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 $k$?

  • $\begingroup$ @CosmasZachos thanks, I might need help unpacking your comments. I can write a Taylor series expansion for each function of $N$: the $O(1)$ term is the same of course, the coefficient of $N$ is $-1-1/2!-(-1)(-2)/3!-\cdots=-\ln 2$ for the former and $i \pi$ for the latter, etc. Are you simply saying that the eigenvalues of each operator are very different ($0,1,2,\cdots$ vs $\pm1$)? $\endgroup$ Commented Jan 6, 2022 at 21:39
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    $\begingroup$ Was just a formal wisecrack, to bypass the noncommutative structures of the right hand side. The lhs projector is, according to Fan, $1-N+N(N-1)/2!+...$ with zeros at n=1,2,3,... but don't imagine it is $\lim_{x\to 1} (1+x)^{-N}$ ... The parity operator is more elegantly written as $\cos \pi N$. The two appear unrelated, but it is an MSE question. If you are willing to sacrifice normal ordering interpretations, you may dismiss the red herring and contrast the left hand sides. $\endgroup$ Commented Jan 6, 2022 at 22:04
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    $\begingroup$ @CosmasZachos that helps. My takeaway from Fan's paper was that normal ordering is extremely useful, but perhaps this must be tempered (or I need to be aware of more mathematical subtleties to fully take advantage thereof) $\endgroup$ Commented Jan 6, 2022 at 22:57
  • $\begingroup$ If you want, I could expound on the idempotent operator function of Fan's, which condenses to the devilish $f(x)=1+\Gamma (1-x)/(x\Gamma (-x))$ by virtue of the residues of the simple poles of the Gamma function! Perhaps surprisingly, it vanishes for not just all integers >1, but even non-integers, not apparent from the series! If you wanted to know more, I could memorialize it in a fake non-answer. This function is unrelated to the cosine representing the parity operator. $\endgroup$ Commented Jan 12, 2022 at 19:05

2 Answers 2

  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). $\Box$


  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.
  • $\begingroup$ Thanks! It was not explicit to me that the CCR was used in the derivations. Is that because defining $|n\rangle\propto a^{\dagger}|0\rangle$ implicitly assumes the CCR to use ladder operators to shift the eigenvalue of $a^\dagger a$? Must be $\endgroup$ Commented Jan 7, 2022 at 14:32
  • $\begingroup$ Or even.. does $|0\rangle\langle 0|^2=|0\rangle\langle 0|$ implicitly assume the CCR? $\endgroup$ Commented Jan 7, 2022 at 15:28
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Commented Jan 7, 2022 at 15:51
  • $\begingroup$ Fantastic, thanks! So we can easily prove $V(-1)^2=V(-1)$, $V(-2)^2=V(0)$, etc., find rotation operators $\exp{i \theta a^\dagger a}=V(-1+e^{i\theta}$ and keep going... Are your latter expressions normalized? I certainly agree that $a$ has eigenstate $V(\beta)|z)$ with eigenvalue $(1+\beta)z$ but that does not imply that $V(\beta)|z)$ is normalized. $\endgroup$ Commented Jan 7, 2022 at 20:03
  • $\begingroup$ The counterexample is that there should be no linear amplification operator that enacts $\alpha|z_1)+\beta|z_2)\to\alpha|\lambda z_1)+\beta|\lambda z_2)$ [equivalently, $\lambda^{a^\dagger a}|z\rangle =\exp(-|\lambda z|^2/2 + |z|^2/2)|\lambda z\rangle$ in my usual notation]. $\endgroup$ Commented Jan 7, 2022 at 20:04

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 $|\lambda|<1$.

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 $${n \brace 2} = 2^{n-2} + 2^{n-1}+...+2+1 = 2^{n-1}-1, \\ {n\brace 3} = 3^{n-3}+{3\brace 2}3^{n-4}+{4\brace 2} 3^{n-5} +\cdots + {n-2\brace 2}3+{n-1\brace 2}. $$

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 $$(-1)^{a^\dagger a}=N(e^{-2a^\dagger a}).$$

  • $\begingroup$ This is great, thank you. The other way to prove it is to take the expectation value with respect to a coherent state $|\alpha\rangle$, because the LHS gives $e^{-|\alpha|^2}\sum_n (\lambda |\alpha|^2)^n/n!=e^{-|\alpha|^2}e^{\lambda|\alpha|^2}$ and the RHS gives $e^{(\lambda-1)|\alpha|^2}$ $\endgroup$ Commented Jun 17 at 15:10
  • $\begingroup$ @QuantumMechanic ah! you're right, that's a much simpler proof, thanks! $\endgroup$
    – glS
    Commented Jun 17 at 17:32
  • $\begingroup$ Not as fancy as using identities with Stirling numbers! $\endgroup$ Commented Jun 18 at 20:08

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