Relationship between normal-ordered vacuum state and parity operator

Question

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?

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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.

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Bonus: should I expect the normally ordered operator $:e^{-m a^\dagger a}:$ to give something familiar for other integer values of $k$?

Answer 1

1. 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.

2. It turns out that the CCR (B) is used in the derivation of eqs. (17) & (46). Hence OP's last calculation is not valid.

3. 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). $\Box$

References:

1. Hong-yi Fan, Operator ordering in quantum optics theory and the development of Dirac's symbolic method, J. Opt. B: Quantum Semiclass. Opt. 5 (2003) R147.

Answer 2

Another possible approach is to leverage the identity: $$F(\lambda)\equiv\lambda^{a^\dagger a} = N(e^{(\lambda-1)a^\dagger a}),$$ which is found for example in [CG1969] (see Eqs. 4.31 and 4.35).

The LHS is of course to be understood as $\lambda^{a^\dagger a}\equiv e^{a^\dagger a\ln\lambda}$. The operator $F(\lambda)$ is (quoting from the paper) finite for $|\lambda|\le1$ bounded and trace-class for $|\lambda|<1$.

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Proof of the identity

This relation is shown in the paper reasoning in terms of its derivative wrt $\lambda$ and the initial condition $F(1)=1$. Another possible approach, more explicit albeit perhaps more involved, is to directly perform the Taylor expansion of the LHS, and use the known normally-ordered expansion for $(a^\dagger a)^n$ terms: $$F(\lambda)=\sum_{n=0}^\infty \frac{(\ln\lambda)^n}{n!} (a^\dagger a)^n, \\ (a^\dagger a)^n = \sum_{k=0}^n {n\brace k} a^{\dagger k}a^k,$$ where ${n\brace k}$ are the Stirling numbers of the second kind (i.e. the number of partitions of $n$ elements in $k$ subsets). Putting the above two together we get $$F(\lambda) = \sum_{k=0}^\infty \left( \sum_{n=k}^\infty {n\brace k} \frac{(\ln\lambda)^n}{n!} \right) a^{\dagger k}a^k,$$ and then the conclusion using the identity $$ \sum_{n=k}^\infty {n\brace k} \frac{(\ln\lambda)^n}{n!} = \frac{(\lambda-1)^k}{k!} \tag X. $$ For example, for $k=0$ (X) holds because ${n\brace 0}=\delta_{n,0}$. While for $k=1$ we have ${n\brace 1}=1$, and thus (X) amounts to $\sum_{n=1}^\infty \frac{(\ln\lambda)^n}{n!} = \lambda-1$. Larger $k$ can be similarly proved (albeit less obviously so) using recursive combinatorial identities such as $${n \brace 2} = 2^{n-2} + 2^{n-1}+...+2+1 = 2^{n-1}-1, \\ {n\brace 3} = 3^{n-3}+{3\brace 2}3^{n-4}+{4\brace 2} 3^{n-5} +\cdots + {n-2\brace 2}3+{n-1\brace 2}. $$

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Use the identity to prove statements at hand

In the paper the authors then also argue that $F(\lambda=0)=|0\rangle\langle0|$ (see below Eq. 4.41). I think one can simply see this being the case from $\lambda^{a^\dagger a}|n\rangle=\lambda^n|n\rangle$, which in the limit $\lambda\to0$ gives $F(0)|n\rangle=\delta_{n,0}|n\rangle$. So knowing this we get one of the statements at hand: $$F(0)=|0\rangle\langle0| = N(e^{-a^\dagger a}).$$ On the other hand, for $\lambda=-1$, we get $$(-1)^{a^\dagger a}=N(e^{-2a^\dagger a}).$$