# Bell's Theorem with Easy Math - Stuck!

1. May 8, 2013

Hi, I hope there is no issue with me posting this here but I'm stuck trying to get my head around DrChinese's page on Bell's Theorem.

"ASSUME that a photon has 3 simultaneously real Hidden Variables A, B and C at the angles 0 degrees, 120 degrees and 240 degrees per the diagram above. These 3 Hidden Variables, if they exist, would correspond to simultaneous elements of reality associated with the photon's measurable polarization attributes at measurement settings A, B and C. In other words, each hidden variable gives us the answer to the question "will this photon pass through a polarizer lens set at a specific angle?" "

However,

"Once any photon passes through a polarizer lens, its polarization will be aligned exactly with the lens thereafter (even if it wasn't previously)."

then

"According to malus, when completely plane polarized light is incident on the analyzer, the intensity I of the light transmitted by the analyzer is directly proportional to the square of the cosine of angle between the transmission axes of the analyzer and the polarizer." (http://www.physicshandbook.com/laws/maluslaw.htm)

so according to Malus's law, only a certain quantity of photons polarized at a given angle will pass through another polarizer set at another random angle, and this quantity is proportional to cos$^{2}$θ, where θ is the difference in the polarization angles. My assumption here is that there is no way of reducing the intensity of the light other than by some photons being absorbed by the polarizer.

So it seems to me that it is not possible to have a definite yes/no answer to the questions A, B and C for any individual photon.

As such I can't actually make it past DrChinese's statement: "If you are not sure of this point, then please review the table above until you are sure.". It seems to me that the table should contain probabilities based on cos$^{2}$θ or have I got totally confused?

2. May 8, 2013

### DrChinese

1. It is certainly possible that there is no simultaneously definite answers to the polarization "questions" of angles A, B and C. That would be the usual perspective anyway, UNLESS you are a local realist. So the issue begins with assuming local realism to see if you can refute it.

What evidence is there for local realism with entangled particle pairs? As EPR (1935) points out, there is a lot: you can predict the answer for Alice's A by first reading Bob's A. Ditto for Alice's B and C (by reading Bob's B or C). Therefore, EPR says there is an "element of reality" associated with A. And by inference, B and C. They acknowledge that no experiment can answer the question of the values simultaneously. But they claim that it is "unreasonable" to quibble over the simultaneous requirement. They say you should ASSUME that to be the case, ie that A B and C all exist simultaneously. So their key assumption counters your conclusion. (They didn't know about Bell though, you do!)

I can explain this further if that would help.

2. The table shows the possible outcomes ASSUMING EPR is correct. In other words, there is a deduction which can be made from their assumption. That is because each run should produce a value somewhere in the table (2^n where n=number of angles, in this case n=3).

Keep in mind that EPR is essentially a local realist position statement. So Bell is questioning that, as you do.

3. May 8, 2013

Many thanks for answering, I think I get that, the local realism that EPR assumes is incorrect because it assumes that A, B and C have definite values.

I guess my thinking was more that local realism meant that any individual photon would have a 'reality' that is actually its polarization angle, and so couldn't have a definite value for being detected by an analyzer set at any given angle. I'm left with being unsure why EPR makes the assumption it does.

4. May 8, 2013

### DrChinese

Their position made a lot of intuitive sense. If you can predict something in advance with certainty, it must have some element of reality. So if I predict the next ball will be "blue" and it is in fact blue, then its "blueness" must be real. Ditto for polarization, which then allows for the idea that polarization itself may not be a fundamental property of the particle. It might represent something, for example, that also involves a measurement device. But in their example, you always get the expected results when you measure the same thing (A, B or C in our example).

5. May 8, 2013

DrC’s example is superb, however if you want something even more related to polarizers and probabilities, Nick Herbert has an example known as one of the simplest proofs of Bell's Inequality (i.e. in serious competition with DrC ;). This example was also used by John Bell himself, giving lectures.

In the setup we have one source of entangled pair of photons and two polarizers that we can position independently at different angles. The entangled source is of that kind that if both polarizers are set to 0º we will get perfect agreement, i.e. if one photon gets thru the other photon gets thru and if one is stopped the other is also stopped, i.e. 100% match and 0% discordance.

To start, we set first polarizer at +30º and the second polarizer at . If we calculate the discordance (i.e. the number of measurements where we get a mismatching outcome thru/stop or stop/thru), we get:

sin^2(30º) = 25%
(according to QM theory & experiments)

Next, we set first polarizer to and the second polarizer to -30º, and the discordance we will naturally be 25% this time also.

Now it’s time for John Bell’s brilliant logic:

– What will the discordance be if we set the polarizers to +30º and -30º...?

If we assume a local reality – nothing we do to one polarizer can possible affect the outcome of the other polarizer – we can formulate this Bell Inequality:

N(+30°, -30°) ≤ N(+30°, 0°) + N(0°, -30°)
(the symbol N represents the number of discordance)

Of course we can make this Bell Inequality even simpler, and say:

50% = 25% + 25%
25 + 25 = 50
1 + 1 = 2

(i.e. all the way down to the “kindergarten version” ;)

This is the obvious local realistic assumption – one banana plus one banana makes two bananas.

Now comes the shock (that makes some people go bananas) – this local realistic assumption is not compatible with neither QM theory predictions nor all performed experiment this far, which is:

sin^2(60º) = 75%

1 + 1 = 3 !!
<-- he got one banana too many

Enjoy
DA

Last edited: May 8, 2013
6. May 9, 2013

Is the source polarized at 0° ?

If not then I don't get why an assumption of local realism would predict a different result for rotating both polarizaers in opposite directions to that which would be obtained by rotating one polarizer twice as far? i.e. if the source is randomly polarized and the universe has no reference plane, the only way of calculating the expected discordance would be by measuring the difference in the angles which then gives you the correct answer.

7. May 9, 2013

No worries, this is a good question that also puzzled me for some time. We all know that for unpolarized light there’s 50/50 chance to get thru a polarizer. However, once the photon has passed through a polarizer then its polarization is known, and the photon will be in a definite state (definitely polarized).

Entanglement requires the photons to be in an indefinite state (unknown polarization), however since they are entangled (and share the same wavefunction) as soon as you measure one photon the other will instantly be in a definite state (definitely polarized), and the entanglement is then broken (i.e. decohered).

The really ‘weird’ thing about entanglement is that you can set the polarizers at any angle and the exact same thing will happen! I.e. it doesn’t matter if the polarizers are set to 0º, 17º or 243º. If they are parallel you will always know the outcome in advance, so called prefect correlation.

This is an example of a Bell state, using proper notation (Type I):

$\frac{1}{\sqrt{2}} \left( | \uparrow \rangle_A |\uparrow \rangle_B + |\rightarrow \rangle_A |\rightarrow \rangle_B \right)$

Now, what does it say? It says that there’s a 50/50 chance for measuring both photon A & B as vertical polarized (0º) or both as horizontal polarized (90º). In the Nick Herbert example above vertical polarized (0º) would mean thru and horizontal polarized (90º) would mean stop.

However, as I mentioned you could as well choose to measure the Bell state by setting both polarizers at for example 45º.

This is an example of a diagonal Bell state (Type I):

$\frac{1}{\sqrt{2}} \left( | \nwarrow \rangle_A | \nwarrow \rangle_B + | \nearrow \rangle_A | \nearrow \rangle_B \right)$

This shows that the Bell state will yield the same result for both photons when measured diagonally!

(It would be interesting if DrC could elaborate around this. How are we supposed to interpret this ‘flexible’ Bell state? Should it be regarded as superposition of all possible states in an “orthogonal linked correlation”?)

The reason for not turning one polarizer all the way to 60º and stop there is that it then would be much more difficult (or impossible) to rule out LHV/local realism. If I’m not mistaken, this is the original 1935 EPR thought which resulted in the 20 year Bohr–Einstein debates ...

EDIT:
Local realism does not predict a different result for rotating both polarizers in opposite directions. Local realism only predicts that you add the two results together 1 + 1 = 2. That’s the key of it all.

Think of it like this; if we are going to rule out 'Spooky action at a distance' first we will need to do a measurement in one end and check the result. Then we have to do the same thing in the other end. Now logically, if we repeat these two measurements at once, we must obtain the same result in each end, right? If we don’t, there must be some 'Spooky' influence between them.

If you first measure a car leaving to the left at 100 km/h, and then measure a car leaving to the right at 100 km/h, you don’t expect to measure them leaving at 150 km/h respectively, if measured simultaneous...

Last edited: May 9, 2013
8. May 9, 2013

### lugita15

Here is the Nick Herbert explanation that DevilsAvacado is referring to.

9. May 9, 2013

Thanks lugita.

Badvok, I think I know better what you mean regarding the assumption of local realism, but I don’t have the time now, try to find some tomorrow.

10. May 10, 2013

Thanks, no rush, I'm just trying to learn this stuff in some of my idle time at work :)

11. May 10, 2013

Thanks all, I think I've got it now.

The bit I was stumbling over in Nick Herbert's proof is this: "Starting with two completely identical binary messages". Where do these messages come from? So far as I could see the system has a stream of randomly polarized photons, no actual binary message. Thus the only mismatch that could be measured is by comparing the results obtained at A & B, which obviously links the two detectors and makes the mismatch 75% based on the mutual misalignment angle.

I now see that the 'binary message' he refers to is the one that would have been produced by both detectors had they remained in alignment. If it was possible to still determine this once both detectors had been rotated then I see that the mismatch between what each detector produced and that impossible message would be 25% and so the mismatch between the two actual messages could only be at most 50%.

It all still feels a bit like one of those maths puzzles that are designed to trip you up and make you do the maths wrong, but I guess that is a common feeling when dealing with QM.

Last edited: May 10, 2013
12. May 11, 2013

Okay Badvok, are you there? Time to work!

I think you are almost there; it’s just that last part in the puzzle that we haven’t discussed yet, so your question about where these ‘messages’ comes from is well-founded.

We have to remember that from a local realist’s point of view – entanglement doesn’t really exist, it’s just an illusion, and behind QM entanglement is an unknown property of nature that has not yet been discovered, but whom will save us from the “mysterious” QM stuff. ;) Normally, to construct a model that saves local realism we usually talk about a Local Hidden Variable theory (LHV).

So, let’s start with the absolute simplest LHV possible, and say we are using two fixed polarizers at 0º. How would one construct local realism in this setup? Well, it’s very simple. We just claim the photons obtain their (random) “identical twin values” at the source. Piece of cake!

Now, let’s make it a little bit harder and introduce the option of rotating the polarizers.

What do you do? Well, you look at the principle for Malus law, which tell you what happens when a polarized beam of light passes through a polarizer, i.e. cos2(θ) (in my example I use sin), and construct a LHV model that exploit this nice ‘facility’.

You can now put one polarizer at any random angle – and everything works just fine! And of course you can do the same thing with the other polarizer, and things still looks great, changing them one at a time.

However, here comes the harsh aftermath...

As we now have constructed a local realism theory, we are not allowed to account for any changes at the other end. What happens at one polarizer cannot possible influence the result of the other.

And right there we run into a dead end/closed door/mission impossible.

Because it’s the relative angle between A & B that gives the correct result cos2(A-B)!

Our LHV has failed...

Last edited: May 11, 2013
13. May 13, 2013

I guess this is the bit I still feel uncomfortable with.

After a polarizer has been turned it becomes impossible to measure what the result would have been if the polarizer had not been turned, therefore the only measurement that can be made is the difference between the outputs. This seems to bring both polarizers into the equation, or in other words the only results that actually can be measured take into account the changes at both ends.

14. May 13, 2013

### DrChinese

Don't forget that no matter what angles you select - 1 degree, 19 degrees, 246 degrees, etc - if you measure both at the SAME angle you get the same result. So if Alice and Bob agree to independently turn to that same angle, and get the same result, the local realist would say that was a pre-determined outcome.

If you think that the polarizers - even when set independently and too far away for a signal to travel between - form a context for a measurement, then you are in favor of observer dependence (as opposed to observer independence). And then you are not a local realist.

15. May 13, 2013

This is correct, but from a local realist’s point of view we must maintain the value (or ‘the hidden function’ to obtain a value) that is present at the source, i.e. once the photon has left its partner at the source your are not allowed to ‘change your mind’, because there are experiments where the polarizers are placed so far away that not even a message sent at the speed of light would have the time to send this ‘changing message’ to the other, and to make it even more difficult for LHV – these polarizers are rotating randomly at very high speeds.

(There are non-local hidden variable theories that work, but let’s skip this for now)

Correct, and this is exactly what Bell proved. We can try to construct all sort of fancy LHV with all the mathematical tricks you can think of, but we are not allow to account for the state of the other polarizer, or the outcome of the other photon. The best LHV we can generate is the red triangle wave in the diagram below:

[Credit: Alain Aspect]

And QM theory/experiments always produce the blue sinusoidal curve = an impossible nut to crack for the local realist’s...

16. May 13, 2013

Isn't that the only context that can actually be measured though?

(I don't think I favour any position - still early days for me with this stuff. My formal education only took me to English GCE 'A' Level Physics and that was very much what I guess could be called 'realist' physics, so I guess I may have a tendency towards that view. Hence my desire to understand how local realism is ruled out.)

17. May 13, 2013

### DrChinese

That's the entire point! The local realist asserts something that *seemingly* cannot be tested. That was the position in 1935 after EPR. But Bell later shows that is not entirely true. There is a consequence of the assertion that something exists that cannot be directly measured.

18. May 13, 2013

### DrChinese

Here are some key experimental results ruling out local realism, which admittedly are relatively newer (last 10-15 years):

http://arxiv.org/abs/quant-ph/0201134
http://arxiv.org/abs/quant-ph/0609135
http://arxiv.org/abs/1209.4191

You can entangle photons (via swapping) which have never been in contact with each other, and also have never existed in an overlapping spacetime region. That means they are polarization clones of each other. The context of a measurement of one is causally related to the result of a measurement of the other (although the causal direction is technically ambiguous).

How can local realism ever be expected to describe these experiments? They violate strict locality immediately!! On the other hand, this is predicted by standard QM.

19. May 13, 2013

Nice papers DrC, of course I got hooked on the graphics in the last.

[my bolding]

I googled “projecting polarizing beam-splitter” and got 2(!) results. One to PF (your thread) and the other to the paper...

What the **** is it??

It’s right there the “Entangled Love Affair” takes place, right? What’s going on? Just because you run two photons thru a PBS this ‘magic’ happens...?? How does one do the timing to get them exactly “in phase”...?? I don’t get it...?? :uhh:

Weird...

Last edited by a moderator: May 6, 2017
20. May 14, 2013

No, I don't expect local realism can describe them, but I will study them because I'd like to fully understand why that is the case. I'm certainly not proficient enough to even begin to consider that QM doesn't work or what flaws there may be in it, I bet there will be some but it will take a better mind than mine to find them (or at least many, many years of study).

For fun here is an old trick/puzzle that I've seen confuse many:
"You see a shirt for £97. You don't have enough cash. You borrow £50 from Fred and £50 from George and so get £100. You buy the shirt and get £3 in change. You give both Fred and George £1 back and keep the odd £1 for yourself. Now you owe Fred £49 and George £49.
Now £49+£49=£98 add on your £1 and you get £99. Where is the missing £1?"

In the local realism scenarios described previously Malus is used to predict some of the correlations and then those predictions are summed and shown to not equal what Malus would have predicted if both ends of the system are taken into account. I understand that this is a logical assertion that is made to show how local realism is not valid by ensuring that we don't take into account changes to a part of the system that is non-local. However, the use of Malus for some parts of the calculation and not for others seems awkward and feels like a deliberate misdirection, a bit like the fun puzzle.

Consider the following scenario: Set up the above experiment with a source on the equator between two polarizers, one to the North and one to the South. Now you measure the correlation at (0°,30°), (-30°,0°) and finally you measure the correlation for (-30°,30°) and continue to measure that last correlation which will show that Malus (and QM) are correct. Now we know from Malus (and QM) that we wouldn't expect to get a change in correlation with regards to time because the correlation is determined solely by the difference in the angles. However, using the local realism assertion we would expect a change because the whole system is rotating in space relative to that original 0° plane - this is obviously nonsense, and what I don't get is why this is even considered as a valid representation of local reality.

I know that you will simply say that of course local realism is nonsense, that was Bell's point, but I've not quite reached that level of understanding and so I'll go away and look at these further examples.

21. May 14, 2013

Ahhh! :surprised The bank stole it! I knew it!!

... no, it’s a very simple Italian medieval ‘thing’ called double-entry bookkeeping:

Code (Text):

[B]Assets                Liabilities[/B]
------------          ------------
97                    49
1                     49
------------          ------------
98                    98

22. May 14, 2013

### DrChinese

I have always loved this little example, although it was told to me about a guy getting a hotel room...

The thing to remember about the QM application of Malus is two-fold:

a) It doesn't really matter if the accounting is right or wrong. The QM prediction is experimentally verified, and there are NO local realistic solutions that match. So even if QM's formula was a cheat, it wouldn't matter.

b) Although it looks as if Malus is the rule being applied, that is not precisely correct. It turns out the math is more complicated than that, but it eventually reduces to cos^2(theta) - which is the same of course. Keep in mind that Malus applies to photons, but there are lots of things that can be entangled, and lots of bases that things can be entangled on. So each has its own border between local realism vs QM. And those have nothing to do with Malus.

So the lesson is: I wouldn't hang my hat on that line of thinking, it is too limiting. Can I have my £1 change now?

23. May 14, 2013

Okay, I’ll give it a last try.

The main thing about entangled photons is that they are in an indefinite state = no definite polarization, however immediately as one of entangled photons passes thru a polarizer its state become definite = fixed polarization, and the amazing thing is that other photon somehow also obtains this definite fixed polarization state, instantaneous with the first photon.

Now, what happens in practice? Photon A has no definite polarization and is traveling towards Alice’s polarizer. For a unpolarized photon there’s a 50/50 chance to get thru Alice’s polarizer. Photon A now enters Alice’s polarizer and gets thru. In this very instant photon A is now aligned with Alice’s polarizer, so if Alice has put here polarizer at +30º photon A has no idea what that ‘means’. Photon A is aligned with Alice and this means 0º to photon A.

Now, photon B on its side is traveling towards Bob’s polarizer, and photon B is in indefinite state = no definite polarization at this stage, but just as photon B is about to enter Bob’s polarizer he gets an ‘abrupt message’ from photon A – “You are polarized at 0º!” – and photon B wonder for a sec what that the heck 0º means... but does not pay not too much attention to this and continue the travel. As Photon B now enters Bob’s polarizer, which is set at -30º, Photon B has only to follow Malus law, which says that it has a cos2(θ) probability to get thru. But what’s theta (θ)!? Well, it comes automatically, right? Photon B has the same polarization as Photon A, which is 0º in reference to Alice, but Alice polarizer set at +30º, and Bob’s polarizer is set at -30º, which means that if both photons A & B agrees that Alice is alignment 0º polarization, then Bob is 60º off from this alignment. And the rest basically come automagically from classical calculations for polarization.

Now, for LHV this option to send FTL messages like – “You are polarized at 0º!” – does not exist, so the best one could do is to agree that photons A & B are polarized at 0º before they leave the source, and then use Malus law to do the best you can with randomly rotating polarizers.

Get it?

(DrC will go nuts on this simple explanation – that A sets the alignment first – and start talking about ‘backward causation’ and stuff, but that’s Doctor Graduate 9.0 which will drive you crazy if you try it now already... )

24. May 14, 2013

### DrChinese

Yes, it is weird. I am not the expert on this, but I think of it this way: 2 otherwise identical particles can be distinguished by their phase. So if they have identical phase, they cannot be distinguished if they arrive at the PBS at the same time.. That makes the 2 systems candidates for entanglement (swapping in this case).

The 2 source lasers are pulsed, and they are given triggers that are somehow synchronous. If the path lengths are adjusted suitably, they will arrive in phase even though they were not created at the same time. Some pairs will arrive too far apart to allow swapping to occur, since you generally cannot command a specific individual photon to split into 2 entangled ones. Sounds difficult to achieve to me! But that's what makes these experiments so great!

25. May 14, 2013