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?


2 Answers 2


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.

  • $\begingroup$ Wouldn't the Trotter formula be relevant in the special case where the Hamiltonian decomposes to kinetic and potential parts? I am looking for a way to obtain some results without any special assumption about the Hamiltonian, I want it as general as it can be $\endgroup$ Jan 16, 2017 at 16:24
  • 2
    $\begingroup$ @user3257624 The Trotter decomposition is a two-step process: first, you divide a time evolution operator $e^{-i H t}$ for a finite time into infinitesimal time steps $\prod_t e^{-i H(t) dt}$, which is essentially the expression for the time-ordered exponential. This step can be done for any Hamiltonian. Rigorously justifying this step essentially proves the statement in your OP. The second step involves using the fact that infinitesimal operators (almost) commute to break up the exponential, and that step often requires specific assumptions about the Hamiltonian, but you don't need it here. $\endgroup$
    – tparker
    Jan 16, 2017 at 21:03

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.

  • $\begingroup$ Yeah I am aware of that. I am just looking for a way to justify that the time evolution for time-dependent Hamiltonian will be given by the exponential of the hamiltonian (as in the time-independent case, without time-ordering) for sufficiently small time lapse $\endgroup$ Jan 16, 2017 at 16:20

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