- #1
Geremy Holly
- 1
- 0
Homework Statement
Given the spinors:
[tex]\Psi_{1}=\frac{1}{\sqrt{2}}\left(\psi-\psi^{c}\right)[/tex]
[tex]\Psi_{2}=\frac{1}{\sqrt{2}}\left(\psi+\psi^{c}\right)[/tex]
Where c denotes charge conjugation, show that for a vector boson #A_{\mu}#;
[tex]
A_{\mu}\overline{\Psi_{1}}\gamma^{\mu}\Psi_{2}
+
A_{\mu}\overline{\Psi_{2}}\gamma^{\mu}\Psi_{1}
=
2
A_{\mu}\overline{\psi}\gamma^{\mu}\psi
[/tex]
Homework Equations
##\psi^{c}=-i\gamma^{2}\psi^{*}##
##\overline{\psi}=\psi^{\dagger}\gamma^{0}##
##\{\gamma^{\mu},\gamma^{\nu}\}=2\eta^{\mu\nu}##
##\gamma^{2}\gamma^{\mu}\gamma^{2}=(\gamma^{\mu})^{*}##
The Attempt at a Solution
Plugging in ##\Psi_{1,2}## it is easy to show that
[tex]
A_{\mu}\overline{\Psi_{1}}\gamma^{\mu}\Psi_{2}
+
A_{\mu}\overline{\Psi_{2}}\gamma^{\mu}\Psi_{1}
=
A_{\mu}(\overline{\psi}\gamma^{\mu}\psi-\overline{\psi^{c}}\gamma^{\mu}\psi^{c})
[/tex]
So for the identity I want to prove to be true I need to prove that
[tex]\overline{\psi^{c}}\gamma^{\mu}\psi^{c}=-\overline{\psi}\gamma^{\mu}\psi[/tex]
Plugging in the definition of ##\psi^{c}## gives
\begin{align*}
\overline{\psi^{c}}\gamma^{\mu}\psi^{c}
&=
(-i\gamma^{2}\psi^{*})^{\dagger}\gamma^{0}\gamma^{\mu}(-i\gamma^{2}\psi^{*})\\
&=
(i\psi^{T}(\gamma^{2})^{\dagger})\gamma^{0}\gamma^{\mu}(-i\gamma^{2}\psi^{*})\\
&=
\psi^{T}\gamma^{0}\gamma^{2}\gamma^{\mu}\gamma^{2}\psi^{*}\\
&=
\psi^{T}\gamma^{0}(\gamma^{\mu})^{*}\psi^{*}\\
&=
(\psi^{\dagger}\gamma^{0}(\gamma^{\mu})\psi)^{*}\\
&=
(\overline{\psi}\gamma^{\mu}\psi)^{*}\\
\end{align*}
Which disagress with the required expression unless it is purely imaginary! I have absolutely no idea where I've gone wrong and would really appreciate some help spotting my error.