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The helicity operator

  1. Apr 30, 2015 #1
    Hi there,

    The question about the helicity operator ## h= S . \bf{p} ## ( where S is 2 by 2 matrix, with ##\sigma^i ## on the diagonal ), that as mentioned in a reference as [arXiv:1006.1718], it commutes with the Dirac Hamiltonian ## H = \gamma^0 ( \gamma^i p^i + m ) ## equ. (3.3), due to Gamma matrices anticommutation relation, but this isn't clear for me ..

    Bests,
     
  2. jcsd
  3. Apr 30, 2015 #2

    ChrisVer

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    S is obviously not a 2x2 matrix if it has the sigma^i on the diagonal... it's a 4x4 (but you write it in block form)...

    Let's see in the most "primitive" way whether the statement is correct:
    [itex] S \cdot p = S_1 p_1 + S_2 p_2 + S_3 p_3[/itex]

    [itex] S \cdot p =\begin{pmatrix} p_3 & p_1 -ip_2& 0 & 0 \\ p_1+ip_2 & -p_3 & 0 & 0 \\ 0& 0 & p_3 & p_1-ip_2 \\ 0&0&p_1+ip_2& -p_3 \end{pmatrix}[/itex]

    On the other hand:

    [itex]H=\begin{pmatrix} m & 0 & p_3 & p_1-ip_2 \\ 0& m & p_1+ip_2 & -p_3 \\ p_3 &p_1-ip_2&-m &0 \\ p_1+ip_2&-p_3&0& -m\end{pmatrix}[/itex]

    Then it's easy to take:
    [itex][H,h]=Hh -hH=?[/itex] I did it roughly and I guess I was able to make it vanish.

    I have to think a little for the "gamma matrix" reasoning...
     
  4. Apr 30, 2015 #3
    I also checked out: Hh-hH and found it vanish ..

    So thanx and in all casses may be Gamma matrices anticommution do something here..
     
    Last edited: Apr 30, 2015
  5. Apr 30, 2015 #4

    Hepth

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    I'm pretty sure you can write ## S ## in terms of gamma matrices. Depending on your basis (dirac or weyl) I think its something like ##S^i = \gamma^0 \gamma_5 \gamma^i##, and then use commutation relations from there.
     
  6. Apr 30, 2015 #5

    ChrisVer

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    yup I was looking for how to write S in term of the gamma matrices :sorry: but I was unsuccessful in finding an expression.
     
  7. Apr 30, 2015 #6

    Hepth

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    ##
    S^i =\begin{pmatrix} \sigma^i & 0 \\ 0 & \sigma^i \end{pmatrix}
    ##

    Then you can in the dirac basis use:

    $$\gamma^0 = \begin{pmatrix} I_2 & 0 \\ 0 & -I_2 \end{pmatrix},\quad \gamma^k = \begin{pmatrix} 0 & \sigma^k \\ -\sigma^k & 0 \end{pmatrix},\quad \gamma^5 = \begin{pmatrix} 0 & I_2 \\ I_2 & 0 \end{pmatrix}.$$ (from wikipedia)

    And you just can use

    $$ S^i =\begin{pmatrix} \sigma^i & 0 \\ 0 & \sigma^i \end{pmatrix} = \gamma^0 \gamma_5 \gamma^i $$

    you can just do the matrix algebra by hand, i think it works. If you use weyl basis the gamma0 and gamma5 are switched.
     
  8. Apr 30, 2015 #7
    So now we have the right formula ..
     
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