Proving the Poisson summation formula (like a physicist)

  • #1
Hi! I'n my quantum mechanics homework I've been asked to proved the Poisson summation formula. The mathematicians seem to use abstract and confusing notation when proving this kind of thing so I'm hoping for some help from physicists in standard notation ;)

I'm starting with a function
[tex] f(x) = \sum_{k = - \infty}^\infty g(x + 2\pi k) [/tex]

which means that f is periodic in periods of two pi and thus can be explanded as a complex fourier series

[tex] f(x) = \sum_{n = -\infty}^\infty c_n e^{inx}[/tex]

with coefficients

[tex]c_n = \frac{1}{2\pi } \int_0^{2\pi} f(x) e^{-inx} dx[/tex]

This can manipulated further into
[tex]c_n = \frac{1}{2\pi } \int_0^{2\pi} f(x) e^{-inx} dx =\frac{1}{2\pi } \sum_{k=-\infty}^{\infty} \int_0^{2\pi} g(x + 2 \pi k) e^{-inx} dx = \frac{1}{2\pi } \sum_{k=-\infty}^{\infty} \int_{2\pi k}^{2\pi(k+1)}g(x) e^{-inx} dx = \frac{1}{2\pi} \int_{-\infty} ^\infty g(x) e^{-i nx} dx [/tex]
and that is all nice. However, how does one take the step from this and to the resulting

[tex] \sum_{n = -\infty}^\infty g(n) = \sum_{k = - \infty}^\infty \int_{-\infty}^\infty g(x)e^{-2\pi i k x} dx?[/tex]
 

Answers and Replies

  • #2
vanhees71
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Just plug your result into your ansatz:
[tex]f(x)=\sum_{n \in \mathbb{Z}} c_n \exp(\mathrm{i} n x)=\sum_n \frac{1}{2 \pi} \int_{\mathbb{R}} \mathrm{d} y g(y) \exp[\mathrm{i} n(x-y).[/tex]
If you set [itex]x=0[/itex] in this formula you get
[tex]f(x)=\sum_n g(n)=\sum_n \frac{1}{2 \pi} \int_{\mathbb{R}} \mathrm{d} y \mathrm g(y) \exp(-\mathrm{i} n y).[/tex]
Finally make a substitution [itex]y=2 \pi x[/itex], leading to the equation you want to prove.
 

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