Hilbert space of a spinless fermion? What is the Hilbert space of a spinless fermion?
Take for instance the following simple example, a 1D free fermion chain:
$$
H = \sum_i^N E_i f_i^{\dagger}f_i
$$
where $f$ are canonical anti-commuting fermionic creation and annihilation operators (i.e. $\{f_i,f_i^{\dagger}\}=\delta_{ij}$ etc). Now, $f_i := 1\otimes \cdots \otimes\underbrace{f}_i\otimes\cdots1$ where $f:\mathcal{H} \rightarrow \mathcal{H}$.
But what is $\mathcal{H}$? is it $\mathbb{C}^2$, so that the full Hilbert space is $\bigotimes_i^N\mathbb{C}^{2}$ ?
Further more, what would be a suitable basis for $\mathcal{H}$?
 A: Yes and yes. What other options could there be?
A single fermionic orbital can be occupied or unoccupied. It clearly has dimension $2$ and thus has Hilbert space $\mathbb{C}^2$. A basis is $|\text{empty}\rangle,|\text{full}\rangle$ or more compactly, $|0\rangle,|1\rangle=f^\dagger|0\rangle$.
Joining $N$ of these together corresponds to a tensor product. Again, what else could it correspond to? Note that you don't need to anti-symmetrise since the orbitals are obviously distinct and that is what matters.
(As a quick note: this scenario of having spinless fermions in a chain comes up a lot, for instance after performing a Jordan-Wigner transform on an XY chain.)
A: Yes, it's ${\mathbb C}^2$ and the basis can be taken as $|0\rangle$ (no particle) and $|1\rangle \equiv f^\dagger |0\rangle$ (one particle). In this basis, and identifying $|0\rangle \to (0,1)^T$ and $|1\rangle \to (1,0)^T$, we have
$$
f\mapsto \left(\matrix{0&0\cr 1&0}\right), \quad f^\dagger \mapsto \left(\matrix{0&1\cr 0&0}\right)
$$
