= Solution
Use the $X$ eigenbasis $|s_i\rangle_X$, $s_i\in\mathbb Z_2$, and put a dual qubit on every edge with $t_i=s_i+s_{i+1}\pmod2$. The <Kramers--Wannier intertwiner> is
$$
D=\sum_{s_1,\ldots,s_L}|s_1+s_2,\ldots,s_L+s_1\rangle_Z\langle s_1,\ldots,s_L|_X.
$$
An explicit <matrix product operator> tensor is
$$
\boxed{A^{t,s}_{\alpha\beta}=\delta_{\alpha,s}\,\delta_{t,\alpha+\beta\ ({\rm mod}\ 2)}}.
$$
Contracting neighboring virtual indices $\beta_i=\alpha_{i+1}$ around the ring is the graphical MPO: each tensor copies its input bit to the left virtual leg and outputs the XOR of its two virtual legs.
Directly from the domain-wall definition,
$$
D(X_iX_{i+1})=\widetilde Z_{i+1/2}D,
\qquad
DZ_i=(\widetilde X_{i-1/2}\widetilde X_{i+1/2})D.
$$
Therefore $DH(\lambda)=\widetilde H(\lambda)D$, where
$$
\widetilde H(\lambda)=-\sum_i\widetilde Z_{i+1/2}+\lambda\sum_i\widetilde X_{i-1/2}\widetilde X_{i+1/2}.
$$
After exchanging $X$ and $Z$ and rescaling, this is the same Ising family at reciprocal coupling, so corresponding symmetry sectors have the same spectrum and $H(\lambda)$ is dual to $|\lambda|H(1/\lambda)$ up to the elementary sign conventions. In particular the Ising spectrum is self-dual at $|\lambda|=1$.
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