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Isospin Rotation

  1. May 29, 2007 #1
    1. The problem statement, all variables and given/known data

    Show that
    [tex] R_2 \left| I \; 0 \right> = (-1)^I \left| I \; 0 \right> [/tex]
    [tex] R_2 = e^{i \pi I_2} [/tex]

    N.B. the isospin states are represented by the usual notation of
    [tex] \left| \textrm{(total isospin}) \qquad \textrm{(isospin component in the z direction)} \right> [/tex]

    3. The attempt at a solution

    I guess I may be using the angular momentum analogy a bit too strongly here but my understanding is that we are asked for the eigenvalues of the operator representing a [tex] \pi [/tex] rotation about the [tex] I_2 [/tex] axis.

    The states we are working with have no component of isospin in the z (or if you rather [tex] I_3 [/tex] ) direction, so it's all in the 1-2 plane. My main conceptual problem is that I thought (after knowing total and z component of isospin) we can't tell anything else about the orientation. I guess this could be where I'm taking the angular momentum analogy too far but I'll continue... So if we rotate it about the 2 axis how are we to know it's not all in the 2 direction anyway (giving us a 1 eigenvalue) or all in the 1 direction (a -1 eigenstate) or neither?
    Clearly the correct answer doesn't bode well with my thinking.

    I have tried treating this purely mathematically by the way but haven't got anywhere and probably isn't worth showing. I'd appreciate any help with this - be it conceptual or mathematical.
    Last edited: May 29, 2007
  2. jcsd
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