A Fermionic quantum field theory $\psi$ has half integer spin, and thus, $\psi$ is a spinor taking values in $\mathbb{C}^m$. On the other hand, the spin-statistics theorem tells us that these fields are anti-commuting and thus take values in a Grassmann algebra. This suggests that the most general form of a spinor field is $$x \mapsto \psi(x) = \psi^i(x) (v_1\wedge\cdots\wedge v_n) \tag{1}$$ where $i$ is the spinor index. In index free notation, (1) can be written as a tensor product: $$x \mapsto \tilde{\psi}(x) \otimes v \tag{2}$$ where $\tilde{\psi}(x)$ is the vector with components $\psi^i$ as in (1) and $v$ is an arbitrary element of the Grassmann algebra.
Is this correct? If so then I have some further questions:
- Why is the Grassmann algebra part of the field $(v_1\wedge\cdots\wedge v_n)$ often omitted in physics textbooks when discussing Fermionic theories and is only brought up when the path integral is introduced? It seems too fundamental to leave out.
- How are Lorentz transformations defined for these fields? If $\Lambda$ is an arbitrary Lorentz transformation, then, I suspect, using the index free version (2), that $\Lambda$ should actually be written as the tensor product of representations $\Lambda \otimes I$ where $I$ is the trivial representation on the Grassmann algebra and $\Lambda$ transforms the spinor part $\psi$ as spinors usually do under Lorentz transformations.
- How is the vector space and its dimension $n$ in (1) chosen when defining the Grassmann algebra? Will any vector space work?
For my third question, I think the dimension should always be infinite, but I'm not sure why. I have made a separate post about this that can be found here. I have also taken a look at this related question about Grassmann algebras and spinors.
