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Hong-Ou-Mandel effect

by phonon44145
Tags: effect, hongoumandel
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phonon44145
#19
Feb20-12, 04:28 PM
P: 53
Quote Quote by Cthugha View Post
Yes, sure. Outcomes of measurements are no problems.
Just for curiosity, what are those probabilities equal to? Are they 25% each as one would expect if the photons were totally independent?

Quote Quote by Cthugha View Post
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.
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
#20
Feb20-12, 04:39 PM
Sci Advisor
P: 1,627
Quote Quote by phonon44145 View Post
Just for curiosity, what are those probabilities equal to? Are they 25% each as one would expect if the photons were totally independent?
That may depend on how you measure it. However, a "simple" and straightforward measurement would typically give classically expected values.

Quote Quote by phonon44145 View Post
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).
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
#21
Feb20-12, 05:13 PM
P: 53
Quote Quote by Cthugha View Post
That may depend on how you measure it. However, a "simple" and straightforward measurement would typically give classically expected values.
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?


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