Does $i\hbar \frac{d \hat A }{d t} = [\hat A (t_0), \hat H]$ hold when $H$ is time-dependent, but $[H(t_0), H(t'_0)] = 0$?

Question

It is known that - given in Sakurai, ch2.2, p83 - in Heisenberg's picture, for a Hamiltonian, $H$, independent of time, the time evolution of any operator $\hat A$ is given by

$$i\hbar \frac{d \hat A }{d t} = [\hat A (t_0), \hat H].$$

Is this equality true when Hamiltonian is dependent on time, but $[H(t_0), H(t'_0)] = 0$ ? I have given an argument why it should be true, but it is correct, or is there any flaw ?

I tried the following argument;

Since $$A (t) = U(t, t_0)^\dagger A(t_0) U(t,t_0),$$ we have $$\frac{d \hat A (t)}{ dt } = D_t [U(t, t_0)^\dagger]AU(t, t_0) + U(t, t_0)^\dagger A D_t[U(t,t_0)] $$ , but since $\frac{i \hbar d U_t }{ dt } = H(t_0) U_t$, $$=\frac{ 1}{ i \hbar} [U_t A H(t_0) U_t - U_t^\dagger H(t_0) A U_t].$$

In this case $U_t := U(t, t_0) = \exp{(\frac{-i}{\hbar } \int_{t_0}^tH(t') dt')}$, but since $[H(t_0), H(t'_0)]$, when we expand the exponential function as a Taylor series, and consider $H(t_0) U_t$, the $H(t_0)$ term will commute with every integral in the expansion, so $[H(t_0), U_t] = 0$, hence

$$\frac{d \hat A (t)}{ dt } = \frac{ 1}{i \hbar } [U_t^\dagger A U_t, H(t_0)] = \frac{1}{ i \hbar} [A(t), H(t_0)],$$ as desired.

Answer 1

Your argument is completely correct.

Consider an operator $A$ which is not explicitly time-dependent (it is constant in the Schrödinger picture) and the time-evolution with a possibly time-dependent Hamiltonian $H(t)$. Then, the time-evolution is $$ \frac{\mathrm d}{\mathrm dt} A_H(t) = \frac{1}{\mathrm i\hbar} [A_H(t), H_H(t)] $$ in general, where $H_H(t) = U(t,t_0)^\dagger H(t) U(t,t_0)$ is the Hamiltonian in the Heisenberg picture. As you correctly noticed, $H_H(t) = H(t)$ in the case where $[H(t_1), H(t_2)] = 0$ for all times.

For completeness, let us also consider the situation where $A$ is explicitly time-dependent, i.e., $A = A(t)$ already in the Schrödinger picture. The Heisenberg picture is defined as $$ A_H(t) = U(t, t_0)^\dagger A(t) U(t, t_0) $$ and note that this is not the same as $U(t, t_0)^\dagger A_H(t_0) U(t, t_0)$! Taking the derivative, we get an extra term from the product rule: $$ \frac{\mathrm d}{\mathrm dt} A_H(t) = \frac{1}{\mathrm i\hbar} [A_H(t), H_H(t)] + \left( \frac{\partial A}{\partial t} \right)_H . $$ Here, $\bigl( \frac{\partial A}{\partial t} \bigr)_H = U(t,t_0)^\dagger \frac{\partial A}{\partial t} U(t,t_0)$ as you would expect. Compare this with $\frac{\mathrm df}{\mathrm dt} = \{ f, H \} + \frac{\partial f}{\partial t}$ in classical mechanics ($\{\bullet,\circ\}$ is the Poisson bracket).

For reference, see e.g. Cohen-Tannoudji: Quantum Mechanics (complement G-III).

Answer 2

If you plug in your time dependent $H(t)$ for $A(t)$ in the evolution equation, you get $\frac{dH(t)}{dt}=0$, so the set of equations

$i\hbar \frac{d \hat A }{d t} = [\hat A (t_0), \hat H]$

$[H(t_0), H(t'_0)] = 0$

$\frac{d \hat H }{d t} \neq 0$

is inconsistent.