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A non-invertible defect on a three-site Ising ring

Physical scope

An exact finite three-site periodic quantum Ising chain at the zero-field self-dual coupling, with an eight-dimensional Hilbert space; no infinite-volume, continuum, or empirical claim is made.

Example

Let $H_{3} = ℓ^{2} (Z_{2}^{3})$ with orthonormal basis $∣ h_{1}, h_{2}, h_{3} ⟩$ , and define $Z_{j} ∣ h ⟩ = (- 1)^{h_{j}} ∣ h ⟩$ and $X_{j} ∣ h ⟩ = ∣ h + e_{j} ⟩$ . Use periodic indices and the finite Hamiltonian $H_{3} = - J \sum_{j = 1}^{3} (Z_{j} Z_{j + 1} + X_{j})$ , where $J > 0$ has energy units and $ℏ = 1$ . Euclidean propagation through $s \geq 0$ is $e^{- s H_{3}}$ . Define $K_{3}$ from this ring to its half-shifted dual by

⟨ \hat{h} ∣ K_{3} ∣ h ⟩ = 2^{- 3 / 2} (- 1)^{\sum_{j = 1}^{3} {\hat{h}}_{j - 1 / 2} (h_{j - 1} + h_{j})}, h_{0} = h_{3},

and use the supplier's reflection identification $R_{3} ∣ \hat{h} ⟩ = ∣ r ⟩$ , $r_{j} = {\hat{h}}_{1 / 2 - j}$ , to set $D_{3} = R_{3} K_{3}$ . Its complete admissible defect-bearing correlators are exactly the finite decorated tensor contractions, ordinary insertions and external boundary tensors of the supplier at $L = 3$ ; allowed slides keep the rest of that complete input fixed. Let $P_{3} = X_{1} X_{2} X_{3}$ and $v = ∣ 000 ⟩ - ∣ 111 ⟩$ . The defect inserted on a Euclidean time slice is topological under the declared slides within gaps of fixed insertions, satisfies $D_{3}^{2} = I + P_{3}$ , and annihilates $v$ after two fusions. Therefore this exact finite-model defect is non-invertible. No continuum or empirical claim is made.

Example references

Aasen, Mong, and Fendley, §3.2

Fröhlich, Fuchs, Runkel, and Schweigert, critical-Ising defect discussion

Facts & Assumptions

Given: Adopt the three-site Hilbert space, Hamiltonian, original-to-dual kernel and reflection identification, Euclidean transfer model and global spin flip in the Example. The defect line is inserted only in a gap that contains no other insertion.

[F1]

Finite-lattice realization of the Ising non-invertible defect: for every finite periodic length $L \geq 2$ in the stated model, $D_{L}$ is topological under the declared slides and $D_{L}^{2} = I + P_{L}$ .

Verification

1.1givenalgebra

The vector $v$ is nonzero because $∣ 000 ⟩$ and $∣ 111 ⟩$ are distinct orthonormal basis vectors. The global spin flip exchanges them, so $P_{3} v = ∣ 111 ⟩ - ∣ 000 ⟩ = - v$ .

2.1F1step 1.1algebra∎

By [F1] at $L = 3$ , $D_{3}^{2} v = (I + P_{3}) v = v - v = 0$ . If $D_{3}$ had a two-sided inverse, then $D_{3}^{2}$ would be invertible and could not annihilate the nonzero vector $v$ ; thus the defect is non-invertible. Its topologicality in every allowed gap follows from the same exact finite transfer-model result.

Depends on

Finite-lattice realization of the Ising non-invertible defect