I have been reading this paper ("Operator ordering in quantum optics theory and the development of Dirac’s symbolic method" by Hong-yi Fan), and on page 3 (right-hand column) the author writes that $:A:B::\;=\; :AB:$. They then go on to imply that if $:AB: \; =\; 1\!\! 1$, then $:B: \; = \; :A^{-1}:$.

Is this generally the case? If it is, how can one show that it is true?

Essentially I want to know how this holds in order to understand why $:e^{a^{\dagger}a}W: \; =\; :e^{a^{\dagger}a}:W:: \; = \; 1\!\! 1$ implies $:W: \; =\; e^{-a^{\dagger}a}$?

  • $\begingroup$ The :'s to the left of A and B are opening the normal ordering, so the second opening (to the left of B) is not doing anything because it is inside the first. So your last sentence holds because there is a trivial ordering. If the ordering was :A : :B:, then each operator is normal ordered separately, and the inverse relations would not hold. $\endgroup$ – matrp Mar 11 at 17:15
  • $\begingroup$ @matrp Is it always true that normal-ordering within normal-ordering is trivial? Also, is it always true that $:AB:\; =\; 1\!\! 1\;\Rightarrow\; :B: \; = \; :A^{-1}:$? $\endgroup$ – Will Mar 11 at 18:03
  • $\begingroup$ For the first question, yes. Because you will order the one inside, and then that will get re-ordered, so the first will have no effect on the final result. For the second question, I am not sure, I cannot see why this should be true. If I think of something, I'll submit as an answer $\endgroup$ – matrp Mar 12 at 18:00
  • $\begingroup$ I delete my answer since I realize that your question is different than what I was understanding $\endgroup$ – Nogueira Mar 13 at 18:26

Let ${\cal A}$ be the freely$^1$ generated algebra of $\hat{a}$, $\hat{a}^{\dagger}$ and $\hat{\mathbb{1}}$ (modulo the rule that $\mathbb{1}$ can be removed from any term different from $\mathbb{1}$).

Let ${\cal I}$ be the 2-sided ideal generated by the commutator $[a,\hat{a}^{\dagger}]$. Consider the quotient algebra ${\cal B}:={\cal A}/{\cal I}$. Then

$$ \forall \hat{C}\in {\cal A}:\qquad :\hat{C}:~=~0 \qquad\Leftrightarrow\qquad \hat{C}~\in~{\cal I}.$$

OP's sought-for relation follows: $$~:\hat{A}\hat{B}:~=~\hat{\mathbb{1}} \qquad\Leftrightarrow\qquad \hat{A}\hat{B}-\hat{\mathbb{1}}~\in~{\cal I}$$ $$\qquad\Leftrightarrow\qquad \hat{B}-\hat{A}^{-1}~\in~{\cal I} \qquad\Leftrightarrow\qquad :\hat{B}:~=~:\hat{A}^{-1}:~.$$


$^1$ In particular, one is not allowed to use the CCR $[\hat{a},\hat{a}^{\dagger}]=\mathbb{1}$, cf. e.g. this Phys.SE post.

  • $\begingroup$ Thanks for your answer. Is normal ordering linear then? Or is it only true for the identity operator, i.e., $:AB:\; -\; 1\!\!1 \; =\; :AB-1\!\!1:$? I’m confused though, as on the Wikipedia page it says it’s not linear. The one on normal ordering (en.wikipedia.org/wiki/Normal_order), under the “Notation” section. $\endgroup$ – Will Mar 14 at 19:13
  • $\begingroup$ Normal order: ${\cal A}\to {\cal A}$ is linear. Seems like Wikipedia is referring to the issues mentioned in footnote 1, i.e. Wikipedia's algebra is not freely generated. $\endgroup$ – Qmechanic Mar 14 at 19:33
  • $\begingroup$ Why is it ok in this case then? To be honest, I’m still not quite sure why $:C:\; =0\iff C=0$? $\endgroup$ – Will Mar 14 at 19:50
  • $\begingroup$ Concerning operator algebra within normal-ordered products, is it also true that, if $A$ is some operator, and $:U\cdots V: \; =\; :W\cdots X:$, then $:AU\cdots V: \; =\; :AW\cdots X:$? If this is correct, then one has that for $:AB: \; =\; 1\!\!1 \; =\; :1\!\!1:$, and the inverse operator $A^{-1}$: $$:A^{-1}AB: \; =\; :A^{-1}1\!\!1: \qquad\Rightarrow\qquad :B: \; = \; :A^{-1}:\;.$$ $\endgroup$ – Will Mar 17 at 15:44
  • $\begingroup$ $\uparrow$ Yes. $\endgroup$ – Qmechanic Mar 17 at 21:06

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