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Tensor spin composition

Statement

Let $s, t \in \frac{1}{2} N$ , write $L_{s} : = {- s, - s + 1, \dots, s}$ and $L_{t} : = {- t, - t + 1, \dots, t}$ , and let $(V_{s}, J_{x}^{(s)}, J_{y}^{(s)}, J_{z}^{(s)})$ and $(V_{t}, J_{x}^{(t)}, J_{y}^{(t)}, J_{z}^{(t)})$ be the spin models of Finite spin matrices with orthonormal bases $(e_{m}^{(s)})_{m \in L_{s}}$ and $(e_{n}^{(t)})_{n \in L_{t}}$ and the ladders $J_{\pm}^{(s)} : = J_{x}^{(s)} \pm i J_{y}^{(s)}$ , $J_{\pm}^{(t)} : = J_{x}^{(t)} \pm i J_{y}^{(t)}$ and Casimirs $J^{(s) 2}$ , $J^{(t) 2}$ (The elementary tensors of two bases form the product basis of the tensor product). On the tensor product $V_{s} \otimes V_{t}$ put

$J_{i}^{t o t} : = J_{i}^{(s)} \otimes I_{V_{t}} + I_{V_{s}} \otimes J_{i}^{(t)} (i = x, y, z), J_{\pm}^{t o t} : = J_{x}^{t o t} \pm i J_{y}^{t o t}, J^{t o t 2} : = (J_{x}^{t o t})^{2} + (J_{y}^{t o t})^{2} + (J_{z}^{t o t})^{2} .$

Then:

(a) each $J_{i}^{t o t}$ is self-adjoint, the triple $J_{x}^{t o t}, J_{y}^{t o t}, J_{z}^{t o t}$ satisfies the commutation relations of Finite spin-block decomposition, and $J^{t o t 2} = J^{(s) 2} \otimes I_{V_{t}} + I_{V_{s}} \otimes J^{(t) 2} + 2 \sum_{i} J_{i}^{(s)} \otimes J_{i}^{(t)}$ is self-adjoint and commutes with $J_{x}^{t o t}, J_{y}^{t o t}, J_{z}^{t o t}$ ;

(b) there is an orthogonal decomposition $V_{s} \otimes V_{t} = ⨁_{j = ∣ s - t ∣}^{s + t} W_{j}$ into $J^{t o t}$ -invariant subspaces, one for each $j \in {∣ s - t ∣, ∣ s - t ∣ + 1, \dots, s + t}$ , with $\dim W_{j} = 2 j + 1$ , on each of which $(J_{x}^{t o t}, J_{y}^{t o t}, J_{z}^{t o t})$ is unitarily equivalent to the spin- $j$ model of Finite spin matrices; consequently $\sum_{j = ∣ s - t ∣}^{s + t} (2 j + 1) = (2 s + 1) (2 t + 1)$ ;

(c) for every $M \in \frac{1}{2} Z$ the $M$ -eigenspace of $J_{z}^{t o t}$ has dimension $N (M) : = # {m \in L_{s} : M - m \in L_{t}} = {\begin{cases} \max (0, \min (s, M + t) - \max (- s, M - t) + 1), & M - (s + t) \in Z, \\ 0, & M - (s + t) \notin Z . \end{cases},$ and the identity $(e_{m}^{(s)} \otimes e_{n}^{(t)})_{m, n}$ is an orthonormal basis of $V_{s} \otimes V_{t}$ consisting of eigenvectors of $J_{z}^{t o t}$ with eigenvalues $m + n$ ;

(d) the number of summands with parameter equal to $M$ in the decomposition of (b) is $N (M) - N (M + 1)$ if $M \geq 0$ , and this number equals $1$ when $M - (s + t) \in Z$ and $∣ s - t ∣ \leq M \leq s + t$ , and equals $0$ otherwise; hence the multiset of parameters in (b) is exactly ${∣ s - t ∣, ∣ s - t ∣ + 1, \dots, s + t}$ , each occurring once;

(e) for each $j \in {∣ s - t ∣, \dots, s + t}$ the space of vectors of $J_{z}^{t o t}$ -eigenvalue $j$ orthogonal to all summands $W_{j^{'}}$ with $j^{'} > j$ is one-dimensional; for a unit vector $u_{j}$ in it the vectors $v_{j, k} : = (k! (2 j) (2 j - 1) \dots (2 j - k + 1))^{- 1 / 2} (J_{-}^{t o t})^{k} u_{j} (0 \leq k \leq 2 j)$ form an orthonormal basis of $W_{j}$ with $J_{z}^{t o t} v_{j, k} = (j - k) v_{j, k}$ , $J_{+}^{t o t} v_{j, k} = \sqrt{k (2 j - k + 1)} v_{j, k - 1}$ and $J_{-}^{t o t} v_{j, k} = \sqrt{(k + 1) (2 j - k)} v_{j, k + 1}$ ; the coefficients of $v_{j, k}$ in the product basis are therefore determined by the phase of $u_{j}$ , and choosing that phase fixes the Clebsch-Gordan numbers explicitly; in particular $u_{s + t} = e_{s}^{(s)} \otimes e_{t}^{(t)}$ is a permissible choice.

The theorem is a purely mathematical statement about tensor products of the explicit matrices of Finite spin matrices; it asserts no physical coupling, conservation law or measurement fact.

Facts & Assumptions

Given: $s, t \in \frac{1}{2} N$ ; the spin models $V_{s}, V_{t}$ with their operators and orthonormal bases; the tensor product $V_{s} \otimes V_{t}$ ; the operators $J_{i}^{t o t}$ , $J_{\pm}^{t o t}$ , $J^{t o t 2}$ .

[F1]

By The elementary tensors of two bases form the product basis of the tensor product, the algebraic tensor product $V_{s} \otimes V_{t}$ has the finite basis $e_{m}^{(s)} \otimes e_{n}^{(t)}$ . Give its coordinate vectors the inner product $⟨ \sum c_{m n} e_{m} \otimes e_{n}, \sum d_{m n} e_{m} \otimes e_{n} ⟩ = \sum c_{m n} \overline{d_{m n}}$ . This is positive definite, makes that basis orthonormal, and expansion of two elementary tensors gives $⟨ u \otimes v, u^{'} \otimes v^{'} ⟩ = ⟨ u, u^{'} ⟩ ⟨ v, v^{'} ⟩$ . This is a finite construction with explicitly supplied bases; no completion or countable choice is invoked.

[F2]

Define $S \otimes T$ by its values $S e_{m} \otimes T e_{n}$ on the finite product basis. Bilinear expansion gives $(S \otimes T) (u \otimes v) = S u \otimes T v$ , and hence $(S \otimes T) (S^{'} \otimes T^{'}) = (S S^{'}) \otimes (T T^{'})$ . The inner product of [F1] gives $⟨ (S \otimes T) (u \otimes v), u^{'} \otimes v^{'} ⟩ = ⟨ u \otimes v, (S^{*} \otimes T^{*}) (u^{'} \otimes v^{'}) ⟩$ , proving $(S \otimes T)^{*} = S^{*} \otimes T^{*}$ on the spanning product basis. Also $I \otimes I = I$ , so $[S \otimes I, S^{'} \otimes I] = [S, S^{'}] \otimes I$ , $[I \otimes T, I \otimes T^{'}] = I \otimes [T, T^{'}]$ , and operators acting on different factors commute.

[F3]

On $V_{s}$ and $V_{t}$ the operators are self-adjoint, satisfy $[J_{x}, J_{y}] = i J_{z}$ and its cyclic companions, and have $J^{2} = s (s + 1) I_{V_{s}}$ , respectively $t (t + 1) I_{V_{t}}$ ; the ladder actions on the bases are $J_{\pm}^{(s)} e_{m}^{(s)} = \sqrt{(s \mp m) (s \pm m + 1)} e_{m \pm 1}^{(s)}$ and $J_{z}^{(s)} e_{m}^{(s)} = m e_{m}^{(s)}$ (Finite spin matrices).

[F4]

Every finite-dimensional triple of self-adjoint operators satisfying the commutation relations decomposes orthogonally into spin blocks with $\sum_{ν} (2 j_{ν} + 1) = \dim V$ , the block actions being the spin- $j_{ν}$ actions; the number of blocks with parameter $M \geq 0$ equals $\dim \ker (J_{z} - M I) - \dim \ker (J_{z} - (M + 1) I)$ , and $M$ must lie in $\frac{1}{2} N$ (Finite spin-block decomposition (b), (d)).

[F5]

$\dim (V_{s} \otimes V_{t}) = \dim V_{s} \dim V_{t} = (2 s + 1) (2 t + 1)$ , and dimensions add over orthogonal direct sums (Hilbert tensor construction: inner product, completion and product bases, Finite-dimensional vector space, and its dimension $\dim_{F} V$ ; infinite-dimensional means having no finite basis, Orthogonality and the orthogonal complement).

[F6]

For a self-adjoint operator the eigenspaces for distinct eigenvalues are orthogonal, and for a unit vector $w$ and self-adjoint $T$ , $⟨ T w, w ⟩ = ⟨ w, T w ⟩$ is real; $∥ w ∥^{2} = ⟨ w, w ⟩$ (Self-adjoint and normal endomorphisms of a finite-dimensional real or complex inner product space, Orthogonality and the orthogonal complement, The induced length is a norm).

[F7]

Factorials and products: $k! = \prod_{i = 1}^{k} i \geq 1$ for $k \geq 0$ , and a finite product of positive numbers is positive (The factorial $n!$ and the falling factorial $n^{\underline{k}}$ , defined by recursion in $N$ , The natural numbers $N$ (von Neumann)).

Proof

1.1F2F3algebra

[Self-adjointness and the algebra] For $i = x, y, z$ , [F2] and [F3] give $(J_{i}^{(s)} \otimes I)^{*} = J_{i}^{(s)} \otimes I$ and $(I \otimes J_{i}^{(t)})^{*} = I \otimes J_{i}^{(t)}$ , so $J_{i}^{t o t}$ is self-adjoint. Adding the two-sided commutators, $[J_{x}^{t o t}, J_{y}^{t o t}] = [J_{x}^{(s)}, J_{y}^{(s)}] \otimes I + I \otimes [J_{x}^{(t)}, J_{y}^{(t)}] = i J_{z}^{(s)} \otimes I + i I \otimes J_{z}^{(t)} = i J_{z}^{t o t}$ , and cyclically; so the triple satisfies the commutation relations.

1.2F1F3algebra

[Weights] By [F1] and [F3], $J_{z}^{t o t} (e_{m}^{(s)} \otimes e_{n}^{(t)}) = (m + n) e_{m}^{(s)} \otimes e_{n}^{(t)}$ , so the elementary tensors form an orthonormal eigenbasis of $J_{z}^{t o t}$ with eigenvalues $m + n$ , and the number of basis vectors of eigenvalue $M$ is the number of pairs $(m, n) \in L_{s} \times L_{t}$ with $m + n = M$ , which is $# {m \in L_{s} : M - m \in L_{t}} = N (M)$ ; all sums $m + n$ lie in $s + t + Z$ , so $N (M) = 0$ off that lattice. On the lattice, the interval endpoints lie in the same unit-spaced lattice $L_{s}$ ; the displayed closed form follows by counting the interval $[\max (- s, M - t), \min (s, M + t)]$ , which is empty exactly when the maximum exceeds the minimum, and otherwise contributes one element per unit lattice step.

2.1step 1.1F2F3algebra

[The Casimir] Expanding $J^{t o t 2} = \sum_{i} (J_{i}^{(s)} \otimes I + I \otimes J_{i}^{(t)})^{2}$ gives $J^{(s) 2} \otimes I + I \otimes J^{(t) 2} + 2 C$ with $C : = \sum_{i} J_{i}^{(s)} \otimes J_{i}^{(t)}$ , which is self-adjoint by [F2] and [F3]; moreover $[J^{t o t 2}, J_{i}^{t o t}] = [J^{(s) 2}, J_{i}^{(s)}] \otimes I + I \otimes [J^{(t) 2}, J_{i}^{(t)}] + 2 [C, J_{i}^{(s)} \otimes I + I \otimes J_{i}^{(t)}]$ vanishes because the first two commutators are $0$ by [F3] and $2 [C, J_{i}^{(s)} \otimes I + I \otimes J_{i}^{(t)}] = 2 i \sum_{k, l} ε_{k i l} (J_{l}^{(s)} \otimes J_{k}^{(t)} + J_{k}^{(s)} \otimes J_{l}^{(t)}) = 0$ , the parenthesised tensor being symmetric in $k, l$ while $ε_{k i l}$ is antisymmetric.

2.2step 1.2F1algebra

[The parity-qualified difference] Restrict now to $M \geq 0$ . If $M - (s + t) \notin Z$ , neither $M$ nor $M + 1$ is a weight and both counts are zero by step 1.2. On the compatible lattice, exchange the factors if necessary to suppose $s \geq t$ . Counting the interval in step 1.2 gives $N (M) = \min (2 t + 1, s + t - M + 1)$ for $0 \leq M \leq s + t$ , and $N (M) = 0$ for $M > s + t$ . Therefore $N (M) - N (M + 1) = 0$ on the plateau $0 \leq M < s - t$ , equals one for $s - t \leq M \leq s + t$ , and is zero above $s + t$ , including the upper endpoint where $N (s + t) = 1$ and $N (s + t + 1) = 0$ . Restoring factor symmetry gives precisely the admissible lattice points in $[∣ s - t ∣, s + t]$ .

3.1step 1.1step 2.1step 2.2F4F5algebra

[Existence and parameter list] By steps 1.1, 2.1 and [F4] the triple $(J_{x}^{t o t}, J_{y}^{t o t}, J_{z}^{t o t})$ decomposes $V_{s} \otimes V_{t}$ into an orthogonal direct sum of spin blocks with parameters $j_{ν} \in \frac{1}{2} N$ , and by F4 the number of blocks with parameter $M$ equals $\dim \ker (J_{z}^{t o t} - M I) - \dim \ker (J_{z}^{t o t} - (M + 1) I) = N (M) - N (M + 1)$ for $M \geq 0$ , which by step 2.2 is $1$ precisely for $M \in {∣ s - t ∣, ∣ s - t ∣ + 1, \dots, s + t}$ and $0$ otherwise. Hence the block parameters are exactly the numbers $∣ s - t ∣, ∣ s - t ∣ + 1, \dots, s + t$ , each once; each block has dimension $2 j + 1$ and its operators are unitarily equivalent to the spin- $j$ model by F4. The dimension identity $\sum_{j = ∣ s - t ∣}^{s + t} (2 j + 1) = (2 s + 1) (2 t + 1)$ follows from $\sum_{ν} (2 j_{ν} + 1) = \dim (V_{s} \otimes V_{t})$ and [F5]. This proves (b) and (d).

4.1step 3.1F1F3F4F6F7algebra∎

[The constructive bases and Clebsch-Gordan phase] Fix $j$ in the parameter list and let $E_{j}$ be the space of $J_{z}^{t o t}$ -eigenvectors of eigenvalue $j$ orthogonal to all blocks $W_{j^{'}}$ with $j^{'} > j$ . Counting as in step 3.1, $\dim \ker (J_{z}^{t o t} - j I) = # {ν : j_{ν} - j \in Z, j_{ν} \geq j}$ (a block with parameter $j_{ν} \geq j \geq 0$ and $j_{ν} - j \in Z$ contributes exactly one eigenvector of eigenvalue $j$ ), and the higher blocks with parameters $j^{'} > j$ contribute one each, so $\dim E_{j} = # {ν : j_{ν} - j \in Z, j_{ν} \geq j} - # {ν : j_{ν} - j \in Z, j_{ν} > j} = 1$ . Pick a unit $u_{j} \in E_{j}$ , which is determined up to multiplication by a phase; applying the ladder identities of F4 in the block normal form, the vectors $v_{j, k}$ of the Statement are well defined and form the orthonormal basis of $W_{j}$ , with the stated actions, because $∥ (J_{-}^{t o t})^{k} u_{j} ∥^{2} = k! (2 j) (2 j - 1) \dots (2 j - k + 1)$ by [F7] and the recursion of F4; expanding $v_{j, k}$ in the product basis [F1] exhibits the Clebsch-Gordan coefficients, and replacing $u_{j}$ by $e^{i θ} u_{j}$ multiplies them all by the same phase. For $j = s + t$ , $e_{s}^{(s)} \otimes e_{t}^{(t)}$ has weight $s + t$ and is annihilated by $J_{+}^{t o t}$ because $J_{+}^{(s)} e_{s}^{(s)} = J_{+}^{(t)} e_{t}^{(t)} = 0$ by [F3]; this is the top vector of the unique block of parameter $s + t$ , so it is a permissible choice. This proves (e) and completes the proof.

Tensor spin composition

Statement

Facts & Assumptions

Proof

Depends on

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