I'm sure this question is a bit gauche for this site, but I'm just a mathematician trying to piece together some physical intuition.

*Question:*Is the statistical interpretation of Quantum Mechanics still, in any sense, viable? Namely, is it completely ridiculous to regard the theory as follows: Every system corresponds to a Hilbert space, to each class of preparations of a system corresponds to a state functional and to every class of measurement procedure there is a self-adjoint operator, and finally, a state functional evaluated at one of these self-adjoint operators yields the expected value of numerical outcomes of measurements from the class of measurement procedures, taken over the preparations represented by the state?

I am aware of Bell's inequalities and the fact that the statistical interpretation can survive in the absence of locality, and I am aware of the recent work (2012) which establishes that the psi-epistemic picture of quantum mechanics is inconsistent with quantum predictions (so the quantum state must describe an actual underlying physical state and not just information about nature). Nevertheless, I would really like a short summary of the state of the art with regard to the statistical interpretation of QM, against the agnostic (Copenhagen interpretation) of QM, at present.

Is the statistical interpretation dead, and if it isn't...where precisely does it stand?

An expert word on this from a physicist would be very, very much appreciated. Thanks, in advance.

EDIT: I have changed the word "mean" to "expected" above, and have linked to the papers that spurred this question. Note, in particular, that the basic thing in question here is whether the statistical properties prescribed by QM can be applied to an individual quantum state, or necessarily to an ensemble of preparations. As an outsider, it seems silly to attach statistical properties to an individual state, as is discussed in my first link. Does the physics community share this opinion?

EDIT: Emilio has further suggested that I replace the word "statistical" by "operational" in this question. Feel free to answer this question with such a substitution assumed (please indicate that you have done this, though).

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    $\begingroup$ What is this "recent work (2012) which establishes that the psi-epistemic picture of quantum mechanics is inconsistent" of which you speak? $\endgroup$ – Nathaniel Oct 22 '12 at 1:20
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    $\begingroup$ Why "mean value"? Aren't you allowed to say anything about the distribution of measurement results? $\endgroup$ – Peter Shor Oct 22 '12 at 1:35
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    $\begingroup$ @PeterShor: The quote asserts that in a state $\omega$, $\omega(A)$ is the mean value of observations of $A$ (that of course take values in the spectrum). This is directly taken from the standard definition of states in the algebraic approach to quantum mechanics and QFT; see e.g., Vol. 3 of the mathematical physics treatise by Thirring. Assertions about the distribution of measurement results are simply assertions about means of the selfadjoint projection operators associated with the spectral decomposition of $A$. $\endgroup$ – Arnold Neumaier Oct 22 '12 at 7:28
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    $\begingroup$ You may want to distinguish between a "statistical" interpretation and an "operational" interpretation, which is probably closer to what your quote is describing. $\endgroup$ – Emilio Pisanty Oct 22 '12 at 14:55
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    $\begingroup$ The distinction, as I understand it, is that an operational approach does not by itself assert any postulate about the ontological nature of the system under description. People usually understand "statistical" interpretations to make some such assumption. Anyway - they're probably good keywords when googling. $\endgroup$ – Emilio Pisanty Oct 22 '12 at 15:47

The statistical interpretation of quantum mechanics is alive, healthy, and very robust against attacks.

The statistical interpretation is precisely that part of the foundations of quantum mechanics where all physicists agree. In the foundations, everything beyond that is controversial.

In particular, the Copenhagen interpretation implies the statistical interpretation, hence is fully compatible with it.

Whether a state can be assigned to an individual quantum system is still regarded as controversial, although nowadays people work routinely with single quantum systems. The statistical interpretation is silent about properties of single systems, one of the reasons why it can be the common denominator of all interpretations.

[Added May 2016:] Instead of interpreting expectations as a concept meaningful only for frequent repetition under similar conditions, my thermal interpretation of quantum mechanics interprets it for a single system in the following way, consistent with the practice of thermal statistical mechanics, with the Ehrenfest theorem in quantum mechanics, and with the obvious need to ascribe to particles created in the lab an approximate position even though it is not in a position eigenstate (which doesn't exist).

The basic thermal interpretation rule says:

Upon measuring a Hermitian operator $A$, the measured result will be approximately $\bar A=\langle A\rangle$ with an uncertainty at least of the order of $\sigma_A=\sqrt{\langle(A−\bar A)^2\rangle}$. If the measurement can be sufficiently often repeated (on an object with the same or sufficiently similar state) then $\sigma_A$ will be a lower bound on the standard deviation of the measurement results.

Compared to the Born rule (which follows in special cases), this completely changes the ontology: The interpretation applies now to a single system, has a good classical limit for macroscopic observables, and obviates the quantum-classical Heisenberg cut. Thus the main problems in the interpretation of quantum mechanics are neatly resolved without the need to introduce a more fundamental classical description.

  • $\begingroup$ Arnold, if I had registered for this site, I'd upvote your answer. I'll look into the implication here. (If it is trivial why Copenhagen implies statistical, I'd really love to see a short outline of the proof!) $\endgroup$ – Jon Bannon Oct 22 '12 at 11:53
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    $\begingroup$ @JonBannon: Part of the Copenhagen interpretation is Born's rule, from which one can easily derive the statistical properties of the states. This is done at the beginning of any text on statistical mechanics. cf. en.wikipedia.org/wiki/Density_matrix $\endgroup$ – Arnold Neumaier Oct 22 '12 at 12:33
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    $\begingroup$ Arnold is right. The problems with locality begin only when you try to take the statistical interpretation further and insist that all outcomes of all observables must have a joint probability distribution. Say if for a pair of particles you require that $p(x_1, x_2, p_1, p_2)$ exists, where $x_i$, $p_i$ are position and momentum, then you can show that Bell inequality is always satisfied when observables correspond to momentum and position. If you extend this requirement to all observables, you find Bell inequality is always satisfied. This is how far precisely you can go with statistics. $\endgroup$ – SMeznaric Oct 22 '12 at 14:10
  • $\begingroup$ @SMeznaric: Your ideas about joint probabilities are interesting. Can you provide a reference discussing this in depth? It seems to me that the trouble may be that we are applying classical probability to something that requires a noncommutative probability theory. (I'll go out on a limb here and say that assuming classical independence of multiple systems, i.e. tensor product relation in the usual QM postulates, seems awfully optimistic!) $\endgroup$ – Jon Bannon Oct 22 '12 at 14:46
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    $\begingroup$ Yeah, Bell inequalities basically reduce to the marginal problem, where given a number of marginal probability distributions you cannot construct a joint distribution. If you could Bell inequalities would be satisfied. We had a discussion about this on mathoverflow (mathoverflow.net/questions/107007/…) and I am also currently writing a doctoral thesis which will contain a discussion of this. If you are interested I can send it to you by email. I have not found any other reference where this is proven directly. $\endgroup$ – SMeznaric Oct 23 '12 at 15:10

As Peter Shor correctly says, what is actually being measured in one experiment/measurement isn't the mean value – as the OP wrote – but one of the eigenvalues, and the probabilities of individual eigenvalues are given by the squared absolute values of the amplitudes. This has been true since the 1920s, Max Born got a well-deserved Nobel prize for the probabilistic interpretation, and there's no doubt it's the framework in which our world works. Otherwise the blockquoted text in the original question is right.

Quantum mechanics is epistemic because the wave functions etc. express observers' subjective knowledge, but in the quantum sense. It's also "as ontic as you can get" because there fundamentally don't exist any "objective variables" whose values would be agreed upon by everyone. Despite the fundamentally subjective nature of the wave function in quantum mechanics, the theory also guarantees the agreement between observers etc. whenever it is required by logic or experience.

All the recent and not-so-recent papers claiming that they have "ruled out" an ontic or epistemic picture have only ruled out naive classical hidden-variable models that have nothing to do with the reality, so none of the papers has any relevance for physics as the science that studies how Nature works and not how it doesn't work. Both ontic and epistemic hidden-variable models have been known to be inapplicable to the Universe for something like half a century.

Ontic and epistemic hidden-variable models differ by assuming that there exist some extra hidden variables behind the wave function etc. but they still assume that the world is fundamentally classical and that's the problem with both of these classes. To describe Nature, one has to use the correct quantum mechanical framework which is neither "classical and ontic" nor "classical and epistemic" because it is not classical! Whether quantum mechanics itself is labeled "epistemic" is a matter of terminology and I would say Yes.

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    $\begingroup$ There is nothing subjective in quantum mechanics. In contrast to what you write, the variables associated with measurement pointers or counters are objective variables whose values are agreed upon by everyone as being the objective measurement results. Otherwise we couldn't perform even a single scientific experiment. $\endgroup$ – Arnold Neumaier Oct 22 '12 at 7:22
  • $\begingroup$ Dear Lubos, my claim was not that an individual measurement yields a mean value. What I am looking for here is not a statement like "Both ontic and epistemic hidden-variable models have been known to be inapplicable to the Universe for something like half a century", but a succinct summary of the details of this fact. Thanks for the answer! $\endgroup$ – Jon Bannon Oct 22 '12 at 11:50
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    $\begingroup$ This proferred «answer» is a mere tissue of assertions, and is not useful. There seems to be something about this topic that brings out the dogmatic on all sides... $\endgroup$ – joseph f. johnson Feb 12 '13 at 0:05

Quantum mechanics is a deterministic theory. Lagrange's thought experiment still holds within it's framework: should you be able to specify the exact state of the universe, you would know precisely the state for all future time (to the extent non-relativist QM is right.)

The difficulty in interpretation arises in these processes of "measurement" and "preparation" where 2 difficult (and related) things happen.

The first is that we'd like to describe the system isolated from the rest of the universe and there's no guarantee that the state space of the system and the state space of the universe decompose into a direct sum nicely. They certainly do not during preparation and measurement.

Second, the macroscopic measurement system is a statistically described thermodynamic system. We know very little about it's state except the extensive and average quantities (Energy, Entropy etc.) that define it's thermodynamic quantities. So when particles couple to this complicated thermodynamic system, (during both measurement and preparation) the combined wave-function can only be understood in a statistical way. The combined apparatus/particle state is an element from the direct product of the two state spaces, and moves chaotically (ergodically) through this state space.

Measuring systems are peculiar and interesting because they have Hamiltonians that (to the extent that they are good measurement apparatuses) decompose into nearly a direct sum of the particle's eigenstates. The combined state vector moves from the thin bridge of state space between eigenstates into the vastly bigger state space dominated by a single eigenstate. When we project this state out into a direct sum of the the macroscopic and microscopic system, we find (unsurprisingly) that the wavefunction has "collapsed" into an eigenstate of that particular measurement operator. It's statistical, but deterministic.

At least that's how it seems to me. If I'm completely off base here, somebody correct me please.

  • $\begingroup$ On further reflection...While thermalization with the measurement apparatus explains decoherence and why the different eigenstates that composed our original state no longer interfere, it doesn't really explain why macroscopic systems can't partake of multiple macrostates at once. Our composite state should diffuse into all subspaces of the measurement Hamiltonian, just there will be no interference between states... I think you see where this is going. Thinking about this has converted me to a many-worlds-er. Sorry I couldn't be more hard headed. :( $\endgroup$ – Colin LaMont Oct 24 '12 at 7:54

I was beginning to write a rather long entry that said some things that are probably best left unsaid about your referenced paper. However, for the sake of constructive argument and saving of time, I would refer you to this paper by Hoffman and the related web article. The point is clear that it is nonsensical to discuss the measurement probability of things that can not be simultaneously measured, however such things have a clear statistical interpretation. This appears to be a popular conceptual problem that is now at the roots of multiple debates.

  • $\begingroup$ I appreciate your abstaining from the polemical. (My little boat has landed in quite a storm, here, it appears.) $\endgroup$ – Jon Bannon Oct 22 '12 at 18:13

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