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

<br /> \sum_{k=-\infty}^{+\infty}e^{ikx}=2\pi\sum_{n=-\infty}^{+\infty}\delta(x-2\pi n)<br />

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

For n an integer,

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

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.