Square of modified Dirac equation

  • Thread starter cbetanco
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  • #1
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If I take a modified Dirac Eq. of the form [itex](i\gamma^\mu \partial_\mu -M)\psi=0[/itex] where [itex]M=m+im_5 \gamma_5[/itex], and whish to square it to get a Klein-Gordon like equation would I multiply on the left with [itex](i\gamma^\nu \partial_\nu +m+im_5\gamma_5)[/itex] or [itex](i\gamma^\nu \partial_\nu +m-im_5\gamma_5)[/itex]?
I was under the impression that to take the square, you put a minus sign on the mass term and multiply with that expression on the left, but I am unsure if the [itex]im_5\gamma_5[/itex] term should also get the appropriate sign change, since its not the tradition mass term. Any thoughts?
 

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  • #2
Bill_K
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I say use iγμμ + m - imγ5. This will eliminate the cross terms.
 
  • #3
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Thanks. I was getting a little worried about the [itex]m_5 \gamma^\nu\partial_\nu\gamma_5+m_5\gamma_5 \gamma^\mu \partial_\mu[/itex] in my expression but [itex]\gamma_5[/itex] anti commutes with [itex]\gamma^\mu[/itex], so it goes away in the end. Thanks again.
 

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