This equation $\bar n_i=\frac{g_i}{e^{\frac{\epsilon_i-\mu}{kT}}-1}$ is Fermi-Dirac statistics where variables are defined here. The classical equation i.e. the Maxwell Boltzman equation is Gaussian function where its Fourier transform with respect to location is also Gaussian. I need to somehow express the Fermi-Dirac statistics in terms of $\bar x$ in order to calculate the Fourier transform with respect to $\bar x$, somehow perhaps with energies or with $g_i$ -- not clear yet how.

What is the Fourier transform with respect to location such as coordinate $\bar x$ with Fermi-Dirac statistics?


According to my tutor: when you do the Fourier-transform $$\phi(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} f(x)e^{-ikt}dx$$ with Maxwell Boltzman, you get the frequency distribution. The inverse Fourier transform is just $$f(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \phi(k)e^{-ikx}dk.$$

I expected that I would get it here also but it is not Gaussian -- a problem?

According to Wolfram, Fourier Transform of the exponential function $e^{-k_0\left| \bar x \right|}$ is a damped exponential cosine integral more here -- a Lorentzian function. They look similar to Gaussian function, more here, but not satisfying the Gaussian.

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I don't know whether the interperation as frequency distribution is still valid even though we get a different function as with the Fourier Transform of the Gaussian.

Gaussian-Gaussian change is bijective but I cannot immediately see it with Fermi-Dirac-Lorentzian change. How can I be sure to get a nice bell-shaped distribution when I do a Fourier transform on arbitrary function? I know Fourier transform of a single sine wave is just a peak but I don't know how it really works with arbitrary functions such as Fermi-Dirac or Bose-Einstein statistics.

  • $\begingroup$ @AlexanderGruber good point, I need to express the energies or the degenerazy $g_i$ somehow in terms of $\bar x$ -- I don't know yet how to do it, have to think over it... $\endgroup$ – hhh Dec 20 '12 at 22:28
  • $\begingroup$ ...I may be integrating with respect to wrong coordinate, chat here. The goal is to get frequency distribution. $\endgroup$ – hhh Dec 20 '12 at 22:32
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    $\begingroup$ It's been a long time. I don't think you need it anymore. But hey, for other people like me that got here. I don't know if this could help lambda.phys.tohoku.ac.jp/~sirotori/URAitems/Report/… $\endgroup$ – mwoua Aug 30 '16 at 16:16

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