# quantum optics of a polarizing beam splitter

I would like to use the Heisenberg picture in quantum field theory to model a polarizing beam splitter. Is there an easy way for someone to show me how the field operators ($a^\dagger_{input1}$, $a^\dagger_{input2}$etc) transform across a polarizing beam-splitter?

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Leonhardt's Quantum Physics of Simple Optical Instruments may contain some useful information for you. – twistor59 Feb 9 '12 at 20:25

Much depends on how "deep" do you want the description to be. Let's assume the beam-splitter works as an external potential (in QM) and does the scattering. The 1-particle scattering operator is the unitary operator of the "Hadamard gate",

$u=\begin{pmatrix} 1& 1\\ 1& -1 \end{pmatrix}\frac{1}{\sqrt2}.$

This is generated in time $t=-\pi/4$ by the Hamiltonian $h=\sigma_2$ (2nd Pauli matrix). Promoting these to many-body by the rule $G=\sum_{ij} a^*_i \langle \psi_i|g\, \psi_j \rangle a_j$ (for some 1-particle operator $g$) you should get the many-particle Hamiltonian

$H=\frac{i}{\sqrt{2}}\left[a^*b -b^*a \right]$

which in the same time $t=-\pi/4$ should produce the unitary Beam-Splitter-transformation

$U=\frac{1}{\sqrt{2}}\left[(a^*+b^*)a+ (a^*-b^*)b \right]$

(here $a,b$ stand for $a_{input\ 1,2}$). The transformations of any many-particle operators can now be computed by the Heisenberg transformation $G(t)=exp(iHt)G(0)exp(-iHt)$.

If you wanted, however, to know about the interaction with some quantum model of the beam splitter, then it becomes more complicated (note, however, that no entanglement with the state of the beam splitter is apparently being produced in experiments, which is interesting in its own right).

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