When constructing the Lagrangian for a two-component left-handed Weyl field $\psi$, in e.g. Srednicki, one rejects the choice of $\partial^\mu \psi \partial_\mu \psi+\partial^\mu \psi^\dagger \partial_\mu \psi^\dagger$ because it will lead to a Hamiltonian that is unbounded below: the Hamiltonian will look like $\dot{\psi} \dot{\psi}+\dot{\psi}^\dagger \dot{\psi}^\dagger+\partial_i \psi \partial_i \psi+\partial_i \psi^\dagger \partial_i \psi^\dagger$. I struggle to see that this is unbounded below. I usually try to understand Lagrangians in terms of classical fields to check for boundedness issues, but any of my attempts to look at complex-number-valued fields are foiled by the fact that we need Grassmann-valued numbers to write a nontrivial kinetic term of the form above, as we're using the 2x2 antisymmetric invariant symbol $\epsilon$ to tie up the indices.

Thus we're led to consider a kinetic term involving both $\psi$ and $\psi^\dagger$. However, for the sake of Lorentz covariance of the Lagrangian density, we can't use the 2x2 antisymmetric invariant symbol $\epsilon$ anymore to couple the fields, since $\psi^\dagger$ is in the complex conjugate representation. This requires the use of $\bar{\sigma}^\mu$ to get a Lorentz covariant Lagrangian. However, the resulting kinetic term of $i\psi^\dagger \bar{\sigma}^\mu \partial_\mu \psi$ is hard for me to understand in terms of whether the Hamiltonian is bounded below.

How can I tell that for a two-component anticommuting Weyl field $\psi$ that a Lagrangian containing $\partial^\mu \psi \partial_\mu \psi+\partial^\mu \psi^\dagger \partial_\mu \psi^\dagger$ gives an unbounded-below Hamiltonian, but a Lagrangian containing $i\psi^\dagger \bar{\sigma}^\mu \partial_\mu \psi$ gives me a Hamiltonian that is bounded below?


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