I'm trying to find how does this quantity $$\psi^\dagger\psi$$ transforms under a Lorentz transformation. Where $\psi$ is a Dirac spinor.

What I've tried so far:

It is known that a Dirac spinor transforms as $$\psi' = S\psi.$$ The matrix $S$ satisfy certain properties. I calculated $\psi'^\dagger=\psi^\dagger S^\dagger$, then: $$\psi'^\dagger\psi'=\psi^\dagger S^\dagger S\psi.$$

I don't known how to continue from here.

Another way I have tried was rewrite $\psi'^\dagger$ this way: $$\psi'^\dagger = \bar{\psi'}\gamma^0$$ Where $\bar{\psi'}=\psi'^\dagger \gamma^0$. Then, using $\bar{\psi'}=\bar{\psi}S^{-1}$ $$\psi'^\dagger\psi'=\bar{\psi}S^{-1}\gamma^0 S\psi.$$ Using this property of the $S$ matrix $(S^{-1})_{\alpha\beta}(\gamma^\lambda)_{\rho\sigma}(S)_{\sigma\beta}=a^{\lambda}_{\mu}(\gamma^{\mu})_{\alpha\beta}$ with $\lambda=0$: $$\psi'^\dagger\psi' = a^{0}_{\mu}\bar{\psi}\gamma^\mu\psi.$$

The coefficients $a^{\mu}_{\nu}$ are a general Lorentz transformation (proper Lorentz transformation, rotations,etc). I don't know if this is the correct way of approaching this problem. I know that $\psi^\dagger\psi$ is not a Lorentz scalar.

Any help is appreciated.

  • 2
    $\begingroup$ Well, the whole point is that it's not a tensor, it doesn't obey any simple transformation law. $\endgroup$
    – Javier
    Aug 12, 2020 at 16:28
  • 1
    $\begingroup$ @Javier. Can you elaborate a little bit more on that? $\endgroup$ Aug 12, 2020 at 16:35
  • 3
    $\begingroup$ Do know how $\bar \psi \gamma^\mu \psi$ transforms? And what is the connection between that transformation and your $\psi^\dagger\psi$? $\endgroup$
    – mike stone
    Aug 12, 2020 at 17:35
  • 1
    $\begingroup$ This is nicely explained in Tong's notes $\endgroup$
    – bolbteppa
    Aug 12, 2020 at 17:47
  • $\begingroup$ related: physics.stackexchange.com/q/219950/226902 $\endgroup$
    – Quillo
    Jun 9, 2022 at 9:33

2 Answers 2


It's the zeroth component $$ \bar \psi \gamma^0 \psi = \psi^\dagger\gamma^0\gamma^0\psi = \psi^\dagger \psi $$ of Lorentz vector $$ \bar \psi \gamma^\mu \psi. $$

Nothing more, nothing less.

$\psi^\dagger \psi$ is not invariant under general Lorentz transformations, albeit it's invariant under the spacial rotation subset of the Lorentz transformations.

An interesting observation is that $\psi^\dagger \psi$ is invariant under the axial/chiral transformation $$ \psi \rightarrow e^{\theta i \gamma^5}\psi, $$ while $\bar \psi \psi$ breaks the chiral symmetry.


To be explicit, $\psi^\dagger = (\psi^*)^T$. We have that $S[\Lambda] = \exp(\frac12 \Omega_{\mu\nu}S^{\mu\nu})$ and you can check that,

$$S^\dagger = -\frac14 [{\gamma^{\mu\dagger}}, \gamma^{\nu\dagger}]$$

is not anti-hermitian, implying that $S[\Lambda]$ is not unitary. Using that $\psi(x) \to S[\Lambda] \psi(\Lambda^{-1}x)$, we have that the quantity $\psi^\dagger \psi$ transforms as,

$$(\psi^\dagger \psi)(x) \to \psi^\dagger(\Lambda^{-1}x)S[\Lambda]^\dagger S[\Lambda] \psi(\Lambda^{-1}x).$$

I encourage you to check that this is the case, for some explicit Lorentz transformations. As you can see, it is certainly not a Lorentz scalar, and does not transform in a "nice" way that we can neatly describe compared to quantities such as $\bar\psi \psi$ which is a scalar, or $\bar\psi \gamma^5 \psi$ which is a pseudo-scalar.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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