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While reading the Wikipedia page on the P function, I came across the following consideration (paraphrasing from there):

Given a state $\rho$, if we write it in anti-normal order as $\rho_A=\sum_{jk}c_{jk}a^j a^{\dagger k}$, and $$\rho_A(a,a^\dagger) = \frac1\pi\int \rho_A(\alpha,\alpha^*)|\alpha\rangle\!\langle\alpha| d^2\alpha,$$ then we can formally assign $P(\alpha)=\frac1\pi \rho_A(\alpha,\alpha^*)$.

What I don't quite understand is what $\rho_A$ is supposed to be representing here. If I write the state $\rho$ in terms of creation and annihilation operators I get something of the form $$\rho = \sum_{jk} \rho_{jk} a^{\dagger j}|0\rangle\!\langle 0| a^k,$$ which differs from the above $\rho_A$ not only in the ordering of the operators, but also crucially in the presence of the vacuum state between them, so going from this to the "anti-normally ordered expression" $\rho_A$ does not look so straightforward.

What's a better way to understand this?

glS
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    Can't you just write $|0\rangle\langle0$ -- which is a projection -- in terms of the $a$ and $a^\dagger$, and insert it there? This certainly gives an explicit form of $\rho$ in terms only of creation and annihilation operators. (Admittedly, this is a bit more awkward for bosons than for fermions.) – Norbert Schuch May 25 '20 at 12:29
  • @NorbertSchuch you can do that? That would be some function of the operators $f(a,a^\dagger)$ such that $f(a,a^\dagger)|0\rangle=|0\rangle$ but $f(a,a^\dagger)|n\rangle=0$ for all $n>0$... I'd have said that at least no polynomial expression of $a,a^\dagger$ can satisfy these – glS May 25 '20 at 14:51
  • Might find this and this one useful. – Cosmas Zachos May 25 '20 at 16:20
  • @CosmasZachos that's where I came from with this actually. I understand decomposing the state as $\rho=\int d^2\alpha P(\alpha)|\alpha\rangle\langle\alpha|$. What I'm missing is what the expressions cited in that wiki page are supposed to mean (assuming they are not just badly written notation). Or is $\rho_A$ here just a formal way to rewrite the matrix components of $\rho$ as a "matrix"? – glS May 25 '20 at 16:57
  • You might be having change of bases and representation problems. Try simple examples of operators. For example, take their matrix elements between vacuum states, or N=1 states... – Cosmas Zachos May 25 '20 at 17:50
  • @CosmasZachos I don't quite follow. My problem is understanding the given definition of $\rho_A$. If I don't know what the expression is supposed to mean, how am I to work out simple examples? – glS May 25 '20 at 18:06
  • The definition is the one given! An operator in terms of creation and annihilation operators, properly anti normal ordered. For example, $\hat N= a^\dagger a = a a^\dagger - 1!!1= \sum_{n=1} a^{\dagger (n+1) }|0\rangle \langle 0| a^{n+1}$. – Cosmas Zachos May 25 '20 at 18:14
  • @glS Any $f(a^\dagger a)$ with $f(n)=\delta_{n,0}$ will do - so there are plenty of choices. – Norbert Schuch May 25 '20 at 19:14
  • Related. – Cosmas Zachos May 25 '20 at 21:56

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You are probably having problems with bases and representations. For example, take your spectacularly unnormalized formal "state" $$ \rho = a^\dagger a = \hat N = a a^\dagger -1\!\! 1= \rho_A, $$ so the subscript A means we have rewritten the operator in the anti-normal-ordered representation, with the creators on the right. The two forms are strictly equal: quite unlike in QFT, "ordering" does not entail throwing away terms! Verify its vacuum matrix element $\langle 0| \rho|0\rangle=0 $, in both equivalent orderings.

Further develop comfort with sticking in vacua or not, $$ 1\!\!1 = \sum_{n=0} |n\rangle \langle n| = \sum_{n=0} a^{\dagger ~n}|0\rangle \langle 0| a^n \\ a^\dagger = \sum_{n=0} a^{\dagger ~(n+1)}|0\rangle \langle 0| a^n = \sum_{n=1} a^{\dagger ~n}|0\rangle \langle 0| a^{n-1} n\\ a= \sum_{n=1}n a^{\dagger ~(n-1)}|0\rangle \langle 0| a^n= \sum_{n=0} a^{\dagger ~n}|0\rangle \langle 0| a^{n+1} \\ \hat N = \sum_{n=0} n |n\rangle \langle n| = \sum_{n=0} a^{\dagger ~(n+1)}|0\rangle \langle 0| a^{n+1}, $$ etc...

I'm illustrating the idea here, and not fussing about normalizations and states. That will have to be done separately and meticulously, and further transferred to the coherent state basis, also involving infinities of such operators.

Edit added in response to comment/question. Here is a formal wisecrack, in response to the question about the projection operator $|1\rangle \langle 1|$. The formal operator $$ P= \frac{N(2-N)(3-N)...}{ 2\cdot 3\cdot 4\cdot ...}$$ is left/right orthogonal to $|n\rangle \langle n|$ for all n except 1, and $$ P|1\rangle \langle 1|= |1\rangle \langle 1| P = |1\rangle \langle 1|,$$ and thus, properly anti-normal ordered (!), stands a good chance of providing an A representation for the original projector you wrote. Nobody said it would be nice...

Cosmas Zachos
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    I see. But is there a general procedure to do that? For example, how would you bring a Fock state, say $|1\rangle\langle1|=a^\dagger|0\rangle\langle0|a$, in that form? It just seems like using only $a,a^\dagger$ there is no way to have something that only operates on a finite number of Fock states (also, I guess there should be some additional factors in those sums arising from things like $a^{\dagger n}|0\rangle=\sqrt{n!}|n\rangle$?) – glS May 25 '20 at 18:40
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    Generic antinormal ordering is an art with lots of tricks, but clearly outside the range of this question. Here, one is supposed to have achieved it, and moved on to translate into the coherent sate basis... – Cosmas Zachos May 25 '20 at 18:56
  • ok, so the answer is really that $\rho_A$ is nothing but $\rho$ rewritten in a specific ordering. I just assumed that wasn't the case (even if it is indeed the most straightforward interpretation of the way the article was written) because I really couldn't see how you would translate a generic state that way. I guess I was mistaken. So, let me just ask this: there is a way to write something like $|1\rangle\langle 1|$ in anti-normally ordered (or normally ordered for that matter) form, correct? I'm tempted to ask a separate question about that if there is a sensible answer for it – glS May 25 '20 at 19:09
  • In principle there should be, however impractical. $a$ and $a^\dagger$ are a complete operator basis, like $\hat x$ and $\hat p$, and adequate to describe any operator, even density matrices. The details might be far afield. They really are a different question. And, yes, this is why they use coherent states. This P function has well known singularities... – Cosmas Zachos May 25 '20 at 19:15