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Coupled triplet and singlet

Example

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $V = C^{2}$ carry the orthonormal basis $(e_{+}, e_{-})$ and let $(J_{x}^{(1 / 2)}, J_{y}^{(1 / 2)}, J_{z}^{(1 / 2)})$ be the spin- $\frac{1}{2}$ operators of Finite spin matrices, so that $J_{i}^{(1 / 2)} = \frac{1}{2} σ_{i}$ and $J_{\pm}^{(1 / 2)} = J_{x}^{(1 / 2)} \pm i J_{y}^{(1 / 2)}$ satisfy $J_{+}^{(1 / 2)} e_{+} = 0$ , $J_{+}^{(1 / 2)} e_{-} = e_{+}$ , $J_{-}^{(1 / 2)} e_{+} = e_{-}$ and $J_{-}^{(1 / 2)} e_{-} = 0$ . On $V \hat{\otimes} V$ (inner product of Hilbert tensor construction: inner product, completion and product bases) put, as in Tensor spin composition, $J_{i}^{t o t} : = J_{i}^{(1 / 2)} \otimes I_{2} + I_{2} \otimes J_{i}^{(1 / 2)}, J_{\pm}^{t o t} : = J_{x}^{t o t} \pm i J_{y}^{t o t}, J^{t o t 2} : = (J_{x}^{t o t})^{2} + (J_{y}^{t o t})^{2} + (J_{z}^{t o t})^{2},$ and define $w_{1, 1} : = e_{+} \otimes e_{+}, w_{1, 0} : = \frac{e_{+} \otimes e_{-} + e_{-} \otimes e_{+}}{\sqrt{2}}, w_{1, - 1} : = e_{-} \otimes e_{-}, w_{0, 0} : = \frac{e_{+} \otimes e_{-} - e_{-} \otimes e_{+}}{\sqrt{2}} .$ Then:

(a) $(w_{1, 1}, w_{1, 0}, w_{1, - 1}, w_{0, 0})$ is an orthonormal basis of $V \hat{\otimes} V$ ;

(b) the triplet vectors satisfy $J_{z}^{t o t} w_{1, m} = m w_{1, m}$ and $J^{t o t 2} w_{1, m} = 2 w_{1, m}$ for $m \in {1, 0, - 1}$ , while the singlet satisfies $J_{i}^{t o t} w_{0, 0} = 0$ for each $i \in {x, y, z}$ and $J^{t o t 2} w_{0, 0} = 0$ ;

(c) the ladder actions on the triplet are $J_{+}^{t o t} w_{1, 1} = 0$ , $J_{+}^{t o t} w_{1, 0} = \sqrt{2} w_{1, 1}$ and $J_{+}^{t o t} w_{1, - 1} = \sqrt{2} w_{1, 0}$ , together with the adjoint relations $J_{-}^{t o t} w_{1, - 1} = 0$ , $J_{-}^{t o t} w_{1, 0} = \sqrt{2} w_{1, - 1}$ and $J_{-}^{t o t} w_{1, 1} = \sqrt{2} w_{1, 0}$ ; hence the correspondence $e_{m} \leftrightarrow w_{1, m}$ is a unitary isomorphism of the spin- $1$ model of Finite spin matrices onto the triplet span, so the orthogonal decomposition of Tensor spin composition (b) is here $V \hat{\otimes} V = span {w_{1, 1}, w_{1, 0}, w_{1, - 1}} \oplus span {w_{0, 0}}$ with parameters $j = 1$ and $j = 0$ and dimensions $3 + 1 = 4$ , summing to $(2 \cdot \frac{1}{2} + 1)^{2}$ .

All claims are finite-dimensional linear-algebra statements; no physical interpretation of the vectors as particle states is asserted.

Facts & Assumptions

Given: the spin-half operators on $V = C^{2}$ with basis $(e_{+}, e_{-})$ , the tensor product and the total operators $J_{i}^{t o t}, J_{\pm}^{t o t}$ , and the four vectors $w_{1, 1}, w_{1, 0}, w_{1, - 1}, w_{0, 0}$ .

[F1]

Spin-half data (Finite spin matrices): $J_{z}^{(1 / 2)} e_{\pm} = \pm \frac{1}{2} e_{\pm}$ , $J_{+}^{(1 / 2)} e_{+} = 0$ , $J_{+}^{(1 / 2)} e_{-} = e_{+}$ , $J_{-}^{(1 / 2)} e_{+} = e_{-}$ , $J_{-}^{(1 / 2)} e_{-} = 0$ , $J_{\pm}^{(1 / 2)} = J_{x}^{(1 / 2)} \pm i J_{y}^{(1 / 2)}$ , the operators $J_{i}^{(1 / 2)}$ are self-adjoint, and $J^{(1 / 2) 2} = J_{x}^{(1 / 2) 2} + J_{y}^{(1 / 2) 2} + J_{z}^{(1 / 2) 2} = \frac{3}{4} I_{2}$ .

[F2]

Tensor calculus (Hilbert tensor construction: inner product, completion and product bases, Linear map between vector spaces over the same field): $(S \otimes T) (u \otimes v) = (S u) \otimes (T v)$ on elementary tensors, operators are extended linearly, and $⟨ u \otimes v, u^{'} \otimes v^{'} ⟩ = ⟨ u, u^{'} ⟩ ⟨ v, v^{'} ⟩$ ; hence the four displayed vectors are pairwise orthogonal with $∥ w_{1, 1} ∥ = ∥ w_{1, - 1} ∥ = 1$ and $∥ w_{1, 0} ∥ = ∥ w_{0, 0} ∥ = 1$ by direct expansion of the two-term sums; four orthonormal vectors in the four-dimensional space $V \hat{\otimes} V$ form an orthonormal basis (The induced length is a norm, Cauchy–Schwarz: $∣ ⟨ u, v ⟩ ∣ \leq ∥ u ∥ ∥ v ∥$ , with equality exactly for linearly dependent vectors).

[F3]

The total operators satisfy $J_{z}^{t o t} = J_{z}^{(1 / 2)} \otimes I_{2} + I_{2} \otimes J_{z}^{(1 / 2)}$ and $J_{\pm}^{t o t} = J_{\pm}^{(1 / 2)} \otimes I_{2} + I_{2} \otimes J_{\pm}^{(1 / 2)}$ by bilinearity of the tensor product, and $J^{t o t 2} = J_{-}^{t o t} J_{+}^{t o t} + (J_{z}^{t o t})^{2} + J_{z}^{t o t} = J_{+}^{t o t} J_{-}^{t o t} + (J_{z}^{t o t})^{2} - J_{z}^{t o t}$ by the algebraic identities of Tensor spin composition (a) (or, directly, of Finite spin matrices (b),(c) applied to the sums).

[F4]

Composition theorem (Tensor spin composition (b),(e)): for $s = t = \frac{1}{2}$ the space $V \hat{\otimes} V$ decomposes orthogonally into $J^{t o t}$ -invariant summands of dimensions $2 j + 1$ for $j = 0, 1$ , on each of which the total operators are unitarily equivalent to the spin- $j$ model; the summand with $j = 1$ contains the unique (up to phase) unit vector of $J_{z}^{t o t}$ -eigenvalue $1$ , and the summand with $j = 0$ is the kernel of all three $J_{i}^{t o t}$ .

Verification

1.1F2algebra

[(a) orthonormality] By [F2], $⟨ w_{1, 1}, w_{1, 0} ⟩ = (⟨ e_{+}, e_{+} ⟩ ⟨ e_{+}, e_{-} ⟩ + ⟨ e_{+}, e_{-} ⟩ ⟨ e_{+}, e_{+} ⟩) / \sqrt{2} = 0$ , and similarly for all pairs: the tensors $e_{+} \otimes e_{+}$ , $e_{-} \otimes e_{-}$ , $(e_{+} \otimes e_{-} + e_{-} \otimes e_{+}) / \sqrt{2}$ and $(e_{+} \otimes e_{-} - e_{-} \otimes e_{+}) / \sqrt{2}$ are mutually orthogonal of norm $1$ because the four tensors $e_{\pm} \otimes e_{\pm}$ are orthonormal and the two combined vectors are orthogonal to them and to each other. Since $\dim (V \hat{\otimes} V) = 4$ (Hilbert tensor construction: inner product, completion and product bases), these four vectors form an orthonormal basis.

1.2F1F3algebra

[(b) triplet] By [F1] and [F3]: $J_{z}^{t o t} w_{1, 1} = (J_{z}^{(1 / 2)} e_{+}) \otimes e_{+} + e_{+} \otimes (J_{z}^{(1 / 2)} e_{+}) = \frac{1}{2} w_{1, 1} + \frac{1}{2} w_{1, 1} = w_{1, 1}$ ; $J_{z}^{t o t} (e_{+} \otimes e_{-}) = (\frac{1}{2} - \frac{1}{2}) e_{+} \otimes e_{-} = 0$ and $J_{z}^{t o t} (e_{-} \otimes e_{+}) = (- \frac{1}{2} + \frac{1}{2}) e_{-} \otimes e_{+} = 0$ , so $J_{z}^{t o t} w_{1, 0} = 0$ ; and $J_{z}^{t o t} w_{1, - 1} = - w_{1, - 1}$ . Moreover $J_{+}^{t o t} w_{1, 1} = 0$ (both terms vanish by [F1]), so $J_{-}^{t o t} J_{+}^{t o t} w_{1, 1} = 0$ ; hence $J^{t o t 2} w_{1, 1} = (0 + 1^{2} + 1) w_{1, 1} = 2 w_{1, 1}$ by [F3].

1.3F1F3algebra

[(b) singlet] $J_{z}^{t o t} w_{0, 0} = (\frac{1}{2} e_{+} \otimes e_{-} + \frac{1}{2} e_{-} \otimes e_{+} - \frac{1}{2} e_{+} \otimes e_{-} - \frac{1}{2} e_{-} \otimes e_{+}) / \sqrt{2} = 0$ ; moreover $J_{+}^{t o t} w_{0, 0} = (J_{+}^{(1 / 2)} e_{+} \otimes e_{-} + e_{+} \otimes J_{+}^{(1 / 2)} e_{-} - J_{+}^{(1 / 2)} e_{-} \otimes e_{+} - e_{-} \otimes J_{+}^{(1 / 2)} e_{+}) / \sqrt{2} = (0 + e_{+} \otimes e_{+} - e_{+} \otimes e_{+} - 0) / \sqrt{2} = 0$ and $J_{-}^{t o t} w_{0, 0} = (J_{-}^{(1 / 2)} e_{+} \otimes e_{-} + e_{+} \otimes J_{-}^{(1 / 2)} e_{-} - J_{-}^{(1 / 2)} e_{-} \otimes e_{+} - e_{-} \otimes J_{-}^{(1 / 2)} e_{+}) / \sqrt{2} = (e_{-} \otimes e_{-} + 0 - 0 - e_{-} \otimes e_{-}) / \sqrt{2} = 0$ by [F1] and [F3]. Hence $J_{i}^{t o t} w_{0, 0} = 0$ for all $i$ , since $J_{x}^{t o t} = \frac{1}{2} (J_{+}^{t o t} + J_{-}^{t o t})$ and $J_{y}^{t o t} = \frac{1}{2 i} (J_{+}^{t o t} - J_{-}^{t o t})$ , and consequently $J^{t o t 2} w_{0, 0} = 0$ .

2.1step 1.1step 1.2step 1.3F1F2F3F4algebra∎

[(c) ladder] Using [F1] and [F3]: $J_{+}^{t o t} w_{1, 1} = (J_{+}^{(1 / 2)} e_{+}) \otimes e_{+} + e_{+} \otimes (J_{+}^{(1 / 2)} e_{+}) = 0$ ; $J_{+}^{t o t} w_{1, 0} = [(J_{+}^{(1 / 2)} e_{+}) \otimes e_{-} + e_{+} \otimes (J_{+}^{(1 / 2)} e_{-}) + (J_{+}^{(1 / 2)} e_{-}) \otimes e_{+} + e_{-} \otimes (J_{+}^{(1 / 2)} e_{+})] / \sqrt{2} = (0 + e_{+} \otimes e_{+} + e_{+} \otimes e_{+} + 0) / \sqrt{2} = \sqrt{2} w_{1, 1}$ ; and $J_{+}^{t o t} w_{1, - 1} = (J_{+}^{(1 / 2)} e_{-}) \otimes e_{-} + e_{-} \otimes (J_{+}^{(1 / 2)} e_{-}) = e_{+} \otimes e_{-} + e_{-} \otimes e_{+} = \sqrt{2} w_{1, 0}$ . Since $J_{-}^{t o t} = (J_{+}^{t o t})^{*}$ ([F1] self-adjointness and [F3]), the adjoint relations follow: $J_{-}^{t o t} w_{1, - 1} = 0$ , $J_{-}^{t o t} w_{1, 0} = \sqrt{2} w_{1, - 1}$ , $J_{-}^{t o t} w_{1, 1} = \sqrt{2} w_{1, 0}$ (each verified by pairing with the orthonormal basis vectors of step 1.1). Together with step 1.2 this shows that sending $e_{m} \mapsto w_{1, m}$ carries the spin- $1$ operators of Finite spin matrices to $J_{x}^{t o t}, J_{y}^{t o t}, J_{z}^{t o t}$ on the triplet: the spin- $1$ model has $J_{+} e_{0} = \sqrt{2} e_{1}$ , $J_{+} e_{- 1} = \sqrt{2} e_{0}$ , $J_{+} e_{1} = 0$ and $J_{z} e_{m} = m e_{m}$ , matching steps 1.2 and this step, and $J_{x}, J_{y}$ are recovered from $J_{\pm}$ ; the singlet span is orthogonal to the triplet by step 1.1 and is annihilated by all $J_{i}^{t o t}$ by step 1.3, so this is exactly the decomposition of [F4] for $s = t = \frac{1}{2}$ .

Coupled triplet and singlet

Example

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

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