- #1
FranzDiCoccio
- 342
- 41
Fourier series vs Integral: just one "coherent" notation?
Hi all,
notations for Fourier analysis always confuse me.
I understand that the "volume" normalization factor can be disposed of either in the definition of the Fourier coefficient or in the definition of the Fourier series (or split among the two).
However, I have the feeling that the only choice which does not need "coefficient rearrangement" in taking the infinite volume limit is the following
[tex] F_k = \int dx e^{-i k x} F(x), \qquad F(x) = \frac{1}{L}\sum_k e^{i k x} F_k [/tex]
Indeed in the limit [tex]L\to\infty[/tex] one has
[tex]\sum_k \rightarrow \frac{L}{2 \pi}\int dk [/tex]
which neatly cancels out the [tex]L[/tex] coefficient in [tex]F(x)[/tex].
It seems to me that any other choice making sense in the finite volume limit, like e.g.
[tex] F_k = \frac{1}{L} \int dx e^{-i k x} F(x), \qquad F(x) = \sum_k e^{i k x} F_k [/tex]
gives rise to awkward volume dependent coefficients.
Does this make sense?
Thanks a lot
Hi all,
notations for Fourier analysis always confuse me.
I understand that the "volume" normalization factor can be disposed of either in the definition of the Fourier coefficient or in the definition of the Fourier series (or split among the two).
However, I have the feeling that the only choice which does not need "coefficient rearrangement" in taking the infinite volume limit is the following
[tex] F_k = \int dx e^{-i k x} F(x), \qquad F(x) = \frac{1}{L}\sum_k e^{i k x} F_k [/tex]
Indeed in the limit [tex]L\to\infty[/tex] one has
[tex]\sum_k \rightarrow \frac{L}{2 \pi}\int dk [/tex]
which neatly cancels out the [tex]L[/tex] coefficient in [tex]F(x)[/tex].
It seems to me that any other choice making sense in the finite volume limit, like e.g.
[tex] F_k = \frac{1}{L} \int dx e^{-i k x} F(x), \qquad F(x) = \sum_k e^{i k x} F_k [/tex]
gives rise to awkward volume dependent coefficients.
Does this make sense?
Thanks a lot