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Equality in 7 might be wrong. We should distinguish between formal, $|0\rangle$, and physical, $|0\rangle_{phys}$, vacuum vectors. The formal vacuum vector is defined by its main property $$ \hat{b}^-|0\rangle = 0. $$ The physical vacuum vector is the least energy stationary state of the system's Hamiltonian. If the Hamiltonian contains "anomalous terms" ...


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For an ideal system (with no loss or scattering in any of the elements), $D_0$ will indeed see the number of photons equal to those detected in the other half of the system. But changing the beamsplitter coefficients doesn’t change the number of photons, it merely redistributes them among the various detectors. So the number of photons measured by $D_0$ will ...


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Well, here is a seat-of-the pants "lark" formal answer, going to "rigged" Fock spaces and places you (or anybody else) shouldn't really be at; except you may already have been there, when you learned about bras and kets, if you looked in Dirac's QM book ― actually, the section virtually all modern texts skip! I will cover your Fock space question, but ...


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The photon wavefunction has an essentially infinite number of degrees of freedom. Proof: the shape of a wave front describing a photon wavefunction can be described by Zernike polynomials. The terms are orthogonal: change the amplitude of one, and it does not affect the amplitudes of the other terms (other than the necessity of normalizing the sum of all ...


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