Video: Spooky Actions At A Distance?: Oppenheimer Lecture – David Mermin – and Rhetorical Homework Problem Solution

If you review the Rhetorical Homework Problem and have a different method of handling the math please leave a comment on how and if possible a link.

Research Links

• Bringing Home the Atomic World – Quantum Mysteries for Anybody
• Physics 6335 – Quantum Mechanics I  Course Information Fall 2021 -Cornell – Scalise

 

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

• Related video presentation on quantum computation uses the 3 entangled photons
• PDF: Mermin > Spooky Action at a Distance 
• spukhafte Fermwarkunge – spooky!
•  All the figures of the video talk:  Spooky action at a distance figures  Local Copy
• Mermin home page
• Quantum Computer Science – an introduction
• Solid State Physics
• Original EPR paper  NonLocal Copy
• Test setup = emitter / tester / detector
• use brick wall to verify the detector is actually detecting a property of the particle and not just a random output unrelated at all
• does the tester impose properties on the particle?
•  Can a particle have 2 different real properties? Appears it can not.

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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

Type-1 detectors measure polarization flashing red if a photon is polarized horizontally and blue if it is polarized vertically.  This detector is on the V,H axes.

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. This detector is on the R,L axes.

The source produces three photons in the polarization state: 

         

Thus if all the detectors are type-1 an odd number must flash red.  You can see that by counting the number of Hs in the state.  Going sequentially through the states you see HHH or HVV or VHV or VVH and counting reds results in 3 or 1 or 1 or 1 reds.

HOMEWORK PROBLEM: 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.

The single Type-1 detector needs to be fed H or V for us to know if it flashes Red or Blue.  The following factoring appears it will work.

Given we see Type 1 detector flash = Red  the left term is in effect

 

Substituting the axis transforms:

Inserted Note:  If you look at the states HH and VV as you do single particles the particle is interferring with itself just as with a 2 slit experiment with single particle.  HH and VV interfer with one another to eliminated the RR and LL terms in the result.

 

Given we see Type 1 detector flash = Blue  the right term is in effect

Substituting the axis transforms:

   Leave the factored V out in V,H reference frame for Type 1 detector

Thus the detectors flash a count of blue = 1 or 3.

 

Further Thoughts

It seems like the sum over paths approach makes this more understandable.  The following diagram demonstrates that cancellation only occurs when the input wave function is applied as a superposition.  That is to say we're in wave mode.  Only by taking the two outputs and summing do you see the cancellation.

When an H polarized wave enters a type 2 detector it can break R(ight) or L(eft).  The diagrams below detail each permutation of what can happen. But then you have to treat the input as a superposition of HH and VV and thus all the outputs of the HH and vV input diagrams have to be summed.

You only see cancellations by summing the output of both.  This emphasizes that "it" was a ghost going into the apparatus and it was a ghost when it came out.  Thus spooky actions at a distance should not be surprising.

The results seen here can be used on the simpler "basic case" used in introductory presentations like Quantum Entanglement and Spooky Action at a distance,   You can see the derivation here  The math is very similar and is derived here.

 

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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.