On 06-04-2022 08:29, Brent Meeker wrote:

On 4/5/2022 5:14 PM, smitra wrote:On 06-04-2022 01:14, Bruce Kellett wrote:On Wed, Apr 6, 2022 at 9:03 AM smitra <smi...@zonnet.nl> wrote:On 06-04-2022 00:21, Bruce Kellett wrote:On Wed, Apr 6, 2022 at 1:13 AM smitra <smi...@zonnet.nl> wrote:The conclusion that local hidden variables are rules out doesdepend onan argument about what would have happened had differentpolarizersetting been used than the ones that were actually used.That is false. Where in the Aspect experiments is reference madetonon-performed measurements?It's in the argument that local hidden variable theories must satisfy Bell's inequalities.You are stretching things quite a bit. All Aspect needed to do was show that his measured results violated the inequalities. No counterfactual reasoning involved. If you then want to argue that Aspect's results extend to all possible such experiments, then of course, not all those experiments have been done. But that does not impact Bell's theorem. Bell, himself, did not have to make any measurements.How does one conclude that a set of measurement results cannot beexplained by a local hidden variable theory if you don't invokecounterfactual reasoning to the performed experiment? Of course,Aspect could simply refer to Bell's theorem, but Bell's theorem reliesheavily on counterfactual reasoning. This is also why you can havesuperderminisitic loopholes to the argument against localdeterministic theories.SaibalAttached are my lecture notes on Aspect's test of the Bell inequalities. I don't see anywhere I invoke counterfactual reasoning, but I'd be happy to have it pointed out to me if I have missed it.

`The counterfactual reasoning is in the derivations of the inequalities.`

`Take e.g. the Bell inequality involving the angle and the double angle.`

`If you flip the sign of Bob's results relative to Alice's results then`

`for the same polarizer settings they get the same results. For a nonzero`

`relative angle we can consider the number of differences. Then we need`

`to invoke a counterfactual reasoning to argue that the number of`

`differences at twice the angle must be less than or equal to twice the`

`number if differences as the original angle. Suppose Bob keeps his`

`polarizer setting fixed and Alice changes it from theta to 2 theta. Then`

`the argument goes that her 2 theta results are what she would have`

`gotten at theta plus on average he same number of changes relative to`

`what she got relative to Bob at theta, because the relative angle`

`between 2 theta and theta is theta.`

`In general, if one has concluded via some reasoning that the value of a`

`correlation at some angle is constrained by the values at some other`

`angles (assuming local determinism) then that begs the question of how`

`the different experiments at the different angles can tell you something`

`about the results at that particular angle, if not for counterfactual`

`reasoning.`

`The GHZ experiment is another simple example, this involves 4 entangled`

`spins.`

We have an entangled state: |psi> = 1/sqrt(2) [|up, up, up> + |down, down, down>]

`Alice, Bob and Charlie receive one spin and then choose to measure`

`either the x or y-component. Using:`

sigma_x|up> = |down> sigma_x|down> = |up> sigma_y|up> = i |down> sigma_y|down> = -i |up>

`and with the notation: (A) (B) (C) |state> for tensor product of A, B,`

`and C so that A acts on the first component of |state> and B on the`

`second component, and C on the third component, we see:`

(sigma_x) (sigma_x) (sigma_x) |psi> = |psi>

`So, of Alice, Bob and Charlie measure the x-components, then despite`

`their results being random, the product of their results will always`

`equal 1.`

We also have: (sigma_x) (sigma_y) (sigma_y) |psi> = -|psi>

`So, if one person measures the x-component and the others measure the`

`y-component then the product of the results will always equal minus 1.`

`Argument against local hidden variables: We assume that if the ith`

`person chooses to measure the X component then the result will be Xi,`

`while it will be Yi if the Y component is chosen. This does not depend`

`on the choices made by the others.`

`For the cases where two people choose to measure the Y-component, we`

`then have the results:`

(X1, Y2, Y3): product = -1 (Y1, X2, Y3): product = -1 (Y1. Y2, X3): product = -1 Taking the product of these results then yields: -1 = Y1^2 Y2^2 Y3^2 X1 X2 X3 = X1 X2 X3

`But as shown above, if all 3 measure the x-components and multiply their`

`results, the result will always be equal to 1. This contradiction with`

`local hidden variables clearly follows from invoking counterfactual`

`measurement results.`

Saibal

Brent

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