Interference seen in a member of an entangled pair

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SUMMARY

The discussion centers on the implications of measuring the spin of entangled electrons in a Stern-Gerlach and double-slit experiment setup. It is established that measuring the spin of one electron affects the probability distribution of the other, but does not allow for faster-than-light (FTL) communication. The interference pattern observed in the right electron is contingent upon whether the spin of the left electron is measured. If the left electron's spin is measured, the right electron will not display an interference pattern due to the collapse of the wave function, which eliminates the superposition necessary for interference.

PREREQUISITES
  • Understanding of quantum entanglement and its implications.
  • Familiarity with the Stern-Gerlach experiment and its principles.
  • Knowledge of wave function collapse and superposition in quantum mechanics.
  • Basic concepts of interference patterns in quantum experiments.
NEXT STEPS
  • Study the Quantum Eraser experiment to understand the relationship between measurement and interference.
  • Explore the implications of wave function collapse in quantum mechanics.
  • Investigate the double-slit experiment with entangled particles for deeper insights into interference.
  • Read Anton Zeilinger's works on quantum entanglement and its experimental setups.
USEFUL FOR

Quantum physicists, researchers in quantum mechanics, and students studying the principles of entanglement and interference in quantum experiments.

  • #61
Have you studied at the actual mathematical analysis of how erasing the information in the apparatus is supposed to restore interference

Please see my post in the thread How do you determine that a particle is/was entangled?
 
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  • #62
One more try at realistic explanation.

Say the same experiment http://arxiv.org/abs/quant-ph/9810080
There they say "The coincidence peak was nearly noise-free (SNR > 100) with approximately Gaussian shape and a width (FWHM) of about 2 ns. All data reported here were calculated with a window of 6 ns."
If we draw coincidence graph not for all data but only for 45° coincidences and if it has not one peak but two peaks with equal height (conditional +- and -+ peaks) then ... well, case solved.
 

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