# Charge operator for Dirac spinor

In QED, the gauge transformation which acts upon a fermionic field $\psi$ is $$\psi'(x)= e^{i \alpha(x) Q}\psi(x)$$ where $Q$ is the charge operator. Most of the time it's just written as $$\psi'(x)= e^{-i \alpha(x)}\psi(x)$$ However, if the general solution of the Dirac equation is $$\psi(x) = \begin{pmatrix} u(x) \\ v(x) \end{pmatrix}$$ where $u$ is the vector with electron fields and $v$ with positron fields, then how can the second formula be true, if $Q u = + u$ and $Q v = -v$? In that case we would get $$Q \psi(x) = \begin{pmatrix} u(x) \\ -v(x) \end{pmatrix}\neq -\psi$$

• what if either $u=0$ or $v=0$? – Phoenix87 Jan 19 '15 at 23:51
• I don't believe that's the case when you have the most general QED lagrangian – user38680 Jan 20 '15 at 0:37
• Can you give some reference? I have never seen a charge operator in the exponential... – Rexcirus Jan 20 '15 at 23:00
• Halzen & Martin, "Quarks and leptons", formula (15.2) – user38680 Jan 21 '15 at 12:04

The point is that a Dirac field isn't really a solution for an electronand positron but instead an electron and the conjugate of the positron. This means that the Dirac electron field does indeed obey, $$e ^{ i \alpha (x) Q } \psi = e ^{ - i \alpha (x) } \psi$$
Such misconceptions are dissolved if one works in the Weyl representation. There the fundamental (2-component) fields which make the Lagrangian are, $e _L , e _R ^c$ (where $e _L$ is negatively charged and $e _R ^c$ is positive). Then a Dirac fermion is, $$\psi = \left( \begin{array}{c} e _L \\ ( e _R ^c ) ^\dagger \end{array} \right)$$