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Let $\mathscr{H}$ be a Hilbert space and $\mathscr{H}^{n}$ be the associated $n$-fold tensor product of this Hilbert space. I'll skip the mathematical details in what follows, but my approach follows Reed & Simon's book. We can define an operator $A_{n}$, called anti-symmetric operator, which is an orthogonal projection on $\mathscr{H}^{n}$ to its anti-symmetric subspace $A_{n}\mathscr{H}^{n}$. Mathematically, we can define what is usually called a fermionic Fock space $\mathcal{F}_\text{fer}$ by: $$\mathcal{F}_\text{fer} := \bigoplus_{n=0}^{\infty}A_{n}\mathscr{H}^{n}$$ where $A_{0}\mathscr{H}^{0} := \mathbb{C}$ and $A_{1}\mathscr{H}^{1} := \mathscr{H}$. We can proceed and define operators on $\mathcal{F}_\text{fer}$, such as the Hamiltonian, creation and annihilation operators and so on. The procedure to do that is basically as follows: we have a given operator $T$ on $\mathscr{H}$ and we extend it to $\mathscr{H}^{n}$ by using a procedure called second quantization and then define $T$ on $\mathcal{F}_\text{per}$ by letting it act on each 'component' $A_{n}\mathscr{H}^{n}$.

The above scenario is pretty much general, and it is basically mathematics. However, the name 'fermionic' always caught my attention. I was sure that this was the mathematical formulation of a system of many fermions in some sense, and this would justify the name. However, as time went by I've found some very different models which seems to be described by the above formalism. Let me give you some quick informal examples:

(1) We can take $\mathscr{H} = L^{2}(\mathbb{R}^{d};\mathbb{C})$ the space of complex-valued square integrable functions defined on $\mathbb{R}^{d}$.

(2) If we are in a discretized space, we can replace $\mathbb{R}^{d}$ in the above example by some finite set $\Lambda$. This finite set could be a finite lattice or even the cartesian product of a finite lattice and some finite 'spin' set.

(3) If we are thinking about Dirac fields, it seems that the proper Hilbert $\mathscr{H}$ is $\mathbb{C}^{2}$, which accounts to spins line ${1}\choose{0}$ and ${0}\choose{1}$.

These three examples (and possibly other ones) are found in many different books under the same name of 'fermionic system' or 'fermionic field theory'.

Question: I don't know if all these three examples are just different realizations of just one physical system, but I believe it is not. So, it seems that by a fermionic field theory, one usually thinks of a fermionic Fock space $\mathcal{F}_\text{fer}$ together with some second quantized Hamiltonian $H$ acting on this Fock space, so there are actually a lot of different realizations of fermionic field theories rather than just one. Is this accurate?

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A fermionic field theory isn't characterized by its Hilbert space. It's characterized by the algebraic relationships among its field operators.

A fermionic field theory is one whose observables are constructed from field operators $\psi_n(x)$ that satisfy the equal-time anticommutation relations \begin{align} \{\psi_n(x,t),\psi_m(y,t)\} &= 0 \\ \{\psi_n(x,t),\psi_m^\dagger(y,t)\} &= 0\text{ if }x\neq y\text{ or }n\neq m, \end{align} where $\{A,B\}\equiv AB+BA$. The Pauli exclusion principle is a consequence of this. The various constructions described in the question can be used as different ways of representing such operators on a Hilbert space.

My answer https://physics.stackexchange.com/a/582216 gives some general comments about representations. The central message is that the Hilbert space itself tells us almost nothing about a theory. The pattern of observables is what matters. Different ways of representing the Hilbert space may be more or less convenient for different patterns of observables, but the observables are what matters.

Yes, there are many different fermionic field theories. The Hamiltonian $H$ itself may be sufficient to distinguish between some of them, but the Hamiltonian is just one operator. The relationship between the Hamiltonian and other observables is more discerning, and such a relationship is implied whenever we express $H$ in terms of the field operators, because other observables are also expressed in terms of the field operators. Note that observables must commute with each other at spacelike separation (this is one of the causality principles), so an observable necessarily involves a product of an even number of anticommuting field operators.

By the way, once in a while, I see the name "fermion" used as a synonym for something whose spin is an odd multiple of $\hbar/2$. In relativistic quantum field theory, the spin-statistics connection justifies this language. But more generally, that connection doesn't hold: a fermion (something whose field operators anticommute) does not necessarily have such a spin, or conversely.

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  • $\begingroup$ Thanks for your answer! Such an amaing answer in fact. I think thi is very enlightening! The only point that is not entirely clear to me is how to distinguish these constructions. You said that the Hamiltonian is not always enough. Maybe a possible approach is to verify the anti-commutation relations of the creation and annihilation operators on the one-particle system? Does the anti-comutation on $\mathcal{F}_{fer}$ is inherited by the one in this one-particle system in some way? It is not clear to me, tho. $\endgroup$
    – IamWill
    Oct 7, 2020 at 14:36
  • $\begingroup$ @IamWill I'm not sure I understand what you're asking, but what I meant by "the Hamiltonian is not enough" is that a single self-adjoint operator by itself has only one distinguishing feature: its spectrum. But if we have an algebra of field operators at time $t=0$ and a Hamiltonian that is expressed in terms of those field operators, then we know everything: we know what the field operators are at $t-0$, and we know how to use the Hamiltonian to determine the field operators at any other time, because the Hamiltonian is expressed in terms of them. $\endgroup$ Oct 7, 2020 at 23:10
  • $\begingroup$ @IamWill Sometimes when people refer to the Hamiltonian $H$, they don't just mean that they've specified that one operator $H$ on the Hilbert space. They often really mean that they've expressed $H$ in terms of other operators, such as the field operators, which is much more informative. $\endgroup$ Oct 7, 2020 at 23:10
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    $\begingroup$ @AccidentalFourierTransform Good point! I wasn't thinking about QFTs that don't depend on a metric structure but still depend on a spin structure. Even if we only consider QFTs that are covered by my definition, it's probably still not the best definition because the field-operator formulation is not necessarily intrinsic to the theory (bosonization in 2d, etc). So my answer falls short both in generality and in intrinsic-ness. $\endgroup$ Oct 8, 2020 at 23:29

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