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I was seeing the proof that $\exp(A)\exp(B)=\exp(A+B)$ on link

Show that $ e^{A+B}=e^A e^B$

where uses the hypothesis $AB=BA$?

Thanks!

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  • $\begingroup$ The use of the Binomial Theorem. $\endgroup$
    – Hayden
    Commented Apr 15, 2014 at 20:54
  • $\begingroup$ Thanks Hayden, but I can't understand. To use the Binomial Theorem I don't need change the position A and B, so I could not use AB=BA. $\endgroup$
    – yemino
    Commented Apr 15, 2014 at 20:57
  • $\begingroup$ Now I see, Thanks Hayden and all of you!! $\endgroup$
    – yemino
    Commented Apr 15, 2014 at 21:00

3 Answers 3

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The step that claims $$\sum_{m=0}^l{\frac{l!}{m!(l-m)!}A^{m}B^{l-m}}=(A+B)^l$$ For example, we would usually say that $$(A+B)^2=A^2+AB+BA+B^2$$ Only because $$AB=BA$$ can this be simplified.

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The commutativity of $A$ and $B$ is used where the double summation becomes a single summation. Consider as an example the term in that sum with $\ell =2$: $(A+B)^2 = A^2+AB+BA+B^2$.

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The binomial formula $$(A+B)^l=\sum_{k=0}^l{l\choose k}A^kB^{l-k}$$ is valid if $AB=BA$.

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  • $\begingroup$ Not quite: $(A+B)^{1}=A+B$ is valid even if $AB\ne BA$. I think you meant "if" instead of "only if." (sorry to be picky!) $\endgroup$
    – user142299
    Commented Apr 15, 2014 at 20:57
  • $\begingroup$ In the proof $l$ take the values from $0$ to $\infty$. $\endgroup$
    – user63181
    Commented Apr 15, 2014 at 21:03

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