Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Jones vector of circularly polarized light

  1. Nov 10, 2011 #1
    Why is the jones vector of circularly polarized light <1,i> ?

    Things like <1,0> and <0,1> make perfect sense for linearly polarized light along the x and y axes but what exactly is that i doing there that makes the vector represent a circular polarization?

    I never really intuitively understood that.

    Thanks for any help guys
  2. jcsd
  3. Nov 10, 2011 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The x and y components of the E-field here are being represented by complex numbers. The reason for doing so is because complex numbers have a magnitude and a phase, and as a result they are a natural way of compactly representing the amplitude and phase of sinusoidal, time-varying quantities. In the case of circular polarization, if the x component of the E-field (after normalizing by dividing out the magnitude) is equal to 1, then it is entirely real (its phase angle is 0 in the complex plane). In contrast, if the y component is represented by an i, then although the magnitude of the y component is the same, it is pi/2 out of phase with the x-component. It is therefore entirely imaginary (being rotated by 90 degrees from the x component in the complex plane) Recall that 'i' can be represented in polar form as:

    [tex] i = 1e^{i(\pi/2)} [/tex]

    So the amplitude and phase of the y component of the E-field are 1 (after normalization) and pi/2 respectively.

    If you think about it, having the x-component and the y-component be 90 degrees out of phase with each other will lead to a total (resultant) E-field vector that rotates in a circle in the plane of polarization.

    For example, at t = 0, let's say that Ex is at a maximum in its cycle and Ey is at the point in its cycle where it is 0. Then the resultant E field vector points entirely in the x-direction. But an eighth of an oscillation period later, the Ex vector is now only 1/root(2) of its initial (max) value, and the Ey vector has increased from 0 length to 1/root(2) of the max value. Therefore, the two vectors have the same magnitude, and their resultant points at 45 degrees to the x direction. So, the vector has rotated by this amount. After about a quarter of an oscillation period, the Ex vector has now lessened down to 0 length, and the Ey vector has increased all the way to its max value. The resultant E vector is therefore now entirely in the y direction. It has rotated 90 degrees since it started. If you look at other sample points in the oscillation, you'll find that over a full period, the vector will have traced out a full circle in space.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook