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Operator that interchanges variables

  1. Sep 3, 2009 #1
    1. The problem statement, all variables and given/known data

    I suppose that this is not directly a quantum mechanical problem, but this have been assigned as homework for the Quantum Mechanics course.

    Let be an operator L and eigenvalue equation [tex]Lf=\lambda f[/tex]. This operator, applied to a function [tex]f(x,y)[/tex], interchanges the variables i.e. [tex]Lf(x,y)=f(y,x)[/tex]. What's the general property of the eigenfunctions of this problem? Get the possible eigenvalues.

    2. Relevant equations



    3. The attempt at a solution

    Well. I think that if [tex]Lf(x,y)=f(y,x)[/tex] then if [tex]f[/tex] is an eigenfunction, obviously, [tex]\lambda f(x,y)=f(y,x)[/tex]. One possible kind of [tex]f[/tex] that fills conditions is one that is symmetric, that is [tex]f(x,y)=f(y,x)[/tex] then [tex]\lambda[/tex] for this kind of eigenfunctions will be [tex]\lambda=1[/tex]. Others are the antisymmetric ones, those for is true [tex]f(y,x)=-f(x,y)[/tex] and then the eigenvalue for this kind is [tex]\lambda=-1[/tex]

    But I'm sure that there are more conditions that generate other kinds of eigenfunctions, not only symmetric nor antisymmetric. My question is: there are more or the antisymmetric and symmetric ones are the only ones, and if there are more how I get them and their eigenvalues?
     
  2. jcsd
  3. Sep 3, 2009 #2

    gabbagabbahey

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    If [itex]f(y,x)=\lambda f(x,y)[/itex] is to be true for all [itex](x,y)[/itex], then surely it must be true for [itex]x=y[/itex]....what does that tell you?:wink:
     
  4. Sep 3, 2009 #3
    it tells me [tex]\lambda=1[/tex]?
     
  5. Sep 3, 2009 #4

    Avodyne

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    Here's another hint: what happens if you act on f with L twice? What does that tell you about [itex]\lambda^2[/itex]?
     
  6. Sep 3, 2009 #5

    Avodyne

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    Or that f(x,x)=0.
     
  7. Sep 3, 2009 #6
    Or that operator [tex]L[/tex] is an involution, that's, it's its own inverse.
     
  8. Sep 3, 2009 #7
    and then [tex]\lambda=\pm 1[/tex] necessarily and takes all values only if [tex]f\equiv 0[/tex]. In fact zero function is symmetric and antisymmetric function at the same time.
     
  9. Sep 3, 2009 #8

    gabbagabbahey

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    f(x,y)=0 is a trivial solution, and not really an eigenfunction...so it should be discarded.

    That leaves you with [itex]\lambda=1[/itex] and symmetric eigenfunctions, or [itex]\lambda=-1[/itex] and antisymmetric eigenfunctions.
     
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