Quantum probability and quantum measure theory

Question

Do quantum probability and free probability mean the same thing - that is, they deal with noncommutative random variables? What about quantum measure theory? Is quantum measure theory the foundation of quantum probability (kind of like measure theory is the foundation of the classical probability theory)?

Answer 1

Quantum probability and free probability do not mean the same thing, the latter is a special case of the former.

In quantum probability there exist several fundamentally different notions of independence, including free independence and tensor independence. See for example these two articles: http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.6.6129&rep=rep1&type=pdf and http://repository.kulib.kyoto-u.ac.jp/dspace/bitstream/2433/42337/1/1278_09.pdf

Tensor independence is the noncommutative generalization of independence in classical (i.e., Kolmogorovian) probability theory. When in classical probability theory two random variables $X_1$ and $X_2$ are independent, then this can be described by the usual product of two classical probability spaces. The quantum probabilistic translation of this is to interpret the random variable $X_i$, $i=1,2$, as an element of an algebra $A_i$, and expectation value as a state $\varphi_i$ on $A_i$, the pair $(A_i,\varphi_i)$ then is called a quantum probability space, and the independence of $X_1$ and $X_2$ is described by the tensor product of $(A_1,\varphi_1)$ and $(A_2,\varphi_2)$.

Free independence involves the free product of quantum probability spaces, rather than the tensor product, but otherwise also satisfies certain characteristics which define independence. It has no classical counterpart.