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Three spin-half factors: the quartet and the two exchange-coupled doublets

Example

Use the standard Pauli matrices $σ^{x} = (\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}), σ^{y} = (\begin{array}{cc} 0 & - i \\ i & 0 \end{array}), σ^{z} = (\begin{array}{cc} 1 & 0 \\ 0 & - 1 \end{array}),$ and let $σ_{j}^{a}$ act by $σ^{a}$ on tensor factor $j$ and identity on the other factors. Let $∣ a b c ⟩ : = ∣ a ⟩ \otimes ∣ b ⟩ \otimes ∣ c ⟩$ for $a, b, c \in {0, 1}$ . In $H_{3} = (C^{2})^{\otimes 3}$ define $\begin{aligned} q_{3 / 2} & = ∣ 000 ⟩, & q_{1 / 2} & = (∣ 100 ⟩ + ∣ 010 ⟩ + ∣ 001 ⟩) / \sqrt{3}, \\ q_{- 1 / 2} & = (∣ 110 ⟩ + ∣ 101 ⟩ + ∣ 011 ⟩) / \sqrt{3}, & q_{- 3 / 2} & = ∣ 111 ⟩, \\ a_{+} & = (∣ 010 ⟩ - ∣ 100 ⟩) / \sqrt{2}, & a_{-} & = (∣ 011 ⟩ - ∣ 101 ⟩) / \sqrt{2}, \\ b_{+} & = (2 ∣ 001 ⟩ - ∣ 010 ⟩ - ∣ 100 ⟩) / \sqrt{6}, & b_{-} & = (∣ 101 ⟩ + ∣ 011 ⟩ - 2 ∣ 110 ⟩) / \sqrt{6} . \end{aligned}$ These eight vectors form an orthonormal basis. With $K_{z} : = \frac{1}{2} \sum_{j = 1}^{3} σ_{j}^{z}$ and $K_{+} : = \frac{1}{2} \sum_{j = 1}^{3} (σ_{j}^{x} + i σ_{j}^{y})$ , the $q$ -vectors form the spin- $3 / 2$ ladder and the $a$ - and $b$ -pairs form two spin- $1 / 2$ ladders. For $K_{-} : = \frac{1}{2} \sum_{j = 1}^{3} (σ_{j}^{x} - i σ_{j}^{y})$ , their lowering actions are the adjoints of the raising actions displayed below: $K_{+} q_{1 / 2} = \sqrt{3} q_{3 / 2}, K_{+} q_{- 1 / 2} = 2 q_{1 / 2}, K_{+} q_{- 3 / 2} = \sqrt{3} q_{- 1 / 2}, K_{+} a_{-} = a_{+}, K_{+} b_{-} = b_{+},$ with the highest vectors killed by $K_{+}$ and the stated subscripts giving the $K_{z}$ eigenvalues. The adjacent swaps act as identity on the quartet and on each magnetic pair $(a_{+}, b_{+})$ and $(a_{-}, b_{-})$ by $s_{12} = (\begin{array}{cc} - 1 & 0 \\ 0 & 1 \end{array}), s_{23} = (\begin{array}{cc} 1 / 2 & \sqrt{3} / 2 \\ \sqrt{3} / 2 & - 1 / 2 \end{array}) .$ For $ℏ > 0$ , set $S_{j}^{a} = ℏ σ_{j}^{a} / 2$ , and for real couplings $J_{12}, J_{23}$ define $H_{3} : = \sum_{j = 1}^{2} J_{j, j + 1} ℏ^{- 2} \sum_{a = x, y, z} S_{j}^{a} S_{j + 1}^{a} .$ This finite-chain exchange operator is the scalar $(J_{12} + J_{23}) / 4$ on the quartet and, on each doublet multiplicity basis $(a_{\pm}, b_{\pm})$ , has matrix $\frac{1}{4} (\begin{array}{cc} - 3 J_{12} & \sqrt{3} J_{23} \\ \sqrt{3} J_{23} & J_{12} - 2 J_{23} \end{array}) .$ The latter matrix has eigenvalues $- \frac{J_{12} + J_{23}}{4} \pm \frac{1}{2} \sqrt{J_{12}^{2} - J_{12} J_{23} + J_{23}^{2}},$ each repeated for the two magnetic states.

Example references

Finite spin-half tensor chains

Spin-half swaps realize the Temperley–Lieb algebra at loop weight two

The finite isotropic spin-half exchange Hamiltonian is a Temperley–Lieb element

Unitary exchange blocks for collective spins

Facts & Assumptions

Given: The ordered tensor basis of $H_{3}$ , the Pauli matrices, and real couplings $J_{12}, J_{23}$ .

[F1]

Finite spin-half tensor chains, The elementary tensors of two bases form the product basis of the tensor product: the eight binary tensor words form a basis of $H_{3}$ and are orthonormal for the product inner product linear in its first argument, whose elementary-tensor formula is $⟨ v_{1} \otimes v_{2} \otimes v_{3}, w_{1} \otimes w_{2} \otimes w_{3} ⟩ = \prod_{j = 1}^{3} ⟨ v_{j}, w_{j} ⟩_{C^{2}}$ with $⟨ (a, b), (c, d) ⟩_{C^{2}} = a \overset{ˉ}{c} + b \overset{ˉ}{d}$ ; $σ_{j}^{a}$ acts by the Pauli matrix on factor $j$ .

[F2]

Finite spin matrices: in the ordered basis $∣ 0 ⟩, ∣ 1 ⟩$ , $σ^{z} ∣ 0 ⟩ = ∣ 0 ⟩$ , $σ^{z} ∣ 1 ⟩ = - ∣ 1 ⟩$ , and $\frac{1}{2} (σ^{x} + i σ^{y}) ∣ 1 ⟩ = ∣ 0 ⟩$ , $\frac{1}{2} (σ^{x} + i σ^{y}) ∣ 0 ⟩ = 0$ , $\frac{1}{2} (σ^{x} - i σ^{y}) ∣ 0 ⟩ = ∣ 1 ⟩$ , and $\frac{1}{2} (σ^{x} - i σ^{y}) ∣ 1 ⟩ = 0$ ; these raising and lowering matrices are adjoints.

[F3]

Spin-half swaps realize the Temperley–Lieb algebra at loop weight two: on three sites, $e_{12} = I - s_{12}$ and $e_{23} = I - s_{23}$ ; these are the loop-weight-two Temperley–Lieb generators.

[F4]

The finite isotropic spin-half exchange Hamiltonian is a Temperley–Lieb element: a real-coupled finite chain has the exact bond identity $S_{i} \cdot S_{i + 1} / ℏ^{2} = I / 4 - e_{i} / 2$ .

[F5]

Unitary exchange blocks for collective spins: permutation operators act on Specht factors and collective unitary operators act on multiplicity factors.

Verification

1.1F1algebra

The coefficient vectors of $a_{+}$ and $b_{+}$ in the ordered one-down subspace $(∣ 001 ⟩, ∣ 010 ⟩, ∣ 100 ⟩)$ are $(0, 1, - 1) / \sqrt{2}$ and $(2, - 1, - 1) / \sqrt{6}$ ; their squared norms are $1$ and their inner product is $0$ . The one-up pair $a_{-}, b_{-}$ has coefficient vectors $(0, - 1, 1) / \sqrt{2}$ and $(- 2, 1, 1) / \sqrt{6}$ in the ordered basis $(∣ 110 ⟩, ∣ 101 ⟩, ∣ 011 ⟩)$ , again with norm $1$ and inner product $0$ . Each $q$ -vector has norm $1$ ; the two three-term $q$ -vectors are orthogonal to the corresponding $a, b$ vectors because their coefficient vectors are constant while the latter have coefficient sum zero. Vectors with different numbers of entries equal to $1$ are orthogonal. Thus the eight vectors are orthonormal, and [F1] says the ambient space has dimension eight; they are a basis.

2.1F2step 1.1algebra

Applying $K_{z}$ counts $+ 1 / 2$ for each $0$ and $- 1 / 2$ for each $1$ , which gives the subscripts in the statement. By [F2], $K_{+}$ changes one $1$ to $0$ with coefficient $1$ . Summing over positions gives $K_{+} q_{1 / 2} = \sqrt{3} q_{3 / 2}, K_{+} q_{- 1 / 2} = 2 q_{1 / 2}, K_{+} q_{- 3 / 2} = \sqrt{3} q_{- 1 / 2}, K_{+} a_{-} = a_{+}, K_{+} b_{-} = b_{+};$ the top states $q_{3 / 2}, a_{+}, b_{+}$ are killed because their coefficients cancel or they contain no $1$ . The lowering actions are the adjoints of these raising actions because the displayed basis is orthonormal and each local lowering matrix is the adjoint of its raising matrix by [F2]. This verifies both ladder directions.

2.2F1step 1.1algebra

The swaps fix every symmetric $q$ -vector. On both magnetic pairs, $s_{12}$ changes the sign of $a_{\pm}$ and fixes $b_{\pm}$ , by their displayed coefficients. For the upper pair, $s_{23} a_{+} = (∣ 001 ⟩ - ∣ 100 ⟩) / \sqrt{2} = \frac{1}{2} a_{+} + \frac{\sqrt{3}}{2} b_{+},$ and direct substitution gives $s_{23} b_{+} = (\sqrt{3} / 2) a_{+} - (1 / 2) b_{+}$ . On the lower pair the same substitution gives $s_{23} a_{-} = (∣ 011 ⟩ - ∣ 110 ⟩) / \sqrt{2} = \frac{1}{2} a_{-} + \frac{\sqrt{3}}{2} b_{-}, s_{23} b_{-} = (\sqrt{3} / 2) a_{-} - (1 / 2) b_{-} .$ These are exactly the two matrices stated.

3.1F5step 2.1step 2.2algebra

The two matrices act on the same coordinate pair $(a_{+}, b_{+})$ and $(a_{-}, b_{-})$ , while $K_{z}$ and $K_{\pm}$ act on the magnetic labels with the same coefficients for either exchange coordinate. The two-dimensional exchange representation is irreducible: the only invariant lines of $s_{12} = diag (- 1, 1)$ are the two coordinate axes, and $s_{23}$ mixes both axes because its off-diagonal entries are nonzero. Thus this basis displays a spin- $3 / 2$ block with trivial exchange and a two-dimensional exchange factor repeated over the two spin- $1 / 2$ magnetic states, in the factor placement of [F5].

4.1F3F4step 2.2algebra∎

By [F3]–[F4], each bond contributes $J (I / 4 - e / 2)$ . On the quartet both swaps are $I$ , so both $e$ 's vanish and the total is $(J_{12} + J_{23}) I / 4$ . On either doublet pair, $e_{12} = (\begin{array}{cc} 2 & 0 \\ 0 & 0 \end{array}), e_{23} = (\begin{array}{cc} 1 / 2 & - \sqrt{3} / 2 \\ - \sqrt{3} / 2 & 3 / 2 \end{array}) .$ Therefore the two bond matrices are $(\begin{array}{cc} - 3 / 4 & 0 \\ 0 & 1 / 4 \end{array}), (\begin{array}{cc} 0 & \sqrt{3} / 4 \\ \sqrt{3} / 4 & - 1 / 2 \end{array}),$ and their real-coupled sum is the displayed doublet matrix, identical for $+$ and $-$ . For $a = J_{12}$ and $b = J_{23}$ , its trace is $- (a + b) / 2$ and determinant is $- 3 (a - b)^{2} / 16$ ; solving the quadratic characteristic equation gives $- (a + b) / 4 \pm \frac{1}{2} \sqrt{a^{2} - a b + b^{2}}$ . This proves the exact block decomposition and multiplicity of these eigenvalues.

Remarks

In the English convention, the parts of a shape give the left-aligned row lengths from top to bottom. A standard tableau fills these three boxes with $1, 2, 3$ , each exactly once, with entries strictly increasing from left to right along rows and from top to bottom down columns. The standard tableaux below label the Specht factors of shapes $(3)$ and $(2, 1)$ . The spin-coupling basis $(a_{\pm}, b_{\pm})$ is a singlet/triplet coupling basis for the two-dimensional multiplicity space; it is not asserted to be the displayed tableau basis. The diagram is an index convention, not part of the proof.

Three spin-half factors: the quartet and the two exchange-coupled doublets

Example

Example references

Facts & Assumptions

Verification

Remarks

Depends on

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Sources