In the paper "Operator ordering in quantum optics
theory and the development of Dirac’s
symbolic method" by Hong-yi Fan, as referenced in this question, the authors mention the property
$$:A:B::\;=\; :AB:$$ for normal-ordering operation $:\circ:$. This means that one can delete a normal-ordering symbol within another normal-ordering symbol (seemingly at odds with the answer to this question). The paper then goes on to prove the two relations defining the vacuum state
$$|0\rangle\langle 0|=:e^{-a^\dagger a}:\tag{17}$$ and the parity operator
$$e^{i\pi a^\dagger a}=:e^{-2a^\dagger a}:\tag{46}$$ for bosonic operators $a$. Are these two related in any useful way?
Making copious use of $:A:B::\;=\; :AB:$, I seem to be able to choose $A=1$ and set $::B::=:B:$ etc. to achieve
\begin{aligned}
|0\rangle\langle 0|\quad=\quad(|0\rangle\langle 0|)^2 \quad&\Rightarrow\quad :|0\rangle\langle 0|:\quad=\quad:(|0\rangle\langle 0|)^2:\\&
\Rightarrow\quad :\quad:e^{-a^\dagger a}:\quad:\quad=\quad:\quad:e^{-a^\dagger a}:\quad:e^{-a^\dagger a}:\quad:\\
&
\Rightarrow \quad:e^{-a^\dagger a}:\quad=\quad:e^{-a^\dagger a}e^{-a^\dagger a}:\\
&
\Rightarrow \quad|0\rangle\langle 0|\quad=\quad:e^{-2a^\dagger a}:\\
&
\Rightarrow \quad|0\rangle\langle 0|\quad=\quad e^{i \pi a^\dagger a}=(-1)^{a^\dagger a}.
\end{aligned}
Obviously this makes no sense, which leads me to suspect the relationship $:A:B::\;=\; :AB:$ and wonder if there is some "freshman's dream" problem in these calculations. It would be nice to know why this is incorrect, but my main question is still whether there is a useful relationship between the vacuum and the parity operator.
Bonus: should I expect the normally ordered operator $:e^{-m a^\dagger a}:$ to give something familiar for other integer values of $k$?
Best Answer
Let the 3 quantities $f$, $g$ and $h$ depend on $a$ and $a^{\dagger}$. The nested property $$N(f N(g) h)= N(fgh)\tag{A}$$ of the normal order symbol $N$ is valid as long as one does not apply the CCR $$[a,a^{\dagger}]~=~{\bf 1}\tag{B}$$ under the normal-order symbol $N$, cf. e.g. this Phys.SE post.
It turns out that the CCR (B) is used in the derivation of eqs. (17) & (46). Hence OP's last calculation is not valid.
Let us for completeness sketch an independent proof of eqs. (17) & (46). If we define a vertex operator $$V(\beta)~=~N(e^{\beta a^{\dagger}a})~=~\sum_{n\in\mathbb{N}_0}\frac{\beta^n}{n!} (a^{\dagger})^n a^n, \tag{C}$$ and coherent state $$ |z)~=~e^{za^{\dagger}}|0\rangle, \qquad z~\in~\mathbb{C}, \qquad a|z)~\stackrel{(B)}{=}~z|z),\tag{D}$$ then we calculate $$V(\beta)|z)~\stackrel{(C)+(D)}{=}~|(1\!+\!\beta)z).\tag{E} $$ It is not hard to see from eq. (E) that $$\begin{align} V(\beta)V(\beta^{\prime})~=~&V\left((\beta\!+\!1)(\beta^{\prime}\!+\!1)\!-\!1\right), \cr V(0)~=~&{\bf 1}, \cr V(-1)~=~&|0\rangle\langle 0|,\cr V(-2)~=~&e^{i\pi a^{\dagger}a},\end{align} \tag{F} $$ which confirm eqs. (17) & (46). Note the implicit use of the CCR (B). $\Box$
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