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Find Fourier coefficients - M. Chester text

  1. Jun 24, 2017 #1

    GreyNoise

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    1. The problem statement, all variables and given/known data
    I am self studying an introductory quantum physics text by Marvin Chester Primer of Quantum Mechanics. I am stumped at a problem (1.10) on page 11. We are given

    [itex] f(x) = \sqrt{ \frac{8}{3L} } cos^2 \left ( \frac {\pi}{L} x \right ) [/itex]

    and asked to find its Fourier coefficients [itex] C_0 [/itex] and [itex] C_1 [/itex] from

    [itex] C_k = \frac{1}{L} \displaystyle {\int_0^L} e^{-ikx} f( x ) dx [/itex]

    I found [itex] C_0 = \sqrt{ \frac{2}{3L} } [/itex]; it was easy as k=0 greatly simplified the integral. But I am completely baffled at an answer for the [itex] C_1 [/itex] coefficient. The text has the answer as [itex] \frac{1}{ \sqrt{6L} } [/itex], and I do not get anything remotely close that; I keep getting answers with nonzero imaginary parts and [itex] \pi [/itex] in the terms.

    2. Relevant equations

    [itex] f(x) = \sqrt{ \frac{8}{3L} } cos^2 \left (\frac{\pi}{L} x \right) [/itex]

    [itex] C_k = \frac{1}{L} \displaystyle {\int_0^L} e^{-ikx} f( x ) dx [/itex]

    [itex] C_1 = \frac{1}{ \sqrt{6L} } [/itex]

    3. The attempt at a solution

    To begin, I substituted f(x) into the equation for [itex] C_k [/itex] for k = 1

    [itex] C_1 = \frac{1}{L} \displaystyle {\int_0^L e^{-ix} \sqrt{ \frac {8}{3L} }} cos^2 \left ( \frac {\pi}{L} x \right ) dx [/itex]

    Then I factored the constants out of the integral

    [itex] C_1 = \frac{1}{L} \sqrt{ \frac {8}{3L} } \displaystyle {\int_0^L e^{-ix}} cos^2 \left ( \frac {\pi}{L} x \right ) dx [/itex]

    Then I attempted to solve the integral

    [itex] \displaystyle {\int_0^L e^{-ix}} cos^2 \left ( \frac {\pi}{L} x \right ) dx [/itex]

    I tried doing it by parts, then I tried by substituting [itex] e^{-ix} = cos (x) - i sin (x) [/itex] and distributing [itex] cos^2 \left ( \frac {\pi}{L} x \right ) [/itex] into that and integrating the result, but I only generated 4 pages of blundering chicken scratch that gave me things like [itex] \textstyle {\frac {2i}{3L} \sqrt{ \frac{8}{3L}} } [/itex]; another attempt led to [itex] \textstyle {\frac {1}{\pi} \frac{2i}{3} } [/itex]. I even went to my TI-89 and the Wolfram website and got an analytical solution to the integral, but the definite integrals they returned gave complex numbers that did NOT look anything like [itex] \frac{1}{ \sqrt{6L} } [/itex]. Can anyone tell me; am I even setting up the integral correctly?
     
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  3. Jun 25, 2017 #2

    Dick

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    Double check your form defining the Fourier coefficents. The ##e^{-ikx}## doesn't have period ##L## for ##k=1##. Shouldn't it be something more like ##e^{-i\frac{2k\pi}{L}x}##? ##e^{-ikx}## is used in the transform over the whole line.
     
    Last edited: Jun 25, 2017
  4. Jun 25, 2017 #3

    GreyNoise

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    Thnx much for that Dick! That was the fix I needed. Now it brings to my mind a question about the form of the exponential argument. The author
    wrote
    $$ C_k = \frac{1}{L} \int_0^L e^{-ikx} f(x) dx $$
    and I used the form literally, but now I am concerned that I missed a basic Fourier series concept. The author had written (before the above equation) that
    $$ kL = 2 \pi n $$
    So the form I should have used would appear as
    $$ C_n = \frac{1}{L} \int_0^L e^{-i \frac{2 \pi n}{L} x} f(x) dx $$
    where I substituted ## k = \textstyle \frac{2 \pi n}{L} ## in the exponent's argument. Am I correct on this point?
     
  5. Jun 25, 2017 #4

    Ray Vickson

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    Not quite, but it depends in part on how you plan to use the ##C_n##.

    Basically, it is always easiest to start with an orthonormal set of functions ##\{ u_k(x) \}##, which have the inner-product properties that
    ## \langle u_j, u_k \rangle = 0 ## if ##k \neq j## and ##\langle u_k,u_k \rangle = 1## for all ##k##. Here, ##\langle \cdot , \cdot \rangle## denotes the inner product for functions on ##[0,L]##:
    $$ \langle f,g \rangle = \int_0^L f^*(x) g(x) \, dx, $$
    where ##f^*## is the complex congugate of ##f##.

    You can easily verify that the functions ##u_k(x) =(1/\sqrt{L}) \: e^{i 2 \pi k x/L}, \; k = 0, \pm 1, \pm 2, \ldots ## constitute an orthonormal set. So, if we expand a function ##f(x)## as
    $$f(x) = \sum_{k=-\infty}^{\infty} c_k u_k(x), $$
    we get (with almost no work) that the coefficients are
    $$c_k = \langle u_k , f \rangle = \int_0^L u^*(x) f(x) \, dx = \frac{1}{\sqrt{L}} \int_0^L e^{-i 2 \pi k x/L} f(x) \; dx.$$

    In other words, we can either use the expansion ##\sum c_k u_k## or use the expansion ##\sum_k \sqrt{L} C_k e^{i 2 \pi k x/L}## with your ##C_k## (because ##\text{my} \: c_k = \sqrt{L} \times \text{your} \: C_k##). The factor ##\sqrt{L}## needs to go somewhere, and IMHO it is easier to put it into the definition of the basis functions ##\{ u_k \}##.

    Note added in edit: it turns out that what YOU wrote is also correct! You are using expansion functions ##f_k(x) = e^{i 2 \pi x k/L}, \; k = 0, \pm 1, \pm 2, \ldots## and coefficients ##C_k = (1/L) \int_0^L f^*_k(x) f(x) \, dx,##, so your ##k##th term of the expansion has the form
    $$\text{term}\; k = \frac{1}{L} \int_0^L f^*_k(t) f(t) \, dt \times f_k(x),$$
    and this is the same as the ##k##th term ##c_k u_k(x)## given above.
     
    Last edited: Jun 25, 2017
  6. Jun 29, 2017 #5

    GreyNoise

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    Sorry for delayed response; thnx are in order; I was away on vacation. Thank you Dick and Ray. The posts got me back on top of the curve (my Fourier forays are ten years ago now), so I committed them to my notes. I might be back for more regarding the M. Chester text I am reading. Thnx again.
     
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