# Issue in deriving Ehrenfest's theorem

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

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.

• thanks for your answer. Do you know an example of an operator that has an 'implicit' time dependence? Aug 24, 2021 at 7:57
• 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. Aug 24, 2021 at 8:45

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

• 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. Aug 24, 2021 at 11:42
• @Cream. I agree in priciple, but I stil think the partial is better practice. Aug 24, 2021 at 11:53
• 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. Aug 24, 2021 at 12:57