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

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 $$A(t)$$.

• Are you assuming $[A(t), B(t')]=0$ for all $t,t'$, or only $[A(t), B(t)]=0$ (when $t=t'$)? For the former, then I think the answer is yet and your derivation is basically correct. In the latter case, the time evolution operator does not necessarily factorize. Commented Apr 24, 2022 at 18:31

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 $$[A(t),B(t)]=0$$.