambda}{dt} \langle \frac{\partial}{\partial \lambda} \Phi(\textbf{r},\textbf{R},t)| \frac{\partial}{\partial \lambda} \Phi(\textbf{r},\textbf{R},t)\rangle + \nonumber \\ &+ \langle \Phi(\textbf{r},\textbf{R},t)|\frac{\partial}{\partial \lambda}\hat{H}|\Phi(\textbf{r},\textbf{R},t)\rangle -i\hbar\frac{d\lambda}{dt} \langle \frac{\partial}{\partial \lambda} \Phi(\textbf{r},\textbf{R},t)| \frac{\partial}{\partial \lambda} \Phi(\textbf{r},\textbf{R},t)\rangle = \nonumber \\ &= \langle\Phi(\textbf{r},\textbf{R},t)|\frac{\partial\hat{H}}{\partial\la mbda}|\Phi(\textbf{r},\textbf{R},t)\rangle \end{align} I cannot understand the step in which the total derivatives appear, why? could somebody help me? Thanks in advance
 PF Gold P: 375 I think lambda is not supposed to depend on the other parameters (time, position) so $$d/d\lambda = \partial_\lambda$$