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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.

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    $\begingroup$ 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. $\endgroup$
    – Ivo Terek
    Commented Jul 27, 2020 at 5:42
  • $\begingroup$ @IvoTerek well, I will confirm again about this with my professor. Thank you too $\endgroup$ Commented Jul 27, 2020 at 7:25
  • $\begingroup$ 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 $\endgroup$ Commented Jul 27, 2020 at 22:25

1 Answer 1

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The second displayed formula you write is correct and standard in exponentiating Lie algebra coproducts into tensor products of group elements: you use it in composing rotation group representations all the time.

The first one is nonsense, and you should desist from mindless general manipulations until you are comfortable with what they actually mean. I am surprised your teacher did not demonstrate this with a trivial counterexample. ("Spectral decomposition"? What?)

Take $X=Y=i\pi \sigma_1 /2$, the obvious symmetric real Pauli matrix.

You then know, on the one hand, that
$$ e^{X}\otimes e^{Y}= e^{i\pi\sigma_1 /2 }\otimes e^{i\pi\sigma_1 /2 } = i\sigma_1 \otimes ~~i\sigma_1 =- \begin{pmatrix}0&0&0&1\\0&0&1&0\\0&1&0&0\\ 1&0&0&0 \end{pmatrix}\equiv -M. $$

On the other hand, $$e^{X\otimes Y}= e^{-\pi^2 M/4}= \cosh (\pi^2/4)~ 1\!\!1 -\sinh (\pi^2/4)~ M.$$

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