How to propagate a planar e/m field in free space using plane waves? I read this great answer to this question:
Numerical software to manipulate a light beam in its plane wave representation?
The main thing that I am trying to clear in my head is the following:
Given that plane waves are eigenmodes of maxwell's equations in free space in the absence of sources, shouldn't I be able to use this to propagate a planar e/m field exactly (ignoring aliasing effects)?
For example I know the field that exits in the x, y plane, can't I talk about what it will look like at +/- z0? From the answer I would assume you cannot and the closer that comes to using this type of thinking is the "angular spectrum propagation" as the poster describes it, which seems to me to be the same as the "beam propagation method" but not sure about this either.
If the question requires further clarification please let me know.
Thank you
 A: The propagation of electromagnetic radiation is governed by diffraction.  The exact form of the field at any point can be calculated by solving the diffraction integral
$$
  E(r)\propto\int\int E_0(r^\prime)\frac{e^{-ik|\mathbf{r}-\mathbf{r^\prime}|}}
    {4\pi|\mathbf{r}-\mathbf{r^\prime}|}dx^\prime dy^\prime.
$$
Solving this integral analytically is impossible except for a very small number of initial fields.  Solving it numerically on a computer is very computationally costly, but it can certainly be done.  Fortunately, in the far field the integral can be approximated as
$$
  E(r)\propto\frac{e^{-ikr}}{4\pi r}\int\int E_0(r^\prime)
    e^{ik\mathbf{r^\prime}\cdot\hat{\mathbf{r}}}dx^\prime dy^\prime.
$$
Note that that is a dot product in the exponential with the unit vector $\hat{\mathcal{r}}$.  Writing this out in rectangular coordinates yields
$$
  E(r)\propto\frac{e^{-ikr}}{4\pi r}\int\int E_0(r^\prime)e^{-i(k_x x^\prime+k_y y^\prime)}
    dx^\prime dy^\prime,
$$
which is just the two dimensional Fourier transform.  So, the FFT numerical propagation techniques are used when working in the far field because they are significantly less computationally costly even though an exact solution is possible. 
