# QM newbie (me) asks <sigh> yet another question about Bell inequality experiments

1. Nov 17, 2009

### Neo_Anderson

Alain Aspect's experiment:

In the Alain Aspect experiment, the two entangled photons will indeed both go through their respective polarizers. I'm hip with that. I'm also hip with Aspect's diligence in making it a "two-channel" testing apparatus.

What I don't understand is why should the fact that each entangled photon going through it's respective polarizer a matter of violating Bell's inequality. Shouldn't it be common sense that an entangled pair of particles share a common state of polarization (even if it's circular polarization)?
I mean, when I visualize the Aspect experiment, I'm visualising the simultaneous creation of a correlated pair of photons--photons that are indistinguishable in all respects including polarization state--going through their polarisers as they should. After all, I'd think that one photon that goes through the polarizer, then its exact replica should likewise go through a different polarizer of the same orientation. Don't you think it would be freaky if one entangled photon goes through it's polarizer, while it's exact replica does not?

Knowing that, what makes the Aspect experiment so special? Why should it be considered evidence that favors something as far-fetched as non-locality? Shouldn't Aspect's result simply be considered merely the observation of two twin photons, created at the same time and with the exact same properties, and nothing more?

I just can't see the connection between photons created with the same polarization and non-locality. Help!

2. Nov 17, 2009

### JesseM

Re: QM newbie (me) asks <sigh> yet another question about Bell inequality experiments

The idea in all Bell inequalities is that you assume that the reason the particles give identical results whenever both experimenters measure the same property, even if they have a choice of which property to measure on each trial, is that the particles were created with identical "hidden" values for each possible property. Then, based on this assumption you can come up with some statistical statements about what you should see on trials where experimenters measure different properties, giving a Bell inequality that tells you something about the statistics...then you can show that in QM the Bell inequality is violated, meaning that your original explanation for how the particles always give correlated results when the same property is measured must have been wrong, it can't simply be a matter of the two particles being created with identical hidden states.

The inequality used in the experiment above is known as the CHSH inequality. I showed the reasoning behind a different Bell inequality, in which there are three possible properties you can measure for each particle rather than two, in this post:
And you can modify this example to show some different Bell inequalities, see post #8 of this thread for one example. As for the CHSH inequality, it's explained in the wikipedia article, but the proof may be a bit hard to follow there so I'll try to explain it in a step-by-step manner here. Instead of Alice and Bob having three boxes to scratch on their respective lotto cards, imagine they only have two boxes on their cards, and that we label Alice's two boxes a and a' while we label Bob's two boxes b and b'. When scratched, any given box will reveal either a cherry or a lemon. Once Alice and Bob have both found the fruit behind the box they choose, they can adopt the convention that a cherry is represented by a +1 and a lemon is represented by a -1, and multiply their respective numbers together to produce a single number for each trial (and that single number will itself be +1 if they both got the same fruit, and -1 if they got different fruits). Then we are interested in the "expectation value" for a given choice of boxes--for example, E(a,b') means the average result Alice and Bob will get after multiplying their numbers together on the subset of trials where Alice chose to scratch box a and Bob chose to scratch box b'. The CHSH inequality then states that if we define the value S by S=E(a,b) - E(a,b') + E(a',b) + E(a',b'), then $$-2 \leq S \leq 2$$.

As for the hidden states, there are 16 different possibilities (and here I am replacing each fruit with the number they've chosen to represent it, so a=+1 means that the hidden fruit in box a on Alice's card is a cherry):

1: a=+1, a'=+1, b=+1, b'=+1
2: a=+1, a'=+1, b=+1, b'=-1
3: a=+1, a'=+1, b=-1, b'=+1
4: a=+1, a'=+1, b=-1, b'=-1
5: a=+1, a'=-1, b=+1, b'=+1
6: a=+1, a'=-1, b=+1, b'=-1
7: a=+1, a'=-1, b=-1, b'=+1
8: a=+1, a'=-1, b=-1, b'=-1
9: a=-1, a'=+1, b=+1, b'=+1
10: a=-1, a'=+1, b=+1, b'=-1
11: a=-1, a'=+1, b=-1, b'=+1
12: a=-1, a'=+1, b=-1, b'=-1
13: a=-1, a'=-1, b=+1, b'=+1
14: a=-1, a'=-1, b=+1, b'=-1
15: a=-1, a'=-1, b=-1, b'=+1
16: a=-1, a'=-1, b=-1, b'=-1

In this case, define something like A(a,12) to mean "the value Alice gets if she picks a and the hidden state of the two cards is 12", so going by the above we'd have A(a,12)=-1. Similarly B(b',7)=+1, and so forth. And we can assume that there must be well-defined probabilities for each of the possible hidden states, which can be represented with notation like p(8) and p(15) etc.

Since the expectation value E(a,b) is the average value Alice and Bob get when they multiply their results together in the subset of trials where Alice picks box a and Bob picks box b, we should have: $$E(a,b) = \sum_{N=1}^{16} A(a,N)*B(b,N)*p(N)$$. Likewise, we should also have $$E(a,b') = \sum_{N=1}^{16} A(a,N)*B(b',N)*p(N)$$. Combining these gives:

$$E(a,b) - E(a,b') = \sum_{N=1}^{16} [A(a,N)*B(b,N) - A(a,N)*B(b',N)]*p(N)$$

With a little creative algebra you can see the above can be rewritten as:

$$E(a,b) - E(a,b') = \sum_{N=1}^{16} A(a,N)*B(b,N)*[1 \pm A(a',N)*B(b',N)]*p(N) - \sum_{N=1}^{16} A(a,N)*B(b',N)*[1 \pm A(a',N)*B(b,N)]*p(N)$$

The triangle inequality says that for any real numbers x and y, |x + y| $$\leq$$ |x| + |y|, so applying this to the above gives:

$$|E(a,b) - E(a,b')| \leq | \sum_{N=1}^{16} A(a,N)*B(b,N)*[1 \pm A(a',N)*B(b',N)]*p(N) | + | \sum_{N=1}^{16} A(a,N)*B(b',N)*[1 \pm A(a',N)*B(b,N)]*p(N) |$$

The triangle inequality also implies that the absolute value of a sum of terms is less than or equal to the sum of the absolute value of each term, so this gives:

$$|E(a,b) - E(a,b')| \leq \sum_{N=1}^{16} |A(a,N)*B(b,N)*[1 \pm A(a',N)*B(b',N)]*p(N)| + \sum_{N=1}^{16} |A(a,N)*B(b',N)*[1 \pm A(a',N)*B(b,N)]*p(N)|$$

Since A(a,N)*B(b,N) is either +1 or -1, and so is A(a,N)*B(b',N), then we can remove them without affecting the absolute value of each term, giving:

$$|E(a,b) - E(a,b')| \leq \sum_{N=1}^{16} |[1 \pm A(a',N)*B(b',N)]*p(N)| + \sum_{N=1}^{16} |[1 \pm A(a',N)*B(b,N)]*p(N)|$$

Since A(a',N)*B(b',N) must be either +1 or -1, [1 ± A(a',N)*B(b',N)] must always be non-negative, and likewise for [1 ± A(a',N)*B(b,N)]. And each p(N) is a probability, so p(N) must always be non-negative too. This means the absolute values on the right side of the equation are unnecessary, so we have:

$$|E(a,b) - E(a,b')| \leq \sum_{N=1}^{16} [1 \pm A(a',N)*B(b',N)]*p(N) + \sum_{N=1}^{16} [1 \pm A(a',N)*B(b,N)]*p(N)$$

Since $$\sum$$ (x ± y) = $$\sum$$ x + $$\sum$$ ± y, we can rewrite that as:

$$|E(a,b) - E(a,b')| \leq \sum_{N=1}^{16} p(N) + \sum_{N=1}^{16} \pm [A(a',N)*B(b',N)]*p(N) + \sum_{N=1}^{16} p(N) + \sum_{N=1}^{16} \pm [A(a',N)*B(b,N)]*p(N)$$

And since there are 16 possible hidden states, it must be true that $$\sum_{N=1}^{16} p(N) = 1$$, so:

$$|E(a,b) - E(a,b')| \leq 2 + \sum_{N=1}^{16} \pm [A(a',N)*B(b',N)]*p(N) + \sum_{N=1}^{16} \pm [A(a',N)*B(b,N)]*p(N)$$

And by the definition of the expectation values, this reduces to:

|E(a,b) - E(a,b')| $$\leq$$ 2 ± [E(a',b') + E(a',b)]

Or:

|E(a,b) - E(a,b')| ± [E(a',b') + E(a',b)] $$\leq$$ 2

which implies

|E(a,b) - E(a,b')| + |E(a',b') + E(a',b)| $$\leq$$ 2

And again using the triangle inequality we get:

|E(a,b) - E(a,b') + E(a',b') + E(a',b)| $$\leq$$ 2

...which is just the CHSH inequality. So this shows that if we assume there is a definite hidden state for the four boxes on the two cards given to Alice and Bob on each trial, then no matter what the probabilities of different possible hidden states (different possible combinations of fruits behind the four boxes), we'd expect this inequality to be respected. If the inequality is violated as in quantum mechanics, that means that picturing their measurements as just revealing preexisting hidden states cannot be correct.

Last edited: Nov 17, 2009
3. Nov 17, 2009

### Fredrik

Staff Emeritus
Re: QM newbie (me) asks <sigh> yet another question about Bell inequality experiments

I like the derivation in Isham's book: 215, 216. But your proof might be better for people who don't have the math skills of a typical QM student.

4. Nov 17, 2009

### Haelfix

Re: QM newbie (me) asks <sigh> yet another question about Bell inequality experiments

As an aside..

Bells theorem (and the inequalities) is so fundamentally simple, I hate all these long proofs of something that can be done by drawing a single picture.

The inequality is basically the following (completely equivalent to the other forms):

The number of things with property A but not B, plus the number of things with B but not C is greater than or equal to the number of things with A but not C. or N(A, ~B) + N(B, ~C) >= N(A, ~C)

Draw three mutually intersecting circles (representing the set of A, the set of B, the set of C). According to classical set theory, this inequality trivially holds true (simply denote regions in the circles and add them up according to the formula above, and check that it must be true).

Then to check that quantum mechanics violates this inequality, you simply take a 2 state spin system in the singlet state and plug and chug the above statement.

5. Nov 17, 2009

### zonde

Re: QM newbie (me) asks <sigh> yet another question about Bell inequality experiments

In short intuitive picture of two photons sharing the same polarization works fine when polarizers have angle differences 0°, 90°, 180° and 270° - exact positive and negative correlations. It works fine as well for no correlation angles 45°, 135°, 225° and 315°.
But it does not work well for all intermediate angles the most popular being 22.5° and 67.5°.
From that intuitive picture for these angles one would expect correlations like 0.5 and -0.5 (exactly between 0 and +-1) but they are more like 0.7 and -0.7.

6. Nov 17, 2009

### DrChinese

Re: QM newbie (me) asks <sigh> yet another question about Bell inequality experiments

The above replies cover the spectrum well. I would add the following:

The EPR paper took the same approach as you have. But that was long before Bell. As mentioned, the Bell Inequalities operate at certain specific angles; not all angles demonstrate the issue. Nobody noticed this problem prior to Bell, even though some of the greatest minds of history looked at it!

There are 2 elements to the issue to remember:

a) There must be agreement when Alice and Bob are set to the same angles; i.e. this is the so-called perfect correlations that occur when entangled particles are measured in the same manner. This demonstrates you have a system generating entangled particles! It also shows the upper limits of experimental efficiencies.

b) You test a Bell inequality at the specified angles which have the potential to violate the local realistic expectation value. (The CHSH inequality is commonly used.) In essense, this result will show that the correlation is not possible unless something "weird" is happening. The weird thing that is happening is that the Heisenberg Uncertainty Principle is operating on both particles as a combined system, not as if they are separate particles.

7. Nov 17, 2009

### Count Iblis

Re: QM newbie (me) asks <sigh> yet another question about Bell inequality experiments

Are there also proofs of Bell's inequality that address at least some physically reasonable aspects of superdeterminism?

8. Nov 17, 2009

### JesseM

Re: QM newbie (me) asks <sigh> yet another question about Bell inequality experiments

There are numerous Bell inequalities, in what sense can they all be considered "completely equivalent"? The one you refer to was the first one I believe, but it wouldn't be the one used for the experimental setup in the OP (and unlike the CHSH inequality I think it assumes a situation where we find that the experimenters always get perfectly correlated results on trials where they both measure the same property). In my post I showed the reasoning behind a simpler inequality before going into the proof of the CHSH inequality, and I also linked to post #8 on this thread where I showed the reasoning behind the inequality you mention.

9. Nov 17, 2009

### Fredrik

Staff Emeritus
Re: QM newbie (me) asks <sigh> yet another question about Bell inequality experiments

How do you use this to prove (9.34) on page 216 of the book I linked to in #3?

10. Nov 18, 2009

### Neo_Anderson

Re: QM newbie (me) asks <sigh> yet another question about Bell inequality experiments

However--and I understand how useful anologies can be in physics (trains/lightning flashes instead of the Lorentz transformations, for example)--I still can't wrap my mind around this Aspect experiment, as dull-witted as I may be.
You showed the equations, but they're (you'll have to admit) somewhat laborious to pour through. The triangle identity seems to be key for this Bell inequality (but that doesn't mean I see how the triangle identity or its expectation value fits into non-locality for the Aspect experiment...).
Looking at the equations again--and please understand my apprehension--a number may be dropped; a sign may be a - when it should be a +, etc, because there are so many of them, and so many summation symbols! For this reason, I'm hoping you can give me a more heuristic explination of the Aspect experiment.
Since photons do not have spin, I'm hoping we can keep the heuristic discussion restricted to the Aspect experiment only (for the time being; that is, until I get my understanding of non-locality up to speed).

We can express Special Relativity in terms of trains as refrence frames, as well as the lightning flashes (and the observer on the ground with his mirror as a different frame) and pretty much make this heuristic explination of Relativity fit in nicely with the Lorentz transformations. I am hoping we can do the same with the Aspect experiment, meaning Bob and Alice are out of the question.

Last edited: Nov 18, 2009
11. Nov 18, 2009

### Neo_Anderson

Re: QM newbie (me) asks <sigh> yet another question about Bell inequality experiments

Is this true you guys? If so, I have a very, very important question: Will the violation of Bell's inequality in the Aspect experiment hold up when only the polarzers are oriented in these four, distinct angles? Or does the violation hold up no matter what the orientation of the polarizers may be (provided the orientation angle of one poarizer equals the orientation angle of the other)?

12. Nov 18, 2009

### Count Iblis

Re: QM newbie (me) asks <sigh> yet another question about Bell inequality experiments

This is a derivation of a Bell's inequality for primary school children. We consider the usual experiment with entangled photons and count the fraction of events where the outcome on both sides differs (photon passes through one polarizer, does not move through the other polarizer). We can denote the outcome on both side as a string of bits, a one if the photon passes through the polarizer, a zero if it doesn't. So, we could have the following outcome:

Polarizer 1: 0,0,1,1,1,0,1,0,0,1,1,0

Polarizer 2: 1,0,1,0,1,0,1,0,1,1,0,0

And we see that 4 out of the 12 are different. Polarizer 2 is set to some angle alpha w.r.t. polarizer 1. Had polarizer 1 been oriented in the same way as polarizer 2, you would have had the same bitstring as recorded at polarizer 2 at polarizer 1. Had polarizer 2 been oriented at twice the angle and polarizer 1 oriented in the way polarizer 2 originally was, then the relative angle is still alpha, so you would have had the same fraction of disagreements. But at polarizer 1 you would have had exactly the same bitstring as recorded at polarizer 2 in its original setting.

We can think of the bitstring at polarizer 2 at angle alpha as obtained from the bitstring recorded at polarizer 1 in its original setting by adding some fraction of mutations. The bittering at polarizer 2 when it is at an angle of 2 alpha is then obtained by adding the same fraction of mutations again to the mutated bittering. Since adding a mutation to a mutation will undo the mutation, you get the inequality that says that the fraction of mutations at 2*alpha is equal or less than twice the fraction of mutations at alpha.

13. Nov 18, 2009

### JesseM

Re: QM newbie (me) asks <sigh> yet another question about Bell inequality experiments

But here you are not deriving the CHSH inequality which Neo_Anderson was interested in, but rather a different Bell inequality--in the CHSH inequality no assumption is being made that the two photons have identical hidden states so that if you measure both with the same polarizer angle you'll get the same result.

14. Nov 18, 2009

### DrChinese

Re: QM newbie (me) asks <sigh> yet another question about Bell inequality experiments

My I suggest the following if you are not already familiar with it:

http://drchinese.com/David/Bell_Theorem_Easy_Math.htm

This is based on Mermin's derivation and I think it is a fairly straight-forward complete proof - although I am biased because I wrote it.

Another trick based on the same proof you can play with a friend:

a. Label 3 coins as A, B and C. Your friend flips them, and you don't look at the result. Now pick any 2 to look at randomly (as you are best able). Write down a 1 if they are the same (both heads or both tails). Repeat enough times to see a trend: you should have a 1 at least 1/3 of the cases. The more times you do it, the more certain you will be that the result is at least 1/3.

b. Now, repeat this process and tell your friend to cheat by choosing the coins as being heads or tails in advance. But you still continue to pick the 2 to look at randomly as before. You will quickly see that the results are the same: 1/3 of the time, at a minimum, you will have a 1.

c. Suppose your friend now cheats even more: as soon as you pick which 2 coins to look at, your friend changes 1 of them so they no longer match (and you end up with a 0 instead of a 1). If your friend does that often enough, your percentage will drop below 33%.

As long as you select the 2 coins to look at randomly, without knowledge of whether they are heads or tails, you will ALWAYS get a 1 at least 33% of the time. The only kind of cheating that gets you a lower value is that of case c. In that case, your friend gets to change one of the coins AFTER you select which 2 to look at.

Now, what does this have to do with Bell? When the same experiment is done with entangled quantum particles, the result can be as little as 25%. And yet the setup is really no different than our coins. So the only way you can explain that is by assuming that your friend is "cheating" as in case c. In this case, the role of your friend is being played by Mother Nature. But if the 2 particles (coins) are separated by distances too far for Mother Nature to send a signal at light speed - which has been done in numerous experiments - then something really strange must be going on. That is Bell's Theorem in a nutshell. The solution is one of the following:

1. Mother Nature can act faster than light (FTL). This is also called quantum non-locality.
2. There is an observer dependence to the situation, and the results are valid only in the "context" of the actual observation performed. This is in keeping with the Heisenberg Uncertainty Principle. This is also called "contextuality", "non-realism" or "no hidden variables".

15. Nov 18, 2009

### DrChinese

Re: QM newbie (me) asks <sigh> yet another question about Bell inequality experiments

There are all kinds of angles in which the inequality is demonstrated. But there are other angles at which the effect does not appear. You can imagine that the angle settings usually selected are ones which violate the inequality by a large amount (rather than small or not at all).

An analogous situation occurs in tests of relativity. You don't see relativistic effects at low speeds. So if you want to see the effect, you need to look at something traveling an appreciable percentage of c.

16. Nov 18, 2009

### DougW

Re: QM newbie (me) asks <sigh> yet another question about Bell inequality experiments

Dr. Chinese.... I read the page you referenced, and have a few questions/comments.

At the end of the page, you say "Reality is somehow dependent upon how we observe it.", but I wonder if the word 'reality' is really correct in this context. Isn't what has really been proven that "Our future observations are somehow dependent upon prior observations."?

I don't think we can or should assume that our observations are all there is of reality. Has there been a distinct proof that more than one value of a measurement can not exist at one time? (I know this is drifting into a 'multiple universe' theory, but again, until someone has a distinct proof against this, I don't see why it is so quickly discarded).

When I first read EPR, I scratched my head and wondered why it didn't mention the possibility that the 'spooky' effect could indeed be 'local' if one considers that the effect is not on the measured particle, but on the observer. You can think of it as the observer's existence encompassing multiple measurement values prior to the actual observation, and then after the observation, the observer's existence is now limited to a single measurement value. Multiple Universe theory might speculate that there would be other 'versions' of the observer created by the measurement, one for each 'possible' value of the measurement taken. This concept would then satisfy Einstein's worry that the moon exists whether someone is looking at it or not. The restatement of this issue would be "By observing the moon, you limit your existence to a universe where the moon must have existed prior to your observation".

Anyway, I just think any discussion of Bell's or EPR should include something about the fact that Multiple Universe theory could eliminate the problem between QM and Einstein's classical approach to physics...

17. Nov 18, 2009

### DrChinese

Re: QM newbie (me) asks <sigh> yet another question about Bell inequality experiments

Sure, a few observations:

1. It is possible that "our future observations are somehow dependent on our prior..." so I wouldn't discard that as an acceptable interpretation. There are time symmetric interpretations, including a recent one called Relational Blockworld, which have this as an explicit feature.

2. Actually, there are such! In fact, quite a number (and I am excluding Bell). The group of such theorems is call "No-go". These demonstrate directly that realism is violated by QM. So either QM is wrong or realism is. The best is the GHZ Theorem, which has been experimentally verified. It explicitly assumes that there is simultaneous existence to non-commuting observables (spin components of 3 photons).

Zeilinger discusses this in sections 16.2 and 16.3 of:
http://www.drchinese.com/David/Bell-MultiPhotonGHZ.pdf

Other no-go theorems (some of which have been verified and some which have yet to be): Leggett, Hardy, Cabello. There are others but that all feature the same basic assumptions: that there is a reality beyond what the Heisenberg Uncertainty Principle (HUP) describes. This assumption turns out to be directly contradicted by QM in case after case. Locality is not an assumption in any of these, although there are locality issues in the actual experimental setups.

3. MWI (your multiple universes) is an interpretation which is generally considered compatible with Bell (and is usually considered local non-realistic). There are other interpretations that are considered compatible too, such are the Bohmian types (which are explicitly non-local) and mentioned in the article. My conclusion being on the web page that QM is supported and local realism (a la Einstein) is not. I was not trying to take a stand among the QM interpretations.

18. Nov 18, 2009

### Neo_Anderson

Re: QM newbie (me) asks <sigh> yet another question about Bell inequality experiments

An excellent reply, Dr. Chinese! Thanks. Your Easy Math link was of immense help in my understanding.
Isn't this entanglement teleportation similar in mechanics to the electron action of stimulated emission radiation? That the mechanism for both (entanglement teleportatiion and the raising of the ruby electron to the higher energy level), the mechanism is similar? And if this is the case, doesn't the mere act of the electron "jumping" to the higher energy level, then "jumping" back to the lower level, somewhat violate Special Relativity? I mean if an electron instantly "jumps" up and down energy levels in space, and instantly does so, then this violates Relativity on its own, no?

Last edited: Nov 19, 2009
19. Nov 19, 2009

### Neo_Anderson

Re: QM newbie (me) asks <sigh> yet another question about Bell inequality experiments

Also, I see this all the time. Can someone tell me why the photon states are always divided by the square root of 2? An example:

http://www.sciencedirect.com/cache/MiamiImageURL/B6TVM-45HFHW9-7-1K/0?wchp=dGLzVtz-zSkWb [Broken]​
[/URL]

Occasionally, they're divided by just 2, but mostly by the square root of two.

Last edited by a moderator: May 4, 2017
20. Nov 19, 2009

### JesseM

Re: QM newbie (me) asks <sigh> yet another question about Bell inequality experiments

Neo, just to be clear, are you interested specifically in the CHSH inequality or are you fine with looking at other Bell inequalities? DrChinese's "easy math" link was not about the CHSH inequality, and it is indeed a bit easier to show why the other inequalities are implied by local hidden variables theories (that's what I did in the first section of my post before I got into all the equations which were specifically about proving the CHSH inequality)