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Fourier series vs Integral: just one coherent notation?

  1. Sep 24, 2008 #1
    Fourier series vs Integral: just one "coherent" notation?

    Hi all,

    notations for Fourier analysis always confuse me.
    I understand that the "volume" normalization factor can be disposed of either in the definition of the Fourier coefficient or in the definition of the Fourier series (or split among the two).

    However, I have the feeling that the only choice which does not need "coefficient rearrangement" in taking the infinite volume limit is the following

    [tex] F_k = \int dx e^{-i k x} F(x), \qquad F(x) = \frac{1}{L}\sum_k e^{i k x} F_k [/tex]

    Indeed in the limit [tex]L\to\infty[/tex] one has

    [tex]\sum_k \rightarrow \frac{L}{2 \pi}\int dk [/tex]

    which neatly cancels out the [tex]L[/tex] coefficient in [tex]F(x)[/tex].

    It seems to me that any other choice making sense in the finite volume limit, like e.g.

    [tex] F_k = \frac{1}{L} \int dx e^{-i k x} F(x), \qquad F(x) = \sum_k e^{i k x} F_k [/tex]

    gives rise to awkward volume dependent coefficients.

    Does this make sense?

    Thanks a lot
  2. jcsd
  3. Sep 24, 2008 #2


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    Re: Fourier series vs Integral: just one "coherent" notation?

    The factor out front of the integral when you "convert" the sum to an integral comes about because you have to take account of the fact that the variable you're integrating over does not have uniform spacing when you sum over it.

    Let me explain in the context of statistical mechanics: In statistical mechanics, you often find yourself evaluating sums such as

    [tex]\sum_{n=0}^{\infty} e^{-\beta E_n}[/tex]

    where the energy [itex]E_n[/itex] depends on the index n. Note that the indices, n, are equally spaced; this does NOT need to be true for E_n! If we have a LARGE number of terms in this sum, we might approximate it by an integral. For instance, we could approximate this sum by

    [tex]\int_{0}^{\infty}dn~e^{-\beta E(n)}[/tex]

    Note that the variable I'm integrating over is still the index n, and the energy is now approximated as a continuous function of n. However, integrating over n isn't really desirable - we'd much rather integrate over the energy. The problem is that the energy isn't evenly spaced out - the values of [itex]E_n[/itex] have some distribution - a "density of states", typically labelled [itex]g(E)[/itex]. If you want to integrate over energy instead of the index n, you have to include this density of states in your integral:

    [tex]\int_{E_0}^{E_{max}}dE~g(E)e^{-\beta E}[/tex]

    where E is now the variable of integration.

    This is what's happening when you convert your sum over k. Your sum is written as being over "k", but really you have to order the k you are summing over by some index n. Then, when you convert your sum into an integral and you integrate over k, you have to include a density of states:

    [tex]\sum_{n=-\infty}^{\infty}f(k_n) \rightarrow \int dk~g(k) f(k)[/tex]

    For your specific example, [itex]g(k) = L/2\pi[/itex]. This is derived in some model by treating particles/waves in a 1d box. The boundary conditions give [itex]kL = n\pi[/itex], where n runs from [itex]-\infty[/itex] to [itex]\infty[/itex], so when you convert the sum to an integral over n, you get the relation [itex]dn = dk L/\pi[/itex]. Now, the factor of 1/2 comes about because you're integrating over is in some sense a "1 dimensional circle" of the index n, and you only want the positive half of that 'circle', so you take divide [itex]L/\pi[/itex] by 2.

    (The analog of this in 2d is that you have two indices, m and n, and you're integrating the indices over a circle [itex]N_max^2 = m^2 + n^2[/itex], but m and n are >0, so you take 1/4 the area of the circle, which relates to your density of states.)
  4. Sep 25, 2008 #3
    Re: Fourier series vs Integral: just one "coherent" notation?

    Hi Mute,

    and thanks for replying.

    Yes, of course. The origin of the factor is clear to me.

    What slightly confuses me is the following: as far as one is interested in Fourier sums, the two notations I have written in the previous post are equivalently good, right? I mean, the 1/L factor can be either in front of the sum or in front of the integral. It is a matter of taste, I guess, and I think have seen both choices in solid state books.

    When one lets L go to infinity, the sum over k becomes an integral to within a constant proportional to L. If the above 1/L factor is originally in front of the sum over k this constant is canceled out, and the result does not contain L any more (explicitly). The original 1/L factor is replaced by a [tex]1/2 \pi[/tex] factor in the integral over k.

    Starting with the other choice yields a 1/L factor in front of the dx integral and a L factor in front of the dk integral, which seems odd in view of the fact that [tex]L\to\infty[/tex].

    I see that one could write

    F_k = \int dx e^{-i k x} \frac{1}{L}\sum_k e^{i k x} F_k \to F(k)=\int dx e^{-i k x} \frac{1}{2 \pi}\int dq e^{i q x} F(q)

    and then "re-define" the relation between [tex]F(k)[/tex] and [tex]F(x)[/tex] by assigning the [tex]1/2 \pi[/tex] factor at will. This is the "coefficient rearrangement" I was referring to.

    What I was pointing out is that performing the limit on the two relations separately appears to make sense only when the 1/L factor is in front of the sum over k.
    Which, b.t.w., is not the usual notation in solid state.
    Last edited: Sep 25, 2008
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