7
$\begingroup$

Working in Schrodinger picture, while deriving Ehrenfest's theorem, we go - $$ \frac{d}{d t}\langle A\rangle=\frac{d}{d t}\langle\psi|\hat{A}| \psi\rangle $$ $A$ is an operator. Expanding RHS- $$ \frac{d}{d t}\langle A\rangle=\left\langle\frac{d}{d t} \psi|\hat{A}| \psi\right\rangle+\left\langle\psi\left|\frac{\partial}{\partial t} \hat{A}\right| \psi\right\rangle+\left\langle\psi|\hat{A}| \frac{d}{d t} \psi\right\rangle $$ My doubt is regarding the second term. Why do we write $\frac{\partial}{\partial t}\hat{A}$ and not $\frac{d}{d t}A$? Of course, this notation wouldn't matter incase there is only an explicit dependence of $t$, if there's any $t$ dependence at all.

What if $A$ were composed of other time dependent operator $\hat{O}(t)$, i.e. $\hat{A}(t)=A(\hat{O}(t),t)$. Can we have such operators? In that case $\frac{\partial}{\partial t}\hat{A} \neq\frac{d}{d t}A$.

$\endgroup$

2 Answers 2

8
$\begingroup$

When the Ehrenfest theorem is derived in the Heisenberg picture, the operator $A$ can have two different kinds of time-dependence. An "inherent" (or explicit) time dependence (in red) and the one due to the time evolution (shifting from the Schrödinger picture):

$$ A_H(t) = e^{+i Ht/\hbar} A(\color{red}{t}) e^{-iHt/\hbar}. $$ In that case, one has to emphasize that the derivative in the Ehrenfest theorem is with respect to the inherent time-dependence. To avoid any confusion, one would/should write: $$ \frac{d}{dt} \langle A_H(t) \rangle = \frac{i}{\hbar} [H_H, A_H] + \Big( \frac{\partial}{\partial t} A(t) \Big)_H. $$ But being a bit sloppy, you could also write $\frac{\partial}{\partial t} A_H(t)$ and mean $\Big( \frac{\partial}{\partial t} A(t) \Big)_H$. This notation is adopted in your case, although it is not strictly needed in the Schrödinger picture.

What if A were composed of other time-dependent operator $\hat O(t)$, i.e. $\hat A(t)=A(\hat O(t),t)$. Can we have such operators?

In QM, you usually work with a limited number of different operators, which are all time-independent in the Schrödinger picture. To get an explicit time-dependence, you really have to add a $t$ there.
The $\partial$ is meant strictly for this case, and not (as in different context) for functions like $f(g(t),t)$.

A common example is the Hamiltonian of a spin particle in a magnetic field. If the field is oscillating (i.e., $B(t) = B_0 \sin t$), then the Hamiltonian is explicitly time-dependent: $\hat H \propto B(t) \hat S_z = B_0 \sin t ~ \hat S_z$, where $\hat S_z$ is the spin operator.

$\endgroup$
2
  • 2
    $\begingroup$ thanks for your answer. Do you know an example of an operator that has an 'implicit' time dependence? $\endgroup$ Commented Aug 24, 2021 at 7:57
  • 2
    $\begingroup$ Not that I know of. I edited the answer to clarify the example a bit. The explicit time-dependence usually means that the system is not closed, but that there is an external cause for the change (like someone turning on a magnetic field). An "implicit" time-dependence of a Schrödinger operator would imply that even in a closed system the measurement is time-dependent. Or put differently, that the physics tomorrow is different from the physics today. $\endgroup$
    – Cream
    Commented Aug 24, 2021 at 8:45
1
$\begingroup$

We us the partial derivative because there are other variables in play --- such as $x$ and $p$, both of which may be time dependent. The partial derivative symbol is used because it implies that we are keeping all the other variables fixed when we vary $t$.

Using the "$d$" derivative would imply that $$ \frac{d}{dt}F(x(t),p(t),t)= \frac{\partial F}{\partial t}+ \dot x \frac{\partial F}{\partial x}+\dot p\frac{\partial F}{\partial p}. $$

$\endgroup$
3
  • 3
    $\begingroup$ This is the usual definition in classical mechanics but I think it is missing the point here. In the Schrödinger picture, $\hat x$ and $\hat p$ are both time-independent. On the other hand, any observable $A$ might depend on operators other than $\hat x$ and $\hat p$ (e.g. the pauli matrices). You seldomly see an equation as you wrote it in quantum mechanics. Instead, you'd see the Heisenberg equations. $\endgroup$
    – Cream
    Commented Aug 24, 2021 at 11:42
  • 1
    $\begingroup$ @Cream. I agree in priciple, but I stil think the partial is better practice. $\endgroup$
    – mike stone
    Commented Aug 24, 2021 at 11:53
  • 1
    $\begingroup$ I agree that using the partial here (in this question) is "good practice". But I think that one should use it for a different reason than you. $\endgroup$
    – Cream
    Commented Aug 24, 2021 at 12:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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