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I learned that a unitary matrix generated by time-dependent Hamiltonians is written down as

\begin{equation} U(t) = \mathcal{T}\exp\Big(-i\int_0^t H(t') dt' \Big),\tag{1} \end{equation} where $\mathcal{T}\exp$ is the time-ordered exponential operator.

My question is: is the $k$th power of the unitary operator $$U^k(t) = \mathcal{T}\exp\Big(-i\int_0^t kH(t') dt' \Big)~?\tag{2}$$

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    $\begingroup$ Have you tried a simple example with three time points? $\endgroup$ Aug 3, 2023 at 0:27
  • $\begingroup$ Also, don't forget that the notation of a "time-ordered exponential" is just a notational convenience. It must be taken to mean the power series of time-ordered operators. The "time-ordering operator" is not a linear operator and we are not "applying it" linearly to each term in an expansion of an exponential. $\endgroup$
    – hft
    Aug 5, 2023 at 15:53

2 Answers 2

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TL;DR: No, OP's eq. (2) does not necessarily hold.

Counterexample. Take e.g. $$H(t)~=~A~ \delta(t-\frac{1}{3})~+~B~ \delta(t-\frac{2}{3}), $$ where the operators $A$ and $B$ do not commute: $[A,B]\neq 0$. Then $$ T\exp\left[2\int_0^1\! dt~H(t)\right]~=~e^{2B}e^{2A},$$ while $$ T\left[\exp\int_0^1\! dt~H(t)\right]~ T\left[\exp\int_0^1\! dt~H(t)\right]~=~e^Be^Ae^Be^A.$$

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  • $\begingroup$ Thanks for the great answer. Actually I confronted a small confusion while rethinking about this problem after looking at your answer. Since the superoperator $\text{ad}_{H(t)}$ commutes with itself for any choice of $H(t)$, isn't the following equality hold: $T \exp(-i \int_0^T \text{ad}_{2H(t)} dt) = T\exp(-i \int_0^T \text{ad}_{H(t)} dt) T\exp(-i \int_0^T \text{ad}_{H(t)} dt) = \mathcal{U} \circ \mathcal{U}$, where $\mathcal{U}$ is the superoperator, i.e. $\mathcal{U}(\rho) = U \rho U^\dagger$?. This contradicts to your answer. What am I missing here? $\endgroup$
    – Hailey Han
    Aug 6, 2023 at 21:47
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For time dependent Hamiltonian $U$ does not necessarily have the time translation symmetry. You should write $U(t,t')$ and then I don't really know what $U^k$ represents.

For time independent Hamiltonian $U^k(t) = U(kt)$ is just a time propagation by $kt$ due to the group structure of $U$ operators. On the other hand $ U(kt) = \mathcal T \exp\left(-i \int_0^{kt} H dt'\right) = \mathcal T \exp\left(-i \int_0^{t} k H dt''\right)$

which is what you asked for.

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    $\begingroup$ If the Hamiltonian is not time dependent the time ordering is the identity and everything is trivial…. $\endgroup$ Aug 3, 2023 at 7:16
  • $\begingroup$ Yeah, you're right $\endgroup$ Aug 3, 2023 at 7:22

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