# Why can't noncontextual ontological theories have stronger correlations than commutative theories?

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EDIT: I found both answers to my question to be unsatisfactory. But I think this is because the question itself is unsatisfactory, so I reworded it in order to allow a good answer.

One take on contextuality is to develop an inequality on measurement outcomes that is satisfied for any ontological noncontextual theory, and see that it is violated by quantum mechanics.

Another take would be to assume an algebraic structure and see that if one restricts the observable algebra to be commutative, the expected values of certain operators are restricted to lie in a given range, whereas if we allow non-commutativity the range is greater.

These approaches coincide? I've seen plenty of works that assume that it does, but without discussing it; in particular there is this paper by Tsirelson that states (in the particular case of Bell inequalities) that both approaches are equivalent, but without proving it. Is it too obvious?

At first sight, an ontological noncontextual theory is much more general than some theory embedded in a C*-algebraic framework. Why then can't it generate stronger correlations than theories with commuting algebras of observables?

Can one find a more direct connection between non-commutativity and the violation of a contextual inequality?

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Well, a reasonable approach to local hidden variables is to require commutativity of operators on different space-like separated systems. This is pretty straight forward to motivate, since otherwise you are essentially using non-local operators, which can be seen via the relevant transform instead as a non-local hidden variable theory with local operators.

You might want to check out Scott Aaronson's paper exploring the consequence of this when taken as an axiom together with some other desirable properties of hidden variable theories (Phys. Rev. A 71, 032325).

I'm not too sure it makes sense to talk about commutativity beyond this, since we care about post-measurement outcomes it is not clear that the domain of the operator should contain its image, and so multiplication, and hence the commutator, isn't necessarily defined.

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answered Sep 20, 2011 by (3,575 points)
I'm afraid you didn't understand my question. Commutativity of operators on different space-like separated system is respected by QM (or AQFT, for that matter). I'm talking about commutativity of the whole algebra. In fact it may not make sense to talk about commutativity in the strict sense; I've never seen anyone construct an operator algebra to model a hidden-variable theory. But if we only care about the measurement outcome, not the post-measurement state, we have commutativity in a trivial sense, by substituting the observable by its predefined outcome, which is just a real number.

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@Mateus: I think I did understand, which is what my last paragraph attempts to address.

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Non-commutativity of operators ensures that in general we can't construct a joint probability distribution over the observables that we model using those operators. In some states and for some choices of non-commuting operators we can construct joint probability distributions, for example the vacuum state and coherent states of a quantized simple harmonic oscillator generate a positive-definite Wigner function for position and momentum, which can be taken to be a probability distribution. Of course that possibility falls apart when one considers almost any superposition of coherent states, say. The Wigner function is not positive-definite in the general case, making the interpretation of the Wigner function as a probability distribution in the special cases quite tendentious.

Conversely, if we have a commutative algebra of operators we can construct a joint probability distribution over the any subset of the observables in any state over the algebra. One could take this property as a somewhat plausible definition of classicality.

For the technical basis of this, I like best two short papers, John Baez, Letters in Mathematical Physics 13 (1987) 135-136, and Lawrence J. LANDAU, PHYSICS LETTERS A, Volume 120, number 2 (1987), which put remarkably little interpretation in the way of the mathematics, but there is a substantial literature that has tried to get at this relationship in some sort of clear way.

A literature that gives an alternative way into the relationship between non-commutativity and measurement, and that focuses on the relationship between quantum theory and classical probability theory in a way that I find helpful, albeit not conclusive, is the positive-operator valued measure approach, which is well represented by the book by Paul Busch, Marian Grabowski, and Pekka J. Lahti, Operational Quantum Physics, Springer, 1995. Searching the literature or the ArXiv for anything more recent by any of these three authors will give you something enlightening to read. To my taste, Paul Busch is always worth reading.

As far as physicality is concerned, classical physics models measurements as not affecting other measurements, so that joint probability distributions over multiple measurements are possible. In the presence of any finite level of noise ---there always is noise, everywhere (only the thermal component of the noise goes away when one is close to absolute zero, the Lorentz invariant quantum component of the noise is not controllable)--- the uncontrolled nature of the noise is something that has to be accommodated by our models of our measurements. Quantum theory accommodates the non-trivial effects of joint measurements on each other by introducing non-commutativity of the operators that are used to model the measurements, whereas classical physics models the non-trivial effects of joint measurements on each other by modeling the measurement apparatus. Contextual models are precisely models that include the measurement apparatus, or the complete experimental apparatus, in the extreme case the whole universe, not just a putative measured system.

That's somewhat bashed out. Hope someone finds it congenial.

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answered Sep 20, 2011 by (1,220 points)
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I don't think there is a more direct connection between noncontextuality and noncommutativity for a couple of reasons. Firstly, there are noncommutaive sets of observables and states that can be simulated by a noncontextual model. Think of the Kochen-Specker model for a qubit for example. Secondly, to really answer the question of what a violation of some inequality means, you should not assume that the data you collect in the experiment is necessarily produced by quantum theory (in particular, we don't do this for Bell's inequalities). Now, there are plenty of operational theories that are contextual (in Rob Spekkens sense) but that do not have a C*-algebraic structure, e.g. the theory wherein the state space is a square. Unless you can define what it means for measurements in these theories to be "commutative", which seems unlikely because they do not have an algebraic structure, then it is clear that the relationship between commutativity and noncontextuality breaks down in this context.

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answered Nov 5, 2011 by (130 points)
Good point. It would be interesting to find an operational definition of commutativity. But allow me to be picky: I think my answer is clear on relating commutative theories and noncontextual models; what it lacks is a relation between non-commutativity and contextuality.

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They can't, because ontological noncontextual theories are not more general than commuting subsets of quantum mechanics. In a nuthsell, commutative quantum mechanics is just classical probability theory, and the question whether there exists an ontological noncontextual model for quantum mechanics is precisely the question if it can be reduced to classical probability theory.

To see this, one needs Spekkens' operational definition of contextuality: an ontological noncontextual model is one where each state $\rho$ is represented by a probability distribution $\mu_\rho(\lambda)$ on an ontological space $\Lambda$, and each POVM $\{E^k\}$ by a probability distribution $\xi_{E^k}(λ)$. Then the probability of outcome $k$ will be given by $$\int \mathrm{d}λ\, \mu_\rho(λ) \xi_{E^k}(λ).$$

Now, if all $E^k$ commute, I can write $ρ$ in a basis in which they are diagonal. Then $$\mathrm{tr} (\rho E^k) = \sum_n \rho_{nn} E^k_{nn},$$ that is, the only part of $\rho$ who'll play a part in the expected value is its main diagonal, and that's just a probability distribution. If we do the identifications $\lambda \mapsto n$, $μ_ρ(λ) \mapsto ρ_{nn}$, and $\xi_{E^k}(λ) \mapsto E^k_{nn}$, we have an embedding of an ontological noncontextual model into commutative quantum mechanics.

In the particular case of nonlocality, one could see directly that a diagonal $\rho$ is always separable, and thus admits a local ontological model. The converse path, to construct a separable $\rho$ and commutative algebra from a local ontological model is essentially the same as above.

That said, I don't think my last question is answered; I'd still like to see a more direct, physical connection between non-commutativity and contextuality.

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answered Nov 2, 2011 by (270 points)

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