Cascaded Polarizers Problem

Question

I am going through the MIT OCW course on Quantum Physics and while looking at the recitations slide on polarizers I got stuck on the problem given in the second last slide (the extra-credit problem):

The squares in the above image are polarizing beam splitters (splits the incoming beam into two beams with perpendicular polarization). The vectors in red indicate the direction of the E-field and not the direction of propagation of the wave.

I understand that they have used mathematical sleight of hand. Essentially by writing the zero-valued horizontal component as $+\frac{1}{2}$ and $-\frac{1}{2}$ then selectively filtering out one component they are able to get a non-zero horizontal component in the end. Hence, output power is more than input power.

But what is the unphysical thing that is going on? I can't for the life of me figure out that part.

EDIT: Added description about the diagram.

Answer 1

The unphysical thing, is the 'adding' of amplitudes. Those things being added are oscillatory, a sum can be addition, or subtraction, or anything inbetween, depending on the phase.

One of the two branches is taking an extra reflection that the other does not, and a 45 degree reflector changes the sense (from positive to negative) of one component (half the light) as it reflects the light. So, that component cannot add to the final sum unless the other component (reflected without inversion of direction) is subtracting from the final sum.

I'm uncertain what the symbols indicate, but an additional difficulty is that light is a transverse wave, in a two-dimensional plot it looks like some direction-of-travel is not orthogonal to the polarization?

Answer 2

Oh, this question had me going for some time. By the principle of simple superposition $I_0 ={I_1} +{I_2}+2\sqrt{I_1}\sqrt{I_2}\cos\phi$ where $\phi$ is the angle between the two waves. As you know, the factor bothering you is that extra $2\sqrt{I_1}\sqrt{I_2}\cos\phi$ as the energy of the total waves is proportional to the intensity and the initial energy was thus proportional to $I_1+I_2$. Then, how and from where is this extra energy being created? Well, the answer is that the frequency of the wave also changes with intensity. Now, the frequency is actually less, which will compensate the apparent energy gain from the change in intensity.