# Entanglement of initially unentangled state of two non-interacting systems

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 $$N \to \infty$$?

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. $$\tag*{\blacksquare}$$