Is there any relationship between Gleason's theorem and Riesz representation theorem?

Question

Gleason's Theorem Section 2.3.3 states that if a mapping $p$ satisfies several conditions, $p$ can be written as $p(E)=\text{tr}(\rho E)$ which is the inner product with a given density matrix. The third condition if $E_1E_2=0$, then $p(E_1+E_2)=p(E_1)+p(E_2)$ looks like linearity condition. So I think it is very similiar to Riesz representation theorem in functional analysis. Is there any further relationship between the two theorems?

Answer 1

The two results you mentioned are indeed related in the sense that Gleason's theorem is a generalization of (something which in finite dimensions boils down to) the Riesz representation theorem. In order to understand why that is we have to introduce two sets: the "effects" $$ \mathbb E(\mathcal H):=\{E\in\mathcal B(\mathcal H):E=E^*\text{ and }0\leq E\leq{\bf 1}\} $$ and the orthogonal projections $$ \mathcal P(\mathcal H):=\{P\in\mathcal B(\mathcal H):P^*=P=P^2\}\,. $$ (Here $\mathcal B(\mathcal H)$ is the collection of all bounded linear operators on the Hilbert space $\mathcal H$; if you're not too familiar with infinite dimensions you can think of this as square matrices.)

Having introduced these sets we can define generalized probability measures, and our goal is to show that these have to be of the form $E\mapsto{\rm tr}(\rho E)$ for some quantum state $\rho$. (NB: when we say quantum state we mean a trace-class operator $\rho\in\mathcal B^1(\mathcal H)$ which is positive semi-definite $\rho\geq 0$ and has unit trace, i.e. ${\rm tr}(\rho)=1$. Again if you're unfamiliar with infinite dimensions think of $\mathcal B^1(\mathcal H)$ as square matrices). The following result can, e.g., be found in "The Mathematical Language of Quantum Theory" (2012) by Heinosaari & Ziman:

Proposition 2.38. Let $f:\mathbb E(\mathcal H)\to[0,1]$ be a generalized probability measure, that is, $f(0)=0$, $f({\bf 1})=1$, and for all $E_1,E_2\in\mathbb E(\mathcal H)$ with $E_1+E_2\leq{\bf 1}$ one has $f(E_1+E_2)=f(E_1)+f(E_2)$. Then there exists a unique quantum state $\rho$ such that $$ f\equiv{\rm tr}(\rho(\cdot))\,. $$

As Heinosaari & Ziman state in their proof: "the main idea [...] is to show that $f$ extends to a positive normal linear functional on [...] the real vector space of bounded self-adjoint operators." Once this is shown one can use the well-known fact that every normal linear $f:\mathcal B(\mathcal H)\to\mathbb C$ is of the form $f(B)={\rm tr}(\rho B)$ for some $\rho\in\mathcal B^1(\mathcal H)$. The latter establishes a duality between (a class of functionals on) $\mathcal B(\mathcal H)$ and the trace class; in finite dimensions this is the same as the Riesz representation theorem.

Now Gleason's theorem generalizes the above result as it is about functionals not on $\mathbb E(\mathcal H)$, but rather on the smaller set $\mathcal P(\mathcal H)$:

Theorem 2.44. Suppose that $\dim\mathcal H\geq 3$. Let $f:\mathcal P(\mathcal H)\to[0,1]$ be a generalized probability measure, i.e. $f(0)=0$, $f({\bf 1})=1$, and for all $P_1,P_2\in\mathcal P(\mathcal H)$ with $P_1P_2=0$ one has $f(P_1)+f(P_2)=0$. Then there exists a unique quantum state $\rho\in\mathcal B^1(\mathcal H)$ such that $$ f(\cdot)={\rm tr}(\rho(\cdot))\,. $$

Notice the condition $\dim\mathcal H\geq 3$ which in fact is necessary (cf. Example 2.45 in Heinosaari & Ziman's book)! Therefore Gleason's theorem cannot be a direct consequence of Riesz or really any result on linear functionals which is independent of the Hilbert space dimension. In a way going from the convex set $\mathbb E(\mathcal H)$ to the set of its extreme points $\mathcal P(\mathcal H)$ gets rid of any of the linear structure which we used to get from functionals on $\mathcal B(\mathcal H)$ to states. Sure, under certain conditions on $P_1,P_2\in\mathcal P(\mathcal H)$ we know that $f(P_1+P_2)=f(P_1)+f(P_2)$, but from this one cannot deduce that $f$ is linear: due to the lack of "linearity" of $\mathcal P(\mathcal H)$ there is no way of establishing (or making sense of, for that matter) $f(\lambda P)=\lambda f(P)$.