In optics, we often come across complex wave vectors that describe absorption, dispersion, etc. given as:

$\textbf{k} = \textbf{k}_{real} + i\textbf{k}_{imag}$

The electric field in phasor notation is:

$\textbf{E} = \textbf{E}_o e^{i(\textbf{k}.\textbf{r} – \omega t)}

= \textbf{E}_o e^{-\textbf{k}_{imag}. \textbf{r}} e^{i(\textbf{k}_{real}.\textbf{r} – \omega t)}$

The question is, how would we define the direction of the wave or rather how would we define the unit vector of the complex wave vector (I am thinking of something like $\hat{k}$).

Because in the case of a real wave vector, there is a physically conceivable concept of a norm and we, therefore, divide the vector by its norm to arrive at a unit vector.

## Best Answer

It's important to point out that $\mathbf{k}\in\mathbb C^3$, here $\mathbb C$ is the field of complex numbers. So, $\mathbf{k}$ is a

three-dimensional vector over $\mathbb C$. However, you can also think about it as asix-dimensional vector over $\mathbb R$. See also this discussion on Mathematics SE for reference about isomorphism of $\mathbb C^n \cong \mathbb R^{2n}$, where $n=3$ for this particular case.Now, a unit vector in a normed vector space - is a vector of length 1. This definition totally applies to our case of wavevector $\mathbf{k}$. So, according to that:

$$ \hat{\mathbf{k}}=\frac{\mathbf{k}}{||\mathbf{k}||} $$ where $||\mathbf{k}||$ is

somenorm in our normed vector space. You can equip your normed vector space with a lot of norms, depending on your purpose, provided the norm you define is actually a norm. However, I would point out that it isa choice you make. Your choice of the norm is dictated by what do you intend to do with your quantities of interest.So, a unit vector without the knowledge

what norm has been used to normalize itprovides incomplete information.It is very common to use a 2-norm, then for vectors in $\mathbb C^3$:

$$ \hat{\mathbf{k}}=\frac{\mathbf{k}}{||\mathbf{k}||_2}=\frac{\mathbf{k}}{\sqrt{\sum\limits_{j=1}^{3}{|k_j|^2}}}=\frac{\mathbf{k}}{\sqrt{\sum\limits_{j=1}^{3}{k_j\bar{k_j}}}}, $$ where $k_j$ is the $j$th element of $\mathbf{k} \in \mathbb C^3$, and $\bar{k_j}$ denotes the conjugation.

Notice, that the unit vector $\hat{\mathbf{k}}\in \mathbb{C}^3$, so is still a complex vector in three-dimensional space over the field of complex numbers.

How do you interpret this unit vector? It depends. I would not say there is an obvious interpretation and usefulness of it in this form. You certainly cannot just say about

the direction(at least in the common understanding of the Cartesian XYZ coordinate system, where $x,y,z\in \mathbb R$).You could ask about the direction of propagation, which would be given by the imaginary part of $\mathbf k$. However, you would have to separate the concepts. Otherwise, you are working in a too complicated vector space.