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Entangled photons in bell experiment: transfer phase or angular momentum?

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jtbell
#37
Nov13-12, 03:25 PM
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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.
San K
#38
Nov13-12, 04:35 PM
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Thanks Dr. Chinese and Cthugha for being patient and taking the time to answer questions and providing good knowledge & information.
Cthugha
#39
Nov14-12, 07:59 AM
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Quote Quote by Iforgot View Post
Ah!!! Here's the meat of our disagreement. Transfer of photon angular momentum with the GT polarizer!

My claim:
1) Angular momentum transfer happens at the GT-prism. Proof: Well defined incident circularly polarized light becomes linearly polarized after passing through. Loss of hbar (helicity reversal by a 1/2 wave plate would be 2hbar)
This is not really exact. A GT-prism is more or less a polarizing beam splitter. You have s-polarized light at one exit port and p-polarized (or mostly p-polarized, depends on the prism design) at the other exit port. 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. You will of course know that a photon passing through will have some well defined polarization, but you will not know whether some photon went through the prism without performing a measurement. At that time you get angular momentum transfer (depending on what kind of light you used possibly also to the prism). However, as the prism is certainly not in an eigenstate of angular momentum, that change will most likely drown in uncertainty and you will not get the prism to another orthogonal eigenstate. Things would differ if you managed to create some very light polarizing beam splitter, where such a tiny change in angular momentum can change the state of the prism. That might actually be a measurement and momentum transfer might happen immediately.


Quote Quote by Iforgot View Post
(I'm trying to take a rigid Bohm approach here. I.e. wavefunctions evolve in a deterministic fashion determined by dirac schrodiner. Measurements never completely collapse the wavefunction. E.g. Optical elements (polarizers, q-waveplates, etc...) force a known previous wave-function into a new wavefunction in a calculable way (the Dirac Schrodinger eq))
Bohmian mechanics is deterministic, but the initial conditions are necessarily unknown to some degree, so you still have uncertainty and stuff. For example for a single photon state, phase is pretty much undefined. Bohmian mechanics is in some respects elegant as it distinguishes the particle and the guiding wave. Nevertheless, it also obeys the fundamental physical law for interpretations of physics: conservation of frustration. ;)
By the way wave plates and all the stuff acting solely on phase without probabilistic elements are also deterministic in other interpretations.

Quote Quote by Iforgot View Post
2) continued... Linear polarized photon on a semiconductor at the bandgap energy would excite 1 electron hole pair with a superposition of opposite spins. I.e. the spins would be precessing.

The linearly polarized photon would be annihilated without collapsing to a circular polarized state.
Annihilated? Yes. Detected? No!

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 e-h pair or maybe exciton and collapsing it to some eigenstate.
Iforgot
#40
Nov14-12, 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 e-h pair or maybe exciton and collapsing it to some eigenstate.

I thought that's how PN junction photo-detectors 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/pre-amplifier.

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.
Cthugha
#41
Nov14-12, 12:31 PM
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Quote Quote by Iforgot View Post

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)
Hmm, I do not think we are on the same page here. Let me get one step back and start at something easier. Do you agree that any circular polarization state can also easily be expressed in a linear basis? This is possible in terms of a superposition of the two linear states with complex coefficients, which is simply equivalent to a well defined phase shift of +/- pi/2 between these two basis states. I just wanted to emphasize that it makes a huge difference whether you add the two linear polarization states coherently or incoherently, the former is more than just the sum of its parts.
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.

Quote Quote by Iforgot View Post
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?
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.

Quote Quote by Iforgot View Post
I thought that's how PN junction photo-detectors 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/pre-amplifier.
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).

Quote Quote by Iforgot View Post
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.
Yes, I do not think we disagree here. Linearly polarized excitation can excite such a superposition.
Iforgot
#42
Nov14-12, 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)
Cthugha
#43
Nov15-12, 07:54 AM
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Quote Quote by Iforgot View Post
1) Ahh, I forgot the "i"s in the equation. Other than that, I don't see what's wrong with the equation.
Much, see below.

Quote Quote by Iforgot View Post
"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?
Yes, but that was just a general remark.

Quote Quote by Iforgot View Post
Could you write ψ for the photon explicitly for after the GT prism photon interaction. That would clear things up.
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.

Quote Quote by Iforgot View Post
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.
Well, if you can be sure of the angular momentum beforehand... The problem with the missing wavefunction of photons makes things complicated again.

Quote Quote by Iforgot View Post
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.
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 non-local, right?

Quote Quote by Iforgot View Post
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)
This puzzles me. For one you said beforehand that you want to follow a Bohmian approach. However, Bohmian mechanics does not even have superpositions. Anyway, please come up with some design for detecting the presence of photoexcited carriers (in terms of a strong measurement) which does not destroy the superposition. You might achieve something similar with weak measurements (and all their drawbacks), but it is basic first semester qm stuff that typical measurements spoil superpositions.
Iforgot
#44
Nov15-12, 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 q-wave 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 non-local, 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 schrod-eq.

E.g. I can semi-quantitatively 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 non-locality arise out of the photon wave-function 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 kick-in :-)


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