touqra
Aug4-07, 06:39 PM
I have a hard time trying to derive the equation in 6.4 . This is what I've done, and as you will see I have a sign difference compared to his.
http://scitation.aip.org/getpdf/servlet/GetPDFServlet?filetype=pdf&id=PRVDAQ000070000001015012000001&idtype=cvips&prog=normal
Starting with page 8, equation 6.2, considering the second term in the action, i.e.
D_{\mu}\overline{\psi}\gamma^{\mu}\psi
note: the author used the four-component Dirac spinor as \left(
\begin{array}{cc}
\chi\\
\overline{\psi}
\end{array}
\right)
I also calculated some identities:
\psi* \gamma^0 \gamma^{\mu} = \gamma^0 \gamma^{\mu} \psi*
\gamma^{\mu}\gamma_{\mu} = 4
\{\gamma^{\mu} , \gamma^{\nu}\} = 2g^{\mu\nu}
With these,
I obtain for \frac{1}{4z}\gamma_{\mu}\gamma_{5}\overline{\Psi}\ gamma^{\mu}\Psi = \frac{-2}{z}(\chi\psi - \overline{\chi}\overline{\psi})
I have a sign difference with the author's in equation 6.4
I calculated for the first term too. Still a sign difference. In fact, I get something like \frac{3}{z} (\chi\psi - \overline{\chi}\overline{\psi})
http://scitation.aip.org/getpdf/servlet/GetPDFServlet?filetype=pdf&id=PRVDAQ000070000001015012000001&idtype=cvips&prog=normal
Starting with page 8, equation 6.2, considering the second term in the action, i.e.
D_{\mu}\overline{\psi}\gamma^{\mu}\psi
note: the author used the four-component Dirac spinor as \left(
\begin{array}{cc}
\chi\\
\overline{\psi}
\end{array}
\right)
I also calculated some identities:
\psi* \gamma^0 \gamma^{\mu} = \gamma^0 \gamma^{\mu} \psi*
\gamma^{\mu}\gamma_{\mu} = 4
\{\gamma^{\mu} , \gamma^{\nu}\} = 2g^{\mu\nu}
With these,
I obtain for \frac{1}{4z}\gamma_{\mu}\gamma_{5}\overline{\Psi}\ gamma^{\mu}\Psi = \frac{-2}{z}(\chi\psi - \overline{\chi}\overline{\psi})
I have a sign difference with the author's in equation 6.4
I calculated for the first term too. Still a sign difference. In fact, I get something like \frac{3}{z} (\chi\psi - \overline{\chi}\overline{\psi})