Dear Venus Kakdarvishi

In a Hanbury Brown and Twiss (HBT) experiment, the second-order correlation function g(2)(τ)g^{(2)}(\tau) is used to measure the statistical properties of light. For a single-photon source like a quantum dot (QD), the function g(2)(τ)g^{(2)}(\tau) provides insights into the photon emission characteristics.

If a single-photon source emits two perfectly indistinguishable photons simultaneously, the result of the second-order correlation function g(2)(τ)g^{(2)}(\tau) at zero delay (τ=0\tau = 0) would be:

g(2)(0)=0g^{(2)}(0) = 0

This result indicates perfect photon antibunching, meaning that the probability of detecting two photons at the same time is zero. In other words, the source emits photons one at a time, and there is no simultaneous emission of multiple photons2.

This behavior is a hallmark of a true single-photon source and is a key feature in quantum optics experiments.

See for instance:

https://www.kth.se/social/files/5cb1833856be5bf03c8165fa/Lecture%203%20Second-order%20Intensity%20Correlation%20Function.pdf

https://en.wikipedia.org/wiki/Higher_order_coherence

https://saaubi.people.wm.edu/TeachingWebPages/Physics404_690_Spring2012/Week2/SecondOrderCoherence_3Feb2012.pdf

Len Leonid Mizrah

Thank you for your response, I appreciate your explanation. However, based on the result of the second-order correlation function, which shows g(2)(0)=0, this suggests that we cannot correctly observe the result. In this case, the emission is not truly single-photon, yet from the second-order correlation function, it does not indicate multi-photon emission either. I will check the references you mentioned to explore this further. Thanks again for your insights.

I guess it will be (in the first approximation) the product of detectors quantum efficiencies, e.g. it will be 1x1 =1 for perfectly efficient noiseless detectors.

Nikolay Pavlov

Thank you for your response. However, in this case, the detector efficiency is not the main concern. What we care about is the nature of the detected photons and their correlation. Since in my question two photons are emitted simultaneously and are perfectly indistinguishable, one might expect the second-order correlation function g(2)(0) to be zero, similar to an ideal single-photon source. However, in reality for this case, g(2)(0) should not be zero, and I am interested in understanding why this happens and how we can address this issue. Any insights on this would be appreciated.

Venus Kakdarvishi Well, single photon, when dealing single photon emitter can only be absorbed=detected into one of the detectors. Yes, there is some kind of conterintuitive fact here: even so the mesaurable electromagnetic field is present simultaneously in both detectors the absorption always happens in only one of them . It shows fundamental non-locality of the photon - and it is even more impressive in Bell's inequality violation experiments (quantum entaglement).