# Bell's theorem proof. Does it really proofs anything?

by 0xDEAD BEEF
Tags: bell, proof, proofs, theorem
 P: 35 here picture. - goes down, + goes up So - does this theory says, that after Bobs measurement (taking place in time between A and A1/A2) we could get A1- or A2+ or we allways will see A1+ or A2- EDIT: it would make more sense to call A1 A+ and A2 A- :) Attached Thumbnails
PF Gold
P: 1,642
0xDEAD BEEF: "Bubuubibeebuu. Bububeee! Babubabubee?"
DevilsAvocado: "Eat this dead beef son, it is good for your teeth!"

 Quote by 0xDEAD BEEF However - would that happen?
With two photons that have a fixed polarization from start, this will happen (at best! ):

Alice polarized 0º
Bob polarized 0º

Alice hit the detector at 0º, and she passes thru 100% of the time.
Bob hit the detector at 30º, and he passes thru 75% of the time.

The relative angle between Alice & Bob is 30º and the mismatch is 25% which seems OK, right?

Next test:

Alice hit the detector at 30º, and she passes thru 75% of the time.
Bob hit the detector at 60º, and he passes thru 25% of the time.

The relative angle between Alice & Bob is now also 30º... but the mismatch is now 50%!? This is not OK!

Babubabubee? Get it?
 P: 35 Thats because you were firing photons at angle about 0. To be honest - i dont get it. If i would - would i start this wierd (nonsense) topic from very begining? By the way - what do you think of my experiment setup with 3 polarizers and 4 detectors at Anna's place? Beef Edit: Nope - i got your experiment! That was exactly what my program outputed. Only - i tried out all different options of photon polarization. Now the question is - where is the catch!
 PF Gold P: 1,642 What is it?? Julian Assange’s contact network!? (terribly sorry, extremely bad joke! )
PF Gold
P: 1,642
 Quote by 0xDEAD BEEF Edit: Nope - i got your experiment! That was exactly what my program outputed. Only - i tried out all different options of photon polarization. Now the question is - where is the catch!
The catch (that I have tried to inform you for TWO WHOLE DAYS) is – QM does not behave in the expected classical way, it’s weird; 1 + 1 = 3!! ()

Seriously, I have to leave now. You've got some reading to do. I’m sure it will come to you.

Good luck and Cheers!
 P: 35 Ha ha! (irony, but maybe not). So, how i see this- Anna got polarization filter and Bob has. Standart setup. What if we further extend this setup, so that Anna has 3 polarization filters all set at same angle. So, Anna would have 4 detectors. 2 for each "second" filter like this - dA++ ........ pA+ dA+- ........ ........................ pA ........ dA-+ ........ pA- dA-- So pA (first polarization filter) decides polarization and redirects photon to either pA+ or pA- polarization filters. In the mean time twin-photon hits Bob's polarizer, which is set at 45 degree angle. If we forget about this QM wierd stuf, no mater what/when/ever second twin-photon hit Bob's polarization filter, Anna should always get either dA-- or dA++, because, if photon once choose to go through Anna's 0 polarizer in + direction (up), why shoul it choose differnetly, when facing second one pA+ (or vice versa - facing pA-). In this case standart physics predict, that Anna should only get dA-- or dA++ output! How ever - if this QM stuff is real and Bob's detector has given new properties to photon, while it was traveling through Bob's polarizer set at very different angle than Anna's was - we should see Anna's dA-+ and dA+- detectors fire as well! This really would prove everything! Scienitifcal brekathrough - i would say. Otherwise... sorry guys, but we are just creating our own virtual reality to play with (wich ain't that bad, since we can learn new thing from it as well). Beef
 PF Gold P: 1,642 @0xDEAD BEEF Maybe unfair to leave in an "outburst" like that... you did ask me a specific question... The "catch" (that your predefined photon polarization will never ever work), is because the relative angle between Alice & Bob can be anything between 0-360, but "your" photons doesn’t care one bit about this relative angle. The only thing that they will "respond" to is their own polarization and the setting of their own polarizer. I.e. there is no "communication" or "link" between Alice & Bob in your "design", and this is crucial. Don’t ask what this "communication" or "link" really is and how this works – no one knows. There are two main "explanations" for this paradox. One is the so called non-locality, meaning some sort of "communication" between Alice & Bob is present (that we don’t know what it is). The other is non-separability, which means that the physical reality is not what we think; one object can be at two places at once, or something like that (I think it involves holism as well). This is why I won’t get into details in your program; I know it will never work they way you hope. Yes, it’s "cocky and rough", but it’s the truth. You will win some substantial time if you first study the problem in detail, to learn everything, and then try to build something on your own. Just by guessing, you will not get anywhere. Good luck & Take care!
Sci Advisor
PF Gold
P: 5,146
 Quote by 0xDEAD BEEF Ha ha! (irony, but maybe not). So, how i see this- Anna got polarization filter and Bob has. Standart setup. What if we further extend this setup, so that Anna has 3 polarization filters all set at same angle...
If they are all set the same way, you can pretty well predict the results. Same all around, as any polarized photon can be sent through a series of similarly oriented splitters and nothing changes.

It would be helpful if you would ask a specific question. You are wandering all over the place and it seems as you ignore the rules I keep trying to explain. There is difference between entangled photons (emerging from a PDC crystal) and a pair of photons of known and identical polarization, such as emerge directly from a laser beam. You should learn this difference first, and understand their statistics.
PF Gold
P: 1,376
 Quote by 0xDEAD BEEF as i said, i would use black-box, so, whatever "polarization axis of respective polarizer" is, i can no know that! So - what output would i get?
I suppose that you mean you don't know polarization axis of photons (and they have different polarizations) not polarizers i.e. black-box is source not measurement equipment.
If that is so then you will have classical output like that:
N(a,b)=1/4+1/2*cos^2(a-b)

If we want to look at real life example we can take PDC source without walkoff compensators.
In that case we have only H and V polarized non-entangled photons. And result is like this:
N(a,b)=1/2*cos^2(a)*cos^2(b)+1/2*sin^2(a)*sin^2(b)
 P: 35 What is PDC source with with walkoff compensators? Yes - my black box device would be photon source which would output two photons to Anna and Bob. These two photons would have same polarization (photon a polarization == photon b polarization), but those polarizations would change on random (photon a == photon b == random value). Beef
 P: 35 So QM states, that particle can have only one "property" at time? For example - i send p (particle) through 0 angle polarizer. It goes either up or down and now it has its 0 angle property set to up or down. Then (p) travels through another polarizer, this time set at 90 angle, so particle now forgets? its 0 angle and decides either to go up or down at 90 angle. And now if i send this particle through 0 angle polarizer again, then it could choose different direction to go this time, because previously it was measured against 90 angle detector??? Example: p --> (0) -- up/down --> (90) -- up/down --> (0) -- same as first time or different??? --> (detector). EDIT: What i am aiming at is: a) photons have some invisible quantum bound between them. In such case (time line): Alisa measures photon polarization at angle 0 Bob measures twin-photon polarization at angle 90 Alisa again measures photon polarization at angle 0, but gets different result, since Bob's measurement on his photon has made Alisa's photon to "forget" it's 0 angle polarization value. Result - faster than light information exchange. Setup (Alisa) - BOB0 \ p(0) \ BOB90 / p(0) <----- photon BOB90 \ p(0) / BOB0 / where BOB90 has 50% chance of firing, if Bob's polarizer is set at 90 degrees and 0% percent chance, when Bob's polarizer is at 0 degrees. b) we are just measuring different properties of photon. In such case (time line): Alisa measures photon polarization at angle 0 Bob measures photon polarization at angle 90 Alisa measures photon polarization at angle 90 Bob measures photon polarization at angle 0 -- Bob's and Alisa's all 4 measurements match, so there is actually no reason to have Bob at all and Alisa could have measured photon both on 0 and 90 angle bu her self. Setup would be - D0+90+ \ p(90) \ D0+90- / p(0) <----- photon D0-90+ \ p(90) / D0-90- / So - if Alisa's D0+90- fires, that means, that photon had 0+ polarization and 90-. If we have Bob in this scenario, then D0+90- or D0-90- should fire, when Bobs (90 angle) - fires and vice versa. EDIT2: I guess, there is also c) option. c) Alisa can measure photon's polarization at one angle (and that is it), and Bob can also do so, so they both can measure different polarization values of "same" photon, which is cool, since we get more information about that photon, but that is it . EDIT3: There might also be option d). Alisa measures her photon at 0 angle, so Bob's photon now can not be longer measured against 90 angle and vice versa. Twin-photon hits Bob's polarizer always first. In this case Alisa could have multiple detector chain BOB0 \ p(0) \ BOB1 / p(0) <----- photon ------> p(0/90) BOB1 \ p(0) / BOB0 / so - if Bob DOES NOT measure his twin-photon polarization angle at 90, then Alisa always gets BOB0. If, however, Bob does measure his twin-photon's polarization before Alisa's photon enters Alisa's first polarizer p(0), then Alisa's photon starts giving random data when measured against p(0), so it should start hitting BOB1 detectors. In this case we would again clearly see at Alisa's end, that Bob has measured his photon (at 90) or has not, thus, we could make conclusion about Bob's setup -> faster than light information exchange. ------ So is it a), b), c) or d) ??? Sincerely, Beef
PF Gold
P: 1,376
 Quote by 0xDEAD BEEF What is PDC source with with walkoff compensators?
It is usual source of polarization entangle photons.
 Quote by 0xDEAD BEEF Yes - my black box device would be photon source which would output two photons to Anna and Bob. These two photons would have same polarization (photon a polarization == photon b polarization), but those polarizations would change on random (photon a == photon b == random value).
Then I understood you correctly and my answers hold.
PF Gold
P: 1,376
 Quote by 0xDEAD BEEF So QM states, that particle can have only one "property" at time? For example - i send p (particle) through 0 angle polarizer. It goes either up or down and now it has its 0 angle property set to up or down. Then (p) travels through another polarizer, this time set at 90 angle, so particle now forgets? its 0 angle and decides either to go up or down at 90 angle. And now if i send this particle through 0 angle polarizer again, then it could choose different direction to go this time, because previously it was measured against 90 angle detector??? Example: p --> (0) -- up/down --> (90) -- up/down --> (0) -- same as first time or different??? --> (detector).
Different
Only if you talk about photons and polarizers then it's 0° and 45° not 0° and 90°. Photons that pass 0° polarizer are completely blocked by 90° polarizer.

 Quote by 0xDEAD BEEF a) photons have some invisible quantum bound between them. ...
No
 Quote by 0xDEAD BEEF b) we are just measuring different properties of photon. ...
No, because after first measurement photons are not entangled any more.
 Quote by 0xDEAD BEEF c) Alisa can measure photon's polarization at one angle (and that is it), and Bob can also do so, so they both can measure different polarization values of "same" photon, which is cool, since we get more information about that photon, but that is it.
I would rather say no. It's not photon that behaves the same way. It's the wavefunction that behaves the same way. So you don't get more information about the "same" photon.
 Quote by 0xDEAD BEEF There might also be option d). Alisa measures her photon at 0 angle, so Bob's photon now can not be longer measured against 90 angle and vice versa. ...
No, because after first measurement photons are not entangled any more.
And you always get BOB0 irrespective of Bob's measurement.
 P: 35 So you are saying, that after first measurement photons are not entangled any more. If so - does it matter at all that they were entangled from very beginning. Or maybe i am just getting this wrong, but - does entanglement gives any other extra properties to photons than just that they have all same properties? In this experiment they use that crystal to create entangled photons with same polarization and send them to Alisa and Bob. Why would it make any difference if i replace "twin-photon crystal" with "black box", which also outputs same photons, only with difference, that they are "manually created" (two "light bulbs" and bunch of polarization filters) . ?
PF Gold
P: 1,376
 Quote by 0xDEAD BEEF So you are saying, that after first measurement photons are not entangled any more. If so - does it matter at all that they were entangled from very beginning.
Apparently it matters. After first measurement they are not entangled but they are correlated nevertheless i.e. if you detect them results are correlated.

 Quote by 0xDEAD BEEF Or maybe i am just getting this wrong, but - does entanglement gives any other extra properties to photons than just that they have all same properties?
Well maybe it's better to take it as speculation but anyways I would say that entanglement is specific correlation of photon phase not only for paired photons but for whole ensemble.

 Quote by 0xDEAD BEEF In this experiment they use that crystal to create entangled photons with same polarization and send them to Alisa and Bob. Why would it make any difference if i replace "twin-photon crystal" with "black box", which also outputs same photons, only with difference, that they are "manually created" (two "light bulbs" and bunch of polarization filters) .
In your "black box" setup you can't control phase of created photons.
 PF Gold P: 1,642 Well, this is not an ensemble of personal speculations; it’s the current professional mainstream scientific view. Trust me, or Dr. Robert Nemiroff:
Sci Advisor
PF Gold
P: 5,146
 Quote by 0xDEAD BEEF So you are saying, that after first measurement photons are not entangled any more. If so - does it matter at all that they were entangled from very beginning. Or maybe i am just getting this wrong, but - does entanglement gives any other extra properties to photons than just that they have all same properties? In this experiment they use that crystal to create entangled photons with same polarization and send them to Alisa and Bob. Why would it make any difference if i replace "twin-photon crystal" with "black box", which also outputs same photons, only with difference, that they are "manually created" (two "light bulbs" and bunch of polarization filters) . ?
There is a difference, and you can tell by an experiment. If you observe Alice and Bob at the same but *random* angles, the following pattern will emerge over a series of trials:

If Alice and Bob ARE entangled (PDC source), they will be 100% correlated.
If Alice and Bob are NOT entangled (black box source), they will be about 75% correlated.

Does that help? In both cases, the photon pairs are clones of each other in the sense that that have the same quantum properties. The difference is that entangled particles are in a superposition of states because the value of one or more of those quantum properties is not known. That leads to some rather unusual experimental situations as compared to particle pairs which are not in a superposition.
 P: 35 DrChinese regarding - If Alice and Bob ARE entangled (PDC source), they will be 100% correlated. If Alice and Bob are NOT entangled (black box source), they will be about 75% correlated. But if i have blackbox, which outputs two photons (to Anna and Bob) having same polarization +---------------- BLACKBOX -------------------| |laser -> beam splitter -> polarizer for anna | -> anna's setup | \-> polarizer for bob | -> bob's setup +---------------------------------------------| Should not hey both get 100% correlated, if "polarizer for anna" is set to same value as "polarizer for bob"? Beef P.S. I did not like that Dr. Robert Nemiroff stuff. It was too easy, too non chalenging. Too much facts, without proof/idea behind them. And I thought education quality is bad only at my country... :)

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