Vacuum projection operator and normal ordering

["Don't panic!"]

I've been reading this book, in which the author expresses the vacuum projection operator $\vert 0\rangle\langle 0\vert$ in terms of the number operator $\hat{N}=\hat{a}^{\dagger}\hat{a}$, where $\hat{a}^{\dagger}$ and $\hat{a}$ are the usual creation and annihilation operators, respectively. I can follow most of the derivation, however, I don't quite understand the following step: $$:\text{exp}\bigg\lbrace\hat{a}^{\dagger}\frac{\mathrm{d}}{\mathrm{d}Z^{\ast}}\bigg\rbrace\,\vert 0\rangle\langle 0\vert\,\text{exp}\big\lbrace\hat{a}Z^{\ast}\big\rbrace :\bigg\vert_{Z^{\ast}=0} \ = \ :\text{exp}\big\lbrace\hat{a}^{\dagger}\hat{a}\big\rbrace\,\vert 0\rangle\langle 0\vert : \qquad (1)$$ How does one get from the left-hand side to the right-hand side of this equation? I'm assuming there are several steps that have been missed out, or am I missing something trivial?

Just in case the link is not viewable, let me elaborate on the details of the calculation a little further, in particular, how one arrives at eq. (1). Using the completeness relation for the basis of number-operator eigenstates, we have $$1\!\!1 \ = \ \sum_{n,m=0}^{\infty}\vert n\rangle\langle m\vert\,\delta_{n,m} \ = \ \sum_{n,m=0}^{\infty}\vert n\rangle\langle m\vert\,\frac{1}{\sqrt{n!m!}}\bigg(\frac{\mathrm{d}}{\mathrm{d}Z^{\ast}}\bigg)^{n}\big(Z^{\ast}\big)^{m}\bigg\vert_{Z^{\ast}=0}\qquad (2)$$ where we make use of the identity $$\frac{1}{\sqrt{n!m!}}\bigg(\frac{\mathrm{d}}{\mathrm{d}Z^{\ast}}\bigg)^{n}\big(Z^{\ast}\big)^{m}\bigg\vert_{Z^{\ast}=0} \ = \ \delta_{n,m}$$ Next, using that $\vert n\rangle =\frac{(\hat{a}^{\dagger})^{n}}{\sqrt{n!}}\,\vert 0\rangle$ we can rewrite (2) as $$\sum_{n,m=0}^{\infty}\frac{(\hat{a}^{\dagger})^{n}}{n!}\,\vert 0\rangle\langle 0\vert\,\frac{(\hat{a})^{m}}{m!}\bigg(\frac{\mathrm{d}}{\mathrm{d}Z^{\ast}}\bigg)^{n}\big(Z^{\ast}\big)^{m}\bigg\vert_{Z^{\ast}=0} \ = \ \sum_{n,m=0}^{\infty}\frac{(\hat{a}^{\dagger})^{n}\big(\frac{\mathrm{d}}{\mathrm{d}Z^{\ast}}\big)^{n}}{n!}\,\vert 0\rangle\langle 0\vert\,\frac{(\hat{a})^{m}\big(Z^{\ast}\big)^{m}}{m!}\bigg\vert_{Z^{\ast}=0} \\[1cm] \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\,\, = \ \text{exp}\bigg\lbrace\hat{a}^{\dagger}\frac{\mathrm{d}}{\mathrm{d}Z^{\ast}}\bigg\rbrace\,\vert 0\rangle\langle 0\vert\,\text{exp}\big\lbrace\hat{a}Z^{\ast}\big\rbrace\bigg\vert_{Z^{\ast}=0}$$ This final expression is already normal-ordered as all creation operators are placed to the left of all annihilation operators. As such we can express this last line as given in eq. (1).

I understand this derivation up to the left-hand side of eq. (1), I just don't understand how the author then gets to the right-hand side of eq. (1).

 

[king vitamin]

So we would like to study the object
$$
\exp\left( \hat{a}^{\dagger} \frac{d}{dZ^{\ast}}\right) \exp\left( Z^{\ast} \hat{a} \right) \Bigg|_{Z^{\ast} = 0} = \sum_{k,k' = 0}^{\infty} \frac{1}{k! k'!} \left( \hat{a}^{\dagger} \frac{d}{dZ^{\ast}}\right)^k \left( Z^{\ast} \hat{a} \right)^{k'} \Bigg|_{Z^{\ast} = 0}
$$
Now the trick is to think about what terms in this sum could possibly survive setting $Z^{\ast} = 0$. We have a set of terms $(Z^{\ast})^{k'}$, and we have derivatives acting on them. If a derivative acts less than $k'$ times on a $(Z^{\ast})^{k'}$ term, it will vanish when you take$Z^{\ast} = 0$. But we also clearly have
$$
\frac{d^k }{dZ^{\ast k}}(Z^{\ast})^{k'} = 0, \quad k>k'.
$$
Therefore, the only terms which can survive the double sum are precisely the $k=k'$ ones. In this case we use
$$
\frac{d^k }{dZ^{\ast k}}(Z^{\ast})^{k} = k!
$$
and find
$$
\exp\left( \hat{a}^{\dagger} \frac{d}{dZ^{\ast}}\right) \exp\left( Z^{\ast} \hat{a} \right) \Bigg|_{Z^{\ast} = 0} = \sum_{k=0}^{\infty} \frac{1}{k!} \left( \hat{a}^{\dagger} \right)^k \left( \hat{a} \right)^{k}
$$

 

["Don't panic!"]

So we would like to study the object
$$
\exp\left( \hat{a}^{\dagger} \frac{d}{dZ^{\ast}}\right) \exp\left( Z^{\ast} \hat{a} \right) \Bigg|_{Z^{\ast} = 0} = \sum_{k,k' = 0}^{\infty} \frac{1}{k! k'!} \left( \hat{a}^{\dagger} \frac{d}{dZ^{\ast}}\right)^k \left( Z^{\ast} \hat{a} \right)^{k'} \Bigg|_{Z^{\ast} = 0}
$$
Now the trick is to think about what terms in this sum could possibly survive setting $Z^{\ast} = 0$. We have a set of terms $(Z^{\ast})^{k'}$, and we have derivatives acting on them. If a derivative acts less than $k'$ times on a $(Z^{\ast})^{k'}$ term, it will vanish when you take$Z^{\ast} = 0$. But we also clearly have
$$
\frac{d^k }{dZ^{\ast k}}(Z^{\ast})^{k'} = 0, \quad k>k'.
$$
Therefore, the only terms which can survive the double sum are precisely the $k=k'$ ones. In this case we use
$$
\frac{d^k }{dZ^{\ast k}}(Z^{\ast})^{k} = k!
$$
and find
$$
\exp\left( \hat{a}^{\dagger} \frac{d}{dZ^{\ast}}\right) \exp\left( Z^{\ast} \hat{a} \right) \Bigg|_{Z^{\ast} = 0} = \sum_{k=0}^{\infty} \frac{1}{k!} \left( \hat{a}^{\dagger} \right)^k \left( \hat{a} \right)^{k}
$$

Thanks for your response. I understand this part though. What I don't understand is how one goes from: $$ : \text{exp}\bigg\lbrace\hat{a}^{\dagger}\frac{\mathrm{d}}{\mathrm{d}Z^{\ast}}\bigg\rbrace\,\vert 0\rangle\langle 0\vert\,\text{exp}\big\lbrace\hat{a}Z^{\ast}\big\rbrace : \bigg\vert_{Z^{\ast}=0}$$ to $$ : \text{exp}\big\lbrace\hat{a}^{\dagger}\hat{a}\big\rbrace\,\vert 0\rangle\langle 0\vert :$$ That is, why is one allowed to freely commute $\text{exp}\big\lbrace\hat{a}Z^{\ast}\big\rbrace$ past $\vert 0\rangle\langle 0\vert$? Is it because the whole thing is normal ordered?

 

[stevendaryl]

Thanks for your response. I understand this part though. What I don't understand is how one goes from: $$ : \text{exp}\bigg\lbrace\hat{a}^{\dagger}\frac{\mathrm{d}}{\mathrm{d}Z^{\ast}}\bigg\rbrace\,\vert 0\rangle\langle 0\vert\,\text{exp}\big\lbrace\hat{a}Z^{\ast}\big\rbrace : \bigg\vert_{Z^{\ast}=0}$$ to $$ : \text{exp}\big\lbrace\hat{a}^{\dagger}\hat{a}\big\rbrace\,\vert 0\rangle\langle 0\vert :$$ That is, why is one allowed to freely commute $\text{exp}\big\lbrace\hat{a}Z^{\ast}\big\rbrace$ past $\vert 0\rangle\langle 0\vert$? Is it because the whole thing is normal ordered?

Yes, if you assume that $|0\rangle \langle 0|$ is expressible in terms of $a$ and $a^\dagger$, then you can freely commute it past any $a$ or $a^\dagger$ within a normal ordering.

The line that I don't understand is how they go from:

$1 = : e^{\hat{a}^\dagger \hat{a}}: W ::$

to

$|0\rangle\langle 0|\ =\ :W:\ =\ : e^{- \hat{a}^\dagger \hat{a}}:$

They know that $: e^{\hat{a}^\dagger \hat{a}}: W :: = : e^{\hat{a}^\dagger \hat{a}}: e^{-\hat{a}^\dagger \hat{a}}::$. But I don't see how can cancel off the $: e^{\hat{a}^\dagger \hat{a}}:$.

It's not true in general that $:A:B::\ =\ :A:C::$ implies that $:B:\ =\ :C:$, is it?

And the other thing is that $W$ is defined to be$|0\rangle \langle 0|$, while the book is writing $:W: = |0\rangle \langle 0|$.

 

["Don't panic!"]

Yes, if you assume that |0⟩⟨0||0⟩⟨0||0\rangle \langle 0| is expressible in terms of aaa and a†a†a^\dagger, then you can freely commute it past any aaa or a†a†a^\dagger within a normal ordering.

But I thought the point of this exercise to show precisely that $\vert 0\rangle\langle 0\vert$ is expressible in terms of $a$ and $a^{\dagger}$? It is not a priori assumed, at least as far as I can tell.

It's not true in general that :A:B:: = :A:C:::A:B:: = :A:C:::A:B::\ =\ :A:C:: implies that :B: = :C::B: = :C::B:\ =\ :C:, is it?

I think it possibly is. If one operates from the left by $:A^{-1}:$, then we get $:A^{-1}::A::B: =:A^{-1}A::B:=1\!\!1 :B:$. Similarly, $:A^{-1}::A::C: =:A^{-1}A::C:=1\!\!1 :C:$, and so $:B: = :C:$. That's legitimate, right? At least it is if $:A::B:=:AB:$ holds, which I'm not sure is true.

I find the last few steps of the authors derivation very confusing. It seems to be of some importance that the author introduces the $Z^{\ast}$ identity too (otherwise, one can essentially arrive at the same point without the need for introducing the $Z^{\ast}$ pieces.

 

[stevendaryl]

But I thought the point of this exercise to show precisely that $\vert 0\rangle\langle 0\vert$ is expressible in terms of $a$ and $a^{\dagger}$? It is not a priori assumed, at least as far as I can tell.

Well, there is a more direct argument to that effect.

$|0\rangle \langle 0|$ is the operator $\hat{A}$ defined by its action on the basis states:

$\hat{A} |0\rangle = |0\rangle$
$\hat{A} |n\rangle = 0$ if $n \neq 0$

So we try to "implement" these rules in terms of creation and annihilation operators:

$\hat{A} = \sum_j c_j (\hat{a}^\dagger)^j (\hat{a})^j$

It's obvious that $c_0 = 1$. Then we can get a recursive definition of $c_j$ for $j > 0$. We want for $n > 0$,

$\sum_{j=0}^{n} c_j (\hat{a}^\dagger)^j (\hat{a})^j |n\rangle = 0$

Using the actions of the creation and annihilation operators, this becomes:

$\sum_{j=0}^{n} c_j\ \frac{n!}{(n-j)!}= 0$

which we can write as:

$\sum_{j=0}^{n-1} c_j \frac{n!}{(n-j)!} + c_{n} n!$

So we have:

$c_{n} = - \sum_{j=0}^{n-1} \frac{c_j}{(n-j)!}$

Obviously, that let's us solve for each $c_n$ in terms of $c_0, c_1, ..., c_{n-1}$.

So clearly, there is a way to write $|0\rangle\langle 0|$ in terms of creation and annihilation operators.

 

["Don't panic!"]

Well, there is a more direct argument to that effect.

$|0\rangle \langle 0|$ is the operator $\hat{A}$ defined by its action on the basis states:

$\hat{A} |0\rangle = |0\rangle$
$\hat{A} |n\rangle = 0$ if $n \neq 0$

So we try to "implement" these rules in terms of creation and annihilation operators:

$\hat{A} = \sum_j c_j (\hat{a}^\dagger)^j (\hat{a})^j$

It's obvious that $c_0 = 1$. Then we can get a recursive definition of $c_j$ for $j > 0$. We want for $n > 0$,

$\sum_{j=0}^{n} c_j (\hat{a}^\dagger)^j (\hat{a})^j |n\rangle = 0$

Using the actions of the creation and annihilation operators, this becomes:

$\sum_{j=0}^{n} c_j\ \frac{n!}{(n-j)!}= 0$

which we can write as:

$\sum_{j=0}^{n-1} c_j \frac{n!}{(n-j)!} + c_{n} n!$

So we have:

$c_{n} = - \sum_{j=0}^{n-1} \frac{c_j}{(n-j)!}$

Obviously, that let's us solve for each $c_n$ in terms of $c_0, c_1, ..., c_{n-1}$.

So clearly, there is a way to write $|0\rangle\langle 0|$ in terms of creation and annihilation operators.

Okay, so do you think the author is assuming this then? What they are not assuming is that it can be written in the simple form of an exponential of the number operator.

What is the point in introducing the $Z^{\ast}$ identity then? One could equally well just use a kronecker delta to write it in terms of two exponentials, and then the last few steps will be the same. Maybe I'm missing something?

Also, is the choice of $Z^{\ast}$ significant? Is it implying something about the complex conjugate of some $Z$? The identity holds regardless of my choice of variable.

 

[A. Neumaier]

It's not true in general that $:A:B::\ =\ :A:C::$ implies that $:B:\ =\ :C:$, is it?

It depends how $A$ is quantified.
With ''$A$ exists such that your formula holds'', your question has a positive answer, since you can take $A=0$.
With ''for all $A$, your formula holds'', your question has a negative answer, since you can take $A=1$.

 

[stevendaryl]

I don't understand the derivation, but the results I can convince myself of in a different way.

Let $n^{(j)}$ be shorthand for $n (n-1) (n-2) ... (n+1-j)$. Then look at the expression:

$S(n) = \sum_{j=0}^n (-1)^j n^{(j)}/j!$

For $n=0$, only the first term survives, so $S_0 = 1$. For $n > 0$, this is the binomial expansion of $(1-1)^n$, so it equals 0.

Now, let's look at a related expression: $\hat{S} = \sum_{j=0}^\infty (-1)^j N^{(j)}/j!$ where $N$ is the number operator. If we let it act on a basis state $|n\rangle$, we get the following:

$\hat{S} |n\rangle = \sum_{j=0}^n (-1)^j n^{(j)} |n\rangle$ (Because for $j > n$, $N^{(j)} |n\rangle = 0$)

So $\hat{S} |n\rangle = S(n) |n\rangle$. Therefore, $\hat{S}$ is our operator that has eigenvalue 1 when acting on $|0\rangle$ and has eigenvalue $0$ when acting on any other basis state. So therefore it's equal to $|0\rangle \langle 0|$.

So $\hat{S}$ is one expression for $|0\rangle \langle 0|$.

Now, here's an amazing fact about normal ordering that you can prove by induction:

$: N^j :\ =\ N^{(j)}$

Just to show it in one case, with $j=2$:

$: N^2 :$
$\ = \ (\hat{a}^\dagger)^2 (\hat{a})^2$
$\ =\ \hat{a}^\dagger (\hat{a}^\dagger \hat{a}) \hat{a}$
$\ =\ \hat{a}^\dagger (-1 + \hat{a} \hat{a}^\dagger) \hat{a}$
$\ =\ - \hat{a}^\dagger \hat{a} + (\hat{a}^\dagger \hat{a})^2$
$\ =\ -N + N^2$
$\ =\ N (N-1)$
$\ =\ N^{(2)}$

So
$\hat{S} = \sum_{j=0}^\infty (-1)^j N^{(j)}/j!$
$\ =\ : \sum_{j=0}^\infty (-1)^j N^{j}/j! :$
$\ = \ : e^{-N} :$

 

[stevendaryl]

Thinking about it some more, I just don't think that the derivation is correct. What the author proves is:

$: e^{\hat{N}} |0\rangle \langle 0| : = 1$

Without further assumptions, I don't see how that implies that $|0\rangle \langle 0| = : e^{- \hat{N}} :$

 

["Don't panic!"]

Thinking about it some more, I just don't think that the derivation is correct. What the author proves is:

$: e^{\hat{N}} |0\rangle \langle 0| : = 1$

Without further assumptions, I don't see how that implies that $|0\rangle \langle 0| = : e^{- \hat{N}} :$

Ah, that's frustrating. I've seen the same argument as the one this author gives in an academic paper too. It would be helpful if they actually explained things in further detail. I can't find anywhere else that gives a detailed derivation.

 

[Avodyne]

The simplest thing to do is to start with $A = :\!\exp(-N){:}\$ and then prove that $A=|0\rangle\langle 0|$. What follows is essentially the same as stevendaryl's proof, but IMO is a little simpler.

First, Taylor expand:

$A=\sum_{j=0}^\infty {(-1)^j\over j!} {:}N^j{:}$

Then use

${:}N^j{:} = (a^\dagger)^j a^j$

Then use

$a^j|n\rangle = \sqrt{n!/(n{-}j)!}\,|n{-}j\rangle$

$(a^\dagger)^j|n{-}j\rangle = \sqrt{n!/(n-j)!}\,|n\rangle$

Note that these hold even for $j>n$ if we take $1/(n{-}j)!=0$ for $j>n$. We now have

$(a^\dagger)^j a^j|n\rangle = {n!\over(n{-}j)!}\,|n\rangle$.

Therefore

$A|n\rangle = \left[\sum_{j=0}^\infty {(-1)^j n!\over j!(n{-}j)!}\right]|n\rangle$

The factor in brackets is the binomial expansion of $(1-1)^n$, and equals $\delta_{n0}$, QED.

 

["Don't panic!"]

The simplest thing to do is to start with $A = :\!\exp(-N){:}\$ and then prove that $A=|0\rangle\langle 0|$. What follows is essentially the same as stevendaryl's proof, but IMO is a little simpler.

First, Taylor expand:

$A=\sum_{j=0}^\infty {(-1)^j\over j!} {:}N^j{:}$

Then use

${:}N^j{:} = (a^\dagger)^j a^j$

Then use

$a^j|n\rangle = \sqrt{n!/(n{-}j)!}\,|n{-}j\rangle$

$(a^\dagger)^j|n{-}j\rangle = \sqrt{n!/(n-j)!}\,|n\rangle$

Note that these hold even for $j>n$ if we take $1/(n{-}j)!=0$ for $n>j$. We now have

$(a^\dagger)^j a^j|n\rangle = {n!\over(n{-}j)!}\,|n\rangle$.

Therefore

$A|n\rangle = \left[\sum_{j=0}^\infty {(-1)^j n!\over j!(n{-}j)!}\right]|n\rangle$

The factor in brackets is the binomial expansion of $(1-1)^n$, and equals $\delta_{n0}$, QED.

Thanks, this derivation (and stevendaryl’s) makes a lot of sense. It is reasonable to take $0^0=1$ (there doesn’t seem to be a general consensus)?

Also, it would interesting to see why the author is able to write: $$: \text{exp}\Big\lbrace\hat{a}^{\dagger}\frac{d}{dZ^{\ast}}\Big\rbrace\vert 0\rangle\langle 0\vert\text{exp}\Big\lbrace\hat{a}Z^{\ast}\Big\rbrace\bigg\vert_{Z^{\ast}=0}: \ = \ : \text{exp}\big\lbrace\hat{a}^{\dagger}\hat{a}\big\rbrace\vert 0\rangle\langle 0\vert :$$ That is, why is it ok to commute the $\hat{a}$ past $\vert 0\rangle\langle 0\vert$?

 

[A. Neumaier]

It is reasonable to take 00=100=10^0=1 (there doesn’t seem to be a general consensus)?

In a power series expansion, $x^0$ is always 1, by definition.

 

["Don't panic!"]

In a power series expansion, $x^0$ is always 1, by definition.

Good point, fair enough.

 

[Avodyne]

Fixed a typo in my post, "for $n>j$" should be (and now is) "for $j>n$".

Re the author's derivation: inside the normal-ordering colons, the factors can be written in any order, since, whatever order they are written in, they will be re-ordered by the normal-ordering. That said, I still can't follow every step. Perhaps some of the manipulations were explained earlier in the book.

 

[stevendaryl]

Fixed a typo in my post, "for $n>j$" should be (and now is) "for $j>n$".

Re the author's derivation: inside the normal-ordering colons, the factors can be written in any order, since, whatever order they are written in, they will be re-ordered by the normal-ordering. That said, I still can't follow every step. Perhaps some of the manipulations were explained earlier in the book.

I'm convinced that the argument in the reference proves that

1. $:e^{\hat{a}^\dagger \hat{a}} |0\rangle \langle 0|: \ = \ 1$

What I don't understand is how that implies that

2. $|0\rangle \langle 0| \ = \ :e^{-\hat{a}^\dagger \hat{a}}:$

I believe that 2. is true but I don't see how it is implied by 1. (I can see that 1. follows from 2., but not vice-versa)

 

["Don't panic!"]

Perhaps some of the manipulations were explained earlier in the book.

Possibly, but unfortunately I don't have access to the rest of the book :(

I'm convinced that the argument in the reference proves that

1. $:e^{\hat{a}^\dagger \hat{a}} |0\rangle \langle 0|: \ = \ 1$

What I don't understand is how that implies that

2. $|0\rangle \langle 0| \ = \ :e^{-\hat{a}^\dagger \hat{a}}:$

I believe that 2. is true but I don't see how it is implied by 1. (I can see that 1. follows from 2., but not vice-versa)

This is what leaves me confused too. It is discussed in this paper (right-hand side of page 3) and this paper (right-hand side of page 6) on the arXiv also (admittedly using a different approach), but without further explanation.

 

[stevendaryl]

The paper https://arxiv.org/pdf/0704.3116.pdf sort of explains it.

One representation of $|0\rangle \langle 0|$ is $lim_{\lambda \rightarrow \infty} e^{-\lambda \hat{N}}$. Of course, that limit doesn't actually exist, but the meaning is this: for any state $|\psi\rangle$,

$|0\rangle \langle 0|\psi\rangle = lim_{\lambda \rightarrow \infty} e^{-\lambda \hat{N}} |\psi\rangle$

The operator $e^{-\lambda \hat{N}}$ acting on $|0\rangle$ returns $|0\rangle$, and when acting on any other basis state $|n\rangle$ returns something close to zero. So that's the same behavior as the operator $|0\rangle \langle 0|$.

Now, we show an unexpected fact about the number operator and normal ordering:

$e^{-\lambda \hat{N}} = \ : e^{- (1 - e^{-\lambda}) \hat{N}}:$ (equation 30, slightly rewritten, from that paper).

So taking the limit as $\lambda \rightarrow \infty$ gives:

$|0\rangle \langle 0| = \ :e^{- \hat{N}}:$

The way that the paper proves equation 30 makes a detour through "Bell numbers", but it seems to me that it can be proved more directly if you use the fact that I derived earlier:

$: N^n : \ = \ N(N-1)...(N+1-n)$

We can use the taylor expansion of $(1+\alpha)^y$:

$(1+\alpha)^y = \sum_{j=0}^{\infty} \frac{\alpha^j}{j!} y (y-1) ... (y+1-j)$

So letting $\alpha = e^{-\lambda} - 1$, we have:

$e^{-\lambda y} = \sum_{j=0}^{\infty} \frac{e^{(-\lambda} - 1)^j}{j!} y (y-1) ... (y+1-j)$

Pluggine $N$ in for $y$:

$e^{-\lambda N} = \sum_{j=0}^{\infty} \frac{(e^{-\lambda} - 1)^j}{j!} N (N-1) ... (N+1-j)$
$= \sum_{j=0}^{\infty} \frac{(e^{-\lambda} - 1)^j}{j!}\ :N^j:$
$= : (\sum_{j=0}^{\infty} \frac{(e^{-\lambda} - 1)^j}{j!}\ N^j):$
$= :e^{(e^{-\lambda} - 1) N}:$

 

["Don't panic!"]

The paper https://arxiv.org/pdf/0704.3116.pdf sort of explains it.

One representation of $|0\rangle \langle 0|$ is $lim_{\lambda \rightarrow \infty} e^{-\lambda \hat{N}}$. Of course, that limit doesn't actually exist, but the meaning is this: for any state $|\psi\rangle$,

$|0\rangle \langle 0|\psi\rangle = lim_{\lambda \rightarrow \infty} e^{-\lambda \hat{N}} |\psi\rangle$

The operator $e^{-\lambda \hat{N}}$ acting on $|0\rangle$ returns $|0\rangle$, and when acting on any other basis state $|n\rangle$ returns something close to zero. So that's the same behavior as the operator $|0\rangle \langle 0|$.

Now, we show an unexpected fact about the number operator and normal ordering:

$e^{-\lambda \hat{N}} = \ : e^{- (1 - e^{-\lambda}) \hat{N}}:$ (equation 30, slightly rewritten, from that paper).

So taking the limit as $\lambda \rightarrow \infty$ gives:

$|0\rangle \langle 0| = \ :e^{- \hat{N}}:$

The way that the paper proves equation 30 makes a detour through "Bell numbers", but it seems to me that it can be proved more directly if you use the fact that I derived earlier:

$: N^n : \ = \ N(N-1)...(N+1-n)$

We can use the taylor expansion of $(1+\alpha)^y$:

$(1+\alpha)^y = \sum_{j=0}^{\infty} \frac{\alpha^j}{j!} y (y-1) ... (y+1-j)$

So letting $\alpha = e^{-\lambda} - 1$, we have:

$e^{-\lambda y} = \sum_{j=0}^{\infty} \frac{e^{(-\lambda} - 1)^j}{j!} y (y-1) ... (y+1-j)$

Pluggine $N$ in for $y$:

$e^{-\lambda N} = \sum_{j=0}^{\infty} \frac{(e^{-\lambda} - 1)^j}{j!} N (N-1) ... (N+1-j)$
$= \sum_{j=0}^{\infty} \frac{(e^{-\lambda} - 1)^j}{j!}\ :N^j:$
$= : (\sum_{j=0}^{\infty} \frac{(e^{-\lambda} - 1)^j}{j!}\ N^j):$
$= :e^{(e^{-\lambda} - 1) N}:$

Thanks for this.

Going back to the normal-ordered stuff, I managed to get hold of a copy of the book I was mentioning in my OP ("Nonequilibrium Quantum Transport Physics in Nanosystems: Foundation of Computational Nonequilibrium Physics in Nanoscience and Nanotechnology" by F. A. Buot). In it, the author states that normal-ordered products satisfy the following property: If $A$ is some operator, and $:U\cdots V: \; =\; :W\cdots X:$, then $$:AU\cdots V: \; =\; :AW\cdots X:$$. Given this property, it is then clear that, for $:AB: \; = 1\!\!1 =\; :1\!\!1:$, it follows that $$:A^{-1}AB: \; =\; :A^{-1}1\!\!1:\qquad \Rightarrow\qquad :B: \; =\; :A^{-1}: \;.$$