User:MRFS/Photons

The poor man's Bell tests

Needless to say the conduct of a Bell test is a very costly business indeed. But the good news is that it's possible to run an accurate simulation in the comfort of one's own home at minimal expense. The only equipment required consists of two coins and an icosahedral die, commonly known as a D20 because it has 20 faces rather than the usual 6. The reason for using a D20 is that it can mimic the quantum probabilities to 2 decimal places, and thereby provide a reasonably faithful representation of the real Bell test.

So what are these probabilities and how is the test to be done? As described above the detector settings $a=0$ °, $a'=45$ °, $b=22$ ½°, $b'=67$ ½° are chosen because they give the greatest violation of Bell's inequality. According to quantum theory if the nominal angle between two settings is γ then the probability of agreement between two detectors at these settings is cos²γ. Thus the probabilities of mutual agreement are as follows :-
P( $a$ and $b$ agree) = cos²(22½°) = P( $a'$ and $b$ agree) = P( $a'$ and $b'$ agree) = 0.85 which is an 85% chance of agreement (1-17)
P( $a$ and $b'$ agree) = cos²(67½°) = 0.15 which is a 15% chance of agreement (18,19,20)
The D20 can exactly emulate these probabilities. 85% of the time it will roll a number from 1 to 17 inclusive. 15% of the time it will roll 18, 19 or 20.

As for the coins, instead of heads and tails one is marked $a$ and $a'$ for Alice's detectors. The other bears $b$ and $b'$ to show Bob's possible choices. A 'trial' will consist of tossing the coins and rolling the D20. First decide on the number of trials and create a blank table like the one below to accommodate the results. The results of 100 trials are displayed below.

Once the preliminaries are over the simulation can begin. Simply run a series of trials and for each trial record the value shown by the D20 in the appropriate cell.

D20 rolls for each detector pairing
$a$ and $b$	4	14	5	2	4	19	8	17	17	12	9	7	18	4	20	13	6	10	1	2	17	19	11	1
$a$ and $b'$	14	17	3	2	19	9	5	7	3	12	3	18	15	4	3	5	17	11	4	4	12	11	18	11	20
$a'$ and $b$	12	15	15	19	16	8	6	3	14	9	15	16	9	5	18	2	2	8	13	16	8	20	3	14
$a'$ and $b'$	11	7	2	7	3	12	6	1	18	1	13	14	17	7	19	7	20	10	1	20	8	12	15	18	8	11	6

The next stage is to replicate the table changing every entry to either an 'A' or a 'D' according to its value. In rows 1, 3 and 4 the probability of agreement is 85% so values from 1 to 17 inclusive qualify for an 'A'; values 18, 19, 20 are given a 'D'. But in row 2 the probability of agreement is only 15% so here 18, 19, 20 receive an 'A' and 1 - 17 get a 'D'.

Agreements and disagreements for each detector pairing
$a$ and $b$	A	A	A	A	A	D	A	A	A	A	A	A	D	A	D	A	A	A	A	A	A	D	A	A				$E(a,b)=16/24=0.67$
$a$ and $b'$	D	D	D	D	A	D	D	D	D	D	D	A	D	D	D	D	D	D	D	D	D	D	A	D	A			$E(a,b')=-17/25=-0.68$
$a'$ and $b$	A	A	A	D	A	A	A	A	A	A	A	A	A	A	D	A	A	A	A	A	A	D	A	A				$E(a',b)=18/24=0.75$
$a'$ and $b'$	A	A	A	A	A	A	A	A	D	A	A	A	A	A	D	A	D	A	A	D	A	A	A	D	A	A	A	$E(a',b')=17/27=0.63$

The final column of each row shows the average (or expectation) of the row. It is given by (N_A − N_D)/(N_A + N_D) where N_A and N_D respectively are the number of 'A's and the number of 'D's in the row. In terms of what has gone before N_A = N₊₊+ N_– – and N_D = N_+–+ N_–+ . It only remains to calculate the Bell signal

$S=E(a,b)-E(a,b')+E(a',b)+E(a',b')=0.67$ – ( – $0.68)+0.75+0.63=2.73$

which shows that the Bell/CHSH inequality has been well and truly violated. What a surprise!

However one rather puzzling aspect of this experiment remains. It has faithfully simulated the actual Bell test and has violated the Bell/CHSH inequality, yet it has somehow managed to avoid any mention of loopholes or hidden variables or nonlocality.

This simulation has concentrated on photons because they are the commonest experiments performed nowadays. With minor modifications it applies equally well to tests with electrons and Stern-Gerlach detectors, say Q, R, S, T replacing $a$ , $a'$ , $b$ , $b'$ . The Bell test angles are then 0°, 90°, 225°, and 315° and the probability of agreement between detectors nominally δ apart is sin²(δ/2). The end results are exactly the same as before.

Interested readers are encouraged to run their own simulations and draw their own conclusions.

The detection loophole

Much of the emphasis of the real test has centred on a possible loophole if the detectors aren't randomly chosen. As stated above, Rauch et al used signals from distant quasars to ensure complete independence but was this really necessary? Sure enough, after 45 minutes and 40 seconds of "Einstein's quantum riddle" a Bell signal of S = 2.6734 flashes briefly onto the screen showing a clear violation of the Bell/CHSH inequality just as predicted by both quantum theory and the poor man's test.

Timed Bell tests

Signals from quasars at opposite ends of the universe have been used in an attempt to eliminate this loophole, but is this really necessary? Suppose the test is run in four quarters of roughly equal duration. In the first quarter $a$ and $b$ will be active. For the second one Bob will deactivate $b$ and activate $b'$ . At half time Alice will switch from $a$ to $a'$ , and for the final quarter Bob will switch back from $b'$ to $b$ . Consider 100 tests in each of quarters 1, 3, 4. For each quarter the expectation is that there will be 85 agreements and 15 disagreements, leading to an expectation of 0.7 in each case. In quarter 2 the position is reversed. The expectation is 15 agreements and 85 disagreements leading to an expectation of –0.7. However $E(a,b')$ has a negative sign in the Bell signal so its contribution is actually 0.7. Adding the four terms in the Bell signal together gives 2.8, and once again it's clear that the Bell/CHSH inequality has been seriously violated.