electromagnetism - Determining wave vector

If I have an $\mathbf{E}$ field:

$$\mathbf{E}_1 = (\hat x 2e^{j\pi/2} + \hat y5)e^{-j4z}$$

How do I find the wave vector $\vec{k}$?

If I multiply through I get $\mathbf{E}_1 = \hat x 2e^{-j(4z-\pi/2)} + \hat y5e^{-j4z}$, but then what? If it was just in the $\hat y$ direction I think I could just read $\vec{k}$ = 4$\vec{z}$ right?

Answer

Think about what the wave vector represents, and what kind of wave your equation describes.

The most general equation for the spatial variation of the electric field for a plane wave is

$$\mathbf{E}(\mathbf{r}) = \mathbf{E_{0}} e^{-i\mathbf{k} \bullet \mathbf{r}}$$

where $\mathbf{E_{0}}$ is some vector with no dependence on $\mathbf{r}$. This can be written in terms of $x$, $y$, and $z$ as

$$\mathbf{E}(x, y, z) = (E_{0,x}\mathbf{\hat{x}} + E_{0,y}\mathbf{\hat{y}} + E_{0,z}\mathbf{\hat{z}})exp[-i(k_{0,x}x + k_{0,y}y + k_{0,z}z)]$$

Let's compare your equation to the one above. Notice that the $\mathbf{\hat{z}}$ term is not present, meaning that $\mathbf{E}$ has no $z$-dependence, and must lie in the $xy$-plane. Now matching terms, you can see that $E_{0,x} = 2e^{j \pi/2}$ and $E_{0,y} = 5$. I'll leave you to do the comparison with the exponential factor, but hopefully it's clear how to find $\mathbf{k}$ at this point.

It's always important to try and visualize the wave as best you can. All plane waves are expressible in this form, so it's just a matter of figuring out in which plane the $\mathbf{E}$-field oscillations are occurring (the $xy$-plane here), and if it's a transverse wave, the direction of propagation and the direction of $\mathbf{k}$ is normal to that plane.