Automorphism Action on String Operators

Set-up

\(\beta\) is an LP *-automorphism, \(\beta:\mathcal{A}_{sym}\vert_I\) \(\rightarrow\mathcal{A}_{sym}\vert_I\), where \(\mathcal{A}_{sym}\vert_I\) is the symmetric subalgebra restricted to a finite interval \(I\) of lattice sites on a 1D infinite spin chain.

In our \(\text{sym} = \mathbb{Z}_2\) case, our total algebra \(\mathcal{A} = \langle X_i, Z_i \rangle\) with the \(\mathbb{Z}_2\) symmetric subalgebra \(\mathcal{A}_{\mathbb{Z}_2} = \langle X_i, Z_{i,i+1} \rangle\) with \(i \in \mathbb{Z}\).

We define the string-like operators:

\[\bar{S}_x = \prod_{i\in[-n,n]} X_{i} \qquad \bar{S}_{zz} = \prod_{i\in[-n,n]} Z_{i, i+1}\]

where the endpoints of the strings lie far outside the interval \(I\). We claim:

\[\beta(\bar{S}_x)=\pm\bar{S}_x \qquad \beta(\bar{S}_{zz})=\pm\bar{S}_{zz}\]

Approach. We use the fact that *-automorphisms preserve centers and the identity.

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Lemma 1

Lemma 1. Let \(\bar{S}_x\vert_I = \prod_{i\in I} X_i\). The center of our \(\mathbb{Z}_2\)-symmetric subalgebra \(\mathcal{A}_{\mathbb{Z}_2}\vert_I = \langle X_i, Z_{j,j+1} \mid i \in [1,n] = I,\; j \in [1,n-1] \rangle\) is generated by this restricted X string operator, i.e. \(Z(\mathcal{A}_{\mathbb{Z}_2}\vert_I) = \text{span}\{\mathbf{e},\, \bar{S}_X\vert_I\}\)

Proof

Step 1. With \(\mathcal{A}_{\mathbb{Z}_2} = \langle X_i, Z_{j,j+1} \rangle\), we know for \(a \in \mathcal{A}_{\mathbb{Z}_2}\):

\[a = \sum_{\alpha} c_{\alpha}W_{\alpha}\]

it is a linear combination of words from the presentation (\(W_{\alpha} = s_1 s_2 \cdots s_m\) with \(s_k \in \{X_i, Z_{i,i+1}\}\)) and weights (\(c_{\alpha} \in \mathbb{C}\)). Now, since each \(Z_{i,i+1} = Z_i Z_{i+1}\), we can further say that

\[W_{\alpha} = s_1 s_2 \cdots s_m \quad \text{with } s_k \in \{X_i, Z_i\}\]

(Since we are dealing with algebras over a field, this is just saying that the words are all possible operators formed by generators of the algebra, and any element in the algebra is some superposition of these words with some complex coefficient.)

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Step 2. By commuting generators and using \(X_i^2 = Z_i^2 = \mathbf{1}\), we can rewrite each word \(W_\alpha\) in reduced form:

\[\widetilde{W}_\alpha = \prod_{i\in S_\alpha} X_i \prod_{i\in T_\alpha} Z_i\]

where \(S, T\) denote the subsets of lattice sites with \(X_i\) and \(Z_i\) generators on them. In doing so we pick up at most a sign, i.e. \(W_\alpha = \pm \widetilde{W}_\alpha\), after applying the relations: \(X_i Z_i = -Z_i X_i, \quad X_i^2 = Z_i^2 = 1, \quad [X_i, X_j] = [Z_i, Z_j] = 0\)

In sum,

\[a = \sum_{\alpha} c_{\alpha}W_{\alpha} = \sum_{\alpha} c_{\alpha}'\,\widetilde{W}_{\alpha}, \qquad c_{\alpha}' = \pm c_{\alpha}\]

Example. Let \(a = c_1(X_2 X_4 Z_2 Z_3 X_5 Z_3 Z_4) + c_2(Z_4 Z_5 X_4)\).

First word: \(\begin{aligned} X_2 X_4 Z_2 Z_3 X_5 Z_3 Z_4 &= X_2 Z_2 X_4 Z_3 X_5 Z_3 Z_4 && \text{(commute $X_4$ past $Z_2$)}\\ &= -Z_2 X_2 X_4 Z_3 X_5 Z_3 Z_4 && \text{($X_i Z_i = -Z_i X_i$)}\\ &= -Z_2 X_2 X_4 X_5 Z_3 Z_3 Z_4 && \text{(commute $Z_3$ past $X_5$)}\\ &= -Z_2 X_2 X_4 X_5 Z_4 && \text{($Z_3^2 = \mathbf{1}$)}\\ &= X_2 Z_2 X_4 X_5 Z_4 && \text{($X_i Z_i = -Z_i X_i$)}\\ &= X_2 X_4 X_5 Z_2 Z_4 && \text{(commute $Z_2$ past $X_4, X_5$)} \end{aligned}\)

Second word: \(Z_4 Z_5 X_4 = Z_4 X_4 Z_5 = -X_4 Z_4 Z_5\)

So in reduced form: \(a = c_1\, X_2 X_4 X_5 Z_2 Z_4 - c_2\, X_4 Z_4 Z_5\)

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Step 3. Since our total algebra on the interval is \(\mathcal{A}\vert_I \subset \bigotimes_{i \in I} M_2(\mathbb{C})\), and the Pauli operators \(\{\mathbf{e}, X_i, Z_i, Y_i = iX_iZ_i\}\) form a basis of \(M_2(\mathbb{C})\) at each site, their tensor products form a basis of \(\bigotimes_{i\in I} M_2(\mathbb{C})\).

Each reduced word \(\widetilde{W}_\alpha\) is a tensor product:

\[W = \bigotimes_{i \in I} O_i, \qquad O_i = \begin{cases} \mathbf{1} & i \notin S \text{ and } i \notin T \\ X_i & i \in S \text{ and } i \notin T \\ Z_i & i \notin S \text{ and } i \in T \\ X_i Z_i & i \in S \text{ and } i \in T \end{cases}\]

Since these reduced words exhaust all combinations of single-site basis elements and are all distinct, they are linearly independent.

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Step 4. Since \(\widetilde{W}_\alpha\) form a basis, \(a\) is central if and only if every \(\widetilde{W}_\alpha\) with \(c'_\alpha \neq 0\) is central. So it suffices to determine for which \(S, T \subset I\) a single word

\[W = \prod_{i\in S} X_i \prod_{i\in T} Z_i\]

satisfies \([W, X_j] = 0\) and \([W, Z_{j,j+1}] = 0\) for all \(j\) with \(\{j, j+1\} \subset I\).

Commutator with \(X_j\): Using that \(X_j\) commutes with \(X_i, Z_i\) for \(i \neq j\) and anticommutes with \(Z_j\),

\[X_j W = (-1)^{\mathbf{1}_{j\in T}}\, W X_j\]

Thus \([W, X_j] = 0\) for all \(j \in I\) iff \(T \cap I = \varnothing\), i.e. \(T = \varnothing\). So any central word must be of the form:

\[W = \prod_{i\in S} X_i, \qquad S \subset I\]

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Step 5. Commutator with \(Z_{j,j+1} = Z_j Z_{j+1}\): Each of \(Z_j\) and \(Z_{j+1}\) anticommutes with \(X_j, X_{j+1}\) and commutes with all other \(X_i\), so

\[Z_{j,j+1}\, W = (-1)^{\mathbf{1}_{j\in S} + \mathbf{1}_{j+1\in S}}\, W Z_{j,j+1}\]

Therefore \([W, Z_{j,j+1}] = 0\) iff \(\mathbf{1}_{j\in S} + \mathbf{1}_{j+1\in S} \equiv 0 \pmod{2}\), i.e. either both \(j, j+1\) lie in \(S\) or both lie in \(S^c\). This forces either \(S = \varnothing\) or \(S = I\).

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Thus the only words commuting with all generators are

\[W = \mathbf{e} \quad\text{and}\quad W = \prod_{i\in I} X_i = \bar{S}_X\vert_I\]

and we obtain:

\[\boxed{Z\!\left(\mathcal{A}_{\mathbb{Z}_2}\vert_I\right) = \text{span}\{\mathbf{e},\, \bar{S}_X\vert_I\}} \qquad \square\]

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Claim: \(\beta(\bar{S}_x) = \pm\bar{S}_x\)

By Lemma 1, \(Z(\mathcal{A}_{\mathbb{Z}_2}\vert_I) = \langle \mathbf{1}, \bar{S}_x\vert_I \rangle\).

A *-automorphism preserves the center and the identity, so

\[\beta(\bar{S}_x\vert_I) = a\,\mathbf{1} + b\,\bar{S}_x\vert_I\]

for some \(a, b \in \mathbb{C}\). Set \(S := \bar{S}_x\vert_I\). Since \(S^* = S\) and \(S^2 = \mathbf{1}\),

\[\beta(S)^2 = \beta(S^2) = \beta(\mathbf{1}) = \mathbf{1}\]

Thus

\[\beta(S)^2 = (a\mathbf{1} + bS)^2 = (a^2 + b^2)\mathbf{1} + 2ab\,S = \mathbf{1}\]

So \(2ab = 0\) and \(a^2 + b^2 = 1\). Since \(S\) is not a scalar multiple of the identity and \(\beta\) is an automorphism, we must have \(b \neq 0\), forcing \(a = 0\) and \(b = \pm 1\). Therefore:

\[\boxed{\beta(\bar{S}_x\vert_I) = \pm\,\bar{S}_x\vert_I} \qquad \square\]

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Lemma 2

To approach the effect of \(\beta\) on \(\bar{S}_{zz}\), we consider a slightly different \(\mathbb{Z}_2\) subalgebra:

\[\mathcal{A}_{\mathbb{Z}_2}'\vert_I = \langle X_i, Z_{i,i+1} \mid i \in [2,n-1],\; j \in [1,n-1] \rangle\]

Lemma 2. The center of \(\mathcal{A}_{\mathbb{Z}_2}'\vert_I\) is generated by the restricted Z string operator: \(Z(\mathcal{A}_{\mathbb{Z}_2}'\vert_I) = \text{span}\{\mathbf{e},\, \bar{S}_{zz}\vert_I\}\)

Proof

For any \(a \in \mathcal{A}_{\mathbb{Z}_2}'\) we have the reduced form

\[\widetilde{W}_\alpha = \prod_{i\in S_\alpha} X_i \prod_{i\in T_\alpha} Z_i\]

with \(S_\alpha \subset \{2, \ldots, n-1\}\) and \(T_\alpha \subset \{1, \ldots, n\}\). By the same argument as before, these reduced words are linearly independent.

X commutation condition. For \(j \in \{2, \ldots, n-1\}\):

\[X_j W = (-1)^{\mathbf{1}_{j\in T}}\, W X_j\]

So \([W, X_j] = 0\) for all \(j = 2, \ldots, n-1\) iff \(T \subset \{1, n\}\).

Z bond commutation condition. By the same analysis:

\[Z_{j,j+1} W = (-1)^{\mathbf{1}_{j\in S} + \mathbf{1}_{j+1\in S}}\, W Z_{j,j+1}\]

So \([W, Z_{j,j+1}] = 0\) for all \(j = 1, \ldots, n-1\) iff \(\mathbf{1}_{j\in S} + \mathbf{1}_{j+1\in S} \equiv 0 \pmod{2}\) for all \(j\).

But \(S \subset \{2, \ldots, n-1\}\), so:

• For \(j = 1\): \(\mathbf{1}_{1\in S} + \mathbf{1}_{2\in S} = 0\). Since \(1 \notin S\), this forces \(2 \notin S\).
• For \(j = 2\): now \(2 \notin S\) forces \(3 \notin S\).
• Continuing inductively: \(S = \varnothing\).

Hence no \(X_i\) can appear in a central word, and any central \(W\) has the form \(W = \prod_{i\in T} Z_i\) with \(T \subset \{1, n\}\) and \(\vert T \vert\) even (since the generators are Z-bonds, not on-site \(Z_i\)). The only possibilities are:

\[W = \mathbf{e} \quad\text{or}\quad W = Z_1 Z_n\]

Now \(Z_1 Z_n \in \mathcal{A}_{\mathbb{Z}_2}'\vert_I\) since:

\[\prod_{j=1}^{n-1} Z_{j,j+1} = \prod_{j=1}^{n-1}(Z_j Z_{j+1}) = Z_1 Z_n\]

as each interior \(Z_2, \ldots, Z_{n-1}\) appears twice and squares to \(1\). So

\[Z_1 Z_n = \bar{S}_{zz}\vert_I = \prod_{j=1}^{n-1} Z_{j,j+1} \in \mathcal{A}_{\mathbb{Z}_2}'\vert_I\]

Hence:

\[\boxed{Z(\mathcal{A}_{\mathbb{Z}_2}'\vert_I) = \text{span}\{\mathbf{e},\, \bar{S}_{zz}\vert_I\}} \qquad \square\]

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Claim: \(\beta(\bar{S}_{zz}) = \pm\bar{S}_{zz}\)

By Lemma 2, \(Z(\mathcal{A}_{\mathbb{Z}_2}'\vert_I) = \text{span}\{\mathbf{e},\, \bar{S}_{zz}\vert_I\}\).

A *-automorphism preserves the center and the identity, so

\[\beta(\bar{S}_{zz}\vert_I) = a\,\mathbf{e} + b\,\bar{S}_{zz}\vert_I\]

for some \(a, b \in \mathbb{C}\). Set \(S := \bar{S}_{zz}\vert_I\). Since \(S^* = S\) and \(S^2 = \mathbf{1}\),

\[\beta(S)^2 = \beta(S^2) = \beta(\mathbf{e}) = \mathbf{e}\]

Thus

\[\beta(S)^2 = (a\,\mathbf{e} + bS)^2 = (a^2 + b^2)\,\mathbf{e} + 2ab\,S = \mathbf{e}\]

So \(2ab = 0\) and \(a^2 + b^2 = 1\). Since \(S\) is not a scalar multiple of the identity and \(\beta\) is an automorphism, \(b \neq 0\), forcing \(a = 0\) and \(b = \pm 1\). Therefore:

\[\boxed{\beta(\bar{S}_{zz}\vert_I) = \pm\,\bar{S}_{zz}\vert_I} \qquad \square\]

By Lemma 2, \(Z(\mathcal{A}_{\mathbb{Z}_2}'\vert_I)\) = \(\text{span}\{\mathbf{e}, \bar{S}_{zz_{I}}\}\).

A *-automorphism preserves the center and the identity, so

\[\beta(\bar{S}_{zz}\vert_I) = a\,\mathbf{e} + b\,\bar{S}_{zz}\vert_I\]

for some \(a, b \in \mathbb{C}\). Set \(S := \bar{S}_{zz}\vert_I\). Since \(S^* = S\) and \(S^2 = \mathbf{1}\),

\[\beta(S)^2 = \beta(S^2) = \beta(\mathbf{e}) = \mathbf{e}\]

Thus

\[\beta(S)^2 = (a\,\mathbf{e} + bS)^2 = (a^2 + b^2)\,\mathbf{e} + 2ab\,S = \mathbf{e}\]

So \(2ab = 0\) and \(a^2 + b^2 = 1\). Since \(S\) is not a scalar multiple of the identity and \(\beta\) is an automorphism, \(b \neq 0\), forcing \(a = 0\) and \(b = \pm 1\). Therefore:

\[\boxed{\beta(\bar{S}_{zz}\vert_I) = \pm\,\bar{S}_{zz}\vert_I} \qquad \square\]