Evaluating the sum over all n for e^{ikna}

Physics Q&A
1

Hi, in the book (and lecture notes) the Fourier transform of the density of lattice points in one dimension is calculated and eventually the following sum is obtained:

\sum_n e^{ikna}.

I have some problems of obtaining the same result as in the lecture notes (see end).

Intuitively, I would want to approach this sum as follows:

\sum_n e^{ikna} = \sum_{-\infty}^{\infty} (e^{ika})^n = \sum_{0}^{\infty} (e^{ika})^n + \sum_{-\infty}^{-1} (e^{ika})^n
= \sum_{0}^{\infty} (e^{ika})^n + \sum_{-\infty}^{0} (e^{ika})^n - 1
= \sum_{0}^{\infty} (e^{ika})^n + \sum_{0}^{\infty} (e^{-ika})^n - 1
= \frac{1}{1-e^{ika}} + \frac{1}{1-e^{-ika}} - 1
= \frac{1-e^{-ika}+1-e^{ika}}{(1-e^{ika})(1-e^{-ika})} - \frac{(1-e^{ika})(1-e^{-ika})}{(1-e^{ika})(1-e^{-ika})}
= \frac{1-e^{-ika}+1-e^{ika}-1+e^{-ika}+e^{ika}-1}{(1-e^{ika})(1-e^{-ika})}

Now we see that for all k, except for when k is a multiple of 2 \pi/a, the sum is 0.

(I assumed that

\sum_{0}^{\infty} (e^{ika})^n = \frac{1}{1-e^{ika}}

even though I guess this is only true when |e^{ika}| < 1? So I am not sure if this is allowed.)

So now we are left with the possible values of k being a multiple of 2 \pi/a and so we are left with the sum:

\sum_n e^{i2\pi m n } = \sum_n (e^{i2\pi m})^n = \sum_n 1.

What mathematical steps are made to rewrite this to (specifically, where does that the factor in front come from)

\frac{2\pi}{|a|} \sum_m \delta(k-2\pi m/a)

?

2

You may recognize the sum \sum_n e^{ikna} as a Fourier series of a signal f(k) that has period 2\pi/a and Fourier coefficients c_n=1 for all n.

This series is indeed the Fourier series of f(k) = \frac{2\pi}{a}\sum_m \delta(k-2\pi m/a). We can see that by writing f(k) as a Fourier series f(k) = \sum_n c_n e^{ikna} and then calculating the coefficients using c_n= \frac{1}{2\pi/a}\int_{1 period} f(k) e^{-inka}. (Note here that we need to integrate over one period of the signal and divide by the period of the signal). Plugging in f(k) we indeed see that we get c_m=1 for all m

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