Square of modified Dirac equation

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To derive a Klein-Gordon-like equation from a modified Dirac equation, the appropriate expression to multiply on the left is (iγν∂ν + m - im5γ5). This choice ensures that the cross terms are eliminated during the squaring process. Concerns about the sign change for the im5γ5 term are addressed, as it does not require alteration due to its distinct nature from the traditional mass term. The anti-commutation property of γ5 with γμ ultimately simplifies the expression, confirming that the additional terms vanish. The discussion clarifies the correct approach for squaring the modified Dirac equation effectively.
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If I take a modified Dirac Eq. of the form (i\gamma^\mu \partial_\mu -M)\psi=0 where M=m+im_5 \gamma_5, and whish to square it to get a Klein-Gordon like equation would I multiply on the left with (i\gamma^\nu \partial_\nu +m+im_5\gamma_5) or (i\gamma^\nu \partial_\nu +m-im_5\gamma_5)?
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 im_5\gamma_5 term should also get the appropriate sign change, since its not the tradition mass term. Any thoughts?
 
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I say use iγμμ + m - imγ5. This will eliminate the cross terms.
 
Thanks. I was getting a little worried about the m_5 \gamma^\nu\partial_\nu\gamma_5+m_5\gamma_5 \gamma^\mu \partial_\mu in my expression but \gamma_5 anti commutes with \gamma^\mu, so it goes away in the end. Thanks again.
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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