# Tensor product of exponential operators

My quantum mechanics professor asked to show a demonstration of the following mathematical result:

$$e^{X\otimes Y}=e^{X}\otimes e^{Y}$$

When $$X$$ and $$Y$$ are some normal operators. But I think that this is wrong, the only thing that I can achieve is: $$e^{X\otimes I_{b}+I_{a}\otimes Y}=e^{X}\otimes e^{Y}$$

Using this material. Have some way to show that $$e^{X\otimes Y}=e^{X\otimes I_{b}+I_{a}\otimes Y}$$ in general? He spoke to me to use spectral decomposition, but I don't know-how.

• Make sure $X\otimes Y$ is not actually an abuse of notation for $X\otimes I + I\otimes Y$, this is somewhat common depending on the context. You might be already done without realizing it. Commented Jul 27, 2020 at 5:42
• @IvoTerek well, I will confirm again about this with my professor. Thank you too Commented Jul 27, 2020 at 7:25
• PhysMath using this I can reach the second equality $e^{X\otimes I_{b}+I_{a}\otimes Y}=e^{X}\otimes e^{Y}$, but thank you Commented Jul 27, 2020 at 22:25

Take $$X=Y=i\pi \sigma_1 /2$$, the obvious symmetric real Pauli matrix.
$$e^{X}\otimes e^{Y}= e^{i\pi\sigma_1 /2 }\otimes e^{i\pi\sigma_1 /2 } = i\sigma_1 \otimes ~~i\sigma_1 =- \begin{pmatrix}0&0&0&1\\0&0&1&0\\0&1&0&0\\ 1&0&0&0 \end{pmatrix}\equiv -M.$$
On the other hand, $$e^{X\otimes Y}= e^{-\pi^2 M/4}= \cosh (\pi^2/4)~ 1\!\!1 -\sinh (\pi^2/4)~ M.$$