It maybe a stupid question, but from the Ehrenfest's theorem, we have
\begin{eqnarray\*}
\frac{d\langle A\rangle}{dt} &=& \left\langle\frac{\partial A}{\partial t}\right\rangle + \frac{1}{i\hbar}\left\langle[A,H]\right\rangle
\end{eqnarray\*}
The if we apply it to the Hamiltonian,
\begin{eqnarray\*}
\frac{d\langle H\rangle}{dt} &=& \left\langle\frac{\partial H}{\partial t}\right\rangle + \frac{1}{i\hbar}\left\langle[H,H]\right\rangle
\end{eqnarray\*}
But since the last term vanishes
\begin{eqnarray\*}
\frac{d\langle H\rangle}{dt} &=& \left\langle\frac{\partial H}{\partial t}\right\rangle
\end{eqnarray\*}
But in general cases, the expectation value of the time derivative of the Hamiltonian is not zero, i.e. in the infinite potential well.
$$ \left\langle\frac{\partial H}{\partial t}\right\rangle=\int\Psi^\*\frac{\partial H}{\partial t}\Psi dx=\int\sum\_n c\_n^\* \psi\_n^\*e^{iE\_n t/\hbar}\frac{\partial H}{\partial t}\sum\_m c\_m \psi\_m e^{-iE\_m t/\hbar}dx$$
$$ =\int\sum\_n c\_n^\* \psi\_n^\*e^{iE\_n t/\hbar}\sum\_m c\_m (H \psi\_m) \frac{\partial }{\partial t}e^{-iE\_m t/\hbar}dx$$
$$=\int\sum\_n c\_n^\* \psi\_n^\*e^{iE\_n t/\hbar}\sum\_m c\_m{1\over{i\hbar}}E\_m^2\psi\_m e^{-iE\_m t/\hbar}dx$$
$$={1\over{i\hbar}}\sum\_n\sum\_m e^{i(E\_n -E\_m) t/\hbar}c\_n^\*c\_m\int E\_m^2 \psi\_n^\*\psi\_m dx$$
$$={1\over{i\hbar}}\sum\_n |c\_n|^2E\_n^2 $$
But since the expectation value of the Hamiltonian in the infinite well is a constant, it is obviously a contradiction. Is it impossible to apply the Ehrenfest's theorem to the Hamiltonian, or is there any mistake in my calculation?