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edited Dec 20, 2019 at 14:35

Given a sequence of real symmetric $(2^N\times 2^N)$- matrices $(H_N)_{N\in\mathbb{N}}$ on the $N$-fold tensor product of $\mathbb{C}^2$ with itself, such that \begin{align} \lim_{N\to\infty}||[H_N,S_N]||_N=0, \end{align} where $||\cdot||_N$ denotes the operator norm on $B(\bigotimes_{n=1}^N\mathbb{C}^2)$ (the space of bounded liner operators on $\bigotimes_{n=1}^N\mathbb{C}^2$), $[\cdot,\cdot]$ the usual commutator, and the matrix $S_N$ denotes the symmetrization (projection) operator given by linear extension of the following map on elementary tensors: \begin{align}S_N (v_1 \otimes \cdots \otimes v_N) = \frac{1}{N!} \sum_{\sigma \in {\cal P}(N)} v_{\sigma(1)} \otimes \cdots \otimes v_{\sigma(N)}. \end{align}\begin{align}S_N (v_1 \otimes \cdots \otimes v_N) = \frac{1}{N!} \sum_{\sigma \in {\cal P}(N)} v_{\sigma(1)} \otimes \cdots \otimes v_{\sigma(N)}, \end{align} where ${\cal P}(N)$ denotes the symmetric group. It is known that for all $N\in\mathbb{N}$ the ground state eigenvector $\psi_N$ of $H_N$ (i.e. the eigenvector corresponding to two lowest eigenvalue of $H_N$) is unique and has strictly positive components. My question is if we can conclude that \begin{align} \lim_{N\to\infty}||S_N\psi_N-\psi_N||_N=0, \end{align} where the latter norm is considered on the space $\bigotimes_{n=1}^N\mathbb{C}^2$.

Given a sequence of real symmetric $(2^N\times 2^N)$- matrices $(H_N)_{N\in\mathbb{N}}$ on the $N$-fold tensor product of $\mathbb{C}^2$ with itself, such that \begin{align} \lim_{N\to\infty}||[H_N,S_N]||_N=0, \end{align} where $||\cdot||_N$ denotes the operator norm on $B(\bigotimes_{n=1}^N\mathbb{C}^2)$ (the space of bounded liner operators on $\bigotimes_{n=1}^N\mathbb{C}^2$), $[\cdot,\cdot]$ the usual commutator, and the matrix $S_N$ denotes the symmetrization (projection) operator given by linear extension of the following map on elementary tensors: \begin{align}S_N (v_1 \otimes \cdots \otimes v_N) = \frac{1}{N!} \sum_{\sigma \in {\cal P}(N)} v_{\sigma(1)} \otimes \cdots \otimes v_{\sigma(N)}. \end{align} It is known that for all $N\in\mathbb{N}$ the ground state eigenvector $\psi_N$ of $H_N$ is unique and has strictly positive components. My question is if we can conclude that \begin{align} \lim_{N\to\infty}||S_N\psi_N-\psi_N||_N=0, \end{align} where the latter norm is considered on the space $\bigotimes_{n=1}^N\mathbb{C}^2$.

Given a sequence of real symmetric $(2^N\times 2^N)$- matrices $(H_N)_{N\in\mathbb{N}}$ on the $N$-fold tensor product of $\mathbb{C}^2$ with itself, such that \begin{align} \lim_{N\to\infty}||[H_N,S_N]||_N=0, \end{align} where $||\cdot||_N$ denotes the operator norm on $B(\bigotimes_{n=1}^N\mathbb{C}^2)$ (the space of bounded liner operators on $\bigotimes_{n=1}^N\mathbb{C}^2$), $[\cdot,\cdot]$ the usual commutator, and the matrix $S_N$ denotes the symmetrization (projection) operator given by linear extension of the following map on elementary tensors: \begin{align}S_N (v_1 \otimes \cdots \otimes v_N) = \frac{1}{N!} \sum_{\sigma \in {\cal P}(N)} v_{\sigma(1)} \otimes \cdots \otimes v_{\sigma(N)}, \end{align} where ${\cal P}(N)$ denotes the symmetric group. It is known that for all $N\in\mathbb{N}$ the ground state eigenvector $\psi_N$ of $H_N$ (i.e. the eigenvector corresponding to two lowest eigenvalue of $H_N$) is unique and has strictly positive components. My question is if we can conclude that \begin{align} \lim_{N\to\infty}||S_N\psi_N-\psi_N||_N=0, \end{align} where the latter norm is considered on the space $\bigotimes_{n=1}^N\mathbb{C}^2$.

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asked Dec 20, 2019 at 13:19

Kris

On norm convergence of a commutator and implications

Given a sequence of real symmetric $(2^N\times 2^N)$- matrices $(H_N)_{N\in\mathbb{N}}$ on the $N$-fold tensor product of $\mathbb{C}^2$ with itself, such that \begin{align} \lim_{N\to\infty}||[H_N,S_N]||_N=0, \end{align} where $||\cdot||_N$ denotes the operator norm on $B(\bigotimes_{n=1}^N\mathbb{C}^2)$ (the space of bounded liner operators on $\bigotimes_{n=1}^N\mathbb{C}^2$), $[\cdot,\cdot]$ the usual commutator, and the matrix $S_N$ denotes the symmetrization (projection) operator given by linear extension of the following map on elementary tensors: \begin{align}S_N (v_1 \otimes \cdots \otimes v_N) = \frac{1}{N!} \sum_{\sigma \in {\cal P}(N)} v_{\sigma(1)} \otimes \cdots \otimes v_{\sigma(N)}. \end{align} It is known that for all $N\in\mathbb{N}$ the ground state eigenvector $\psi_N$ of $H_N$ is unique and has strictly positive components. My question is if we can conclude that \begin{align} \lim_{N\to\infty}||S_N\psi_N-\psi_N||_N=0, \end{align} where the latter norm is considered on the space $\bigotimes_{n=1}^N\mathbb{C}^2$.