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Define a time-dependent Hamiltonian $$H(t) = H_1(t) + H_2(t),\tag{1}$$ where $$[H_1(t), H_2(t)] = 0 ~ \forall t \in [0,T].\tag{2}$$ Is it true that the unitary operator generated by $H(t)$ is a product of two unitaries generated by $H_1(t)$ and $H_2(t)$, i.e.

\begin{equation} U(T) = \mathcal{T}\exp\Big(-i \int_0^T dt H(t) \Big) = \mathcal{T}\exp\Big(-i \int_0^T dt H_1(t) \Big)\mathcal{T}\exp\Big(-i \int_0^T dt H_2(t) \Big) = U_1(T) U_2(T)~?\tag{3} \end{equation}

(Essentially, I'm curious whether the BCH formula works in time-ordered exponentials.)

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  • $\begingroup$ One idea to check: $U$ must obey a differential equation (time-dependent Schrödinger equation with $H$), and likewise must $U_1$ and $U_2$ obey their respective equations. Then, in particular, if $U=U_1 U_2$, the LHS and RHS must obey the SE for $H$, with the corresponding initial condition. And since the solution is unique, if $U_1 U_2$ obeys the Schrödinger equation for $H$ (and the same initial condition as $U$), it is equal to $U$. [of course, this is a, a priori, not very rigorous way] $\endgroup$ Aug 9, 2023 at 7:12
  • $\begingroup$ For an even less rigorous take, we can treat everything at the level of formal power series. In that case I believe the LHS can be rearranged into $\mathcal{T}\left[\exp\left(-i\int dt H_1(t)\right)\exp\left(-i\int dt H_2(t)\right)\right]$ (the ordering of symbols under $\mathcal{T}$ does not matter, the operator order is fixed by time ordering, so the proof should go through as if the symbols were numbers). I would be really surprised if this was equal to the right hand side, which has all the $H_1$s to the left of the $H_2$s for all times, while they are mixed on the left hand side. $\endgroup$ Aug 9, 2023 at 9:34
  • $\begingroup$ Related: physics.stackexchange.com/q/774543/2451 $\endgroup$
    – Qmechanic
    Aug 9, 2023 at 15:13

1 Answer 1

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Surely, your instructor must have drilled you to illustrate such expressions for a finite number of time points, first; here, take them to be just two. Incorporate the -i into the Hamiltonian pieces, and use lower case for the former time point and upper case for the latter point. So, define $$ -iH_1(t_1)\equiv a, \qquad -iH_2(t_1)\equiv b,\qquad -iH_1(t_2)\equiv A, \qquad -iH_2(t_2)\equiv B,\\ [a,b]=[A,B]=0, $$ but all other commutators need not vanish, in general, so, here, they are taken to be non-vanishing and uncorrelated.

Then, $$ -i \int_0^T dt H_1(t) = a+A, \leadsto \qquad U_1(T)=e^A e^a,\\ -i \int_0^T dt H_2(t) =b+B, \leadsto \qquad U_2(T)=e^B e^b ,\\ -i \int_0^T dt H(t) = a+b+A+B, \leadsto \qquad U(T)=e^{A+B} e^{a+b}=e^{A} e^{B} e^{a} e^b, $$ which manifestly violates your wrong conjecture, in general.

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  • $\begingroup$ So the sufficient condition for $U(T) = U_1(T)U_2(T)$ seems to be like $[H_1(t), H_2(t')] = 0 ~\forall t, t' \in [0,T]$, is it correct? For example, if $H_1 = H_2$, then does $U(T) = U_1(T)^2 = U_2(T)^2$? $\endgroup$
    – Mohan
    Aug 9, 2023 at 20:18
  • $\begingroup$ Yes to the first. Your second question is linked in a comment. Note lower cases do not commute with upper cases, in general, but you made them do? $\endgroup$ Aug 9, 2023 at 20:29
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    $\begingroup$ Ah, gotcha, thank you for your clear explanations! $\endgroup$
    – Mohan
    Aug 9, 2023 at 20:37

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