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:

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


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

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...

  • 1
    $\begingroup$ 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".... $\endgroup$
    – lkcl
    Commented Jan 16, 2017 at 9:48
  • 1
    $\begingroup$ 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 :) $\endgroup$
    – lkcl
    Commented Jan 16, 2017 at 11:52

1 Answer 1


just for the records, i found the answer and it is part-described here with a link to the paper, it turns out that a phase shift by 180 degrees prior to superposition results in exactly the rotation of 90 degrees required, however the only waves that may superimpose are those described in the paper, exactly as expected from the observations made. fascinating stuff (if you like this sort of thing... which i do :) ) superposition of elliptically polarised light


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.