How to find wave vector of superposition of electromagnetic waves Suppose I have two arbitrary electric fields (vector fields), $\mathbf{E}_1 (\mathbf{r})$ and $\mathbf{E}_2 (\mathbf{r})$, which are a function of position $\mathbf{r}$ (the $e^{i\omega t}$ is implicit). The wave vectors of these fields are $\mathbf{k}_1 (\mathbf{r})$ and $\mathbf{k}_2 (\mathbf{r})$. I understand that the superposition of these electric fields, $\mathbf{E} (\mathbf{r})$, at a point $\mathbf{r}$ is given by:
$$ \mathbf{E} (\mathbf{r}) = \mathbf{E}_1 (\mathbf{r}) + \mathbf{E}_2 (\mathbf{r}) $$
How do I find the wavevector of the superposition at the point $\mathbf{r}$? Is it as simple as:
$$ \mathbf{k} (\mathbf{r}) = \mathbf{k}_1 (\mathbf{r}) + \mathbf{k}_2 (\mathbf{r}) $$
Or is there another way to calculate it?
 A: A way to define the wave vector of an arbitrary field is in terms of the expectation value. First, consider an electric field that is described by an angular spectrum $H(\mathbf{k})$. Then one electric field is given by
$$ E(\mathbf{r}) = \int H(\mathbf{k}) \exp(-i \mathbf{k}\cdot\mathbf{r})\ d\mathbf{k} . $$
The expectation value would then be
$$ \langle k \rangle = \frac{1}{P} \int \mathbf{k} |H(\mathbf{k})|^2\ d\mathbf{k} , $$
where
$$ P = \| E(\mathbf{r}) \|^2 = \int |H(\mathbf{k})|^2\ d\mathbf{k} . $$
In the example of the superposition of two electric fields, where each is (presumably) a plane wave with wave vector $\mathbf{k}_1$ and $\mathbf{k}_2$, respectively, the expectation value would simplify to
$$ \langle k \rangle = \frac{\mathbf{k}_1 \| E_1(\mathbf{r}) \|^2 + \mathbf{k}_2\| E_2(\mathbf{r}) \|^2}{\| E_1(\mathbf{r}) \|^2 + \| E_2(\mathbf{r}) \|^2} . $$
Hope this explanation is clear enough.
A: As you correctly wrote, the superposition of waves is
$$\mathbf{E}(\mathbf{r}) = \mathbf{E}_1(\mathbf{r}) + \mathbf{E}_2(\mathbf{r}).$$ This is true for any form of electric field (not encessarily plane waves) and in no way translates in a superposition rule for wave vectors.
If $\mathbf{E}_{1,2}(\mathbf{r})$ are plane waves,
$$\mathbf{E}_{1,2}(\mathbf{r}) = \mathbf{E}_{1,2}e^{i\mathbf{k}_{1,2}\mathbf{r}} + c.c.$$ there superposition is just
$$\mathbf{E}(\mathbf{r}) = \mathbf{E}_{1}e^{i\mathbf{k}_{1}\mathbf{r}} + \mathbf{E}_{2}e^{i\mathbf{k}_{2}\mathbf{r}} + c.c.$$
A: I'm assuming that the waves in your question are plane waves $\mathbf{E}_i(r) = \mathbf{E}_i \, \mathrm{e}^{i \mathbf{k}_i \cdot \mathbf{r}}$.
In general there is no way to write $\mathbf{E}_1(r) + \mathbf{E}_2(r)$ in the form $\mathbf{E} \, \mathrm{e}^{i \mathbf{k} \cdot \mathbf{r}}$. In the special case where $\mathbf{E}_1 = \mathbf{E}_2$, you can use trigonometric sum-to-product identities to write
$$\mathbf{E}(r) = \mathbf{E}_1 \cos\left[\frac{(\mathbf{k}_1 - \mathbf{k}_2)\cdot \mathbf{r}}{2} \right] \exp\left[i \frac{(\mathbf{k}_1 + \mathbf{k}_2) \cdot \mathbf{r}}{2} \right],$$
but there is no formula for the general case.
A: TL,DR: k-vectors don’t interact during superposition.
Your electric fields my look like $\vec{E}_m =\vec{E}_{m,0}\exp(i\vec{k}_m\,\vec{r})$. If you superimpose two electric fields of this kind, the k-vectors are not interacting with ich other: The product $\vec{k}_m\cdot\vec{r}$ is a scalar and argument of the e-function.
$$
\vec{E}_1+\vec{E}_2=\left.\vec{E}_{1,0}\,\exp(i\vec{k}_1\,\vec{r})+\vec{E}_{2,0}\,\exp(i\vec{k}_2\,\vec{r})\right|_{r=r_0} = \vec{E}_{1,0}\,c_1+\vec{E}_{2,0}\,c_2
$$
with $c_m$ denoting to a constant, $|c_m|\leq 1$.
k denotes to the direction of propagation of the electromagnetic wave. Unless something weird is going on (like Photon-Photon-scattering), each wave keeps its original direction during superposition.
