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Proof of Hellmann Feynman Theorem for TD wavefunctions

by Pablolopez
Tags: hellmann-feynman, physics, proof, quantum, theorem
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Pablolopez
#1
Dec20-10, 09:54 AM
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:

[tex]
\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}
[/tex]

I cannot understand the step in which the total derivatives appear, why? could somebody help me?

Thanks in advance
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naima
#2
Dec20-10, 10:46 AM
PF Gold
P: 375
I think lambda is not supposed to depend on the other parameters (time, position)
so
[tex]d/d\lambda = \partial_\lambda[/tex]
Pablolopez
#3
Dec21-10, 01:23 AM
P: 2
Thanks naima,

I agree with that, however the step to transform:
[tex]
\begin{equation}
i\hbar \langle \frac{\partial}{\partial \lambda} \Phi(\textbf{r},\textbf{R},t)| \frac{\partial}{\partial t} \Phi(\textbf{r},\textbf{R},t)\rangle
\end{equation}
[\tex]

and:

[tex]
\begin{equation}
- i\hbar \langle \frac{\partial}{\partial t} \Phi(\textbf{r},\textbf{R},t)|\frac{\partial}{\par tial \lambda} \Phi(\textbf{r},\textbf{R},t)\rangle
\end{equation}
[\tex]

into:
[tex]
\begin{equation}
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
\end{equation}
[\tex]

and:
\begin{equation}
[tex]
-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
\end{equation}
[\tex]

it is still not clear.

Thanks for your help!

naima
#4
Dec21-10, 07:47 AM
PF Gold
P: 375
Proof of Hellmann Feynman Theorem for TD wavefunctions

A problem with tex?
Why do you keep using [tex]\frac{d\lambda}{dt}[\tex] ?


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