vectors - superposition of elliptically polarised light

i've done a google search "elliptical polarisation superposition" and also included the terms "jones vector" but have not found the exact answer that i seek.

i have some very specific criteria and suspect that the answer is straightforward but do not know how to prove it: help greatly appreciated.

in the following dissertation it provides a description of polarization by a jones vector http://www.diss.fu-berlin.de/diss/servlets/MCRFileNodeServlet/FUDISS_derivate_000000002688/04_chapter2.pdf

what i need to know is: if you superimpose two polarisations at:

• exactly the same position


• exactly the same size in both X and Y


• exactly the same orientation in both X and Y


• in exactly the same plane (XY) about Z


• with exactly the same phase



• with the SOLE EXCLUSIVE difference being their angle of rotation about Z

then what is the result?

intuitively i am guessing: is the end result simply a new polarisation field with exactly the same position, exactly the same size, still in the same XY plane about Z, still with exactly the same phase, but with the angle now being the SUM of the two polarisation's angles of rotation about Z?

or, is it more complex than that?

also, what happens in the case where the phase between the two is inverted by exactly 180 degrees?

in essence: given two jones vectors of same magnitude, frequency and location, how do i add them up?

Answer

You seem to be:

1. Describing what happens at the recombiner of an interferometer;


2. Describing the recombination of mutually coherent plane wave fields of exactly the same frequency propagating in the $\hat{Z}$ direction;

In that case, the polarizations are wholly described by the $2\times1$ Jones vectors and, when the fields recombine, the superposition is modelled by the complex vector summation of the two Jones vectors. Thus if the two vectors are:

$$E_1=\left(\begin{array}{c}x_1\\y_1\end{array}\right)\quad\text{and}\quad E_2=\left(\begin{array}{c}x_2\\y_2\end{array}\right)$$

where the $x_i,\,y_i$ are all complex scalars then the superposition is simply:

$$E_1+E_2=\left(\begin{array}{c}x_1+x_2\\y_1+y_2\end{array}\right)$$

Note that this operation does not of course generally conserve energy. One field can indeed completely cancel another if in the same direction and in anti-phase. What's happenning is that you're modelling one output port of the interferometer (a wave incident on a beamsplitter, for example, is split along two paths) so that, in the lossless case, the sum of the powers output from the two ports equals that of the two incoming waves. I analyze this in more detail in this answer here.

So, your question kind of implies that the two Jones vectors are of the form:

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

so that you'll find it's more complicated than simply adding the rotation angles $\theta_1$ and $\theta_2$, even if the phases $\varphi_1,\,\varphi_2,\,\phi_1,\,\phi_2$ are all nought.