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A many-body model must specify what its factors represent before it specifies a Hamiltonian. Tensor factors can label distinguishable subsystems, while symmetric or exterior powers describe identical particles. These choices lead to different local observables and different infinite-system limits. This page explains the operator consequences of those choices and gives a domain-correct construction of independent-particle energy.

Helpful background. Hilbert Space Tensor Products supplies the completed tensor product. Symmetric and Antisymmetric Tensor Products proves the permutation-projection construction.

Particle slots, sites, and modes are different structures

Section titled “Particle slots, sites, and modes are different structures”

For NN distinguishable subsystems, the Hilbert space is ⨂j=1Nhj\bigotimes_{j=1}^N\mathcal h_j. For identical particles with one-particle space h\mathcal h, the physical fixed-number space is instead

HN+=Sym⁡Nh,HN−=⋀Nh.\mathcal H_N^+=\operatorname{Sym}^N\mathcal h, \qquad \mathcal H_N^-=\bigwedge\nolimits^N\mathcal h.

A continuum spinful particle can have h=L2(Rd)⊗Cq\mathcal h=L^2(\mathbb R^d)\otimes\mathbb C^q. The spin factor belongs to each one-particle state; exchange symmetry acts on the full spatial-and-spin slot. Antisymmetrizing only spatial coordinates would impose a different restriction unless a spin sector had already been fixed.

For a spin lattice, by contrast, a site is a physically specified subsystem with a local space hx\mathcal h_x. Its finite-volume Hilbert space is ⨂x∈Λhx\bigotimes_{x\in\Lambda}\mathcal h_x. This tensor product does not impose exchange symmetry between lattice sites.

A mode decomposition of a particle space is a further construction. Its Fock factorization does not identify mode labels with particle slots. Fock Space Construction makes that distinction precise.

Local operators must respect the chosen space

Section titled “Local operators must respect the chosen space”

On a distinguishable tensor product, AjA_j acting in one factor is a well-defined local operator. Operators on disjoint factors commute.

For identical particles, an individual formal-slot operator generally does not preserve the symmetric or antisymmetric subspace. A bounded one-particle observable AA instead lifts to the permutation-invariant sum

A(N)=∑j=1NAj,∥A(N)∥≤N∥A∥.A^{(N)}=\sum_{j=1}^N A_j, \qquad \|A^{(N)}\|\leq N\|A\|.

The sum commutes with every permutation, so it restricts to either statistics sector. The physical occupation-number expression and reduced-density expectations belong to One-Body Operators.

This restriction is not a ban on spatially local measurements of identical particles. A one-particle spatial projector can be summed to count particles in a region. What fails is the interpretation of an arbitrary labeled slot as a separately addressable identical particle.

Independent-particle energy uses joint spectral calculus

Section titled “Independent-particle energy uses joint spectral calculus”

Let hh be self-adjoint on h\mathcal h. The lifted copies hjh_j strongly commute because their spectral projections act in different tensor factors. Their joint projection measure defines the self-adjoint operator

HN(0)=∫RN(λ1+⋯+λN) d(Eh⊗N)(λ).H_N^{(0)}=\int_{\mathbb R^N} (\lambda_1+\cdots+\lambda_N)\, d(E_h^{\otimes N})(\lambda).

Its domain consists of vectors with

∫RN∣λ1+⋯+λN∣2 dμΨ(λ)<∞.\int_{\mathbb R^N} |\lambda_1+\cdots+\lambda_N|^2\,d\mu_\Psi(\lambda)<\infty.

On algebraic tensors of vectors in D(h)D(h), this agrees with the familiar sum ∑jhj\sum_jh_j. The spectral construction specifies its closure and maximal domain. If hh is not semibounded, cancellation can make this domain larger than the intersection of the separate D(hj)D(h_j).

The joint multiplier is invariant under permutations. Therefore the bosonic and fermionic projections commute with every spectral projection of HN(0)H_N^{(0)}, and its restrictions to HN±\mathcal H_N^\pm are self-adjoint. A merely invariant algebraic test set would not, by itself, establish this reducing-subspace statement.

Interactions require additional control. A symmetric potential may respect permutations while still needing operator or form bounds for self-adjointness; see Domains of Hamiltonians and Quadratic Form Methods.

Take h=ℓ2(Z)\mathcal h=\ell^2(\mathbb Z) and hen=nenhe_n=ne_n. For two distinguishable slots, choose

Ψ=C∑n=1∞1n en⊗e−n.\Psi=C\sum_{n=1}^\infty\frac1n\,e_n\otimes e_{-n}.

It is normalizable. Neither h1Ψh_1\Psi nor h2Ψh_2\Psi exists as a Hilbert vector because its squared norm would contain ∑n≥11\sum_{n\geq1}1.

Nevertheless, the joint sum multiplier vanishes on every displayed basis vector. Hence Ψ∈D(H2(0))\Psi\in D(H_2^{(0)}) and H2(0)Ψ=0H_2^{(0)}\Psi=0. Writing h1Ψ+h2Ψh_1\Psi+h_2\Psi as the sum of two individually defined vectors would miss this state.

The example explains why an exact domain belongs to the resulting self-adjoint operator, rather than being inferred from formal notation.

Increasing the system size changes the comparison problem

Section titled “Increasing the system size changes the comparison problem”

Fixed-NN particle spaces have no unique canonical embedding into the (N+1)(N+1)-particle space. One must specify an added particle, a mode, or another comparison map. Fock space packages all finite sectors orthogonally, but that does not turn a thermodynamic limit into a Hilbert-norm limit of particle vectors.

For a lattice, local observable algebras do have canonical embeddings: tensor with identities on added sites. This motivates the quasi-local algebra and local-state convergence developed in Thermodynamic Limit.

Dimension growth, particle-number growth, spatial-volume growth, and removal of a mode cutoff are different limits. A theorem must say which one is taken and which quantities remain fixed.

  1. Why is A1A_1 not generally an observable acting within the symmetric two-particle space?
Solution

For a swap UU, UA1U∗=A2UA_1U^*=A_2. Unless the action happens to agree on the state under consideration, A1A_1 need not preserve the U=1U=1 subspace. The sum A1+A2A_1+A_2 commutes with UU and therefore preserves both statistics sectors.

  1. Modify the domain-cancellation example to make a symmetric two-particle vector.
Solution

Replace each tensor by (en⊗e−n+e−n⊗en)/2(e_n\otimes e_{-n}+e_{-n}\otimes e_n)/\sqrt2. These vectors remain orthonormal and have zero total energy. The same coefficients yield a vector in the symmetric total-energy domain but outside the separate slot domains.

  • J. Dereziński, “Introduction to representations of the canonical commutation and anticommutation relations,” in Large Coulomb Systems, Lecture Notes in Physics 695, Springer, 2006, pp. 63–143, doi:10.1007/3-540-32579-4_3, author text.
  • G. Teschl, Mathematical Methods in Quantum Mechanics: With Applications to Schrödinger Operators, 2nd ed., Graduate Studies in Mathematics 157, American Mathematical Society, 2014, Sections 4.2 and 4.6, author’s text.