Why does a 45-degree mirror reflection cause a π/2 phase shift in photons?

[LarryS]

I read about one version of the double-slit experiment that uses a light beam from a laser and a combination of fully-silvered and half-silvered mirrors.

The author, Roger Penrose, describes one part of the experimental apparatus in which a light beam from a laser, traveling “north”, encounters a fully-silvered mirror at a 45 degree angle to the beam and is reflected “east”. He said that the wave function for those photons from the laser had to be multiplied by i in order to account for the ¼ phase shift (pi/2) caused by the mirror.

My question is this: Is the pi/2 phase shift caused by the fact that the laser beam changed direction by 90 degrees or is the phase shift due solely to the physical properties of the mirror?

Thanks in advance.

 

[f95toli]

It is just due to the mirror. The phase change is due to the usual boundary conditions for
an EM wave impinging on a reflective surface (i.e is is just classical EM).

 

[LarryS]

It is just due to the mirror. The phase change is due to the usual boundary conditions for
an EM wave impinging on a reflective surface (i.e is is just classical EM).

So, are you saying that if the angle of incidence of the beam on the mirror had been, say, 10 degrees instead of 45 degrees, the phase shift still would have been pi/2?

 

[IttyBittyBit]

Yes.

 

[LarryS]

It is just due to the mirror. The phase change is due to the usual boundary conditions for
an EM wave impinging on a reflective surface (i.e is is just classical EM).

Thanks. How does QED explain, at the individual photon/electron level, the phenomena of the reflection of light?