Does a time-dependent Hamiltonian commute with its self at different times?

Question

Let $H\colon\mathbb{R}\to\operatorname{End}\mathcal{H}$ be a time-dependent hamiltonian operator, where $\mathcal{H}$ is an arbitrary Hilbert space. Does $H$ commutes with itself at different times, i.e. is $\left[H(t_1),H(t_2)\right]=0$ for $t_1\neq t_2$? Since I expect the answer to be negative, I move on to the actual question. Because it commutes with itself at the same time instant, I expect that: $$\lim_{\epsilon\to 0^+} \left[H(t+\epsilon),H(t)\right]=0$$ where $\epsilon>0$. In which limiting procedure would the above limit make sense?

And one final question. The formal solution to the initial value problem: $$i\hbar\tfrac{\mathrm{d}}{\mathrm{d}t}\psi(t)=H(t)\psi(t), \ \ \psi(t_0)=\psi_0$$ is given by the time-ordered exponential: $$\psi(t)=\mathcal{T}\exp\left(\tfrac{1}{i\hbar}\int_{t_0}^{t}H(t')\mathrm{d}t'\right)\psi_0$$ where $t\ge t_0$. Taking into account the above, I intuitively expect that as $t\rightarrow t_0$ then: $$\mathcal{T}\exp\left(\tfrac{1}{i\hbar}\int_{t_0}^{t}H(t')\mathrm{d}t'\right)\rightarrow \exp\left(\tfrac{1}{i\hbar}(t-t_0)H(t_0)\right)$$ If the above is in some sense correct, how would you (attempt to) prove it?

Answer 1

Does $H$ commutes with itself at different times?

In general, no. If it does happen to, then the eigenstates don't change in time and you don't need to time-order the exponential in the time-evolution operator.

In which limiting procedure would the above limit make sense?

Define the operator $A_t(\epsilon) := [H(t + \epsilon), H(t)]$ which depends on the parameter $\epsilon$. If the Hamiltonian depends continuously on time at time $t$, then as $\epsilon \to 0^+$, $A_t(\epsilon)$ will approach the zero operator with respect to the "operator trace norm," which is proportional to the RMS value of the eigenvalues of $A_t(\epsilon)$. (In practice, it will generally approach the zero operator with respect to any reasonable operator norm.)

If the above is in some sense correct, how would you (attempt to) prove it?

The result follows from the Lie product formula. This idea forms the basis of the extremely important Trotter decomposition, which has been studied extensively in the context of almost all areas of quantum mechanics.

Answer 2

1.Very often the Hamiltonian at the different time are not commuting with each other. As we all known, the time evolution is governed by the total Hamiltonian, so if you modify you Hamiltonian (I mean you will add some energy to your system by the physical interaction) at some later time, you will get the different evolution process with different total energy.

2.The time-ordered exponential is just a shorthand. You must realize it is a solution for the original Schrodinger equation and actually represents an infinite series (Dyson series). So if you take $t \rightarrow t_0$, namely your system keep on still and does not evolve.

Hope it helps.