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 $$Q \psi(x) = \begin{pmatrix} u(x) \\ -v(x) \end{pmatrix}\neq -\psi$$

  • $\begingroup$ what if either $u=0$ or $v=0$? $\endgroup$
    – Phoenix87
    Jan 19, 2015 at 23:51
  • $\begingroup$ I don't believe that's the case when you have the most general QED lagrangian $\endgroup$
    – user38680
    Jan 20, 2015 at 0:37
  • $\begingroup$ Can you give some reference? I have never seen a charge operator in the exponential... $\endgroup$
    – Rexcirus
    Jan 20, 2015 at 23:00
  • $\begingroup$ Halzen & Martin, "Quarks and leptons", formula (15.2) $\endgroup$
    – user38680
    Jan 21, 2015 at 12:04

1 Answer 1


The point is that a Dirac field isn't really a solution for an electron and positron but instead an electron and the conjugate of the positron. This means that the Dirac electron field does indeed obey, \begin{equation} e ^{ i \alpha (x) Q } \psi = e ^{ - i \alpha (x) } \psi \end{equation}

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, \begin{equation} \psi = \left( \begin{array}{c} e _L \\ ( e _R ^c ) ^\dagger \end{array} \right) \end{equation}


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