Confusion with the meanings of fermion fields $\hat{\Psi},\hat{\overline{\Psi}},\hat{\Psi}^C$ If $\hat{\Psi}$ is a field that annihilates an electron and creates a positron, $\hat{\overline{\Psi}}$ is a field that annihilates a positron and creates an electron. This takes all possibilities into account. Then, what do we gain by defining a new field $\hat{\Psi}^C=C\gamma^T_0\Psi^*$ where $C=i\gamma_0\gamma_2$? I am confused that how does this field differ from $\hat{\overline{\Psi}}$ in terms of what it does? 
 A: The point of these funny terms in the "conjugate" fields is that they have the same Lorentz transformation properties as the original fields. 
For example, let's forget about the spinors and consider vectors. Suppose a vector $\mathbf{v}$ transforms with a phase $e^{i \theta}$ under some $U(1)$ transformation. Then the conjugate transpose $\mathbf{v}^\dagger$ transforms with the opposite phase. But it's qualitatively different than $\mathbf{v}^\dagger$ because it is now a spatial row vector, not a spatial column vector. We can instead consider $\mathbf{v}^c = (\mathbf{v}^\dagger)^T$. This still transforms under the $U(1)$ like $\mathbf{v}^\dagger$, but it transforms under rotations like $\mathbf{v}$. That's useful and keeps things simple. For example, you know that you can put $\mathbf{v}^c$ in the place of $\mathbf{v}$ in any rotationally invariant expression, and it'll stay rotationally invariant. 
The transition from $\Psi$ to $\bar{\Psi}$ to $\Psi^c$ is precisely analogous to the transition from $\mathbf{v}$ to $\mathbf{v}^\dagger$ to $\mathbf{v}^c$. The quantity $\Psi^c$ is useful for the same reason. For example, you can take any Lorentz invariant expression involving $\Psi$ and replace it with $\Psi^c$, and it will remain Lorentz invariant.
