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So I am reading the book named 'Quantum Optics An Introduction' by Werner Vogel, Dirk-Gunnar Welsch, Sascha Wallentowitz and in the section 'Phase-space representations', I struggle to follow one step in their derivation of the s-parameterized operator-valued Dirac $ \delta $ function. It reads as follow:

$$\hat{\delta}(\hat{a}-\alpha;s) = \frac{2}{\pi(1-s)}:\exp[-\frac{2\hat{n}(\alpha)}{1-s}]:$$

Then it says this can be further evaluated to this:

$$\hat{\delta}(\hat{a}-\alpha;s) = \frac{2}{\pi(1-s)}:\exp[\frac{s+1}{s-1}\hat{n}(\alpha)]\exp[-\hat{n}(\alpha)]:$$

This is where I struggle to follow. If anyone can explain how to get from the first line to the second line, it will be very helpful.

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  • $\begingroup$ Isn't it just $\frac{s+1}{s-1}-1 = -\frac{2}{1-s}$? $\endgroup$ – Prahar Jul 11 at 13:15
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I don't know about the details of the operator $\hat{n}(\alpha)$, but it seems like a standard application of the BCH formula $$[X,Y] = 0 \; \Rightarrow \; \exp(X)\exp(Y) = \exp(X+Y)$$ with $$X = \frac{s+1}{s-1}\hat{n}(\alpha), \quad Y = -\hat{n}(\alpha). $$ We're only using the fact that $\hat{n}(\alpha)$ commutes with itself; the normal ordering is not important for this argument.

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  • $\begingroup$ That's what I thought initially but the $\hat{n}(\alpha)$ is multiply by 2 in the first line while applying the BCH formula in the second line give me a $s$ in the front $\endgroup$ – J L Jul 10 at 13:47
  • $\begingroup$ Open Mathematica and type $\text{Factor[}(s+1)/(s-1)-1\text{]}$. $\endgroup$ – Hans Moleman Jul 10 at 13:53
  • $\begingroup$ LOL I feel so stupid Thanks man $\endgroup$ – J L Jul 10 at 14:08
  • $\begingroup$ You should probably accept the answer, such that nobody else spends more time looking at it. $\endgroup$ – Hans Moleman Jul 10 at 19:03
  • $\begingroup$ @HansMoleman: Perhaps you can just add in your answer that the BCH formula has the form it has because $\hat{n}(\alpha$ commutes with itself. $\endgroup$ – flippiefanus Jul 11 at 4:18

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