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Is this correct? (eigenfunctions)

  1. Jun 23, 2009 #1
    1. The problem statement, all variables and given/known data

    a) Show that the functions [tex]f=sin(ax)[/tex] and [tex]g=cos(ax)[/tex] are eigenfunctions of the operator [tex]\hat{A}=\frac{d^2}{dx^2}[/tex].

    b) What are their corresponding eigenvalues?

    c)For what values of [tex]a[/tex] are these two eigenfunctions orthogonal?

    d) For [tex]a=\frac{1}{3}[/tex] construct a linear operator of [tex]f[/tex] and [tex]g[/tex] which is orthogonal to [tex]f[/tex]

    3. The attempt at a solution

    a) [tex]\hat{A}f=\frac{d^2}{dx^2}sin(ax)=-a^2sin(ax)[/tex]
    [tex]\hat{A}g=\frac{d^2}{dx^2}cos(ax)=-a^2cos(ax)[/tex]

    b) the eigenvalues are [tex]-a^2[/tex]

    c)orthogonality condition is: [tex]\int f^*gdx=0[/tex]

    so to satisfy the above condition [tex]a[/tex] would have to be [tex]\pm\frac{n\pi}{2} \pm\n\pi[/tex] where [tex]n=\pm1,\pm2,\pm3....[/tex]

    d) I have no clue how to do this one.. Any help???
     
  2. jcsd
  3. Jun 23, 2009 #2

    HallsofIvy

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    Okay.

    Okay.

    Integrated over what interval?

    so to satisfy the above condition [tex]a[/tex] would have to be [tex]\pm\frac{n\pi}{2} \pm\n\pi[/tex] where [tex]n=\pm1,\pm2,\pm3....[/tex][/quote]
    How did you get that?

    What does it mean for a linear functional be be orthogonal to a (function). For that matter why are they asking for a linear operator of f and g? Why not just a linear operator on the set of function f and g belong to?
     
  4. Jun 24, 2009 #3
    For part c) the integral is performed under all space.

    When you write out the integral, the integrand is [tex][sin(\frac{x}{3})]^*cos(\frac{x}{3})[/tex] with a constant on the outside of the integral. Now for this to be 0. X would have to be those values that I wrote in my previous post. Does this seem right?

    and for part d) I believe what they are asking is to make a new function (the linear combination of {f and g}) that will then be orthogonal to f. Using the above orthogonality relation. I am not to sure how to construct the linear combination though..
     
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