When a single photon with polarization $\mathbf{a}$ arrives at a linear polarizing filter in the direction $\mathbf{p}$, the photon has a probability of $(\mathbf{a}\cdot\mathbf{p})^2$ to pass through the filter, while it has a probability of $1-(\mathbf{a}\cdot\mathbf{p})^2$ to be absorbed.

What does a typical quantum mechanical model of such a linear polarizing filter look like? In particular, I'm looking for how the quantized EM-field interacts with the many degrees of freedom in the filter. Also, any references to an article discussing such a model for a particular realistic polarizing filter would be appreciated.

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    $\begingroup$ I don't know if this can be useful, but a beam splitter is in some sense a polarization filter (it deflects the radiation differently depending on the polarization), and it is extensively used in quantum optics. An example of its quantum modelization is for example this. $\endgroup$ – yuggib Oct 1 '14 at 15:37
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    $\begingroup$ I guess you can model the system via a quantum operation. You will describe the filter as a collection of harmonic oscillators (the phonon field) and couple it to a mode of your quantum field (in canonical quantization the EM field is a collection of oscillators as well) through a coupling of the form a b^\dagger + a^\dagger b. a is the EM field annihilation operator of the relevant mode, and b is the collective annihilation operator of the phonons modes resonating with your field. $\endgroup$ – giulio bullsaver Oct 1 '14 at 15:47
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    $\begingroup$ Intuitevely tracing over the filter degrees you will be left with a master equation describing the probability of absortion. I do not any reference about this though, maybe Nielsen and Chuang or quantum optics books. $\endgroup$ – giulio bullsaver Oct 1 '14 at 15:47
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    $\begingroup$ The key thing to realize here is that the EM field is harmonic and usually in a coherent state. Because of this, interactions with the environment do not take the system out of the correspondance limit. For this reason, it turns out (not proving this in a comment) that you can just take the classical picture of loss and it's actually correct in the quantum description as well. You can see this because the interaction Hamiltonian between the photon and whatever else is $a$ on the photon, and the coherent state is an eigenstate of that operator. $\endgroup$ – DanielSank Oct 1 '14 at 17:41
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    $\begingroup$ On a broader level, you are essentially asking how to understand dissipation from a quantum mechanical picture. Excellent question! There is much literature about this if you Google "decoherence" or something like that, but maybe this simplified example will help you: physics.stackexchange.com/questions/36475/…. $\endgroup$ – DanielSank Oct 1 '14 at 17:43

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