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I Dirac commutation relations

  1. Jul 6, 2017 #1
    Hello! I am reading Peskin's book on QFT and at a point he wants to show that the Dirac field can't be quantified using this commutation relations: ##[\psi_a(x),\psi_b^\dagger(x)]=\delta^3(x-y)\delta_{ab}## (where ##\psi## is the solution to Dirac equation). I am not sure I understand the math behind the commutation relation (I understand why physically it is wrong) as you have a column and a raw vector, so doing the commutation you have the difference between a number and a 4x4 matrix and I am not sure how does this work. Can someone explain it to me? Thank you!
     
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  3. Jul 6, 2017 #2

    ChrisVer

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    Just to solve your misconception about "physically wrong", what you have on the left-hand-side is again a 4x4 matrix (has indices a,b running from 1 to 4), and similarily for the right-hand-side.
     
  4. Jul 6, 2017 #3
    The in
    The indeces a and b are fixed, they don't run from 1 to 4. They just specify the spin state.
     
  5. Jul 6, 2017 #4

    ChrisVer

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    what do you mean by fixed? the equation has 16 fixed numbers in the left-hand-side (like a 4x4 matrix) and 16 fixed numbers in the right hand side (again like a 4x4 matrix).
    The indices indicate one of the Dirac 4-spinor components:
    [itex] \psi = \begin{pmatrix} \psi_1 \\ \psi_2 \\ \psi_3 \\ \psi_4 \end{pmatrix}[/itex]
    Why would you have vectors then for [itex]\psi_a[/itex] ?
     
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