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I want to show that the property $e^{tA}e^{tB}=e^{t(A+B)}$ implies that $AB=BA$. Here $A,B$ denotes matrices and $t\in \mathbb{R}$.

Im stuck, have tried to expand both sides with their taylor series but not sure if this is the way to go.

Could anyone give me a hint?

Biggiez
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  • It follows from the Baker-Campbell-Hausdorff formula (a bit of a sledgehammer). – Angina Seng Nov 25 '17 at 19:25
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    You don't need BCH; just compute the coefficient of $t^2$ on both sides. – Qiaochu Yuan Nov 25 '17 at 19:25
  • @QiaochuYuan Why is it enough to compute the coefficient of $t^{2}$ and how does one see that? – Biggiez Nov 25 '17 at 19:34
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    If you computed the coefficient of $t^2$ on both sides you would already know the answer to that question! – Qiaochu Yuan Nov 25 '17 at 19:41
  • @QiaochuYuan Is there an easy way to simplify the left hand side into only one sum, so that the coefficient of $t^{2}$ can easily be computed? – Biggiez Nov 25 '17 at 20:09
  • @Biggiez: the LHS is $\left( \sum A^k \frac{t^k}{k!} \right) \left( \sum B^k \frac{t^k}{k!} \right)$, and using standard facts about multiplying power series we get that the coefficient of $\frac{t^n}{n!}$ is $\sum_{k=0}^n {n \choose k} A^k B^{n-k}$, whereas on the RHS the coefficient is $(A + B)^n$. – Qiaochu Yuan Nov 25 '17 at 20:57
  • @QiaochuYuan Im having trouble with the simplification in the LHS leading to only one sum, could you write down the steps? – Biggiez Nov 26 '17 at 09:18

2 Answers2

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Differentiate 2 times with respect to time $t$ the identity $$e^{tA}e^{tB}=e^{t(A+B)}$$ and then set $t=0$.

RTJ
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  • If I differentiate 2 times w.r.t to $t$ I obtain $(A+B)^{2}e^{tA}e^{tB}=(A+B)^{2}e^{t(A+B)}$. Just wondering if I have differentiated correctly? – Biggiez Nov 25 '17 at 19:38
  • in order to write the LHS, you assume A and B commutes. – karakfa Nov 25 '17 at 19:41
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    @Biggiez: you have not differentiated correctly. – Qiaochu Yuan Nov 25 '17 at 19:42
  • @Biggiez No, $\frac{d}{dt}[e^{tA}e^{tB}]=Ae^{tA}e^{tB}+e^{tA}e^{tB}B$. Can you take it from there? – RTJ Nov 25 '17 at 19:42
  • @CTNT Why does the matrix $B$ end up being multiplicated to the far right and not from the far left as in the first term in the sum? – Biggiez Nov 25 '17 at 19:47
  • Since B and $e^{tB}$ commutes $Be^{tB}=e^{tB}B$ – karakfa Nov 25 '17 at 19:49
  • @Biggiez $\frac{d}{dt}[e^{At}e^{Bt}]=\frac{d}{dt}[e^{At}]e^{Bt}+e^{At}\frac{d}{dt}[e^{Bt}]=Ae^{At}e^{Bt}+e^{At}Be^{Bt}=Ae^{At}e^{Bt}+e^{At}e^{Bt}B$ since $B$ and $e^{Bt}$ commute. – RTJ Nov 25 '17 at 19:52
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    If I've differentiated correctly this time: $L.H.S=(A^{2}+2BA+B^{2})e^{tA}e^{tB}=R.H.S=(A+B)^{2}e^{t(A+B)}$. Setting t=0 we obtain that $A^{2}+2BA+B^{2}=A^{2}+AB+BA+B^{2}$ which in turn gives us that $2BA=AB+BA$, so the property holds only if $BA=AB$ ? – Biggiez Nov 25 '17 at 19:53
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    @Biggiez Correct! – RTJ Nov 25 '17 at 19:54
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This is not what has been asked but to show the identity is valid if $A$ and $B$ commutes, you can...

Define $$ f(t) = e^{t(A+B)}e^{-tB}e^{-tA} $$

Then, $$ f'(t) = (A+B)e^{t(A+B)}e^{-tB}e^{-tA} - e^{t(A+B)}Be^{-tB}e^{-tA} - e^{t(A+B)}e^{-tB}Ae^{-tA} $$

if $A$ and $B$ commutes, we can factor out A and B from the last two. $$ f'(t) = (A+B) f(t) - A f(t) - B f(t) = 0 $$

Therefore $f(t)$ is a constant, you can show that it's identity.

karakfa
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  • The question is about the converse: if $f(t)$ is a constant, show that $A$ and $B$ commute. – Qiaochu Yuan Nov 25 '17 at 19:42
  • I may have understood your solution wrong. But aren't you now assuming that $A$ and $B$ commutes? The question was to show that the identity implies that $A,B$ commutes. – Biggiez Nov 25 '17 at 19:42
  • Yes, this shows if $AB=BA$, then the exponential identity holds. – karakfa Nov 25 '17 at 19:50
  • @karakfa Yes I agree with your statement, the only problem is that the exercise is to show that that identity IMPLIES that $A$ and $B$ commutes. As Qiaochu already stated you have shown the converse of that. – Biggiez Nov 25 '17 at 19:58
  • Yes, right now you have both directions proved. – karakfa Nov 25 '17 at 20:03