I was looking through my old, well-known book of David Griffiths, Intrododuction to elementary particles, and couldn't tell if I made a mistake in interpreting an expression, or Griffiths made a little mistake in writing it. So I hope someone can come "to the rescue". It's about an equation on page 223 (7.60). There he constructs the bilinear quantity $\bar {\psi} \psi$, the product of a Dirac spinor and its adjoint. The Dirac spinor itself isn't Lorenz invariant, but transforms as ${\psi}'=S \psi$ when going from one inertial frame to another which move with a speed $v$ relative to each other. $S$ is a 4x4 matrix with the property $$S^{\dagger} {\gamma}^0 S={\gamma}^0 (1)$$ Then he writes:

$$(\bar{\psi} \psi)' ={(\psi)'}^\dagger{\gamma}^0 (\psi)'={\psi}^{\dagger} S^{\dagger}{\gamma}^0 S \psi=\psi^{\dagger}{\gamma}^0 \psi=\bar{\psi} \psi (2)$$ which is Lorentz invariant.

I know it's a little thing, but shouldn't the two $\psi$'s after the second equal sign in $(2)$ stay primed, and let the third part of the calculation become:

$${({\psi})'}^{\dagger}{(S^{\dagger})}^{-1} S^{\dagger}{\gamma}^0 S S^{-1} {\psi}'$$wich you can write [see $(1)$] as

$${(\psi)'}^{\dagger}(S^{\dagger})^{-1}{\gamma}^0S^{-1} \psi'=\bar{\psi}\psi$$

because otherwise you can conclude in going from the third to the fourth part in $(2)$ that $S\psi=\psi$?

By the way, in Appendix C, on the top of page 379, Griffiths writes that $\gamma^0$ is equal to the 4x4 unit matrix, which is obviously wrong. Maybe I should send him an e-mail... But I guess there is not one book (of whatever kind) without one spelling or calculation error at least.

  • 1
    $\begingroup$ Why do you think they should stay primed? You explicitly wrote $\psi' = S\psi$ and that's precisely what is used in that step. $\endgroup$
    – ACuriousMind
    Apr 2, 2017 at 11:58
  • $\begingroup$ No. He doesn't replace $({\psi}')$ by $S\psi$, but ${\gamma}^0$ by $S^{\dagger}{\gamma}^0 S$. If you do the first thing then you'll get (see the transition from the third part to the fourth part of the calculation): $S\psi=\psi$, which is not true. You should get $S{\psi}'=\psi$ to end up with ${\psi}\bar \psi$ $\endgroup$ Apr 2, 2017 at 18:50
  • $\begingroup$ On the other hand, though, you can also say (${\psi}'$ is equal to $S\psi$, that can't be denied), like you do, that he replaces ${\psi}'$ by $S \psi$ to get to the third part, and then uses $S^{\dagger}{\gamma}^0 S={\gamma}^0$ to get the final expression, but doesn't the transition from the third to the fourth part imply that $S\psi=\psi$ (and ${\psi}^{\dagger}S^{\dagger}={\psi}^{\dagger}$)? $\endgroup$ Apr 2, 2017 at 19:29

1 Answer 1


What Griffith's does is right. By replacing $\psi '$ with the equivalent expression $S\psi$, he obtains results in terms of unprimed expressions. However, he started with $\overline{\psi} '\psi '$, so non-trivially equates this with $\overline{\psi}\psi$.

  • $\begingroup$ As I wrote in the previous comment, he doesn't replace ${\psi}'$ by $S\psi$, but ${\gamma}^0$ by $S^{\dagger}{\gamma}^0 S$. That's why he first writes $S^{\dagger}{\gamma}^0 S={\gamma}^0$. $\endgroup$ Apr 2, 2017 at 18:59
  • $\begingroup$ @descheleschilder I thought you were asking about the second $=$ sign, not the third. $\endgroup$
    – J.G.
    Apr 2, 2017 at 19:00
  • $\begingroup$ @J.G.-I did indeed. I'm just not sure how to go from the second part to the third part. Obviously, Griffiths replaces ${\psi}'$ by $S\psi$, but doesn't this imply (see the transition from the third to the fourth part, the third = sign) that $S\psi=\psi$? Can't you just as well replace ${\gamma}^0$ (after the first = sign) by $S^{\dagger}{\gamma}^0 S$ to get the same result? $\endgroup$ Apr 2, 2017 at 19:48
  • $\begingroup$ I see now why you can't do that. I assumed $S{\psi}'=\psi$. This isn't true: $S^{-1}{\psi}'=\psi$. But it still keeps bothering me why you can't conclude in going from the third part to the fourth that $S\psi=\psi$. $\endgroup$ Apr 2, 2017 at 21:04
  • $\begingroup$ In (2) the first, second, third and fourth $=$ signs respectively use the definition of barred spinors (and in particular $\overline{\psi}=S\psi$), the definition of primes in our transformation, (1) and the definition of bars again. The reason you can't jump to $S\psi=\psi$ is because $A^\dagger\gamma^0A=B^\dagger\gamma^0B$ doesn't imply $A=B$. $\endgroup$
    – J.G.
    Apr 2, 2017 at 22:56

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