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), $${\cal L} _{ kin} = \frac{1}{2} \partial _\mu \phi \partial ^\mu \phi$$ 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, $$\bar{\psi} i \partial _\mu \gamma ^\mu \psi \quad ?$$ i.e., would having a negative in front of the Dirac Lagrangian make the Hamiltonian unbounded?

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1 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.

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Great answer, thanks! –  JeffDror Feb 26 at 13:35