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  Interference in "Frustrated Two photon Creation"

+ 4 like - 0 dislike

In this paper it is explained how an SPDC source can have its emission inhibited by destructive interference with emission from a previous time.

I have had a (maybe incorrect?) general intuition that destructive interference of wavefunctions can never make things "disappear." For instance, while destructive interference of a photons' possible paths can cause a particular path to vanish, the photon itself can only have its pathway altered. (Likewise in two-photon interference, like HOM interference, the output possibilities are interfering destructively)

In this example (as illustrated from the picture from the paper below), mirrors are set up such that photon pairs created by SPDC interfere upon reflection with a subsequent spdc event.

It appears as though when two creation events are prepared to have opposite phase-relations, the creation event is inhibited, as shown here: $$|\Psi\rangle = \alpha|1\rangle_{s_1}|1\rangle_{i_1} + e^{\mathrm{i}\phi_p}\alpha|1\rangle_{s_2}|1\rangle_{i_2}, $$

When the photons pairs are made to be identical, then these events are seen to destructively interfere. $$ I_i = I_s = 2I_0 \{ 1 + \cos(\phi_i+\phi_s-\phi_p) \},$$ where $I_0\propto |\alpha|^2$ is the rate of photon emission into either mode without mirrors present.

My question is: Where does the probability "go" in this case? In the case of a quantum mach-zender interferometer, for instance, to have a wavefunction interfere with itself you need to construct a unitary operator in which probability amplitudes can split between different pathways. When there is destructive interference, there's constructive interference in another pathway. So where is the constructive interference that's happening in parallel in this case?

Maybe other "leftover" modes are enhanced (in other spatial modes, for example) enhanced as a result of this inhibition due to interference?

Or maybe the total state could be expressed in a form describing a larger system, such as: $|\text{pair creation}\rangle + |\text{no pairs}\rangle$ And something constructive is happening there?

This post imported from StackExchange Physics at 2020-01-22 12:21 (UTC), posted by SE-user Steven Sagona

asked Sep 18, 2019 in Experimental Physics by Steven Sagona (20 points) [ revision history ]
recategorized Jan 22, 2020 by Dilaton
If you explain SPDC and HOM it would make the question more self-contained.

This post imported from StackExchange Physics at 2020-01-22 12:21 (UTC), posted by SE-user my2cts
'destructive interference of wavefunctions can never make things "disappear." ' Correct but it will prevent things from appearing in the first place.

This post imported from StackExchange Physics at 2020-01-22 12:21 (UTC), posted by SE-user my2cts
I think I found this on arxiv, arxiv.org/abs/1909.03513 ( I do not have easy access to phys rev). It is not my field , it needs a quantum optics specialist, but I have found very useful in accepting the behavior of photons in interference the Mit video youtube.com/watch?v=RRi4dv9KgCg . It explains for the shown experimental setup where the energy goes (let alone the probabilities) : one has to consider the whole setup , the energy goes back to the laser. In this case crystal and mirrors and what ever generates the waves,, should be in one quantum mechanical solution.

This post imported from StackExchange Physics at 2020-01-22 12:21 (UTC), posted by SE-user anna v

I have found useful in understanding interference in light beams this MIT video .. It shows that the whole quantum mechanichal setup has to be used, not only the beams. In case the video does not show search for "MIT video interference where does the light go"

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