# I Bell inequality and linearity

1. Jan 23, 2017

### exponent137

Bell inequality in page 171 of
https://www.scientificamerican.com/media/pdf/197911_0158.pdf
is
$n[A^+B^+] \le n[A^+C^+]+n[B^+C^+]$
In page 174 we can see that this causes linear dependency according to angle. How to derive this?

Let us suppose that angle between $A^+$ and $B^+$ is 30°. Which direction of $C$ to take that the calculation is the most simple.

I intuitively understand why result at 45° is 0.5. But, for instance, at 30°, I do not understand, why it is 0.6667.

2. Jan 24, 2017

### DrChinese

Great question. Technically, there is no one specific local realistic number and therefore no one specific Bell limit (at least that I am aware of). For example, having a function with a constant value of 0.5 at all angle differences would also work, and would therefore be a different curve than in the article. Of course you can already see that wouldn't come remotely close to experiment results.

You start with some hypothetical local realistic function for theta, and you have 3 different theta inputs that are related (the permutations of A, B and C). Perhaps there are many prospective functions, and they do not need to either match QM (just as the article's does not) and they don't need to equal .6667 either.

But you can back into it with a few assumptions. Normally, you assume that the function matches the QM result at 0 and 45 degrees. And as a practical matter you want the delta between it and QM to be as small as possible. And the function must be evaluated independently on both side (if it is to be realistic). Usually the line presented in the article is used as it is meets the requirement that it is "close" to the QM value and easy to model.

3. Jan 25, 2017

### exponent137

I need to understand, by which model linearity came from.
Question: why do hidden variables need to imply a linear variation?
This is not strictly required by Bell, but it is more of a practical consequence. The candidate local hidden variable theory needs a relationship which both works for perfect correlations (which is the requirement of EPR's elements of reality) and needs to yield a result at other angle settings which is proportional to the angle difference so that there aren't anomalies at certain angles (as Bell discovered). A linear relationship solves that instantly. Of course, that won't match experiment.
https://www.physicsforums.com/threa...ly-a-linear-relationship.589923/#post-3834739
in bold: Which anomalies arise at certain angles? Can you explain more precisely or you find me, where it is explained already, maybe in around your link?

4. Jan 25, 2017

### DrChinese

I actually have a page on those "anomalies", specifically they are negative probabilities. Please note those only arise when considering local realistic theories, and do not arise in QM itself. (I also have a spreadsheet that graphs them across 90 degrees.)

Bell's Theorem and Negative Probabilities

5. Jan 26, 2017

### exponent137

I think that your paper is the second level of explanation of Bell theorem, but I wish for now only a answer why linear line (also in your figure 3) exists. You said that
"which is proportional to the angle difference so that there aren't anomalies at certain angles (as Bell discovered)."

6. Jan 26, 2017

### stevendaryl

Staff Emeritus
Bell made some remark about a local hidden variables theory predicting a linear relationship between Alice's and Bob's results for EPR, but I never understood whether he was speaking generally, or about a particular hidden-variables model. The linear relationship can be derived for a particular toy model:

• Assume that the spin of a spin-1/2 particle is just a unit vector $\vec{s}$ that can point anywhere.
• Assume that the measurement of the spin along an axis $\vec{a}$ deterministically produces spin-up if the absolute value of the angle between $\vec{a}$ and $\vec{s}$ is less than 90o, and produces spin-down otherwise.
• Assume that in EPR, when an electron-positron pair is produced, the electron is given a random spin direction $\vec{s}$ (all directions equally likely--or rather, an equal probability density in all directions), and the positron is given the spin direction $-\vec{s}$
• If Alice measures the spin of the electron along axis $\vec{a}$, and Bob measures the spin of the positron along axis $\vec{b}$, then the probability that they will get the same result (both spin-up or both spin-down) would be: $|\frac{\theta}{\pi}|$, where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$.
So this model predicts a linear relationship. I don't know how general the linear result is, though.

Last edited: Jan 26, 2017
7. Jan 26, 2017

### exponent137

stevendaryl,
I agree. But why this your point 4 is linearly dependent on angle, why not of sin of some other function of angle? Probably it is trivially, but I do not see how.

Last edited: Jan 26, 2017
8. Jan 26, 2017

### stevendaryl

Staff Emeritus
That's a fact of spherical geometry. Think of it this way:
Alice picks her axis, $\vec{a}$. Bob picks his axis, $\vec{b}$. Those vectors can be thought of as points on a sphere of radius 1. Like with globes, we can draw lines of latitude and longitude on that sphere, and we're free to choose whichever points we like to be the North Pole and South Pole. Let's choose them so that $\vec{a}$ and $\vec{b}$ are both on the equator. Let $\theta_a$ be the longitude of $\vec{a}$ and let $\theta_b$ be the longitude of $\vec{b}$. By swapping the North and South poles, if necessary, and by picking the point of 0 longitude, we can make sure that $0 \leq \theta_a \leq 90$ and $\theta_a \leq \theta_b \leq 180$. Now, let $\vec{s}$ be the point corresponding to the spin of the electron. We just have to consider its longitude, $\theta_s$ (a number between -180 and +180):
• If $\theta_a - 90< \theta_s < \theta_a + 90$, then Alice will measure spin-up.
• If $\theta_b - 90 < \theta_s < \theta_b + 90$, then Bob will measure spin-down (because he's measuring the positron, which has the opposite spin)
• If $\theta_a +90 < \theta_s < +180$ or $-180 < \theta_s < \theta_a - 90$, then Alice will measure spin-down.
• If $\theta_b +90 < \theta_s < +180$ or $-180 < \theta_s < \theta_b - 90$, then Bob will measure spin-up.
• So they both measure spin-up if $\theta_a - 90 < \theta_s < \theta_b - 90$. That range has size $\theta_b - \theta_a$.
• They both measure spin-down if $\theta_a + 90 < \theta < \theta_b + 90$. That range also has size $\theta_b - \theta_a$
• So the range in which they have the same result has size $2 (\theta_b - \theta_a)$
• So if $\theta_s$ is chosen randomly in the range $-180 < \theta_s < +180$, then the probability it will be in a range where Alice and Bob get the same result is $\frac{2(\theta_b - \theta_a)}{360} = \frac{\theta_b - \theta_a}{180}$
• If we measure the longitude in radians, rather than degrees, that probability would be $\frac{\theta_b - \theta_a}{\pi}$

Last edited: Jan 26, 2017
9. Feb 3, 2017

### edguy99

The above exact model is used here. Dehlinger and Mitchell propose a model where each photon has a polarization angle λ. When a photon meets a polarizer set to an angle γ , it will always register as Vγ if λ is closer to γ than to γ + π/2, i.e.,
• if |γ − λ| ≤ π/4 then vertical
• if |γ − λ| > 3π/4 then vertical
• horizontal otherwise.
The OP has a very good question as to why they pick a linear relationship since it surely fails to predict nature properly. Why dont they try one that works like here?

Which if I understand correctly is this:

Assume the photon is coming at you with an angle of λ, but with a wobble:
• Chance of vertical measurement = (cos((γ − λ)*2)+1)/2
• Chance of horizontal measurement = (cos((γ − λ + π/2)*2)+1)/2

10. Feb 3, 2017

### DrChinese

Surely you must know that there aren't any LR models that predict nature properly (i.e. match QM).

So... the formulae you have are far off. For example, even when measurement angles A=B you would get incorrect statistical results. You should get a certain match every time, but instead it will only be as expected 75% of the time at best. That is because (cos((A − λ)*2)+1)/2 and (cos((B − λ)*2)+1)/2 will not yield the same outcomes since they are evaluated independently.

11. Feb 3, 2017

### edguy99

Yes, you are absolutely correct that you do not get a match every time if they are off their basis vectors as is pointed out in the comments to the insight here. The point here is that if the experiment is only using the "matched" photons and is pre-selecting and throwing out the "un-matched" photons as being noise, then clearly they are left with only "matched" photons and skewed statistics.