What kind of tensor is $\psi^\dagger\psi$?

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.

• Well, the whole point is that it's not a tensor, it doesn't obey any simple transformation law. – Javier Aug 12 '20 at 16:28
• @Javier. Can you elaborate a little bit more on that? – Isai Dávila Cuba Aug 12 '20 at 16:35
• Do know how $\bar \psi \gamma^\mu \psi$ transforms? And what is the connection between that transformation and your $\psi^\dagger\psi$? – mike stone Aug 12 '20 at 17:35
• This is nicely explained in Tong's notes – bolbteppa Aug 12 '20 at 17:47

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.