# $s$-parameterized operator-valued Dirac delta function

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.

• Isn't it just $\frac{s+1}{s-1}-1 = -\frac{2}{1-s}$? – Prahar Jul 11 at 13:15

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.
• 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 – J L Jul 10 at 13:47
• Open Mathematica and type $\text{Factor[}(s+1)/(s-1)-1\text{]}$. – Hans Moleman Jul 10 at 13:53
• @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. – flippiefanus Jul 11 at 4:18