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Pauli matrices

  1. Jun 6, 2006 #1
    Ok, I have a stupid question on pauli matrices here but it is bugging me. In a book i'm reading it gives the equation [tex][\sigma_i , \sigma_j] = 2 I \epsilon_{i,j,k} \sigma_k [/tex] , I understand how it works and everything but I do have a question, when you have k=i/j and i!=j (like 2,1,2) you get a non zero commutator and yet [tex]\epsilon[/tex] gives you 0. What am I missing here? Thanks
     
    Last edited: Jun 6, 2006
  2. jcsd
  3. Jun 6, 2006 #2
    The i and j determine the value of k.

    For instance if i and j are equal then obviously the commutator is zero and the [tex]\epsilon_{ijk} = \epsilon_{iik} = 0[/tex] so that's fine.

    If i and j are different then k is the other value from {1,2,3}. To use your example, if i=2 and j=1 then k MUST be 3 to get a non-zero value of [tex]\epsilon_{ijk}[/tex] (you don't put commas in the subscript by the way). Therefore [tex][\sigma_{2},\sigma_{1}] = \epsilon_{213}\sigma_{3} = -\sigma_{3}[/tex]

    If it makes it easier, think about summation convention, so that the right hand side becomes

    [tex]\epsilon_{ijk}\sigma_{k} = \epsilon_{ij1}\sigma_{1} + \epsilon_{ij2}\sigma_{2} + \epsilon_{ij3}\sigma_{3} [/tex]

    This means that when i isn't equal to j only one of those three terms remains non-zero otherwise k will be equal to either i or j. If i=j, then all three are zero.
     
  4. Jun 6, 2006 #3
    The Levi-Civita symbol is zero if any of the 3 indices are the same. Calculate a few of those commutators and see if they equal the RHS for the kind of indices you picked and see if the relation holds if you want to see this for yourself.
     
  5. Jun 6, 2006 #4
    Thanks, that clears it up.
     
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