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This question is concerned with the last line of these notes by M. Fowler titled "Coherent States of the Simple Harmonic Oscillator."

I understand the things before that but I don't see how the last line comes about.

Thank you, George

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Making the question a little more self-contained is both friendly to potential answerers, and avoids the question becoming unanswerable due to something happening to the link. By adding the name and author of the notes I intend to make them searchable in case the link changes. (If I were more ambitious I might type the last line and a little context to actually make the question self contained.) (This comment was deleted and reposted because of a typo.) –  Jonas Meyer Oct 2 '12 at 2:38
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2 Answers

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You have that $e^{A+B}=e^Ae^Be^{-\frac12[A,B]}$. Exchanging the roles of $A$ and $B$ you get $$ e^Be^Ae^{-\frac12[B,A]}=e^Ae^Be^{-\frac12[A,B]}, $$ so $$ e^Be^A=e^Ae^Be^{-\frac12[A,B]}e^{\frac12[B,A]}=e^Ae^Be^{-\frac12[A,B]}e^{-\frac12[A,B]} =e^Ae^Be^{-[A,B]}. $$ This, of course, provided that the original requirement of $[A,B]$ commuting with both $A$ and $B$ holds (so that the original identities are valid).

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Thank you! I feel so silly now! –  George Sep 29 '12 at 15:22
    
You are welcome, and don't worry: we all have a long story of missing the obvious. –  Martin Argerami Sep 29 '12 at 15:25
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So, assume we know $e^{A+B} = e^A e^B e^{-\frac12[A,B]}$, then using $A+B=B+A$, this also must hold: $$e^A e^B e^{-\frac12[A,B]} = e^{A+B} = e^{B+A} = e^B e^A e^{-\frac12[B,A]} $$

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Thank you very much! –  George Sep 29 '12 at 16:20
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