I am trying to show the following relationships: $\bar{u}_{\pm p\sigma}\gamma^\mu u_{\pm p\sigma'} = 2p^\mu \delta_{\sigma\sigma'}$, $\bar{u}_{\pm p\sigma} u_{\pm p\sigma'} = \pm 2m\delta_{\sigma\sigma'}$, and $\bar{u}_{+ p\sigma}u_{-p\sigma'} = 0$.

Now, for the plane wave solution of the Dirac equation, we have \begin{equation} \psi_{\pm p \sigma} = \frac{1}{\sqrt{2\epsilon_\mathbf{p}V}}u_{\pm p\sigma}e^{\mp ipx}, \end{equation} where $px = p^\mu x_\mu$, and has a normalization of $\psi^\dagger_{\pm p\sigma}\psi_{\pm p\sigma} = 1/V$, then $\frac{1}{2\epsilon_\mathbf{p}V}u^\dagger_{\pm p\sigma}u_{\pm p\sigma} = 1/V\implies u^\dagger_{\pm p\sigma}u_{\pm p\sigma} = \bar{u}_{\pm p\sigma}\gamma^0u_{\pm p\sigma} = 2\epsilon_\mathbf{p}$, but I am not sure how to get the full covariant form with $\gamma^\mu$, or is it just something I argue such as, "due to Lorentz invariance..." and similar for the mass relationship, but I know how to get out the $2m$ by using the Dirac equation, plug in the plane wave solution and simplify, but I don't know how to get out the Delta function above. But for the third relationship, is that doing exactly what I did above, but specify the momentum (I am not sure)?

EDIT: The link provided in the first answer helped with finding the mass relationship, but for the rest I am at a hard wall, without going the route of the general Gordon Identity, which is something I don't want to do (i.e when $p = p'$).

For the mass part, I used the vector representation of of the spinors, aka equation 38.6 and equation 38.10, but I believe this only gets me the $+p$ solution, not the minus, unless I am interpreting something wrong, or wrong in my calculation. So I am very lost admittedly.

  • $\begingroup$ Do you mean $\delta_{\sigma \sigma'}$ rather than $\delta(\sigma-\sigma')$? $\endgroup$
    – mike stone
    Commented Oct 30, 2022 at 17:49
  • $\begingroup$ @mikestone Yes, it wasn't clear was the case, so I assumed the dirac delta. $\endgroup$
    – MathZilla
    Commented Oct 30, 2022 at 18:33

1 Answer 1


The detailed proof of everything you need can be found in section 38. of Srednicki's book, a draft of which can be found on his personal webpage.

  • $\begingroup$ It is definitely of great help for the mass identity, but when it comes to the gamma matrix one, I don't want the Gordon identity, just the momentum term, and it doesn't do that well the way it is derived. $\endgroup$
    – MathZilla
    Commented Oct 31, 2022 at 1:23
  • 1
    $\begingroup$ I think the source of your confusion is the notation you are given. First notice that you can identify your $u_{+}$ with $u$ in the book and $u_{-}$ with $v$ respectively. Then take all possible product combinations of (38.6) and (38.10) and this will yield the last two equations that you want to prove. $\endgroup$
    – Evan
    Commented Oct 31, 2022 at 2:29
  • 1
    $\begingroup$ As for the Gordon identities, I am not sure I understand the issue. Setting $p'=p$ immediately makes the term that is analogous to $S^{\mu\nu}$ vanish and you are left with just the momentum term as you wanted. Then you can use the second identity to get rid of the masses and that's it. From a broader perspective it would be very instructive to also prove the Gordon identities as they are useful in calculations of tree level amplitudes. $\endgroup$
    – Evan
    Commented Oct 31, 2022 at 2:42
  • $\begingroup$ I would honestly want to do this since it seems useful in current computation methods in research (as you say about tree level amplitudes) but our course could be considered, "old school," so who knows. $\endgroup$
    – MathZilla
    Commented Oct 31, 2022 at 12:32

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