What is weak coupling of photon polarization to a pointer?
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Yes.
So the polarization must come from interference effects between multiple photons, is that correct?
That is completely muddled. What makes you think that? The hamiltonian is constant over this space, so every polarization is an eigenstate. Every single photon can have any polarization it wants - linear in any direction, circular, or elliptical - and it does not need other photons to make any linear polarization.
Superposition states like |R⟩=12√(|x⟩+i|y⟩)|R⟩=12(|x⟩+i|y⟩) are always states of a single particle unless explicitly indicated. This doesn't mean that "a right-circular photon is a mixture of xx- and yy-linear photons", which would lead to a pretty paradox since those polarizations are also superpositions of of right- and left-circular states (not photons):
|x⟩|y⟩=12–√(|R⟩+|L⟩),=−i2–√(|R⟩−|L⟩
So the polarization must come from interference effects between multiple photons, is that correct?
That is completely muddled. What makes you think that? The hamiltonian is constant over this space, so every polarization is an eigenstate. Every single photon can have any polarization it wants - linear in any direction, circular, or elliptical - and it does not need other photons to make any linear polarization.
Superposition states like |R⟩=12√(|x⟩+i|y⟩)|R⟩=12(|x⟩+i|y⟩) are always states of a single particle unless explicitly indicated. This doesn't mean that "a right-circular photon is a mixture of xx- and yy-linear photons", which would lead to a pretty paradox since those polarizations are also superpositions of of right- and left-circular states (not photons):
|x⟩|y⟩=12–√(|R⟩+|L⟩),=−i2–√(|R⟩−|L⟩
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Explanation:
This weak coupling entangles the quantum pointer with the system, until the final projective measurement of B correlates the obtained system eigenstate with the quantum state of the weak pointer.
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