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?
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7$\begingroup$ "It is said" by whom? $\endgroup$– AccidentalFourierTransformCommented Jun 8, 2018 at 13:27
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1$\begingroup$ It's a dimension 5 term $\endgroup$– FrodCubeCommented Jun 8, 2018 at 14:00
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$\begingroup$ @AccodentalFourierTransform my lecturer $\endgroup$– KeithCommented Jun 8, 2018 at 14:05
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$\begingroup$ @Keith Perhaps it was a slip of the tongue. The Lagrangian is Lorentz invariant, but not renormalizable (in four dimensions). $\endgroup$– gj255Commented Jun 8, 2018 at 14:08
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$\begingroup$ @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? $\endgroup$– Oktay DoğangünCommented Jun 8, 2018 at 16:11
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1 Answer
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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.
So, I would say it looks like a weird mixture of 4 Klein Gordon fields. If you compute the equations of motion it seems every component of what you called a "Dirac spinor" satisfies a Klein-Gordon equation. If you are given this equation there is nothing forcing them to be Dirac spinors.
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.
