- #1
center o bass
- 560
- 2
Hi. I just wondered why we use a [itex] 1/\sqrt{V}[/itex] in the Fourier expansion of the vector potential. A regular 3 dimensional Fourier expansion is just
[tex] f(\vec r) = \sum_{\vec k} c_\vec{k} e^{i \vec k \cdot \vec r}[/tex]
but as the solution to the equation
[tex] (\frac{\partial ^2}{\partial t^2} - \nabla^2 ) \vec A(\vec r,t) = 0[/tex]
one usually writes
[tex] \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}.[/tex]
What is the reason for this? Doesn't this also screw up the dimension of the euation when [itex]\vec A_{0 \vec k}[/itex] already has the same dimension as the vector potential since the plane wave solutions are
[tex] \vec A_{0 \vec k}(t) e^{i \vec k \cdot \vec r}?[/tex]
[tex] f(\vec r) = \sum_{\vec k} c_\vec{k} e^{i \vec k \cdot \vec r}[/tex]
but as the solution to the equation
[tex] (\frac{\partial ^2}{\partial t^2} - \nabla^2 ) \vec A(\vec r,t) = 0[/tex]
one usually writes
[tex] \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}.[/tex]
What is the reason for this? Doesn't this also screw up the dimension of the euation when [itex]\vec A_{0 \vec k}[/itex] already has the same dimension as the vector potential since the plane wave solutions are
[tex] \vec A_{0 \vec k}(t) e^{i \vec k \cdot \vec r}?[/tex]