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: この文書について... : lecture_10 : The Born-Oppenheimer Approximation

Proof of the Hellman-Feynman Theorem

Consider a system with a Hamiltonian $H(\lambda)$ that depends on some parameters $\lambda$. Let $\vert\psi(\lambda)\rangle$ be an eigenvector of $H(\lambda)$ with eigenvalue $E(\lambda)$

\begin{displaymath}
H(\lambda)\vert\psi(\lambda)\rangle = E(\lambda)\vert\psi(\lambda)\rangle
\end{displaymath} (16)

We further assume that $\vert\psi(\lambda)\rangle$ is normalized so that
\begin{displaymath}
\langle\psi(\lambda)\vert\psi(\lambda)\rangle = 1
\end{displaymath} (17)

The Hellman-Feynman theorem states that
\begin{displaymath}
{dE \over d\lambda} = \left<\psi(\lambda)\left\vert{dH \over d\lambda}\right\vert\psi(\lambda)\right>
\end{displaymath} (18)

The proof of the Hellman-Feynman theorem is straightforward. We begin with the fact that
\begin{displaymath}
E(\lambda)= \langle \psi(\lambda)\vert H(\lambda)\vert\psi(\lambda)\rangle
\end{displaymath} (19)

Differentiating both sides yields
\begin{displaymath}
{dE \over d\lambda} = \left<\psi(\lambda)\left\vert{dH \over...
...i\left\vert H(\lambda)\right\vert{d\psi \over d\lambda}\right>
\end{displaymath} (20)

Since $\vert\psi(\lambda)\rangle$ is an eigenvector of $H(\lambda)$, this can be written as
$\displaystyle {dE \over d\lambda}$ $\textstyle =$ $\displaystyle \left<\psi(\lambda)\left\vert{dH \over d\lambda}\right\vert\psi(\...
...da)\right>
+ E(\lambda)\left<\psi\left\vert\right.{d\psi \over d\lambda}\right>$  
$\displaystyle \left<\psi(\lambda)\left\vert{dH \over d\lambda}\right\vert\psi(\...
...ambda)\right>
+ \left<\psi\left\vert\right.{d\psi \over d\lambda}\right>\right]$      

However, since $\vert\psi(\lambda)\rangle$ is normalized, we have, from the normalization condition:
$\displaystyle \langle\psi(\lambda)\vert\psi(\lambda)\rangle$ $\textstyle =$ $\displaystyle 1$  
$\displaystyle \left<{d\psi \over d\lambda}\left\vert\right.\psi(\lambda)\right>
+ \left<\psi\left\vert\right.{d\psi \over d\lambda}\right>$ $\textstyle =$ $\displaystyle 0$  

Hence, the term in square brackets vanishes, and we have
\begin{displaymath}
{dE \over d\lambda} = \left<\psi(\lambda)\left\vert{dH \over d\lambda}\right\vert\psi(\lambda)\right>
\end{displaymath} (21)

which is just the Hellman-Feynman theorem.


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: この文書について... : lecture_10 : The Born-Oppenheimer Approximation
Mark Tuckerman 平成17年3月8日