# Understanding Bell's mathematics

1. May 28, 2010

### Gordon Watson

From: Is action at a distance possible as envisaged by the EPR Paradox.

I thought this needed a new thread (to stop a hi-jack), with an emphasis on Bell's mathematics please.

Last edited: May 28, 2010
2. May 29, 2010

### DrChinese

OOO goooood.

Bell covers plenty of ground in his paper. As I said, he wrote it for a certain audience. He believed they would be able to follow his line of reasoning. So it is necessary to follow Bell using the strongest arguments. He could see that the basic idea was important: that the EPR result (QM was not complete) was now incompatible with the kind of world they envisioned. We call that world Local Realistic. So we should talk about the requirements - per Bell - for a local realistic world.

Bell used 2 central ideas:

a) that the setting used for Alice did not change the result for Bob (and vice versa). This is often called the separability requirement. Bell said said this was to restore locality. This is his (2).
b) that there should be definite outcomes possible for counterfactual measurement settings. This is introduced after his (14). He says "It follows that c is another unit vector..." but what he really means is: "Assume c is another unit vector...".

Both of the above are needed to get the result, but they can be expressed a variety of ways.

3. Jun 1, 2010

### Gordon Watson

DrC, ThomasT.

You both appear to agree that Bell uses P(AB|H) = P(A|H).P(B|H) in his work.

I cannot see how EPR studies using that formula could be serious. If H includes a hidden variable for each particle, that formula gives P(AB|H) = P(A|H).P(B|H) = (1/2).(1/2) = 1/4.

Can you direct me to an example where Bell uses P(AB|H) = P(A|H).P(B|H) in his work, please?

4. Jun 1, 2010

### alxm

I just read the Bell paper for the first time (woo!), and the way it looks to me, his [2] does not imply "P(AB|H) = P(A|H).P(B|H)" at all.

That would imply that A and B weren't correlated and as DrChinese said, they can be as correlated as they want, through their mutual dependence on lambda - any number of hidden variables you're free to pick. And yes, Bell does explicitly address this.

You can only get P(AB|H) = P(A|H).P(B|H) from that if you throw out the hidden-variable dependence. They'd naturally be uncorrelated then. The point of [2] is that A is independent of B's detector settings. Apart from that, they can be as mutually interdependent as one chooses.

5. Jun 1, 2010

### Gordon Watson

Thank you alxm.

So Bell needs to do something like this -

P(AB|Hab$$\lambda$$) = P(A|Hab$$\lambda$$).P(B|Hab$$\lambda$$A) = P(A|Ha$$\lambda$$).P(B|Hab$$\lambda$$A).

Does he need to? Does he do it?

Thank you.

6. Jun 1, 2010

### DrChinese

Again, Bell was writing for a small audience who could be counted on to understand his points. So if it is read as a general piece for a general audience, it will end up having a somewhat different meaning than intended.

There has been a lot of discussion around Bell's (2) over the years. Some try to read it literally. He is attempting to express the idea that the detector setting for Alice does not affect the result at Bob, and vice versa. He says those words. Now, my question is: how might you express it better mathematically than his (2)?

Second, does he need it? Well, I don't think so. You can express everything you need - in my opinion - with something akin to the following:

a) P(A, B) = P(A, B, C) + P(A, B, ~C)
b) P(A, B, C) + P(~A, B, C) + ...(the other 6 permutations) = 1

So I get this directly from Bell, although I realize some don't see these points. I mean, really, what does it matter? Once you see the line of reasoning, you can express it different ways. I have a page where I show that, based on the above, some cases should have a -10% chance of occurance - an absurd result. Mermin has a great way of expressing it too, I have a page on that approach as well. Or you can follow the master, Bell.

So when I see someone trying to say that P(AB) = P(A)*P(B) I usually think they have already missed the starting line (I am talking to ThomasT here). What he is really saying is that if you consider A and B as settings, P(AB) must equal the quantum mechanical value AND it must be able to be decomposed into P(A) and P(B). This is pretty much the same thing as my a) and b) above.

Last edited: Jun 1, 2010
7. Jun 1, 2010

### DrChinese

It's an interesting read, isn't it? There is a lot hidden in the paper, and because of the intended audience it sometimes takes an extra read (or three) to appreciate the full value. He really circled things from a lot of perspectives. If I had had the great insight Bell had, I would never have been able to dot the i's and cross the z's like he did.

8. Jun 1, 2010

### Gordon Watson

Thank you DrC, but this seems confusing.

A and B considered "settings"? Aren't they specific outcomes? P(two settings) is not making sense to me.

Agreed that detector setting a should not influence outcome B, and b ... A. Agreed that P(AB|H) must equal the QM value. We agree on the essentials.

Then your requirement that P(AB|H) be decomposed into P(A|H) and P(B|H) puts you and your mathematics outside the EPR context (because EPR outcomes are correlated).

Granted there may be other ways of equating with Bell's conclusion, can you help me stay with Bell's mathematics please? This thread is about understanding Bell's mathematics.

9. Jun 2, 2010

### zonde

It is assumed that correlation appears because of H (LHV assumption). And then it is correct that P(AB|H) can be decomposed into P(A|H) and P(B|H).

Where you deviate from experimental situation is that P(A|H)+P(~A|H)=1 is not correct. To get that you have to normalize P(A|H)+P(~A|H) to 1. Like that k*P(A|H)+k*P(~A|H)=1.

Now you can write k3*P(AB|H)=k1*P(A|H)*k2*P(B|H). And there you invoke fair sampling assumption by equating k1=k2=k and k3=k^2.
But that's my line about unfair sampling and it might be a bit to the side from your question.

10. Jun 2, 2010

### Gordon Watson

Thank you zonde.

This information appears to be to the side of my question on Bell's mathematics.

H stands for the conditions under which we are analyzing AB.

11. Jun 2, 2010

### zonde

I am quite sure this is upside down.
It's H that we are analyzing under conditions A and B.
H stands for photons (I prefer to keep closer to real experiments) but A and B stands for polarizer or PBS settings.

12. Jun 2, 2010

### Gordon Watson

Isn't A an outcome from Alice's detector which Alice had set at angle a? B an outcome from Bob's detector which Bob had set at angle b? With photons A can be G (green light) or R (red light), B can be G' (green) or R' (red). So combined outcomes can be GG', GR', RG', RR' and we are interested in their probability under condition H?

So the settings are a and b which are angles?

13. Jun 2, 2010

### ThomasT

Right. This is what the probability analog (in which A and B are specific outcomes) to Bell's ansatz is saying. The EPR context comprises only two joint settings, or two values for |a-b|. These are the only joint settings where there's a correlation between A and B . For these two joint settings the form P(AB|H) = P(A|H) P(B|AH) doesn't reduce to P(AB|H) = P(A|H) P(B|H). (And this doesn't imply ftl transmission, because the information that allows us to write P(AB|) = P(A|H) P(B|AH) contingent on certain joint settings is in the experimental preparation which is in the past light cones of both observers.) For all other values of |a-b|, P(AB|H) = P(A|H) P(B|AH) reduces to P(AB|H) = P(A|H) P(B|H). (That is, when there's no correlation between A and B, then P(AB|H) = P(A|H) P(B|H) obtains.)

The requirement (Bell's task), per EPR, was to model the joint, entangled, state in terms of parameters which determine individual results. That is, in order to model separable predetermination, per EPR, Bell had to model the joint measurement situation in a separable form as a combination of the individual situations which are determined by the hidden parameter(s), H. The use of the probability analogs is an attempt to support Bell's result by showing that the EPR requirement entails a contradiction between the reality of the experimental situation and the form that any, EPR constrained, local realistic model has to be rendered in.

The 'physical' reason why it can't be done is because the joint measurement context involves different parameters than the individual contexts. The RELATIONSHIP between counter-propagating, entangled, photons is NOT what determines the individual results -- but it IS what determines the joint results. Hence, there's a dilemma if it's required that local realistic models of joint, entanglement, situations be rendered in terms of individual results.

Thus, it can be understood that inequalities based on certain (EPR) modelling constraints don't (and can't) represent the actual joint experimental situations or preparations designed to produce entanglement. That's why the experimental results (and qm predictions) exceed the limits set by inequalities so constrained -- and not because Nature is nonlocal or because the results which determine individual results aren't predetermined and separately determining those individual results. (Keeping in mind that separable predetermination IS compatible with the qm description of the individual measurement situations.)

Further, it's been shown that the joint, entangled, situation CAN be viably modelled as a nonseparable situation involving predetermined (eg., via emission), realistic (but not in the EPR sense of determining individual results), global hidden parameters.

14. Jun 2, 2010

### Gordon Watson

Looking at the mathematics only, this seems confused.

In EPR-Bell settings, the formula P(AB|H) = P(A|H).P(B|HA) holds for any value of a and any value of b. So it holds for any value of |a - b| or (a - b). In my opinion.

Last edited: Jun 2, 2010
15. Jun 3, 2010

### zonde

If we talk about photons and analyze them with polarizers then event A would be that Alice's photon with hidden variable H was detected after passing polarizer that is set at an angle $$\vec{a}$$.
So if we write $$P(A\cap B|H\vec{a}\vec{b})=P(A|H\vec{a})P(B|H\vec{b})$$ would it be unambiguous now?

In case of Bell A is two different events i.e. A=+1 when spin up was detected and A=-1 when spin down was detected. So it messes up things a bit if you refer to A as an event with some probability while meaning it in context of Bell paper. Maybe that's the reason of confusion? Were you taking H as either +1 or -1 condition?

16. Jun 4, 2010

### Gordon Watson

Thank you zonde for your attempts at clarification. I totally support them. However (IMHO) there a some hidden subtleties and confusions (that need amendment) in what you wrote. My cuts and pastes and amendments and extensions are --

1. If we talk about photons and analyze them with polarizers then event A would be that, under condition H, Alice's photon was detected after passing Alice's polarizer that is oriented $$\vec{a}$$.

1a. A will be signaled by an outcome in Alice's analyzer, either G or R; that's how we know the photon is detected. A = {G, R}? So we can discuss P(G|H$$\vec{a}$$) = 0.5, and P(R|H$$\vec{a}$$) = 0.5; H defining the conditions.

1b. B will similarly be a signaled outcome, either G' or R' at Bob's analyzer. B = {G', R'}? So we can discuss P(G'|H$$\vec{b}$$) = 0.5, and P(R'|H$$\vec{b}$$) = 0.5.

2. So if zonde wrote

$$P(GG'|H\vec{a}\vec{b})=P(G|H\vec{a})P(G'|H\vec{b})$$,

that would be unambiguous.

2a. But it would equal 0.25 and be unhelpful. Because H specifies the conditions and the conditions are EPR-Bell conditions and under such conditions G and G' are correlated.

2b. $$P(GG'|H\vec{a}\vec{b}) = P(G|H\vec{a}\vec{b})P(G'|H\vec{a}\vec{b}G) = P(G|H\vec{a})P(G'|H\vec{a}\vec{b}G)$$ would be OK.

3. Was I taking H as either +1 or -1 condition? No. H specifies the general overall total conditions.

17. Jun 4, 2010

### zonde

No, analyzer consisting of polarizer and detector does not give you G and R. It just gives you "clicks" time after time.
If you want G and R case then we have to switch to analyzer consisting of PBS and two detectors. Then we can have "click" in G detector or we can have "click" in R detector.

Next H is defining certain but unknown conditions that are shared between Alice and Bob. So these are conditions that have causal connection with source. Conditions that are causally related only to either Alice or Bob are not included in H.
P(G|H$$\vec{a}$$) does not have to be 0.5 because we talk about certain (but unknown) value of H.
If H is polarization of photon clearly in case where $$\vec{a}$$ is perfectly aligned with that polarization of photon we should have probability of 1 for G signal (or 0 depending which output we define as G).

You can say that P(G|$$\vec{a}$$)=0.5 but that too only after normalization.

That's not good either.
H is only shared conditions or more trivially speaking it's polarization of individual photon.
If you want to include still something else this must be described as additional variable individually for Alice and Bob (like $$\vec{a}$$ and $$\vec{b}$$).

18. Jun 4, 2010

### Gordon Watson

I see your NO-s and NOT-s et cetera and suspect you are wrong or confused in each case.

1. You say NO ... DOES NOT ... IF ... ? My G/R polarizer-analyzers use pure Iceland spar so G or R for Alice, G' or R' for Bob, works quite OK.

2. Aren't I the one that introduced H? My H includes $$\vec{a}$$ and $$\vec{b}$$, but there's no problem pulling them out of H (if you wish and when it helps).

3. You talk about that if you wish. I choose not to. Makes no sense (to me).

4. Makes no sense with my H. Time to bring in your own Z, maybe?

5. Aren't probabilities normalized by definition? They're not the same as raw experimental frequencies.

6. See 2 above.

I think you are making too many assumptions about my notation and approach. Time to bring in your own for me to follow?

All we want is an agreed notation that leads us to agree on Bell's mathematics.

19. Jun 4, 2010

### zonde

I am not sure I understand how your analyzer works. Can you describe it a bit more? Where is photon when your analyzer gives G and where is photon when your analyzer gives R?

If you mean that you introduced H when you wrote
P(AB|H)=P(A|H)P(B|H)
then my objections still hold. It's because you use the same H for both Alice and Bob. But they are spatially separated so they can't be described with the same conditions.
If you write something like P(AB|HH')=P(A|H)P(B|H') then yes you can introduce H and H' as you wish.

20. Jun 4, 2010

### Gordon Watson

For G (or +1), photon is absorbed in the ordinary ray detector. For R (or -1), photon is absorbed in extraordinary ray detector. (Easy to make. I supply DrC, Mermin, Clauser, Aspect, Zeilinger. You want some

No problem to make you happy?

1. With EPR-Bell common condition H, Alice controls orientation a, sees R or G, assumes z has arrived. Bob controls orientation b, sees R' or G', assumes z has arrived.

2. z is Bell's lambda for your photon example.

3. I write formula. You give answer:

P(G|H) = ?

P(G|Ha) = ?

P(G|Haz) = ?

P(G|Hazb) = ?

P(G|HazbG') = ?

Repeat for R replacing G ... ... ... ...

P(G'|H) = ?

P(G'|Hb) = ?

P(G'|Hbz) = ?

P(G'|Hbza) =

P(G'|HazbG) =

Repeat for R' replacing G' ... ... ... ...

P(GG'|H) = ?

P(GG'|Ha) = ?

P(GG'|Haz) = ?

P(GG'|Hazb) = ?

P(GG'|HazbR') = ?

Repeat for R' replacing G' ... ... ... ...

et cetera

You happy?

Last edited: Jun 4, 2010