This post (together with the previous one) left me in a quite a bit of a confusion. How does this model with polarization vectors correspond to the old “amplitude distribution over a configuration space of «a photon here and a photon there»”? What are the configurations here, and when are they distinct? (And it seems I am not the only one who got confused by this.)
I think, I found the solution; the photons have a distinguishing property: spin. So, if configurations are more like “a photon with a +1 spin here, a photon with a −1 spin there...”, then it all fits nicely in the same model. And the amplitude distribution corresponding to the situation described in the article would be:
|a+> |b-> - |a-> |b+>
(modulo a constant factor). Where |p+> (a photon with a +1 spin at P) corresponds to the P=(1 ; i) and |p-> to the P=(1 ; -i) in the article’s notation. Of course, the math remains the same, but now I can see a bit more clearly the amplitude distribution and what are the distinct configurations.
Hmm, it’s nice that there is this pretty compact formulation for two
coupled but separately “unpolarized” photons. But, this still leaves
me with a question of how does one “unpolarized” photon (a photon
for which half of the squared amplitude would pass any polarized
filter) looks like?
I would guess that there is no such thing. We might be ignorant about
the photon’s polarization, but it does have some definite polarization
even before it passes any filter. Otherwise, it has to be in a
similarly tangled state with something (eg. its source). Hmm, how would I check this?..
You can construct a photon that will pass any linear polarization filter 50% of the time, by constructing a circular polarization photon. If you include arbitrary polarizing filters including elliptical polarization, then yes, you will need to have a 2-or-more-particle entangled state to get 50% regardless of filter.
Having 2 or more particles entangled is of course the overwhelmingly normal case. If you take photons from an incandescent lightbulb and attenuate the signal until you’re counting photons, then half of them will pass any polarizing filter you can construct (not counting inefficiencies in the polarizer, obviously).
You’re right. Such a thing is not expressible as a wavefunction, as an unentangled pure state. Entanglement with something else that you are ignorant of (unentangled with) is one way of getting the right statistics. So too is expressing it as an impure state in the density matrix formalism. Some people appeal to the “church of the larger Hilbert space”, saying that only pure states exist, and that system is entangled with other, unobservable ones.
This post (together with the previous one) left me in a quite a bit of a confusion. How does this model with polarization vectors correspond to the old “amplitude distribution over a configuration space of «a photon here and a photon there»”? What are the configurations here, and when are they distinct? (And it seems I am not the only one who got confused by this.)
I think, I found the solution; the photons have a distinguishing property: spin. So, if configurations are more like “a photon with a +1 spin here, a photon with a −1 spin there...”, then it all fits nicely in the same model. And the amplitude distribution corresponding to the situation described in the article would be:
(modulo a constant factor). Where |p+> (a photon with a +1 spin at P) corresponds to the P=(1 ; i) and |p-> to the P=(1 ; -i) in the article’s notation. Of course, the math remains the same, but now I can see a bit more clearly the amplitude distribution and what are the distinct configurations.
Hmm, it’s nice that there is this pretty compact formulation for two coupled but separately “unpolarized” photons. But, this still leaves me with a question of how does one “unpolarized” photon (a photon for which half of the squared amplitude would pass any polarized filter) looks like?
I would guess that there is no such thing. We might be ignorant about the photon’s polarization, but it does have some definite polarization even before it passes any filter. Otherwise, it has to be in a similarly tangled state with something (eg. its source).
Hmm, how would I check this?..
You can construct a photon that will pass any linear polarization filter 50% of the time, by constructing a circular polarization photon. If you include arbitrary polarizing filters including elliptical polarization, then yes, you will need to have a 2-or-more-particle entangled state to get 50% regardless of filter.
Having 2 or more particles entangled is of course the overwhelmingly normal case. If you take photons from an incandescent lightbulb and attenuate the signal until you’re counting photons, then half of them will pass any polarizing filter you can construct (not counting inefficiencies in the polarizer, obviously).
You’re right. Such a thing is not expressible as a wavefunction, as an unentangled pure state. Entanglement with something else that you are ignorant of (unentangled with) is one way of getting the right statistics. So too is expressing it as an impure state in the density matrix formalism. Some people appeal to the “church of the larger Hilbert space”, saying that only pure states exist, and that system is entangled with other, unobservable ones.