Square root of volume in fourier expansion of the vector potential

[center o bass]

Hi. I just wondered why we use a $1/\sqrt{V}$ in the Fourier expansion of the vector potential. A regular 3 dimensional Fourier expansion is just

$$f(\vec r) = \sum_{\vec k} c_\vec{k} e^{i \vec k \cdot \vec r}$$

but as the solution to the equation

$$(\frac{\partial ^2}{\partial t^2} - \nabla^2 ) \vec A(\vec r,t) = 0$$

one usually writes

$$\vec A(\vec r,t) = \frac{1}{\sqrt{V}}\sum_{\vec k} \vec A_{0 \vec{k}}(t) e^{i \vec k \cdot \vec r}.$$

What is the reason for this? Doesn't this also screw up the dimension of the euation when $\vec A_{0 \vec k}$ already has the same dimension as the vector potential since the plane wave solutions are

$$\vec A_{0 \vec k}(t) e^{i \vec k \cdot \vec r}?$$

 

[Jano L.]

why we use a 1/V−−√ in the Fourier expansion of the vector potential

It is just a convention. Perhaps you came across it in a book on quantum theory of radiation, where it is sometimes used because it leads to expression of the Poynting energy of the field in the box V in which V does not appear.

If we used the standard convention

$$\vec A(\vec r,t) = \sum_{\vec k} \vec A_{0 \vec{k}}(t) e^{i \vec k \cdot \vec r},$$

the integration over volume would introduce the prefactor V in the total Poynting energy.

...Doesn't this also screw up the dimension of the euation when A⃗ 0k⃗ already has the same dimension as the vector potential...

The dimension of $\vec A(\vec r,t)$ stays the same as in the standard convention, but the dimension of $\vec A_{0\vec k}$ is $[ A(\vec r,t) ] \mathrm{m}^{3/2}$.