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 ?

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    $\begingroup$ There is a Bayesian interpretation of quantum mechanics, see for example this nlab post, I don't know about a frequentist one (probably there is also). However, I think these interpretations are only a philosophical point of view on something that "exists independently", i.e. a mathematical model of physical reality, that works quite well independently of the interpretation. $\endgroup$ – yuggib May 11 '15 at 14:17
  • $\begingroup$ Any non-frequentist interpretation of physics dies when it comes in contact with the fact that we can only measure frequencies. Having said that, some derivations in Bayesian probability are cute, so there is little reason for us not to steal them, strip them to their true frequentist self and use them as we wish. $\endgroup$ – CuriousOne May 12 '15 at 10:50
  • $\begingroup$ @CuriousOne that's a strange comment. what do you mean we can only measure frequencies? even in the most fundamental experiments e.g. collider experiments, we measure numbers of events. whatever you mean by it, would that be a problem for bayesian probability? $\endgroup$ – innisfree May 12 '15 at 14:13
  • $\begingroup$ It's not a problem at all. Probability theory always has to live with the distinction between finite sample data and mathematical distributions. Bayesians are merely taking that to the next level, and as long as you ignore some of the more outrageous statements (like cosmology based on a sample of one), then you are good. $\endgroup$ – CuriousOne May 12 '15 at 19:49
  • $\begingroup$ if only being able to measure frequencies (whatever that means) isn't a problem for bayesian probability, why do you write that "Any non-frequentist interpretation of physics dies when it comes in contact with the fact that we can only measure frequencies"? $\endgroup$ – innisfree May 13 '15 at 7:46

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.

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    $\begingroup$ A hidden deterministic variable might cause problems but would a hidden random variable cause problems? If anything, I would imagine it would imply Bayesian is more correct than frequentist... $\endgroup$ – Mehrdad May 11 '15 at 17:35
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    $\begingroup$ @Mehrdad It depends on what you mean by "problems". We know that any hidden-variable theory must 1) be non-local (by Bell's theorem), 2) be contextual (by the Kochen-Specker theorem), and 3) describe orthogonal quantum states by non-overlapping probability distributions over the hidden-variable state space (by the Pusey-Barrett-Rudolph theorem). These constraints are so severe (and experimentally verified to great precision, modulo several rather outlandish "loopholes" which one could imagine), that very few viable alternatives remain. $\endgroup$ – Mark Mitchison May 11 '15 at 17:45
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    $\begingroup$ @Mehrdad No, all of those theorems allow for hidden random variables. $\endgroup$ – Mark Mitchison May 11 '15 at 17:51
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    $\begingroup$ @Mehrdad Any competent source should make this point clear. The proofs for all of these theorems start from the assumption of random hidden variables, of which deterministic variables are obviously a subset. The Wikipedia pages for the various theorems should be a good place to start. $\endgroup$ – Mark Mitchison May 11 '15 at 17:55
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    $\begingroup$ @Mehrdad For me the best reference on Bell's theorem has to be Bell's original paper. If you read and understand this, it should be clear. It's difficult for me to know what other references to give you, since I don't really understand why you thought the hidden variables had to be deterministic. Where did you get this idea from, or what made you assume that this would be true? $\endgroup$ – Mark Mitchison May 11 '15 at 18:56

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.


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).

  • $\begingroup$ this is still quite misleading (even without the math). the math of QM does not look at all the same as that in Bayesian probability - QM is based on 2-norms $\endgroup$ – innisfree May 12 '15 at 10:49

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