Proving the Poisson summation formula (like a physicist)

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SUMMARY

The discussion focuses on proving the Poisson summation formula using standard notation familiar to physicists. The function defined is f(x) = ∑_{k=-∞}^∞ g(x + 2πk), which is periodic and can be expressed as a complex Fourier series. The coefficients are derived as c_n = (1/2π) ∫_0^{2π} f(x)e^{-inx} dx, leading to the transformation of the summation into the desired form. The final steps involve substituting values and manipulating the integrals to achieve the proof of the formula.

PREREQUISITES
  • Understanding of complex Fourier series
  • Familiarity with periodic functions and their properties
  • Knowledge of integral calculus, particularly in the context of Fourier transforms
  • Basic concepts of quantum mechanics and mathematical notation
NEXT STEPS
  • Study the derivation of the Fourier series coefficients in detail
  • Explore the applications of the Poisson summation formula in quantum mechanics
  • Learn about the properties of periodic functions and their Fourier expansions
  • Investigate the implications of the Poisson summation formula in signal processing
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Students and professionals in physics, particularly those studying quantum mechanics, as well as mathematicians interested in Fourier analysis and summation techniques.

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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]
 
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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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