1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Hermiticity of permutation operator

  1. Aug 11, 2011 #1
    Hi

    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]

    and

    [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

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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...........
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Hermiticity of permutation operator
  1. Permutation operator (Replies: 1)

Loading...