We’ve all had it happen: in a frenzied attempt to prove some major keystone result, we get stuck on a seemingly necessary technical lemma only to, after successful proof, discover it was unnecessary all along! In order to at least feel like the effort wasn’t wasted, I’ve converted one such lemma into the following blog post.
Let \(\mathfrak{D}(\mathcal{H})\) be the set of density matrices on Hilbert space \(\mathcal{H}.\) Let \(T\in U(\mathcal{H}^{\otimes N})\) be defined by \(T(\bigotimes_{i\in \mathbb{Z}_N} \ket{\psi_{i}}) := \bigotimes_{i\in\mathbb{Z}_N}\ket{\psi_{i-1}}.\) Note that \(T\) furnishes \(\mathcal{H}^{\otimes N}\) with a unitary representation of \(\mathbb{Z}_N.\)
We wish to investigate the following scenario: Let \(\rho\in \mathfrak{D}(\mathcal{H}^{\otimes N})\) be a translationally invariant state, i.e., one such that \(T\rho T^\dagger = \rho.\) Let \(\rho^{(1)}\) be the reduced density matrix of \(\rho\) to any given site (this is well-defined due to translation invariance). Further, suppose that \((\ker\rho^{(1)})^\perp = V\subset \mathcal{H},\) i.e., that \(\rho^{(1)}\) is supported on some strict subspace of \(\mathcal{H}.\) Another way to think of this is that all of the positive-eigenvalue eigenvectors of \(\rho^{(1)}\) are contained in \(V.\)
What we wish to show is that \(\rho\) is supported on \(V^{\otimes N}.\) To begin, we diagonalize \(\rho:\)
\(\rho = \sum_{i} p_i \ket{p_i} \bra{p_i}.\) Note that by Schur’s lemma each \(\ket{p_i}\) is an \(\omega_N^{k_i}\)-eigenstate of \(T\) for some \(k_i \in \mathbb{Z}_N,\) where \(\omega = e^{\frac{2\pi i}{N}}.\) This in turn implies that \(\ket{p_i} \bra{p_i}\) is translation-invariant. Now, choose some site \(m^*\) and Schmidt decompose \(\ket{p_i}\) over that site
$$\begin{aligned}\ket{p_i} &= \sum_{j} \alpha^{(i)}_j\ket{u^{(i)}_j}\otimes \ket{v^{(i)}_j}\\ \ket{p_i}\bra{p_i}&=\sum_{j,k}\alpha^{(i)}_j\alpha^{(i)}_k\ket{u^{(i)}_j}\bra{u^{(i)}_k}\otimes \ket{v^{(i)}_j}\bra{v^{(i)}_k}\\ \rho &= \sum_{i,j,k} p_i\alpha^{(i)}_j\alpha_k^{(i)}\ket{u^{(i)}_j}\bra{u^{(i)}_k}\otimes \ket{v^{(i)}_j}\bra{v^{(i)}_k}\\ \rho^{(1)}&= \sum_{i, j} p_i \left(\alpha_j^{(i)}\right)^2 \ket{v_j^{(i)}}\bra{v_j^{(i)}}\end{aligned}$$
Now, suppose that \(\rho\) has support outside of \(V^{\otimes N}.\) This means that \((\ker\rho)^\perp \not\subset V^{\otimes N}.\) In other words, there is some subspace \(W\subseteq \left(V^{\otimes N}\right)^\perp\) such that \(W\subseteq (\ker \rho)^\perp.\) Let
$$S_b:=\begin{cases}V & b=0\\
V^\perp & b=1\end{cases}.$$
For \(\mathbf{b}\in\{0,1\}^N,\) let \(\Pi_{\mathbf{b}}\) be the orthogonal projector onto \(\bigotimes_{i=1}^{N} S_{b_i}\subset \mathcal{H}^{\otimes N}\) Then, we have that \(I=\sum_{\mathbf{b}\in\{0,1\}^N} \Pi_{\mathbf{b}}.\) Then, asserting that \(\rho\) has support outside of \(V^{\otimes N}\) is equivalent to asserting that \(\rho(I-\Pi_{\mathbf{0}})\neq 0.\) This, in turn, is equivalent to asserting that there exists some bit string \(\mathbf{b}\) of Hamming weight \(\geq 1\) such that \(\rho\Pi_{\mathbf{b}}\neq 0.\)
Now, note that by the translation-invariance of \(\rho,\) \(\rho \Pi_{\mathbf{b}}\neq 0\) implies that \(\rho\Pi_{T^{q}\mathbf{b}}\neq 0\) for all \(q\in \mathbb{Z}_N,\) where \(T\mathbf{b}\) is the bitwise cyclic shift of \(\mathbf{b}.\) Redefine \(\mathbf{b}\) to be some \(T^q\mathbf{b}\) of itself such that \((T^{q}\mathbf{b})_{m^*}=1.\)
Thus, we have some \(\mathbf{b}\) with \(b_{m^*}=1\) such that \(\rho \Pi_{\mathbf{b}} \neq 0.\) This in turn implies that for at least one \(i,j,\) \(\Pi_{\mathbf{b}}(\ket{u^{(i)}_j}\otimes \ket{v^{(i)}_j})\neq 0.\) As \(\Pi_{\mathbf{b}} = \bigotimes_i P_{S_{b_i}},\) where \(P_{S}\) is the orthogonal projector onto \(S,\) we see that this implies that \(P_{V^\perp} \ket{v_j^{(i)}}\neq 0.\) This would in turn imply that \(\rho^{(1)}P_{V^\perp}\neq 0,\) contradicting the assumption that \(\rho^{(1)}\) is completely supported in \(V.\)