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I am currently taking a class on Quantum Field Theory. The propagator was defined as:

$$K(x,t;x',t') = \langle x|\hat{T}e^{{\frac{-i}{\hbar}\int_{t}^{t'}dtH(t)}}|x\rangle$$

where, $\hat{T}$ is the time ordering operator. I fully well agree with this so far. However, we next considered '$n$' time slices (which is later considered to approach infinity) and equated:

$$\hat{T}e^{{\frac{-i}{\hbar}\int_{t}^{t'}dtH(t)}} = e^{{\frac{-i}{\hbar}\int_{t_{n-1}}^{t'}dtH(t)}} e^{{\frac{-i}{\hbar}\int_{t_{n-2}}^{t_{n-1}}dtH(t)}}... e^{{\frac{-i}{\hbar}\int_{t}^{t_1}dtH(t)}} \tag1$$

The argument given for this was, "Due to time ordering we now have all operators at later times to the left of earlier times". I do not fully agree with this because I believe in order to use the identity $e^{A+B} = e^Ae^B$, we need to have $[A,B]=0$. Here, our Hamiltonian does not necessarily commute with itself at different times.

However, texts I have looked up, seem to fully ignore time-dependent Hamiltonians and instead use time-independent Hamiltonians.

So my question is: How do I properly deal with the derivation in the case of a time-dependent Hamiltonian? Is eq(1) after all correct? What Am I missing?

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    $\begingroup$ See e.g. this QCSE post. $\endgroup$
    – Tobias Fünke
    Commented May 29 at 15:10
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    $\begingroup$ I believe you are right. As it is written, that expression cannot be correct for finite $n$. All versions of Trotter formula are correct only in the limit $n\to \infty$. $\endgroup$
    – lcv
    Commented May 29 at 16:28
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    $\begingroup$ The error in that formula comes exactly from what you say and is of the order of $(t'-t)/n$ $\endgroup$
    – lcv
    Commented May 29 at 16:33
  • $\begingroup$ You are indeed correct. eq(1) is derived at the limit $n \to \infty$. Although, I can see the confusion in notation. This and the fact that operators commute under the time ordering operator, like @11zaq showed in their answer, was the source of my confusion. $\endgroup$
    – ofbrackets
    Commented May 29 at 17:05
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    $\begingroup$ If you're happy I'm happy. $\endgroup$
    – lcv
    Commented May 29 at 17:41

2 Answers 2

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They do commute at different times under the time ordering symbol. To see this, $$ T[H(t_1),H(t_2)] \\= T[H(t_1)H(t_2)-H(t_2)H(t_1)]\\=\begin{cases} H(t_1)H(t_2)-H(t_1)H(t_2) & (t_1 >t_2) \\ H(t_2)H(t_1)-H(t_2)H(t_1) & (t_2 > t_1) \end{cases} \\=0$$

So you can ignore the extra terms from the BCH formula and split the exponentials. After you do this (and your $\Delta t$ intervals are small enough) you can drop the time ordering because you've already written this expression in a manifestly time ordered way. This gets you all the way to your Equation 1.

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  • $\begingroup$ So you are essentially saying, $\hat{T}e^{{\frac{-i}{\hbar}\int_{t}^{t'}dtH(t)}} = \hat{T} e^{{\frac{-i}{\hbar}\int_{t_{n-1}}^{t'}dtH(t)}} ... e^{{\frac{-i}{\hbar}\int_{t}^{t_1}dtH(t)}} = e^{{\frac{-i}{\hbar}\int_{t_{n-1}}^{t'}dtH(t)}} ... e^{{\frac{-i}{\hbar}\int_{t}^{t_1}dtH(t)}}$ right? Essentially, everything commutes until the second step and then we simply order it correctly in the last step (given the time step is small enough it is already ordered correctly)... It seems I was missing the middle step which clarifies things for me. $\endgroup$
    – ofbrackets
    Commented May 29 at 14:50
  • $\begingroup$ Yes, that is what I'm saying! And the time step being small enough is what let's you assume H(t) is roughly time independent (at least over that interval) which is why dropping the time order symbol makes sense in the last step. $\endgroup$
    – 11zaq
    Commented May 29 at 18:50
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This is because by definition the operator-ordering (in this case time-ordering $T$) takes symbols/functions to operators; not operators to operators.

Before its evaluation, symbols/functions within its argument supercommute. After the operator-ordering procedure has been applied, the result consists of possible non-commutative operators.

See also this and this related Phys.SE posts.

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  • $\begingroup$ Frankly I find it hard to believe. How can the answer be correct? Consider, for example the limiting case with $n=1$. I would say the expression is true only modulo an error $O((t'-t)/n)$ which goes to zero only in the limit $n\to \infty$, no? $\endgroup$
    – lcv
    Commented May 29 at 16:18
  • $\begingroup$ Otherwise you managed to write a Texp as a simple product of standard exponential. $\endgroup$
    – lcv
    Commented May 29 at 16:19
  • $\begingroup$ Sorry maybe I wasn't too clear. What I was trying to say is that the expression that the OP wrote cannot be exact for finite $n$. Any version of Trotter formula has a small error for finite $n$ $\endgroup$
    – lcv
    Commented May 29 at 16:54
  • $\begingroup$ Yes, because the time-ordered exponential depends on infinitely many instants of time, it is formally evaluated to an infinite product of exponentials. $\endgroup$
    – Qmechanic ♦
    Commented May 29 at 16:57

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