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20h
comment Loss of interference in single-photon Mach–Zehnder interferometer with detector in only one arm
@David No, in your other question you assumed the existence of an auxiliary system $A$ which does not appear here. The correct expression is $\langle \Psi_{P_i}\rvert T \exp \left (\int_0^t\mathrm{d}s\, H(s)\right) \lvert \Psi_{P_i}\rangle$, with $P$ the detector system and $H$ a Hamiltonian that switches the detector on. Given that here $\lvert \Psi_{P_i}\rangle = \lvert D_{off}\rangle$, in the end this reduces to $\langle D_{off}\lvert D_{on}\rangle = 0$.
20h
comment Loss of interference in single-photon Mach–Zehnder interferometer with detector in only one arm
The state of the system at $t_2$ is $|\Psi_{t_2}\rangle = (1/\sqrt 2)(|D_{on}\rangle |A_x\rangle + |D_{off}\rangle |A_y\rangle. $ Your second expression for $|\Psi_{t_2}\rangle $ is only true if the detector doesn't work (i.e. if its state is unchanged by the presence of the photon). The interference visibility, as defined in my other answer (physics.stackexchange.com/questions/233190/…), is $\langle D_{on}|D_{off}\rangle = 0$.
21h
comment How bizarre can the Quantum two slit experiment get?
No, your understanding is not correct. What matters is whether the "which-path" information is available in principle, not whether it is actually accessed by the observer. In order for this information to be available in principle, you need to introduce a detector which interacts with the quantum particle and completely changes the nature of the experiment, so that the interference pattern disappears.
1d
comment Single quantum particle in beam splitter, with different systems located in each channel
@David BTW there was a bit of a typo in my answer which is corrected now, specifically the explicit expression for $\langle C_a \rvert \rho_C(t)\rvert C_b\rangle$.
1d
revised Single quantum particle in beam splitter, with different systems located in each channel
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2d
comment Single quantum particle in beam splitter, with different systems located in each channel
@David You seem to already understand the tensor product of vectors $\lvert \phi\rangle \otimes \lvert \psi \rangle$, a way of composing vectors living in different Hilbert spaces. The notation $A\otimes B$ means a tensor product of operators, meaning that the operators act on different Hilbert spaces. So, the notation $\otimes$ just means the following: $$ (A\otimes B) (\lvert \phi\rangle \otimes \lvert \psi \rangle) = (A\lvert \phi\rangle) \otimes (B\lvert \psi \rangle) $$ Here (and in general) it is meaningless to write something like $U(t) \otimes \lvert \psi\rangle$.
2d
comment Single quantum particle in beam splitter, with different systems located in each channel
@David The zeroth order term in the exponential is the identity operator 1. This is the only term which survives, the higher order terms are killed by the $\langle C_a \rvert C_b \rangle$.
Feb
7
comment Single quantum particle in beam splitter, with different systems located in each channel
@David Indeed, the visibility is just given by the overlap between the initial and final states of A and P. I agree, interferometry is a very rich and fascinating subject!
Feb
6
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Feb
6
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Feb
6
revised Single quantum particle in beam splitter, with different systems located in each channel
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Feb
6
revised Single quantum particle in beam splitter, with different systems located in each channel
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Feb
6
answered Single quantum particle in beam splitter, with different systems located in each channel
Feb
5
comment Single quantum particle in beam splitter, with different systems located in each channel
@David It's a result of decoherence theory. I'll write up an answer soon which should clarify where it comes from.
Feb
4
comment Single quantum particle in beam splitter, with different systems located in each channel
@David That's what I just said in a slightly more formal way. Such a scenario corresponds precisely to the interaction I wrote down above. If you work through the calculation you will see that this leads to your idea of "two different Hamiltonians". Actually there is just one Hamiltonian but it depends on the states $C_{a/b}$.
Feb
4
comment Single quantum particle in beam splitter, with different systems located in each channel
Note that your assumption is essentially that the full interaction Hamiltonian is $H_{AP} \otimes \lvert C_a\rangle \langle C_a\rvert$, i.e. the interaction energy vanishes in state $\lvert C_b\rangle$.
Feb
4
comment Single quantum particle in beam splitter, with different systems located in each channel
This is mostly fine although you have neglected the free evolution of the particle and auxiliary system. In general your propagator should be (schematically) $U(t) =T\exp (\int H_{AP}(t))$, where $T$ is time-ordering and $H_{AP}(t)$ is the interaction Hamiltonian in the interaction-picture representation. Then you find the interference fringe visibility will be proportional to $\langle \Psi_{A_i}\lvert \langle \Psi_{P_i} \rvert U(t) \lvert \Psi_{P_i}\rangle \lvert \Psi_{A_i} \rangle$. So how much the coherence is lost depends on the interaction $H_{AP}$.
Feb
4
comment Quantized light-atom Hamiltonian
@Frank All modes do couple to the atom. It is just that the coupling is most important for modes near resonance with $\omega_{12}$. So really only modes within a range $\gamma$ around $\omega_{12}$ contribute significantly to the emitted photon (technically the frequency distribution in the rest frame of a single free atom is a Lorentzian of width $\gamma$ centred on $\omega_{12}$). This is only true because the photon-atom coupling is relatively weak, so that the interaction energy contributes just a small level-broadening $\gamma \ll \omega_{12}$.
Feb
4
comment Quantized light-atom Hamiltonian
@Frank Only the total energy, including the interaction energy, is conserved. So it is not strictly true that the photon must be the same energy as the two-level atom $\omega_{12}$. Actually the interaction leads to broadening of the electronic level. The photon will therefore have a frequency uncertainty $\gamma$, where $\gamma^{-1}$ is the lifetime of the electronic transition. Emission cannot suddenly create a plane-wave photon state of single frequency (which would extend infinitely far in space), it creates a wave packet which flies away and doesn't return.
Feb
4
comment Quantized light-atom Hamiltonian
Picking out a single mode of the field will only give you the right physics in certain scenarios, for example when the atom is inside an optical cavity. If you want something irreversible to happen, e.g. spontaneous emission, you need to include all of the modes.