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Annoying sign error in gamma matrix calculation

+ 2 like - 0 dislike
137 views

I am trying to show that

\[\bar{\psi}\gamma^{\mu}\psi =\bar{\psi}_L\gamma^{\mu}\psi_L + \bar{\psi}_R\gamma^{\mu}\psi_R\]

For $\mu = 0$, this is using $\gamma^0 \gamma^0 = 1$ rather straightforward.

However, for the case $\mu \neq 0$ I obtain an annoying sign error (using the Weyl representation for the gamma matrices:

\[\bar{\psi}\gamma^1\psi = \left( \begin{array}{cc} \psi_L^*,\psi_R^*\\ \end{array} \right) \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \\ \end{array} \right) \left( \begin{array}{cc} 0 & -\sigma_1 \\ \sigma_1 & 0 \\ \end{array} \right) \left( \begin{array}{c} \psi_L \\ \psi_R \\ \end{array} \right)\]

which when multiplying out stuff gives me the wrong result

\[\bar{\psi}\gamma^{1}\psi =\bar{\psi}_L\gamma^{1}\psi_L - \bar{\psi}_R\gamma^{1}\psi_R\]

For the other two components, the same would happen. How can I get rid of this annoying minus sign in front of the second term? Or is my way to show this identity not a good one in the first place?

asked Nov 24, 2015 in Theoretical Physics by anonymous [ revision history ]
edited Nov 29, 2015 by Arnold Neumaier

Just a note: I fixed your equations to what I think was your intention, considering that your "wrong result" evaluated to zero, and the "right one" was \(2\bar\psi\gamma^\mu\psi\), which probably wasn't what you intended. Feel free to rollback.

1 Answer

+ 4 like - 0 dislike

Start noticing that the identity operator $\textbf{1}$ can be written as $\textbf{1} = P_L + P_R$, where $\psi_{R,L} = P_{R,L}\psi$, respectively. As such we have:
$$\bar{\psi} = \bar{\psi}(P_L+P_R)$$ (and likewise for $\psi$). From there one can derive

$$\bar{\psi}\gamma^{\mu}\psi = \bar{\psi}(P_L+P_R)\gamma^{\mu}(P_L+P_R)\psi$$ the trick being to make the projection operators come into play. Expanding the above equation one gets: $$\bar{\psi}P_R \gamma^{\mu}P_L\psi + \bar{\psi}P_L \gamma^{\mu}P_R\psi + \bar{\psi}P_L\gamma^{\mu}P_L\psi + \bar{\psi}P_R\gamma^{\mu}P_R\psi. $$

Notice now that $\bar{\psi}P_R = \bar{\psi}_L$ (and viceversa), so that the previous equation becomes $$\bar{\psi}_L\gamma^{\mu}\psi_L + \bar{\psi}_R\gamma^{\mu}\psi_R + \bar{\psi}\left(P_L\gamma^{\mu}P_L + P_R\gamma^{\mu}P_R\right)\psi$$

the last contribution being zero due to the anti-commutation relations $\left\{\gamma^5,\gamma^{\mu}\right\}=0$.

answered Nov 28, 2015 by GennaroTedesco (80 points) [ revision history ]
edited Nov 30, 2015 by GennaroTedesco

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