2

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
  • 15,488
  • 5
  • 42
  • 114

1 Answers1

2

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
  • 67,623