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I have a problem in Quantum mechanics 1 with Operators. I have to prove the following equation. I tried it for about 4 hours without any result:

Condition: $[[\hat A,\hat B],\hat A]=[[\hat A,\hat B],\hat B]=0$

$$ e^{\hat A} \hat B = (\hat B + [\hat A,\hat B]) e^{\hat A} $$

Info: $e^{\hat A}=\sum\limits_{n=0}^\infty \frac{(\hat A) ^n}{n!}$

Maybe you could help me?

I've done these steps:

$$ [e^{\hat A}, \hat B] = [\hat A,\hat B] e^{\hat A} = e^{\hat A} [\hat A,\hat B] $$

$$ \sum\limits_{n=0}^\infty \left( \frac{(\hat A)^n}{n!} \hat B - \hat B \frac{(\hat A)^n}{n!}\right) = \sum\limits_{n=0}^\infty \left( \frac{(\hat A)^n}{n!} \hat A \hat B - \frac{(\hat A)^n}{n!} \hat B \hat A\right) $$

But now I don't know how to go on...

bjs
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1 Answers1

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$[A^n,B]=n[A,B]A^{n-1}$ (prove by induction) $[\exp(A),B]=\Sigma\frac{1}{n!}[A^n,B]=\Sigma\frac{1}{n!}n[A,B]A^{n-1}=[A,B]\exp(A)$

akhmeteli
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