Posts Tagged ‘Quantum-Entanglement’

Video: Spooky Actions At A Distance?: Oppenheimer Lecture – David Mermin

Saturday, August 1st, 2009

Einstein's real complaint about the quantum theory was not that it required God to play dice, but that it failed to "represent a reality in time and space, free from spooky actions at a distance." I shall use the rhetorical device of a computer-simulated lecture demonstration (a cartoon version of recent experiments in Vienna) to explain both the appeal of Einstein's criticism and the remarkable act that the "reality" he insisted upon is nevertheless unattainable.

Global Notes

  • After doing the "Rhetorical Homework" problem I watch Ron Garrett's Google Techtalk again: Quantum Conspiracy again.  It is a very good build / second chapter to this David Mermin "Stuff Left Behind" presentation.  He explains the stuff left behind in terms of Von Neuman entropy and it is really helpful for understanding.

Video Notes


Rhetorical Homework Problem:  

Let H and V be polarization eigenstates for horizontally and vertically polarized photons, and let R and L be polarization eigenstates for two orthogonal  right and left directions at 45 degrees so that

H = 1/sqrt 2  (R + L)

V = 1/sqrt 2  (R - L)

Type-1 detectors measure polarization flashing red if a photon is polarized horizontally and blue if it is polarized vertically. Type-2 detectors behave the same way with respect to diagonal photons — I.E. they are given by simply rotating a type-1 detector through 45 degrees about the line joining it to the source of the three photons. 

The source produces three photons in the polarization state: 

       1/2  (HHH - VVH - VHV - HVV )  

so if all the detectors are type-1 an odd number must flash red.  Show by appropriate substitutions of (1) into (2) that if only one detector is type-1 then an odd number must flash blue.

Rhetorical Homework Solution

  • My Interpretation: R causes type-2 detector to flash blue.  L causes type-2 detector to flash red.
  • Assume position 1 is the has a type-1 detector and positions 2,3 have a type-2 detector
  • "Factor" position one out.  It's not really factoring but rather more like a probability statement saying given position 1 = something statement.

1/2  (HHH-VVH-VHV-HVV )

1/2[  H(HH-VV )-V(VH + HV) ]

In order to get things to work out correctly I have to deduce that: 

  -VV = +  -V,V     ……. I am not absolutely sure of this one.  I do know it was the only place to look for a color switch that I needed to arrive at a single blue in the final expression below.  See note % below.

  1/2[H(H,H+ -V,V )-V(V,H+H,V)] 

1/2[H(1/sqrt 2(R + L),1/sqrt 2  (R + L) + -1/sqrt 2  (R - L),1/sqrt 2  (R - L) )  - V(1/sqrt 2  (R - L),1/sqrt 2  (R + L) + 1/sqrt 2  (R + L),1/sqrt 2(R - L))]    

                                                                        1/2[  H(1/sqrt 2  2L,1/sqrt 2  2R)  - V(1/sqrt 2  2R,1/sqrt 2  2R) ]

With the two measurement sets reading:             (Red    Red    Blue)               (Blue      Blue     Blue)

You get the same result for shifting the type-1 detector to any of the positions 1,2,3. 


Notes

  • each one of the 4 terms is an eigenvector. So if all detectors are type 1, an odd number of boxes must flash red. 
  • QUESTION:  Why must an odd number flash red?  ANSWER: Because when measured the system collapses in to one of the eigenstates.  The eigenstates shown all have an odd number of H  – horizontal polarizations.  Thus they pass through the red flashing direction of the crystal.
  • You can see they are entangle because not all possible states are represented.  If the photons were independent you should see HHH, HHV, HVH, VHH, HVV,VHV, VVH, VVV  
  • %- A vector [a,b] is perpendicular to [b,-a] & [-b,a].  I assume that in this notation -vv in this manner means a 90 rotation.