Register to reply

Proof of Hellmann Feynman Theorem for TD wavefunctions

by Pablolopez
Tags: hellmann-feynman, physics, proof, quantum, theorem
Share this thread:
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
Phys.Org News Partner Physics news on Phys.org
Mapping the optimal route between two quantum states
Spin-based electronics: New material successfully tested
Verifying the future of quantum computing
naima
#2
Dec20-10, 10:46 AM
PF Gold
P: 354
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: 354
Proof of Hellmann Feynman Theorem for TD wavefunctions

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


Register to reply

Related Discussions
Role of mean value theorem in fundamental theorem of calculus proof Calculus & Beyond Homework 5
Discrete hellmann-feynman theorem ? Linear & Abstract Algebra 5
Wavefunctions and probability-Proof Introductory Physics Homework 2
Feynman-Hellmann Theorem Advanced Physics Homework 7
What is Hellmann-Feynman forces? Quantum Physics 0