# Applying Ehrenfest's theorem to Hamiltonian

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?

• As Robin Ekman noted, $\frac{\partial H}{\partial t}=0$. Apr 9, 2014 at 12:24
• You mean that $\frac{\partial H}{\partial t}=-{\hbar^2\over{2m}}\frac{\partial^3 }{\partial t\partial x^2}$ (in the well) is zero. Am I right? But I don't get it. Why does the non-applied operator become zero? Apr 9, 2014 at 12:27
• See comment I wrote below Robin's post. Basically, you're taking the derivative of a matrix $\frac{p^2}{2m}$, rather than a matrix-vector product $\frac{p^2}{2m}\cdot\psi$. The matrix is time-independent, even though the matrix-vector product isn't. The idea that operators like $\frac{\partial^2}{\partial x^2}$ can be represented as matrices that are applied to functions (vectors) may seem strange, but it can be made moderately rigorous. Apr 9, 2014 at 12:34

For the infinite potential well, do we not have $H = \frac{p^2}{2m}$ inside the well? Then $\frac{\partial H}{\partial t} = 0$.

I think you have misinterpreted $\frac{\partial H}{\partial t}$. You seem to be applying $\frac{\partial }{\partial t}$ to $(\psi^* H\psi)$, but you should be applying $\frac{\partial H}{\partial t}$ to $\psi$, and then multiplying that by $\psi^*$.

Ehrenfest's theorem applied to the Hamiltonian is the analogue to the classical mechanics theorem that $H$ is conserved unless it depends explicitly on time.

• I did apply the operators in the right order as you said in my calcaulation above. As far as I know, $p$ is the momentum operator, so, by applying $\frac{\partial }{\partial t}$, we cannot make it zero since it is not a variable. Is it right? since $\frac{\partial^3 }{\partial t\partial x^2}$ is not zero but just an operator... Apr 9, 2014 at 12:05
• @dielectric: $\frac{\partial^3 }{\partial t\partial x^2}$ is not the same as $\frac{\partial H}{\partial t}$ (ignoring the extra factors of $i,\hbar$). It is basically the difference between taking $\frac{\partial}{\partial t}(\mathbf{A}(t)\cdot\mathbf{x}(t))$ where $\mathbf{A}$ is a matrix and $\mathbf{x}$ is a vector, versus $\frac{\partial\mathbf{A}}{\partial t}\cdot\mathbf{x}(t)$. Think of $p^2$ as a matrix (after all, with the appropriate choice of basis it is a matrix). Apr 9, 2014 at 12:27
• $p$ is the momentum operator, yes. But $p$ is a constant function of time. Your third order differential operator is $\frac{\partial}{\partial t} \circ p^2$, as composition, not $\frac{\partial p^2}{\partial t}$ as in the time derivative of the function that assigns to every time $t$ the operator $p^2$. Since this function is constant, the derivative is 0. Apr 9, 2014 at 12:34
• @DumpsterDoofus: Uh... I still don't get it clearly, but you mean that I have to calculate the time derivative of the operator $H$ itself, not just to compose two operators? But how can I define the partial derivative of the partial derivative operator itself? Apr 9, 2014 at 12:34
• It is the same operator for every time, right? The derivative of a constant is 0. Apr 9, 2014 at 12:35

There is indeed a contradiction. The nonzero partial derivative of the Hamiltonian implies that it contains time-dependent terms meaning that the wave function can not be dependent on time just like $e^{iE_nt/\hbar}$ The dependence should be more complicated. You assume that partial derivative is non-zero and, at the same time, you take the wave functions corresponding to a stationary solution with time-independent coefficients $c_n$.

• Thank you for your answer, but can the parital derivative of $H$ w.r.t. time itself be 0? I mean, as far as I know, the Hamiltonian in QM is not merely a variable but an operator, so it cannot be 0 just by applying partial derivative. Am I right? And also in the stationary infinite potential well, the wave functions indeed have the time dependence like $e^{iE_nt/\hbar}$, so I think that's not the main reason that I got the contradiction... Apr 9, 2014 at 12:15