Bell’s inequality

A simple proof of Bell’s inequality

(as posted in sci.physics.research)

Ok, so here’s the version of the theorem which I find the easiest to fit inside my head all at once (if anybody knows a simpler one I’d be pleased to hear about it):

There are two experimenters separated by a distance so large that a light signal cannot reach one experimenter from the other during the time in which they are performing measurements. One could imagine that each can see the other on a television screen, and are broadcasting while they experiment, so that each experimenter performs his measurements, makes his notes, and then watches the television to see the other experimenter doing the same procedure, knowing that the television signals which he is now watching were travelling towards him across the vastness of space while he was doing his experiment a few moments before.

Each experimenter recieves N particles and performs one of three experiments, A, B or C, on each particle. Each of the experiments gives either +1 or -1 as an answer. The experimenter decides for himself which experiment he will perform on which particle, and deliberately doesn’t decide which to do until just before the particle arrives.

We have fairly good reasons to believe that, if the two experimenters perform the same experiment (A, B or C), they will get opposite results (one will get +1 and the other will get -1). The reasons are both theoretical (conservation laws) and experimental (whenever it has been tried it has in fact turned out to be true, and it has been tried millions of times).

The condition of locality which is used here is the condition that, whatever the results of a given experiment depends upon, it does not depend upon the choice made by the distant experimenter about which experiment (A, B, or C) he will perform. That is, locality means that neither experimenter should find that the results which he gets have been affected by the choices of the other experimenter, which he sees on the television a few minutes after he does the experiment.

We say that an experimental result, X, “depended upon” a choice from the set {A, B, C} by an experimenter if it can be shown (or statistically inferred) that, had the experimenter made a different choice, X would have had a different value. In such a circumstance we can draw up a table like this:

    choice | A  B  C 
           |---------
value of X | +1 -1 -1

or at least we could do so if we knew what the various results would have been in the various circumstances.

When we say that X did not depend on a choice, it means that X would have had the same value that it actually did regardless of what choice was made, and we can draw a table that looks like:

    choice | A  B  C 
           |---------
value of X | +1 +1 +1

where it has been assumed that the result was +1. Drawing such a table isn’t very interesting or worthwhile, but it is important to realise that the statement that a result doesn’t depend on a choice means that such a table can be drawn up and is accurate.

Now, back to the experimenters. If experimenter 1 is to declare that one of his results depended on his own choice of which experiment to do, but did not depend on the experimenter 2’s choice of which to do, then that means that he can draw up a table that looks like:

 choice of experimenter 1  | A  B  C 
---------------------------|---------
       choice             A| +1 -1 -1
       of                 B| +1 -1 -1
       experimenter 2     C| +1 -1 -1

or at least he could if he knew what the other results would have been had he made the other choices. Since the choice of experimenter 2 doesn’t affect the result, the table can be considered to be specified by any one of its rows.

Now we add in the bit about the experimenters getting opposite results when they choose the same experiment, and we realise that if we suppose that the result of an experimenter depends on his own choice of what experiment to do, but doesn’t depend on the choice of the other experimenter, then we can draw a pair of tables like:

 1's choice| A  B  C    2's choice| A  B  C 
           |---------             |---------
   result  | +1 -1 -1     result  | -1 +1 +1

or at least we could etc. At least, the statements above about what depends on what are equivalent to the statement that such tables can be drawn up.

Then we can categorize the N particles according to the results that would be obtained upon a measurement of each of A, B and C, by experimenter 1, knowing that that fixes the result that experimenter 2 would get. For an individual particle, each experimenter can, having watched the other experimenter on the television, partially reconstruct the tables for each particle. That is, suppose experimenter 1 performed experiment A and got +1 and experimenter 2 performed experiment B and got +1. Then the tables can be reconstructed thus far:

 1's choice| A  B  C    2's choice| A  B  C 
           |---------             |---------
   result  | +1 -1 ?      result  | -1 +1 ?

That is, experimenter 1 can say to himself: “Well he did experiment B and got +1, so if I had done experiment B I would have gotten -1.” For him to be wrong on this point, it would have to be the case that, had he done experiment B instead of A, he would have gotten +1 (obviously). That would mean that, had he done experiment B, experimenter 2 would have to have gotten -1, because the two experimenters have to get opposite results when they do the same experiment.

So, either experimenter 1 is correct in drawing up his table (the table on the left), *or* the result of experimenter 2 depended on the choice of experimenter 1. Similarly, experimenter 2 has the same justification for drawing up the table on the right.

So the two experimenters draw up partially reconstructed tables for each pair of particles. Now, we don’t know what was in column C in the tables above, but what we do know is that it was +1 in one table and -1 in the other. That is, the full table for experimenter 1 was either:

 1's choice| A  B  C  
           |---------  
   result  | +1 -1 +1 

or

 1's choice| A  B  C  
           |---------  
   result  | +1 -1 -1 

There were N pairs of particles altogether. We can call the number of pairs for which the first table above was correct N_3, and draw a table which assigns similar symbols to the numbers of pairs of particles with each table:

 1's choice| A  B  C    2's choice| A  B  C 
           |---------             |---------
      N_1  | +1 +1 +1             | -1 -1 -1
      N_2  | +1 +1 -1             | -1 -1 +1
      N_3  | +1 -1 +1             | -1 +1 -1
      N_4  | +1 -1 -1             | -1 +1 +1
      N_5  | -1 +1 +1             | +1 -1 -1
      N_6  | -1 +1 -1             | +1 -1 +1
      N_7  | -1 -1 +1             | +1 +1 -1
      N_8  | -1 -1 -1             | +1 +1 +1

where N_1 + … + N_8 = N

The example given above, where experimenter 1 performed experiment A and got +1 for a result, and experimenter 2 performed experiment B and got +1 as a result, would be a contribution to either N_3 or N_4 (we don’t know which because we don’t know what result would have been obtained upon a measurement of C, though we know it would have been either +1 or -1).

Bell’s inequality is: N_3+N_4 <= N1+N3 + N4+N7

which is trivial because all of these are non negative integers, denoting as they do numbers of particles involved in the experiments.

The numbers N_3 + N_4 and so on, cannot be measured directly, since each experimenter can only make one measurement on the particles which arrive, but they can be estimated in the following way. Take the number of particles for which the situation above arose (A +1, B +1), and divide it by the fraction of experiments for which experiments A and B were performed by experimenters 1 and 2 respectively. If the choice of which experiment to do does not affect which table is associated with that particle, then this will provide a fair sampling – that is, if the pair of experiments (A,B) was done one nineth of the time, and 500 of those pairs of experiments yielded +1 for both, then we infer that, had the pair (A,B) been done all of the time, there would have been about 9*500 incidents, and so N_3 + N_4 is about 9*500.

We can make estimates of N1+N3 in a similar way by counting how often experimenter 1 performs experiment A and gets +1 while experimenter 2 performs C and gets -1. Something similar works for N4+N7.

Notice that this inference can only systematically fail (as opposed to failing because of a statistical fluctuation) if the choice of experimenter B affects the appropriate table for experimenter A, since the choice of experimenter A is already taken into account in his table. That is, it can only fail if experimenter B’s choice affects the result of experimenter A, or vice versa.

For the sake of completeness, I’ll add that some people have historically objected to the use of “counterfactuals”, claiming that this argument makes too much use of “What would have happened if such and such an experimenter had chosen something else.” In each case, where such “facts” were used in the proof, there was a good reason. For example, it was supposed that, had C been performed, either +1 or -1 would have been obtained. This statement was made because the experiment C has been done many times before and either +1 or -1 have been obtained in each case.

Such inferences are the stuff of science. That is, this objection, while self-consistent, is rather too powerful since it rejects the kind of reasoning used in science all the time. For example, a person who refuses to accept arguments involving counterfactuals would not accept as evidence of faster than light signalling an experiment in which the message “This is a faster than light signal” was sent at one event and received at a distant destination, because they would have to object that it is not known, or reasonable to discuss, what would have been received if a different message had been sent, and so could not accept the conclusion that the contents of the sent message “caused” the contents of the received message.

Also notice that during the entire proof of the inequalities, the things referred to were the experimenters, their choices and experimental results, they tables they draw up and so on. No mention is made of any theoretical framework or even of any theory making the predictions. Essentially, the theorem is not about theories at all, but says that if certain experimental results are obtained, namely counts of +1’s and -1’s which violate the inequalities, then the choices of one experimenter affect the results of a distant one.

The experiments were done to look for a evidence that the setting of one device can affect the results obtained on a distant one, and the evidence was certainly found. I’ll mention quantum mechanics for the first time in this post to point out that the role it had was to suggest what the experiments A, B and C should be.

R.

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11 Responses to “Bell’s inequality”

  1. physics musings » Blog Archive » Bell’s theorem Says:

    […] I just found a beautiful, elementary proof of Bell’s inequality at sci.physics.research. It’s the more transparent and intutive i’ve ever read, and, as an additional plus, no quantum mechanics is involved in the argument. In fact, i’ve liked it so much that have reproduced it as the first entry in my new ‘Bits and bolts’ section. […]

  2. Mike Says:

    I don’t really call something that takes that long to explain “beautiful, elementary and ituitive”!

    Bell’s Theorem (in its discrete binary form) is simply an inequality on set sizes:

    |Ac| + |Cb| >= |Ab|

    where sets A, B, C have complements a, b, c.

    The proof really is elementary Boolean algebra:

    The LHS is >= |Ac U Cb| (actually = because those sets are disjoint). Now intersect Ab with Ac U Cb to find that Ac U Cb contains Ab, whence the result.

    This models the physical situation by using simultaneous measurements on pairs of systems to deduce properties of single systems (the underlying sample set in the above) by appealing to EPR.

    Bob’s your uncle :)

    • David M Says:

      To follow up to Mike, the last step can be made a bit more explicit. The set Ab can be expanded to |AbC U Abc|. The former is a subset of Cb and the latter a subset of Ac.

    • Valentin Tihomirov Says:

      How do you associate Ab with (N3,N4)? I see that N3 differs from N4 by C property. Similarly, N1 differs from N3 by property B. But, N4=”+–” differs from N7=”–+” in every property. This says like we have experimenter 1 measured “+” or “-” regardless which property he chosen to measure. This falls out the pattern “experimenter 1 measures one property, experimenter 2 measures one another and third property is left undefined” which works for N3 with N4 and N1 with N3.

    • Valentin Tihomirov Says:

      I guess that what you are doing is N2 +N3 + N4 + N7 = 110 + 101 + 100 + 001 = 110 + (101 + 100) + 001 = 110 + 10- + 001 = |ABc| + |Ab| + |abC| >= |Ab|. You also say that |Ab| = N3 + N4 <= (N2 + N4) + (N3 + N7) = (110 + 100) + (101 + 001) = 1-0 + -01 = |Ac| + |bC|. I wonder why did author chose N1 +N3 + N4 + N7 instead of N2 +N3 + N4 + N7.

  3. Khalid Says:

    Hi,

    Well, I think there is a problem at the hypothesis!
    How can each experimenter see the other “live” while performing their respective experiment, if even the light cannot reach one another before they have finished their experiments???
    What form of communication signal their receivers will use to receive signals faster than the speed of light?

    Khalid

  4. David Kemp Says:

    I hope you are still listening given that it is years since you wrote this. Unless some of the numbers are negative, the inequality (N3+N4 <= N1+N3 + N4+N7) will surely always hold given that N3 and N4 are on both sides. Perhaps you meant N3+N4 <= N1+N3 + N3+N7… that would also make more sense as N3 and N7 correspond to how often experimenter 1 performs experiment B and gets -1 while experimenter 2 performs C and gets +1. N4 and N7 do not seem to correspond to any particular scenario.

  5. Lou Jagerman Says:

    Excellent. But since I’m a compulsive stickler on clarity in scientific writing, I have a few questions:

    No doubt “answer” which each experiment “gives” is the same as “value of X” is the same as “result.” Are these assumptions correct?

    Moreover, it seems that “N_3” is the number of correct pairs in row 3; is that correct? Then what is the numeric value of N_3 in line 3?

    Please define the terms N1, N3, N4 and N7, and give their numeric values in the “Bell’s inequality.”

    By “inference” do you mean the inequality?

  6. The d0 parameter | jfoadi Says:

    […] interested in accessible (but long!) explanations of this subject can have a look at another blog or at the following web […]

  7. Ed Unverricht Says:

    You may be interested in a local realistic model of a photon being used in the Dehlinger and Mitchell setup.

    http://www.animatedphysics.com/photons/bells_inequality.htm

    The linear photon is modelled as not only having a specific “average” polarization direction (as is modelled by Dehlinger and Mitchell), but also as having a “wobble” or “instantaneous” polarization direction. This “wobble” of the linear polarization introduces all the right properties required to be able to properly model photon polarization and have it match reality.

    The sample 24° photon, with a 30° wobble, will most of the time be picked up as a vertical, but sometimes when the angle is over 45°, it will be picked up as a horizontal. This invalidates the grouping of N_1, N_2, etc. since a photon measured at an angle to its polarization, has a probability of being measured different.

    Not sure if you would have any comments or questions, but I would appreciate hearing them if you do.

  8. Fred Johnson Says:

    An interesting paper with new research called Randomness, Nonlocality And Information In Entangled Correlations may be found at http://arxiv.org/abs/quant-ph/9505020v1

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