# I Concrete contrast between local and nonlocal hidden variable

1. Jun 10, 2016

### ObjectivelyRational

Can someone concretize the difference between a "hypothetical reality" where hidden variables are local and hidden variables are nonlocal in the following context:

Some physical event occurs involving particles A and B (which are/become entangled and will exhibit correlations) at time T and location X. Insofar as any hidden variables exist at this time, they let us assume they are local (whatever that means) as A and B are local to one another at X as must also be any attributes or properties thereof.

At another time or times TA and TB (same or different depending on the reference frame) each particle is respectively measured at locations XA and XB (for example at locations at distances in opposite directions).

Regardless of who is correct about reality, what does one mean when one claims "hidden variables are local" and consequently what would one mean when one says "No, hidden variables are not local, they are nonlocal"?

2. Jun 10, 2016

### Heinera

When we talk about a "local hidden variable theory", it should be parsed as "local [hidden variable theory]", not "[local hidden variable] theory". In other words, it is the model that is local, not the hidden variables. In a local model, signals cannot travel at superluminal speeds. In the context of Bell's theorem, this means that we assume that Alice's detector settings cannot influence the results at Bob's detector, and vice versa (because space-like separation of events is assumed). In a non-local model, such an influence is allowed.

3. Jun 10, 2016

### naima

In another thread i recall that Bell wrote the conditionof locality
$P(a,b,\lambda) = P(a,\lambda) P(b,\lambda)$
Here a and b are the result of measurements by distant devices and lambda is the hidden variable. here $P(a,\lambda)$ is not the probabilité to get "a" knowing lamda (if it were a hidden variable it would be 1) but the probability that in the measurement we get a and $\lambda$
I asked how this property is related with the speed of light or with the past light cones. Have you an idea?
Thanks.

4. Jun 10, 2016

### Jilang

The probability is different to the wavefunction, right? Shouldn't the condition for locality be written in a different way if the hidden variable is variable of the wavefunction?

5. Jun 11, 2016

### naima

The "problem", if any, is that a probability is associated to the hidden variables $\lambda$. When you see probabilities $P(a,b,\lambda)$
The first thing which comes to you is: how could i sum them to get P(a,b)?
It is the same thing with rhe Young slits. If I give you the probability for each slit you will wonder how to mix them to get the interference pattern. And you will get formulas violated by experiments.
Things might be different if a complex number $C(a,b,\lambda)$ was associated. It might also be a quaternion (proposed by Joy Christian)
What Bell's theorem precludes is to use local hidden variables with probabilities not local variables.

Last edited: Jun 11, 2016
6. Jun 11, 2016

### Jilang

Yes, that is what I was wondering. It's not the hidden variables per se that are disallowed, but the assignment of such variables to a classical probability.

7. Jun 11, 2016

### naima

There is a disproof of Bell's theorem by Joy Christian. He is not mainstream.
He is mainly criticised because he associates quaternions to the hidden variables instead of positive numbers (probabilities).
When he does so he find another inequality which is not violated by the experiments.
You can see here how he is appreciated by the physics community

8. Jun 11, 2016

### Heinera

Indeed he is not mainstream.
No, he is mainly critizised because his (unpublished) papers are complete gibberish. Since you linked to Motl's blog, let me link to an even more entertaining blog post by MIT professsor Scott Aaronson: http://www.scottaaronson.com/blog/?p=1028

9. Jun 11, 2016

### Jilang

Oh dear, that does not read well. Are here are any other formulations that assign lambda to the wavefunction rather than to the probability?

10. Jun 11, 2016

### stevendaryl

Staff Emeritus
Christian's ideas are not rejected because they are not mainstream. They are rejected because they are nonsense. Just my opinion (and that of just about everyone who has looked into it).

11. Jun 11, 2016

### stevendaryl

Staff Emeritus
I don't understand what that means. In the EPR experiment, $\lambda$ is supposed to represent facts about the twin pair that are true at the time they are created. There is no assumption about what $\lambda$ is, whether it's a real number or a quaternion or whatever. But the assumption is that:
1. The probability that Alice measures spin-up depends only on $\lambda$ and facts about Alice's detector.
2. The probability that Bob measures spin-up depends only on $\lambda$ and facts about Bob's detector.
3. There is some probability distribution governing the possible values of $\lambda$.
Now, what is interesting (to me, anyway) is that quantum amplitudes do work exactly like Bell assumed probabilities work. If we let

$\Psi(\alpha, \beta)$ = the probability amplitude that Alice and Bob will both measure spin-up, given that Alice's detector has setting $\alpha$ and Bob's detector has setting $\beta$.

For the spin-1/2 anti-correlated twin pair version of EPR, the quantum prediction for this amplitude is:

$\Psi(\alpha, \beta) = \frac{1}{\sqrt{2}} sin(\frac{\beta- \alpha}{2})$

Then we can find a "hidden variable" explanation for this amplitude:

$\Psi(\alpha, \beta) = \sum_{\lambda} \psi(\lambda) \psi_A(\alpha, \lambda) \psi_B(\beta, \lambda)$

This hidden variable model is just this:
1. $\lambda$ has two possible values: $0$ and $\pi$
2. $\psi(0) = \frac{1}{\sqrt{2}}$
3. $\psi(\pi) = -\frac{1}{\sqrt{2}}$
4. $\psi_A(\alpha, \lambda) = cos(\frac{\alpha - \lambda}{2})$
5. $\psi_B(\beta, \lambda) = -sin(\frac{\beta - \lambda}{2})$
With these choices, we compute:

$\Psi(\alpha, \beta) = \frac{1}{\sqrt{2}} (- cos(\frac{\alpha}{2}) sin(\frac{\beta}{2}) + cos(\frac{\alpha - \pi}{2}) sin(\frac{\beta - \pi}{2}))$
$= \frac{1}{\sqrt{2}} sin(\frac{\beta- \alpha}{2})$

Then we square the amplitude $\Psi(\alpha, \beta)$ to get the appropriate probability: $\frac{1}{2} sin^2(\frac{\beta- \alpha}{2})$

So in a sense, probability amplitudes for QM work like we would expect probabilities to work.

Last edited: Jun 11, 2016
12. Jun 12, 2016

### Jilang

Yes, that's exactly what I was asking. It is interesting to me too!

13. Jun 12, 2016

### naima

You are right but it is not what i said.
It is not suggested that $\lambda$ could be a quaternion or something like that.
In the Bell machinery there is a map p from $\Lambda$ to [0 1]
This map is supposed to sum to 1. to verify the locality condition. More than that, when integrated it is supposed to give the statistial frequencies. And this does not succeed.
It is not surprising. Suppose that these hidden variables are physical. QM
tells us that waves of probabilities are not the good tool. that amplitudes of probabilities are more useful.
So the idea would be to map (i call this map m) $\Lambda$ to another set of numbers.
We could also have $m(a,b,\lambda) = m(a,\lambda) m(b,\lambda)$ that is locality. and so on
Of course this would need a way to get the frequencies at the end.

Edit: in your model what is the value of $\psi(a,b,\lambda)$

Last edited: Jun 12, 2016
14. Jun 12, 2016

### stevendaryl

Staff Emeritus
It's $-\frac{1}{\sqrt{2}} cos(\frac{a- \lambda}{2}) sin(\frac{b- \lambda}{2})$

15. Jun 12, 2016

### naima

I think that a hidden variable model has to explain how the knowledge of the variable gives you the outcomes. As $\lambda$ may be a couple of number, i would have thought that $\lambda = (1,-1)$ would give a = up and b = down and so on.
With this model the hidden variable is exactly the unknown (hidden) outcome that you will observe in the future!
Of course this model is contextual.

16. Jun 12, 2016

### stevendaryl

Staff Emeritus
That's a misconception about Bell's argument. It's not necessary to assume that the hidden variable uniquely determines the outcomes. What Bell assumed was that for Alice's measurement of her particle's spin, the most complete information about the future outcome must come from local variables: Facts about Alice's detector + facts about her particle. So the probability of her measuring spin-up is assumed to be of the form:

$P_A(\sigma_p, \sigma_a)$

where $\sigma_p$ is the state of the particle, and $\sigma_a$ is the state of Alice's detector. It is not necessary to assume that the outcome is determined by these variables. The outcome might be random, but with probabilities affected by $\sigma_p$ and $\sigma_a$. However, in the EPR experiment, there is perfect correlation between Alice's result and Bob's result. Perfectly correlation is not possible unless the outcome is deterministic.

17. Jun 12, 2016

### stevendaryl

Staff Emeritus
I should say that Bell's notion of a local model allows perfect correlation only if the outcome of measurements is deterministic. Nondeterministic nonlocal interactions can bring about perfect correlations.

18. Jun 12, 2016

### naima

Wiki says:
hidden variable theories were espoused by some physicists who argued that the state of a physical system, as formulated by quantum mechanics, does not give a complete description for the system; i.e., that quantum mechanics is ultimately incomplete, and that a complete theory would provide descriptive categories to account for all observable behavior and thus avoid any indeterminism
Maybe you do not agree?
The aim is to avoid indeterminism.
In a hidden variable theory the outcomes have to be determined.

19. Jun 13, 2016

### Demystifier

"It’s as if someone announced his revolutionary discovery that P=NP implies N=1, and then critics soberly replied that, no, the equation P=NP can also be solved by P=0."