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Finite-mode occupation configurations: fermionic exclusion counts, bosonic counts and truncated dimensions

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $H$ be a complex Hilbert space of finite dimension $d \geq 1$ with orthonormal basis $(e_{1}, \dots, e_{d})$ (Hilbert space), let $H^{\otimes n}$ and $P_{\pm}^{(n)}$ be as in Slater and symmetric occupation-number bases of the finite-dimensional permutation sectors (with $P_{\pm}^{(0)} = {i d}_{C}$ ), and let $W (n, d) = {λ \in N^{d} : \sum_{j} λ_{j} = n}$ and $m_{λ}$ be the symmetric vectors of that theorem. Then:

(a) fermionic configurations: the map $J \mapsto 1_{J}$ is a bijection from the set of $n$ -element subsets of ${1, \dots, d}$ onto the set ${λ \in {0, 1}^{d} : \sum_{j} λ_{j} = n}$ ; consequently $\dim ran P_{-}^{(n)} = # {λ \in {0, 1}^{d} : \sum_{j} λ_{j} = n}, \sum_{n = 0}^{d} \dim ran P_{-}^{(n)} = 2^{d};$

(b) bosonic configurations: the assignment $λ \mapsto m_{λ}$ induces a bijection from $W (n, d)$ onto the set of lines spanned by the orthogonal basis $(m_{λ})$ , so the number of bosonic configurations of total occupation exactly $n$ is $(\binom{n + d - 1}{n}) = (\binom{n + d - 1}{d - 1})$ , and for every $N \geq 0$ the number of bosonic configurations of total occupation at most $N$ is $\sum_{n = 0}^{N} \dim ran P_{+}^{(n)} = \sum_{n = 0}^{N} (\binom{n + d - 1}{d - 1}) = (\binom{N + d}{d});$

(c) exclusion counts: no fermionic configuration has an entry $\geq 2$ , so the antisymmetric Slater index set contains no $λ$ with $λ_{j} \geq 2$ for some $j$ ; for $n > d$ there is no fermionic configuration at all and $ran P_{-}^{(n)} = {0}$ , while for every $n$ and every $j$ the bosonic configuration with $λ_{j} = n$ and all other entries $0$ occurs, so no occupation bound holds in the symmetric sector;

(d) consistency with the total occupation: for every $λ \in W (n, d)$ one has $\sum_{j} λ_{j} = n$ , and for every fermionic configuration $λ \in {0, 1}^{d} \cap {\sum_{j} λ_{j} = n}$ likewise; in particular the total occupation equals the particle number $n$ in both sectors.

Facts & Assumptions

Given: The finite-dimensional complex Hilbert space $H$ of dimension $d \geq 1$ with orthonormal basis $(e_{1}, \dots, e_{d})$ , the sectors $ran P_{\pm}^{(n)}$ of the cited theorem, and $n, N \geq 0$ .

[F0]

Empty configuration, directly. At $n = 0$ , the Statement's conventions give $H^{\otimes 0} = C$ and $P_{\pm}^{(0)} = {i d}_{C}$ . The empty tensor is the scalar $1$ , so the empty symmetric vector is $m_{0} = 1$ , where $0$ is the unique element of $W (0, d)$ . Thus both sector ranges are one-dimensional and have this single normalized basis vector; the unique fermionic configuration is the indicator of the empty subset and is also $0$ . These are direct calculations in $C$ , not consequences of the cited theorem.

[F1]

Occupation basis for positive particle number. For $n \geq 1$ , the Slater vectors $e_{J}^{\land}$ , $J \subseteq {1, \dots, d}$ with $∣ J ∣ = n$ , form an orthonormal basis of $ran P_{-}^{(n)}$ , and $\dim ran P_{-}^{(n)} = (\binom{d}{n})$ with $ran P_{-}^{(n)} = {0}$ for $n > d$ ; the symmetric vectors $(m_{λ})_{λ \in W (n, d)}$ form an orthogonal basis of $ran P_{+}^{(n)}$ with $\dim ran P_{+}^{(n)} = (\binom{n + d - 1}{n})$ ; each $e_{J}^{\land}$ has occupation vector $1_{J}$ and each $m_{λ}$ has occupation vector $λ$ (Slater and symmetric occupation-number bases of the finite-dimensional permutation sectors clauses (b),(c),(d)).

[F2]

Binomial counts and symmetry. $(\binom{d}{n})$ is the number of $n$ -element subsets of ${1, \dots, d}$ , $(\binom{n}{0}) = (\binom{n}{n}) = 1$ , and $(\binom{n}{k}) = 0$ for $k > n$ (The set $[A]^{k}$ of $k$ -element subsets and the binomial coefficient $(\binom{n}{k}) : = ∣ [n]^{k} ∣$ ); also $(\binom{n + d - 1}{d - 1}) = (\binom{n + d - 1}{n})$ ( $(\binom{n}{k}) k! (n - k)! = n!$ for $k \leq n$ ; hence $(\binom{n}{k}) k! = n^{\underline{k}}$ , the quotient $n! / (k! (n - k)!)$ is a natural number, and $(\binom{n}{k}) = (\binom{n}{n - k})$ clause 3, with $N = n + d - 1$ and $k = d - 1$ ).

[F3]

Weak compositions. For $m \geq 1$ and $n \in N$ , $∣ W (n, m) ∣ = (\binom{n + m - 1}{m - 1})$ (For $m \geq 1$ the number of weak compositions of $n$ into $m$ parts is $(\binom{n + m - 1}{m - 1})$ , and the number of compositions is $(\binom{n - 1}{m - 1})$ for $n \geq 1$ ).

[F4]

Pascal rule and hockey stick. For all $n, k \in N$ , $(\binom{n + 1}{k + 1}) = (\binom{n}{k}) + (\binom{n}{k + 1})$ and $\sum_{i < n + 1} (\binom{i}{k}) = (\binom{n + 1}{k + 1})$ as an $N$ -valued finite sum (Pascal's rule $(\binom{n + 1}{k + 1}) = (\binom{n}{k}) + (\binom{n}{k + 1})$ , and the hockey-stick identity $\sum_{i \leq n} (\binom{i}{k}) = (\binom{n + 1}{k + 1})$ ); the sector formula of the theorem is stated for the finite sums there defined.

[F5]

Row sum. $\sum_{k < n + 1} (\binom{n}{k}) = 2^{n}$ in $N$ ( $\sum_{k < n + 1} (\binom{n}{k}) = 2^{n}$ , and $\sum_{k < n + 1} (- 1)^{k} ι ⁣ (\binom{n}{k}) = 0$ for $n \geq 1$ clause 1).

Proof

technique · direct; compare the two index sets of the occupation basis with the configuration sets, count them with the cited finite-counting facts, and read the vanishing statement off the antisymmetric sector

Given: The data of the statement and [F0]-[F5].

1.1F0F1F2algebraconstruct

[Fermionic configurations] Let $J \subseteq {1, \dots, d}$ have $∣ J ∣ = n$ ; its indicator $1_{J}$ lies in ${0, 1}^{d}$ with $\sum_{j} 1_{J} (j) = n$ . Conversely, if $λ \in {0, 1}^{d}$ has $\sum_{j} λ_{j} = n$ , then $J_{λ} : = {j : λ_{j} = 1}$ satisfies $∣ J_{λ} ∣ = n$ and $1_{J_{λ}} = λ$ . The two assignments are inverse bijections, so the number of fermionic configurations of total occupation $n$ equals the number of $n$ -element subsets, which is $(\binom{d}{n})$ by [F2]. For $n \geq 1$ , this is $\dim ran P_{-}^{(n)}$ by [F1]; for $n = 0$ , the dimension is $1 = (\binom{d}{0})$ by [F0] and the boundary value in [F2]. This proves (a)'s dimension assertion in all cases.

1.2F0F1F2F3algebra

[Bosonic configurations] If $n = 0$ , then $W (0, d) = {0}$ and $m_{0} = 1$ by [F0], so its assignment is a bijection onto the single basis line of $ran P_{+}^{(0)} = C$ . If $n \geq 1$ , the assignment $λ \mapsto m_{λ}$ is injective by the linear independence of the orthogonal family [F1], and its image spans $ran P_{+}^{(n)}$ , so it is a bijection onto the set of basis lines. In either case, the number of its elements is $∣ W (n, d) ∣ = (\binom{n + d - 1}{d - 1}) = (\binom{n + d - 1}{n})$ by [F3] and [F2], the first assertion of (b).

1.3F0F1F2F4algebra

[Truncated bosonic dimension] For $n \geq 1$ the dimension $\dim ran P_{+}^{(n)} = (\binom{n + d - 1}{n}) = (\binom{n + d - 1}{d - 1})$ is given by [F1] and [F2]; at $n = 0$ it is $1 = (\binom{d - 1}{d - 1})$ by [F0] and the boundary value of the binomial coefficient. Thus $\sum_{n = 0}^{N} \dim ran P_{+}^{(n)} = \sum_{n = 0}^{N} (\binom{n + d - 1}{d - 1})$ . Substituting $i : = n + d - 1$ , the terms with $i < d - 1$ vanish by the boundary values of the binomial coefficient, so the hockey stick [F4] with $k = d - 1$ and upper index $N + d - 1$ gives $\sum_{i = 0}^{N + d - 1} (\binom{i}{d - 1}) = (\binom{N + d}{d})$ ; this completes (b).

2.1F0F1F5step 1.1algebra

[Fermionic row sum] Summing the dimensions over all particle numbers, using [F0] at $n = 0$ and [F1] for $1 \leq n \leq d$ and the vanishing for $n > d$ , gives $\sum_{n = 0}^{d} \dim ran P_{-}^{(n)} = \sum_{n = 0}^{d} (\binom{d}{n}) = 2^{d}$ by the row sum [F5], which completes (a).

2.2F1step 1.1step 1.2algebra

[Exclusion counts] By construction every fermionic configuration lies in ${0, 1}^{d}$ , so no entry is $\geq 2$ ; for $n > d$ the set ${0, 1}^{d} \cap {\sum_{j} λ_{j} = n}$ is empty because at most $d$ entries can be equal to $1$ , and then $ran P_{-}^{(n)} = {0}$ by [F1], while for the bosonic sector the configuration $λ^{(j)}$ with $λ_{j}^{(j)} = n$ and all other entries $0$ lies in $W (n, d)$ for every $j$ , so no occupation bound holds there; this is (c).

3.1step 1.1step 1.2step 2.1step 1.3step 2.2algebra∎

[Consistency and assembly] Each $λ \in W (n, d)$ satisfies $\sum_{j} λ_{j} = n$ by definition, and each fermionic configuration in ${0, 1}^{d} \cap {\sum_{j} λ_{j} = n}$ does so as well, which is (d); step 1.1 and 2.1 prove (a), steps 1.2 and 1.3 prove (b), and step 2.2 proves (c). Only the cited spectral basis and finite-counting facts were used.

Remarks

Reading. The theorem is the finite-mode bookkeeping behind the occupation-number description: for a $d$ -mode one-particle space there are $(\binom{d}{n})$ antisymmetric $n$ -particle sector dimensions and $(\binom{n + d - 1}{n})$ symmetric ones, and the antisymmetric configurations have all occupations in ${0, 1}$ while the symmetric ones do not. The physical reading of the occupations as a readout of the framework is fixed on The declared finite-mode model and its occupation readouts and Pauli exclusion in the declared finite-mode occupation readout, with the bosonic contrast.
Truncation. The sums over $n \leq N$ are finite sums of finite dimensions; no infinite Fock space and no convergence statement is made here.

Finite-mode occupation configurations: fermionic exclusion counts, bosonic counts and truncated dimensions

Statement

Facts & Assumptions

Proof

Remarks

Depends on

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