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Proof of Hellmann Feynman Theorem for TD wavefunctions |
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| Dec20-10, 09:54 AM | #1 |
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Proof of Hellmann Feynman Theorem for TD wavefunctions
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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| Dec20-10, 10:46 AM | #2 |
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I think lambda is not supposed to depend on the other parameters (time, position)
so [tex]d/d\lambda = \partial_\lambda[/tex] |
| Dec21-10, 01:23 AM | #3 |
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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! |
| Dec21-10, 07:47 AM | #4 |
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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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