# Fourier components of vector potential

1. Jun 18, 2011

### dkin

Hi all,

I was reading the http://en.wikipedia.org/wiki/Quanti...d#Electromagnetic_field_and_vector_potential" and I am a little bit confused.

In this equation defining the vector potential

$\mathbf{A}(\mathbf{r}, t) = \sum_\mathbf{k}\sum_{\mu=-1,1} \left( \mathbf{e}^{(\mu)}(\mathbf{k}) a^{(\mu)}_\mathbf{k}(t) \, e^{i\mathbf{k}\cdot\mathbf{r}} + \bar{\mathbf{e}}^{(\mu)}(\mathbf{k}) \bar{a}^{(\mu)}_\mathbf{k}(t) \, e^{-i\mathbf{k}\cdot\mathbf{r}} \right)$

are the $a^{(\mu)}_\mathbf{k}(t)$ and $\bar{a}^{(\mu)}_\mathbf{k}(t)$ complex or real?

On my first reading I assumed complex as the text says the bar indicates complex conjugates but the vector potential is later defined as a linear addition with complex $\mathbf{e}^{(1)}(\mathbf{k})$ and $\mathbf{e}^{(-1)}(\mathbf{k})$ basis vectors so why would these vectors have complex multipliers? They are also not in bold font which would indicate scalers.

Is the bar in this case simply a label for the scaler?

Last edited by a moderator: Apr 26, 2017
2. Jun 18, 2011

### atyy

Fourier coefficients of a function are generally complex, because they are the Fourier transform of the function.

3. Jun 18, 2011

### dkin

Thanks for the reply, I think the fourier coefficients will be complex regardless as the $\mathbf{e}^{(1)}(\mathbf{k})$ and $\mathbf{e}^{(-1)}(\mathbf{k})$ vectors are complex therefore $\mathbf{e}^{(1)}(\mathbf{k}) a^{(1)}_\mathbf{k}(t)$ should be complex regardless of whether $a^{(1)}_\mathbf{k}(t)$ is.. I think.

Last edited: Jun 18, 2011