Entanglement of initially unentangled state of two non-interacting systems

Question

Suppose we have two subsystems $A$ and $B$, and the hamilitonian $H$ for the system separates into a sum $H_A(t) + H_B(t)$ of two time dependent hamiltonians $H_A(t)$ and $H_B(t)$ that act only on subsystems $A$ and $B$ respectively.

Suppose the collective system begins in a non-entangled (product) state $\psi_o \equiv |a\rangle \otimes |b\rangle$ and the system evolves under the action of $H$ for some period of time so that the state is now some $\psi'$.

Will $\psi'$ also be a product state? If not wouldn't this be surprising that correlations would develop in non-interacting systems?

A short time $dt$ later the Schrodinger equation tells us that $$ |a\rangle \otimes |b\rangle \to |a\rangle \otimes |b\rangle + \frac{dt}{i\hbar} \left( H_A | a \rangle \otimes | b \rangle + | a \rangle \otimes H_B | b \rangle \right) $$ so that to first order in $dt$ we could write $$ |a\rangle \otimes |b\rangle \to \left( 1 + \frac{dt}{i\hbar} H_A \right) |a\rangle \otimes \left( 1 + \frac{dt}{i\hbar} H_B \right) |b\rangle $$ so that it appears that the state remains in a product state. I do not know how to show this for finite $dt$.

Could we add a term $-\frac{\Delta t^2}{\hbar^2}$ in the Dyson series, i.e. \begin{align} U(t_o,t) & \equiv \Pi_{n=0}^N \left(1 + \frac{\Delta t}{i\hbar}\left( H_A(t_o + n \Delta t ) + H_B(t_o + n \Delta t) \right) \right) \\ & \to \Pi_{n=0}^N \left[ \left(1 + \frac{\Delta t}{i\hbar} H_A(t_o + n \Delta t ) \right)\left(1 + \frac{\Delta t}{i\hbar} H_B(t_o + n \Delta t ) \right) \right] \equiv \tilde{U}(t_o,t) \end{align} where $\Delta t \equiv \frac{t-t_o}{N}$ so that $U=\tilde{U}$ in the limit $N \to \infty$?

Answer 1

Define $|\alpha(t)\rangle$ and $|\beta(t)\rangle$ as the solutions of the IVPs \begin{align} i\hbar \frac{\mathrm d}{\mathrm dt}|\alpha(t)\rangle & = H_A(t) |\alpha(t)\rangle \quad\text{under}\quad |\alpha(0)\rangle = |a\rangle, \\ i\hbar \frac{\mathrm d}{\mathrm dt}|\beta(t)\rangle & = H_B(t) |\beta(t)\rangle \quad\text{under}\quad |\beta(0)\rangle = |b\rangle. \end{align} Then you can show that \begin{align} i\hbar \frac{\mathrm d}{\mathrm dt}|\alpha(t)\rangle\otimes|\beta(t)\rangle & = (H_A(t)+H_B(t)) |\alpha(t)\rangle\otimes|\beta(t)\rangle \\ \text{under}\quad |\alpha(0)\rangle\otimes|\beta(0)\rangle & = |a\rangle\otimes|b\rangle, \end{align} i.e. the product state $|\alpha(t)\rangle\otimes|\beta(t)\rangle$ is a solution of the TDSE you're interested in; since that solution is unique, it follows that it is the solution you're interested in. $$\tag*{$\blacksquare$}$$