2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ Let us continue this discussion in chat. $\endgroup$ – Noiralef Jan 24 at 19:49
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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