# Entanglement and polarization

1. Aug 7, 2008

### daniel29208

In photon polarization placing a horizontal polarizer in a beam of randomly polarized photons allows only half the photons through.

Question 1: Say, you have twin beams of entangled photons orthagonally polarized. If you place a horizontal polarizer in a one beam does it cut the intensity of the beam in half? Or, does it maintain its intensity and only "sets" the polarization of both photons.

Question 2: If you then place a vertical polarizer in the same beam does it eliminate the beam (beams?) altogether?

Thanks

2. Aug 8, 2008

### wawenspop

The thing about these polarizers is to remember that when the photon is between polarizers - (just travelling along in space - not being observed) then it has not got an exact polarization, its got a probabilty of having one -if it were to be observed or filtered. This is where things differ from classical behaviour. In fact all the states are probabilties only, until observed, when a value reveals itself (when its 'forced' to give a single answer).
In the case of two polarizers at 90 degrees to each other, then there is zero probabilty of a photon going through - because no photon has a probabilty of being at the orthogonal angle (90 degrees). But change the polarizer angles just a little and there is an easily calculable probability that some will pass through. (see Leonard Susskind, Stanford uNIVERSITY, on youtube for the mathematics of this - I forgotten which video- he's done lots, sorry).

These polaroid filters (eg sunglasses do not alter the polarization angle, they just filter). But slip a piece of transparent tape between the orthogonally set filters and the light passes once more because clear sellotape actually rotates the polarization of all photons by 90 degrees.

3. Aug 8, 2008

### clem

1. The other beam's intensity and polarization are unchanged. It's vertical photons are still correlated with the continuing horizontal photons in the first beam.

2. A vertical polarizer following a horizontal polarizer in the same beam will eliminate any beam. If the crossed polarizers are in different beams, each beam will have half intensity and be polarized. They will still be correlated.

4. Aug 8, 2008

### DrChinese

1. Once a measurement is performed on a photon - such as determining polarization - it continues to act "as if" it has that polarization until any subsequent observation. (Not sure if that is different than stated above or not.) But if you determine a photon is vertically polarized, all subsequent observations will yield the same results with certainty. This is true even if the measurement is on an entangled twin.

2. Polaroid filters definitely change the polarization of light which passes through. Obviously any filter has a frequency range at which it operates efficiently, and sunlight operates over a range. Even with a cheap polaroid filter (say from sunglasses): If you shine a flashlight between 2 that are crossed (i.e. perpendicular), little light will emerge. Put a third filter in between at 45 degrees to both, and the light beam will significantly increase. That is due to the effect of changing the polarization of the light as it passes from filter to filter.

5. Aug 9, 2008

### moving_on

I suppose this is kind of what I was getting it on the other thread...

6. Aug 10, 2008

### wawenspop

Hello DrChinese, nice to meet you, I've seen your website - its great

I am assuming that polarization is a state vector and then use QM mathematics (this is debatable because the spin of a photon is definitely a state vector but the polarization is only related to that, so some say, its not a state vector - any clues here anyone?).

Assuming it behaves quantumly, then when a polarized wave is 'observed' then QM calculates (very well) its probabilties of passing through another filter.
The result is: P = cos squared (alpha - theta) where alpha is the polaraization angle of the wave and theta the orientation of the polaroid filter.

If the second filter is aligned exactly then a photon has a probability of 1 of passing through, if the second filter is 90 degrees then it has P of 0 of passing. If its 45 degrees, then the photons has a .5 probability of passing.

Hence for orthogonally placed filters with a third filter between at 45 degrees?: This combination will pass 0.5 times 0.5 which is 0.25 of the photons. (The middle filter passes 0.5 then the third filter passes 0.5 of that).

If anyone wants the QM derivation of this I am only too pleased to add it here and discuss it.

A polaroid filter can be thought of as a 'grating' which can only pass one polarisation angle of EM waves. So anything getting through must be aligned with it. But a wave polarised at say, 45 degrees, has a probability of having photons aligned to the filter - 50% at 45 degrees. AFIK it does not rotate the polarisation plain. E.g. polaroid sun glasses. But many plastics especially under stress do really rotate the angle. The best one I know is clear sticky tape (you know it, its the very very clear type!). Try interposing that between two orthogonally crossed polarisers and its amazing and beautiful to see what it does. Pure QM at work!

Pls argue with this, - I am not using the classical Malus stuff, purely QM, as I believe that's what polarizers need to explain them correctly - but am willing and looking forward to be 'educated' about what's really correct.

7. Aug 10, 2008

### DrChinese

Thanks for the compliment about the web site!

I am not sure there is any disagreement here. The polaroid filter causes the photon to take on a now known alignment. If it was known to be aligned at a specific angle previously, it will have changed once it passes through the filter. So I would not call it rotating so much as resetting.

I guess it makes sense to state that the polarization exists as a probability (which can be 1) once it has been observed. I.e. once it passes a filter, its behavior at a similarly aligned filter can be predicted with near certainty. Any other angle relative to that would be a lower probability per Malus' Law.

I guess the point I was trying to make is that once a photon passes a filter, any previous information about polarization of that photon is lost and replaced with information from the alignment of the most recent filter. There is no "memory" of prior quantum observables once a subsequent non-commuting observation takes place.

8. Aug 10, 2008

### wawenspop

There's an important point here:
After a photon leaves a filter, its angle of polarization is a 'ray in a wave
packet' or a 'probability line'. If we observe it then a value is revealed,
but as its travelling its a combination of many angles at once.

Say, if it were an electron travelling along between two devices. Then we
cannot say that the electron has a position exactly, or a momemtum or indeed
a spin. It has probabilities of them but no actual real values, or, said another
way, it has all values at once.

I see so many references to photons having an exact angle of polarization, that
sometimes I doubt the QM version of events. Yet Bell's Theorem, Alain Apspect's
experiment and especially Quantum Computing reinforce the QM side of it. QBits
being 0 and 1 simultaneously is crucial to QC. And thus the angle of polarization
would be many at once too for the same reason.
Or have I slipped up here and indeed the angle is completely fixed on its path
from one filter to another?

9. Aug 10, 2008

### moving_on

This agrees with my understanding, a probability density function centred on the
alignment of the filter, i.e. a 'bell curve' with the top centred on the alignment of the
filter, tailing off to zero at the orthogonal alignment.
However, if the photon (say) passes a filter at 45 degrees to the first, it 'resets' this function to align to the filter it just passed.
This is why placing a filter at 45 degrees between two orthogonal filters allows photons
(say) to pass whereas they would not without it. Is this correct?

10. Aug 11, 2008

### DrChinese

Any non-destructive observation of a photon - say its polarization or wavelength, etc. - should yield identical values on repeated measurements consistent with the HUP. If the first measurement is done with great precision, then the uncertainty (deviation) in subsequent readings should be negligible and almost entirely due to experimental setup. In that respect, I think of the value as being fixed and well-defined. But other non-commuting observables would be completely undetermined and ill-defined.

So I guess the question - to me - comes down to what we know at point in time T1 after the first observation. If we knew something then, that should also be true at T2 upon second observation - assuming an identical observation. Change what you observe, and all bets are off if it does not commute with the first observation.

Now what happens in between T1 and T2? I guess you can say anything depending on your (philosophical) perspective. I think this is referred to as unitary evolution, not sure if this is the appropriate application of the term.