Understanding the Relationship Between Kronecker Delta and Dirac Delta

[andrewm]

Is it true that $$\sum_x e^{i(k-k')x} = \delta_{k-k'}$$, where $$\delta$$ is the Kronecker delta? I've come across a similar relation for the Dirac Delta (when the sum is an integral). I do not understand why $$k-k' \neq 0$$ implies the sum is zero.

Edit: In fact, I'm really confused, since it seems that when the $$x=0...\inf$$ and k=k' the sum is infinite. So is it a Dirac delta?

 

[Meir Achuz]

In that sum, x should be an integer, so it is usually written as n.
The sum should be from n=-N to n=+N, and divided by 2N+1.
Then in the limit N-->infinity, it is the Kronecker delta.

 

[Avodyne]

There are really two different formulas here. For $x$ real (not necessarily an integer), we have

$$\sum_{k=-\infty}^{+\infty}e^{ikx}=2\pi\sum_{n=-\infty}^{+\infty}\delta(x-2\pi n)$$

where $\delta(x)$ is the Dirac delta function.

For $n$ an integer,

$$\sum_{k=1}^{N}e^{2\pi ikn/N}=N\delta_{n0}$$

where $\delta_{nm}$ is the Kronecker delta. To get an idea of why this sums to zero when $n\ne 0$, consider the case $N=4$; then the four numbers being summed are $i^n, (-1)^n, (-i)^n, 1^n$.

 

[andrewm]

Thank you both, I've written these handy formulas for my future reference.

 

[Avodyne]

$$N\delta_{n0}$$ should have been $$N\delta_{n\mathop{\rm mod}N,0}$$
That is, $N$ if $n=0,\pm N,\pm 2N, \ldots,$ and zero otherwise.