Can the time-evolution operator be factorised if the Hamiltonian is a sum of two commuting operators?

Question

Let the time-dependent Hamiltonian $H(t) = A(t) + B(t)$ for some quantum system be given as the sum of two time-dependent operators $A(t)$ and $B(t)$. Further, assume that $A(t)$ and $B(t)$ commute, so $[A(t), B(t)] = 0$ for all $t$. The general solution to the time-dependent Schrödinger equation $$ i \hbar \frac{d |\psi(t)\rangle}{dt} = H(t) |\psi(t)\rangle, \qquad |\psi(t_0)\rangle = |\psi_0\rangle $$ can be obtained once the time-evolution operator $U(t;t_0)$ has been found. Now let $U_A(t;t_0)$ (or $U_B(t;t_0)$ respectively) denote the time-evolution operator that would govern the evolution if only $A(t)$ (or $B(t)$) was present in the Hamiltonian, so $$ i \hbar \frac{d U_A(t;t_0)}{dt} = A(t) U_A(t;t_0), \qquad U(t_0;t_0) = \mathbb{I} $$ and similarly for $U_B(t)$.

Is the overall time-evolution operator for the case when $A(t)$ and $B(t)$ commute given by $U(t;t_0) = U_A(t;t_0) U_B(t;t_0) = U_B(t;t_0) U_A(t;t_0)$?

My feeling is that the answer is yes, because one can write $$ i \hbar \frac{d}{dt} (U_A(t) U_B(t)) = i \hbar \frac{d U_A(t)}{dt} U_B(t) + i \hbar U_A(t) \frac{d U_B(t)}{dt} = A(t) U_A(t) U_B(t) + U_A(t) B(t) U_B(t). $$ I think that since the equation for $U_A(t)$ only depends on $A(t)$, it can only be a function of that operator, so $U_A(t) = f(A(t),t)$. Since $A(t)$ commutes with $B(t)$ for all times, it seems reasonable that $[B(t), f(A(t))] = 0$ and therefore $[B(t), U_A(t)] = 0$ (and similarly if the roles of the operators are reversed). Thus, $$ i \hbar \frac{d}{dt} (U_A(t) U_B(t)) = A(t) U_A(t) U_B(t) + U_A(t) B(t) U_B(t) = (A(t) + B(t)) U_A(t) U_B(t) = H(t) U_A(t) U_B(t), $$ which means that $U_A(t) U_B(t)$ solves the equation required for $U(t)$. Since the time-evolution operator is unique, $U(t) = U_A(t) U_B(t)$ is the only solution. The part that I am not sure about is whether just because the equation for $U_A(t)$ only depends on $A(t)$ it follows that $B(t)$ and $U_A(t)$ always commute. In general, $U_A(t)$ will involve some integral of $A(t)$ and I am not sure whether $B(t)$ would always commute with an arbitrary integral of $A(t)$.

Answer 1

Your worry is correct. We have the nice property that $$[A,B]=0 \quad \Rightarrow \quad [A,f(B)]=0,$$ which can be seen from a power series expansion of $f(B)$, but $U_A(t)$ depends on more than just $A(t)$, it depends on the whole history of $A(t^\prime)$, so we cannot immediately make your conclusion.

Let's formally solve the Schrödinger equation from some initial condition at $t_0$ to $t>t_0$ by using the Magnus expansion: \begin{align}|\psi(t)\rangle=U(t;t_0)|\psi(t_0)\rangle&=\exp\left[-\frac{i}{\hbar}\Omega(t)\right]|\psi(t_0)\rangle, \end{align} where the phase takes the following form \begin{align} \Omega(t;t_0)=&\int_{t_0}^t H(t^\prime)dt^\prime+\frac{1}{2!}\int_{t_0}^t dt^\prime \int_{t_0}^{t^\prime}dt^{\prime\prime}[H(t^\prime),H(t^{\prime\prime})]\\ &+\frac{1}{3!}\int_{t_0}^t dt^\prime \int_{t_0}^{t^\prime}dt^{\prime\prime}\int_{t_0}^{t^{\prime\prime}}dt^{\prime\prime\prime}\left\{[H(t^\prime),[H(t^{\prime\prime}),H(t^{\prime\prime\prime})]]+[H(t^{\prime\prime\prime}),[H(t^{\prime\prime}),H(t^{\prime})]]\right\}\\&+\cdots. \end{align} Expanding, for example, the second integrand $[A(t^\prime)+B(t^\prime),A(t^{\prime\prime})+B(t^{\prime\prime})]$, we immediately notice that we don't just have to worry about the commutators between $A$ and $B$ or only wory about commutators at equal times. If we extend OP's assumption to be that $[A(t),B(t^\prime)]=0$ for all $t,t^\prime\geq t_0$, then we find $$\Omega(t;t_0)=\Omega_A(t;t_0)+\Omega_B(t;t_0),$$ with \begin{align} \Omega_A(t;t_0)=&\int_{t_0}^t A(t^\prime)dt^\prime+\frac{1}{2!}\int_{t_0}^t dt^\prime \int_{t_0}^{t^\prime}dt^{\prime\prime}[A(t^\prime),A(t^{\prime\prime})]\\ &+\frac{1}{3!}\int_{t_0}^t dt^\prime \int_{t_0}^{t^\prime}dt^{\prime\prime}\int_{t_0}^{t^{\prime\prime}}dt^{\prime\prime\prime}\left\{[A(t^\prime),[A(t^{\prime\prime}),A(t^{\prime\prime\prime})]]+[A(t^{\prime\prime\prime}),[A(t^{\prime\prime}),A(t^{\prime})]]\right\}\\&+\cdots, \end{align}and similarly for $\Omega_B$, as all of the commutators involving both $A$ and $B$ will always vanish. Then, because each $\Omega_A$ only contains terms with $A$ and likewise for $\Omega_B$, our more stringent assumption allows us to realize that $[\Omega_A,\Omega_B]=0$ for all $t\geq t_0$, and so the unitary can be factorized into $U=U_A U_B=U_B U_A$.

For this, a sufficient condition is that $$[A(t),B(t^\prime)]=0\qquad \forall \quad t,t^\prime \geq t_0.$$ In general, we can expand to ask for \begin{align} \int_{t_0}^{t^\prime} [A(t^\prime),B(t^{\prime\prime})] dt^{\prime\prime}=0, \end{align} now for all $t^\prime\geq t_0$. We know the sufficient condition, and this can hold without the sufficient condition for a particular $t^\prime$; can we find a different sufficient condition to show our first one to not be necessary? Very artificially, we can consider a situation where $[A(t^\prime),B(t^{\prime\prime})]\sim \sin [\omega(t^\prime -t^{\prime\prime})]$. In the limit of infinitely large oscillation frequencies $\omega$, the integrated commutators will vanish for all times $t$. This is a forced example, and in general we should not expect to find $U=U_A U_B$ with only the guarantee that $[A(t),B(t)]=0$, but it shows that there may be a number of disparate conditions for which the factorizability of the unitaries may hold. Incidentally, the unitaries may factor by having $[A(t^\prime),B(t^{\prime\prime})]\sim \cos [\omega(t^\prime -t^{\prime\prime})]$ in the limit of large $\omega$, devoid of the property that $[A(t),B(t)]=0$.