Simplifying expression with gamma matrix and slashes

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The discussion centers on simplifying the expression \not p \gamma^\mu \not p. The proposed simplification is \frac{1}{2} \gamma^\mu p^2, though there is uncertainty about its correctness. Tom provides a relevant identity involving gamma matrices, stating that \gamma^{\nu}\gamma^{\mu}\gamma^{\lambda} can be expressed in terms of the metric tensor and other gamma matrices. He concludes that \not p \gamma^{\mu}\not p simplifies to 2p^{\mu}\not p - \gamma^{\mu} p^2. The conversation highlights the complexities of manipulating gamma matrices in quantum field theory.
besprnt
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I am trying to simplify the expression
\not p \gamma^\mu \not p.
I believe the answer should be
- \frac{1}{2} \gamma^\mu p^2,
but I am not sure.

Tom
 
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##\gamma^{\nu}\gamma^{\mu}\gamma^{\lambda} = g^{\mu\nu}\gamma^{\lambda} + g^{\mu\lambda}\gamma^{\nu} - g^{\nu \lambda}\gamma^{\mu} - i\epsilon^{\delta\nu\mu\lambda}\gamma_{\delta}\gamma^5 \\ \Rightarrow \not p \gamma^{\mu}\not p = 2p^{\mu}\not p - \gamma^{\mu} p^2 ##.
 
Thanks.
 
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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