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 $$ \frac d {da^\dagger} \sqrt{a^\dagger a} = a \left(2\sqrt{a^\dagger a}\right)^{-1} + \lim_{\delta \to 0} {\left[\sqrt{(a^\dagger + \delta) a}, \sqrt{a^\dagger a}\right]} \left( \delta\sqrt{(a^\dagger + \delta) a} + \delta\sqrt{a^\dagger a} \right)^{-1}. $$

  • $\begingroup$ Please show more effort. What have you tried to do before posting the question? What do you think the answer is? $\endgroup$ – Abhimanyu Pallavi Sudhir Jun 19 '13 at 8:46
  • $\begingroup$ @dimension10 Is it really that simple? I don't even understand how one would define the square root. $\endgroup$ – user10001 Jun 19 '13 at 9:29
  • $\begingroup$ @user10001: I know, when the OP starts trying to solve it, he will realise that himself. $\endgroup$ – Abhimanyu Pallavi Sudhir Jun 19 '13 at 9:31
  • $\begingroup$ $\sqrt A$ is an operator such as $\sqrt A \sqrt A$ = A. What's wrong? $\endgroup$ – Andrii Jun 19 '13 at 9:59
  • $\begingroup$ Good, you added more detail so I retracted my -1. $\endgroup$ – Abhimanyu Pallavi Sudhir Jun 19 '13 at 11:50

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 $$\left[a, \sqrt{\hat{n}}\right] = \left(\sqrt{\hat{n}+1} - \sqrt{\hat{n}}\right)~ a.$$

  • $\begingroup$ Thanks, it is simple. I believe it was in undergrad QM course I've completely forgotten. BTW, for some reason I can't vote for your answer -- it says 'You can't vote for your own post.' $\endgroup$ – Andrii Jun 19 '13 at 13:15
  • $\begingroup$ @Andrii : Bug ? Normally, you click on the check box. Were you logged in ? Use the meta channel (see menu at the top of the S.E.page) if something does not work. $\endgroup$ – Trimok Jun 19 '13 at 16:45

We are given

$$[\hat{a},\hat{a}^{\dagger}]~=~{\bf 1}.$$




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

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

  3. Argue that the commutator $[\hat{a},\sqrt{\hat{n}}]$ (at the physical level of rigor) should have the following (partially) normal-ordered form $$[\hat{a},\sqrt{\hat{n}}]= (\sqrt{\hat{n}+{\bf 1}}- \sqrt{\hat{n}})\hat{a}.$$

  • $\begingroup$ Thanks, it is simple. I believe it was in undergrad QM course I've completely forgotten. $\endgroup$ – Andrii Jun 19 '13 at 13:15
  • $\begingroup$ Sorry, am I missing something? I don't believe $f(x) = \sqrt{x}$ qualifies for step 2 because step 2 requires that you should expand the (bounded) operator as a Taylor series: the Taylor series doesn't exist around nought and moreover $\hat{n}$ is not bounded: so clearly there is some physical argument I'm missing? $\endgroup$ – WetSavannaAnimal Aug 19 '13 at 7:56
  • $\begingroup$ Property 2 also holds for the square root function. In fact, property 2 holds for a much wider class of functions $f$ than just the class of (real) analytic functions, cf. generalizations of the Weierstrass approximation theorem. In particular, the square root function can be view as a limit of a sequence $(f_n)_{n\in\mathbb{N}}$ of real analytic functions, where the property 2 holds, and therefore property 2 also holds for the square root function itself. $\endgroup$ – Qmechanic Aug 19 '13 at 8:35
  • $\begingroup$ Thanks, I must admit I had not thought of the more general, Weierstrass-approximated class. I'll have to think about this - non-boundedness might pose a hitch to a strict mathematical proof grounded on the Stone-Weierstrass theorem. But, given that at least one square root certainly exists, I agree it is a physically perfectly reasonable thing to assume that observables behave in this way. $\endgroup$ – WetSavannaAnimal Aug 20 '13 at 0:37

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 $n \in \mathbb{Z}$.

  • $\begingroup$ One can easily define $A^q$ for any q rational and A operator. I think the only problem here occures when operator acts in space with infinite dimensions. For example, in this case matrix elements of inverse operator can be singular, as it is with the of boson annihilation operator $a^{-1}$. $\endgroup$ – Andrii Jun 19 '13 at 12:25
  • $\begingroup$ Oops they are not singular but close to zero (in Fock representation), becase the determinant of matrix of operator $a$ is $1\cdot2\cdot3\ldots$ -- infinite =) $\endgroup$ – Andrii Jun 19 '13 at 12:39
  • $\begingroup$ as for your answer, 1) I doubt the descent of the binomial theorem for $n$ not natural; 2) your final formula should be understood as $.5 \sqrt a a^{\dagger-1/2}$ or $.5 a^{\dagger-1/2} \sqrt a $ or is it the same? $\endgroup$ – Andrii Jun 19 '13 at 12:43
  • 1
    $\begingroup$ It should be $\sqrt{a}$ first, but my answer is wrong, please look those given by Qmechanic and Trimok $\endgroup$ – Jorge Lavín Jun 19 '13 at 12:59

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