2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ what if either $u=0$ or $v=0$? $\endgroup$ – Phoenix87 Jan 19 '15 at 23:51
  • $\begingroup$ I don't believe that's the case when you have the most general QED lagrangian $\endgroup$ – user38680 Jan 20 '15 at 0:37
  • $\begingroup$ Can you give some reference? I have never seen a charge operator in the exponential... $\endgroup$ – Rexcirus Jan 20 '15 at 23:00
  • $\begingroup$ Halzen & Martin, "Quarks and leptons", formula (15.2) $\endgroup$ – user38680 Jan 21 '15 at 12:04
1
$\begingroup$

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, \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}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.