2
$\begingroup$

Consider the equation of motion for the expectation value of an operator $A$ $$\frac{d\langle A\rangle}{dt} = \frac{1}{i\hbar}\langle [A,H]\rangle + \left \langle \frac{\partial A}{\partial t} \right \rangle \, .$$ I am confused with the second term, $\langle \partial A / \partial t \rangle$. Why does $\langle \partial A / \partial t \rangle$ vanish for observables?

if the observable $A$ does not depend explicitly on time, the term $\langle \frac{\partial A}{\partial t}\rangle$ will vanish

-- Zettili's Quantum Mechanics Book

What does it really mean when $A$ is $X$ or $P_x$ for example? If $$\hat P_x= -i\hbar\frac{\partial }{\partial x}$$ then $$ \left \langle \frac{\partial \hat P_x}{\partial t} \right \rangle = \left \langle -i\hbar\frac{\partial}{\partial t} \left( \frac{\partial }{\partial x} \right) \right \rangle = ? $$ Why does it vanish and what does it really mean?

[I am very confused with the concept of the expectation value of an operator. I have checked these:

$\endgroup$
  • $\begingroup$ I think you may be confusing an observable operator on Hilbert space, e.g. $X$, with the result of the observation on a state. Typically (in the Schrodinger picture) the state is changing in time, but the operator is not. $\endgroup$ – Charles Francis Mar 15 at 8:07
  • $\begingroup$ Sometimes they do explicitly depend on time - e.g., the electric field is often included via time-dependent vector potential. $\endgroup$ – Vadim Mar 15 at 9:09
  • $\begingroup$ I think the OP's question is pretty clear: how can an operator, like $-i\hbar \frac{\partial}{\partial x} $ depend on time? To anyone with the usual math background before QM, they are only used to functions depending on time. And an explicit example of an operator which depends on time is hard to find in the opening chapters of textbooks. So what's the issue with the question? @Qmechanic . I think this is a case where a bit of "putting yourself in the OP's shoes" makes it clear what sort of answer is needed.I don't have the time to write one today, but I don't see why thequestion is closed. $\endgroup$ – doublefelix Mar 15 at 13:49
  • $\begingroup$ I am still looking for an answer! But I don't know how to reask and request here! $\endgroup$ – raf Mar 28 at 6:44

Your Answer

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

Browse other questions tagged or ask your own question.