# Why is the Lagrangian $-\partial^u\overline\Psi\partial_u\Psi-m^2\overline\Psi\Psi$ not Lorentz invariant?

Let $-\partial^u\overline\Psi\partial_u\Psi-m^2\overline\Psi\Psi$ be a Lagrangian density. Here $\Psi$ is the Dirac spinor and $\overline \Psi$ is defined to be $\Psi^\dagger \gamma^0$. It is said that this Lagrangian is not Lorentz invariant. However, I cannot see why. For me, it has perfect form of being Lorentz invariant. Could anyone please explain why this Lagrangian is not Lorentz invariant?

• "It is said" by whom? Jun 8 '18 at 13:27
• It's a dimension 5 term Jun 8 '18 at 14:00
• @AccodentalFourierTransform my lecturer Jun 8 '18 at 14:05
• @Keith Perhaps it was a slip of the tongue. The Lagrangian is Lorentz invariant, but not renormalizable (in four dimensions). Jun 8 '18 at 14:08
• @gj255 It is renormalizable in any dimensions. The first term in N dimensions would imply $\Psi$ to be in dimensions $GeV^{N/2-1}$ which is the same for the second term because of the squared mass. It looks like a Klein-Gordon equation but has a four-component field with mixed energy eigenstates? Jun 8 '18 at 16:11

Let us first agree on the fact that $\bar{\psi}\psi$ is indeed a Lorentz invariant, so the mass term obviously is. For the second term, you could integrate by parts arrive to: $$\bar{\psi}(\Box - m^2)\psi$$ which is again Lorentz invariant since $\Box$ also is. Here the $\gamma^0$ is causing kinetic mixing originally if you want, but it looks Lorentz invariant.
EDIT: What people are saying in the comments about 5th dimension operators is also true if you assume beforehand that those fields have the mass dimensions of a Dirac field, namely 3/2. But then again it makes no sense to declare them Dirac fields if there is no $\bar{\psi}\gamma^\mu\partial_\mu \psi$ term.