Proof of Hellmann Feynman Theorem for TD wavefunctions

In summary, the proof of the Hellman Feynman-theorem states that the wavefunction given by the equation:[tex]\begin{equation}&\frac{\partial}{\partial \lambda}\langle\Phi(\textbf{r},\textbf{R},t)|\hat{H}|\Phi(\textbf{r},\textbf{R},t)\rangle=\nonumber\\&=i\hbar \langle \frac{\partial}{\partial \lambda} \Phi(\textbf{r},\textbf{R},t)| \frac{\partial}{\partial t} \Phi(\
  • #1
Pablolopez
2
0
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/Hellmann–Feynman_theorem#Proof_2)

I got stack:

[tex]
\begin{align}
&\frac{\partial}{\partial \lambda}\langle\Phi(\textbf{r},\textbf{R},t)|\hat{H}|\Phi(\textbf{r},\textbf{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\lambda}|\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
 
Last edited:
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  • #2
I think lambda is not supposed to depend on the other parameters (time, position)
so
[tex]d/d\lambda = \partial_\lambda[/tex]
 
  • #3
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!
 
Last edited:
  • #4
A problem with tex?
Why do you keep using [tex]\frac{d\lambda}{dt}[\tex] ?
 

What is the Hellmann Feynman Theorem for TD wavefunctions?

The Hellmann Feynman Theorem for TD wavefunctions is a mathematical theorem that relates the change in energy of a quantum system to the change in its wavefunction. It states that the derivative of the energy with respect to a parameter is equal to the expectation value of the derivative of the Hamiltonian with respect to that same parameter.

Why is the Hellmann Feynman Theorem important in quantum mechanics?

The Hellmann Feynman Theorem is important in quantum mechanics because it allows us to calculate the change in energy of a quantum system without explicitly solving the Schrodinger equation. This makes it a powerful tool for understanding the behavior of quantum systems and making predictions about their behavior.

What is the proof of the Hellmann Feynman Theorem for TD wavefunctions?

The proof of the Hellmann Feynman Theorem for TD wavefunctions involves using the time-dependent Schrodinger equation and the variational principle to derive the theorem. It also involves using the properties of Hermitian operators and the commutator to simplify the equations and arrive at the final result.

Can the Hellmann Feynman Theorem be applied to any quantum system?

Yes, the Hellmann Feynman Theorem can be applied to any quantum system as long as the Hamiltonian is time-independent and the wavefunction is a solution to the Schrodinger equation. It is a general theorem that applies to all quantum systems, regardless of their complexity.

What are the practical applications of the Hellmann Feynman Theorem?

The Hellmann Feynman Theorem has many practical applications in quantum mechanics, including in the calculation of atomic and molecular energies, as well as in the study of chemical reactions and solid state physics. It is also useful in the development of quantum computing algorithms and in the design of new materials with specific properties.

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