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.

  • 2
    $\begingroup$ The solution is that the exponential map does not distribute over tensor products, since sums don't distribute over tensor products. $\sum_k A^k\otimes B^k\neq (\sum_k A^k)\otimes(\sum_k B^k)$. $\endgroup$ Oct 7, 2020 at 11:03
  • $\begingroup$ 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? $\endgroup$ Oct 7, 2020 at 11:05
  • $\begingroup$ 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. $\endgroup$ Oct 7, 2020 at 11:06
  • $\begingroup$ Are my expansion up to the point when I mistakenly identify it as distributive? $\endgroup$ Oct 7, 2020 at 11:31
  • 1
    $\begingroup$ Yes, that's right. $\endgroup$ Oct 8, 2020 at 11:02

1 Answer 1


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.


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