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


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    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.
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