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Homework Help: Hermiticity of permutation operator

  1. Aug 11, 2011 #1

    here's a problem I am having.

    Consider the hilbert space of two-variable complex functions [tex]\psi (x,y)[/tex].

    A permutation operator is defined by its action on [tex]\psi (x,y)[/tex] as follows.

    [tex]\hat{\pi} \psi (x,y) = \psi (y,x) [/tex]

    a) Verify that operator is linear and hermitian.

    b) Show that

    [tex]\hat{\pi}^2 = \hat{I}[/tex]

    find the eigenvalues and show that the eigenfunctions of [tex]\hat{\pi}[/tex] are given by

    [tex]\psi_{+} (x,y)= \frac{1}{2}\left[ \psi (x,y) +\psi (y,x) \right] [/tex]


    [tex]\psi_{-} (x,y)= \frac{1}{2}\left[ \psi (x,y) -\psi (y,x) \right] [/tex]

    I could show that the operator is linear and also that its square is unity operator I . I did
    find out the eigenvalues too. I am having trouble showing that its hermitian and the
    part b about its eigenfunctions.

    Any help ?
  2. jcsd
  3. Aug 11, 2011 #2
    After you already show that [itex]\hat{\pi}[/itex] is linear. Take a look at [itex]\hat{\pi}^2[/itex]. What is it? From this what can you infer about the possible eigenvalues for [itex]\hat{pi}[/itex]? What properties for the eigenvalues for an operator implies Hermiticity?
  4. Aug 11, 2011 #3
    Hi , eigenvalues are [tex]\pm 1[/tex] they are real. I know that hermitian operators have real eigenvalues but does it mean that if operator has real eigenvalues then its a hermitian
    operator ( this is converse statement , so not necessarily true) ?
  5. Aug 11, 2011 #4
    If all the eigenvalues of an operators are real.

    More explicitly, you are trying to show that:
    [tex]\int_{-\infty}^{\infty} \psi(r)^{*} \phi(-r) dr = \int_{-\infty}^{\infty} \psi(-r)^{*} \phi(r) dr[/tex]
    which is easy by simple change of variable.
    Last edited: Aug 11, 2011
  6. Aug 11, 2011 #5
    So since all eigenvalues of [itex]\hat{\pi}[/itex] are real , its hermitian. fine thats solved.
    what about the second part about the eigenfunctions ? can you help ?
  7. Aug 11, 2011 #6


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    Just apply the operator to the functions and show that you get an eigenvalue times the function.
  8. Aug 12, 2011 #7
    vela, yes that's one approach. but how do I derive these eigenfunctions in the first place ?
  9. Aug 12, 2011 #8
    There are really no general approach for arbitrary operator. A good guess is a start.
  10. Aug 12, 2011 #9


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    I suppose you could make a hand-waving argument that the Hilbert space can be written as a direct sum of subspaces, where each subspace is the span of Φ(x,y) and Φ(y,x) for a particular Φ(x,y). Then find the matrix representing [itex]\hat{\pi}[/itex] in one of these subspaces and diagonalize it. Blah, blah, blah...

    It's more useful, though, to recognize this pattern of combining elements to achieve a certain symmetry.
  11. Aug 12, 2011 #10
    thanks fellas. makes some sense now...........
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