Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question
    
As Robin Ekman noted, $\frac{\partial H}{\partial t}=0$. –  DumpsterDoofus Apr 9 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? –  dielectric Apr 9 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. –  DumpsterDoofus Apr 9 at 12:34
add comment

2 Answers 2

up vote 4 down vote accepted

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.

share|improve this answer
    
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... –  dielectric Apr 9 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). –  DumpsterDoofus Apr 9 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. –  Robin Ekman Apr 9 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? –  dielectric Apr 9 at 12:34
    
It is the same operator for every time, right? The derivative of a constant is 0. –  Robin Ekman Apr 9 at 12:35
show 5 more comments

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 $c_n$.

share|improve this answer
    
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... –  dielectric Apr 9 at 12:15
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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