# Can quantum randomness be somehow explained by classical uncertainty? [closed]

In quantum mechanics, the outcome of each measurement is random, distributed according to the squared amplitude of the wave function obtained from the Schrodinger's equation. Now, can someone suggest that QM measurement outcomes are just produced by a deterministic, real, local process (or field) in the 3-D space, varying because of uncontrollable phenomena like noise?

(An example of such a classical field can be a real 3-D coherently rotating vector field accounting for the Schrödinger equation and electron spin in a consistent manner, which carries distributed angular momentum and energy in the same way as a circularly polarized electromagnetic wave).

Setting aside Bell's theorem which forbids such an explanation and any other local realistic theory, will such a classic explanation ever be able to reproduce the probabilities obtained from QM? I have seen that deterministic QM interpretations like Bohmian mechanics introduce concepts like infinite dimensional configuration space. Why is this necessary for reproducing QM?

## closed as off-topic by WillO, Aaron Stevens, ZeroTheHero, Jon Custer, ChairFeb 8 at 15:29

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• square root of the wavefunction (which is complex)? Do you mean modulus square? – ZeroTheHero Feb 5 at 4:48
• @ZeroTheHero Oh yes, I corrected it. Thanks for reminding. – Ali Lavasani Feb 5 at 4:48
• If you “set aside Bell’s theorem”, you’re no longer in the realm of mainstream physics, and therefore outside the realm of this site. – G. Smith Feb 5 at 5:40
• @G. Smith I want to know why classical deterministic explanations cannot reproduce QM predictions. This is not about Bell's theorem, which says that local realism doesn't exist even in a random way. Prior to Bell's theorem, many people like Einstein and Bohm tried to offer some "hidden variable" explanation, but they never considered a simple classic probability theory. I want to know what strictly rules such theories out. I mentioned that in order not to get the answer "Because of Bell's theorem". – Ali Lavasani Feb 5 at 6:03
• You should look into Stochastic Electrodynamics (SE), which reproduces most (if not all) Quantum Mechanics predictions from a random classical field theory. Contrary to what many people may say here, SE is a main stream physical theory (maybe not very well known), and it's still under development today. It's a kind of hidden variables theory (but not of the usual type). I don't know its current status relative to Bell's theorem. See this article: en.wikipedia.org/wiki/Stochastic_electrodynamics – Cham Feb 5 at 15:20

There are several serious obstructions. One is provided by Bell's analysis concerning the conflict between realism and locality. However that obstruction concerns a very peculiar situation, referring to a bipartite system, with parts causally separated, and quantum entangled states. There is another no-go result, usually called Kochen-Specker theorem, leading to a very severe obstruction against any completely classical interpretation of Quantum Mechanics based on hidden variables and epistemic randomness (however Bohmian quantum mechanics is untouched by it).

Actually this theorem exists into a number of versions and its origin can be traced back to the celebrated Gleason's theorem as observed by Bell himself in his second famous paper of 1966 preceding the paper by Kochen and Specker of 1967.

The basic idea underpinning the no-go result is that quantum observables $$A$$ (selfadjoint operators on the Hilbert space of the system) are actually classical variables and there is a classical hidden state $$\lambda$$ (a set of hidden classical variables $$\lambda \in \Lambda$$) which fixes the values $$v_\lambda(A) \in \mathbb R$$ of every observable $$A$$.

In this view, randomness of values attained by measurements of quantum observables is explained by assuming that $$\lambda$$ is unknown, but we know only a probability distribution $$\mu$$ over $$\Lambda$$ describing the probability that $$\lambda$$ attains some value (discrete distribution) or stay in some "continuous" set. This is what happens, for instance, in classical statistical mechanics. Here quantum probability becomes epistemic instead of ontic as in the standard interpretation of QM. In other words there must exist some correspondence $$\mu \leftrightarrow |\psi \rangle$$ such that $$\langle \psi| A \psi \rangle = \int_{\Lambda} v_\lambda(A) d\mu(\lambda)\:.$$

It remains to fix general rules to associate sharp values $$v_\lambda(A)$$ to observables $$A$$. The problem is how one should deal with functional relations as $$C=A+B$$. The naive idea to always assume that $$v_\lambda(C) = v_\lambda(A) + v_\lambda(B)$$ turns out to be untenable when $$A$$ and $$B$$ are quantistically described as incompatible observables as explained by Bell analysing an earlier no-go theorem by von Neumann in 1966.

A fair set of assumptions for $$A \mapsto v_\lambda(A)$$, which avoids to tackle any classical interpretation of quantum incompatibility, was proposed by Kochen and Specker referring to the algebra of observables $$B(\cal H)_{sa}$$ over a finite-dimensional Hilbert space $$\cal H$$ (finite dimensionality requirement can be relaxed by assuming some suitable continuity requirement on $$v_\lambda$$).

(1) The map $$v_\lambda : B({\cal H})_{sa} \ni A \mapsto v_\lambda(A) \in \mathbb R$$ is non-trivial (not all values are $$0$$).

(2) If $$A,B \in B(\cal H)_{sa}$$ are compatible observables (i.e. they commute), then $$v_{\lambda}(A+B) = v_\lambda(A)+ v_\lambda(B)$$.

(2) If $$A,B \in B(\cal H)_{sa}$$ are compatible observables (i.e. they commute), then $$v_{\lambda}(AB) = v_\lambda(A)v_\lambda(B)$$.

A more precise theory would also fix how the map $$v_\lambda$$ deals with incompatible observables. This thechnical specification is not necessary for producing the no-go result I go to state and this fact also shows how KS' result is powerful.

Kochen-Specker Theorem

If $$3\leq \dim(\cal H) < +\infty$$, then there is no map $$v_\lambda : B(\cal H)_{sa} \ni A \mapsto v_\lambda(A) \in \mathbb R$$ satisfying requirements (1), (2), (3).

This theorem rules out from scratch every classical interpretation of QM where the realism hypothesis, i.e. every quantum observable is actually classical and always has an (unknown) sharp value. All that before any attempt to explain quantum randomness in terms of some classical uncertainty.

Actually, a closer scrutiny shows that there is a way out when assuming the contextuality requirement: that the always existing values $$v_\lambda(A)$$ depend also on which observable $$B$$ I measure together with $$A$$ ($$B$$ is therefore assumed to be compatible with $$A$$). It may happens that $$v_\lambda(A|B)\neq v_\lambda(A|B')$$ if $$B$$ and $$B'$$ are incompatible (compatibility is not a transitive relation!). This impervious approach seems to be logically consistent even if it requires a big revision of our classical ideas on the physical world (personally I definitely prefer the standard interpretation of QM!).

The result of Kochen and Specker rules out realistic non-contextual hidden-variable interpretations of quantum theory.

There is an equivalent formulation of K-S theorem which is more suitable for experiments. It is based on the notion of test. A test is an observable which can assume only the value $$0$$ or $$1$$, in the standard formalism tests are all of orthogonal projectors $$P \in B(\cal H)_{sa}$$.

Kochen-Specker Theorem

If $$3\leq \dim(\cal H) < +\infty$$, then there is a set $$\cal P$$ of tests such that there is no map $$v_\lambda: {\cal P} \ni P \to \{0,1\}$$ satisfying the following requirements

(1) If $$P,P' \in \cal P$$ are compatible mutually exclusive tests ($$PP'=0$$ as orthogonal prjectors), then at most one of $$v_\lambda(P)$$, $$v_\lambda(P')$$ does not vanish.

(2) If $$P_1,\ldots, P_n \in \cal P$$ is a set of pairwise compatible and mutually exclusive tests such that $$P_1+\ldots + P_n =I$$, then one of $$v_\lambda(P_k)$$ does not vanish.

The original proof of KS theorem in 1967 proved that if $$\dim(\cal H)=3$$ there is a set of 117 tests satisfying the theorem. Actually a general proof valid for every dimension (also infinite when assuming some continuity hypothesis on $$v_\lambda$$) easily arises from Gleason's theorem as already noticed by Bell.

I have seen that deterministic QM interpretations like Bohmian mechanics introduce concepts like infinite dimensional configuration space.

I do not think so. Bohmian mechanics for particles is formulated in the standard $$3N$$ dimensional configuration space of a system of $$N$$ particle. Maybe you are considering the system of a quantum field. I am not an expert on this subject however.

As recent references I would like to mention various entries of Stanford Encyclopedia of Philosophy, Landsman's book on foundations of quantum theory, a book consisting of a wide collections of recent papers on Bell's analysis and further foundational issues.

(I am publishing a book on fundamental mathematical structures in quantum theoryand chapter 5 is completely devoted to study these issues including Bell's inequality and its interplay with locality and contextuality).

My answer here could be of interest

• Thanks for your answer and additional references (including your very excellent new book). Note that the Landsman book you cite is open access, and Springer explicitly says so on its copyright page. A free and perfectly legal pdf is downloadable from researchgate.net/publication/… – John Forkosh Feb 5 at 13:06
• Thank you for the piece of information about Klaas's book. I do not know if my book is excellent! Thank you however :) – Valter Moretti Feb 5 at 13:47

Bell's theorem is a "no-go theorem" that draws an important distinction between quantum mechanics and the world as described by classical mechanics, particularly concerning quantum entanglement where two or more particles in a quantum state continue to be mutually dependent, even at large physical separations.

.....

Bell's theorem states that any physical theory that incorporates local realism cannot reproduce all the predictions of quantum mechanical theory. Because numerous experiments agree with the predictions of quantum mechanical theory, and show differences between correlations that could not be explained by local hidden variables, the experimental results have been taken by many as refuting the concept of local realism as an explanation of the physical phenomena under test. For a hidden variable theory, if Bell's conditions are correct, the results that agree with quantum mechanical theory appear to indicate superluminal (faster-than-light) effects, in contradiction to the principle of locality.

....

(Currently accepted quantum field theories are local in the terminology of the Lagrangian formalism and axiomatic approach.)

The "no go" means that all the data fitted with quantum mechanical models and thus validate quantum mechanics, cannot be fitted with classical theories if locality is assumed in the mathematical model.

Locality is a principle in both classical and quantum physics, principles are axioms for physics models.

n physics, the principle of locality states that an object is directly influenced only by its immediate surroundings. A theory which includes the principle of locality is said to be a "local theory". This is an alternative to the older concept of instantaneous "action at a distance". Locality evolved out of the field theories of classical physics. The concept is that for an action at one point to have an influence at another point, something in the space between those points such as a field must mediate the action. To exert an influence, something, such as a wave or particle, must travel through the space between the two points, carrying the influence.

So the answer is no, you cannot ignore Bell's theorem within main stream physics which is what this site discusses.

• What I have heard is that the quantum probability is obtained from Born's rule, which is psi-squared. A classical uncontrollable (and unpredictable in practice, not in principle) factor such as noise or chaos cannot produce probability distributions we get from Born's rule, for example such a classical distribution would be Gaussian. Is this correct? This doesn't have anything to do with Bell's theorem. – Ali Lavasani Feb 5 at 6:56
• QM probability is $Ψ*Ψ$ , the complex conjugate $Ψ*$ squared with the $Ψ$.. Bohm's model does reproduce non relativistic QM, that is why it is called an interpretation of quantum mechanics, because it is local.. The statistical argument does not suffice for complicated models aiming at finding a classical physics explanation of quantum mechanics. locality does. look at deterministic proposals en.wikipedia.org/wiki/… – anna v Feb 5 at 7:52
• en.wikipedia.org/wiki/Hidden-variable_theory . Bells theorem constrains them to be non local – anna v Feb 5 at 7:54

Setting aside Bell's theorem which forbids such an explanation and any other local realistic theory, will such a classic explanation ever be able to reproduce the probabilities obtained from QM?

No. See Mermin's article, "Is the Moon Really There When Nobody Looks?", reprint at http://www.theory.caltech.edu/classes/ph125a/istmt.pdf That satisfies your "setting aside Bell's theorem" requirement.

Pay particular attention to the (unnumbered) section "a gedanken demonstration" on pages 4-9 of that reprint. Note the table of trial runs on page 6, and Mermin's subsequent analysis leading to his conclusion on page 9 that  ...  "There is no conceivable way to assign such instruction sets to the particles from one run to the next that can account for the fact that in all runs taken together, without regard to how the switches are set, the same colors flash half the time."

By "such instruction sets", Mermin means (see the details on reprint pages 4-9) the classical notion that once a system has been prepared, the outcome of any subsequent measurement is deterministically predictable. And that's precisely what you're suggesting in your first paragraph, "...that QM measurement outcomes are just produced by a deterministic, real, local process..."

Mermin's very simple analysis of his gedanken experiment's numerical outcomes unambiguously demonstrates the absolute impossibility of your suggestion. Bell's theorem not required (though you might say Mermin's argument is another approach to it). Again, read that reprint for the full details.