Does the Hong-ou-Mandel effect happen within a delay equal to the coherence time of the two photons?

Question

I read that photon bunching happens for thermal light within the coherence time of the thermal light. That is, if the time for detection is greater than the coherence time of the thermal light, no bunching is observed.

I also read that atoms of Bose-Einstein condensates are randomly bunched, but they become bunched if they are heated above the critical temperature, and they are bunched only within the coherence length of the matter wave.

Sources: http://www.sciencealert.com/cold-atoms-act-like-laser-beam https://physics.anu.edu.au/news.php?NewsID=13 http://www.abc.net.au/science/articles/2011/02/25/3149175.htm?site=indepthfeature&topic=latest http://phys.org/news/2011-02-scientists-atoms-lasers.html

Does this also hold for the Hong-ou-Mandel effect? Do the photons bunch at the same output of the beam splitter only if they arrive together at the beamsplitter within the coherence time?

Answer 1

Yes, the photons need to arrive within the coherence time. The reason, as I understand it, is that if the photons arrive at times that differ by more than the coherence time then the photons are effectively tagged by the time. As a result they become distinguishable and the HOM effect does not work anymore. For this reaosn one always have a trombone in the optical setup to change the path length difference. The HOM dip manifests as a function of this pathlength difference. The function of the dip is directly related to the spectrum of the light.

Answer 2

This is all treated within the multimode description of quantum optics. You end up with photon creation, destruction operators that correspond to a photon in a specific pulse, i.e. operators like

$\hat{a}_\alpha=\int d\omega \alpha\left(\omega\right)\hat{a}\left(\omega\right)$

$\hat{a}^\dagger_\alpha=\int d\omega \alpha\left(\omega\right)^*\hat{a}\left(\omega\right)^\dagger$

Which desribe the single-photon within a very specific wave-packet given by $\alpha$. The spectrum of the amplitude of light (i.e. electric field in some normalized form) due to this wave-packet is $\alpha=\alpha\left(\omega\right)$. It follows that the spectral density of power due to this pulse is $P\left(\omega\right)=\mathcal{F}\left(\alpha\left(t\right)\alpha^*\left(t\right)\right)=\int d\omega' \alpha\left(\omega'-\omega\right)\alpha^*\left(\omega'\right)$.

Also, above we use the single-frequency creation/destruction operators that commute as $\left[\hat{a}\left(\omega\right),\hat{a}\left(\omega'\right)^\dagger\right]=\delta\left(\omega-\omega'\right)$

Now, the HOM effect arrises as a result of the non-vanishing commutator, which in case of multimode light will be:

$I=\left[\hat{a}_\alpha,\hat{a}_\beta^\dagger\right]=\int d\omega \alpha\left(\omega\right)\beta\left(\omega\right)^*$

The above expression is basically the overlap of the pulse $\alpha$ with pulse $\beta$. I have written it in frequency domain, but you could also do it in time domain. Now, what happens in case of incoherent light is that overlap of the light with its delayed version goes to zero (if delay time exceeds coherence time). In classical optics this leads to dissaperance of interference pattern. In quantum optics, you get dissapearance of anti-bunching ($I=0$)

Answer 3

As far as I know, the answer is no. Coherence time is a classical concept, whereas Hong-ou-Mandel effect is a quantum interference phenomenon. That's all I can explain with my limited knowledge!