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!

Quantum physics - Symmetrizer operator

  1. Dec 13, 2013 #1
    Hi everyone.

    I am studying 'identical particles' in quantum mechanics, and I have a problem with the properties of the Symmetrizer (S) and Antisymmetrizer (A) operators.

    S and A are hermitian operators. Therefore, for what I know, their set of eigenkets must constitute a basis of the space ket.
    However, the set of eigenkets of S (A) contains only symmetric (antisymmetric) kets. Given an arbitrary ket (not necessarily symmetrical or antisymmetrical), I don't think it is always possible to write it as linear combination of only symmetrical (antisymmetrical) kets. Therefore, I can't understand how the eigenkets of S (A) can constitute a basis of the space ket.
     
  2. jcsd
  3. Dec 13, 2013 #2

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Try a 4-dimensional example: Consider the state space V spanned by {|1>|1>, |1>|2>, |2>|1>, |2>|2> }. What are the eigenvalues of S? What are the corresponding eigenspaces?
     
  4. Dec 13, 2013 #3
    I think in this case the eingevalue is 1 (4 times degenerate), and the eigenkets
    |1>|1>
    (1/√2) (|1>|2>+|2>|1>)
    (1/√2) (|2>|1>+|1>|2>)
    |2>|2>
    So.. 4 eigenkets for a 4D space. Does this mean that every ket can be obtained as a linear combination of symmetric (antisymmetric) kets?
     
  5. Dec 13, 2013 #4

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Addition is commutative, so your middle two kets are the same. :smile:
     
  6. Dec 13, 2013 #5
    Err... right. I did not solve the eigenvalue equation, but I can't see a 4th eigenket of S...

    I guess the missing one to form a basis of the ket space is:
    (1/√2) (|1>|2>-|2>|1>)
    that is an eigenket of A (and thus not of S).
     
  7. Dec 13, 2013 #6

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Are you sure this isn't an eigenket of S?
     
  8. Dec 13, 2013 #7
    S [(1/√2) (|1>|2>-|2>|1>)]=
    (1/2)(P12+P21) [(1/√2) (|1>|2>-|2>|1>)] =
    (1/2) {[(1/√2) (|1>|2>-|2>|1>)] + [(1/√2) (|2>|1>-|1>|2>)]} =
    0
    ...

    At this point, the only thing I can imagine is that
    (1/√2) (|1>|2>-|2>|1>)
    is an eigenket of S with eigenvalue 0.
     
  9. Dec 13, 2013 #8

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Yup.

    S [(1/√2) (|1>|2>-|2>|1>)] = 0 [(1/√2) (|1>|2>-|2>|1>)].

    Things are sort of reversed for A.
     
  10. Dec 13, 2013 #9
    Great: thanks!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Quantum physics - Symmetrizer operator
  1. Quantum operators (Replies: 2)

Loading...