Entangled photons in bell experiment: transfer phase or angular momentum?by Iforgot Tags: bell, entangle, quantum information, quantum optics 

#37
Nov1312, 03:25 PM

Mentor
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To everybody to whom this might apply: please knock it off with the personal remarks and cattiness. If you have a personal dispute with someone, carry it on privately and not in public.




#38
Nov1312, 04:35 PM

P: 915

Thanks Dr. Chinese and Cthugha for being patient and taking the time to answer questions and providing good knowledge & information.




#39
Nov1412, 07:59 AM

Sci Advisor
P: 1,563

By the way wave plates and all the stuff acting solely on phase without probabilistic elements are also deterministic in other interpretations. You can create such a superposition of say +3/2 heavy hole /1/2 electron and 3/2 heavy hole/ +1/2 electron spin for example. These might precess in a magnetic field. However, there still is no measurement. You cannot know whether such an absorption process took place without measuring the eh pair or maybe exciton and collapsing it to some eigenstate. 



#40
Nov1412, 11:31 AM

P: 101

1) "What you get if you shoot a single photon at the prism is a superposition state of having one linear polarization and the photon being transmitted and having the other polarization and the photon being reflected. "
To make sure we are on the same page you are saying the photon state after GT prism interaction is some like ψ = (1>+1>)e^k1*x +constant*(1>1>)e^k2*y where the 1> and 1> are spin eigenstates), k2 and k1 are the wave vectors along the x and y directions respectively. (I know, I know, the reflected and transmitted rays have a colinear component) We could perform calculations to determine if that is true. If the CP photon doesn't exchange angular momentum with the GT prism, the super position state you describe would still have angular momentum. And for consistency, I would also show that calculations on a 45 degree polarized photon that is split into the state you described doesn't have any angular momentum. Let me know if you agree with the above before I start going about doing the tedious equations. I lose your train of thought starting at "the prism is certainly not in an eigenstate of angular momentum".... What's this very light polarizer you speak of? 2) Detected. No! You cannot know whether such an absorption process took place without measuring the eh pair or maybe exciton and collapsing it to some eigenstate. I thought that's how PN junction photodetectors work? Once the carriers are excited, the field across the PN junction (or the bias) sweeps the carriers to the electrodes for subsequent detection stages, e.g. a current amplifier/preamplifier. Yes, precession requires a field. I meant to say a linear polarized photon (with wave vector along Z), would excite an electron with spin in a super position of z up and z down. We can show that an electron in such a super position is an eigenstate of the Pauli spin matrix along x or y. 



#41
Nov1412, 12:31 PM

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As you seem to be interested in semiconductor physics: This can also be used in coherent control schemes. If you shine two short laser pulses (polarized horizontally and vertically, respectively) on e.g. a semiconductor quantum well, you will notice a huge increase in the degree of circular polarization of the excitons in the quantum well. This results from the coherent interaction of the second pulse with the coherent excitonic population created by the first pulse. Anyway: relative phase matters. 



#42
Nov1412, 02:32 PM

P: 101

1) Ahh, I forgot the "i"s in the equation. Other than that, I don't see what's wrong with the equation.
Yes I agree circularly polarized light can be expressed in a linear basis with a pi/2 phase shift. "whether you add the two linear polarization states coherently or incoherently" you mean putting the "i"s in the correct place/ properly accounting for the phases? Could you write ψ for the photon explicitly for after the GT prism photon interaction. That would clear things up. Regardless how we choose to write the photon ψ for the post GT prism interaction, do we agree that without angular momentum transfer, it must conserve its angular momentum? I.e. no change in angular momentum from before and after GT interaction. This can be verified evaluating the angular momentum operator on ψ. 2) I just wanted to emphasize that it does not make much sense to speak of momentum transfer before performing any measurement on the system. A VERY light polarizing beam splitter might actually end up in a different state when a single quantum of angular momentum is transfered to it, so that might constitute a measurement. For any real polarizing beam splitter, this will not be the case. Ah! Another difference in our assumptions! I am of the opinion that we don't have to make a measurement or require the wavefunction to collapse into an eigenstate for there to be transfer of angular momentum. 3) Yes, that is part of how they work. But you need to detect the current somehow to know that there was some light. This will in any case break the superposition state and put the electrons and holes into some eigenstates. While the superposition still exists you cannot even know whether a photon was detected (and also you will not know that the superposition exists). I don't agree that current detection would break the electron carrier spin super position. a) I thought we agreed the electron is in an spin eigenstate along the xy plane. b) And even if it was in a super position, detecting the charge (say using a transistor based amplifier) wouldn't break the spin superposition. (I'm thinking as to how I can support these assertions) 



#43
Nov1512, 07:54 AM

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#44
Nov1512, 12:20 PM

P: 101

1)No, I obviously cannot do that because there is no generally accepted formalism for wavefunctions of photons. There have been some efforts in that direction, but the results are far from giving a complete picture. In quantum optics the quantities of interest are therefore probability amplitudes for certain events.
Ahh! I did not know this. I thought the dirac equation for a massless particle was generally accepted as providing the photon wavefunction. 2) Well, if you can be sure of the angular momentum beforehand... The problem with the missing wavefunction of photons makes things complicated again. ??? Making a photon with a definite angular momentum can readily be done with a polarizer and a qwave plate! 3) Hmm, so you insist on some kind of realistic interpretation of qm? Ok, but you are aware that every tenable realistic interpretation like the Bohmian one is nonlocal, right? I was under the impression that the Bohm approach was a strict adherence to the (dirac) schrodinger equations. That is, any modelling of a measurement had to be introduced in a rigorous fashion by introducing the potential contributed from the detector into the schrodeq. E.g. I can semiquantitatively show how an electron in the double slit experiment is localized to one of the slits when one introduces the detector as a potential term in the Schrodinger equation. (Have you seen this mathematical derivation before?) I have wanted to see the nonlocality arise out of the photon wavefunction by applying the appropriate operators. But in light of your comment in 1) this is not a trivial task. 4) Anyway: relative phase matters. I couldn't agree more. My original post was an attempt to draw an analogy between a) super luminous phase velocity of light in medium with refractive index less than 1 and the fact that it doesn't transmit information faster than light and b) relative phase determining polarization states in bell experiments that also can't be used to transmit information faster than light. Both of these strange properties (a & b) arise from some sort of phase term. This analogy led me to speculate that the underlying mathematical structure of both these phenomena might share other similarities. (Thanks for listening and responding. I've really appreciated having some one to talk to about this.) Oooh. My head is starting to spin. Conversation (meant to say conservation) of frustration is starting to kickin :) 


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