# How to check whether Weyl field Hamiltonian is bounded below?

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?