PropositionStatement: AI-adaptedProof: AI-adaptedPipeline-generated

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Finite spin matrices

Statement

Let $s \in \frac{1}{2} N = {0, \frac{1}{2}, 1, \frac{3}{2}, \dots}$ , put $N : = 2 s \in N$ and let $V_{s} : = C^{N + 1}$ carry the standard inner product $⟨ z, w ⟩ = \sum_{k = 0}^{N} z_{k} \overline{w_{k}}$ , linear in the first argument (Real and complex inner product spaces, with the inner product linear in the first argument). Write $(e_{m})_{- s \leq m \leq s}$ for the standard basis, indexed so that $e_{m}$ is the vector with entry $1$ in position $k = s - m$ and entry $0$ elsewhere; thus $⟨ e_{m}, e_{n} ⟩ = 1$ if $m = n$ and $⟨ e_{m}, e_{n} ⟩ = 0$ otherwise. Define $C$ -linear maps $J_{z}, J_{+}, J_{-} \in {End}_{C} (V_{s})$ by their values on this basis (Linear map between vector spaces over the same field):

$J_{z} e_{m} : = m e_{m}, J_{+} e_{m} : = \sqrt{(s - m) (s + m + 1)} e_{m + 1}, J_{-} e_{m} : = \sqrt{(s + m) (s - m + 1)} e_{m - 1},$

where $e_{s + 1} : = 0$ and $e_{- s - 1} : = 0$ , and put

$J_{x} : = \frac{1}{2} (J_{+} + J_{-}), J_{y} : = \frac{1}{2 i} (J_{+} - J_{-}), J^{2} : = J_{x}^{2} + J_{y}^{2} + J_{z}^{2} .$

Then:

(a) $J_{z}$ , $J_{x}$ and $J_{y}$ are self-adjoint endomorphisms of $V_{s}$ (Self-adjoint and normal endomorphisms of a finite-dimensional real or complex inner product space), and $J_{+}^{*} = J_{-}$ , $J_{-}^{*} = J_{+}$ (The adjoint $T^{*} : W \to V$ is characterised by $⟨ T v, w ⟩_{W} = ⟨ v, T^{*} w ⟩_{V}$ );

(b) on $V_{s}$ the commutators are

$[J_{z}, J_{+}] = J_{+}, [J_{z}, J_{-}] = - J_{-}, [J_{+}, J_{-}] = 2 J_{z}, [J_{x}, J_{y}] = i J_{z}, [J_{y}, J_{z}] = i J_{x}, [J_{z}, J_{x}] = i J_{y};$

(c) $J^{2} = J_{-} J_{+} + J_{z}^{2} + J_{z} = J_{+} J_{-} + J_{z}^{2} - J_{z} = s (s + 1) I$ , and $[J^{2}, J_{z}] = [J^{2}, J_{\pm}] = 0$ ;

(d) the matrix elements of the raising operator are $∥ J_{+} e_{m} ∥^{2} = (s - m) (s + m + 1)$ and $∥ J_{-} e_{m} ∥^{2} = (s + m) (s - m + 1)$ , and $⟨ J_{+} e_{m}, e_{m + 1} ⟩ = \sqrt{(s - m) (s + m + 1)}$ , which is nonzero exactly when $m < s$ ;

(e) for $s = \frac{1}{2}$ , identify $V_{\frac{1}{2}} = C^{2}$ with the basis ordered as $(e_{\frac{1}{2}}, e_{- \frac{1}{2}})$ and let

$σ_{x} : = (\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}), σ_{y} : = (\begin{array}{cc} 0 & - i \\ i & 0 \end{array}), σ_{z} : = (\begin{array}{cc} 1 & 0 \\ 0 & - 1 \end{array})$

be the Pauli matrices. Then $J_{i} = \frac{1}{2} σ_{i}$ for $i = x, y, z$ , $σ_{x}^{2} = σ_{y}^{2} = σ_{z}^{2} = I$ , $σ_{x} σ_{y} = i σ_{z}$ , $σ_{y} σ_{z} = i σ_{x}$ , $σ_{z} σ_{x} = i σ_{y}$ , and consequently $σ_{i} σ_{j} + σ_{j} σ_{i} = 2 δ_{i j} I$ and $σ_{i} σ_{j} = δ_{i j} I + i \sum_{k} ε_{i j k} σ_{k}$ for all $i, j$ , where $δ_{i j} = 1$ if $i = j$ and $0$ otherwise, and $ε_{i j k} = 0$ when two of $i, j, k$ coincide, and otherwise $ε_{i j k}$ is the sign of the permutation $(x, y, z) \mapsto (i, j, k)$ ; the sum ranges over $k \in {x, y, z}$ . The same convention with indices $1, 2, 3$ uses the order-preserving relabelling $(1, 2, 3) = (x, y, z)$ .

All statements are finite-dimensional linear algebra: no physical interpretation of $J_{x}, J_{y}, J_{z}$ as angular momentum, and no measurement postulate, is asserted here.

Facts & Assumptions

Given: $s = \frac{N}{2}$ with $N \in N$ ; the space $V_{s} = C^{N + 1}$ with its standard inner product; the standard basis $(e_{m})_{- s \leq m \leq s}$ ; the maps $J_{z}, J_{\pm}, J_{x}, J_{y}, J^{2}$ displayed in the Statement.

[F1]

The standard basis is orthonormal: $⟨ e_{m}, e_{n} ⟩ = δ_{m n}$ in the sense of the Statement, and every $z \in V_{s}$ has the unique expansion $z = \sum_{m} ⟨ z, e_{m} ⟩ e_{m}$ with coefficients $⟨ z, e_{m} ⟩$ ; a linear map is determined by, and may be prescribed by, its values on a basis (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis, Linear combination of a finite list, and the span $span (S)$ as the smallest linear subspace containing $S$ , Linear map between vector spaces over the same field).

[F2]

For $T \in End (V_{s})$ the adjoint $T^{*}$ is the unique endomorphism with $⟨ T z, w ⟩ = ⟨ z, T^{*} w ⟩$ for all $z, w$ ; $T$ is self-adjoint when $T = T^{*}$ , equivalently when $⟨ T z, w ⟩ = ⟨ z, T w ⟩$ for all $z, w$ (The adjoint $T^{*} : W \to V$ is characterised by $⟨ T v, w ⟩_{W} = ⟨ v, T^{*} w ⟩_{V}$ , Self-adjoint and normal endomorphisms of a finite-dimensional real or complex inner product space).

[F3]

Complex conjugation is additive, satisfies $\overline{z w} = \overline{z} \overline{w}$ , $\overline{\overline{z}} = z$ and $z \overline{z} = ∣ z ∣^{2} \geq 0$ , and $z \overline{z} = 0$ only for $z = 0$ ; also $i^{2} = - 1$ and $\overline{i} = - i$ (The complex numbers as $R [x] / (x^{2} + 1)$ , with the real embedding and imaginary unit $i$ , Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive).

[F4]

For real $t \geq 0$ the square root symbol denotes the unique nonnegative $r$ with $r^{2} = t$ ; the numbers $(s - m) (s + m + 1)$ and $(s + m) (s - m + 1)$ are nonnegative integers for every $m \in {- s, - s + 1, \dots, s}$ , and they vanish exactly at the endpoints $m = s$ , respectively $m = - s$ , where an endpoint vector is killed (The natural numbers $N$ (von Neumann)).

[F5]

Matrix multiplication of the displayed $2 \times 2$ matrices is entrywise-finite multiplication; the identities $σ_{x}^{2} = σ_{y}^{2} = σ_{z}^{2} = I$ , $σ_{x} σ_{y} = i σ_{z}$ , $σ_{y} σ_{z} = i σ_{x}$ , $σ_{z} σ_{x} = i σ_{y}$ are checked by direct multiplication using [F3].

Proof

1.1F1F2F3F4algebra

[Well-definedness; (a)] By [F4] the coefficients are nonnegative reals and the endpoint conventions are compatible, so prescribing the values $J_{z} e_{m}, J_{\pm} e_{m}$ on the basis defines unique linear maps $J_{z}, J_{+}, J_{-} \in End (V_{s})$ by [F1]; $J_{x}, J_{y}, J^{2}$ are then defined by the displayed formulas and are linear. For all basis vectors, $⟨ J_{z} e_{m}, e_{n} ⟩ = m δ_{m n} = ⟨ e_{m}, J_{z} e_{n} ⟩$ , and $⟨ J_{+} e_{m}, e_{n} ⟩ = \sqrt{(s - m) (s + m + 1)} δ_{n, m + 1}$ , while $⟨ e_{m}, J_{-} e_{n} ⟩ = \sqrt{(s + n) (s - n + 1)} δ_{m, n - 1}$ , which at $n = m + 1$ equals $\sqrt{(s + m + 1) (s - m)} = \sqrt{(s - m) (s + m + 1)}$ , and both sides vanish otherwise; hence $⟨ J_{+} e_{m}, e_{n} ⟩ = ⟨ e_{m}, J_{-} e_{n} ⟩$ and $⟨ J_{-} e_{m}, e_{n} ⟩ = ⟨ e_{m}, J_{+} e_{n} ⟩$ for all $m, n$ , so $J_{+}^{*} = J_{-}$ and $J_{-}^{*} = J_{+}$ by [F1], [F2]. Taking adjoints in the definitions of $J_{x}, J_{y}$ and using $(J_{+} + J_{-})^{*} = J_{-} + J_{+} = J_{+} + J_{-}$ and $(\frac{1}{2 i} (J_{+} - J_{-}))^{*} = \frac{1}{\overline{2 i}} (J_{+} - J_{-})^{*} = \frac{1}{- 2 i} (J_{-} - J_{+}) = \frac{1}{2 i} (J_{+} - J_{-})$ gives $J_{x}^{*} = J_{x}$ and $J_{y}^{*} = J_{y}$ . This is (a).

1.2F1F4algebra

[(b), ladder relations] For every basis vector, $J_{z} (J_{+} e_{m}) = J_{z} \sqrt{(s - m) (s + m + 1)} e_{m + 1} = (m + 1) J_{+} e_{m} = J_{+} (J_{z} e_{m}) + J_{+} e_{m}$ , so acting on the basis and using linearity, $[J_{z}, J_{+}] = J_{+}$ ; similarly $J_{z} (J_{-} e_{m}) = (m - 1) J_{-} e_{m} = J_{-} (J_{z} e_{m}) - J_{-} e_{m}$ gives $[J_{z}, J_{-}] = - J_{-}$ . Moreover, using the endpoint conventions, $J_{+} J_{-} e_{m} = (s + m) (s - m + 1) e_{m}$ and $J_{-} J_{+} e_{m} = (s - m) (s + m + 1) e_{m}$ , so $[J_{+}, J_{-}] e_{m} = [(s + m) (s - m + 1) - (s - m) (s + m + 1)] e_{m} = 2 m e_{m} = 2 J_{z} e_{m}$ , and by linearity $[J_{+}, J_{-}] = 2 J_{z}$ .

1.3F1F4algebra

[(d)] For $- s \leq m \leq s$ , orthonormality [F1] and the definitions give $∥ J_{+} e_{m} ∥^{2} = (s - m) (s + m + 1)$ and $∥ J_{-} e_{m} ∥^{2} = (s + m) (s - m + 1)$ , and $⟨ J_{+} e_{m}, e_{m + 1} ⟩ = \sqrt{(s - m) (s + m + 1)}$ ; the coefficient vanishes exactly when $m = s$ by [F4].

2.1step 1.2F3algebra

[(b), remaining commutators] From $J_{x} = \frac{1}{2} (J_{+} + J_{-})$ , $J_{y} = \frac{1}{2 i} (J_{+} - J_{-})$ and the relations $[J_{+}, J_{-}] = 2 J_{z}$ , $[J_{-}, J_{+}] = - 2 J_{z}$ of step 1.2, expand the products bilinearly: $[J_{x}, J_{y}] = \frac{1}{4 i} ([J_{+}, J_{+}] - [J_{+}, J_{-}] + [J_{-}, J_{+}] - [J_{-}, J_{-}]) = \frac{1}{4 i} (0 - 2 J_{z} - 2 J_{z} - 0) = \frac{- 4 J_{z}}{4 i} = i J_{z}$ , because $1 / i = - i$ by [F3]. Likewise $[J_{z}, J_{x}] = \frac{1}{2} ([J_{z}, J_{+}] + [J_{z}, J_{-}]) = \frac{1}{2} (J_{+} - J_{-}) = i J_{y}$ , and $[J_{z}, J_{y}] = \frac{1}{2 i} ([J_{z}, J_{+}] - [J_{z}, J_{-}]) = \frac{1}{2 i} (J_{+} + J_{-}) = - i J_{x}$ , which is $[J_{y}, J_{z}] = i J_{x}$ . This completes (b).

2.2step 1.2F1algebra

[(c)] The definitions give $J_{x}^{2} + J_{y}^{2} = \frac{1}{2} (J_{+} J_{-} + J_{-} J_{+})$ ; combining this with $[J_{+}, J_{-}] = 2 J_{z}$ from step 1.2 yields $J^{2} = J_{-} J_{+} + J_{z}^{2} + J_{z} = J_{+} J_{-} + J_{z}^{2} - J_{z}$ . Direct expansion on each basis vector gives $(J_{-} J_{+} + J_{z}^{2} + J_{z}) e_{m} = [(s - m) (s + m + 1) + m^{2} + m] e_{m} = s (s + 1) e_{m}$ and $(J_{+} J_{-} + J_{z}^{2} - J_{z}) e_{m} = [(s + m) (s - m + 1) + m^{2} - m] e_{m} = s (s + 1) e_{m}$ , using $(s \mp m) (s \pm m + 1) = s^{2} + s - m^{2} \mp m$ ; hence $J^{2} = s (s + 1) I$ by [F1]. Since $J^{2}$ is a scalar multiple of the identity, it commutes with $J_{x}, J_{y}, J_{z}$ and with $J_{\pm} = J_{x} \pm i J_{y}$ , giving (c).

3.1F3F5algebra∎

[(e), Pauli case] Let $s = \frac{1}{2}$ , so $N = 1$ and the basis is $(e_{\frac{1}{2}}, e_{- \frac{1}{2}})$ . From the definitions, $J_{z} e_{\pm \frac{1}{2}} = \pm \frac{1}{2} e_{\pm \frac{1}{2}}$ , $J_{+} e_{- \frac{1}{2}} = e_{\frac{1}{2}}$ , $J_{+} e_{\frac{1}{2}} = 0$ , $J_{-} e_{\frac{1}{2}} = e_{- \frac{1}{2}}$ , $J_{-} e_{- \frac{1}{2}} = 0$ ; reading these actions as matrices in the ordered basis $(e_{\frac{1}{2}}, e_{- \frac{1}{2}})$ gives $J_{z} = \frac{1}{2} σ_{z}$ , $J_{+} = \frac{1}{2} (σ_{x} + i σ_{y})$ and $J_{-} = \frac{1}{2} (σ_{x} - i σ_{y})$ , whence $J_{x} = \frac{1}{2} (J_{+} + J_{-}) = \frac{1}{2} σ_{x}$ and $J_{y} = \frac{1}{2 i} (J_{+} - J_{-}) = \frac{1}{2} σ_{y}$ . The displayed matrices satisfy $σ_{i}^{*} = σ_{i}$ by inspection. Direct multiplication using [F3] and [F5] gives $σ_{x}^{2} = σ_{y}^{2} = σ_{z}^{2} = I$ , $σ_{x} σ_{y} = i σ_{z}$ , $σ_{y} σ_{z} = i σ_{x}$ , $σ_{z} σ_{x} = i σ_{y}$ ; taking adjoints of the three off-diagonal products and using that the $σ_{i}$ are Hermitian with $\overline{i} = - i$ gives $σ_{j} σ_{i} = - σ_{i} σ_{j}$ for $i \neq j$ , so $σ_{i} σ_{j} + σ_{j} σ_{i} = 2 δ_{i j} I$ and $σ_{i} σ_{j} = δ_{i j} I + i \sum_{k} ε_{i j k} σ_{k}$ for all ordered pairs $i, j \in {x, y, z}$ . For $i = j$ , every $ε_{i i k}$ is zero, and the product formula reduces to $σ_{i}^{2} = I$ ; for $i \neq j$ , only the remaining coordinate $k$ has nonzero $ε_{i j k}$ , with sign positive in the three cyclic orders and negative in their reversals, exactly matching the products just computed. This is (e).

The displayed relations are those of the standard $(2 s + 1)$ -dimensional spin representation; the proposition alone asserts only the finite matrix identities written in its Statement.

Finite spin matrices

Statement

Facts & Assumptions

Proof

Depends on

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