1
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ As Robin Ekman noted, $\frac{\partial H}{\partial t}=0$. $\endgroup$ Apr 9, 2014 at 12:24
  • $\begingroup$ 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? $\endgroup$ Apr 9, 2014 at 12:27
  • $\begingroup$ 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. $\endgroup$ Apr 9, 2014 at 12:34

2 Answers 2

4
$\begingroup$

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.

$\endgroup$
10
  • $\begingroup$ 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... $\endgroup$ Apr 9, 2014 at 12:05
  • $\begingroup$ @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). $\endgroup$ Apr 9, 2014 at 12:27
  • $\begingroup$ $p$ 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. $\endgroup$ Apr 9, 2014 at 12:34
  • $\begingroup$ @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? $\endgroup$ Apr 9, 2014 at 12:34
  • $\begingroup$ It is the same operator for every time, right? The derivative of a constant is 0. $\endgroup$ Apr 9, 2014 at 12:35
0
$\begingroup$

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 $c_n$.

$\endgroup$
1
  • $\begingroup$ 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... $\endgroup$ Apr 9, 2014 at 12:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.