Hilbert space of a spinless fermion?

Question

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, $f_i := 1otimes cdots otimesunderbrace{f}_iotimescdots1$ where $f:mathcal{H} rightarrow mathcal{H}$.
But what is $mathcal{H}$? is it $mathbb{C}^2$, so that the full Hilbert space is $bigotimes_i^Nmathbb{C}^{2}$ ?
Further more, what would be a suitable basis for $mathcal{H}$?

jacob1729 · Accepted Answer

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, $|0rangle,|1rangle=f^dagger|0rangle$.
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.)

mike stone · Answer

Yes, it's ${mathbb C}^2$ and the basis can be taken as $|0rangle$ (no particle) and $|1rangle equiv f^dagger |0rangle$ (one particle). In this basis, and identifying $|0rangle to (0,1)^T$ and $|1rangle to (1,0)^T$, we have
$$
fmapsto left(matrix{0&0cr 1&0}right), quad f^dagger mapsto left(matrix{0&1cr 0&0}right)
$$

Hilbert space of a spinless fermion?

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