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?

  • $\begingroup$ As far as I understand things, saying "exactly the same phase" doesn't even make sense for polarization vectors in different direction. $\endgroup$ Commented Jan 13, 2017 at 16:28
  • $\begingroup$ hi henning: this is where i lack knowledge of the subject to be able to comment (my area of expertise is software engineering). i do however have A-Level maths. looking at equation 2.1 of the paper i referenced, it has - omega t as part of the equation. in superimposing two such Jones vectors, i am in effect saying "omega is exactly the same value in both vectors to be superimposed" however i could be wrong here, hence why i am asking the question. i should imagine that i would be also saying that they have the exact same E0x and E0y. as for Phix and Phiy, that i don't know. $\endgroup$
    – lkcl
    Commented Jan 13, 2017 at 17:51

2 Answers 2


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:


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.

  • $\begingroup$ rod that is absolutely beautiful - yes i believe you are correct in that it is just the addition of the vectors E1 and E2. what i didn't explain was that i am actually rotating the two polarizations about the Z axis, see the recent "mobius light" experiments from 2015 or look up isaac freund's work. by aligning the Z-axis rotation angle of the two polarization fields at all times, the very VERY important result comes out that yes you just add E1 and E2, which is a highly significant result and a major breakthrough in the particle physics theory i am working on so i am extremely grateful $\endgroup$
    – lkcl
    Commented Jan 14, 2017 at 7:30
  • $\begingroup$ rod, comments are limited to 300 characters (and also cannot put in paragraphs! total pain....) so here is the reference to a draft: lkcl.net/reports/rishon_group_analysis basically by aligning the Z-axes of the elliptical polarisation i entirely avoid the problem of having to mix the angles, and, highly significantly, the Rishon Model drops out of the vector summation. $\endgroup$
    – lkcl
    Commented Jan 14, 2017 at 7:41

hi okay so for anyone interested, i followed a lead from a paper by castillo published in 2008, where he remaps jones vectors onto pauli matrices (at a tangent to the vector) on a poincare sphere. his paper contains a very important equation which shows the conditions under which constructive or destructive interference does not occur: it aligns with the other answers kindly given above.

what i then did was to work out the additional conditions under which superposition would occur, and it's when one of the waves is phase-shifted by 180. that results in the elliptical axis being rotated by half that angle (90 degrees), and at that point the superposition is successful when the other wave is at 90 degrees at all times to the other.

as this is electro-magnetism we're referring to, intuitively it would make sense that when one wave's field is always at right-angles to the other they would not interfere with each other.

i wrote this up in a paper where i show both dual and triple superposition, if anyone is interested: http://rxiv.org/abs/1702.0131


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.