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When two optical fibers A and B are brought close enough that there can be quantum tunneling between the two, we can write the "beam splitter" Hamiltonian as

$$H = a^\dagger b + b^\dagger a$$

If, instead, the two fibers are spliced together, such that (travelling from left to right) photons in B continue in B, but photons in A merge into B, how do I write the Hamiltonian for this?

My guess is

$$H = a b^\dagger.$$

It's not Hermitian, but maybe it's correct. I would be interested in references to any material that discusses something like this in the context of quantum optics and second quantisation representation. Thanks!

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    $\begingroup$ Non-hermitian hamiltonians are never correct, unless you have losses (i.e. actual absorption) in your system. $\endgroup$ – Emilio Pisanty Aug 8 at 14:52
  • $\begingroup$ Related: Phase added on reflection at a beam splitter? $\endgroup$ – Emilio Pisanty Aug 8 at 14:53
  • $\begingroup$ Hmm, ok I read the links, but still have no real sense of how to construct my Hamiltonian. Any ideas? $\endgroup$ – Tom Aug 8 at 15:10
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    $\begingroup$ Hint: start with the classical optics, build a hamiltonian formulation in terms of modes, and then quantize. There's nothing about your question that cannot be figured out first in the classical-optics arena and then brought back to the QM side, and then you can just do old-school classical optics instead of wondering about non-hermitian hamiltonians. $\endgroup$ – Emilio Pisanty Aug 8 at 15:15
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    $\begingroup$ @Emilio Pisanty I know that you deal with this stuff, so I am curious to understand how a beam-splitter Hamiltonian may work. Up to now, I only read descriptions of the BS device in terms of unitary matrices accounting for transition processes. A Hamiltonian $H$ would instead imply a true temporal process: is there a sort of time evolution? (Yes obviously!) If Yes, how is the evolutor operator $e^{-itH}$ related with the standard BS description? $\endgroup$ – Valter Moretti Aug 9 at 9:43

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