2
$\begingroup$

As I understand quantum mechanics, a particle does not have a particular value for a property until it's been measured. So for instance a photon does not have a polarisation, that is an electric field direction, until then.

Once it's been measured, further up/down measurements along the same axis will yield the same answer. This is consistent with a parallel polarised filter passing 100% of an incident polarised beam. It's also consistent with a slightly tilted filter passing essentially 100% of the beam, but rotating the plane of polarisation.

If the beam is incident on a crossed polar filter, then one expectation I might have is that each photon will be measured randomly as being left or right, and so produce a beam with no net polarisation, but full power.

However, what happens is that there is no power transmitted.

In the classical description of a polarised EM beam, the electric fields of the beam may be parallel or normal to some anisotropic conductivity of the polarising filter, which inhibits the transmission of waves with an electric field in the 'wrong' direction.

Questions

When a dipole emits a polarised EM wave, what is the 'measurement' that establishes the polarisation of the constituent photons? Or does the emission itself constitute a measurement?

At the photon level, what's happening at a polarising filter, where is my (obviously) wrong expectation wrong?

$\endgroup$
  • $\begingroup$ I believe the measurement is whether it makes it through the slit or not. If I'm understanding your question. It has been shown that no photons travel through a perpendicular slit compared to 100% through A parallel slit. It's also interesting that the percentages change by increments of 25% as the angle rotates. 90° =0%, 60° = 25%, 45° =50%, 30° = 75%, and 0° equals 100% $\endgroup$ – Bill Alsept Jul 4 '17 at 19:15
2
$\begingroup$

In the classical picture, when a polarized light beam falls onto a non-parallel polarizer we calc the projection of the E-field onto the direction of the polarizer. The square yields the transmitted power, $P\propto I \propto E^2$.

In qm we also take the projection, however, we need to change the interpretation: Consider a single photon which falls onto the polarizer behind which a photo detector is placed. The photon will be either transmitted or absorbed. Therefore, the projection must yield the probability amplitude for the transmission and not the E-field amplitude.

Regarding your second question: If the set-up is such, that the emitted photon must be polarized along one direction, then the probability amplitude to get a different polarization vanishes. Hence, there is no need to assume a measurement at the emission, your "just" have to insure that all other prob amplitudes vanish. However, it's much simpler to assume a measurement at the start.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.