A Identity with Gamma matrices and four-vector contractions

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The discussion centers on verifying the identity involving gamma matrices and a generic four-vector q, specifically whether the expression qα qβ γα γμ γβ simplifies correctly. The defining property of the gamma matrices, {γμ, γν} = 2ημν, is crucial for the proof. The participants derive the expression step-by-step, ultimately showing that it simplifies to 2qα γα qμ - q² γμ. Additionally, there is a query about the LaTeX package for typesetting Feynman slashes, indicating a need for clarity in notation. The conversation highlights the importance of understanding gamma matrix properties in quantum field theory.
Francisco Alegria
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I want to determine a specific identity involving gamma matrix and four vectors
Is the fowwowin identity correct for a generic four-vector"q"? What is the proof? Thank you.
Quantum Physics Identity with Gamma matrices.png
 
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That's "Diracology". You need the defining property of the ##\gamma## matrices,
$$\{\gamma^{\mu},\gamma^{\nu} \}=2 \eta^{\mu \nu}.$$
Your expression can be written as (with the Einstein summation convention in action)
$$\begin{split}
q_{\alpha} q_{\beta} \gamma^{\alpha} \gamma^{\mu} \gamma^{\beta}&=q_{\alpha} q^{\beta} \left (\gamma^{\alpha} \{\gamma^{\mu},\gamma^{\beta} \} - \gamma^{\alpha} \gamma^{\beta} \gamma^{\mu} \right) \\ &= q_{\alpha} q_{\beta} \left (2\gamma^{\alpha} \eta^{\mu \beta} - \frac{1}{2} \{\gamma^{\alpha} ,\gamma^{\beta} \} \gamma^{\mu} \right)\\
&=2q_{\alpha} \gamma^{\alpha} q^{\mu} - q^2 \gamma^{\mu}.
\end{split}$$
[Note: Is there some equivalent of "slashed" from the LaTeX package slashed.sty to type "Feynman slashes"?]
 
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vanhees71 said:
[Note: Is there some equivalent of "slashed" from the LaTeX package slashed.sty to type "Feynman slashes"?]
##\not{\!p}## is produced by typing "\not{\!p}".
 
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For the quantum state ##|l,m\rangle= |2,0\rangle## the z-component of angular momentum is zero and ##|L^2|=6 \hbar^2##. According to uncertainty it is impossible to determine the values of ##L_x, L_y, L_z## simultaneously. However, we know that ##L_x## and ## L_y##, like ##L_z##, get the values ##(-2,-1,0,1,2) \hbar##. In other words, for the state ##|2,0\rangle## we have ##\vec{L}=(L_x, L_y,0)## with ##L_x## and ## L_y## one of the values ##(-2,-1,0,1,2) \hbar##. But none of these...