Polarisation Rotators and Measurement

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SUMMARY

This discussion focuses on the behavior of entangled photons when subjected to polarization rotators and polarizing beam splitters (PBS). It establishes that upon detection in the |H> or |V> basis, the photon is indeed in that eigenstate, resulting in a collapsed wave function. The conversation highlights that a circularly polarized photon has a 50% chance of passing through a |H> eigenstate. Additionally, it raises questions about the output polarization of entangled photons when one photon is measured, affecting the state of the other photon before it reaches a PBS.

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  • Understanding of quantum mechanics and wave function collapse
  • Familiarity with polarization states |H> and |V>
  • Knowledge of polarizing beam splitters (PBS) and their function
  • Experience with entangled photon pairs and their measurement
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  • Study the principles of quantum entanglement and measurement theory
  • Learn about the operation and applications of polarizing beam splitters (PBS)
  • Explore the effects of polarization rotators on photon states
  • Investigate the implications of wave function collapse in quantum mechanics
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Quantum physicists, optical engineers, and researchers working with entangled photons and polarization measurement techniques.

StevieTNZ
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When we use polarisation rotators to measure certain polarisations for entangled photons, the photon usually meets a polarised beam splitter after.

When we detect in the |H> or |V> basis, does that mean the photon is in |H> or |V>? What has happened to the polarisation we were measuring? Shouldn't it pass/fail into that polarisation?
 
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Any photon that comes out of a polarisation filter, splitter or other polarising device will be in an eigenstate of that basis, i.e. it will have a collapsed wave function. That does not have to be true for the photon/wave function going into the polariser.

For example, a circularly polarized photon contains both |H> and |V> components. It will have a 50% chance of passing a |H><H|filter and coming out as |H> eigenstate.

But please don`t ask me what happens to the entangled photon...
 
(1) I wonder: If we have a pair of entangled photons (entangled as HH+VV), we send photon #2 to a polarising rotator orientated at 22.5 degrees. Prior to photon #2 reaching the subsequent PBS (in the H/V basis) - so all it has done is gone through the polarising rotator - we measure photon #1, also with a polarising rotator orientated at 22.5 degrees. Photon #1 also reaches the PBS and we detect the photon as H polarised.

Because the input of photon #2 into the polarising rotator was undefined at the time (photon #1 hasn't been detected as H), we get an undefined output. Now that photon #1 has acquired H polarisation, does the output at the polarising rotator that photon #2 went into now become: input H polarisation -> output 45 degree polarisation?

(2) In another case, if we had two photons entangled as HH+VV, and we send both through quarter wave plates, then only one goes onto PBS in H/V, do we write the outcome as H(output) + V(output) OR HH + VV, where (output) is the output polarisation of the 2nd photon not yet reached a subsequent PBS? Or because photon #1 takes on H polarisation, we write photon #2 as taking on polarisation H as well (even though it hasn't reached a PBS)?
 

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