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The electric field in phasor notation is often written \begin{align} \mathbf{E}(x,y,z,t)&=\Re\{\mathbf{E}_0\mathrm{e}^{j\phi}\mathrm{e}^{j\omega t}\}\\ &=\Re\{\tilde{\mathbf{E}}\mathrm{e}^{j\omega t}\}. \end{align} Does it mean $\mathbf E_0$ is a constant vector, i.e. $$\mathbf E_0=E_{0x}\mathbf{\hat{x}}+E_{0y}\mathbf{\hat{y}}+E_{0z}\mathbf{\hat{z}}?$$

Or a vector field, $\mathbf E_0:\mathbb R^3\rightarrow \mathbb R^3$, i.e. $$ \mathbf{E}_0(x,y,z)=E_{0x}(x,y,z)\mathbf{\hat x}+E_{0y}(x,y,z)\mathbf{\hat y}+E_{0z}(x,y,z)\mathbf{\hat z}? $$

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  • $\begingroup$ Regarding the two extra questions in v5 of this post: yes, and yes. $\endgroup$ – Emilio Pisanty Nov 11 '17 at 15:02
  • $\begingroup$ @EmilioPisanty Great! I thought my edit was more of a math-question so I posted it at math.stackexchange instead. $\endgroup$ – JDoeDoe Nov 12 '17 at 7:33
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It means a combination of the two: the complex amplitude can depend on space, i.e. it is a complex-valued vector field $\tilde{\mathbf{E}}: \mathbb R^3\to\mathbb C^3$, which gives you a space-dependent polarization, \begin{align} \mathbf{E}(x,y,z,t) &=\mathrm{Re}\mathopen{}\bigg[\tilde{\mathbf{E}}(x,y,z)\mathrm{e}^{-i\omega t}\bigg]\mathclose{}, \end{align} including spatial dependence of everything from intensity and ellipticity through the orientation of the polarization plane and of the polarization ellipse within that plane.

For examples of this in action, see e.g. the references in this answer or, say, this paper or this one or this one.

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  • $\begingroup$ I think the direct answer to the question then is that JDoeDoe's second suggestion, that $\textbf{E}_0$ is a complex vector field, is correct. Each component can be a function of x, y, and z. But not time, time dependence has been separated to the $e^{-i \omega t}$ as we are assuming the time dependence is sinusoidal. $\endgroup$ – Kthaxt Nov 8 '17 at 19:20
  • $\begingroup$ @Kthaxt JDoeDoe's second suggestion hat $\mathbf E_0$ as a real vector field. $\endgroup$ – Emilio Pisanty Nov 8 '17 at 20:43

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