In the Chapter/Section 36 of Srednicki's Quantum Field Theory, he argued that the terms of $\partial^\mu\psi\partial_\mu\psi + h.c.$ cannot be included in the Lagrangians for spinor fields because of that these terms would yield a Hamiltonian with unbounded below. I cannot understand this argument, so I try to write down the Lagrangian of $$ \mathscr L = -\frac12 (\partial^\mu \psi \partial_\mu \psi + h.c.), $$ in which I ignore the quadratic terms like $\psi\psi=\psi^a\psi_a=\varepsilon^{ab}\psi_a\psi_b$ and its Hermitian conjugate, which can be interpreted as the mass terms.
Then, I calculate the conjugate momentum of $\psi_a$ and its Hermitian conjugate $\psi_{\dot a}^\dagger$ like that $$ \begin{aligned} &\Pi^a(x) = \frac{\partial\mathscr L}{\partial(\partial_0\psi_a)} = \varepsilon^{ab}\partial_0\psi_b =: \dot\psi^a(x), \\ &\Pi^{\dagger\dot a}(x) = \frac{\partial\mathscr L}{\partial(\partial_0\psi^\dagger_\dot{a})} = \varepsilon^{\dot a \dot b}\partial_0\psi^\dagger_\dot{b} =: \dot{\psi}^{\dagger\dot a}(x). \end{aligned} $$ Therefore, we can get the Hamiltonian from Legendre transformation $$ \begin{aligned} \mathscr H & = \Pi\dot{\psi} + \Pi^\dagger\dot{\psi}^\dagger - \mathscr L \\ & = \dot\psi\dot\psi + \dot\psi^\dagger\dot\psi^\dagger - \frac12 \left[ \dot\psi\dot\psi - (\nabla\psi)\cdot(\nabla\psi) + \dot\psi^\dagger\dot\psi^\dagger - (\nabla\psi^\dagger)\cdot(\nabla\psi^\dagger) \right] \\ & = \frac12 \left[ \dot\psi\dot\psi + (\nabla\psi)\cdot(\nabla\psi) + \dot\psi^\dagger\dot\psi^\dagger + (\nabla\psi^\dagger)\cdot(\nabla\psi^\dagger) \right]. \end{aligned} $$ Then, how can I argue that Hamiltonian is unbounded below? I'd like to explain from $$ \psi \psi = \psi^a \psi_a = \varepsilon^{ba} \psi_a \psi_b = \psi_2 \psi_1 - \psi_1 \psi_2, \\ \psi^\dagger \psi^\dagger = \psi^\dagger_\dot{a} \psi^{\dagger\dot a} = \varepsilon^{\dot a\dot b} \psi^\dagger_\dot{a} \psi^\dagger_\dot{b} = \psi^\dagger_{\dot1} \psi^\dagger_{\dot2} - \psi^\dagger_{\dot2} \psi^\dagger_{\dot1}, $$ in which the minus signs will make the $\mathscr H \ge 0$ not correct. Is that right?
Finally, Srednicki chose the term of $i \psi^\dagger \bar\sigma^\mu \partial_\mu\psi$ included in the Lagrangian, whose hermicity is shown in $(36.1)$. But I have the same question about it that does this term of $i \psi^\dagger \bar\sigma^\mu \partial_\mu\psi$ yield a Hamiltonian with bounded below?
New Argument: The conjugate momentum of the spinor fields may be wrong, and they should be $$ \begin{aligned} \Pi^a & = \frac{\partial\mathscr L}{\partial (\partial_0\psi_a)} \\ & = \frac12 \left[ \delta_{ac}\varepsilon^{bc}\partial_0\psi_b - \delta_{ab}\varepsilon^{bc}\partial_0\psi_c \right] \\ & = \epsilon^{ba}\partial_0\psi_b = -\dot\psi^a, \end{aligned} $$ and $$ \Pi^{\dagger\dot a}=-\dot\psi^{\dagger\dot a}. $$ Therefore the Hamiltonian of this theory is $$ \begin{aligned} \mathscr H & = \Pi\dot{\psi} + \Pi^\dagger\dot{\psi}^\dagger - \mathscr L \\ & = -\dot\psi\dot\psi - \dot\psi^\dagger\dot\psi^\dagger - \frac12 \left[ \dot\psi\dot\psi - (\nabla\psi)\cdot(\nabla\psi) + \dot\psi^\dagger\dot\psi^\dagger - (\nabla\psi^\dagger)\cdot(\nabla\psi^\dagger) \right] \\ & = \frac12 \left[ -3\dot\psi\dot\psi + (\nabla\psi)\cdot(\nabla\psi) - 3\dot\psi^\dagger\dot\psi^\dagger + (\nabla\psi^\dagger)\cdot(\nabla\psi^\dagger) \right], \end{aligned} $$ in which the Hamiltonian, I think, is no positive definite. Some useful arguments may be that put this Hamiltonian as the production of annihilation operators and creation operators.
