Is Interpretation of state vectors and density matrices according to Frequentist or Bayesian interpretation of probability? I asked a question on math stack exchange what does probability mean. I did not know about Frequentist and Bayesian interpretation of probability previously. So according to which interpretation are the density operators and amplitude squares of state vector defined ? I am reading the introduction to quantum information from Nielsen and Chuang. For measurement operators $\{M_m \}$ such that $\sum_m M_m^{\dagger}M_m=I$ the book defines probability that the outcome is $m$ as $\langle \psi|M^{\dagger}M_m|\psi\rangle$ when measuring state  $|\psi\rangle$. They also define similarly in terms of density operator.
So is all this according to Frequentist or Bayesian interpretation of probability or are both interpretations used whenever one seems more suitable ?
 A: It's pretty hard to be Bayesian about quantum mechanics without believing in some sort of underlying hidden-variable theory. Such theories are highly unpopular in modern culture (not to mention experimentally falsified in the majority of cases) and so the overwhelming interpretation amongst physicists is an operationalist/frequentist one.
A: Heisenberg discusses probability in his Physics and Philosophy. He first stresses that quantum mechanics contains objective, frequentist probabilities:

Probability in mathematics or in statistical mechanics means a
  statement about our degree of knowledge of the actual situation. In
  throwing dice we do not know the fine details of the motion of our
  hands which determine the fall of the dice and therefore we say that
  the probability for throwing a special number is just one in six. The
  probability wave of Bohr, Kramers, Slater, however, meant more than
  that; it meant a tendency for something. It was a quantitative version
  of the old concept of 'potentia' in Aristotelian philosophy. It
  introduced something standing in the middle between the idea of an
  event and the actual event, a strange kind of physical reality just
  in the middle between possibility and reality.

He continues, arguing that in addition to the objective probability, there is a subjective, Bayesian probability, related to our incomplete knowledge about the initial wave-function:

This probability function represents a mixture of two things, partly a
  fact and partly our knowledge of a fact. It represents a fact in so
  far as it assigns at the initial time the probability unity (i.e.,
  complete certainty) to the initial situation: the electron moving with
  the observed velocity at the observed position; 'observed' means
  observed within the accuracy of the experiment. It represents our
  knowledge in so far as another observer could perhaps know the
  position of the electron more accurately. The error in the experiment
  does - at least to some extent - not represent a property of the
  electron but a deficiency in our knowledge of the electron...the
  probability function contains the objective element of tendency and
  the subjective element of incomplete knowledge

The key point is that even with maximum information about the wave-function, we cannot make concrete predictions  for all observables in quantum mechanics - there is an intrinsic, objective uncertainty, unrelated to our degree of knowledge. For that reason, it is difficult to understand quantum mechanics with purely Bayesian probability.
In spite of that QBism (quantum Bayesianism) - an interpretation of quantum mechanics that aims to do just that. In my opinion, however, the interpretation is inconsistent. Furthermore, it is of course conceivable that probabilities in quantum mechanics result from our incomplete knowledge of so-called "hidden variables"; if that were so, probability in quantum mechanics would indeed be Bayesian, much like Heisenberg's example of throwing a dice.
A: At a fundamental level, if you are asking whether the universe has unknown variables (which Bayesian statistics often teases out in other sciences), the answer is provably not. 
If you are asking whether any of the math looks the same, the answer is an unequivocal yes, though you must remember that calculating a probability in Bayesian statistics is fundamentally different from QM. Instead, we calculate any number of intermediate quantities with math that looks almost identical to Bayesian statistics. Doing so allows us to understand a variety of intermediate processes too difficult to measure directly.
For example, a collection of known states interacts with an unknown potential. When we measure the outcomes, we use test potentials to fit the data. Next, we quantify how well those potentials fit. This is a categorically Bayesian approach and we know how a variety of potentials which are far too small to measure directly are shaped because of this math. Mathematically, a purely Frequentist approach would be to simply report the data and venture no guess about the shape of the potential as it is a pre-condition. 
As far as interpreting what anything means, when it comes to Bayesian vs. Frequentist discussions, they talk about different things - Bayesians will say the probability is $a$ given $b$ and they are $c$ uncertain about $a$ or $b$ whereas Frequentists say the probability is $b$ and the uncertainty of that is $d$. When it comes to your example, if $|\psi \rangle$ has been prepared in some way, you're doing a Bayesian calculation, and if it has not been prepared, you are doing a Frequentist approach (it appears $| \psi \rangle$ is unconstrained, so I'd guess Frequentist).
