lukephysics said:
Could you please explain that a little more? As far as I know there is no such thing as quantum probability. It is simply the born rule which gives you a pdf which is classical.
That right and wrong at the same time ;-). Born's rule gives usual probabilities a la Kolmogorov at best for the measurement of a complete compatible (!!!) set of observables, ##A_1,\ldots,A_n##, because then the possible outcomes of such a measurement are the n-tupels of eigenvalues ##(a_1,\ldots,a_n)## of the corresponding operators, and the eigenstates are unique up to an irrelevant phase factor and build a complete orthonormal (!!!) basis. That means here you have to usual notion of a probability space a la Kolmogorov, i.e., with the set of "elementary outcomes" given by the ##n##-tupels of eigenvalues, and the corresponding probabilities, given by Born's rule,
$$P(a_1,\ldots,a_n) = \langle a_1,\ldots a_n| \hat{\rho}|a_1,\ldots,a_n \rangle,$$
where ##\hat{\rho}## is the statistical operator describing the quantum state the system is prepared in. In this sense the quantum probabilities form a standard realization of Kolmogorov's axioms.
However, there are almose more observables, which are not compatible with all the observables in any complete set. So the probabilities for the entire set of possible observables doesn't form such a Kolmogorovian probability scheme. You can even extent the description of probabilities for measurements involving incompatible observables, where you do "weak measurements", i.e., you gain some probabilistic information about the incompatible observables measured, but it can never be a complete information. That's formalised with positive operator-values measures (POVMs).