Hong-Ou-Mandel effect

Cthugha

Sure, probability amplitudes just work fine. However, these just give you probabilities for processes and outcomes of measurements, but no eigenstates.

Yes, sure. Outcomes of measurements are no problems.

Just one more issue I just noticed where I am not sure whether the point is clear or not. The photons in a Fock state are not entangled, but that does not mean that they are statistically independent. Even photons in classical states are not necessarily statistically independent.

phonon44145

Just for curiosity, what are those probabilities equal to? Are they 25% each as one would expect if the photons were totally independent?

I think I can understand the bunching of photons (at least in a thermal source) and antibunching (e.g. resonance fluorescence) and can see that it does not involve any entanglement. I just can't see what would cause such behavior in a Fock state (whether single-mode or multimode).

Cthugha

That may depend on how you measure it. However, a "simple" and straightforward measurement would typically give classically expected values.

Hmm, but the case of a general Fock state is not that different from resonance fluorescence. Resonance fluorescence gives you the n=1 Fock state and g2=g3=...=gn=0. The n=2 Fock state will give you g2=0.5 and g3=...=gn=0. The n=3 Fock state will give you g2=0.6666666, g3=2/9, g4=0,...gn=0 and so on.

There is not really a conceptual difference between resonance fluorescence and Fock states of higher photon number.

phonon44145

Then can we write |2V,0H> biphoton state in an orthogonal basis according to

|2V,0H> = 1*|2V,0H> + 0*|1V,1H> + 0*|0V,2H>

and

|2V,0H> = (1/2) * |2R,0L> + (1/sqrt 2) * |1R,1L> + (1/2) * |0R, 2L>

so that light will behave "classically"? Also, by "simple and straightforward" you mean ordinary von Neumann measurements?