Hong–Ou–Mandel effect

Question

The Hong–Ou–Mandel effect is a two-photon interference effect in quantum optics that was demonstrated in 1987 by three physicists from the University of Rochester: Chung Ki Hong, Zheyu Ou, and Leonard Mandel. The effect occurs when two identical single-photon waves enter a 1:1 beam splitter, one in each input port. When the temporal overlap of the photons on the beam splitter is perfect, the two photons will always exit the beam splitter together in the same output mode, meaning that there is zero chance that they will exit separately with one photon in each of the two outputs giving a coincidence event. The photons have a 50:50 chance of exiting (together) in either output mode. If they become more distinguishable (e.g. because they arrive at different times or with different wavelength), the probability of them each going to a different detector will increase. This effect can be used to test the degree of indistinguishability of the two photons and to implement logic gates in linear optical quantum computing.

The following picture shows the four possible outcome categories:

If a detector is set up on each of the outputs (in bottom and top directions) then coincidences can never be observed, while both photons can appear together in either one of the two detectors with equal probability. While Quantum mechanics agree with empirical results, a classical prediction of the intensities of the output beams for the same beam splitter and identical coherent input beams would suggest that all of the light should go to one of the outputs (the one with the positive phase).

Why does classical prediction demand that only situation 1 is possible, even if the two photons are indistinguishable? I mean there cannot be a physical difference between the bottom side and the upside of the beam splitter if it is a symmetrical 50:50 beam splitter.

Answer 1

Why does classical prediction demand that only situation 1 is possible, even if the two photons are indistinguishable?

The classical image is not quantum-mechanical, therefore you are not considering two photons, rather two beams:

a classical prediction of the intensities of the output beams for the same beam splitter and identical coherent input beams would suggest that all of the light should go to one of the outputs

You can see that in the wikipedia image for the phase relation between the incoming waves and the output ports.

https://www.cs.princeton.edu/courses/archive/fall06/cos576/papers/zetie_et_al_mach_zehnder00.pdf

In the above document you can find a description of the classical beamsplitter experiment, which is further explained by the vectorial description of the fields interacting at the boundary of the beamsplitter:

here:

$\vec{E_i} = \vec{E_{oi}} e^{i \vec{k_i}\cdot \vec{r}}$

$\vec{E_r} = \vec{E_{or}} e^{i \vec{k_r}\cdot \vec{r}}$

$\vec{E_t} = \vec{E_{ot}} e^{i \vec{k_t}\cdot \vec{r}}$

I omit the temporal description of the wave for simplicity's sakes but there is an additional temporal phase term in the exponential $\omega_{i,r,t}t$.

Continuity at the boundaries require that the phase is matched at the boundary (for energy conservation purposes):

$\hat{n} \times \vec{E_i} + \hat{n} \times \vec{E_r} = \hat{n} \times \vec{E_t}$

This is generally derived from Maxwell's equations (Here I don't provide proof besides the fundamental equations but this explanation can be found in fundamental optics and photonics books e.g. Saleh/Teich, Born/Wolf, etc.).

Hopefully this clears why the quantum-mechanical description would violate classical rules: For the "non-classical" cases, the phase relation is not retained from the classical pictures, but as could be seen in HOM experiments, it is possible to detect each photon at a different detector but only when dealing with single photons.

Answer 2

Actually, there is a difference between the two sides of the beam splitter, depending on the type and design of the beam splitter. For example, in a polarizing beam splitter, one side reflects one polarization state of the incident light, while the other side transmits the orthogonal polarization state. In a phase-sensitive beam splitter, there is a phase difference of π between the two output ports of the beam splitter. In a colour-sensitive beam splitter, one side reflects one part of the spectrum while the other side transmits the other part.

In the context of the Hong-Ou-Mandel effect, it is important to note that according to the Wikipedia page on the Hong-Ou-Mandel effect reflection from one side of the beam splitter can introduce a relative phase shift between the two interfering photons.

“reflection from the bottom side of the beam splitter introduces a relative phase shift of π, corresponding to a factor of −1 in the associated term in the superposition.”

This means that when two photons enter a beam splitter, one from the top and one from the bottom, the transmitted photon from the top and $-\pi$ phase shifted reflected photon from the bottom will interfere destructively and cancel each other out, resulting in no light being transmitted to the output port on the bottom side.

But empirical results and QM give equal chances for exiting on either output port (but only one output port at a time meaning identical photons should exit together in the same output port) for two identical photons.