We have the following equation.s for the E field of a wave : $$ \dot{E} = \left[ \hat{x} + 0.75 \hat{y} + \left( 2 + j \ 5 \right) \hat{z} \ \right] \ e^{-j \ 2.3 \ \left(-0.6 \ x \ + \ 0.8 \ y \right)} $$

I know that in order to determine the polarization I need to look at the phase difference between the electric field components but I'm having trouble here as I usually have only two components.


1 Answer 1


The electric field is in the plane orthogonal to the wavevector, the latter being proportional to $\vec{k}\propto -0.6\hat{x}+0.8\hat{y}$ (Check this assertion).

So, all you need to do is find an orthonormal basis - any such basis for this plane orthogonal to $\vec{k}$ and resolve the electric field components onto this basis - the complex quantities so gotten are the components of the Jones vector relative to that basis (and polarization is always defined relative to a basis of the plane transverse to $\vec{k}$ that must be chosen for a full specification).

I would suggest that the vectors $e_1 = \hat{x} + 0.75\hat{y}$ and $e_1 = \hat{x} + 0.75\hat{y} + \hat{z}$ are linearly independent and both orthogonal to $\vec{k}$, thus they span the plane you need. They're not orthogonal, though.

So make an orthonormal basis out of these two, using the Gramm-Schmidt procedure to do so - then calculate your Jones vector.

  • $\begingroup$ I think such a calculation is not necessary, the polarisation of the wave is same as the direction of the electric field, and the direction can be obtained from the magnitude part of the field, which can easily be observed to be perpendicular to direction of propagation. $\endgroup$
    – Tejas P
    Jun 5, 2017 at 14:23
  • $\begingroup$ @TejasP Yes, but the direction of the E field is not in general constant, and it is not constant here. You need to find the relative phases of the orthogonal plane components to work these details out - the electric field's head follows an ellipse in this plane and this wave is elliptically polarized. $\endgroup$ Jun 5, 2017 at 14:26
  • $\begingroup$ Thanks for pointing out, I did not observe the derivative. $\endgroup$
    – Tejas P
    Jun 5, 2017 at 14:39

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