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I Hamiltonian for an Optical Phase Shifter?

  1. Dec 1, 2017 #1

    Twigg

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    Gold Member

    Hey all,

    I was reading Efficient Linear Optics Quantum Computation by Knill, Laflamme, and Milburn, when I came across their expression for the Hamiltonian for a phase shifter, given as ##\textbf{n}^{(\ell)} = \textbf{a}^{(\ell)\dagger} \textbf{a}^{(\ell)}##, where ##\ell## indicates the mode. (They ignore polarization and work in the formalism of spinless bosons.) How does one get this?
    I was able to get the beam splitter Hamiltonian by looking at the evolution operator (the unitary operator that takes an input photon in the x direction and spits out a superpostion of x and y photons) and subtracting out identity. But this gives a scalar Hamiltonian of ##e^{i\phi} - 1## when I think about following the same procedure for the phase shifter. Am I looking at this the wrong way? I'm completely new to quantum optics. I'm familiar with the quantization of the EM field, but that's about it.

    Thanks!
     
  2. jcsd
  3. Dec 2, 2017 #2

    vanhees71

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    Science Advisor
    2016 Award

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