# Bell theorem, question

1. Jul 10, 2015

### iamcj

I am just a noob, trying to understand.

I suspect the use of the Bell theorem is wrong. Ik think there is only one hidden variable and not 3 or more. That variable is time. A partical is in motion and has a position, depending on time. When two particles are entangled they have the same position in time. When a particle hits something, time in combination with position decides if the particle goes trough or not, gets absorbed or not. So it’s always a 50% ballgame and gives a 50% * 50% = 25% result not 33%.

2. Jul 10, 2015

### atyy

Bell's theorem says that if quantum mechanics is right, there cannot be hidden variables consistent with a reality that obeys special relativity.

So the number of hidden variables doesn't matter. In particularly, if a real hidden variable is that particles share the same time, that will violate special relativistic reality. In special relativity, distant particles cannot really share the same time, because special relativity forbids absolute simultaneity for distant events.

3. Jul 10, 2015

### stevendaryl

Staff Emeritus
Well, before trying to come with your own model for what's going on, you need to understand what the actual predictions of Quantum Mechanics are for the EPR experiment. There are several different versions, but I'll give you the spin-1/2 version.

You have some source that creates electron-positron pairs. For each pair, one particle is measured by one experimenter, Alice, and the other particle is measured by another experimenter, Bob. The measurement goes like this:

Alice has a detector that can be spun around at various angles. She picks an angle $\alpha$ to orient her detector. So $\alpha$ is any angle between 0 and 360 degrees. (You could actually imagine more options than that, by allowing the detector to rotate in 3 dimensions, but let's just limit it to two dimensions for simplicity) A particle passes through the detector, and is either deflected "up" or "down" (up and down are relative to the orientation of the detector). Let's let her result be a variable $A$, which is +1 if her particle goes up and -1 if her particle goes down.

Similarly, Bob picks an orientation $\beta$ and he also gets a result, either "up" or "down". Let $B$ record his result.

The prediction of quantum mechanics for this experiment is:

1. $P(A=1) = P(B=1) = P(A=-1) = P(B=-1) = 50%$. So no matter what Alice and Bob choose for $\alpha$ and $\beta$, they always have a 50/50 chance of getting +1 (spin-up) or -1 (spin-down).
2. $P(A=B) = sin^2(\frac{\alpha - \beta}{2})$.
So if Alice and Bob choose the same orientation ($\alpha = \beta$), then there is zero chance that they get the same result. If they choose orientations that are 180 degrees apart, then there is 100% chance that they get the same result.

4. Jul 10, 2015

### iamcj

No information is send, the "phase" of the partical is known in time. I am just saying the particals that are entangeld are always (in or?) out of phase. Just like two sound waves that are out of phase (in time), they react contradictory to an object like a filter. Soundwaves ar not in violation of special relativistic reality. The interaction with a filter goes like this in my imagination:

Partical A is contantly moving up and down, so is partical B. Let's say partical A hits the filter and is Up, it will pass, if it's down it will not. 50% chance based on position and time. Now we entangle particle A and B so they are out of phase. When they hit a filter the reaction will contradictory . The angle of a filter is not important, the particles will behave contradictory if the are out of phase. So there is not hidden communication, I don't get the fuzz?

5. Jul 11, 2015

### StevieTNZ

What is your level of understanding on Quantum Mechanics?

6. Jul 11, 2015

### iamcj

I am just a dreamer, not a mathematician.

7. Jul 11, 2015

### Staff: Mentor

First and foremost, are you aware that particles do not constantly 'move up and down'?

8. Jul 11, 2015

### StevieTNZ

That doesn't really answer my question. We are trying our best to assist you. How much Quantum Mechanics do you know? Then we can respond accordingly to clarify your questions in a way you might understand.

(P.S. you also spell particle "partical".)

You could try books such as:
"Dance of the Photons" by Anton Zeilinger (https://www.amazon.com/Dance-Photons-Einstein-Quantum-Teleportation/dp/0374239665) and
"Quantum Chance: Nonlocality, Teleportation and Other Quantum Marvels" by Nicolas Gisin (https://www.amazon.com/Quantum-Chance-Nonlocality-Teleportation-Marvels/dp/3319054724/)

Last edited: Jul 11, 2015
9. Jul 11, 2015

### iamcj

It is just a figure of speach, i think the have a moving state, to eplain the behaviour. I see it as vibrating.

10. Jul 11, 2015

### iamcj

Thanks for that. I am dutch, so please forgive my spelling. I can follow a litte math but please use words. I always think that sometimes a dumb question, like lets assume particles indeed are vibrating, can bring others to new insights. When I read about QM my gut says Einstein is right and the Bell theorem must be wrong in its assumptions. I am just looking for the awnser to rest my mind even if it is undecided. Really understanding the math takes a physics degree and more.

11. Jul 11, 2015

### Staff: Mentor

It's not vibrating. The wavefunction that describes a particle is not a physical wave and the wave-like properties of a particle should not be interpreted as the particle vibrating, moving back and forth, or anything else like that. I think the problem here is that you have some very serious, and also very common, misunderstandings about how things work at the atomic and subatomic scale.

To start with, you need to absolutely forget every possible version of the phrase "i see it as..." or "my gut tells me..." because those will NOT help you understand things. Our gut instincts were never designed for anything but our normal, everyday scale (and they still work poorly in many circumstances here).

12. Jul 11, 2015

### Staff: Mentor

13. Jul 11, 2015

### Staff: Mentor

You're imagining that there is something about the two particles that is completely out of phase so that whatever one particle does when it reaches the detector, the other will do the opposite. That's an example of a hidden variable: if we just knew the exact state of that something, we'd be able to predict the exact result when each particle reaches its detector and they'd always be opposite.

As long as the two detectors are lined up on the same axis so that the results are always exactly opposite, you can explain everything by assuming that there is some hidden variable that acts as you're imagining. But suppose the two detectors are not lined up on the same axis? Then the two particles won't always do the opposite when they arrive at their detector, the probability of getting opposite results will depend on the angle between the detectors.

What Bell proved is that there will be no way that you can assign the phases to your hidden variable so that you will get the quantum mechanical results for all combinations of detector angles. It's easy to do as long as the detectors are both on the same axis so that we always get opposite results, but not possible in general.

14. Jul 11, 2015

### iamcj

Last edited: Jul 11, 2015
15. Jul 11, 2015

### iamcj

So the photon reaches the filter and my assumption is that his polarity is changing 180 degrees lets say every "smallest moment in time". That is my reason for the fact half of the photons are let trough an the other half is absorbed.

"Suppose we consider a single particle (photon) of light. We ask a simple question: does it have a definite polarization at the following three angles: 0 degrees (A), 120 degrees (B), and 240 degrees (C)?"

I say no, it has a 50% chance each in each case. So if a photon reaches an x degrees filter and has a 10% chance to go trough, 20% of photons could go through because they flip polarity. 80% is absorbed because the are always have a wrong polarity no matter if they are 180 degrees flipped or not.

If a phonton is polarized and it flips 180 every smallest moment in time, it will always will of never will pass the second filter. See the next quote.

This can be confirmed via conventional optics: any photon with a known polarization (say 0 degrees) can be checked at a later time by another polarizer at the same angle setting. The predicted result (100% certainty) is that the photon will have the same polarization at that later time, and this has been known to be true for about 200 years (Malus, 1809). By extension, A, B and C must correspond individually to elements of reality, according to the definition of EPR. (In fact, any angle's polarization will meet this test individually.)""

So time decides it a photon goes trough or not. I hope somebody is willing to hold my assumptions to Bell and can tell me if my assumptions can lead to QM predictions.

16. Jul 11, 2015

### stevendaryl

Staff Emeritus
That sort of explanation is exactly what Bell proved cannot be true.

The details for why it can't require some mathematics--you can't just convince yourself using words alone. The mathematics isn't terribly difficult, though. Dr. Chinese gives an explanation in one of the links someone posted.

17. Jul 11, 2015

### stevendaryl

Staff Emeritus
It would be better for you to investigate yourself whether your ideas can reproduce the quantum predictions. Bell already proved that they can't, but you have to convince yourself that that's true.

What I think is easier than proving Bell's theorem and proving the QM violates is just directly proving for a particular experiment that it is impossible to have a hidden-variables type explanation for the results.

If we simplify the description to just the statistics, and ignore the details of how the measurements are performed, the EPR experiment with twin photons of the same polarization (there is an alternate experiment where the two photons have opposite polarizations) can be described this way:

1. Each round of the experiment, Alice chooses a setting $\alpha$, where $\alpha = 0^o$, $120^o$ or $240^o$.
2. Each round, Alice gets a result $A$ that is either +1 (the photon passes the filter) or -1 (the photon does not).
3. Each round, Bob chooses a setting $\beta = 0, 120, 240$
4. Each round, Bob gets a result $B = \pm 1$
You play the game for many many rounds, with Alice and Bob choosing settings randomly. What QM predicts is the following statistics:
• Out of those rounds for which $\alpha = \beta$, $A = B$. If they choose the same setting, they get the same result.
• Out of those rounds for which $\alpha \neq \beta$, 25% of the time $A=B$, and 75% of the time, $A \neq B$.
• Regardless of the settings, 50% of the time, $A = 1$, and 50% of the time $A=-1$.
• Regardless of the settings, 50% of the time $B = 1$ and 50% of the time $B=-1$ .
(The numbers are computed by $cos^2(\alpha - \beta)$)

A deterministic, local, hidden-variable model of this result would be a way of generating a sequence of three numbers
$(R_{0,n}, R_{120,n}, R_{240, n})$
with the statistics:
1. $R_{\alpha, n} = \pm 1$
2. The average over all $n$ of $R_{0, n}$ is 0 (just as many +1 as -1). Similarly for $R_{120, n}$ and $R_{240, n}$
3. 25% of the time, $R_{0,n} = R_{120, n}$
4. 25% of the time, $R_{120, n} = R_{240, n}$
5. 25% of the time, $R_{0, n} = R_{240, n}$
If you could come up with such a sequence, then you could explain Alice's and Bob's results by saying:
• On round number $n$, if Alice chooses setting $\alpha$, then she gets result $R_{\alpha, n}$
• Similarly for Bob.
But it is mathematically impossible to come up with such a sequence.

Your model has the two photons fluctuating in-flight, but exactly in-synch, so that Alice and Bob always get the same result if they choose the same polarization angle. But the fluctuation doesn't change the mathematics. It's not hard to make sure that Alice and Bob always get the same result when they choose the same angle, but you can't ALSO make sure that they get DIFFERENT results 75% of the time when they choose different angles.

18. Jul 11, 2015

### atyy

You need to work out your scenario in full detail, and show that it can violate the Bell inequality. At present, I have no idea what you are talking about.

19. Jul 11, 2015

### Staff: Mentor

What 'polarity' are you talking about?

20. Jul 12, 2015

### iamcj

You will probably laugh at me, but I had some fun J and eventually I came up with this very simple thing. Everything is to be interpreted as a figure of speech. But please prove me wrong.

We take the smallest amount of time *
We assume a photon flips his polarity 180 degrees every 4*
1 means gone trough, 0 means absorbed.

When the angels are not the same QM predicts only 25% equal results for Bob and Alice. With different angels I assume there is a 3* (time) difference in interaction with the filter at the decisive moment (absorb or not) for BOB and then we get these patterns:

The sortcut I take to Bell is that I assume the same timeshift for a difference of 120 degrees as a difference of 240 degrees. On the other hand cos2(120) and cos2(240) are the same?

21. Jul 12, 2015

### stevendaryl

Staff Emeritus
I don't know what that chart means. Let me try to make it more explicit:

You play the game (or run the experiment) for many rounds. Each round has 4 associated numbers:

$\alpha_n, \beta_n, A_n, B_n$

where $\alpha_n$ is Alice's setting for round $n$, either 0, 120 or 240. $A_n$ is Alice's result, either +1 or -1. $\beta_n$ and $B_n$ are Bob's settings and results for round $n$.

So, the question is, for a local deterministic model, what is the rule for calculating $A_n$ and $B_n$? For a local model, $A_n$ cannot depend on $B_n$ or $\beta_n$, and $B_n$ cannot depend on $A_n$ or $\alpha_n$.

What rule are you using for computing $A_n$ and $B_n$?

22. Jul 12, 2015

### atyy

If you look at http://drchinese.com/David/Bell_Theorem_Easy_Math.htm, Bell's theorem is described as "The fact that the matches should occur greater than or equal to 1/3 of the time is called Bell's Inequality."

However, in post #20, there are 3 pairs of settings in which the matches occur only 1/4 of the time, thus violating "Bell's inequality".

The calculation in post #20 looks right to me. Off the top of my head, I don't know whether the 1/3 condition is really equivalent to say CHSH, so I think one should calculate CHSH for the results in post #20.

EDIT: My calculation for CHSH for post #20 is 0, so I think Bell's inequality is not violated. But I have not checked my calculation, so I don't know whether it is correct. Can anyone verify that DrChinese's condition is strictly equivalent to CHSH or some other Bell inequality?

EDIT: I think the CHSH calculation is wrong, see my new attempt in post #25.

Last edited: Jul 12, 2015
23. Jul 12, 2015

### iamcj

I will try to explain the first row:

The first column is a single photon flipping polarity by 180 degrees every 4 time-untis at $\alpha_0$. Thats an assumption I make, to make this work. I reality I think its rotating or something.
The second column represents a photon which is cought up in a filter with an angle 120 or 240 greater than $\alpha_0$. My ussumption is that it takes 3 time units longer to polarize this photon by the filter, because it is a greater angle. Because it takes longer, the moment that determines if the photon goes trough is 3 time units later. At that moment the photon is 3 time-units later flipping his polarity.

The third and fourth column gives the results for the same angels. Here only the flipping of the photons determines if it goes trough. "the are in the same timezone"

AC and AB just use column "flip" and "different angle".

I suspect that when you change the angles the percentages are different? I don't no, but then I say the timeshift will move accordingly.

24. Jul 12, 2015

### iamcj

Actually I think, I don't violate Bell, I just bypass it.

25. Jul 12, 2015

### atyy

Sorry, I think my CHSH calculation in post #22 for the scenario in post #20 is wrong. Here is another attempt. Please correct it if it is wrong!

For AB, AC and BC, the probability of matching is 0.25, so the probability of not matching is 0.75, so E(AB) = E(AC) = E(BC) = 0.25-0.75 = -0.5.

So CHSH = E(A,B) − E(A,C) + E(A,B) + E(B,C) = -0.5 + 0.5 - 0.5 - 0.5 = -1.

So CHSH is not violated.

Could it be that DrChinese's http://drchinese.com/David/Bell_Theorem_Easy_Math.htm condition "The fact that the matches should occur greater than or equal to 1/3 of the time is called Bell's Inequality" is not correct?