# Beam splitters and quantum mechanics

Suppose two incoming photons one coming from the axis x and the other from the y axis, both in the positive direction of this axes. Now suppose we put a beam splitter in the origin, oriented with 45° wrt x axis. The initial state of the photon x is (1,0) and of the photon y is (0,1) After both meet the beam splitter, the photon x states now become (1,i) and the photon y becomes (i,1) (non normalized).

This is what i read, but i am not understand the latter part, how does this change of states occurs? I would guess that the change of phases make the photon x become (1,-1) and the y (-1,1), -1 = i*i = rotation of 180° or, in another word, change of phase of 180°

• You are interested in the case, where at time $t$ only a single photon is split. If both photons arrive at the beam splitter simultaneously, the problem becomes more complex, see quantum optics. – Semoi Dec 8 '20 at 18:57

## 1 Answer

You should think link this. After the beam splitter the light rays will be in superposition of (1,0) and (0,1). It means that if you measure the polarization of photon x after the beam splitter it is 50% of the time (1,0) and 50% (0,1). If you calculate the probability of (1,i) you will find that it is 50% of the time (1,0) and 50% (0,1).

Polarization of type (1,i) i.e. when there is "i" in there, means rotating. You can think of (1,i) and (i,1) as a circular polarization which can be archived by superpositioning of (1,0) and (0,1).

Last note, a photon which comes out of an interaction is always circularly polarized i.e. it carries angular momentum. (1,0) and (0,1) polarization are always superposition of left- and right-handed photons.

• I see what you mean, but why is my interpretation wrong? In the Argand-Gauss plane "i" means rotation of 90°, but the reflection does not imply 180° of rotation? So why not i*i = -1? You can see too that the probability in the case we replace i by -1 remains 50/50 – Gabriela Da Silva Dec 8 '20 at 17:48
• because photons are quantum properties and polarization is not a vector. It is a spinor. You should always be careful when dealing with spinors. Your reasoning is correct in classical point of view but not quantum mechanically. There is a full discussion on this topic here. physics.stackexchange.com/questions/154468/… – Kian Maleki Dec 8 '20 at 17:58