Commutator with a square root How to find the commutator $[a, \sqrt{a^\dagger a}]$? Here $a$ is a usual bosonic annihilation operator, and $[a, a^\dagger] = 1$.
The first thing I tried is
$$
[x,A] = [x, \sqrt{A}]\sqrt{A} + \sqrt{A}[x, \sqrt{A}]
$$
which clearly shows either commutators similarity to derivatives and the difference between them. In general, $[[x,\sqrt A], \sqrt A] \neq 0$, so $[x, \sqrt{A}] \neq \frac{[x,A]}{2\sqrt A}$.
The usual trick (see Mandel and Wolf, Optical Coherence and Quantum Optics)
$$
[a, f(a,a^\dagger)] = \frac{df}{da^\dagger}
$$
is of no use here. Indeed, calculating the derivative defined as
$$
\frac{df(a,a^\dagger)}{da^\dagger} = 
    \lim_{\delta \to 0} \frac {f(a,a^\dagger + \delta) - f(a,a^\dagger)}\delta
$$
for $f = \sqrt{a^\dagger a}$ leads to
$$
    \frac d {da^\dagger} \sqrt{a^\dagger a} 
         = a \left(2\sqrt{a^\dagger a}\right)^{-1} 
         + \lim_{\delta \to 0} 
           {\left[\sqrt{(a^\dagger + \delta) a}, \sqrt{a^\dagger a}\right]} 
           \left(
              \delta\sqrt{(a^\dagger + \delta) a} + \delta\sqrt{a^\dagger a}
           \right)^{-1}.
$$
 A: You have to use the eigenstates $|n\rangle $ of the operator $\hat{n} = a^\dagger a$.
You have, then, that $a \sqrt{\hat{n}} ~|n\rangle    = a \sqrt{n} ~|n\rangle = \sqrt{n} ~ a |n\rangle              = \sqrt{\hat{n}+1} ~ a |n\rangle ,$  where the last equality is because $a |n\rangle \sim |n-1\rangle$.
So, $\left[a, \sqrt{\hat{n}}\right]~ |n\rangle = \left(\sqrt{\hat{n}+1} - \sqrt{\hat{n}}\right)~ a |n \rangle   $, for each $|n\rangle  $, and therefore
$$\left[a, \sqrt{\hat{n}}\right] = \left(\sqrt{\hat{n}+1} - \sqrt{\hat{n}}\right)~ a.$$
A: We are given 
$$[\hat{a},\hat{a}^{\dagger}]~=~{\bf 1}.$$ 
Let 
$$\hat{n}~:=~\hat{a}^{\dagger}\hat{a}.$$
Hints: 


*

*Prove that 
$$\hat{a}\hat{n} = (\hat{n}+{\bf 1}) \hat{a}.$$

*Prove that if $f:\Omega \subseteq \mathbb{C}\to \mathbb{C}$ is a sufficiently well-behaved function, then 
$$\hat{a}f(\hat{n}) = f(\hat{n}+{\bf 1}) \hat{a}.$$

*Argue that the commutator $[\hat{a},\sqrt{\hat{n}}]$ (at the physical level of rigor) should have the following (partially) normal-ordered form
$$[\hat{a},\sqrt{\hat{n}}]= (\sqrt{\hat{n}+{\bf 1}}- \sqrt{\hat{n}})\hat{a}.$$
A: Formally, you can say
$$ \frac{df(a,a^\dagger)}{da^\dagger} = 
    \lim_{\delta \to 0} \frac {f(a,a^\dagger + \delta) - f(a,a^\dagger)}\delta $$
for $f(a,a^{\dagger})=\sqrt{a a^{\dagger}}=\sqrt{a}\sqrt{a^{\dagger}}$
$$ \frac{df(a,a^\dagger)}{da^\dagger} =\sqrt{a}  
    \lim_{\delta \to 0} \frac {\left(\sqrt{a^{\dagger}+\delta}-\sqrt{a^{\dagger}} \right)}\delta $$
Note that $(a^{\dagger}+\delta)^{n}=(a^{\dagger})^n+(a^{\dagger})^{n-1}n\delta+O(\delta^2)$ (binomial theorem), so that
$$ \sqrt{a}  
    \lim_{\delta \to 0} \frac {\left(\sqrt{a^{\dagger}+\delta}-\sqrt{a^{\dagger}} \right)}\delta=\sqrt{a}\lim_{\delta \to 0}\frac{\sqrt{a^{\dagger}}}{\delta}+\frac{\delta}{2\sqrt{a^{\dagger}}\delta}+\frac{O(\delta^2)}{\delta}-\frac{\sqrt{a^{\dagger}}}{\delta}$$
And finally
$$\frac{df(a,a^\dagger)}{da^\dagger} = \frac{\sqrt{a}}{2\sqrt{a^{\dagger}}} $$
However I'm not sure if $\left(a^{\dagger}\right)^{n}$ is defined for $n \in \mathbb{Q}$ or even $n \in \mathbb{Z}$.
