# Problem Proving a Spinor Identity

Geremy Holly

## Homework Statement

Given the spinors:
$$\Psi_{1}=\frac{1}{\sqrt{2}}\left(\psi-\psi^{c}\right)$$
$$\Psi_{2}=\frac{1}{\sqrt{2}}\left(\psi+\psi^{c}\right)$$
Where c denotes charge conjugation, show that for a vector boson #A_{\mu}#;
$$A_{\mu}\overline{\Psi_{1}}\gamma^{\mu}\Psi_{2} + A_{\mu}\overline{\Psi_{2}}\gamma^{\mu}\Psi_{1} = 2 A_{\mu}\overline{\psi}\gamma^{\mu}\psi$$

## 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
$$A_{\mu}\overline{\Psi_{1}}\gamma^{\mu}\Psi_{2} + A_{\mu}\overline{\Psi_{2}}\gamma^{\mu}\Psi_{1} = A_{\mu}(\overline{\psi}\gamma^{\mu}\psi-\overline{\psi^{c}}\gamma^{\mu}\psi^{c})$$
So for the identity I want to prove to be true I need to prove that
$$\overline{\psi^{c}}\gamma^{\mu}\psi^{c}=-\overline{\psi}\gamma^{\mu}\psi$$
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.

Staff Emeritus
So for the identity I want to prove to be true I need to prove that
$$\overline{\psi^{c}}\gamma^{\mu}\psi^{c}=-\overline{\psi}\gamma^{\mu}\psi$$

That doesn't seem like it could be true. Look at the ##\gamma^0## term. You would need:

##\overline{\psi^{c}}\gamma^0 \psi^{c}=-\overline{\psi}\gamma^0\psi##

But ##\overline{\psi} = \psi^\dagger \gamma^0##. So the left-hand side is:

##- \psi^\dagger \gamma^0 \gamma^0 \psi = - \psi^\dagger \psi## (because ##\gamma^0 \gamma^0 = 1##).

But ##\psi^\dagger \psi## is always positive. Similarly, ##(\psi^c)^\dagger (\psi^c)## is always positive. So they can't have opposite signs (unless they're both zero).

Something is screwy here.