(Note - I edited to the question in response to answers)

In the 1935 EPR paper, Einstein, Podolsky and Rosen write that given two entangled particles, one particle can be used to predict with certainty the value of a quantity of the second particle, or in other words the outcome of measurement on the second particle:

Thus, by measuring either A or B we are in a position to predict with certainty, and without in any way disturbing the second system, either the value of the quantity P or the value of the quantity Q.

One way to come to terms with this phenomenon is to stipulate faster than light interaction. This is how John Stewart Bell put it:

In a theory in which parameters are added to quantum mechanics to determine the results of individual measurements, without changing the statistical predictions, there must be a mechanism whereby the setting of one measuring device can influence the reading of another instrument, however remote. Moreover, the signal involved must propagate instantaneously, so that such a theory could not be Lorentz invariant.

However, if I understand correctly, the current view in physics is that the remote particles to not interact upon measurement, and the phenomena is viewed in terms of correlations.

In particular in Quantum Field Theory, which according to the Wikipedia "is brought forward as an unavoidable consequence of the reconciliation of quantum mechanics with special relativity", the remote particles do not interact. For example, see the following answer by the physicist Luboš Motl:

there are no interaction terms operating in between the two particles at all! Because there are no interactions, there is no influence, and the observed correlations clearly can't have anything to do with any non-local interactions.

So Einstein writes that the nearby particle can be used to predict with certainty the outcome of a measurement of the remote particle.

One can ask how the particle does it, and the answer is that we do not know.

Is it possible in principle to build a machine that can predict with certainty the outcome of a measurement on the remote particle?

The answer is yes, the machine could use the nearby particle as an "oracle" to make its prediction.

But is it possible in principle to build a machine that can predict with certainty the outcome of a measurement on the remote particle without using the nearby particle?

As far as we know, the answer is no.

In particular there is no (Turing) computation, not even in principle, that can make the prediction Einstein is writing about.

The question is therefore, why is quantum entanglement not acknowledged as an observable incomputable phenomenon?

  • The "quantum" part is measurement, not computation, and once you performed it you hardly need a Turing machine to compute the other particle's state. If you did not perform it all you have is a correlation, not a state of either particle, so you can not compute it either. And if we are augmenting quantum computer with physical powers to perform measurements it's only fair to do the same for the Turing machine.
    – Conifold
    Apr 23, 2016 at 22:57
  • 1
    @Conifold, you write that "The 'quantum' part is measurement, not computation", but what do you think Seth Lloyd means by "the universe is a quantum computer"? I would guess that he means that the event of measurement can be seen as a computation done by the universe. don't you agree?
    – nir
    Apr 24, 2016 at 0:41
  • I asked this question about Tegmark's similar claim that "universe is a mathematical structure" philosophy.stackexchange.com/questions/32253/… and I am still not sure if it means anything coherent. But whatever it means it does not tell me what you have in mind as the common input for TM and QC in this case if not the output of the measuring apparatus.
    – Conifold
    Apr 24, 2016 at 2:08
  • My point is that the measurement of the first particle can be seen as an incomputable Turing oracle for the measurement of the other particle.
    – nir
    Apr 25, 2016 at 5:40
  • If you see measurement as part of the computation then the only way I can think of interpreting your question is if the entire setup is computable, i.e. if TM can simulate the "universe" of EPR. And of course it can, as Feynman showed in the same paper where he introduced the idea of quantum computer, although it involves exponential blow-up in running time cs.berkeley.edu/~christos/classics/Feynman.pdf That's simply because TM can solve QM equations, and simulated "time" and "space" need not have anything to do with "physical" time and space of TM.
    – Conifold
    Apr 25, 2016 at 20:38

3 Answers 3


You seem to be confused about the distinction between two different tasks you might perform with a computer. You can take some set of bits B with value V(B) as input to a computation and compute some function of those bits F(V(B)). The function F is a rule that maps any specific sequence V(B) to some specific value of those bits. Any computer that obeys the laws of physics is limited to computing only the set of functions that a Turing machine could compute.

Not all of the patterns of information flow that can take place in reality fit the pattern of being a function of a set of bits with specific values. A quantum system is best described by a set of Heisenberg picture observables. For a system prepared in a particular way, some of its observables may be sharp - that is, every system prepared that way gives you the same result when you measure that observable. But in general not all of the observables can have that property. If one observable of the system is sharp, others will not be sharp. There are processes that can take place where there is no explanation of those processes in terms of quantities that have a single value all the time. Entanglement is an example of this kind of process. There is no explanation of what is happening in the EPR experiment in which each of the systems involved has only a single value for each measurable quantity. See




However, it is still possible to simulate a quantum system to any degree of accuracy you like. A suitable simulation will not only give you accurate information about the probabilities of various outcomes once the process has finished, it could also give you the same information about the probabilities at stages intermediate between the initial and final state. All of that information about the probabilities gives you information about other properties of the system like relative phases. For a brief discussion with some references see


It is possible to build a device that can calculate a detailed abstract description of a quantum system, including an entangled system. People who say that quantum computation isn't real computation or something like that have an idea about what sorts of things can be calculated that is arbitrarily limited to a subset of what can be calculated in reality.

  • Why do you say I am confused about computation? it is clear that a computation may yield the probabilities of various outcomes. but this is not what I am asking. EPR suggests using one of the particles to predict with certainty the outcome of measurement of the other particle, but no computation can yield such a prediction.
    – nir
    Apr 24, 2016 at 12:57
  • This "predicting with certainty" thing is shorthand for the following. Suppose you measure suitable observables A and B on two different systems. If the systems are entangled, then A and B can both be unsharp, but some function of both observables can be sharp. For example, A and B may both be unsharp but the product of A and B is sharp. Since the quantum computer predicts the probabilities of the outcomes of measurements of the product of A and B it also predicts the "certainty" of which you speak. In this case the probability happens to be 1.
    – alanf
    Apr 24, 2016 at 13:24
  • I think we have a misunderstanding. EPR suggests using one of the particles, say particle A, to predict with certainty the outcome of a measurement of particle B. Are you saying there exists an alternative method not involving particle A for predicting with certainty the outcome of a measurement of particle B?
    – nir
    Apr 24, 2016 at 14:03
  • No. "Predict the outcome of a measurement of Y on B with certainty given the outcome of a measurement of X on A" means that there is some observable Z that is a function of X and Y such that Z is sharp. There is a particular outcome of the measurement on Z that has probability 1. And a quantum computer can predict that probability just as easily as it can predict any other probability.
    – alanf
    Apr 24, 2016 at 14:48
  • So generally without making the measurement on the first particle A there is no way to predict with certainty the outcome of measurement of the other particle B, right?
    – nir
    Apr 25, 2016 at 5:24

QM accurately predicts the result of the EPR experiment; the problematic that Einstein is high-lighting is 'spooky action at a distance'.

It's worth recalling that this territory has been covered before: Newtonian gravity is explicitly action at a distance which made Newton uneasy; it was one of Einsteins signal contribution to show that there was a propagating influence - curvature in spacetime; perhaps this discovery steered him into fingering this aspect of QM as suspect and in need of an explanation.

As Seth Lloyd points out in his paper this was the achievement of Bohm working from an abandoned theory of Broglie that demonstrated a such a theory, though the propagating field was super-luminal.

Information in physical or mathematical terms is philosophically suspect, in the same way that the word 'selfish' has in the term 'selfish gene'; genes cannot be selfish, they have no sense of self for a start; it's a useful and suggestive shorthand for a biological mechanism with a faint relationship to the actual notion of selfishness; the same goes for information, or more philosophically - knowledge: for something to be known, it must be potentially knowable, and in fact knowable; and there must be something (usually a person) that can potentially know it, and in fact know it. Here we see that consciousness is directly implicated.

Consider the letter 'T'; without knowing that it is a letter, we see only one horizontal line on top of a vertical one; but this only makes sense because we know what a line is; and if we forget that, then what do we see?

  • What do you mean by "QM accurately predicts the result of the EPR experiment"? I apologize but I do not understand how this is an answer to the question.
    – nir
    Apr 23, 2016 at 18:10

If the measurement does not precipitate a decision, then the information about the decision has been made when the particles were together, and is carried with them as state. I think this is the framing of QFT, but that means all kinds of historical information can be bound up in a particle when it coheres with others.

There is not more information here, of the form "as this particle is treated, so all these others will respond", there is less information, the coherent particles somehow actually share a quantum number. So entanglement is not a source of complexity, it is an economy of information.

If the wave-equation is your main point of reference, then the collection of particles does not so much travel, as it spreads. Since, in the dual model, the observation is what makes them back into particles to begin with, they are free to come "back" into being with this encoded information, which has been stored all along in their shared state. Because underlying their state as particles is a continuous potentiality distribution that indirectly indicates the likelihood of so many of them with such-and-such a distribution of quantum numbers, and not a real collection of separable entities, to begin with.

From a computational standpoint, we can look at this as an of object class with inherited properties. Given so straightforward a model, it is hard to see the effect as any more or less incomputable than the case where every particle fends for itself.

As a lame analog, consider a class in an object/class language (like Python) with a class-level cached property. The first time you access it, it has no value, so it finds one, then every instance will have that value for the property. But information has not been transmitted between instances, at faster-than-light or any other speed. So, whatever other aspects of physics this interaction offends, computability is not one of them.


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