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Exchange operator

  1. Aug 25, 2010 #1
    Hello Everybody,

    I am trying to understand the concept of symmetric and antisymmetric wave functions in QM.

    Now the Griffiths and other textbooks I have introduce the exchange operator as an operator that switches two particles in a given two particle wave function.

    But then an eigenvalue equation is introduced for the exchange operator with eigenvalues +1 or -1

    And finally the original wave function is said to equal the one with the switched particles multiplied by a +1 or a -1.

    Now I don't understand how this statement can be made because the switched wave function can not be an eigenfunction of the exchange operator because it is switched and not the original wave function so it does not satisfy the eigenvalue equation for the exchange operator.

    Thanks for your Help!! =)
  2. jcsd
  3. Aug 25, 2010 #2


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    The wave function with two identical particles interchanged can (at most) give a change of phase (multiplication by e^(-i*omega)), since it has to result in the same observables.

    So the switched wave function can only differ by a constant factor, which are the eigenvalues of P.
  4. Aug 25, 2010 #3
    The proof usually given is simply flawed. I wish textbooks would stop using it. Simply swapping labels is not a physical act, and does not change anything. True exchange, by adiabatically moving the particles, picks up a Berry phase due to the intrinsically curved nature of projective complex spaces. For 2 particles in 3D or higher, the phase is restricted to +1 and -1. In 2D, it can be any phase at all. For systems with extra degenerate degrees of freedom, the "phase" can be a matrix which mixes between them. For more particles, more complex patterns can occur (though it may be proved that it is always a combination of the 2 particle possibilities).

    So yeah, I'm not surprised that you find the "proof" strange. *wink*
  5. Aug 26, 2010 #4
    I don't see what you mean by true exchange. How do you implement adiabatically moving one particle around another, and how do you deduce from it interesting facts about the states of two electrons in a 1-d harmonic well?
  6. Aug 31, 2010 #5
    alxm, do you mean by same observables the same expectation values of the observables?
    I can follow your argument.
    If I understood you right the phase IS the eigenvalue of P. And we obtain this equating the first equation derived from the expectation values to the eigenvalue equation of P.

    Genneth, I read something in that regard but didn't delve into it further. But as I still didn't study Berry phases I am quite unsure if your argumentation is good as a starting point although I think it is the most accurate.
  7. Aug 31, 2010 #6


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    Expectation values, then.

    Switching twice must result in the same function, so the phase factor squared must equal unity, hence -1 or 1.
  8. Sep 1, 2010 #7
    Got it! Thank you!!
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