Issue in deriving Ehrenfest's theorem

Question

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 $\frac{\partial}{\partial t}\hat{A} \neq\frac{d}{d t}A$.

Answer 1

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.

Answer 2

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 $$ \frac{d}{dt}F(x(t),p(t),t)= \frac{\partial F}{\partial t}+ \dot x \frac{\partial F}{\partial x}+\dot p\frac{\partial F}{\partial p}. $$