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.

  • $\begingroup$ Are you asking about the direction of $\mathbf k$ or $\mathbf E$? $\endgroup$ Jun 25, 2019 at 17:10
  • $\begingroup$ I am asking about k $\endgroup$
    – Babaji
    Jun 26, 2019 at 8:35

4 Answers 4


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 a six-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 some norm 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 is a 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 it provides 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.

  • 1
    $\begingroup$ I think you mean that the direction of propagation would be given by the real part of k. Maybe the imaginary part of k provides us with the direction in which the field is being attenuated? $\endgroup$
    – Babaji
    Jun 26, 2019 at 8:39

I can't comment so I write it as an answer, but the norm for a complex number $z = a + ib$ is $||z||= \sqrt{z\overline{z}}$, where $\overline{z}= a - ib$. For the case where you have $z = (z_1, z_2, ...,z_n)$ you have $||z|| = \sqrt{\sum_i^nz_i\overline{z_i}}$.

  1. The imaginary part of the complex vector corresponds to the decay magnitude and the decay direction.
  2. The real part of the complex vector corresponds to the spatial oscillation frequency and the direction of oscillation.

In general, the decay and trig oscillation directions can be different.


Your instincts about pulling out the imaginary part is spot on! Here is an explanation as to why, and some educated guesses from there.

Consider an electric wave penetrating from air to metal.

Let $\vec{E}=\vec{E_0}e^{i(\vec{k}\cdot \vec{x}-\omega t)}$

Then taking the curl:

$\nabla \times \vec{E} =-\partial \vec{B}/ \partial t=i\vec{k}\times\vec{E_0}e^{i(\vec{k}\cdot \vec{x}-\omega t)}$

So $\vec{B}=(\vec{k}/\omega) \times \vec{E_0}e^{i(\vec{k}\cdot \vec{x}-\omega t)}$

The curl of $\vec{B}$ by

By Ohm's Law, $\vec{J}=\sigma\vec{E}$. Where $\vec{J}$ is the current density and $\sigma$ is the conductivity. We also assume $\vec{E_0}$ is time independent, then

$\frac{\partial \vec{E}}{\partial t }=-i\omega\vec{E}$

So by Ampere's Law: $\nabla \times \vec{B}=\mu_0\sigma\vec{E}-i\omega\vec{E}/c^2$

And using the identity:

$$\nabla \times (\vec{A}\times \vec{B})=(\nabla \cdot \vec{B})\vec{A}+(\vec{B}\cdot \nabla)|\vec{A}-\vec{B}(\nabla \cdot \vec{A})-(\vec{A} \cdot \nabla)\vec{B}$$

And where the curl of $\vec{k}$ is 0.

So $\nabla \times \vec{B}=(ik^2/\omega)\vec{E}$

So $(ik^2/\omega)=\mu_0\sigma-i\omega/c^2$

Solve for $k$, you get a complex number, say $k=a+bi$. So $ik=ai-b$. So we get $e^{-b}$. The parameter, $b$, measures the decay of the electric field in the metal corresponding to the electric field being zero in the conductor.

So while the imaginary part carries physical meaning regarding decay in intensity, it doesn't seem to imply anything about the location of the particle. Not only does only the real part enter into the expression $i(\vec{a}\cdot \vec{x}-\omega t)$.

The time derivative of the electric field is also suggestive about what to make of the imaginary part of the wave number.

$$\frac{d\vec{E}}{dt}=-\nabla \vec{E}\cdot \vec{u}=-\vec{k}\vec{E}\cdot \vec{u}=-i\omega t \vec{E}$$

Where here $\vec{u}$ represents a velocity , $\vec{u}=\vec{m}+i\vec{n}$.

This leaves us with

$$(\vec{a}+i\vec{b})\cdot (\vec{m}+i\vec{n})=\omega $$

Equating real and imaginary parts we have:

$$\vec{a}\cdot \vec{m}-\vec{b}\cdot \vec{n}=\omega $$ $$\vec{a}\cdot\vec{n}+\vec{b}\cdot \vec{m}=0$$

Now $i\vec{n}$ represents the imaginary part of the velocity, $\vec{u}$. But what does an imaginary velocity mean?

If we assume it's zero, then the imaginary part of our wave number is orthogonal to the velocity.

Also remember the role the wave number plays trigonometrically, its is essentially the argument for $\sin{kx-\omega t}$.

If $k=a+ib$, we have $\sin{(a+ib)x-\omega t}$

But what is the the sine of a purely complex number? We can find this from Euler's Formula and find that its a decaying exponential.

So in terms of both velocity and cycles of constituent sine functions, the imaginary part of the wave number plays no role. It only represents a real exponent.

So it seems the physical meaning of the norm for $(\vec{k}=\vec{a}+i\vec{b}) $is $ norm Re\{\vec{k}\}$

  • $\begingroup$ How does this answer the question? $\endgroup$ Jun 25, 2019 at 17:09
  • $\begingroup$ I think i patched it up. $\endgroup$
    – R. Romero
    Jun 26, 2019 at 19:14
  • $\begingroup$ It still seems like there is a lot of superfluous information that doesn't tackle the main point of the question. $\endgroup$ Jun 26, 2019 at 20:22

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