# Defining a trace-class operator with a Bochner integral

For complex numbers $z$ consider the system of $L^2(\mathbb{R})$-vectors with norm equal to $1$ \begin{equation} \psi_z=e^{-\frac{1}{2}|z|^2}\sum_{n=0}^{\infty}\frac{z^n}{\sqrt{n!}} |n\rangle, \end{equation} where $|n\rangle$ is an orthonormal basis of the hilbert space $L^2(\mathbb{R})$.

(Actually I'm only interested in the special case where the orthonormal basis is given by \begin{equation} |n\rangle (x)= \frac{\pi^{\frac{1}{4}}}{\sqrt{2^n n!}}H_n(x)\exp({-\frac{x^2}{2}}) \end{equation} with the Hermite-Polynomials $H_n$. This is the eigenbasis of the quantum harmonic oscillator with $m\omega =1=\hbar$. To add further background, the system $\psi_z$ are called coherent states and are closely linked to Segal-Bargmann spaces. I don't think the exact choice of basis matters though, in any case it doesn't seem to matter for calculations of the type needed to solve the problem below)

I'm trying to solve the following problem:

Let $P$ be a signed (real) measure on the Borel sets of $\mathbb{C}$, such that $P(\mathbb{C})=1.$ Consider the linear operator defined by $$R:L^2(\mathbb{R})\ni\phi\mapsto \int_{\mathbb{C}} \psi_z \langle \psi_z, \phi\rangle dP(z).$$ What are additional, sufficient conditions to the measure $P$, such that this operator becomes self-adjoint, positive and trace class with trace equal to $1$?

(This representation for a density operator is called "Glauber–Sudarshan P representation" and used in quantum optics) I am a physicist, and (no implication implied) my ability to make rigorous mathematical statements is somewhat lacking, especially when it comes to Bochner-integrals, as used in the definition of $R$ and, even worse, signed measures. Nevertheless I decided to get the math right on this one, which I need a little help for. Here is what I have done so far:

I'm already failing to prove that the operator is even well-defined as a bounded map. The main problem seems to be that any decomposition theorem for the measure will prevent me from using the fact that the measure of the whole plane is one. If the measure were positive, I would say (to prove that $R$ is a well-defined bounded operator)

\begin{equation} \|R\phi\|\leq \int_{\mathbb{C}} |\langle \psi_z, \phi \rangle |dP(z)\leq \int_{\mathbb{C}} \| \phi\| dP(z)=\|\phi\| \end{equation} using Cauchy-Schwarz on the second inequality, but I don't think the first inequality holds with a signed measure...

Next I would naively say that the operator is obviously self-adjoint, since $P$ is real. Also I would compute the trace of $R$ to be the reassuring \begin{align} \operatorname{tr} R=\sum_n \langle n, Rn\rangle&=\int_\mathbb{C}\sum_n \langle n,\psi_z\rangle \langle \psi_z , n\rangle dP(z)=\int_\mathbb{C}e^{-|z|^2}\sum_{n,m,k} \frac{z^m\overline{z}^k}{\sqrt{m!k!}}\langle n,m\rangle\langle k,n\rangle dP(z) \\ &= \int_\mathbb{C} dP(z)=1. \end{align} However none of this is mathematically rigorous. What do I need to assume, so that my arguments work? I'm also unsure what exactly I need to make the resulting operator positive. It seems the continuous part of the measure (w.r.t. Lebesgue) needs to be positive, that isn't enaugh however and I'm also not sure if it's true.

I expect conditions for the measure will include that it can be decomposed into two sigma-finite (positive) measures to even define the Bochner integral; it seems reasonable that both the positive and negative parts should be finite on compact sets and have vanishing "part" which is both absolutely continuous and singluar w.r.t. the Lebesgue measure on $\mathbb{C}$ in their unique decompositions. It is less of a problem to add too many conditions in this case, as I don't expect finding the sufficiant and necessary conditions will be doable. (I would also be very interested in finding conditions for replacing integration against the measure by the action of certain (probably tempered) distributions, that seems even harder though) Thank you all very much in advance for the help!

Since I still don't know the answer several months later, I decided to try asking this question in MathOverflow.

Since I still don't know the answer several months later, I decided to try asking this question in MathOverflow. I created this answer to complete this particular question topic and make sure it doesn't show up as on open question to avoid parallel discussions.

• I would assume that the measure can be decomposed into two positive $\sigma$-finite measures, one of which (and thus both) is finite. I suspect that the operator is not bounded on $L^2$ without this assumption. This makes your boundedness argument work, replacing $dP$ with its variation $\lvert dP \rvert$. – Eric Thoma Aug 26 '16 at 2:12
• Thank you very much for your insights! While I'm familiar with the definition of the variation, I don't really see how to make the inequality work, since I never really had to deal with signed measures before. Could you elaborate on the necessary step or point me towards sources where the necessary result is proven? Maybe something occurs to you that would make the operator positive? My current suspicion is that the negative part of the measure must be singular w.r.t. Lebesgue measure... – Adomas Baliuka Aug 26 '16 at 15:35