Quantum mechanics defines a state of a system as a superposition of 'classical' states with complex coefficients, thus reducing many problems to linear algebra. Can classical statistics be approached to that way? Is this approach useful?
2026-04-02 10:47:19.1775126839
'Quantum' approach to classical probability
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Yes. You can define a(n algebra dual to a) noncommutative probability space as a complex $^{\ast}$-algebra $A$ together with a positive linear $^{\ast}$-functional $\mathbb{E} : A \to \mathbb{C}$ (the expectation). The quantum examples occur when $A$ is a suitable $^{\ast}$-algebra of linear operators on a Hilbert space and the classical examples occur when $A$ is a suitable $^{\ast}$-algebra of integrable functions on a measure space $X$.
I am still thinking about the merits of this approach for classical probability but I think it is better suited for making certain kinds of more algebraic arguments (e.g. it is perfectly suited for thinking about convergence in the sense of moments). Terence Tao's notes on free probability contain some good motivation for this point of view.
Let me give an example. Suppose I want to randomly choose a conjugacy class in a compact Hausdorff group $G$. Classically I would have to construct Haar measure on $G$ to do this. In the algebraic setting, by Peter-Weyl the span of the characters of finite-dimensional unitary irreducible representations of $G$ is dense in the space of class functions, and the integral of any such character over $G$ is uniquely determined by representation-theoretic concerns, so as long as I only care about integrating class functions on $G$ that I know how to express in terms of characters, I can do representation theory instead of constructing the Haar measure!
In this way the problem of, for example, computing moments of various random variables (e.g. the trace in some irreducible representation $V$) reduces to the problem of decomposing tensor products of representations of $G$. For a more specific example see my answer to this MO question.
This observation addresses an interesting question, which is why the category $\text{Rep}(G)$ of, say, finite-dimensional unitary representations of $G$ behaves like a categorified inner product space, with $\text{Hom}_G(V, W)$ categorifying the inner product and so forth. The answer is that what it actually behaves like is a categorified (algebra dual to a) noncommutative probability space, namely Haar measure on the conjugacy classes of $G$ (it more or less categorifies the algebra of class functions), and any such thing is a semi-inner product space with semi-inner product given by
$$\langle a, b \rangle = \mathbb{E}(a^{\ast} b).$$
Here the expectation $\mathbb{E}$ is categorified by taking invariant subspaces $\text{Inv}$, and we have $\text{Hom}_G(V, W) \cong \text{Inv}(V^{\ast} \otimes W)$ naturally.
Let me give a perhaps more accessible example. Suppose I have a random variable $X$ and I want to talk about an independently identically distributed sequence $X_1, X_2, ...$ of random variables all of which are distributed like $X$. Classically I would have to construct an appropriate measure space on which all of the $X_i$ are functions. Algebraically it is enough to start with an algebra $A$ as above in which $X$ sits and consider the "tensor product"
$$A_1 \otimes A_2 \otimes ...$$
of countably many copies of $A$. It is worth emphasizing that this does not agree with the tensor product of underlying vector spaces; its elements are linear combinations of finite tensor products $a_{i_1} \otimes ... \otimes a_{i_n}$ where $a_{i_j} \in A_{i_j}$ and $i_1 < i_2 < ... < i_j$. The expectation is defined multiplicatively on pure tensors (so that the variables coming from different $A_i$ are independent as desired). In fact this strategy works with no modification for an arbitrary collection of i.i.d. variables, whereas classically my impression is that there are some technical issues with defining an arbitrary product of measure spaces (although I could be mistaken here).