To be more specific, say I have the wavefunction for the position of some arbitrary particle, but then I also have the QFT operator (is that the correct term?) modeling the position of that same particle. What's the relationship between those two mathematical objects? Like, could a 1-to-1 correspondence be reasonably created between the wave functions of QM and the operators of QFT?

Or, to put it in more conceptual terms, is it accurate to say that particles are excitations in quantum fields and the PDF's we get from squaring the wavefunctions in QM correspond to the physical distributions of the excitations in the fields? The image I have in my head is of an object being dropped onto a clear sheet and creating ripples, and I imagine the shape of the ripples would correspond to the shape of the probability distribution for the particle, where the particle is just the ripples themselves. I'm guessing the actual relationship is more complicated than that though. Is it? If so, in what way?

Edit: Looking at the answer to this question, as was suggested: Why we use fields instead of wave functions? I get the impression that a quantum field is just the superposition of all the relevant wave functions, like, that the electron field is the superposition of the wavefunctions of all electrons. Is that right?

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    $\begingroup$ Does this answer your question? Why we use fields instead of wave functions? $\endgroup$ Jan 12, 2023 at 0:57
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    $\begingroup$ Also possibly relevant: physics.stackexchange.com/q/734234 $\endgroup$
    – Andrew
    Jan 12, 2023 at 1:39
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    $\begingroup$ "I get the impression that the electron field is the superposition of the wavefunctions of all electrons. Is that right?" No, it is not right. $\endgroup$
    – hft
    Jan 12, 2023 at 2:07
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    $\begingroup$ A decent answer to this question would explain how to translate between regular and "second quantization" descriptions of the same multi-particle system. Anything else is kind of dodging the question. $\endgroup$
    – DanielSank
    Jan 12, 2023 at 5:57
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    $\begingroup$ I voted to re-open. @Quillo the close votes make no sense, given that the referenced questions don't answer or even ask what this one is asking. In my opinion, the accepted answer doesn't answer the question either, though. $\endgroup$
    – DanielSank
    Jan 12, 2023 at 16:29

4 Answers 4


When we do quantum field theory, we still are doing quantum mechanics. We don't normally work a lot with a wave function explicitly because the wavefunction (other than perhaps the vacuum state) is quite complicated (and hard to prepare in many cases).

Nevertheless, all the postulates of Quantum Mechanics still hold, we just usually do not write out a single N-particle Hamiltonian, since we usually want to consider the case where the number of particles is changing.

Another way to understand why we have to introduce all this fancy new formalism is to realize that in "ordinary" N-particle quantum mechanics space and time are treated completely differently. In N-particle QM, Spatial position is an operator, but time is a real number.

If you try to promote time to an operator, you run into trouble. So, instead we demote position to a number. But, we still need operators, and those are the fields; the fields are operator-valued functions of space and time (where both space and time are just real numbers).

What is the mathematical relationship between the wave functions of QM and the fields in QFT?

The fields are operators, so they act on wavefunctions.

For example, when we write a Klein-Gordan field operator like: $$ \hat\phi(\vec r, t)\;, $$ we mean that there is an operator at every point in space-time.

This operator can act on, say, the ground state wave function of the universe $|\Phi_0\rangle$ and create a new wavefunction: $$ |\Psi\rangle(\vec r, t) = \hat\phi(\vec r, t)|\Phi_0\rangle\;. $$

I get the impression that a quantum field is just the superposition of all the relevant wave functions, like, that the electron field is the superposition of the wavefunctions of all electrons. Is that right?

This is not the right way to say this.

It is perhaps helpful to think of particles as localized excitations associated with the field. This is nice, because it help us feel good about why, say, every electron is identical. We can say they are all identical because they are all excitations created by the same field operator $\hat \psi(\vec r,t)$, which itself obeys the Dirac equation.

Update to address one comment:

How do we recover the Schrodinger equation? We don't need to recover it, per se, we just need to write it in terms of the field operators. It looks the same: $$ i\partial_t |\Psi\rangle(t) = \hat H |\Psi\rangle(t)\;, $$ but here the Hamiltonian is written as an integral over all space of the operators. For example, for a free Klein-Gordan field: $$ \hat{H} = \int d^3r \frac{1}{2}\left( \dot{\phi(\vec r)}^2 + \nabla\phi(\vec r)^2+ m^2\phi(\vec r)^2\;, \right) $$ where the operators have no time dependence above since we are working in the Schrodinger picture now. (And I have dropped the "hats" from the operators.)

This Hamiltonian can be rewritten as: $$ H = \int \frac{d^3p}{(2\pi)^3}\omega_p a^\dagger_{\vec p}a_{\vec p} + C\;, $$ where $C$ is a numerical (non-operator) constant. And where: $$ \phi(\vec r) = \int \frac{d^3p}{\sqrt{2\omega_{\vec p}}(2\pi)^3}\left( a_{\vec p}e^{i\vec p\cdot \vec x}+ a_{\vec p}^\dagger e^{-i\vec p\cdot \vec x} \right) $$

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    $\begingroup$ +1 but how do you recover the schrodinger equation for the wave function? Do you know any reference? $\endgroup$
    – Quillo
    Jan 12, 2023 at 2:39
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    $\begingroup$ We do talk about the wavefunction, we just do it in the momentum basis instead of the wavefunctional basis. For example, the $S$ matrix lays out the Fock basis components of the outgoing wavefunction $\endgroup$
    – Ryder Rude
    Jan 12, 2023 at 2:47
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    $\begingroup$ @Quillo I will try to find a reference, but I don't have one immediately accessible $\endgroup$
    – hft
    Jan 12, 2023 at 3:04
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    $\begingroup$ @MikaylaEckelCifrese When I say position is an operator, I mean it is an operator on a Hilbert space in the usual sense, just like the Hamiltonian, the momentum, etc. In QM every observable is a Hermitian operator. For example, position is an operator: $\hat X$. The action of this operator on a single-particle wavefunction $\psi(x)$ is $\hat X\psi(x) = x\psi(x)$. As another example, momentum is an operator $\hat P$. The action of this operator on a single-particle wavefunction $\psi(x)$ is $\hat P\psi(x) = -i\frac{d\psi}{dx}$. $\endgroup$
    – hft
    Jan 12, 2023 at 4:34
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    $\begingroup$ You write $|\Psi\rangle(\vec r, t) = \hat\phi(\vec r, t)|\Phi_0\rangle$. That equation makes me uneasy. For one thing, shouldn't it only work for a one-particle state? But my main concern is that the Hilbert spaces for the two kets are different. One's for a Schrodinger wave function and one's for a QFT wave function. And I would have guessed that $\hat\phi$ maps from the QFT Hilbert space to itself, not to the Schrodinger Hilbert space. Could you explain this a bit more? $\endgroup$ Jan 13, 2023 at 3:07

The problem connecting QFT and $N$-body quantum mechanics isn't so much with QFT but rather with relativistic field theories. For a non-relativistic theory the connection is actually quite straight forward, if we define the one-particle momentum states by $$a^{\dagger}_\mathbf{p}|0\rangle=|\mathbf{p}\rangle$$ then the position basis from regular QM is simply $$|\mathbf{x}\rangle=\frac{1}{(2\pi)^{3/2}}\int d^3p \ e^{-i\mathbf{p}\cdot\mathbf{x}}a^{\dagger}_\mathbf{p}|0\rangle.$$ These are delta-normalized because of the commutation relations for $a_\mathbf{p}$ and so the regular QM wavefunction for a one particle state is $$\psi(\mathbf{x})=\langle\mathbf{x}|\psi\rangle=\frac{1}{(2\pi)^{3/2}}\int d^3p \ e^{i\mathbf{p}\cdot\mathbf{x}}\langle0|a_\mathbf{p}|\psi\rangle.$$ The problem comes with relativistic field theories where the measure $d^3p$ has to be replaced with $d^3p/2E_p$ for lorentz invariance and this ruins the orthogonality of the analogous position basis states, more precisely $$\frac{1}{(2\pi)^3}\int \frac{d^3p}{2E_p}\ e^{i(\mathbf{x}-\mathbf{x'})\cdot\mathbf{p}}\neq\delta^3(\mathbf{x-x'}).$$ The way to think about this is that relativistic particles/fields have an inherent amount of spatial spread and so it's impossible to construct a completely localized state. Duncan briefly discusses this in section 6.5 of his textbook The Conceptual Framework of Quantum Field Theory in which he showed that you can't construct a one-particle state that is delta-localized for both the number density operator and the enery density operator.

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    $\begingroup$ But... this answer dodges multiple particles, which is where QFT really differs from introductory quantum mechanics. $\endgroup$
    – DanielSank
    Jan 12, 2023 at 5:56
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    $\begingroup$ Everything I wrote here generalizes readily to any number of particles. $\endgroup$
    – AfterShave
    Jan 12, 2023 at 20:09
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    $\begingroup$ I think not. How do I recover the symmetrization and antrisymmetrization criteria of wave functions based on this answer? $\endgroup$
    – DanielSank
    Jan 13, 2023 at 1:20
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    $\begingroup$ For two particles the position basis is $\psi^{\dagger}(x)\psi^{\dagger}(y)|0\rangle$, then the wavefunction for a state $a^{\dagger}_\mathbf{p}a^{\dagger}_\mathbf{k}|0\rangle$ can be evaluated to $e^{i \mathbf{k \cdot x}} e^{i \mathbf{p \cdot y}} \pm e^{i \mathbf{p \cdot x}} e^{i \mathbf{k \cdot y}}$ depending on whether you two commutators or anti-commutators. $\endgroup$
    – AfterShave
    Jan 13, 2023 at 2:27
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    $\begingroup$ @DanielSank What is your issue with this answer? Is it just that the answer does not explicitly explain that we have operator commutation relations like $\left[a_i, a_j^\dagger \right]_{\pm} = \delta_{ij}$? The formalism of QFT must account for the statistics of multiple particle or else it would not work at all... For example, if it didn't then we would get the wrong answers when we calculated scattering matrix elements, etc. $\endgroup$
    – hft
    Jan 17, 2023 at 15:59

Everything depends on the notion of wavefunction one adopts.

Generally speaking, a wavefunction is expected to be an element of a Hilbert space of functions $\psi_t=\psi(t,\vec{x})$ defining the state of a particle at time $t$. Here, $|\psi(t,\vec{x})| d^3x$ is assumed to be the probability to detect the particle in the small spatial volume $d^3x$ around $\vec{x}$ of a Minkowski reference frame at its time $t$.

This object, in relativistic quantum theory simply does not exist.

As soon as one switches on the Poincar'e symmetry and assumes that energy is positive, there is no triple of selfadjoint operator representing the position observable and thus a corresponding $L^2(\mathbb{R}^3, d^3x)$ space on the joint spectrum describing the position (in a Minkowski reference frame).

This is due to a number of no-go results essentially arising from the popularly called "Hegerfeldt Theorem". In particular, they rule out the Newton-Wigner operator as a selfadjoint operator defining the position observable. In spite of the fact that this is the only possible Projector Valued Measure (PVM) that correctly transform under the action of Poincar'e group (as proved by Wightman).

There are other possibilities to describe the position observable in terms of a POVM (Positive- Operator - Valued measures) and not selfadjoint operators (i.e PVM). POVM are the standard notion of observable in the modern view on quantum theory arising from, for instance, quantum information.

However, the Newton Wigner operator has still some relevance, as is the first moment of the position probability distribution (for all conceivable notion of position described by POVMs)

See my recent paper (starting from a proposal due to D.Terno) on the subject and the even more recent one by D. Castrigiano

There are many issues still open regarding the notion of relativistic position, especially concerning the post-measurement state.

I stress that the problem of providing a meaningful sense to some also approximated notion of position for a relativistic particle is crucial, as experimental physicists can say, with a certain precision, where a particle is detected in the rest space of the laboratory. Hence the kick-off answer "Everything exists is a field so we do not need of that notion" is completely misleading in my view.

It is clear, however, that this notion cannot be given in terms of selfadjoint operators (i.e. PVMs), but POVMs are necessary.

The definition (for scalar bosons) based on th extractio of the positive energy part of the quantum field $$\psi(t,\vec{x}) = \int e^{ipx} \hat{\psi}(p) \frac{d^3p}{2(2\pi)^{3/2}E(p)}$$ does not define a wavefunction in the sense said above, as $|\psi(t, \vec{x})|^2$ is not the wanted proabability density. However, in the non-relativistic limit $\vec{p}^2<< m^2$, then $E\sim m$ and the definition (up to a factor) becomes the standard wavefunction of non relativistic physics with the usual properties.

  • $\begingroup$ Tacitly, the assumption seems to be that the link between the wavefunction and the position operators is $\psi(x,t)=\langle x|\psi\rangle$, where $|x\rangle$ are assumed to be eigenstates of some position operator. Then it is posed that with no position operator there are no position eigenstates and thus no wavefunction. However, why could one not just have a position basis represented by $|x\rangle$ without assuming a position operator? In my view the former does not require the latter. $\endgroup$ Oct 6, 2023 at 10:23
  • $\begingroup$ First of all, the standard formalism based on objects like $|x\rangle$ is too rough to handle these subtleties and it should be replaced for a more rigorous formalsm. However, the short answer is that $X= \int x |x\rangle\langle x| dx$ would be the position operator. $\endgroup$ Oct 6, 2023 at 10:37
  • $\begingroup$ The point is not the position operator, but the spectral measure of it. If a spectral measure (made of orthogonal projectors) existed for the position observable (and this is equivalent to the existence of a selfadjoint position operator), it would imply a superluminal propagation of the spatial detection probability of a particle. $\endgroup$ Oct 6, 2023 at 10:39
  • $\begingroup$ What is a "spectral measure"? $\endgroup$ Oct 6, 2023 at 10:41
  • $\begingroup$ A map associating every set $E$ of the say, x axis, with a corresponding orthogonal projector $P(E)$ in the Hilbert space. In this sense $X = \int x dP(x)$. Roughly speaking $P(E)= \int_E |x\rangle \langle x| dx$ $\endgroup$ Oct 6, 2023 at 10:43

In quantum mechanics, the wave function is a way to represent the state of a system. It need not be just one body, which is what your question is tacitly alluding to, but I'll get to that in a moment. If there are $n$ bodies, each with $3$ coordinates, then it's a function of $3n$ coordinate. It all goes down the same kettle as an undistinguished $3n$-dimensional mess. The wave function for the Helium atom - if treating the nucleus as an external source - would include two electrons in it, not just one, and so would have $6$ coordinates, not $3$.

Frequently, however - typically seen in textbook presentations - the wave function is used as a representation for a single body, as if the body, itself, were a wave in ordinary $3$ dimensional space. These are the one-body states.

However, it's not really a case of a body being "smeared out" in space, which is the usual visualization behind the "wave function" - betrayed by the name, itself - but rather the configurations of an entire system that are smeared out, non-locally.

For two bodies, for instance, the states, themselves, may not be factorable into a product of a state for one body, and the state for a second body, although the state spaces where they reside are factorable. Thus, if you have two electrons whose states are described by a Hilbert space $H$, then the two electron-states reside in $H⊗H$. A wave function would have the form $ψ = \sum_{m,n∈ℤ} ψ_{m,n}|m⟩_0|n⟩_1$, with $\{|m⟩_0: m∈ℕ\}$ being the (countable) basis for the first $H$ for the first electron (its basis indexed by the natural numbers $ℕ = \{0,1,2,⋯\}$, and $\{|n⟩_1: n∈ℕ\}$ the basis for the second $H$ for the second electron. The state, itself, can't generally be factored, e.g $ψ = ψ_0 ψ_1$, since the sum may involve entanglement (e.g. $ψ = |0⟩_0|0⟩_1 + |1⟩_0|1⟩_1$).

The quantum theory of many-body systems, technically, still resides in quantum mechanics, rather than quantum field theory, but starts to cross the bridge that your question is asking about.

Many-body state spaces are called Fock Spaces. Quantum field theory uses many-body spaces - with a twist will be described in a moment. The same Hilbert space $H$ can be used for the one-body state, and as the foundation for the many-body state space, as its the "1-body" sector.

The historical reason quantum field theory fell back into a formulation in terms of many-body states is because of vacuum polarization and its reverse: two or more bodies can come into or or out of existence and transform into or from other bodies. The classical case is an electron and positron cancelling each other out and leaving behind two photons. The reversal of the process would correspond to pumping energy into a vacuum (the two photons) and polarizing the vacuum into an electron and positron.

So, out of necessity, you have to use many-body states.

The following notations and conventions - which are not standard, but are very useful in bridging the gap between the two cases - will help explain where the difference lies and what remains of the "one-body" state in the larger setting of the Fock space. The one-body state is what's left of your original wave function, if thinking of the wave-function as a one-body state.

Maxwell-Boltzmann Fock Spaces In Algebraic Form - "Stack Machines"
So, start with a Hilbert space $H$. Create the following spaces:

  • The "zero body" states: $H^0 = ℂ$. It's a trivial one dimensional Hilbert space. Apart from a normalization factor, there is only one such state. That's the "vacuum".
  • The "one body" states: $H^1 = H$, itself. That's where your wave functions lie. The basis is just $\{|m⟩: m ∈ ℕ\}$.
  • The "two body" states: $H^2 = H⊗H$, consisting of the two-body states $\{|m⟩|n⟩ = |mn⟩: m, n ∈ ℕ\}$.
  • The "three body" states: $H^3 = H⊗H⊗H$, consisting of the three-body states $\{|m⟩|n⟩|p⟩ = |mnp⟩: m, n, p ∈ ℕ\}$.
  • Do the same for $H^n$, for $n > 3$.
Together, we collect all the states to form the Hilbert space that we denote using the Kleene star (just like those seen for regular expressions): $$H^* ≡ H^0 ⊕ H^1 ⊕ H^2 ⊕ H^3 ⊕ ⋯.$$ It has the property that it can be described recursively as the least solution to the following fixed-point inclusion: $$H^* ⊇ ℂ ⊕ \left(H⊗H^*\right).$$ Every state in $ψ ∈ H^*$ has one of the following forms:
  • The $0$-body states $ψ = z ∈ ℂ$,
  • The $n+1$-body states $ψ = v ψ'$, where $v ∈ H$ is a $1$-body state and $ψ'$ an $n$-body state.
The states, themselves, are represented as "words", e.g. $u_1u_2u_3Ω$ for a $3$-body state, where $u_1, u_2, u_3 ∈ H$, where $$Ω = 1 ∈ ℂ = H^0,$$ denotes the vacuum state, normalized to $1$. Specifically, the original "wave functions" are now the one-body states corresponding to $v ∈ H$ which are now $|v⟩Ω = vΩ ∈ H^*$.

Technically, the state space is called the Maxwell-Boltzmann Fock space associated with $H$. In it, each body is tagged (by its word position, e.g. $u_1$ is at position $1$, $u_2$ at position $2$, $u_3$ at position $3$ in $u_1u_2u_3$), and is distinguishable from the other bodies.

Here's the non-standard, algebraic, part: instead of treating $|v⟩$ as synonymous with $v ∈ H$, we will treat it as an operator defined by: $$|v⟩: ψ ∈ H^* ↦ v ψ ∈ H^*.$$ That's analogous to the "push" operation of a stack in a computing machine. Instead of treating $⟨u|$ as the dual of $u ∈ H$ that resides in the dual Hilbert space $\tilde{H}$, we will treat it as an operator, defined recursively by: $$⟨u|: z ∈ ℂ ↦ 0, \quad ⟨u|: v ψ ↦ ⟨u|v⟩ ψ \quad \left(v ∈ H, \quad ψ ∈ H^*\right).$$ This corresponds to a "pop and test for equality to $u$" operation in a computer stack.

Finally, we will introduce the zero-state projection operator, defined recursively by: $$π_0: z ∈ ℂ ↦ z, \quad π_0: v ψ ↦ 0 \quad\left(v ∈ H, \quad ψ ∈ H^*\right).$$ This tests for the "empty stack" condition.

The algebra then satisfies the following relations: $$ {π_0}^2 = π_0, \quad ⟨i| π_0 = 0 = π_0 |j⟩, \quad ⟨i||j⟩ = δ^i_j = \left\{\begin{align}1 \quad (i = j)\\0 \quad (i ≠ j)\end{align}\right\}\quad \left(i, j ∈ ℕ\right),\\ π_0 + \sum_{i ∈ ℕ} |i⟩⟨i| = 1. $$

The Maxwell-Boltzmann State Space $H^*$ is a de facto Hilbert space implementation of a stack memory, formed of "stack words" or "stack configurations", whose elements are the bodies, themselves, represented via their states in $H$.

The one-body sector, within it, is where you will find your (one-body) "wave functions". But the $ψ$'s in $H^*$ are much more far-reaching in scope, as they also include the vacuum and $n$-body states for $n > 1$.

Einstein-Bose Fock Spaces: For Energy
The quanta that make up an energy field, such as the electromagnetic field, are no more distinguishable from one another than waves of the same shape on an ocean are. Unlike Maxwell-Boltzmann Fock Spaces, there is only one way for two photons of the same type to be in two places, with one in each; while for Maxwell-Boltzmann 2-body states, each body is tagged and there are two ways for them to be like so.

Algebraically, the state space has the same structure as $H^*$, except that the elements of the words commute. Define the following relation: $$ρ_{+}: u v = v u \quad (u, v ∈ H).$$ Then the Einstein-Bose Fock Space associated with $H$ can be defined as: $$H^{EB} = H^*/ρ_{+}.$$

The bodies in such a state space are called "bosons" are are typically associated with energy fields.

Correspondingly, we have the following relation between the $|u⟩$ operators: $$|u⟩|v⟩ = |v⟩|u⟩ \quad (u, v ∈ H).$$

To accommodate this change, we redefine the operators $⟨u|$ for $u ∈ H$, recursively by: $$⟨u|: z ∈ ℂ ↦ 0, \quad ⟨u|: v ψ ↦ ⟨u|v⟩ ψ + |v⟩(⟨u|ψ) \quad \left(v ∈ H, \quad ψ ∈ H^*\right).$$

The algebra satisfies the following relations: $$ {π_0}^2 = π_0, \quad ⟨i| π_0 = 0 = π_0 |j⟩ \quad (i, j ∈ ℕ),\\ \quad [⟨i|,|j⟩] = ⟨i||j⟩ - |j⟩⟨i| = δ^i_j, \quad ⟨i|⟨j| = ⟨j|⟨i|, \quad |i⟩|j⟩ = |j⟩|i⟩ \quad (i, j ∈ ℕ),\\ \sum_{i ∈ ℕ} |i⟩⟨i| = n, $$ where $n$ denotes the "number" operator that determines how many bodies are in the state.

Since order no longer matters, the bodies are no longer tagged or distinguishable. Each "word" is now just a monomial over the set $\{|0⟩, |1⟩, |2⟩, ⋯\}$, and states are generated by sums of monomials, i.e. polynomials, e.g. $$|0⟩(|2⟩Ω) + |1⟩(|1⟩Ω) = (|0⟩|2⟩ + |1⟩|1⟩)Ω.$$ The operators $\{⟨0|, ⟨1|, ⟨2|, ⋯\}$ act as derivatives on the polynomials, e.g. $$\begin{align} ⟨1|(|0⟩|2⟩ + |1⟩|1⟩)Ω &= ⟨1|(|0⟩|2⟩Ω) + ⟨1|(|1⟩|1⟩Ω)\\ &= |0⟩⟨1|(|2⟩Ω) + |1⟩Ω + |1⟩(⟨1||1⟩Ω)\\ &= |0⟩|2⟩⟨1|Ω + |1⟩Ω + |1⟩(Ω + |1⟩⟨1|Ω)\\ &= |1⟩Ω + |1⟩Ω\\ &= 2|1⟩Ω\\ &= \frac{∂}{∂|1⟩}(|0⟩|2⟩ + |1⟩|1⟩)Ω \end{align}$$

Fermi-Dirac Fock Spaces: For Matter
In contrast to the Einstein-Bose Fock Spaces, in the Fermi-Dirac Fock Space, only one body may exist in a given state. That's the Pauli Exclusion Principle. These are the states typically used for matter. The difference is that the algebra is now defined by the anti-commuting relation: $$ρ_{-}: u v = -v u \quad (u, v ∈ H),$$ as: $$H^{FD} = H^*/ρ_{-}.$$ The relation implements the exclusion principle, since, under it, $|v⟩|v⟩ = -|v⟩|v⟩$, for any $v ∈ H$, thus $|v⟩|v⟩ = 0$.

This time, to accommodate the relation $ρ_{-}$, the operators $⟨u|$, for $u ∈ H$ are defined, recursively by $$⟨u|: z ∈ ℂ ↦ 0, \quad ⟨u|: v ψ ↦ ⟨u|v⟩ ψ - |v⟩(⟨u|ψ) \quad \left(v ∈ H, \quad ψ ∈ H^*\right).$$ The corresponding identities for the algebra are: $$ {π_0}^2 = π_0, \quad ⟨i| π_0 = 0 = π_0 |j⟩ \quad (i, j ∈ ℕ),\\ \quad \{⟨i|,|j⟩\} = ⟨i||j⟩ + |j⟩⟨i| = δ^i_j, \quad ⟨i|⟨j| = -⟨j|⟨i|, \quad |i⟩|j⟩ = -|j⟩|i⟩ \quad (i, j ∈ ℕ). $$

In place of "words" and "monomials", the elements of $H^{FD}$ are essentially sets, since only one body of a given type may appear in a state.

All of this might have connection to Finkelstein's notion of Quantum Sets, but I only know of his work vaguely and haven't had an opportunity to examine it in any detail.


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