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Qm Math Orbital And Spin — Examples

1 · Prerequisites

2 · Summary

The companion page to qm-math-orbital-and-spin records three explicit finite-dimensional checks behind its spin and orbital results. All three items are mathematics: they carry domain: mathematics, depend only on the items of the parent page and on mathematical suppliers, and make no physical or empirical claim.

Spin singlet CHSH computes the singlet correlations through Finite spin matrices and Tsirelson bound: for $Ψ = (e_{+} \otimes e_{-} - e_{-} \otimes e_{+}) / \sqrt{2}$ one has $⟨ (σ_{i} \otimes σ_{j}) Ψ, Ψ ⟩ = - δ_{i j}$ , hence $⟨ (σ_{a} \otimes σ_{b}) Ψ, Ψ ⟩ = - a \cdot b$ for unit $a, b \in R^{3}$ ; with the planar settings $a_{0} = e_{1}$ , $a_{1} = e_{2}$ , $b_{0} = (e_{1} + e_{2}) / \sqrt{2}$ , $b_{1} = (e_{1} - e_{2}) / \sqrt{2}$ the CHSH combination of the four correlations equals $- 2 \sqrt{2}$ , attaining the bound, and since $2 \sqrt{2} > 2$ these four numbers admit no common-outcome $\pm 1$ realization by the CHSH inequality for that explicitly specified model.

Coupled triplet and singlet writes out the $\frac{1}{2} \otimes \frac{1}{2}$ Clebsch-Gordan basis: $e_{+} \otimes e_{+}$ , $(e_{+} \otimes e_{-} + e_{-} \otimes e_{+}) / \sqrt{2}$ , $e_{-} \otimes e_{-}$ span the triplet with $J_{z}^{t o t}$ -weights $1, 0, - 1$ and $J^{t o t 2} = 2$ , the singlet $(e_{+} \otimes e_{-} - e_{-} \otimes e_{+}) / \sqrt{2}$ is annihilated by all three $J_{i}^{t o t}$ , the four vectors form an orthonormal basis, and the ladders $J_{+}^{t o t} w_{1, 1} = 0$ , $J_{+}^{t o t} w_{1, 0} = \sqrt{2} w_{1, 1}$ , $J_{+}^{t o t} w_{1, - 1} = \sqrt{2} w_{1, 0}$ realize the spin- $1$ model on the triplet, matching the decomposition $V \otimes V = W_{1} \oplus W_{0}$ of Tensor spin composition.

Orbital core commutators verifies on the three-dimensional block $V_{1}$ of Orbital self-adjoint generators that the normalized coordinate harmonics $v_{m} = Y_{1, m}$ satisfy $L_{3} v_{m} = ℏ m v_{m}$ , $L_{+} v_{1} = 0$ , $L_{+} v_{0} = \sqrt{2} ℏ v_{1}$ , $L_{+} v_{- 1} = \sqrt{2} ℏ v_{0}$ , $[L_{1}, L_{2}] = i ℏ L_{3}$ and $L_{1}^{2} + L_{2}^{2} + L_{3}^{2} = 2 ℏ^{2}$ on $V_{1}$ (the last equalling $ℏ^{2} A$ there, with $A ∣_{V_{1}} = 2 i d$ by Sphere Laplacian diagonal domain), and that $e_{m} \mapsto v_{m}$ is a unitary equivalence of the orbital degree-one block with the spin- $1$ model of Finite spin matrices.

All three are finite matrix computations inside two- or three-dimensional spaces; none of them turns these mathematical identities into a statement about nature.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

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Spin singlet CHSH

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 $σ_{x}, σ_{y}, σ_{z}$ be the Pauli matrices of Finite spin matrices (e), with $σ_{a} : = \sum_{i \in {x, y, z}} a_{i} σ_{i}$ for $a \in R^{3}$ ; on the completed tensor product $V \hat{\otimes} V$ use the inner product of Hilbert tensor construction: inner product, completion and product bases. Define the singlet vector $Ψ : = \frac{1}{\sqrt{2}} (e_{+} \otimes e_{-} - e_{-} \otimes e_{+}) \in V \hat{\otimes} V .$ Then:

(a) $∥ Ψ ∥ = 1$ and $⟨ (σ_{i} \otimes σ_{j}) Ψ, Ψ ⟩ = - δ_{i j}$ for all $i, j \in {x, y, z}$ ;

(b) for all unit vectors $a, b \in R^{3}$ the singlet correlation is $⟨ (σ_{a} \otimes σ_{b}) Ψ, Ψ ⟩ = - a \cdot b;$

(c) with the unit vectors $a_{0} : = (1, 0, 0)$ , $a_{1} : = (0, 1, 0)$ , $b_{0} : = (1, 1, 0) / \sqrt{2}$ , $b_{1} : = (1, - 1, 0) / \sqrt{2}$ the CHSH combination of the four correlations $E_{i j} : = ⟨ (σ_{a_{i}} \otimes σ_{b_{j}}) Ψ, Ψ ⟩$ satisfies $E_{00} + E_{01} + E_{10} - E_{11} = - 2 \sqrt{2}, so ∣ E_{00} + E_{01} + E_{10} - E_{11} ∣ = 2 \sqrt{2},$ which is the value of the Tsirelson bound of Tsirelson bound (b),(c); thus the bound is attained at these settings;

(d) since $2 \sqrt{2} > 2$ , the four numbers $E_{i j}$ of (c) cannot be realized as $E [A_{i} B_{j}]$ for four $\pm 1$ -valued random variables $A_{0}, A_{1}, B_{0}, B_{1}$ on one probability space, by the CHSH inequality for that common-outcome model (Tsirelson bound (e)).

All claims are finite-dimensional linear-algebra statements about the indicated matrices and vectors; no preparation, measurement or experimental claim is made.

Facts & Assumptions

Given: the Pauli matrices $σ_{x}, σ_{y}, σ_{z}$ , the basis $(e_{+}, e_{-})$ of $C^{2}$ , the tensor product $V \hat{\otimes} V$ with its inner product, the vector $Ψ$ , and the unit vectors $a_{0}, a_{1}, b_{0}, b_{1}$ .

[F1]

Pauli algebra and its consequences: $σ_{i}^{2} = I_{2}$ , $σ_{x} σ_{y} = i σ_{z}$ , $σ_{y} σ_{z} = i σ_{x}$ , $σ_{z} σ_{x} = i σ_{y}$ (and the reversed products with the opposite sign), $σ_{i} σ_{j} + σ_{j} σ_{i} = 2 δ_{i j} I_{2}$ ; for unit $a$ the matrix $σ_{a}$ is a self-adjoint involution; and the actions on the basis are $σ_{x} e_{+} = e_{-}$ , $σ_{x} e_{-} = e_{+}$ , $σ_{y} e_{+} = i e_{-}$ , $σ_{y} e_{-} = - i e_{+}$ , $σ_{z} e_{+} = e_{+}$ , $σ_{z} e_{-} = - e_{-}$ by the displayed matrices and matrix-vector multiplication (Finite spin matrices (e), Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes).

[F2]

The tensor product: $V \hat{\otimes} V$ is a complex Hilbert space containing the elementary tensors densely, with $⟨ u \otimes v, u^{'} \otimes v^{'} ⟩ = ⟨ u, u^{'} ⟩ ⟨ v, v^{'} ⟩$ ; the four elementary tensors $e_{\pm} \otimes e_{\pm}$ form an orthonormal basis (Hilbert tensor construction: inner product, completion and product bases, Real and complex inner product spaces, with the inner product linear in the first argument, Hilbert space).

[F3]

The operator $S \otimes T$ on $V \hat{\otimes} V$ is determined on elementary tensors by $(S \otimes T) (u \otimes v) = (S u) \otimes (T v)$ ; sums and scalar multiples of such operators act componentwise, so $(σ_{a} \otimes σ_{b}) Ψ = \sum_{i, j} a_{i} b_{j} (σ_{i} \otimes σ_{j}) Ψ$ ; the inner product is linear in the first argument and additive and conjugate-linear in the second (Hilbert tensor construction: inner product, completion and product bases, Linear map between vector spaces over the same field, Real and complex inner product spaces, with the inner product linear in the first argument).

[F4]

The CHSH inequality for the common-outcome model: for $\pm 1$ -valued random variables $A_{0}, A_{1}, B_{0}, B_{1}$ on one probability space, $∣ E [A_{0} B_{0}] + E [A_{0} B_{1}] + E [A_{1} B_{0}] - E [A_{1} B_{1}] ∣ \leq 2$ ; the Tsirelson bound states that the operator $C = A_{0} \otimes (B_{0} + B_{1}) + A_{1} \otimes (B_{0} - B_{1})$ built from self-adjoint involutions satisfies $∣ ⟨ C ψ, ψ ⟩ ∣ \leq 2 \sqrt{2}$ for every unit $ψ$ (Tsirelson bound (b),(c),(e)).

Verification

1.1F1F2F3algebra

[(a), the nine pairings] By [F2], $∥ Ψ ∥^{2} = (∥ e_{+} \otimes e_{-} ∥^{2} + ∥ e_{-} \otimes e_{+} ∥^{2}) / 2 = 1$ . Using [F1] and [F3]: for $i = j = x$ , $(σ_{x} \otimes σ_{x}) Ψ = (e_{-} \otimes e_{+} - e_{+} \otimes e_{-}) / \sqrt{2} = - Ψ$ , so the pairing is $- 1$ ; for $i = j = y$ , $(σ_{y} \otimes σ_{y}) (e_{+} \otimes e_{-}) = (i e_{-}) \otimes (- i e_{+}) = e_{-} \otimes e_{+}$ and $(σ_{y} \otimes σ_{y}) (e_{-} \otimes e_{+}) = (- i e_{+}) \otimes (i e_{-}) = e_{+} \otimes e_{-}$ , so again $(σ_{y} \otimes σ_{y}) Ψ = - Ψ$ ; for $i = j = z$ , $(σ_{z} \otimes σ_{z}) (e_{+} \otimes e_{-}) = e_{+} \otimes (- e_{-}) = - e_{+} \otimes e_{-}$ and $(σ_{z} \otimes σ_{z}) (e_{-} \otimes e_{+}) = (- e_{-}) \otimes e_{+} = - e_{-} \otimes e_{+}$ , so $(σ_{z} \otimes σ_{z}) Ψ = - Ψ$ . For clarity put $q = (e_{+} \otimes e_{-} + e_{-} \otimes e_{+}) / \sqrt{2}$ and $d_{\pm} = (e_{+} \otimes e_{+} \pm e_{-} \otimes e_{-}) / \sqrt{2}$ . Direct Pauli actions give $(σ_{x} \otimes σ_{y}) Ψ = - i q$ , $(σ_{y} \otimes σ_{x}) Ψ = i q$ , $(σ_{x} \otimes σ_{z}) Ψ = - d_{+}$ , $(σ_{z} \otimes σ_{x}) Ψ = d_{+}$ , $(σ_{y} \otimes σ_{z}) Ψ = i d_{-}$ and $(σ_{z} \otimes σ_{y}) Ψ = - i d_{-}$ . Each of $q, d_{+}, d_{-}$ is orthogonal to $Ψ$ by the elementary product-basis pairings, so every off-diagonal correlation is zero. Therefore $⟨ (σ_{i} \otimes σ_{j}) Ψ, Ψ ⟩ = - δ_{i j}$ .

2.1step 1.1F3F4algebra

[(b) and (c)] By [F3], $(σ_{a} \otimes σ_{b}) Ψ = \sum_{i, j} a_{i} b_{j} (σ_{i} \otimes σ_{j}) Ψ$ , so step 1.1 and the linearity and conjugate-linearity of the inner product give $⟨ (σ_{a} \otimes σ_{b}) Ψ, Ψ ⟩ = \sum_{i, j} a_{i} b_{j} (- δ_{i j}) = - a \cdot b$ , using real $a, b$ . With the four unit vectors displayed in (c) the dot products are $a_{0} \cdot b_{0} = a_{0} \cdot b_{1} = a_{1} \cdot b_{0} = 1 / \sqrt{2}$ and $a_{1} \cdot b_{1} = - 1 / \sqrt{2}$ , so $E_{00} = E_{01} = E_{10} = - 1 / \sqrt{2}$ and $E_{11} = + 1 / \sqrt{2}$ , giving $E_{00} + E_{01} + E_{10} - E_{11} = - 4 / \sqrt{2} = - 2 \sqrt{2}$ ; the bound in [F4] forces $2 \sqrt{2}$ to be the largest possible value of this combination, so the bound is attained.

3.1step 2.1F4algebra∎

[(d)] By [F4], any four $\pm 1$ -valued random variables on one probability space satisfy the CHSH bound 2 for the same combination; since the combination computed in step 2.1 has absolute value $2 \sqrt{2} > 2$ , no such common-outcome model reproduces the four numbers $E_{i j}$ .

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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 $s = t = \frac{1}{2}$ .

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Orbital core commutators

Example

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Use the notation of Orbital self-adjoint generators: $S^{2} = {x \in R^{3} : ∣ x ∣ = 1}$ with polar measure $σ$ , the Hilbert space $L^{2} (S^{2}, σ)$ with inner product linear in the first argument, the operators $L_{1}, L_{2}, L_{3}$ with $L_{\pm} : = L_{1} \pm i L_{2}$ , the finite span $H$ , the blocks $V_{l}$ , and the first-order operators $D_{i}$ , so that $L_{i} = - i ℏ D_{i}$ on $H$ . Let $d : = \sqrt{3 / (4 π)}$ and, writing $Ω = (Ω_{1}, Ω_{2}, Ω_{3})$ for the points of $S^{2}$ and restricting the coordinate polynomials, $v_{1} : = - \frac{d}{\sqrt{2}} (Ω_{1} + i Ω_{2}), v_{0} : = d Ω_{3}, v_{- 1} : = \frac{d}{\sqrt{2}} (Ω_{1} - i Ω_{2}) \in V_{1} .$ Then:

(a) $(v_{1}, v_{0}, v_{- 1})$ is an orthonormal basis of $V_{1}$ , and it is the degree-one part of the family $(Y_{l m})$ , in the sense that $v_{m} = Y_{1, m} (m \in {1, 0, - 1})$ as elements of $L^{2} (S^{2}, σ)$ (Associated Legendre function normalization);

(b) the core commutators hold on the whole three-dimensional space $V_{1}$ : $L_{3} v_{m} = ℏ m v_{m}, L_{+} v_{1} = 0, L_{+} v_{0} = \sqrt{2} ℏ v_{1}, L_{+} v_{- 1} = \sqrt{2} ℏ v_{0},$ with $(L_{-} ∣_{V_{1}}) = (L_{+} ∣_{V_{1}})^{*}$ (adjoint in the finite Hilbert space $V_{1}$ ) giving $L_{-} v_{- 1} = 0$ , $L_{-} v_{0} = \sqrt{2} ℏ v_{- 1}$ , $L_{-} v_{1} = \sqrt{2} ℏ v_{0}$ ; and $[L_{1}, L_{2}] v = i ℏ L_{3} v, (L_{1}^{2} + L_{2}^{2} + L_{3}^{2}) v = 2 ℏ^{2} v (v \in V_{1});$

(c) the assignment $e_{m} \mapsto v_{m}$ ( $m = - 1, 0, 1$ ) is a unitary isomorphism of the spin- $1$ model $(V_{1}^{s p i n}, J_{x}, J_{y}, J_{z})$ of Finite spin matrices onto $V_{1}$ carrying $J_{i}$ to $ℏ^{- 1} L_{i}$ for $i = x, y, z$ : the orbital degree- $1$ block is unitarily equivalent to the spin- $1$ block, and $L_{1}^{2} + L_{2}^{2} + L_{3}^{2}$ acts on $V_{1}$ as $ℏ^{2} A$ , where $A$ is the diagonal sphere Laplacian of Sphere Laplacian diagonal domain with $A ∣_{V_{1}} = 2 i d$ .

All claims are finite-dimensional matrix computations inside $V_{1}$ ; no physical interpretation of the operators as angular momentum of a particle is asserted.

Facts & Assumptions

Given: the operators $L_{1}, L_{2}, L_{3}, L_{\pm}$ of the generators theorem, the functions $Ω_{i}$ , the number $d = \sqrt{3 / (4 π)}$ , and the vectors $v_{m}$ .

[F1]

Generators theorem (Orbital self-adjoint generators (a),(b)): for $F \in V_{l}$ with representative $p_{F}$ one has $L_{i} F = {r e s}_{l} (- i ℏ D_{i} p_{F})$ with $D_{i} p = (x \times \nabla p)_{i}$ , so $L_{i}$ maps $V_{l}$ into itself and $L_{i} = - i ℏ D_{i}$ on $H$ ; $L_{1}, L_{2}, L_{3}$ are self-adjoint and $L_{i} L_{j} - L_{j} L_{i} = i ℏ \sum_{k} ε_{i j k} L_{k}$ on $H$ ; and the everywhere-defined finite restrictions $A_{i} : = L_{i} ∣_{V_{1}}$ are Hermitian. Hence $A_{\pm} : = A_{1} \pm i A_{2}$ satisfy $A_{-} = (A_{+})^{*}$ as operators on $V_{1}$ : for $f, g \in V_{1}$ , $⟨ A_{+} f, g ⟩ = ⟨ f, (A_{1} - i A_{2}) g ⟩$ , since the inner product is first-variable linear. This is solely a finite-block adjoint identity, with no assertion about the full unbounded sum adjoint.

[F2]

The $L^{2}$ inner products of the coordinate functions: integrating over $S^{2}$ in angle coordinates, $\int_{S^{2}} Ω_{1}^{2} d σ = \int_{0}^{π} \int_{0}^{2 π} \sin^{2} θ \cos^{2} φ \sin θ d φ d θ = π \cdot \frac{4}{3} = \frac{4 π}{3}$ and likewise for $Ω_{2}^{2}, Ω_{3}^{2}$ , while all mixed integrals vanish, since $\int_{0}^{2 π} \cos φ d φ = \int_{0}^{2 π} \sin φ d φ = \int_{0}^{2 π} \sin φ \cos φ d φ = 0$ and the $θ$ -integrals factor; hence $⟨ Ω_{i}, Ω_{j} ⟩ = \frac{4 π}{3} δ_{i j}$ (Sphere integrals in angle coordinates, The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ , The derivatives of sine and cosine are cosine and minus sine, Sine and cosine defined by their real power series, The complex exponential by its power series).

[F3]

The degree-one harmonics: $Y_{1, 0} = b_{10} w_{10} (\cos θ) / \sqrt{2 π}$ with $P_{1} (t) = t$ and $w_{10} = P_{1}$ , $b_{10} = (\int_{- 1}^{1} t^{2} d t)^{- 1 / 2} = \sqrt{3 / 2}$ , so $Y_{1, 0} = \sqrt{3 / (4 π)} Ω_{3}$ ; and $Y_{1, 1} = b_{11} w_{11} (\cos θ) e^{i φ} / \sqrt{2 π}$ with $w_{11} = - (1 - t^{2})^{1 / 2}$ and $b_{11} = (\int_{- 1}^{1} (1 - t^{2}) d t)^{- 1 / 2} = \sqrt{3 / 4}$ , so $Y_{1, 1} = - \sqrt{3 / (8 π)} (Ω_{1} + i Ω_{2})$ ; moreover $Y_{1, - 1} = - Y_{1, 1}$ -conjugate: $Y_{1, - 1} = (- 1)^{1} \overline{Y_{1, 1}} = \sqrt{3 / (8 π)} (Ω_{1} - i Ω_{2})$ , using $Ω_{1} + i Ω_{2} = \sin θ e^{i φ}$ on $S^{2}$ (Associated Legendre function normalization (a),(c), The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ , The derivatives of sine and cosine are cosine and minus sine, Sine and cosine defined by their real power series, The complex exponential by its power series).

[F4]

The space $V_{1}$ and the diagonal operator: $V_{1}$ consists of the classes of the restrictions of the linear polynomials $Ω_{i}$ , the restriction map on harmonic homogeneous polynomials of degree $1$ is a linear isomorphism onto $V_{1}$ , and $A Y_{1, m} = 2 Y_{1, m}$ , so $A ∣_{V_{1}} = 2 i d$ (Harmonic polynomial dimension (b),(c), Sphere Laplacian diagonal domain (a), The Laplacian of a $C^{2}$ function and of a $C^{2}$ vector field).

[F5]

Spin-one data (Finite spin matrices): in the basis $(e_{m})_{m = 1, 0, - 1}$ one has $J_{z} e_{m} = m e_{m}$ , $J_{+} e_{1} = 0$ , $J_{+} e_{0} = \sqrt{2} e_{1}$ , $J_{+} e_{- 1} = \sqrt{2} e_{0}$ , $J_{-} = (J_{+})^{*}$ , and $J_{x} = \frac{1}{2} (J_{+} + J_{-})$ , $J_{y} = \frac{1}{2 i} (J_{+} - J_{-})$ ; the arrows $e_{m} \mapsto v_{m}$ with orthonormal domain and target bases are unitary ([F2], Real and complex inner product spaces, with the inner product linear in the first argument, The induced length is a norm).

Verification

1.1F2F3F4algebra

[(a) orthonormality and identification] By [F2], $∥ v_{0} ∥^{2} = d^{2} \frac{4 π}{3} = 1$ , $∥ v_{\pm 1} ∥^{2} = \frac{d^{2}}{2} (\frac{4 π}{3} + \frac{4 π}{3}) = 1$ , and $⟨ v_{1}, v_{0} ⟩ = ⟨ v_{1}, v_{- 1} ⟩ = ⟨ v_{0}, v_{- 1} ⟩ = 0$ ; so the $v_{m}$ are orthonormal, and since $\dim V_{1} = 3$ they form an orthonormal basis of $V_{1}$ ([F4]). Comparing with the displayed values of [F3]: $Y_{1, 0} = d Ω_{3} = v_{0}$ , $Y_{1, 1} = - \sqrt{3 / (8 π)} (Ω_{1} + i Ω_{2}) = - \frac{d}{\sqrt{2}} (Ω_{1} + i Ω_{2}) = v_{1}$ (because $\sqrt{3 / (8 π)} = d / \sqrt{2}$ ), and $Y_{1, - 1} = + \frac{d}{\sqrt{2}} (Ω_{1} - i Ω_{2}) = v_{- 1}$ .

2.1step 1.1F1algebra

[(b) $L_{3}$ and ladder] Compute the polynomial actions with $D_{1} = x_{2} \partial_{3} - x_{3} \partial_{2}$ , $D_{2} = x_{3} \partial_{1} - x_{1} \partial_{3}$ , $D_{3} = x_{1} \partial_{2} - x_{2} \partial_{1}$ and [F1]: $D_{3} (Ω_{1} \pm i Ω_{2}) = \pm i (Ω_{1} \pm i Ω_{2})$ (since $D_{3} Ω_{1} = - Ω_{2}$ , $D_{3} Ω_{2} = Ω_{1}$ ), hence $L_{3} (Ω_{1} \pm i Ω_{2}) = \pm ℏ (Ω_{1} \pm i Ω_{2})$ and, after the normalisation $v_{\pm 1} = \mp \frac{d}{\sqrt{2}} (Ω_{1} \pm i Ω_{2})$ , $L_{3} v_{1} = ℏ v_{1}$ , $L_{3} v_{- 1} = - ℏ v_{- 1}$ , and $L_{3} v_{0} = 0$ ; further $(D_{1} + i D_{2}) (Ω_{1} + i Ω_{2}) = 0$ , $(D_{1} + i D_{2}) Ω_{3} = - i (Ω_{1} + i Ω_{2})$ , $(D_{1} + i D_{2}) (Ω_{1} - i Ω_{2}) = 2 i Ω_{3}$ , so $L_{+} = - i ℏ (D_{1} + i D_{2})$ satisfies $L_{+} v_{1} = 0$ , $L_{+} v_{0} = - i ℏ d (- i (Ω_{1} + i Ω_{2})) = - ℏ d (Ω_{1} + i Ω_{2}) = \sqrt{2} ℏ v_{1}$ , and $L_{+} v_{- 1} = - i ℏ d \frac{2 i}{\sqrt{2}} Ω_{3} = \sqrt{2} ℏ d Ω_{3} = \sqrt{2} ℏ v_{0}$ ; the relations for $L_{-} ∣_{V_{1}} = (L_{+} ∣_{V_{1}})^{*}$ follow by the finite-block adjunction of [F1].

3.1step 1.1F1F4algebra

[(b) commutators on the core] With the same operators, $L_{1} x_{1} = 0$ , $L_{1} x_{2} = i ℏ x_{3}$ , $L_{1} x_{3} = - i ℏ x_{2}$ ; $L_{2} x_{1} = - i ℏ x_{3}$ , $L_{2} x_{2} = 0$ , $L_{2} x_{3} = i ℏ x_{1}$ ; $L_{3} x_{1} = i ℏ x_{2}$ , $L_{3} x_{2} = - i ℏ x_{1}$ , $L_{3} x_{3} = 0$ . Evaluating $L_{1} L_{2} - L_{2} L_{1}$ on $x_{1}, x_{2}, x_{3}$ : since $L_{2} x_{1} = - i ℏ x_{3}$ and $L_{1} x_{1} = 0$ one has $L_{1} L_{2} x_{1} = - i ℏ L_{1} x_{3} = - ℏ^{2} x_{2}$ and $L_{2} L_{1} x_{1} = 0$ , so $[L_{1}, L_{2}] x_{1} = - ℏ^{2} x_{2} = i ℏ (i ℏ x_{2}) = i ℏ L_{3} x_{1}$ ; $[L_{1}, L_{2}] x_{2} = + ℏ^{2} x_{1} = i ℏ L_{3} x_{2}$ ; and $[L_{1}, L_{2}] x_{3} = 0 = i ℏ L_{3} x_{3}$ . Hence $[L_{1}, L_{2}] = i ℏ L_{3}$ on $V_{1}$ , and cyclically for the other two pairs, consistent with [F1]. Likewise, squaring the displayed actions: $L_{1}^{2} x_{1} = 0$ , $L_{2}^{2} x_{1} = ℏ^{2} x_{1}$ , $L_{3}^{2} x_{1} = ℏ^{2} x_{1}$ ; $L_{1}^{2} x_{2} = i ℏ L_{1} x_{3} = ℏ^{2} x_{2}$ , $L_{2}^{2} x_{2} = 0$ , $L_{3}^{2} x_{2} = ℏ^{2} x_{2}$ ; $L_{1}^{2} x_{3} = ℏ^{2} x_{3}$ , $L_{2}^{2} x_{3} = ℏ^{2} x_{3}$ , $L_{3}^{2} x_{3} = 0$ ; in each case the sum is $2 ℏ^{2} x_{i}$ , so $L_{1}^{2} + L_{2}^{2} + L_{3}^{2} = 2 ℏ^{2} i d$ on $V_{1}$ , which equals $ℏ^{2} A ∣_{V_{1}}$ by [F4]. Alternatively the sum identity follows from the ladder relations of step 2.1 via $L_{1}^{2} + L_{2}^{2} = \frac{1}{2} (L_{+} L_{-} + L_{-} L_{+})$ and $L_{1}^{2} + L_{2}^{2} + L_{3}^{2} = 2 ℏ^{2}$ on each $v_{m}$ .

4.1step 1.1step 2.1step 3.1F4F5algebra∎

[(c) spin-one equivalence] By steps 1.1 and 2.1 the assignment $e_{m} \mapsto v_{m}$ matches the spin-one relations of [F5]: $J_{z} e_{m} = m e_{m}$ with $L_{3} v_{m} = ℏ m v_{m}$ , and $J_{+} e_{1} = 0$ , $J_{+} e_{0} = \sqrt{2} e_{1}$ , $J_{+} e_{- 1} = \sqrt{2} e_{0}$ with $L_{+} v_{1} = 0$ , $L_{+} v_{0} = \sqrt{2} ℏ v_{1}$ , $L_{+} v_{- 1} = \sqrt{2} ℏ v_{0}$ ; since both bases are orthonormal and $L_{x} = \frac{1}{2} (L_{+} + L_{-})$ , $L_{y} = \frac{1}{2 i} (L_{+} - L_{-})$ mirror $J_{x}, J_{y}$ , the assignment is a unitary isomorphism carrying $J_{i}$ to $ℏ^{- 1} L_{i}$ for $i = x, y, z$ . The last assertion is step 3.1 with [F4].

Qm Math Orbital And Spin — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

5 · Examples, experiments and counterexamples

Spin singlet CHSH

Example

Facts & Assumptions

Verification

Coupled triplet and singlet

Example

Facts & Assumptions

Verification

Orbital core commutators

Example

Facts & Assumptions

Verification

Sources