Does a polarisationfilter do a measurement?

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SUMMARY

A polarization filter does indeed measure a photon by placing it into a polarization eigenstate upon passing through. The measurement is completed with the aid of a detector, confirming that the photon is in the eigenstate. This process is governed by quantum mechanics, where the measurement occurs in a normalized basis, and the photon can be prepared in a superposition of eigenvectors. The principles discussed apply not only to photons but also to spin-1/2 particles, as both are described by a two-dimensional Hilbert space.

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entropy1
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Consider a measurement of a photon after it has passed a polarisationfilter. Does the photon jump in a (polarisation-)eigenstate by passing the filter? Does the filter do a measurement? Is the filter part of the entire measurement?
 
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entropy1 said:
Consider a measurement of a photon after it has passed a polarisationfilter. Does the photon jump in a (polarisation-)eigenstate by passing the filter? Does the filter do a measurement? Is the filter part of the entire measurement?

Ordinarily, the filter places the photon into a polarization eigenstate as you say. You could say that it also takes the detector to complete the full measurement. Technically that would be accurate.

Certainly, a 2nd polarizer (or PBS) after the first would then allow you to see that the 1st did in fact place the photon into the eigenstate.

So to a certain extent, what you call it is a matter of taste in this case.
 
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If I understand correctly, a measurement is made in a normalized basis. The measured particle eventually measures as a projection along one of eigenvectors of that basis. However, the particle can also be prepared in a superposition of eigenvectors of the measurement basis. (correct me if I'm wrong)

If I'm correct: consider a two-dimensional basis. Is a particle that is prepared in a superposition of eigenvectors of the basis in an angle α always measured with a distribution of cos2(α) resp. sin2(α) over several measurements? (such as is the case with polarized photons and polarizers?)
 
Exactly, and a polarization filter indeed does prepare photons of a given (linear) polarization state. Of course, nothing jumps since the time evolution in quantum theory is a smooth process governed by the equations of motion for state (statistical) operators and that of the operators representing observables. There are no quantum jumps according to quantum theory!

Of course, FAPP you can often just idealize a polarization filter as a projection operator and neglect the transition time of the state vector, because it's tiny compared to the time scales we resolve in such experiments.
 
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Does the formula I mentioned also hold for, say, measuring electron spin? In other words: does it hold for any particle? (not just for polarisation and photons?)
 
It's true for spin-1/2 particles and photons or for any system that's described by some two-dimensional Hilbert space like the spin component of spin-1/2 particles and the polarization of massless (relativistic) quanta, which both are described in a two-dimensional Hilbert space.
 
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