Derivation of the CHSH inequality

[Alien8]

For this model,
P_{++} = \frac{1}{\pi} \int d\phi cos^2(\alpha - \phi) cos^2(\beta - \phi) = \frac{1}{8} + \frac{1}{4}cos^2(\alpha - \beta)
P_{+-} = \frac{1}{\pi} \int d\phi cos^2(\alpha - \phi) sin^2(\beta - \phi) = \frac{1}{8} + \frac{1}{4}sin^2(\alpha - \beta)
P_{-+} = \frac{1}{\pi} \int d\phi sin^2(\alpha - \phi) cos^2(\beta - \phi) = \frac{1}{8} + \frac{1}{4}sin^2(\alpha - \beta)
P_{--} = \frac{1}{\pi} \int d\phi sin^2(\alpha - \phi) sin^2(\beta - \phi) = \frac{1}{8} + \frac{1}{4}cos^2(\alpha - \beta)

So
E(\alpha, \beta) = \frac{1}{2}(cos^2(\alpha - \beta) - sin^2(\alpha - \beta)) = \frac{1}{2} cos(2 (\alpha - \beta))

That's exactly 1/2 of the QM prediction.

Ok, I got that here: http://www.wolframalpha.com/input/?...n^2(a+−+phi)+*+cos^2(b+−+phi),++phi=0+to+2Pi+

http://www4a.wolframalpha.com/Calculate/MSP/MSP1161fh633fae46422g300001fc1f12g5d92ih4e?MSPStoreType=image/gif&s=58&w=532.&h=71. Only one question remains, the dotted curve, how does QM arrive to: cos2(a - b)?

 

[atyy]

Post #41 dispersed the thickest fog around this derivation for me, thanks atyy. I'm unhappy about binary states like heads and tails being labeled as integers and subjected to ordinary arithmetic, and I don't see expressions like black - white = 2 and white - black = -2 are meaningful, but at least I know now what's going on and can investigate further on my own.

The expression is not "black - white = 2". Rather, we assign a value to black and a value to white. For example, "value of black = 1" and "value of white = -1". Then "value of black - value of white = 2." We could choose other values, but this is the choice that is made when people refer to CHSH.

The derivation given by Aspect http://arxiv.org/abs/quant-ph/0402001 is different from the one in Wikipedia, but both are correct. If we use Aspect's approach, he explains the reason for assigning these values to particular outcomes in section 3.1. In particular, in Eq 11 of section 3 he relates it to his definition of E(a,b) = P₊₊(a,b) + P_--(a,b) - P_+-(a,b) - P_-+(a,b).

 

[stevendaryl]

I guess it doesn't make any difference, but why are you not integrating over 2pi or full 360 degrees? How do you convert that integral into the right hand side function for each of those probabilities, is there some simple principle behind it or you use some kind of integral calculator?

Well, for the particular integral I was talking about, the integrand is symmetric between 0 < \phi < 180 and 180 < \phi < 360. So you get the same results if you integrate over 360 and divide by 2 \pi, or just integrate over 180 and divide by \pi

 

[stevendaryl]

Ok, I got that here: http://www.wolframalpha.com/input/?i=1/(2Pi) * integral cos^2(a − phi) * cos^2(b − phi) + sin^2(a − phi) * sin^2(b − phi) - cos^2(a − phi) * sin^2(b − phi) - sin^2(a − phi) * cos^2(b − phi), phi=0 to 2Pi

http://www4a.wolframalpha.com/Calculate/MSP/MSP1161fh633fae46422g300001fc1f12g5d92ih4e?MSPStoreType=image/gif&s=58&w=532.&h=71. Only one question remains, the dotted curve, how does QM arrive to: cos2(a - b)?

In QM, for EPR with correlated photons, the probability that Alice detects a photon is 1/2. The probability that Bob detects a photon given that Alice detects a photon is cos^2(a - b). So for QM:

E(a,b) = P_{++} + P_{--} - P_{+-} - P_{-+} = 1/2 cos^2(a-b) + 1/2 cos^2(a-b) - 1/2 sin^2(a-b) -1/2 sin^2(a-b)
= cos^2(a-b) - sin^(a-b) = cos(2(a-b))

It's not an integral in the QM case.

 

[Alien8]

In QM, for EPR with correlated photons, the probability that Alice detects a photon is 1/2. The probability that Bob detects a photon given that Alice detects a photon is cos^2(a - b). So for QM:

E(a,b) = P_{++} + P_{--} - P_{+-} - P_{-+} = 1/2 cos^2(a-b) + 1/2 cos^2(a-b) - 1/2 sin^2(a-b) -1/2 sin^2(a-b)
= cos^2(a-b) - sin^(a-b) = cos(2(a-b))

It's not an integral in the QM case.

Ok, but that only moves the question to the probability that Bob detects a photon given that Alice detects a photon: what is "cos^2(a - b)" given by?

 

[stevendaryl]

Ok, but that only moves the question to the probability that Bob detects a photon given that Alice detects a photon: what is "cos^2(a - b)" given by?

What do you mean "what is it given by"? Do you mean, how is it derived?

 

[Alien8]

What do you mean "what is it given by"? Do you mean, how is it derived?

Yes, how, where from, or based on what it is derived.

 

[atyy]

Yes, how, where from, or based on what it is derived.

You'll have to learn quantum mechanics to understand the quantum prediction. The quantum prediction is derived in chapter 4 of John Preskill's lecture notes. http://www.theory.caltech.edu/people/preskill/ph229/#lecture

To learn some quantum mechanics you can start with Braam Gaasbeek's http://arxiv.org/abs/1007.4184 or a standard text like Zettili https://www.amazon.com/dp/0470026790/?tag=pfamazon01-20 or Griffiths https://www.amazon.com/dp/0131118927/?tag=pfamazon01-20 or Rae https://www.amazon.com/dp/1584889705/?tag=pfamazon01-20.

 

[Alien8]

You'll have to learn quantum mechanics to understand the quantum prediction. The quantum prediction is derived in chapter 4 of John Preskill's lecture notes. http://www.theory.caltech.edu/people/preskill/ph229/#lecture

To learn some quantum mechanics you can start with Braam Gaasbeek's http://arxiv.org/abs/1007.4184 or a standard text like Zettili https://www.amazon.com/dp/0470026790/?tag=pfamazon01-20 or Griffiths https://www.amazon.com/dp/0131118927/?tag=pfamazon01-20 or Rae https://www.amazon.com/dp/1584889705/?tag=pfamazon01-20.

I'm not ready for all that, too many unknowns. Let's get back to the integral for the classical prediction:

Shouldn't we be puzzled by this similarity between 1/2 cos2(a-b) and cos2(a-b)? Local theory ended up with (a-b) term even though a and b should be oblivious to one another. It's interesting, the same function only squashed in half. I mean, whatever non-local magic QM describes the mechanics of it is somehow captured by this Malus' law integral. Are we sure we integrated that right? If it's integrated from 0 to 2Pi but only over 1/Pi then the curve stretches out to cos2(a-b):

http://www5a.wolframalpha.com/Calculate/MSP/MSP140920b44a6h65g2f07h00004d93g0a279789h43?MSPStoreType=image/gif&s=40&w=516.&h=68.

I'm not sure what would that practically mean, but isn't it possible that might actually be the proper way to integrate it?

 

[atyy]

I'm not ready for all that, too many unknowns. Let's get back to the integral for the classical prediction:

I'll let stevendaryl handle that, he's the expert. I'll help out if he doesn't reply, but I'm not so familiar with this particular "classical prediction". I think it is important to keep in mind that this "classical prediction" is just one example of a local variable theory, and there could be many others. A violation of CHSH shows that no local variable theory - even those we haven't explicitly constructed - will not work (except for some bizarre exceptions which we can worry about after you understand the main idea of the inequality).

 

[atyy]

Shouldn't we be puzzled by this similarity between 1/2 cos2(a-b) and cos2(a-b)? Local theory ended up with (a-b) term even though a and b should be oblivious to one another. It's interesting, the same function only squashed in half. I mean, whatever non-local magic QM describes the mechanics of it is somehow captured by this Malus' law integral.

If I understand stevendaryl's model correctly, this is in fact the quantum mechanical prediction for unentangled pairs, ie. without entanglement quantum mechanics does not violate the Bell inequality, and its predictions can be reproduced by a local theory. A local theory, and quantum mechanics without entanglement, is able to produce correlations between distant locations, because of correlations in the source. The famous example is that if I prepare a pair of socks, each pair having a different random colour, but both socks in a pair having the same colour, and send one to Alice and the other to Bob, they will receive socks with random colours, but their colours will always be correlated. In stevendaryl's example, the orientation of each pair of unentangled photons from the source is random from trial to trial, but within one trial both photons always have the same polarization. So the presence of correlation alone is not enough to rule out a local model. It must be correlation that violates a Bell inequality.

Are we sure we integrated that right? If it's integrated from 0 to 2Pi but only over 1/Pi then the curve stretches out to cos2(a-b):

http://www5a.wolframalpha.com/Calculate/MSP/MSP140920b44a6h65g2f07h00004d93g0a279789h43?MSPStoreType=image/gif&s=40&w=516.&h=68.

I'm not sure what would that practically mean, but isn't it possible that might actually be the proper way to integrate it?

Yes, stevendaryl integrated it right. You can change the nomalization to $\frac{1}{\pi}$ but you must also change the upper limit of the integral to $\pi$. In fact, since photon polarizations only vary from $0$ to $\frac{\pi}{2}$, his integral will work if you normalize with $\frac{1}{\pi/2}$ and change the upper limit to $\frac{\pi}{2}$.

 

[Alien8]

If I understand stevendaryl's model correctly, this is in fact the quantum mechanical prediction for unentangled pairs, ie. without entanglement quantum mechanics does not violate the Bell inequality, and its predictions can be reproduced by a local theory. A local theory, and quantum mechanics without entanglement, is able to produce correlations between distant locations, because of correlations in the source.

Yes, plain old unpolarized and nonentangled photons, just every two in each pair having the same polarization. Not sure why you call it QM prediction when it's standard Malus' law integration for unpolarized light vs one polarizer:

http://www5a.wolframalpha.com/Calculate/MSP/MSP40581d9f66hfh79ab24500005e9bi8g5bd9fdgi4?MSPStoreType=image/gif&s=53&w=203.&h=40.

Therefore average photon probability for unpolarized light to mark "+" event is P(+) = 0.5 or 50%.

Individually then per each pair we have:

$P_a(+) = cos^2(a - \phi)$
$P_b(+) = cos^2(b - \phi)$

Two independent events, so their joint probability is: $P_{ab}(++) = P_a(+)P_b(+)$, thus for each combination:

$P_{ab}(++) = cos^2(a - \phi) * cos^2(b - \theta)$
$P_{ab}(--) = sin^2(a - \phi) * sin^2(b - \theta)$
$P_{ab}(+-) = cos^2(a - \phi) * sin^2(b - \theta)$
$P_{ab}(-+) = sin^2(a - \phi) * cos^2(b - \theta)$

...which plugged into that same integral for ordinary unpolarized light yields expectation value based on average probability:

$E_{Malus}(a,b) = P_{ab}(++) + P_{ab}(--) - P_{ab}(+-) - P_{ab}(-+) =$

Yes, stevendaryl integrated it right. You can change the nomalization to $\frac{1}{\pi}$ but you must also change the upper limit of the integral to $\pi$. In fact, since photon polarizations only vary from $0$ to $\frac{\pi}{2}$, his integral will work if you normalize with $\frac{1}{\pi/2}$ and change the upper limit to $\frac{\pi}{2}$.

Yes, we should really normalize over the whole limit, but still, it's the same freaking function... only half of it, but exactly half of it! Originally the impression I had is that local and non-local interpretations would be inherently and utterly incompatible, so I find their direct proportionality quite surprising.

 

[Nugatory]

Originally the impression I had is that local and non-local interpretations would be inherently and utterly incompatible, so I find their direct proportionality quite surprising.

On the contrary, they make the same predictions in most cases. This is why it took decades after the EPR paper before someone (John Bell) discovered the conditions under which they make different predictions.

 

[atyy]

Yes, plain old unpolarized and nonentangled photons, just every two in each pair having the same polarization. Not sure why you call it QM prediction when it's standard Malus' law integration for unpolarized light vs one polarizer:

http://www5a.wolframalpha.com/Calculate/MSP/MSP40581d9f66hfh79ab24500005e9bi8g5bd9fdgi4?MSPStoreType=image/gif&s=53&w=203.&h=40.

Well, QM can also deal with plain old unpolarized and nonentangled photons, for which we do get Malus's law. In fact, we can even say that Malus's law applies in some sense to entangled photons if we use a technical tool in QM called the "reduced density matrix".

Yes, we should really normalize over the whole limit, but still, it's the same freaking function... only half of it, but exactly half of it! Originally the impression I had is that local and non-local interpretations would be inherently and utterly incompatible, so I find their direct proportionality quite surprising.

Me too! But in hindsight, the situation with unpolarized and unentangled photons is such that probabilities for detecting a photon are the same for all polarizer angles, and the probability of detecting a photon on both sides depends only on the difference between the polarizer angles. In other words, the source has rotational symmetry. The entangled state that is used in the quantum calculation is the Bell state, which also has rotational symmetry in the sense that it can be thought of as a pair of photons having the same polarization at some angle $\theta$ and simultaneously a pair of photons having the same polarization at the orthogonal angle $\theta + \frac{\pi}{2}$. This particular quantum state has rotational symmetry because one can use any angle for $\theta$ without changing the quantum state. Perhaps the rotational symmetry of this particular classical example and this particular entangled quantum example makes it plausible that the unentangled and entangled curves the same shape, but with larger correlations for the entangled case.

For reference, the rotational symmetry of the quantum state is seen in Eq 1 and 3 of http://arxiv.org/abs/quant-ph/0205171. I should stress that there are entangled states without this high degree of symmetry.