Photon polarization and 1/2 wave plates

[zincshow]

I have seen this explained but would appreciate comments as to whether I understand it correctly.

#1. A vertically polarized photon hits a 1/2 wave plate set at 90 degrees will not pass through since cos(90)^2 = 0

#2. 50% of vertically polarized photons that hit a 1/2 wave plate set at 45 degrees will pass through since cos(45)^2 = 0.5

#3. following #2, 25% of the photons from #2 will get through a 1/2 wave plate set at 90 degrees since cos(90-45)^2 = 0.5 and 0.5*0.5 = 0.25

The question: does this mean that the photons that get through #2 are polarized at 45 degrees and that the photons that get through #3 are now horizontally polarized with the same energy that they started with?

 

[DrChinese]

I have seen this explained but would appreciate comments as to whether I understand it correctly.

#1. A vertically polarized photon hits a 1/2 wave plate set at 90 degrees will not pass through since cos(90)^2 = 0

#2. 50% of vertically polarized photons that hit a 1/2 wave plate set at 45 degrees will pass through since cos(45)^2 = 0.5

#3. following #2, 25% of the photons from #2 will get through a 1/2 wave plate set at 90 degrees since cos(90-45)^2 = 0.5 and 0.5*0.5 = 0.25

The question: does this mean that the photons that get through #2 are polarized at 45 degrees and that the photons that get through #3 are now horizontally polarized with the same energy that they started with?

First, you should substitute "polarizing filter" for "half wave plate" in the above for it to be correct. A wave plate passes everything, and shifts the polarization by a fixed amount.

Second, and with that in mind, the answer is YES: They are now horizontally polarized.

 

[LostConjugate]

First, you should substitute "polarizing filter" for "half wave plate" in the above for it to be correct. A wave plate passes everything, and shifts the polarization by a fixed amount.

Second, and with that in mind, the answer is YES: They are now horizontally polarized.

I thought wave plates only change the phase, not the polarization.

 

[DrChinese]

I thought wave plates only change the phase, not the polarization.

It has the same effect in a lot of cases, keeping in mind that the entangled photons do not have a well defined polarization to change. A half wave plate should change an H> to a V>.

From Wikipedia:

"A wave plate or retarder is an optical device that alters the polarization state of a light wave traveling through it. A wave plate works by shifting the phase between two perpendicular polarization components of the light wave... "

 

[LostConjugate]

It has the same effect in a lot of cases, keeping in mind that the entangled photons do not have a well defined polarization to change. A half wave plate should change an H> to a V>.

From Wikipedia:

"A wave plate or retarder is an optical device that alters the polarization state of a light wave traveling through it. A wave plate works by shifting the phase between two perpendicular polarization components of the light wave... "

So it looks like according to the wiki article it does only change the phase, but this results in an interference that changes the expectation value of polarity to be perpendicular to the source.

In other words it does not change the polarity of any single photon, only the frequency. The combined effect of many measurements or a macroscopic (classical) measurement being shifted in polarization.

Do I have that right?

"A half-wave plate. Linearly polarized light entering a wave plate can be resolved into two waves, parallel (shown as green) and perpendicular (blue) to the optical axis of the wave plate. In the plate, the parallel wave propagates slightly slower than the perpendicular one. At the far side of the plate, the parallel wave is exactly half of a wavelength delayed relative to the perpendicular wave, and the resulting combination (red) is orthogonally polarized compared to its entrance state."

 

[DrChinese]

In other words it does not change the polarity of any single photon, only the frequency.

Generally, I don't think the frequency (wavelength) is affected by such plate.

 

[LostConjugate]

Generally, I don't think the frequency (wavelength) is affected by such plate.

Oh the phase I mean.