In classical waveguide analysis (e.g. for optical fibers, as in the notes Modal analysis of step-index fibers, ECE 4006/5166 Guided Wave Optics, Robert R. McLeod, University of Colorado), one can find the various supported vector modes, which are typically defined as TE, TM, HE, EH groups etc

For any given mode, expressions for the mode field then exist: $E_r$, $E_\theta$, $E_z$, $H_r$, $H_\theta$, $H_z$ where $E$ is electric field, $H$ is magnetic field and we're in polar coordinates assuming a cylindrical waveguide.

Some of these field components are complex. Can anyone please explain the physical significance of the imaginary part please? I assume this is somehow related to phase?

Following on from this, what part of the field then defines the modal polarization. e.g. on p76 of the lecture notes cited above the field for TE01 is shown to exhibit a 'swirling pattern' as follows:

enter image description here

Is this related to the real part of the modal amplitudes? Or the magnitude (i.e. if we converted to cartesian coordinates, this would be found by adding the square of X and Y terms in quadrature?)


1 Answer 1


As usual, whenever you have a complex-valued amplitude $\tilde{\mathbf{E}}(\mathbf r)$, it's a stand-in for a physical field that's obtained as its real part, i.e. $$ \mathbf E(\mathbf r,t) = \mathrm{Re}\left( \tilde{\mathbf{E}}(\mathbf r)e^{-i\omega t}\right). $$ The presence of nontrivial imaginary parts of $\tilde{\mathbf{E}}(\mathbf r)$ signifies nontrivial phase relations between the electric field when taken at different points. This is obvious with e.g. trivial factors of the form $e^{ik_zz}$, but it's all the same whenever the amplitude is complex. Thus, for example, if $\tilde{\mathbf{E}}(\mathbf r)=E_0(\hat{\mathbf e}_x+i\hat{\mathbf e}_y)$, then $$ \mathbf E(\mathbf r,t) = E_0\left(\cos(\omega t)\hat{\mathbf e}_x+\sin(\omega t)\hat{\mathbf e}_x\right), $$ i.e. with a phase of $i=e^{i\pi/2}$ between the two components. Generally, if two scalar amplitudes at two different locations differ by a phase $e^{i\varphi}$, there will be a relative delay of $\varphi/\omega$ in the oscillations of those two components.

The modal polarization is a bit more complicated, and the polarization diagram you've drawn is special in that the polarization is linear everywhere (where in the general case it will be elliptical almost everywhere). The polarization is linear if and only if the complex mode amplitude vector $\tilde{\mathbf{E}}(\mathbf r)$ is real or it is a real vector times a single complex scalar, and in that case the diagram as drawn just represents that real-valued amplitude.

However, in general, complex valued vectors are not of that form (cf. the field from above, $\tilde{\mathbf{E}}(\mathbf r)=E_0(\hat{\mathbf e}_x+i\hat{\mathbf e}_y)$, as an example that cannot be reduced to a real vector times a complex scalar), so if you want to depict the polarization field you need to extract its elliptical axes (via the method in this answer) which are real-valued vectors that you can then draw.

  • $\begingroup$ Thanks for this insightful answer (and for reformatting my question)! Could you please clarify the statement that the diagram shows linear polarization - it looks like a circular field to me. Isn't linear polarization shown by arrows all pointing in the same direction - e.g. the HE11 diagram? Do you know how (i.e. what operations performed on E_r, E_theta, E_z, H_r, H_theta, H_z terms) that diagram in the OP is plotted please? Is it just plotting the real parts? Thanks $\endgroup$
    – PhysLQ
    Jun 26, 2017 at 13:58
  • 2
    $\begingroup$ @PhysLQ No, polarization is a local property, and at each point the polarization is linear (and therefore depicted as an arrow). What you don't have is a global linear polarization (so you can only talk about a polarization field / polarization map), but that doesn't stop you from having a linear polarization at each point. And, as I said, the diagram is plotted by showing the real part of $(E_r,E_\theta,E_z)$, with the obvious due care paid to the direction of the corresponding unit vectors, which change from point to point. $\endgroup$ Jun 26, 2017 at 14:06

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