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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.

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  • $\begingroup$ What part of the definition of a probability measure do you think is violated? $\endgroup$
    – Mushu Nrek
    Commented Mar 13 at 8:26
  • $\begingroup$ @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? $\endgroup$
    – Anacardium
    Commented Mar 13 at 8:50
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    $\begingroup$ 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. $\endgroup$
    – Mushu Nrek
    Commented Mar 13 at 9:50
  • $\begingroup$ @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? $\endgroup$
    – Anacardium
    Commented Mar 13 at 10:45
  • $\begingroup$ 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 $=1$. $\endgroup$
    – Mushu Nrek
    Commented Mar 14 at 11:22

1 Answer 1

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Over the last few days I have been thinking about your question (so thank you for posting it, it has been fun to think about), and I have multiple times begun typing up an answer, only to realise that I was missing some detail or simply had the wrong idea. I think that I finally am satisfied with my answer, so hopefully you will be as well.

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.

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