# Hong-Ou-Mandel (HOM) effect and symmetric beam splitter

When modelling the photons' behaviour in Hong-Ou-Mandel (HOM) effect, those reaching the beam-splitter from the bottom side are considered to act differently than those coming from the other one:

In addition, reflection off the bottom side of the beam splitter introduces a relative phase shift of −1 in the associated term in the superposition.

Source

Is there an easy way to explain that if the beam splitter is symmetric?

Edit. When the two photons reach the beam splitter, in the HOM (Hong-Ou-Mandel) effect, one would think that four different things could happen:

• Both get the same direction and it is the right one
• Both get the same direction and it is the top one
• Each one gets in a different direction (*2)

The fact is that the two terms "Each one gets in a different direction" get compensated because one gets a positive constant and the other one a negative one. My question is about that fact, that is justified because of the reflection off the bottom side in one of this two situations.

• Are you familiar with the behavior of ordinary waves at a fixed boundary? This is the same type of situation. Jul 28 '17 at 9:45
• @probably_someone, what do you mean? I am familiar with it but I do not find the analogy. Jul 28 '17 at 10:15
• Reflection of a transverse wave from a fixed boundary introduces a phase shift of 180 degrees. The waves reflecting off of the beam splitter acquire an analogous phase shift of -1. Meanwhile, in the limit of a perfectly thin beam splitter, the waves transmitted through the beam splitter acquire no phase shift. Jul 28 '17 at 11:23
• @probably_someone, I have edited my question because I think I was not explaining myself well. Jul 28 '17 at 13:22

The reason why the one term gets a minus sign compared to the other, has to do with the fact that in quantum mechanics beamsplitters is represented by a unitary operator (matrix). If one wants to do that without making the elements of the matrix complex, then one side must get a minus sign while the other side doesn't: $$a \rightarrow (a+b)/\sqrt{2}$$ $$b \rightarrow (b-a)/\sqrt{2}$$ Of course, there is no reason why one cannot have complex entries in the matrix. It would just represent a phase. This allows one to write the transformation in a symmetric way: $$a \rightarrow (a+ib)/\sqrt{2}$$ $$b \rightarrow (b+ia)/\sqrt{2}$$ It turns out that both these ways of writing down the transformation through the beam splitter give the same result: HOM interference.