Klao comments on Entangled Photons - Less Wrong

8 Post author: Eliezer_Yudkowsky 03 May 2008 07:20AM

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Comment author: Klao 15 September 2011 10:41:29PM 0 points [-]

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?..

Comment author: wnoise 15 September 2011 11:44:51PM 0 points [-]

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.

Comment author: Luke_A_Somers 24 April 2012 02:25:46PM 1 point [-]

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).