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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°

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  • $\begingroup$ 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. $\endgroup$ – Semoi Dec 8 '20 at 18:57
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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.

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  • $\begingroup$ 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 $\endgroup$ – Gabriela Da Silva Dec 8 '20 at 17:48
  • $\begingroup$ 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/… $\endgroup$ – Kian Maleki Dec 8 '20 at 17:58

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