# Why the Minkowski and Euclidean spinors need to be fermions?

1. Minkowski spinors are the spinor representations of the spin group $$Spin(1,d)$$ of spacetime rotational symmetry.

2. Euclidean spinors are the spinor representations of the spin group $$Spin(1+d)$$ of spacetime rotational symmetry.

The group elements of the spin group are generated by some Clifford algebra with $$\gamma_j$$ obeying $$\{\gamma_j, \gamma_k\} = 2 g_{jk}$$ where the $$g_{jk}$$ is the metric signature for the Minkowski or Euclidean signatures.

At this stage, we do not yet know whether the Minkowski and Euclidean spinors need to be fermions or bosons, or any other statistics.

But when we dicuss the Minkowski and Euclidean spinors, we regard them in QFT as Grassman variables with fermion anticommutator relations $$\{ \psi_a(x) ,\psi_b(y) \}=\delta_{ab}\delta(x-y),$$ thus they are fermions --- obeying fermions statistics.

question: So, why the Lorentz and Euclidean spinors need to be fermions? Could they have some other statistics, like

• bosons?
• anyons in 2+1d?
• Possible duplicates: physics.stackexchange.com/q/17893/2451 , physics.stackexchange.com/q/134577/2451 and links therein. Jun 22, 2021 at 22:49
• >> Why are anticommutators needed in quantization of Dirac fields? --> No this is not what I am asking. That question is addressed in Chap 3 Dirac fields in Peskin book. But I am asking why the spacetime spinor representation needs to be fermionic statistics Jun 22, 2021 at 23:32
• No, both links do not provide what I am asking. I am asking why the spacetime spinor representation needs to be fermionic statistics. Perhaps for certain SUSY theory or nonrelativistic theory, the spacetime spinor representation needs NOT to be fermionic statistics (my hint). Jun 22, 2021 at 23:34
• Also Dirac spinors are not the only spacetime spinor. Maybe we also need to understand Majorana and Weyl spinors (or others) => do they need to have fermionic statistics for being spacetime spinor? Jun 22, 2021 at 23:36
• (But thanks very much for sharing the links :) ) Jun 22, 2021 at 23:37

There are various levels of generality you can aim for when proving the spin statistics theorem but it's probably best to start with free fields coming from the Dirac action $$$$S = \int i\bar{\psi} \gamma^\mu \partial_\mu \psi - m \bar{\psi} \psi \, dx.$$$$ The canonical quantization procedure then tells us to define a conjugate momentum $$$$\pi_\alpha = \frac{\delta L}{\delta \partial_0 \psi^\alpha} = i \psi^\dagger_\alpha.$$$$ (If the spinors are Majorana there is a $$\frac{1}{2}$$ from Dirac brackets but this doesn't change the conclusion.) Now why do we have to impose $$\left \{ \psi^\alpha(\textbf{x}), i\psi^\dagger_\beta(\textbf{y}) \right \} = i\delta^\alpha_\beta \delta(\textbf{x} - \textbf{y})$$? Well let's try imposing $$\left [ \psi^\alpha(\textbf{x}), i\psi^\dagger_\beta(\textbf{y}) \right ] = i\delta^\alpha_\beta \delta(\textbf{x} - \textbf{y})$$ instead and seeing what this says about the creation an annihilation operators. We know the fundamental field needs to have one $$a$$ and one $$b^\dagger$$ to make the energy bounded from below. $$$$\psi(x) = \int a(\textbf{p})^r u_r(\textbf{p}) e^{-ipx} + b^\dagger(\textbf{p})^r v_r(\textbf{p}) e^{ipx} \, \frac{d\textbf{p}}{(2\pi)^3}$$$$ Along with this expression, we also need to know how to contract the plane wave solutions to the Dirac equation. $$$$\bar{u}_r(\textbf{p}) u_s(\textbf{p}) = 2m \delta_{rs}, \;\; \bar{v}_r(\textbf{p}) v_s(\textbf{p}) = -2m \delta_{rs}$$$$ The algebra then leads to \begin{align} [a(\textbf{p})^r, a^\dagger(\textbf{q})^s] = \delta^{rs} \delta(\textbf{p} - \textbf{q}), \;\; [b(\textbf{p})^r, b^\dagger(\textbf{q})^s] = -\delta^{rs} \delta(\textbf{p} - \textbf{q}) \end{align} from which we see that $$b^\dagger(\textbf{p})_r \left | 0 \right >$$ has negative norm. This problem goes away when we use ant-commutators instead.
• Oh well then the answer is "it doesn't". If all you want to do is write down a classical field theory involving spinors, you can make them commuting to your heart's content. It just won't be the $\hbar \to 0$ limit of a sensible QFT anymore. Jun 22, 2021 at 23:53