Why is the phase of the reflected light important in the Hong-Ou-Mandel (HOM) experiment?

Question

Trying to understand the Hong-Ou-Mandel effect (Wikipedia link) I got a bit lost with regards to what the reflection phases mean for the experiment and the kind of beam splitter required.

Instead of single photons, let's say that there are two coherent beams (A and B) and a beam splitter plate with a dielectric coating on the bottom (as suggested from Wikipedia link). Beam A is reaching the beam splitter from above, beam B from below. If beam A is reflected or refracted, there is no phase shift. If beam B gets reflected it gets a phase of $\pi$, and a phase of $0$ if transmitted.

We have four possibilities:

1. Beam A gets reflected and beam B is transmitted. In that case none of the beams gets a relative phase and both beams constructively interfere.
2. Beam A and B get transmitted. Nothing interesting happens, the beams just exchanged sides.
3. Beams A and B are reflected off the beam splitter. Nothing interesting happens, each went back their way. With the caveat that beam B obtains a relative phase of $\pi$.
4. Beam A is transmitted and beam B gets reflected. This case is confusing me, as beam A goes through the beam splitter, no phase added, and beam B gets reflected with a phase of $\pi$. So in principle the beams cancel as they interfere destructively.

In the Hong-Ou-Mandel experiment, only (1) and (4) contribute, while (2) and (3) cancel out. But that's not what I am getting by looking at the phases. Note that, if (2) and (3) indeed cancel out, the beams should still interfere destructively in (4), so (1) would be the only possible result.

Another possibility is that the beam splitter in this experiment is some kind of symmetrical cube beam splitter, where the reflective surface is inside a crystal. In that case no relative phase is gained by the outgoing beams, but then it is even more difficult to see why relative phases from reflection matter here.

What am I missing in this setup? What can be understood from the nature of the beam splitter and the phases of reflection? Can this effect only be understood using second quantization and the reflection phases do not matter here?

Answer 1

The experiment does not work with beams of light (like in the classical case). It needs to be single photons, because this is a quantum multi-particle interference effect. The phases (or phase-changes) are very important. The process needs to be unitary. Without the correct phases-changes the process won't be unitary. However, there are different ways to represent the unitary process. One can, for example, multiply the two input photons with arbitrary phases and do the same for the two output photons. Nevertheless, such changes do not change the interference effect (only the details of the states).

Note that this is a multi-photon interference effect, which is different from ordinary optical interference. The latter is the result of the superposition of two beam in which the same photon appears. So it is of the form $$ |\phi_a\rangle + |\phi_b\rangle . $$ In the multi photon case, like in HOM, the interference is a superposition of multiple photons: $$ |\phi_a\rangle|\psi_a\rangle| + |\phi_b\rangle|\psi_b\rangle . $$ As a result, it does not produce the same interference effect that we are familiar with from ordinary optical interference. In optical interference "every photon only interferes with itself". That is not the case with multi-photon scenarios.

About the phases, one can always multiply the output photons after the beamsplitter by separate local phase factors without affecting the detections of photons. The same can be done for the input photons, but that could affect the synchronization of the photons, which in turn affects location of the HOM dip. Nevertheless, the affect of these local phase modulations allow a variety of different but equivalent ways to define the unitary matrix for the beamsplitter, all of which produce the same multi-photon interference effect associated with the HOM effect.

Answer 2

The short answer is that the reflection phases of a beam splitter do not matter as long as it is a 50%-50% transmission/reflection beam splitter.

flippiefanus provides some insight in why we cannot consider this as a classical effect. As a direct answer to this question, I will post my own calculation.

The answers to this are in the Wikipedia article as it says clearly that the classical result is different from the quantum one. It is just a bit misleading about why it discusses the reflection phase of a particular beam splitter.

Classical calculation

Note that the equation for a beam splitter is just $$(O_A,O_B)=\tau (I_A,I_B)$$ where $I_A$ and $I_B$ are the incoming outputs from beams A and B, $\tau$ is the transfer matrix and $O_a$ and $O_B$ are the output beams in the directions A and B, respectively. The transfer matrix $\tau$ is a unitary be it quantum or classical description as the beam splitter behaves the same backwards per Helmholtz reciprocity theorem.

For a plate dielectric beam splitter as described first above, we have $$\tau_0=\frac{1}{\sqrt{2}}\begin{pmatrix}1 & 1\\ 1 & -1\end{pmatrix}$$

that means that if $(I_A,I_B)=\frac{1}{\sqrt{2}}(1,1)$ then the output is $(O_A,O_B)=(1,0)$, meaning that all of the emitted light is in one direction (A). This is expected due to unitarity, as a single beam will split into two, a symmetrical input of two beams will converge into 1. Note that the results vary if we have a more general beam splitter,

$$\tau=\frac{1}{\sqrt{2}}\begin{pmatrix} e^{i\phi_R}& e^{-i\phi_T}\\ e^{i\phi_T} & -e^{-i\phi_R}\end{pmatrix}$$.

(where $\phi_R$ and $\phi_T$ are phases of reflection/tranmission) as $\tau (I_A,I_B)= \frac{1}{2}( e^{i\phi_R}+ e^{-i\phi_T}, e^{-i\phi_R}+ e^{i\phi_T})$.

So do the $\phi_R,\phi_T$ influence the HOM experiment?

Quantum calculation

As suggested, the only way to find out is to carry out the quantum calculation.

Let the input state be $|1_A,1_B\rangle=a^\dagger b^\dagger|0_A,0_B \rangle$, where $a^\dagger$ and $b^\dagger$ are bosonic creation operations in channels A and B, and $|n_A,n_B\rangle$ correspond to a state with $n_A$ and $n_B$ photons in channels A and B.

The state is modified by $\tau$, such that the output state is given by $c^\dagger d^\dagger |0_A,0_B\rangle$, where

$$\begin{pmatrix} c^\dagger \\ d^\dagger\end{pmatrix}=\tau \begin{pmatrix} a^\dagger \\ b^\dagger\end{pmatrix}$$

thus the output is given by

$$\frac{1}{\sqrt2}(a^\dagger e^{i\phi_R}+b^\dagger e^{-i\phi_T})\frac{1}{\sqrt2}(a^\dagger e^{i\phi_T}-b^\dagger e^{-i\phi_R})|0_A,0_B\rangle=\frac{1}{2}[((a^\dagger)^2 e^{i(\phi_R+\phi_T)}-(b^\dagger)^2 e^{-i(\phi_T+\phi_R)}+a^\dagger b^\dagger (1-1)]|0_A,0_B\rangle= \frac{1}{2}([(a^\dagger)^2 e^{i(\phi_R+\phi_T)}-(b^\dagger)^2 e^{-i(\phi_T+\phi_R)}]|0_A,0_B\rangle=\frac{1}{\sqrt2}(e^{i(\phi_R+\phi_T)}|2_A,0_B\rangle-e^{-i(\phi_R+\phi_T)}|0_A,2_B\rangle) $$

So in general, the output has no crossed term of the type $a^\dagger b^\dagger$ to create a state with one photon in each channel $|1_A,1_B\rangle$ (this would be the case as long as the unitary matrix $\tau$ has permanent $0$). Also there is no way to find a beam splitter that can cancel the output of one of the channels as in the classical result with light beams.