Burnside's Theorem - Ryder McDowell

The previous showed that the \(\mathbb{Z}_2\) symmetric subalgebra necessarily contains multi-site operators. This post finds them explicitly, first working out the qutrit case concretely and then using Burnside’s theorem to handle general qudits.

The Qutrit Case

For a qutrit (\(d=3\)) spin chain, the \(\mathbb{Z}_2\) symmetry acts as

\[U(a) = \bigotimes_i A_i, \qquad A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix}\]

A matrix \(M\) commutes with \(A\) iff it is block-diagonal of the form

\[M = \begin{pmatrix} * & * & 0 \\ * & * & 0 \\ 0 & 0 & * \end{pmatrix}\]

This is a 5-dimensional complex vector space, decomposing as \(M_2(\mathbb{C}) \oplus \mathbb{C}\). A natural basis for the single-site symmetric algebra is therefore built from the Pauli operators projected into the \(\mathcal{H}_+\) block, plus a projector onto the \(\mathcal{H}_-\) sector:

\[\bar{X} = P_1 X P_1, \quad \bar{Z} = P_1 Z P_1, \qquad P_1 = \mathrm{diag}(1,1,0), \quad P_2 = \mathrm{diag}(0,0,1)\]

So the single-site symmetric algebra is generated by \(\{\bar{X}_i, \bar{Z}_i, P_{2,i} \mid i \in \mathbb{Z}\}\).

But as the previous post showed, this isn’t the whole story on a chain.

General Qudits: Three Types of Generators

For a chain of \(d\)-dimensional qudits with \(\mathbb{Z}_2\) symmetry splitting the on-site Hilbert space as \(\mathcal{H}^{(i)} = \mathcal{H}_+ \oplus \mathcal{H}_-\) (with \(\dim \mathcal{H}_+ = p\), \(\dim \mathcal{H}_- = q\)), we claim the symmetric subalgebra is generated by three types of operators:

On-site operators \(O_i\): operators in \(\mathcal{A}_{\mathbb{Z}_2}^{(i)}\) acting as the identity elsewhere — i.e. \(O_i = \mathbf{1}^{\otimes (i-1)} \otimes O \otimes \mathbf{1}^{\otimes (L-i)}\) with \(O \in \mathcal{A}_{\mathbb{Z}_2}^{single}\).
Charge-hoppers \(C_{i,i+1}\): two-site operators on a bond \((i, i+1)\) that swap which site carries the \(\mathcal{H}_-\) component: \(\mathcal{H}_- \otimes \mathcal{H}_+ \;\leftrightarrow\; \mathcal{H}_+ \otimes \mathcal{H}_-\)
Pair swaps \(S_{i,i+1}\): two-site operators on a bond \((i, i+1)\) that exchange the two parity-even combinations: \(\mathcal{H}_+ \otimes \mathcal{H}_+ \;\leftrightarrow\; \mathcal{H}_- \otimes \mathcal{H}_-\)

Call the subalgebra generated by these \(B\). Since all three types preserve the parity sectors \(\mathcal{H}_{even}\) and \(\mathcal{H}_{odd}\), we have \(B \subset \mathcal{A}_{\mathbb{Z}_2}\). We want to show \(B = \mathcal{A}_{\mathbb{Z}_2}\).

Burnside’s Theorem Does the Work

We use the following version of Burnside’s theorem (see Burnside’s Theorem on Matrix Algebras):

For a finite-dimensional complex vector space \(V\), the only irreducible subalgebra of \(\mathrm{End}(V)\) is \(\mathrm{End}(V)\) itself, where irreducibility is equivalent to transitivity: \(\forall v \in V \setminus \{0\}\), \(Mv = V\).

Since \(\mathcal{A}_{\mathbb{Z}_2} \cong \mathrm{End}(\mathcal{H}_{even}) \oplus \mathrm{End}(\mathcal{H}_{odd})\), it suffices to show \(B\) acts transitively on each sector separately.

Claim: \(\forall \psi \in \mathcal{H}_{even} \setminus \{0\}\), \(B\psi = \mathcal{H}_{even}\), and similarly for \(\mathcal{H}_{odd}\).

Proof sketch (even sector)

Take any nonzero \(\psi \in \mathcal{H}_{even}\). Write it as a superposition of tensor-product basis states, each with an even number of \(\mathcal{H}_-\) sites. Since \(\psi \neq 0\), at least one basis state \(v_k = \bigotimes_j \vert e_j \rangle\) appears with nonzero coefficient \(c_k\).

Using the projectors \(P_k = \bigotimes_j \vert e_j \rangle \langle e_j \vert\) inherited from the on-site generators, we get \(P_k \psi = c_k v_k \neq 0\). Since \(B\) is a vector space and \(\frac{1}{c_k} P_k \in B\), we recover \(v_k \in B\psi\).

Now we show \(v_k\) can reach any other even-sector basis state \(w\) using the generators of \(B\):

If \(v_k\) and \(w\) have the same number of \(\mathcal{H}_-\) sites: use charge-hoppers to slide the \(\mathcal{H}_-\) components where needed, then on-site operators to rearrange within \(\mathcal{H}_+\).
If they have a different number of \(\mathcal{H}_-\) sites (but the same parity, since both are in \(\mathcal{H}_{even}\)): use charge-hoppers to collect \(\mathcal{H}_-\) components onto adjacent bonds, then pair swaps to create or annihilate pairs as needed. Since the counts agree mod 2, this is always achievable.

This constructs an operator \(\tilde{O}_w \in B\) with \(\tilde{O}_w v_k = w\) for every even basis state \(w\). Hence \(\mathcal{H}_{even} \subset Bv_k \subset B\psi\), giving \(B\psi = \mathcal{H}_{even}\).

The odd sector is identical. So \(B\) is transitive on both sectors, hence irreducible in \(\mathcal{A}_{\mathbb{Z}_2}\), and by Burnside’s theorem \(B = \mathcal{A}_{\mathbb{Z}_2}\).

A Note on Unitarity

The on-site operators and charge-hoppers can always be built as unitaries. For the charge-hoppers this works because \(\dim(\mathcal{H}_+ \otimes \mathcal{H}_-) = pq = \dim(\mathcal{H}_- \otimes \mathcal{H}_+)\), so the spaces being swapped have equal dimension.

For the pair swaps, unitarity requires \(\dim(\mathcal{H}_+ \otimes \mathcal{H}_+) = \dim(\mathcal{H}_- \otimes \mathcal{H}_-)\), i.e. \(p^2 = q^2\), i.e. \(p = q\). This is achievable for even-dimensional qudits (choose \(U\) with equal \(\pm 1\) eigenvalues) but not for odd-dimensional qudits like the qutrit, where \(p \neq q\) is unavoidable. The transitivity argument above doesn’t require unitarity, so the generating set still works — but the pair swaps are not unitary in the qutrit case. This is worth keeping in mind when thinking about circuits.

Summary

\(\mathcal{A}_{\mathbb{Z}_2} = \mathrm{span}(O_i, C_j, S_k)\) with \(i \in [1,L]\), \(j \in [1, L-1]\), \(k \in [1, L-2]\).

The argument is dimension-independent: as long as the three types of generators exist (which requires only \(p, q \geq 1\)), they generate the full symmetric subalgebra via Burnside. The only subtlety is that pair swaps can’t be made unitary unless \(p = q\).