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Proof that a Dirac particle has spin 1/2?

  1. Nov 17, 2013 #1
    Hi,

    I am having trouble following the Peskin and Schroeder and their derivations to show that a Dirac particle is a spin 1/2 particle (page 60 and 61). I understand how he gets the first (unnumbered) equation on page 61. However, I don't understand how he gets to the second equation:
    \begin{equation}
    J_z a^{s \dagger}_0|0 \rangle = \frac{1}{2m} \sum_r \left(u^{s \dagger}(0) \frac{\Sigma^3}{2} u^r(0) \right) a_0^{r \dagger} |0 \rangle
    \end{equation}
    In particular I do not understand why there is only one "summation" symbol. I would have thought that the equation would be:
    \begin{equation}
    J_z a^{s \dagger}_0|0 \rangle = \frac{1}{2m} \sum_r \sum_{s} \left(u^{s \dagger}(0) \frac{\Sigma^3}{2} u^r(0) \right) a_0^{r \dagger} |0 \rangle
    \end{equation}
    Does anybody know what they have done with the second "summation" symbol?

    Any help would much appreciated.
     
    Last edited: Nov 17, 2013
  2. jcsd
  3. Nov 17, 2013 #2

    JK423

    User Avatar
    Gold Member

    Hi AlbertEi,

    On the left hand side you have the free index 's', so on the right hand side this index shouldn't be summed. This is just a remark without knowing the actual calculation.
     
  4. Nov 17, 2013 #3
    Well, for starters the second equation you wrote down doesn't make too much sense, since there is a sum over s on the right hand side, but there is still an s on the left hand side.
    Now the actual problem, you should convince yourself that when taking the commutator
    $$[J_{z},a_{0}^{s\dagger}]$$
    the only term that does not (anti-)commute (anti-commutator or commutator does not matter, J_{z} will hit the vacuum and annihilate it regardless), is the term
    $$ [a_{p}^{r\dagger}a_{p}^{r},a_{0}^{s\dagger}].$$
    Next you should convince yourself that this term is exactly what Peskin tells you it is. Now check what happens to the exponentials in the first line of the unnumbered equation when you integrate it against the δ(p-p^{\prime}) over p^{\prime}, (as a hint, the position integral should now be easy, and doing the position integral should allow you to do the final momentum integral).
     
  5. Nov 17, 2013 #4
    Those indices represent the spin, i.e. they do are not tensor indices so I don't think we can use the Einstein summation convention in this sense.
     
  6. Nov 17, 2013 #5
    Hi VoxCaelum, my previous reply was not directed towards you. I will look into your answer and come back if I have any questions. Thanks for your reply.
     
  7. Nov 17, 2013 #6

    Bill_K

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    Science Advisor

    To begin with there were two summations over r and r'. The commutator contains a Kronecker delta, δr's, which collapses the r' summation to a single term s, so just the r summation remains.
     
  8. Nov 17, 2013 #7
    That makes sense; I feel a bit silly now. Thank Bill_K!
     
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