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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!
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!