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Gamma matrices and how they operate

  1. Sep 29, 2012 #1
    1. The problem statement, all variables and given/known data

    Just a matter of convention (question)

    2. Relevant equations

    [tex]\gamma^0 = \begin{pmatrix} 1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & -1 & 0 \\0 & 0 & 0 & -1 \end{pmatrix}[/tex]

    3. The attempt at a solution

    If then,

    [tex]\gamma^0 = \begin{pmatrix} 1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & -1 & 0 \\0 & 0 & 0 & -1 \end{pmatrix}[/tex]

    and [tex]\gamma^0[/tex] is just [tex]\beta[/tex] and [tex]\beta \alpha^k = \gamma^k[/tex] is it true then that

    [tex]\gamma^k = \begin{pmatrix} 1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & -1 & 0 \\0 & 0 & 0 & -1 \end{pmatrix}\alpha^k[/tex]
     
  2. jcsd
  3. Oct 1, 2012 #2
    Can no one confirm I have done this right?
     
  4. Oct 1, 2012 #3

    dextercioby

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    Science Advisor
    Homework Helper

    Isn't the particular form of a Dirac matrix representation dependent ? So then only the general relations will hold, i.e.

    βrepαkrepkrep

    where i/o <rep> one has the Dirac, Majorana or Weyl/chiral representations.
     
  5. Oct 1, 2012 #4
    I think so. I think you have to work with [tex]D(\psi(x,t))[/tex] on the three matrices [tex]\gamma^1,\gamma^2,\gamma^3[/tex] to get back the matrix [tex]i\gamma^0 \begin{pmatrix} 1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1 \end{pmatrix}[/tex]

    which when squared gives you the chirality.
     
  6. Oct 1, 2012 #5
    Now I am really confused: consider the matrix form of [tex]a^k[/tex] and calculate it all out we have

    [tex]\begin{pmatrix} 1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & -1 & 0 \\0 & 0 & 0 & -1 \end{pmatrix}\begin{pmatrix} 0 & 0 & 1 & 0 \\0 & 0 & 0 & -1 \\0 & 1 & 0 & 0 \\1 & 0 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 \end{pmatrix}[/tex]

    A nullified matrix?

    Have I got my [tex]a^k[/tex] matrix right... ?
     
  7. Oct 1, 2012 #6
    [tex]a^k[/tex] is just a submatrix, right? of

    0_2 sigma^k

    \sigma^k 0_2

    k=1,2,3

    in my case, 1 and 3
     
  8. Oct 1, 2012 #7
    I just don't understand why the relationship

    [tex]\beta \alpha^k = \gamma^k[/tex]

    would be important if it spat out a zero matrix, which makes me wonder strongly whether I even have the right conditions down.
     
  9. Oct 1, 2012 #8

    TSny

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    Gold Member

    You will not get the null matrix. For example, check the element in the first row, third column of the resultant matrix.
     
  10. Oct 1, 2012 #9
    I'm sorry, I did it all wrong didn't I? I now get

    [tex]\begin{pmatrix} 1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & -1 & 0 \\0 & 0 & 0 & -1 \end{pmatrix}\begin{pmatrix} 0 & 0 & 1 & 0 \\0 & 0 & 0 & -1 \\0 & 1 & 0 & 0 \\1 & 0 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 1 & 0 \\0 & 0 & 0 & -1 \\0 & -1 & 0 & 0 \\-1 & 0 & 0 & 0 \end{pmatrix}[/tex]


    I am an idiot some times lol
     
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