Exchange symmetry, Schur–Weyl and Temperley–Lieb spin algebra — Examples

The identity commutes with every endomorphism, so in particular $I_{H_2}s=s{I}_{H_2}$. By [F1], $s\mathrm{\mid }00⟩=\mathrm{\mid }00⟩$. Also $s\mathrm{\mid }\omega ⟩=(\mathrm{\mid }10⟩-\mathrm{\mid }01⟩)\mathrm{/}\sqrt{2}=-\mathrm{\mid }\omega ⟩$; the two vectors are nonzero and orthogonal because $\mathrm{\mid }00⟩$ is orthogonal to both one-zero/one-one basis vectors. By [F2], this swap is exactly the left place action of the transposition. Thus both exchange eigenspaces in $H_2$ are nonzero.

The operator $I_{H_2}$ satisfies the claimed permutation invariance, but step 1.1 shows that its underlying space is contained in neither single exchange eigenspace. This is a counterexample to the universal claim. The Schur–Weyl decomposition [F3] records both exchange factors and their multiplicity spaces; it does not add a rule that removes one of them.

The coefficient vectors of $a_{+}$ and $b_{+}$ in the ordered one-down subspace $(\mathrm{\mid }001⟩,\mathrm{\mid }010⟩,\mathrm{\mid }100⟩)$ are $(0,1,-1)\mathrm{/}\sqrt{2}$ and $(2,-1,-1)\mathrm{/}\sqrt{6}$; their squared norms are $1$ and their inner product is $0$. The one-up pair $a_{-},b_{-}$ has coefficient vectors $(0,-1,1)\mathrm{/}\sqrt{2}$ and $(-2,1,1)\mathrm{/}\sqrt{6}$ in the ordered basis $(\mathrm{\mid }110⟩,\mathrm{\mid }101⟩,\mathrm{\mid }011⟩)$, 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.

Applying $K_z$ counts $+1\mathrm{/}2$ for each $0$ and $-1\mathrm{/}2$ 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 $K_{+}{q}_{1\mathrm{/}2}=\sqrt{3}{q}_{3\mathrm{/}2},K_{+}{q}_{-1\mathrm{/}2}=2q_{1\mathrm{/}2},K_{+}{q}_{-3\mathrm{/}2}=\sqrt{3}{q}_{-1\mathrm{/}2},K_{+}{a}_{-}=a_{+},K_{+}{b}_{-}=b_{+};$ the top states $q_{3\mathrm{/}2},a_{+},b_{+}$ 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.

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, $s_{23}{a}_{+}=(\mathrm{\mid }001⟩-\mathrm{\mid }100⟩)\mathrm{/}\sqrt{2}=\frac{1}{2}{a}_{+}+\frac{\sqrt{3}}{2}{b}_{+},$ and direct substitution gives $s_{23}{b}_{+}=(\sqrt{3}\mathrm{/}2)a_{+}-(1\mathrm{/}2)b_{+}$. On the lower pair the same substitution gives $s_{23}{a}_{-}=(\mathrm{\mid }011⟩-\mathrm{\mid }110⟩)\mathrm{/}\sqrt{2}=\frac{1}{2}{a}_{-}+\frac{\sqrt{3}}{2}{b}_{-},s_{23}{b}_{-}=(\sqrt{3}\mathrm{/}2)a_{-}-(1\mathrm{/}2)b_{-}\mathrm{.}$ These are exactly the two matrices stated.

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 $s_{12}=\mathrm{diag}(-1,1)$ are the two coordinate axes, and $s_{23}$ mixes both axes because its off-diagonal entries are nonzero. Thus this basis displays a spin-$3\mathrm{/}2$ block with trivial exchange and a two-dimensional exchange factor repeated over the two spin-$1\mathrm{/}2$ magnetic states, in the factor placement of [F5].

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