quantum field theory - Sign in front of QFT kinetic terms

I'd like to know if the sign in front of a kinetic term in QFT important. For the scalar field we conventionally write (in the $+ ---$ metric),

$$\begin{equation} {\cal L} _{ kin} = \frac{1}{2} \partial _\mu \phi \partial ^\mu \phi \end{equation}$$ Based on the answer given here, this makes perfect sense since we want to have positive kinetic energy $\propto \dot{\phi}^2$. So would the Hamiltonian with a negative in front of the kinetic term be unbounded?

Does this logic extend to the Dirac Lagrangian typically given by,

$$\begin{equation} \bar{\psi} i \partial _\mu \gamma ^\mu \psi \quad ? \end{equation}$$ i.e., would having a negative in front of the Dirac Lagrangian make the Hamiltonian unbounded?

Answer

Yes. Though the energy will not be unbounded, but bounded from above, if my calculation is correct.

For real scalar field under $(+---)$ metric, besides the negative classical kinetic energy for the Lagrangian

$$\mathcal{L}=-\frac{1}{2} \partial^{\mu} \phi \partial_{\mu} \phi - \frac{1}{2} m^2 \phi^2 \tag{1}$$ , the classical equation of motion will be $$(\square - m^2 )\phi=0 . \tag{2}$$ For plane wave $\phi ~\sim e^{ipx}$, it gives $p^2+m^2 = (p^0)^2 - \mathbf{p}^2+m^2=0$ which is inconsistent with relativistic energy momentum relation. I am not sure if it is necessary to quantize it.

Though the energy-momentum-relation argument will not work for the Dirac field, we can quantize it to see the energy will be negative definite.

$$\mathcal{L} = \bar{\psi}( -i \gamma^{\mu} \partial_{\mu} - m ) \psi \tag{3}$$

The classical equation of motion is

$$(i \gamma^{\mu} \partial_{\mu} +m) \psi=0 \tag{4}$$

To preserve all properties of $u(\mathbf{p})$ and $v(\mathbf{p})$, we define

$$\psi =: u(\mathbf{p}) e^{ipx}$$ $$\psi =: v(\mathbf{p}) e^{-ipx}$$
Thus we can replace the $u(\mathbf{p})$ as $v(\mathbf{p})$ and $v(\mathbf{p})$ as $u(\mathbf{p})$ in the expansion of $\psi$ and $\bar{\psi}$. By $$\pi = -i \bar{\psi} \gamma^0$$ then $$H= \int d^3 x \bar{\psi} ( i \gamma^i \partial_i \psi + m ) \psi$$

Plug in expansions of spinors in the Schrodinger picture

$$\psi = \int \frac{ d^3 p }{ (2\pi)^3} \frac{1}{ \sqrt{2 E_{\mathbf{p}}}} \sum_s \left( a_{\mathbf{p}}^s v^s (\mathbf{p}) e^{-i\mathbf{p} \cdot \mathbf{x} } + b_{\mathbf{p}}^{s\dagger} u^s(\mathbf{p}) e^{i \mathbf{p} \cdot \mathbf{x} } \right)$$ $$\bar{\psi} = \int \frac{ d^3 p }{ (2\pi)^3} \frac{1}{ \sqrt{2 E_{\mathbf{p}}}} \sum_s \left( b_{\mathbf{p}}^s \bar{u}^s (\mathbf{p}) e^{-i\mathbf{p} \cdot \mathbf{x}} + a_{\mathbf{p}}^{s\dagger} \bar{v}^s(\mathbf{p}) e^{i\mathbf{p} \cdot \mathbf{x}} \right)$$ we have

$$H = \sum_{ss'} \int \frac{d^3p}{ (2\pi)^3 2E_{\mathbf{p}} } b_{\mathbf{p}}^{s'} b_{\mathbf{p}}^{s\dagger} \bar{u}^{s'}(\mathbf{p}) ( - \gamma^i p_i +m) u^s(\mathbf{p}) + a_{\mathbf{p}}^{s'\dagger} a_{\mathbf{p}}^{s} \bar{v}^{s'}(\mathbf{p}) ( \gamma^i p_i +m) v^s(\mathbf{p})$$ $$= \sum_{ss'} \int \frac{d^3p}{ (2\pi)^3 2E_{\mathbf{p}} } b_{\mathbf{p}}^{s'} b_{\mathbf{p}}^{s\dagger} \bar{u}^{s'}(\mathbf{p}) ( \gamma^0 p_0 ) u^s(\mathbf{p}) + a_{\mathbf{p}}^{s'\dagger} a_{\mathbf{p}}^{s} \bar{v}^{s'}(\mathbf{p}) ( - \gamma^0 p_0 ) v^s(\mathbf{p})$$ $$= \sum_s \int \frac{ d^3p}{ (2\pi)^3} E_{\mathbf{p}} ( b_{\mathbf{p}}^{s} b_{\mathbf{p}}^{s\dagger} - a_{\mathbf{p}}^{s\dagger} a_{\mathbf{p}}^{s} )$$ $$= \sum_s \int \frac{ d^3p}{ (2\pi)^3} - E_{\mathbf{p}} (b_{\mathbf{p}}^{s\dagger} b_{\mathbf{p}}^{s} + a_{\mathbf{p}}^{s\dagger} a_{\mathbf{p}}^{s} ) - \infty$$

Changing anticommutator into commutator will make the spectrum unbounded.