Independence of two noncommutative observables If two observables are free, you can find the joint distribution of these two observables. But, by Heisenberg's Uncertainty Principle it is impossible unless $X$ and $Y$ are such that $XY=YX$. 
Is there any physical significance of free independence? What's the physical interpretation for two observables to be free independent?
 A: Free probability theory was developed precisely to deal with noncommuting operators, represented by matrices $X$, $Y$. If $XY\neq YX$, the eigenvalues $\sigma$ of the sum $X+Y$ are not given by the sum of the eigenvalues $\lambda,\mu$ of $X$ and $Y$, so you cannot use the usual construction of the probability distribution $P_\sigma$ of $\sigma$ as the convolution of $P_\mu$ and $P_\lambda$. Free probability theory tells you how to construct, under certain conditions, $P_\sigma$ in terms of $P_\mu$ and $P_\lambda$. If this is the case, $X$ and $Y$ are called "free independent random variables".
Most of the physics applications I know of involve the quantization of classically chaotic systems. Suppose you have a billiard geometry of a certain shape, and you vary the shape a bit, parameterized by some variable $s$. If the billiard has classically chaotic dynamics, then the Hamiltonians $H(s_1)$ and $H(s_2)$ of that system evaluated at two sufficiently different values of $s$ are free independent.
A: I think some of the confusion in this question comes from the use of the phrase "joint distribution"; the slogan is that freeness of two variables allows to determine the joint distribution of the two variables, given the distribution of each of the variables. This is correct, but "joint distribution" has here to be understood as the non-commutative joint distribution, which is, by definition, the collection of all moments of the two variables. If the variables commute this can be identified with the classical meaning of joint distribution, i.e., a probability measure on $\mathbb{R}^2$; if the variables do not commute then this identification does not work any more. If variables are free, then (apart from trivial cases, where one of them is a constant) they do not commute, hence there is no joint distribution in the classical sense for two free variables. 
The search for a good analytic meaning of the joint distribution of non-commuting variables is at the moment actually a quite active direction in free probability. See, for example, my survey article
Free Probability theory.
The physical significance of freeness usually comes via its link with random matrices; then large $N$-limits of physical theories might have a formulation in terms of free probability theory.
Some literature with a more physics flavor are for example: 


*

*Quantum free Yang-Mills on the plane by Anshelevich and Sengupta

*Zooming in on local level statistics by supersymmetric extension of free probability by Mandt and Zirnbauer,  

*Anomalous transport: a mathematical framework" (in particular, chapter 5) by Schulz-Baldes and Bellisard, 

A: Free probability is a mathematical theory that studies noncommutative random variables. The “freeness” property is the analogue of the classical notion
of independence, and it is connected with free products. This theory was
initiated by Dan Voiculescu around 1986 in order to attack the free group
factors isomorphism problem, an important unsolved problem in the theory
of operator algebras. Typically the random variables lies in a unital algebra $ A$ such as a $C^{\ast}$-algebra or a von Neumann algebra. The algebra comes equipped with a noncommutative expectation, a linear functional $\tau: \mathcal{A}\rightarrow\mathbb{C}$ such that $\tau(1) = 1$. Unital subalgebras $A_{1},\ldots, A_{n}$ are then said to be free if the expectation of the product $a_{1}, \ldots, a_{n}$ is zero whenever each $ a_{1}\ldots , a_{n}$ has zero expectation, lies in an $A_{k}$, and no adjacent $ a_{j}$’s come from the same subalgebra $ A_{k}$. Random variables are free if they generate free unital subalgebras.
A: If two observables $X$ and $Y$ are free, all joint moments in $X$ and $Y$ vanish. Thus, if $X$ and $Y$ are self-adjoints like hermitians matrices, one can show with the help of the spectral theorem, the joint distribution  $\mu_{XY}$ of $X$ and $Y$ is uniquely determined by only the distribution $\mu_{X}$ of $X$ and the distribution $\mu_{Y}$ of $Y$. Thus, Heisenberg's uncertainty principle is not applied to free observables.  
