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3

Suppose $a$ and $a^{+}$ operators satisfy $$\left\{ a,a\right\} =0\mbox{ and }\left[a,a^{+}\right]=1$$ We have basically $a^{2}=0$ and $aa^{+}=a^{+}a+1$. Now consider $aaa^{+}$. $$0=aaa^{+}=a\left(a^{+}a+1\right)=aa^{+}a+a=a^{+}aa+2a=2a.$$ So we get $a=0$.

5

This is because the path integral ${\cal Z}$ is an infinite-dimensional version of a Grassmann-odd Gaussian integral $$\int \!\mathrm{d}^n \bar{\theta} ~\mathrm{d}^n\theta ~e^{\sum_{i,j=1}^n\bar{\theta}_i ~M^i{}_j ~\theta^j}~\propto~\det(M),$$ where the indices $i,j$ can be interpreted as DeWitt's condensed notation.

0

You have discovered the fact that the Dirac spinors form a reducible representation of Spin(3,1) $\simeq$ SL(2,C), the covering group of SO(3,1)$^+$. The left and right Weyl spinors, which have two components, are irreducible representations.

3

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 $$... 0 I believe I'm ready to answer my own question. The pin group can alternately be defined as the set of all invertible elements S_{\Lambda} \in \mathrm{Cl}(p,q) satisfying S_{\Lambda} S_{\Lambda}^{\tau} = \pm 1 and$$ \alpha(S_{\Lambda}) \gamma^a S_{\Lambda}^{-1} = {\Lambda^a}_b \gamma^b  for some element ${\Lambda^a}_b \in \mathrm{O}(p,q)$. The map ...

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