1
$\begingroup$

Adjoint spinors $\bar{\psi}$ satisfy the adjoint Dirac equation $$ \bar{\psi} \left( \gamma^{\mu} \, p_{\mu} - mc \right) = 0 \; $$ (I'm using the terminology and units of Griffiths' "Intro. to Elementary Particles"). What puzzles me here is that $p_{\mu}$ is presumably a (four-) vector operator (as indicated below), and by normal convention one expects the operand (i.e. that which it the operator operates upon) to stand on its right side... but here there is nothing on its right side to operate upon. Is it implicit, then, that here $p_{\mu}$ "acts" to its left?

Secondarily, and under the convention $x^{\mu}=(ct, \mathbf{x})$ with diagonal components of the metric tensor being $$ (g_{00}, g_{11}, g_{22}, g_{33}) = (1,-1,-1,-1) \; , $$ is it correct that since $$ p_{\mu}=i \hbar \partial_{\mu} = i \hbar (\frac{\partial}{\partial (ct)}, \nabla) $$ one must have $$ p^{\mu}=i \hbar (\frac{\partial}{\partial (ct)}, - \nabla) \; ? $$ (Note: in my original post I had unintentionally reversed these two definitions, for the co- and contravariant cases.)

Perhaps the resolution of my question is that $p_{\mu}$ is here to be interpreted here not as an operator, but as $p_{\mu}=\hbar k_{\mu}$, where $k_{\mu}$ is the wavenumber vector?

$\endgroup$
  • 1
    $\begingroup$ It really depends on whether you are working in position or momentum space. In momentum space, your first equation is just a matrix equation, and there is no need to talk about how $p$ acts. In position space one often adds an arrow above $p$ that indicates that $p$ acts to the left. $\endgroup$ – user178876 Sep 18 '18 at 17:34
1
$\begingroup$

The four momentum operator in in the Adjoint Dirac equation do operates right to left. For clarification, we wright it with "backward arrow on top".

Also, the contravariant vector is the one you have written as covariant vector and vice verse. This is very basic, you can verify the Dirac adjoint eq. simply by Eular-Lagrange eq. for Dirac Lagrangian for /psi.

$\endgroup$

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.