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

Since $S_i^a=\mathrm{\hslash }{\sigma }_i^a\mathrm{/}2$, the factors ${\mathrm{\hslash }}^2$ cancel and ${\mathrm{\hslash }}^{-2}{\sum }_a{S}_i^a{S}_{i+1}^a=\frac{1}{4}{\sum }_a{\sigma }_i^a{\sigma }_{i+1}^a\mathrm{.}$ On the ordered basis $\mathrm{\mid }00⟩,\mathrm{\mid }01⟩,\mathrm{\mid }10⟩,\mathrm{\mid }11⟩$, the operator $\frac{1}{2}(I+{\sum }_a{\sigma }_a\otimes {\sigma }_a)$ sends these four vectors respectively to $\mathrm{\mid }00⟩,\mathrm{\mid }10⟩,\mathrm{\mid }01⟩,\mathrm{\mid }11⟩$: for the middle two, the Pauli sum is $2\mathrm{\mid }10⟩-\mathrm{\mid }01⟩$ and $2\mathrm{\mid }01⟩-\mathrm{\mid }10⟩$. This is exactly the swap, so $s_i=\frac{1}{2}(I+{\sum }_a{\sigma }_i^a{\sigma }_{i+1}^a)$. Rearranging gives $\frac{1}{2}{s}_i-\frac{1}{4}I$, and substituting $e_i=I-s_i$ gives $\frac{1}{4}I-\frac{1}{2}{e}_i$. This proves the bond identity including its normalization.

Each $e_i$ is self-adjoint by [F3], $I$ is self-adjoint, and every ${\mathcal{J}}_i$ is real; hence their finite real linear combination $H$ is self-adjoint. By [F3] each $e_i$ lies in the represented Temperley–Lieb algebra, so the displayed expression for $H$ is an element of that algebra's image.

An adjacent swap commutes with the diagonal action $U^{\otimes N}$ for every $U\in U(2)$: on an elementary tensor, swapping the two entries after applying $U$ to each gives the same tensor as applying $U$ after the swap. Thus $e_i=I-s_i$ and $H$ commute with that action. By [F4], the decomposition places the permutation operators on $S^{\lambda }$ and the collective unitary action on $M_{\lambda }$; therefore the formula for $H$ acts on $S^{\lambda }$ and as identity on $M_{\lambda }$. When $N=1$ there are no bonds and all displayed sums are zero.