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I hope you're all doing well.

I'm trying to prove the following identity: $$\frac{1}{N} \sum_{j=1}^{N} e^ {\frac{2i\pi(n-n')j}{N}} = \delta_{nn'}$$

but I'm having some troubles.

This is what I tried:

We know that for a finite geometric series, the entire sum is equal to $$\sum_{j=1}^{N} ar^{bj} = \frac{ar^b(r^{bN} - 1)}{r^b - 1}$$

In your case we have: $$r = e$$ $$a = \frac{1}{N}$$ $$b = \frac{2i\pi(n-n')}{N}$$

Using this I found

$$\frac{1}{N} \sum_{j=1}^{N} e^ {\frac{2i\pi(n-n')j}{N}} = \frac{1}{N} \frac{e^{\frac{2i\pi(n-n')}{N}}(e^{{2i\pi(n-n')}} - 1)}{e^{\frac{2i\pi(n-n')}{N}} - 1}$$

but after this I'm not able to see how these exponentials are gonna to result in the Kronecker's delta.

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3 Answers 3

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Be careful with $\delta_{nn'}$: your formula is true if $n, n'$ are either regarded as integers modulo $n$ or restricted to set $\{ 1,...,N \}$. Can you see why the value of the sum is unchanged if $n$ or $n'$ is shifted by an integer multiple of $N$?

Take the formula you derived. In the numerator you have $e^{2 \pi i (n-n')}-1=1-1=0$. This is not quite right if $n-n'$ is divisible by $N$ because then also denominator vanishes. In fact in this special case your formula is wrong, because geometric series formula is valid only if the ratio of consecutive terms of the series is not equal $1$. So in this special case you have to compute the sum in a different way.

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  • $\begingroup$ this helped me a lot. The case with n=n' is trivial and remembering that e^(2*ipiany_integer) is always 1 gives the other case of the solution. Thank you!! $\endgroup$ Mar 3, 2021 at 23:37
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Hint: Try using the Fourier transform. The discrete Fourier transform of $\delta_{nn'}$ is given by:

$$ \mathcal F(\delta_{nn'}) = \sum_{n=1}^N \delta_{nn'} e^{-2 i\pi j n/N} = e^{-2 i\pi j n'/N} $$

We can simply invert $\mathcal F$ to obtain an expression for $\delta_{nn'}$. The inverse Fourier transform of $e^{-2 i\pi j n'/N}$ is given by,

$$ \delta_{nn'} = \mathcal F^{-1}(e^{-2 i\pi j n'/N}) = \frac{1}{N} \sum_{j=1}^N e^{-i 2\pi j n'/N} e^{2 i\pi j n/N} = \frac{1}{N} \sum_{j=1}^N e^{2 i \pi j (n-n')/N} $$

As desired!

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  • $\begingroup$ OP's identity is usually seen when one introduces discrete Fourier transforms, and it is used to prove that it works (specifically, that it is an orthogonal transformation). So using Fourier transforms to prove it would be tautological. $\endgroup$ Mar 4, 2021 at 9:03
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Let $t= 2 \pi (n-n')$ and $r=e^{\frac {it } N}$. Then the given expression is $\frac 1N \sum\limits_{j=1}^{N} r^{j}$. This is $\frac 1N \frac {r^{N+1}-r} {1-r}$. And $r^{N+1}=r$ if $n \neq n'$ (because $r^{N}=1$). The case $n=n'$ is trivial.

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