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

<br /> \vec A(\vec r,t) = \sum_{\vec k} \vec A_{0 \vec{k}}(t) e^{i \vec k \cdot \vec r},<br />

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}.