Saturday, June 06, 2009

Quantum Entanglement

There are three degrees of spookiness in quantum entanglement:

A ← ●● → B

1) Observer A, called Alice, will measure the spin of incoming particle. If it has spin, say, up then she knows that observer B, called Bob, will measure its counter-part particle’s spin to be down.This is like Alice tossing a coin, heads or tails, and say it comes down heads. Her counterpart Bob who is light years away on the other side of the galaxy, who is also tossing a coin, and now his must come out in this case as tails!

2) Not only that, but Alice can choose whatever axis, an infinity of possibilities, along which the particle spin must quantize as spin up or spin down. Now Bob’s particle, on the other side of the galaxy, must also quantize along the same axis. But how does it know that?!?

3) Furthermore, a third observer C, called Claude, might be moving towards either Alice or Bob. If Claude is moving towards Alice, then according to him, she is measuring first, and she decides along which axis the particle spin will quantize. However if at the same time another observer D, called Donna, is moving towards Bob, then according to Donna, it is Bob who makes the first measurement and he decides along which axis the particle will quantize. How can this be??!?? Which is the cause and which is the effect? Accordingly Quantum physics makes no distinction between the two cases.

Bell’s theorem

Assumptions:

(i) Logic
(ii) A, B, C are independent events (often called locality).Examples: A is up or down, B is head or tail, C is red or green, etc.

Derivation of Bell’s inequality

(1) N (A+, B-) = N (A+, B-, C+) + N (A+, B-, C-); since an object must have the characteristic C or not have it.

(2) So N (A+, B-) >= N (A+, B-, C-); since N (A+, B-, C+) cannot be smaller than zero.

(3) N (B+, C-) = N (A+, B+, C-) + N (A-, B+, C-); similar reasoning to step 1.

(4) So N (B+, C-) >= N (A+, B+, C-); similar reasoning to step 2.

(5) So N (A+, B-) + N (B+, C-) >= N (A+, B-, C-) + N (A+, B+, C-); adding inequalities 2. and 4. together

(6) But N (A+, B-, C-) + N (A+, B+, C-) = N (A+, C-); similar reasoning to steps 1. and 3.

(7) So N (A+, B-) + N (B+, C-) >= N (A+, C-); which completes the proof.

Experiment

A+ = right spin up at 00 ; A- = left spin up at 00
B+ = right spin up at 450 ; B- = left spin up at 450
C+ = right spin up at 900 ; C- = left spin up at 900

You put detectors at L and R (Prisms, polarizers, Stern-Gerlach apparatus, etc.)

L ← ●● → R

Case1: R has orientation 00 and L has orientation 450. Measure: N (A+, B-) = N1

L ← ●● → R

Case2: L has orientation 900 and R has orientation 450. Measure: N (B+, C-) = N2 <

L ← ●● → R

Case3: R has orientation 00 and L has orientation 900. Measure: N (A+, C-) = N3

NOW ACCORDING TO BELL’S INEQUALITY:

N (A+, B-) + N (B+, C-) >= N (A+, C-) (Equation 7 above)

This translates for the three cases as: N1 + N2 >= N3

From the experimental data, it turns out that this inequality is wrong (In Quantum physics language: Bell's theorem is violated). See http://perso.wanadoo.fr/eric.chopin/epr/aspect.htm

The conclusion is that either assumption (i) or (ii) is wrong, or both are wrong.

Wednesday, April 08, 2009