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I am looking at the following problem and I am struggling to follow the steps involved. Consider the non-interacting Hamiltonian $$H_{AB}=H_A\otimes I_B+I_A\otimes H_B$$

So I'm trying to prove that the unitary evolution of the joint state is given by $$|\psi\left(t\right)\rangle_{AB}=e^{-iH_At}\otimes e^{-iH_Bt}|\psi\left(t=0\right)\rangle_{AB}$$

Where $|\psi\rangle_{AB}=|\psi\rangle_A \otimes |\psi\rangle_B$

My working so far is

$$|\psi\left(t\right)\rangle_{AB}=e^{-i\left(H_A\otimes I_B+I_A\otimes H_B\right)t}|\psi\left(t=0\right)\rangle_{AB}$$

$$=e^{-i\left(H_A\otimes I_B\right)t}e^{-i\left(I_A\otimes H_B\right)t}|\psi\left(t=0\right)\rangle_{AB}$$

As the two Hamiltonians for the two system commute, so from here I am a little confused, I know the next step must be

$$=\left(e^{-iH_A t}\otimes I_B\right)\left(I_A\otimes e^{-iH_B t} \right)|\psi\left(t=0\right)\rangle_{AB}$$

But this isn't at all obvious to me why this is the case? I'm also not sure if this is off topic here and would be better suited to maths stack exchange so I apologise in advance.

Throughout I have set $\hbar=1$.

At this point I think I should probably use the definition of the matrix exponential as a Taylor series but I'm not sure.

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It really is obvious if you understand how tensor products work. Essentially, your state has two indices instead of one, and a tensor product of operators means that the first operator acts on the first index, and the second operator acts on the second. The operator that act on the "other" index just goes along for the ride. If it helps, you can write the exponential as a Taylor series. Then you will end up with lots of powers of $I$, which you can just collapse down using $I^2 = I$.

For the sake of completeness, $$ \left(U_A \otimes U_B\right) \; \left|\psi\right\rangle_A \otimes \left|\psi\right\rangle_B \equiv \left(U_A \left|\Psi\right\rangle_A \right) \otimes \left(U_B \left|\Psi\right\rangle_B \right) $$ Thus $$ \begin{align} \left(U_A \otimes U_B\right)^2 \; \left|\psi\right\rangle_A \otimes \left|\psi\right\rangle_B &\equiv \left(U_A \otimes U_B\right) \left(U_A \otimes U_B\right) \; \left|\psi\right\rangle_A \otimes \left|\psi\right\rangle_B\\ &= \left(U_A \otimes U_B\right)\;\left(U_A \left|\Psi\right\rangle_A \right) \otimes \left(U_B \left|\Psi\right\rangle_B \right) \\ &= \left(U_A^2 \left|\Psi\right\rangle_A \right) \otimes \left(U_B^2 \left|\Psi\right\rangle_B \right) .\\ \end{align} $$

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    $\begingroup$ Yeah I think i see it now I will be left with something like $I_B \otimes \left(I_A + H_A +\frac{1}{2!} H_A^2+... \right)$ which is $I_B \otimes e^{H_A}$ (dropping $-i$ and $t$) $\endgroup$ – Chris2807 Feb 8 '15 at 23:40
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I just found this post because I was confused by the same step. But I think I got it now with the help of @lionelbrits post and @Chris2807's comment. Just adding this for completeness and maybe helping someone else struggling with this:

$$ \begin{align} e^{(H_A \otimes I_B)} &= \sum_{n=0}^\infty\dfrac{(H_A \otimes I_B)^n}{n!} \\&= I_A \otimes I_B + H_A \otimes I_B + \dfrac{1}{2}(H_A \otimes I_B)^2 + ... \\&= I_A \otimes I_B + H_A \otimes I_B + \dfrac{1}{2}(H_A \otimes I_B)(H_A \otimes I_B) + ... \\&= I_A \otimes I_B + H_A \otimes I_B + \bigg(\dfrac{1}{2}(H_A)^2 \otimes (I_B)^2\bigg)+ ... \\&= (I_A + H_A + \frac{1}{2}(H_A)^2 + ... ) \otimes I_B \\&= e^{H_A}\otimes I_B \end{align} $$

where I also dropped the -i and t and used that $(I_B)^n = I_B$ with $n\in \mathbb{N}$.

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