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It is well known that if $X,Y$ are commuting matrices, then their exponential commute: $$XY=YX\quad\implies\quad e^Xe^Y=e^Ye^X.$$ Now, I am wondering if the following generalization holds:

Question: If $XYZ=ZXY$, does $e^Xe^Ye^Z=e^Ze^Xe^Y$?

Note that if $Z$ commutes with both $X$ and $Y$, then it is obvious.

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    $\begingroup$ I do not believe that. $XYZ=ZXY$ does not imply $X^2 Y^2 Z^2 = Z^2 X^2 Y^2$, for instance, so the two Cauchy products $$\left(\sum_{n\geq 0}\frac{X^n}{n!}\right)\cdot\left(\sum_{n\geq 0}\frac{Y^n}{n!}\right)\cdot \left(\sum_{n\geq 0}\frac{Z^n}{n!}\right)$$ and $$\left(\sum_{n\geq 0}\frac{Z^n}{n!}\right)\cdot\left(\sum_{n\geq 0}\frac{X^n}{n!}\right)\cdot \left(\sum_{n\geq 0}\frac{Y^n}{n!}\right)$$ may not match. $\endgroup$
    – Jack D'Aurizio
    Dec 14, 2015 at 11:17
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    $\begingroup$ Hint: What if $Y=0$? $\endgroup$
    – user1551
    Dec 14, 2015 at 11:18
  • $\begingroup$ It's fine to speculate like this, but even better is to examine the proof that commuting matrices have commuting exponentials ("the binomial theorem applies"), then to ask yourself if there's any reason to expect the proof generalizes. Here the answer is a clear "no", as Jack D'Aurizio notes. And user1551's comment, perhaps the first or second case one "ought" to try, immediately dispatches the conjecture. :) $\endgroup$
    – Andrew D. Hwang
    Dec 14, 2015 at 11:29

3 Answers 3

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OP is asking if

$$ [XY,Z]~=~0\qquad \stackrel{?}{\Rightarrow}\qquad [e^Xe^Y,e^Z]~=~0~? \tag{1}$$

In a comment above, user1551 has already pointed out obvious counterexamples if $X=0$ xor $Y=0$.

Here we will give a counterexample with invertible $2\times 2$ matrices, namely the Pauli matrices:

$$ X~=~ i\pi \sigma_x, \qquad Y~=~ \frac{i\pi}{2} \sigma_y, \qquad Z~=~ \frac{i\pi}{2} \sigma_z, \tag{2}$$

$$ e^X~=~ -{\bf 1}_{2\times 2}, \qquad e^Y~=~ i\sigma_y, \qquad e^Z~=~ i\sigma_z. \tag{3}$$

Now $XY$ is proportional to $\sigma_z$ and therefore commutes with $Z$; while $e^Xe^Y$ is proportional to $\sigma_y$, and hence anticommutes with $e^Z$ (rather than commutes).

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If $X$ and $Y$ commute, $$e^Y e^X = e^X e^Y = e^{X+Y}$$ So $Z$ needs to commute with $X+Y$ for that to be true in this particular case, and not with $e^{XY}$.

An easy counterexample is $Y=0$, $X$ and $Z$ such that $[X,Z]\neq 0$. The condition is still fulfilled but, obviously, $e^X e^Z \neq e^Z e^X$ in the general case.

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  • $\begingroup$ $e^Z$ and $e^{X+Y}$ commute does not imply that $Z,X+Y$ commute. In the same way $[X,Z]\not= 0$ does not imply that $e^Xe^Z\not= e^Ze^X$. $\endgroup$
    – user91684
    Dec 14, 2015 at 15:57
  • $\begingroup$ And that's why I said "in the general case" $\endgroup$
    – AnalysisStudent0414
    Dec 14, 2015 at 16:07
  • $\begingroup$ For point 1. You write "So $Z$ needs..." that is false. For point 2. "in the general case" implies that "it is true for any matrices". That you must write is: "for generic matrices, one has....". Moreover, this last result is not obvious. $\endgroup$
    – user91684
    Dec 14, 2015 at 16:13
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If $(XY)Z=Z(XY)$ we can only say that $$ e^Ze^{XY}=e^{XY}e^Z=e^{Z+XY} $$

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