For example the Maxwell's equations is a local theory. It's a set of differential equations that describe how should the state at a point change based on its neighbourhood.

Counter example: Newtonian mechanics isn't a local field theory because the state at a point can depend on the rest of the world since interactions can be instantaneous in that framework.

Since special relativity we know no interaction can spread faster than the light. So this would mean that effects should be local: the state of the point should depend on the state of it's direct neighborhood.

But is any local field theory automatically a local hidden variable theory thus sorted out by the Bell's theorem?

  • $\begingroup$ there's a problem with bell's theorem: see this $\endgroup$
    – image357
    Mar 28, 2015 at 13:27

2 Answers 2


You asked:

But is any local field theory automatically a local hidden variable theory thus sorted out by the Bell's theorem?

Bell's theorem has been widely misunderstood. What it says is if you have a theory in which the outcomes of experiments are represented by stochastic variables, and if that theory reproduces the predictions of quantum mechanics, then that theory must be non-local. However, quantum mechanics does not describe physical systems in terms of stochastic variables: the full description of a system is given by its Heisenberg picture observables. And when Bell-type experiments are described in terms of Heisenberg picture observables the correlations can be explained by local interactions:


The Heisenberg picture observables of the system are Hermitian operators that describe many versions of that system. They do not suddenly and magically turn into a single number, i.e. - a stochastic variable, when you do a measurement. This is commonly called the many worlds interpretation of quantum mechanics because for some reason people like to focus on the least interesting part of the theory: the part that involves lots of classical systems evolving in parallel. Rather, quantum theory explains why you only see a single outcome despite the existence of multiple versions of the system you measured. You can't interact with multiple versions of the system because this is incompatible with copying information about the system:


Note that you don't copy the entire state, this would be incompatible with the no-cloning theorem. Rather, on any given experiment you copy only the results of a measurement of a single observable.

In Bell type experiments, quantum information is carried in decoherent systems like detectors. The observables are dependent on this information, but the expectation values of measurements on either system alone are not dependent on this information. As a result this locally inaccessible quantum information can change the result of a measurement that compares the state of two decoherent systems. The correlations only take place when the results of measurements are compared and not before then. This entire process can take place locally provided that the equation of motion for the relevant observables is local.

So if you have a field theory where the field is described by quantum mechanical observables, that field theory can be local without conflicting with Bell's Theorem.

  • $\begingroup$ Are you saying that basically it's possible to describe the world in a form of cellular automation given that we can only partially observe it? I'm yet to understand quantum mechanics... $\endgroup$
    – Calmarius
    Jan 20, 2015 at 12:50
  • $\begingroup$ @Calmarius : I am not sure whether I understand what is the cellular automation. Do you mean a cellular array in which the information flows at each beat of a clock? $\endgroup$
    – Sofia
    Jan 20, 2015 at 14:05
  • $\begingroup$ @Sofia: In a cellular automation the values of the the cells are computed from the neighbour cells at each clock this. Of course I'm thinking of the continuous and Lorentz invariant version of this. $\endgroup$
    – Calmarius
    Jan 20, 2015 at 14:15
  • $\begingroup$ @Calmarius : of course! I absolved the electronics department, it's there where I learnt such things. In which department you learn/learnt? Now, I hope to have some talk with Hypnosifl this evening. It may be interesting for you. Anyway, I will notify you. Meanwhile, if you change the version of the question at least do notify the people that answered already, otherwise their answers may become no more to the point and they won't know that. $\endgroup$
    – Sofia
    Jan 20, 2015 at 14:28
  • $\begingroup$ The observables' equation of motion could be local, Lorentz invariant and all that. And there is no particular reason why they could not be simulated by a computer running an FDTD simulation or something like that, given enough time and memory. In practice having enough time and memory will be a large problem for a complicated system: the size of the matrices representing the observables increases exponentially with the number of systems involved. $\endgroup$
    – alanf
    Jan 20, 2015 at 14:41

Yes, you can say that a theory that describes the world by differential equations, is also a local variable theory.

But in the classical physics, i.e. not quantum, the local variables are not necessarily hidden. The Newtonian mechanics describes the world using the observable variables masses, velocities, positions, accelerations, and/or angles, angular velocities, etc. No need of hidden variables. Electromagnetism also uses electric and magnetic fields, charges, etc.

In statistical mechanics we have some difficulty: we use positions and velocities of particles in a gas. These are not hidden variables, but we cannot keep track of the individual movement of each molecule and molecule of gas.

In QM (quantum mechanics) the situation is more difficult. There are observables, as $x$ and $p_x$, that, according to Heisenberg's uncertainty principle cannot be measured together. We call them incompatible observables. So, people thought that there may exist hidden variables that produce well definite values for both $x$ and $p_x$, even if we cannot measure $x$ and $p_x$ together.

Bell's theorem, more exactly the violation of Bell's inequalities, shows that we have problems with these hidden variables. The problem, however, appears when we perform measurements on entangled particles. Moreover, the problems appear when the particles are sufficiently distant from one another so as to be impossible for them to interact.

Well, Newtonian mechanics doesn't deal with entanglements. Thus, no worry for it. Maxwell's equations don't comprise entanglements. Again, no worry.

As long as we don't have to do with entanglements, it's O.K., the theory may be local. But, as soon as we try to describe entanglements in terms of local hidden, or non-hidden variables, we can have troubles.

  • $\begingroup$ I don't think your first statement is correct--isn't Newtonian gravity a non-local theory, since gravitational influences are transmitted instantaneously? And yet gravitational interactions would be written in terms of differential equations. $\endgroup$
    – Hypnosifl
    Jan 19, 2015 at 23:47
  • $\begingroup$ You ask me something that nobody knows with precision. How can I take responsibility to say whether the Newtonian gravitation is local or non-local? Is vacuum an emptiness? On the other hand differential equations of the sort $v(t_0 + \text d t) = v_0 + a \text d t$ and $x(t_0 + \text d t) = x_0 + v_0 \text d t$ are local, i.e. tell me the data now, and I'll tell you the data after dt. Or, alternatively, tell me the data here and I'll tell you the data, a d$x$ further. $\endgroup$
    – Sofia
    Jan 20, 2015 at 0:20
  • 1
    $\begingroup$ Why do you say "nobody knows"? I'm not talking about any empirical question, just the theoretical question of whether Newtonian gravity, which we know to be an incorrect theory superseded by general relativity, satisfies physicist's definition of "non-local". I think you must be misunderstanding the definition, it has nothing to do with whether or not vacuum is "an emptiness"--the principle of locality wiki page says clearly that any theory containing instantaneous action at a distance, including Newtonian grav, violates locality. $\endgroup$
    – Hypnosifl
    Jan 20, 2015 at 0:30
  • $\begingroup$ (cont.) And p. 152 of Probability in Physics available on google books here says "Locality has two components: it asserts the continuity of causal influence and constrains its speed to forbid instantaneous propagation of impact and information ... In contemporary physics, locality is generally taken to mean Lorentz invariance, and the upper bound on the speed of propagation is the speed of light" The idea you were discussing of being able to "tell you the data after dt" is determinism, not locality (also discussed on p. 152). $\endgroup$
    – Hypnosifl
    Jan 20, 2015 at 0:36
  • $\begingroup$ @Hypnosifl : thank you. Will you look now? But, the fellow didn't ask about the non-local problems in the classical theory, but about the situations in which "the state at a point change based on its neighbourhood, look at his question." An answer should be to the point. Why should I begin to tell him that the Newtonian attraction is maybe non-local? He didn't ask about non-local classical cases, but about the local ones. $\endgroup$
    – Sofia
    Jan 20, 2015 at 1:25

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