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.


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).

  • $\begingroup$ where did that $\frac{\partial A}{ \partial t}_H $ term come from ? $\endgroup$ – onurcanbektas Jan 24 at 12:20
  • $\begingroup$ Ok, I'm confused a little; if $H(t) = U^\dagger (t,t_0) H(t_0) U(t,t_0)$, and $[H(t_1), H(t_2)] = 0$, then $H(t) = H(t_0)$, so $H$ is independent of time. Also, if it is independent of time, trivially, $[H(t_1), H(t_2)] = 0$, so these two conditions are equivalent. This clearly show that there cannot be a case where $[H(t_1), H(t_2)] = 0$, but $H(t)$ is time dependent; however, in Sakurai's book, at the end of the page 70, there is such an explicit assumption. $\endgroup$ – onurcanbektas Jan 24 at 12:30
  • $\begingroup$ @onurcanbektas I fixed a typo that was maybe confusing you. But I don't know where you got $H(t) = U^\dagger H(t_0) U$ from, it is wrong. You have to distinguish the Hamiltonian $H(t)$ in the Schrödinger picture and the Hamiltonian $H_H(t)$ in the Heisenberg picture, that's why I put a subscript "H". $\endgroup$ – Noiralef Jan 24 at 13:27
  • $\begingroup$ even if it is wrong, how can you justify the term $\frac{ \partial A}{ \partial t} $ in that expression; that is what I'm failing to see. $\endgroup$ – onurcanbektas Jan 24 at 15:30
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    $\begingroup$ Let us continue this discussion in chat. $\endgroup$ – Noiralef Jan 24 at 19:49

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.

  • $\begingroup$ Thanks for your answer @tonydo; it is probably because I forget to take the derivative of $H$ wrt time. $\endgroup$ – onurcanbektas Jan 24 at 10:22
  • $\begingroup$ Well, after going over the "proof", there is no need for the derivative of $H$ also, since the Sch. eqn, for the time evolution operator valid regardless of the time dependence of $H$. $\endgroup$ – onurcanbektas Jan 24 at 10:30
  • $\begingroup$ This answer is only half-right. The equation derived by OP only holds for $A$ without explicit time-dependence, but they never claimed otherwise. $\endgroup$ – Noiralef Jan 24 at 11:49
  • $\begingroup$ @Noiralef Are your emphasis on explicit/implicit time-dependence ? If so, regardless of its type, if a function is time dependent, it can be represented as both implicitly and explicitly time-dependent. $\endgroup$ – onurcanbektas Jan 24 at 12:32
  • $\begingroup$ @onurcanbektas An operator has "explicit time-dependence" if it is time-dependent in the Schrödinger picture. In your question, $H$ is explicitly time-dependent and $A$ is not. $\endgroup$ – Noiralef Jan 24 at 13:31

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