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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 $\mathcal{H}$?

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  • $\begingroup$ There are no spinless fermions in QFT, AFAIK. I suppose you may, for purely heuristic reasons, consider a quantity that's a Grassman-odd number, call it "fermionic" in that sense, and not relate it to spin in any way --that would be incompatible with the usual supersymmetric algebras. But I don't know what the interest of that could be. Unless I have misunderstood something in your question. $\endgroup$ – joigus Feb 17 at 14:50
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    $\begingroup$ @joigus Spinless fermion models are used routinely in condensed matter physics. They are typically defined by imposing canonical anticommuation relations on a family of operators with no spin indices. Experimentally they can be probed by imposing a strong magnetic field so that the Zeeman splitting is much greater than the other energy scales in the problem, so there is effectively only one spin species present $\endgroup$ – By Symmetry Feb 17 at 14:54
  • $\begingroup$ Thanks for pointing that out. I'll take a back sit and learn something. $\endgroup$ – joigus Feb 17 at 14:58
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    $\begingroup$ @FriendlyLagrangian If one applies high magnetic field, then all the fermions have spins in one diretcion, the contribution of the other spin state being negligible. The use of spinless fermions however goes far beyond such approximations - simply because in condensed matter we often do not need to care about supersymmetries, and also because in many problems spin can be accounted for by multiplying the result by 2. $\endgroup$ – Vadim Feb 17 at 15:35
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    $\begingroup$ @BySymmetry not just condensed matter. Fadeev-Popov ghosts in gauge theories like the standard model are also spinless fermions. The concept makes complete sense so long as one doesn't restrict the definition of particle unnecessarily. $\endgroup$ – jacob1729 Feb 17 at 15:43
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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.)

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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 $$ f\mapsto \left(\matrix{0&0\cr 1&0}\right), \quad f^\dagger \mapsto \left(\matrix{0&1\cr 0&0}\right) $$

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