# Why is $P_{\rho}$ is a probability measure on the Borel subsets of $H\$?

I am going through a paper on Operator Probability Theory by Stan Gudder. The author introduced the notion of probability distribution of self-adjoint operators on a Hilbert space where the self-adjoint operators are thought of as complex valued random variables relative to a fixed state. Let $$A \in \mathcal S (H)$$ (self-adjoint operator) and $$\rho$$ be a state on $$H.$$ Let $$P^A$$ be the spectral measure corresponding to the self-adjoint operator $$A.$$ Then for a Borel subset $$\Delta \subseteq \sigma (A)$$ (spectrum of $$A$$) we define $$P_{\rho} (A \in \Delta) = \text{tr} \left (\rho P^A (\Delta) \right ).$$ So the expectation of $$A$$ is given as $$:$$ $$E_{\rho} (A) = \int_{\sigma (A)} \lambda\ \text{tr} \left (\rho P^A (d \lambda) \right ) = \text{tr} (\rho A).$$

But I don't understand why $$P_{\rho}$$ is a valid probability measure. Also I don't follow why $$E_{\rho} (A)$$ evaluates to $$\text {tr} (\rho A).$$ Any suggestion in this regard would be warmly appreciated.

Thanks for your time.

• What part of the definition of a probability measure do you think is violated? Commented Mar 13 at 8:26
• @MushuNrek$:$ If we want to check that $P_{\mu}$ is a probability measure then its range should be in $[0,1].$ Right? How do I show that? Commented Mar 13 at 8:50
• Do you agree that $\mathrm{tr}(\rho P^A(\Delta)) \leq \mathrm{tr}(\rho P^A(\mathbb R)) = \mathrm{tr}(\rho) = 1$ (where I used that $P^A(\mathbb R) = Id$ and that $\rho$ is of unit trace) ? The fact that its $\geq 0$ should be clear. Commented Mar 13 at 9:50
• @MushuNrek$:$ $P^A$ is defined on $\sigma (A)$; not on $\mathbb R.$ How can it be said that $P^A (\sigma (A)) = \text {Id}\$? Also $\rho$ is state i.e. a positive linear functional with norm $1.$ Are you considering some different norm than the operator norm? I think there is some property of spectral measure which is used here. I only know that $$A = \int_{\sigma (A)} \lambda\ d P^A (\lambda).$$ Do you have some reference? Commented Mar 13 at 10:45
• As far as I am concerned, $P^A(\Delta) = \sum_{\lambda\in \sigma(A)\cap\Delta} P_\lambda$ is defined for any $\Delta\subseteq \mathbb C$ (which is better than taking $\mathbb R$ I guess). Here, $P_\lambda$ denotes the projection on the eigenspace to the eigenvalue $\lambda$. In particular, $P^A(\mathbb C) = Id$. Next, the text says that states are elements of $\mathcal D(H)$, i.e. operators with "unit trace", i.e. trace $=1$. Commented Mar 14 at 11:22

In Gert K. Petersens book "Analysis Now", we find in Proposition 4.6.11 that a functional $$\varphi$$ on $$\mathcal B(H)$$ is given by $$\varphi(T)=\mathrm{tr}(ST)$$ for some trace class operator $$S$$ if and only if $$\varphi$$ is $$\sigma$$-weakly continuous. Proposition 4.6.14 then says that this is the case if and only if $$\varphi$$ is weakly continuous on the bounded subsets of $$\mathcal B(H)$$.
Now let as in your case $$\rho$$ be a positive trace class operator with unit trace, and define a functional $$\varphi$$ by $$\varphi(T)=\mathrm{tr}(\rho T),\quad (T\in\mathcal B(H)).$$ By the above we get that $$\varphi$$ is weakly continuous on bounded subsets of $$\mathcal B(H)$$. We want to show that it is strongly sequentially continuous, so let $$(T_n)_{n\in\mathbb N}$$ be a sequence of operators such that $$T_n\to 0$$ strongly as $$n\to\infty$$, i.e. $$\lVert T_n x\rVert\to 0$$ as $$n\to\infty$$ for every $$x\in H$$. Then of course $$\sup_{n\in\mathbb N}\lVert T_n x\rVert<\infty$$ for every $$x\in H$$, so by the uniform boundedness principle, $$\sup_{n\in\mathbb N}\lVert T_n\rVert<\infty,$$ and hence $$\{ T\in\mathcal B(H)\colon\lVert T\rVert\leq\sup_{n\in\mathbb N}\lVert T_n\rVert\}$$ is a bounded set, and thus the restriction of $$\varphi$$ to this set is weakly continuous. Next, since $$T_n\to 0$$ strongly as $$n\to\infty$$, it follows that $$T_n\to 0$$ weakly as $$n\to\infty$$ (i.e. $$\langle T_n x,y\rangle\to 0$$ for all $$x,y\in H$$). We thus get $$\varphi(T_n)\to 0$$ as $$n\to\infty$$, so $$\varphi$$ is strongly sequentially continuous as desired. This is important because the Borel functional calculus plays nice with strong convergence.
Now for $$A\in\mathcal S(H)$$ we have the spectral measure $$P^A$$, which is a resolution of the identity (or projection valued measure) on $$\sigma(A)$$, so by definition $$P^A(\emptyset)=0$$ and $$P^A(\sigma(A))=I$$. This immediately yields $$\varphi(P^A(\emptyset))=\mathrm{tr}(0)=0,\quad\varphi(P^A(\sigma(A)))=\mathrm{tr}(\rho)=1.$$ Also, $$\varphi$$ is a positive operator, so in general $$\varphi(P^A(\Delta))\geq 0$$ for Borel subsets $$\Delta\subseteq\sigma(A)$$. Finally, if $$\Delta_1,\Delta_2,\Delta_3,\dots$$ are pairwise disjoint Borel subsets of $$\sigma(A)$$, then $$\sum_{k=1}^n P^A(\Delta_k)\text{ converges strongly to }P^A\bigg(\bigcup_{k\in\mathbb N}\Delta_k\bigg)\text{ as }n\to\infty,$$ so since $$\varphi$$ is (linear and) strongly sequentially continuous, $$\varphi\circ P^A\bigg(\bigcup_{k\in\mathbb N}\Delta_k\bigg)=\sum_{k=1}^\infty\varphi(P^A(\Delta_k)).$$ Hence, $$\varphi\circ P^A=\mathrm{tr}(\rho P^A)=P_\rho(A\in\cdot)$$ is a Borel probability measure on $$\sigma (A)$$.
Now to show that $$E_\rho(A)=\mathrm{tr}(\rho A)$$. Let us prove the more general fact that $$\int_{\sigma(A)}f(\lambda)\,\mathrm{tr}(\rho P^A(\mathrm{d}\lambda))=\mathrm{tr}\bigg(\rho\int_{\sigma(A)}f\,\mathrm{d}P^A\bigg)$$ holds for all essentially bounded Borel-measureable functions $$f$$ on $$\sigma(A)$$, i.e. all $$f\in\mathcal L^\infty(P^A)$$. Let $$\mathcal G\subseteq\mathcal L^\infty(P^A)$$ denote the set of functions for which this holds. If $$f=1_\Delta$$ for some Borel subset $$\Delta\subseteq\sigma(A)$$, then $$\int_{\sigma(A)}1_\Delta(\lambda)\,\mathrm{tr}(\rho P^A(\mathrm{d}\lambda))=\mathrm{tr}(\rho P^A(\Delta))=\mathrm{tr}\bigg(\rho\int_{\sigma(A)}1_\Delta\,\mathrm{d}P^A\bigg)$$ as desired, i.e. $$1_\Delta\in\mathcal G$$. Now, if $$f$$ and $$g$$ are two non-negative functions from $$\mathcal G$$, then \begin{align*} \int_{\sigma(A)}f(\lambda)+g(\lambda)\,\mathrm{tr}(\rho P^A(\mathrm{d}\lambda))&=\mathrm{tr}\bigg(\rho\int_{\sigma(A)}f\,\mathrm{d}P^A\bigg)+\mathrm{tr}\bigg(\rho\int_{\sigma(A)}g\,\mathrm{d}P^A\bigg) \\ &=\mathrm{tr}\bigg(\rho\int_{\sigma(A)}f+g\,\mathrm{d}P^A\bigg), \end{align*} so $$f+g\in\mathcal G$$. Finally, if $$f\in\mathcal L^\infty(P^A)$$ is non-negative and $$f_1,f_2,f_3.\dots$$ is a(-n increasing and) bounded sequence of non-negative functions from $$\mathcal G$$ such that $$f_n\to f$$ $$P^A$$-a.e., then it holds that $$\int_{\sigma(A)}f_n\,\mathrm{d} P^A\text{ converges strongly to }\int_{\sigma(A)}f\,\mathrm{d}P^A\text{ as }n\to\infty,$$ so again since $$T\mapsto\varphi(T)=\mathrm{tr}(\rho T)$$ is strongly sequentially continuous, we get (using dominated/monotone convergence in the first step): \begin{align*} \int_{\sigma(A)}f(\lambda)\,\mathrm{tr}(\rho P^A(\mathrm{d}\lambda))&=\lim_{n\to\infty}\int_{\sigma(A)}f_n(\lambda)\,\mathrm{tr}(\rho P^A(\mathrm{d}\lambda)) \\ &=\lim_{n\to\infty}\mathrm{tr}\bigg(\rho\int_{\sigma(A)}f_n\,\mathrm{d}P^A\bigg) \\ &=\mathrm{tr}\bigg(\rho\int_{\sigma(A)}f\,\mathrm{d}P^A\bigg), \end{align*} so indeed also $$f\in\mathcal G$$. By the "standard proof", $$\mathcal G$$ contains all non-negative functions from $$\mathcal L^\infty(P^A)$$, and for a general $$f\in\mathcal L^\infty(P^A)$$ we simply decompose $$f=\Re(f)^+-\Re(f)^-+\mathrm{i}(\Im(f)^+ -\Im(f)^-),$$ and the full result follows immediately from linearity, so $$\mathcal G=\mathcal L^\infty(P^A)$$ as claimed. Taking $$f(\lambda)=\lambda$$ on $$\sigma(A)$$ yields $$E_\rho(A)=\mathrm{tr}(\rho A)$$.
As a final side note, I would like to mention a different approach to defining the distribution of non-commutative random variable $$A\in\mathcal S(H)$$. Let more generally $$\mathcal A$$ be a von Neumann algebra, and let $$A$$ be a self-adjoint element of $$\mathcal A$$. Then we consider the continuous functional calculus $$\Phi:C(\sigma(A))\to\mathcal A$$ of $$A$$. Consider further a positive linear functional $$\tau$$ on $$\mathcal A$$ with $$\tau(I)=1$$ (also usually called a "state"). Then the composition $$\varphi=\tau\circ\Phi$$ is a positive linear functional on $$C(\sigma(A))$$, so by Riesz' representation theorem we get a (finite) Borel measure $$\mu_A$$ on $$\sigma(A)$$ satisfying $$\varphi(f)=\tau(\Phi(f))=\int_{\sigma(A)}f\,\mathrm{d}\mu_A$$ for all $$f\in C(\sigma(A))$$. In particular $$\mu_A(\sigma(A))=\varphi(\mathrm{id})=\tau(I)=1$$, so $$\mu_A$$ is a probability measure. We call this $$\mu_A$$ the distribution of $$A$$. In your case $$\tau(T)=\mathrm{tr}(\rho T)$$, and if we instead consider the Borel functional calculus in place of $$\Phi$$ (which extends the continuous calculus), we can extend the equality $$\tau(\Phi(f))=\int_{\sigma(A)}f\,\mathrm{d}\mu_A$$ to hold for all $$f=1_\Delta$$ with $$\Delta$$ a Borel subset of $$\sigma(A)$$, so that indeed $$P_\rho(A\in\Delta)=\mu_A(\Delta)$$ in your case.