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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.

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    $\begingroup$ Well, the whole point is that it's not a tensor, it doesn't obey any simple transformation law. $\endgroup$
    – Javier
    Aug 12 '20 at 16:28
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    $\begingroup$ @Javier. Can you elaborate a little bit more on that? $\endgroup$ Aug 12 '20 at 16:35
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    $\begingroup$ 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 '20 at 17:35
  • $\begingroup$ This is nicely explained in Tong's notes $\endgroup$
    – bolbteppa
    Aug 12 '20 at 17:47
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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.

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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.

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