Confusion regarding $e^{I}$ and $e^{A \otimes I}$ and $e^{A \otimes B}$

I am a little confused regarding how the differences between $$e^{I}, e^{A \otimes I}$$ and $$e^{A \otimes B}$$ emerge in regards to how the taylor series acts on them.

$$\displaystyle e^I = \sum_{k=0}^\infty \frac{I^k}{k!}$$

But $$I^{k}$$ for $$k \in N$$ is $$I$$, so $$\displaystyle e^I = \sum_{k=0}^\infty \frac{I^k}{k!}=\displaystyle I \sum_{k=0}^\infty \frac{1}{k!}$$, which ends up as $$Ie$$, with e's along the diagonal.

However, given $$A \otimes I$$: $$\displaystyle e^{A\otimes I} = \sum_{k=0}^\infty \frac{(A \otimes I)^k}{k!} = I_{A} \otimes I + A \otimes I + \frac{1}{2}(A^{2} \otimes I)+.....$$ $$=\displaystyle\sum_{k=0}^\infty (I_{A}+A+\frac{1}{2}A^{2}+...)\otimes I=e^{A}\otimes I$$ Now here I am assuming that the reason we can take the fractions out of $$I$$ completely is because of how the tensor product works, ie, it will still be a factor that multiples $$I$$, but that is an assumption I am making and I am not sure if it is correct.

Furthermore: $$\displaystyle e^{A \otimes B} = \sum_{k=0}^\infty \frac{(A \otimes B)}{k!}=\sum_{k=0}^\infty I_{A} \otimes I_{B}+ A \otimes B + \frac{(A^2 \otimes B^2)}{2}+...$$ I am not as sure in this case, but given there is no common factor to take out, I think its $$e^{A}\otimes e^{B}$$ But if the exponential function acts in this distributive manner on the kronecker product, I am confused as to why it doesn't also do so for the case of $$e^{A\otimes I}$$. I can see in the taylor series why I can be taken out, as it is a common tensor factor, but still, I think I am missing something in my understanding.

• The solution is that the exponential map does not distribute over tensor products, since sums don't distribute over tensor products. $\sum_k A^k\otimes B^k\neq (\sum_k A^k)\otimes(\sum_k B^k)$. Oct 7, 2020 at 11:03
• So can I take it that e^{A \otimes B} would result in an operator that is not decomposable into the tensor product of the exponential function on the individual operators? Oct 7, 2020 at 11:05
• There may be a formula to decompose them, but I don't know any. If there is such a formula, it won't be as simple as the one you proposed. Oct 7, 2020 at 11:06
• Are my expansion up to the point when I mistakenly identify it as distributive? Oct 7, 2020 at 11:31
• Yes, that's right. Oct 8, 2020 at 11:02

We use $$1=I$$ for simplicty. $$e^A\otimes e^B=\sum_n\frac{A^n}{n!}\otimes\sum_m\frac{B^m}{m!}=\sum_{n=0}^\infty\underbrace{\sum_{k=0}^n {n\choose{k}}A^n\otimes B^k}_{L_n}.$$
If $$A$$ and $$B$$ commutes, the binomial theorem holds (kinda) in the following form (use induction), $$(A\otimes 1+1\otimes B)^n=\sum_{k=0}^n {n\choose{k}}A^n\otimes B^k=L_n.$$ Thus, $$e^A\otimes e^B=e^{A\otimes 1+1\otimes B}.$$
Edit: Furthermore, we can use the Baker-Campbell-Hasudorff formula, $$e^A e^B=e^Z,\quad Z=A+B+\frac{1}{2}[x,y]+...$$ and thus, since $$A$$ and $$B$$ commute, $$e^A\otimes e^B=e^{A\otimes 1+1\otimes B}=e^{A\otimes 1}e^{1\otimes B}.$$
If $$B=0$$, $$e^A\otimes 1=e^{A\otimes 1}.$$ If $$B=0$$ and $$A=1$$ $$e^1\otimes 1=e^{1\otimes 1}.$$ The non-commutative case can be regarded in the same fashion together with the results of https://arxiv.org/abs/1707.03861 and using the BCH formula.