superposition at pi/4 phase-offset of elliptically polarised light

in an earlier question, i asked if elliptically-polarised light could be superimposed in a way that allowed the vector(s) E (the jones vectors) to make sense:

superposition of elliptically polarised light

it was very kindly answered by rod as "yes if theta and phi and everything is the same, then the result E1 + E2 is the summation of the major axes".

i note however that yes, if this relation does not hold, all bets are off.

i have an additional case - a scenario where i suspect that the vectors may also similarly be summed, and i believe it's when the two exponent parts of Jones Vectors are offset by pi/4 (90 degrees).

going from the two equations E_1 and E_2 of the former answer (apologies i am not wholly familiar with quora otherwise i would cut/paste them: i will edit if i can work it out... got it!)

$$E_1=\left(\begin{array}{c}e^{i\,\varphi_1}\,\cos\theta_1\\e^{i\,\phi_1}\,\sin\theta_1\end{array}\right)\quad\text{and}\quad E_2=\left(\begin{array}{c}e^{i\,\varphi_2}\,\cos\theta_2\\e^{i\,\phi_2}\,\sin\theta_2\end{array}\right)$$

let us suppose that theta_1 = theta_2 (the cos/sin parts are the same), but that phi_1 = phi_2 + pi/2

my intuition tells me that this results in a simple inversion of the sign of E2 relative to E1, such that the vectors may simply be SUBTRACTED instead of ADDED, because sqrt(-1) squared comes back to -1. but my intuition is also telling me that one of the preconditions may need to be modified, so that theta_1 = theta_2 + pi/2 as well for example. however these are the kinds of things i have difficulty working out.

help greatly appreciated to work out if my suspicions are correct, that there are indeed other special cases (right-angle phase-coherent cases) where Jones Vectors may indeed simply be added (or subtracted).

working on this myself (comments as notes below), so far i have:

$$E_{\hat{x}} = E_{n\hat{x}} e^{-i \left( \psi_{\hat{x}} \right) } e^{-i \left( kz / 2 \right) } e^{-i \left( - \omega t / 2 \right) } e^{-i \left( \theta/2 \right) }$$

where

$$E_{n\hat{x}} = E_{0\hat{x}} e^{-i \left( \theta \right) }, \theta = n\tau/12$$

where of course tau is pi / 2 because i love the tauday manifesto :)

the theta/2 is down to a property of the solution i'm working on (mobius-light), but i think may turn out to be very relevant... still uncertain about all this stuff...

• i'm exploring this myself, i found the following (which i can't quite get to grips with), i may be talking about this case: en.wikipedia.org/wiki/Jones_calculus#cite_ref-3 which is a quarter-wave plate "phase retarder".... – lkcl Jan 16 '17 at 9:48
• ooo exciting, i found this: princeton.edu/~wbialek/intsci_web/dynamics2.2.pdf which according to equation 2.76 and 2.77 mean that mutiplication of two complex exponentials is equivalent to addition of the angles, likewise division equivalent to subtraction. i am therefore guessing that the addition of the e^(i pi/2) onto theta_2 may be factored outside as a static multiplicative term... i am slowly getting there :) – lkcl Jan 16 '17 at 11:52