# Are Bell’s Inequalities really violated by rotating polarizer experiments?

In a previous post on Bell’s Inequalities, I argued that the inequalities could be derived from two premises:

Premise 1: One can devise a test that will give one of two discrete results. For simplicity we label these (+) and (-).

Premise 2: We can carry out such a test under three different sets of conditions, which we label A, B and C.

Since a violation of a mathematical relationship falsifies it, and since tests on entangled particles are alleged to comply with these two premises yet the inequalities were violated, either one of these premises were violated, or a new mathematical relationship is required. In this post I shall present one argument that the experiments that involve rotating polarizing detectors, with the classic experiment of Aspect et al. (Phys. Rev. Lett. 49, 91-94, 1982) as an example, did not, as claimed, show violations of the inequality.

Before proceeding, this argument does not in any way deny entanglement, nor does it say anything about locality/non-locality. I am merely arguing there is a logic mistake in what is generally considered. To proceed, to clarify my definition of entanglement:

An entangled pair of particles is such that certain properties are connected by some rule such that when you know the value of a discrete property of one particle, you know the value of the other particle, even though you have not measured it.

Thus if one particle was found to have a spin clockwise, the spin of its entangled partner MUST be either clockwise or anticlockwise, depending on the rule or the particles are not entangled. That there are only two discrete values for properties such as spin or polarization means you can apply Bell’s Inequality, which for purposes of illustration, we shall test in the form

A(+) B(-) + B(+)C(-) ≧ A(+)C(-)

The Aspect experiment tested this by making a pair of entangled photons, and the way the experiment was set up, the rule was, each photon in an entangled pair would have the same polarization. The reason why they had entangled polarization was that the photons were generated from an excited 4P spin-paired state of calcium, which in sequence decayed to the spin-paired 4S state. The polarization arose from the fact that when decaying from a P state to an S state, each electron loses one quantum of action associated with angular momentum, and since angular momentum must be conserved, the photon associated with each decay must carry it away, and that is observed as polarization. There is nothing magic here; all allowed emissions of photons from electronic states involve a change of angular momentum, and usually of one quantum of action.

What Aspect did was to assign a position for the polarization detectors such that A was assigned to be vertical for the + measurement, say, 12 o’clock, and horizontal, say 3 o’clock for the – measurement. The position B was to rotate the detectors clockwise by 22.5o, and for C, to rotate by 45o. There is nothing magic about these rotations, but they are chosen to maximise the chances of seeing the effects. So you do the experiment and what happens? All detectors count the same number of photons. The reason is, the calcium atoms have no particular orientation so pairs of photons are emitted in all polarizations. What has happened is the first detector has detected half the entangled pairs, and the second the other half. We want only the second photon of the entangled pair detected by the first detector, so, instead at the (-) detector we only count photons that arrived within 19 ns of a photon registered at the (+) detector, then we find, as required, if the first detector is at A(+), then no photons come through at A(-). That was nearly the case.

Given that we count only measured photons, the law of probability requires A(+) =1; A(-) =0. The same will happen at B, and at C. (Some count all photons, so their probabilities are the ones that follow, divided by two.) There would be the occasional equipment failure, but for the present let’s assume all went ideally. This occurs because if we apply the Malus law to polarized photons, if the two filters are at right angles, and if working ideally, and if the two photons have the same polarization, and you only count photons at the second detector that are entangled with the first, there are zero photons going through the second filter. What is so special about the Malus law? It is a statement of the law of conservation of energy for polarized light, or the conservation of probability at 1 per event.

Now, let us put this into Bell’s Inequality, from three independent measurements, because the minus determinations are all zero: {A(+).B(-) + B(+).C(-)} = 0 + 0, while A(+)C(-) = 0. We now have 0 + 0 = 0, in accord with Bell’s inequality.

What Aspect did, however, was to argue that we can do joint tests and measure A(+) and B(-) on a set of entangled pairs. The proposition is, if we leave the first polarizing detector at A(+), but rotate the second we can score B(-) at the same time. Let the difference in clockwise rotations of the detectors be θ, thus in this example θ = 22.5 degrees. Following some turgid manipulations with the state vector formalism, or by simply applying the Malus law, if A(+) = 1, then B(-) = sin2 θ, and if we do the same for the others, we find,

{A(+).B(-) + B(+).C(-)} = 0.146 +0.146 while A(+)C(-) = 0.5 Oops! Since 0.292 is not greater or equal to 0.5, Bell’s inequality appears to be violated. At this point, I believe we should carefully re-examine what the various terms mean. In one of Bell’s examples (washing socks!) the socks undergoing the tests at A, B and C were completely separate subsets of all socks, and if we label these as a, b and c respectively, we can write {a} = ~{b, c}; {b}= ~{a, c}; {c} = ~ {a, b} where the brackets {} indicate sets. What Bell did with the sock washing was to take the result A(+) from the subset {a} and B(-) from the subset {b} and so on. But that is not what happened in the Aspect experiment, because as seen above, when we do that we have the result, 0 + 0 = 0. So, does this variation have an effect? In my opinion, clearly yes.

My first criticism of this is that the photons that give the B(-) determination are not those entangled with the B(+) determination. By manipulating things this way, B(+) + B(-) > 1. Previously, we decided that 1 represented the fact that an entangled pair was detected only once during a B(+) + B(-) determination, because the minus indicates “photons not detected”, but it has grown by rotating the B(-) filter. If we recall the derivation, we used the fact that B(+) + B(-) =1. Our experiment has not only violated Bell’s Inequality, but it violates our derivation of it.

Let us return to the initial position, The first detector vertical, the second horizontal, and we interpret that as A(-) = 0. That means that no photons entangled with those assigned as A(+) are recorded, and all photons actually recorded there are the other half of the photons, i.e. ¬A(+), or 0*A(+). Now, rotate the second detector by 90o. Now it records all the photons that are entangled with the selection chosen by A(+). It is nothing more than an extension of the first detector, or part of the first detector translated in space, but chosen to detect the second photon. Its value is equivalent to that of A(+), or 0*A(-). Because the second photon to be detected is chosen as only those entangled with those detected at A(+), surely what is detected is still in the subset A, and what Aspect labelled as B(-) should more correctly be labelled 0.146*A(+), and what was actually counted includes 0.854*A(-), in accord with the Malus law. What the first detector does is to select a subset of half the available photons, which means A(+) is not a variable, because its value is set by how you account for the selection. The second detector applies the Malus law to that selection.

Now, if that is not bad enough, then consider that the B(+).C(-) determination is an exact replica of the A(+).B(-) determination, but has been rotated by 22.5 degrees. Now, you cannot prove 2[A(+) B(-)]≧ A(+)C(-), so how can you justify simply rotating the experiment? The rotational symmetry of space says that simply rotating the experiment does not change anything. This fact is, from Nöther’s theorem, the source of the law of conservation of angular momentum, and conservation laws to arise from such symmetries. Thus the law of conservation of energy depends on the fact that if I do an experiment today I should get the same result if I do it tomorrow. The law of conservation of momentum depends on the fact if I move the experiment to the other end of the lab, or to another town, the same result arises. Moving the experiment somewhere else does not change anything physically. The law of conservation of angular momentum depends on the fact that if I orient the experiment some other way, I still get the same result. Therefore just rotating the experiment does not generate new variables. So, we have the rather peculiar fact that it is because of the rotational symmetry of space that we get the law of conservation of angular momentum, and that is why we assert that the photons are entangled. We then reject that symmetry in order to generate the required number of variables to get what we want.

Suppose there were no rotational symmetry? This happens with experiments involving compass needles, where the Earth’s magnetic field orients the needle. Now, further assume energy is conserved and the Malus law applies. If a thought experiment is carried out on a polarized source and we correctly measure the number of emitted photons, now we have the required number of variables, but we find, surprise, Bell’s Inequality is followed. Try it and see.

My argument is quite simple: Bell’s Inequality is not violated in rotating polarizer experiments, but logic is. People wanted a weird result and they got it, by hook or by crook.

## 8 thoughts on “Are Bell’s Inequalities really violated by rotating polarizer experiments?”

1. Dear Ian: Hmmmm… It seems unlikely to me the thousands of physicists who looked at Aspect’s experiment before giving him the Wolf Prize in physics (among other prizes) don’t know logic. Moreover, the experiment could be run one photon at a time, insuring B(+) + B(-) = 1.

The essence of the problem of the Quantum Entanglement is that it PROPAGATES. I tried to illustrate this in my drawing in:
https://patriceayme.wordpress.com/2016/05/09/entangled-universe-bell-inequality/
OK, it could be clearer…

Then it’s enough to see it in ONE case. Quantum Physics predicts the propagation, and says it happens in ALL cases.

• Dear Patrice: You cannot run it one photon at a time. The reason is the plane of polarization could be at any angle but the detection is discrete, and what is selected depends on probability. If you do it on three photon pairs you can get any result as each result is either 1 or 0. The photons collected at the second detector have to have the same polarisation as those at the first, otherwise the law of conservation of angular momentum is broken. You can only ensure B(+) + B(-) = 1 is if you have both detectors in the B configuration, but then the (-) has to be zero, as noted in the post. All of this has nothing to do with entanglement, propagation or what; it ONLY depends on the fact that you detect the polarisation of the photons, for whatever reason. And, as I say, simply rotating an experiment does not give you two new variables. If it did, it violates Nöther’s theorem.

• Dear Ian:
1) I produce my 2 entangled photon, a and b of total spin zero: if a(+) then b(-).
2) I rotate polarizer PA in the way of photon a in direction DA.
3) Photon a goes through PA. I find, say, DA(+). Event A.
4) I immediately know that b “has” polarization DA(-). Event B.
5) My gnomic associate rotate polarizer PC in the way of photon b, in direction DC.
6) Photon b runs through PC. I measure DC. Event C.

One photon at a time.

Repeat. And again, and again.

Statistical analysis then shows that that the distribution DC(+) versus DC(-) is dependent upon the angle (DA,DC).

So event A has influenced event C. Not just event B, as in Classical Mechanics.

2. Disagree with (1) You produce 2 entangled photons of total spin zero. Suppose the first one spins clockwise. You are saying the second one spins anticlockwise, which is true to a point, BUT in the Aspect experiment, you look at the second photon from the other direction, so from that perspective it also spins clockwise. If you detect the first one in the A plane, then the second is also in the A plane. Accordingly, If you record A(+), then the second should be A(-), and not detected. Because you are COUNTING only photons with the same polarisation as in the first detector, wherever you put the second detector, you detect the probability projection on the A plane.

Single photons are no use, because they only give a probability. You need enough to average out the odd probabilities. Event A does not influence anything relating to B or C because all it does is demand the probability of a projection into the A plane

• Hmmmm… I am confused. First polarization is detected (with a “plane”!) in a direction selected by Aspect’s assistant on a whim. Aspect does not “make” what I call event B, namely the knowledge of the other photon in the direction DA (D for direction). Because event B is a mental event. B is happening in the location where A happened, that is, somewhere else, where Aspect is not. B happens in Aspect’s assistant mind.

More confusion on my part: To say “single photon are no use” is like saying single photons are no use during the 2-slit interference experiment (the pattern is drawn one photon at a time, like any painting, or script, for that matter!).

BTW, I just made a (rather drastic) observation (purely philosophical) which shots down the Copenhagen Interpretation. I may put it on my site today.

3. Obviously I am not as clear as I hoped. The polarisation was detected by plane polarised filters; I am unclear as to exactly what these were, but if you stuck with circular polarisation, rotating the filters would make no difference at all: it would be either clockwise or anticlockwise. However, the event at Aspect’s B position IS dependent on A, because he only counts photons entangled with A, and entanglement means they must have the same polarisation, otherwise they would not be entangled. There are number of other photons detected at B, but those results were discarded.

They reason that the single photon is of no use is because there is a probabilistic aspect to this. Thus suppose the detectors are A(+) and A(-), i.e. at 90 degrees to each other, and one pair of photons were plane polarised, and at 45 degrees to each. Now, each photon has a 50-50 chance of going each way. So we have detection probabilities (y for yes, n for no) of yy, yn, ny, nn at each detector. It is possible get quite awkward results with one photon pair. Of course you could keep doing that, and statistically they average out, but then you r back to the Aspect situation.

Of course, if you COULD generate such a stream of plane polarised photons, you would have a test for non-locality: if the A(+) and A(-) photons detected one at a time ALWAYS behaved coordinated, this would seem to be non-local; if they were statistical, the entanglement would have to be local. The problem is to generate the plane polarised photon waves. I doubt starting with circular polarised waves and converting them when detecting is adequate.

4. Have a look at the pic in my post today. It’s related to part of the problem we have here.
I still maintain there are 3 events:
A measuring polarization in one direction DA (D for direction)
So polarization at distant event B is known in direction DA. But indeed, does not matter. That’s classical entanglement.
Then event C.
C consists, at distant position B, in measuring polarization along direction DC.
Now it turns out that the polarization measured at C depends upon the angle (DA, DC)!!!!!!!!!!!!!!!!!!!!!

Aspect ran three types of experiment, starting around 1982. What we are discussing here is the third type, around 1988, when he rotated the polarizers in flight, haphazardly.

Since 1988, all reasonable loopholes to Aspect’s third experiment have been closed.

• Now I am unsure what you are querying. There can only be two events for the detection of an entangled pair because there are only two photons. And yes, move the second polariser to C and the COUNT at C depends on the angle ADC (if I have your diagram properly.) As I read the Phys Rev letter, I am unsure what you mean by haphazard – to me he specifies the angles I quoted (or misquoted if I got it wrong, but I don’t think so.) What I understand this experiment usually does (and I have seen it demonstrated on a video clip, is you get a stream of photons in each of the three specified positions, one position at a time.

The important point I am hoping to make is that it does not matter how the photons get to where they are reported to be. To apply the inequality, all we need are the numbers of photons detected at each detector and they do follow the Malus law. After all, the Malus law applies if you have enough of them. If it did not, when does it? If it did not, there would be a violation of the law of conservation of energy, so where did the energy go (assuming you don’t believe it is not a law at all)?

A nice post about the Copenhagen Interpretation, by the way.