# Probability measure implies quantum mechanics?

There is a Hilbert space H (probably of dimension 3 or more, as in Gleason's theorem?) equipped with some logical apparatus (the lattice L?).

Correct, and the lattice $L(H)$ is that of orthogonal projectors/closed subspaces of a separable complex Hilbert space $H$. As a partially ordered set, the partial ordering $P\leq Q$ relation is the inclusion of subspaces: $P(H) \subset Q(H)$.

As a consequence $P \vee Q := \sup\{P,Q\}$ is the projector onto the closure of the sum of $P(H)$ and $Q(H)$ and $P\wedge Q := \inf\{P,Q\}$ is the projector onto the intersection of the said closed subspaces.

This lattice turns out to be orthomodular, bounded, atomic, satisfying the covering law, separable, ($\sigma$-)complete.

You also need **not** assume that the lattice of elementary propositions of a quantum system is $L(H)$ from scratch, but you can *prove* it, assuming some general hypotheses (those I wrote above together with a few further technical requirements). However what you eventually find is that the Hilbert space can be real, complex or quaternionic. This result was obtained by Solèr in 1995.

We have a probability measure that satisfies the Kolmogorov axioms (including countable additivity, but without the connotations of boolean logic).

Correct. The lattice is (orthocomplemented and) **orthomodular** ($A= B \vee (A \wedge B^\perp)$ if $B\leq A$) instead of (orthocomplemented and) **Boolean** ($\vee$ and $\wedge$ are mutually distributive).

However the story is much longer. The elements of $L(H)$ are interpreted as the **elementary propositions/observables** of a quantum system, admitting only the outcomes YES and NOT under measurement.

In an orthomodular lattice, two elements $P,Q$ are said to **commute** if the smallest sublattice including both them is Boolean.

It is possible to prove that, for the lattice of orthogonal projectors $L(H)$, a pair of elements $P$ and $Q$ commute if and only if they *commute as operators*: $PQ=QP$.

*A posteriori*, this is consistent with the idea that these elementary observables can be measured simultaneously.

If $P$ and $Q$ in $L(H)$ commute, it turns out that $$P\wedge Q = PQ =QP\tag{*}$$ and $$P\vee Q = P+Q-PQ\:.\tag{**}$$

A crucial point is the following one. *Having a Boolean sublattice (i.e. made of mutually commuting elements) $\vee$ and $\wedge$ can be equipped with the standard logical meaning of OR and AND respectively. The orthogonal $P^\perp = I-P$ corresponds to the negation NOT $P$*.

This is a way to partially recover classical logic form quantum logic.

Do we also need to assume some form of the law of large numbers?

Actually, at least when you make measurements of observables, you always reduce to a Boolean subalgebra where the probability measure becomes a standard $\sigma$-additive measure of a $\sigma$-algebra and here you can assume standard results on the relation between probabilities - frequencies.

Is the following something like the correct list of results?

The probability measure can be described by a density matrix (Gleason's theorem).

Yes, provided the Hilbert space is separable with dimension $\neq 2$.

In particular, the **extremal elements** of the convex set of probability Gleason measures (the probability measures which cannot be decomposed into non-trivial convex cominations) turn out to be of the form $|\psi\rangle \langle \psi|$ for every possible $\psi\in H$ with unit norm. This way, extremal measures coincides to **pure states**, i.e., unit vectors up to phases.

Observables must be represented by self-adjoint operators.

Yes, this is straightforward to prove if one starts by assuming that an observable $A$ is a collection $\{P^{(A)}(E)\}_{E \in B(\mathbb R)}$ of elements of the lattice $L(H)$, that is, projectors $P(E)$ where $E\subset \mathbb R$ is any real Borel set.

The physical meaning of $P^{(A)}(E)$ is "the outcome of the measurement of $A$ lies in (or is) $E$".

Evidently $P^{(A)}(E)$ and $P^{(A)}(E')$ commute and giving the standard meaning to $\wedge$ (= AND), we have from (*) that $$P^{(A)}(E) P^{(A)}(F) = P^{(A)}(E)\wedge P^{(A)}(F) = P^{(A)}(E\cap F)\:.\tag{1}$$

Using completeness, it is not difficult to justify also the property $$\vee_i P^{(A)}(E_i) = P^{(A)}(\cup_i E_i)$$ where the $E_i$ and a finite or countable class of Borel sets pairwise disjoint. This requirement, making in particular use of (**), is mathematically equivalent to $$\sum_i P^{(A)}(E_i) = P^{(A)}(\cup_i E_i)\tag{2}$$ where the $E_i$ and a finite or countable class of Borel sets pairwise disjoint and the sum is computed in the strong operator topology.

Finally since some outcome must be measured in $\mathbb R$, we conclude that $$P^{(A)}(\mathbb R)=I\tag{3}\:,$$ because the trivial projector $I \in L(H)$ satisfies $\mu(I)=1$ for every Gleason state.

Properties (1), (2) and (3) say that $\{P^{(A)}(E)\}_{E \in B(\mathbb R)}$ is a **projection valued measure** (PVM) so that the **self-adjoint** operator
$$A = \int_{\mathbb R} \lambda P^{(A)}(\lambda) $$
exists.

The spectral theorem proves that the correspondence between observables and self-adjoint operators is one-to-one.

Given a pure state represented by the unit vector up to phases $\psi$ and a PVM $\{P^{(A)}(E)\}_{E\in B(\mathbb R)}$ describing the observable/self-adjoint operator $A$, the map $$B({\mathbb R}) \ni E \mapsto \mu^{(A)}_\psi(E) := tr(|\psi\rangle \langle \psi| P^{(A)}(E)) = \langle \psi|P^{(A)}(E) \psi\rangle$$
is a standard probability measure over $\sigma(A)$, and standard results of QM arise like this ($\psi$ is supposed to belong to the domain of $A$)
$$\langle \psi |A \psi \rangle = \int_{\sigma(A)}\lambda d\mu^{(A)}(\lambda)\:,$$
justifying the interpretation of the left-hand side as **expectation value of $A$** in the state represented by $\psi$, and so on.

It also turns out that the support of a PVM coincides with the spectrum $\sigma(A)$ of the associated observable.

The elements $P$ of $L(H)$ are self-adjoint operators and thus the picture is consistent: $P$ is an elementary observable admitting only two values $0$ (NOT) and $1$ (YES). In fact $\{0,1\} = \sigma(P)$ unless considering the two trivial cases (the contradiction) $P=0$ where $\sigma(P)= \{0\}$ and $P=I$ (the tautology) where $\sigma(P)= \{1\}$.

Time evolution must be unitary.

Here one has to introduce the notion of **symmetry** and **continuous symmetry**.

There are at least 3 possibilities which are equivalent on $L(H)$, one is the well-known Wigner's theorem. The most natural one, in this picture, is however that due to Kadison (one of the two possible versions): *a symmetry can be defined as an isomorphism of the lattice $L(H)$, $h: L(H) \to L(H)$*.

It turns out that (Kadison's theorem) isomorphisms are all of the form $$ L(H)\ni P \to h(P) = UPU^{-1}$$ for some unitary or antiunitary operator $U$, defined up to a phase, and depending on the isomorphism $h$.

**Temporal homogeneity** means that there is no preferred origin of time and all time instants are physically equivalent.

So, in the presence of time homogeneity, there must be a relation between physics at time $0$ and physics at time $t$ preserving physical structures. Time evolution form $0$ to $t$ must therefore be implemented by means of an isomorphism $h_t$ of $L(H)$.

Since no origin of time exists, it is also natural to assume that $h_t\circ h_s = h_{t+s}$.

It is therefore natural to assume that, in the presence of *temporal homogeneity*, time evolution is represented by an one-parameter group of such automorphisms ${\mathbb R} \ni t \mapsto h_t$. (One-parameter group means $h_t\circ h_s = h_{t+s}$ and $h_0= id$.)

It is also natural assuming a continuity hypothesis related to possible measurements and states:

$${\mathbb R} \ni t \mapsto \mu(h_t(P))$$

is *continuous* for every $P\in L(H)$ and every Gleason state $\mu$.

Notice that Kadison theorem associates a unitary $U_t$ to every $h_t$ *up to phases*, so that there is no reason, *a priory*, to have $U_tU_s = U_{t+s}$, since phases
depending on $s$ and $t$ may show up.

Even if one is so clever to fix the phases to prove the composition rule of an *one-parameter group of unitary operators* $U_tU_s = U_{t+s}$ and $U_0=I$, there is no *a priory* reason to find a continuous map $t \mapsto U_t$ in some natural operator topology.

Actually under the said hypotheses on $\{h_t\}_{t\in \mathbb R}$, it is possible to prove that (the simplest example of application of *Bargmann's theorem* since the second co-homoloy group of $\mathbb R$ is trivial) the phases in the correspondence $h_t \to U_t$ via Kadison's theorem can be unambiguously accommodated in order that $h_t(P) = U_t P U_t^{-1}$ where $$\mathbb R \ni t \mapsto U_t$$ is a s*trongly continuous* *one-parameter group* of unitary operators.

**Stone's theorem** immediately implies that
$U_t = e^{-itH}$ for some self-adjoint operator $H$ (defined up to an additive constant in view of arbitrariness of the phase of $U_t$).

This procedure extended to other one-parameter groups of unitary operators $e^{-isA}$ describing continuous symmetries gives rise to the well-known quantum version of Noether theorem. The continuous symmetry preserves time evolution, i.e., $$e^{-isA} e^{-itH}= e^{-itH}e^{-isA}$$ for all $t,s \in \mathbb R$, if and only if the observable $A$generating the continuous symmetry is a constant of motion: $$e^{itH}Ae^{-itH}=A\:.$$