Suppose we have $N$ spin-1/2 particles and let $\vec{S} = \frac{\hbar}{2}\sum_{n=1}^N\vec{\sigma_n}$ be the total spin vector operator, with $\vec{\sigma}_n = (\sigma_n^x,\sigma_n^y,\sigma_n^z)$ the Pauli spin matrices of the $n$th spin. It's well known that the squared spin operator $S^2 = \vec{S}\cdot \vec{S}$ takes eigenvalues $s(s+1)\hbar$ for $s=0,1,\ldots,N/2$ if $N$ is even and $s=1/2,3/2,\ldots,N/2$ if $N$ is odd.

My question: Which of these eigenspaces, labeled by $s$, are connected by a single Pauli matrix for a single site? That is, if $|\psi_1\rangle$ and $|\psi_2\rangle$ are eigenstates of $S^2$ with eigenvalues $s_1(s_1+1)\hbar$ and $s_2(s_2+1)\hbar$, for what values of $s_1$ and $s_2$ can $\langle\psi_1|\sigma_n^x|\psi_2\rangle$ be non-zero?

Small values of $s$ correspond to states where roughly as many spins are pointing in one direction as another, whereas large values of $s$ correspond to states where the spins are aligned in the same direction, so intuitively $\sigma_n^x$ (which flips the $n$th spin in the basis of $\sigma_n^z$ eigenstates) should only connect $s_1$ to $s_2=s_1-1,s_1,s_1+1$. I'm not seeing a quick proof though, and, unlike for $S^z$, it does not seem easy to construct raising and lowering operators for $S^2$.

Suppose we have $N$ spin-1/2 particles and let $\vec{S} = \frac{\hbar}{2}\sum_{n=1}^N\vec{\sigma_n}$ be the total spin vector operator, with $\vec{\sigma}_n = (\sigma_n^x,\sigma_n^y,\sigma_n^z)$ the Pauli spin matrices of the $n$th spin. It's well known that the squared spin operator $S^2 = \vec{S}\cdot \vec{S}$ takes eigenvalues $s(s+1)\hbar$ for $s=0,1,\ldots,N/2$ if $N$ is even and $s=1/2,3/2,\ldots,N/2$ if $N$ is odd.

My question: Which of these eigenspaces, labeled by $s$, are connected by a single Pauli matrix for a single site? That is, if $|\psi_1\rangle$ and $|\psi_2\rangle$ are arbitrary eigenstates of $S^2$ with eigenvalues $s_1(s_1+1)\hbar$ and $s_2(s_2+1)\hbar$, for what values of $s_1$ and $s_2$ can $\langle\psi_1|\sigma_n^x|\psi_2\rangle$ ever be non-zero?

Small values of $s$ correspond to states where roughly as many spins are pointing in one direction as another, whereas large values of $s$ correspond to states where the spins are aligned in the same direction, so intuitively $\sigma_n^x$ (which flips the $n$th spin in the basis of $\sigma_n^z$ eigenstates) should only connect $s_1$ to $s_2=s_1-1,s_1,s_1+1$. I'm not seeing a quick proof though, and, unlike for $S^z$, it does not seem easy to construct raising and lowering operators for $S^2$.

Suppose we have $N$ spin-1/2 particles and let $\vec{S} = \frac{\hbar}{2}\sum_{n=1}^N\vec{\sigma_n}$ be the total spin vector operator, with $\vec{\sigma}_n = (\sigma_n^x,\sigma_n^y,\sigma_n^z)$ the Pauli spin matrices of the $n$th spin. It's well known that the squared spin operator $S^2 = \vec{S}\cdot \vec{S}$ takes eigenvalues $s(s+1)\hbar$ for $s=0,1,\ldots,N/2$ if $N$ is even and $s=1/2,3/2,\ldots,N/2$ if $N$ is odd.

My question: Which of these eigenspaces, labeled by $s$, are connected by a single Pauli matrix for a single site? That is, if $|\psi_1\rangle$ and $|\psi_2\rangle$ are eigenstates of $S^2$ with eigenvalues $s_1(s_1+1)\hbar$ and $s_2(s_2+1)\hbar$, for what values of $s_1$ and $s_2$ can $\langle\psi_1|\sigma_n^x|\psi_2\rangle$ be non-zero?

Small values of $s$ correspond to states where roughly as many spins are pointing in one direction as another, whereas large values of $s$ correspond to states where the spins are aligned in the same direction, so intuitively $\sigma_n^x$ (which flips the $n$th spin in the basis of $\sigma_n^z$ eigenstates) should only connect $s$ to $s'=s-1,s,s+1$. I'm not seeing a quick proof though, and, unlike for $S^z$, it does not seem easy to construct raising and lowering operators for $S^2$.

Suppose we have $N$ spin-1/2 particles and let $\vec{S} = \frac{\hbar}{2}\sum_{n=1}^N\vec{\sigma_n}$ be the total spin vector operator, with $\vec{\sigma}_n = (\sigma_n^x,\sigma_n^y,\sigma_n^z)$ the Pauli spin matrices of the $n$th spin. It's well known that the squared spin operator $S^2 = \vec{S}\cdot \vec{S}$ takes eigenvalues $s(s+1)\hbar$ for $s=0,1,\ldots,N/2$ if $N$ is even and $s=1/2,3/2,\ldots,N/2$ if $N$ is odd.

My question: Which of these eigenspaces, labeled by $s$, are connected by a single Pauli matrix for a single site? That is, if $|\psi_1\rangle$ and $|\psi_2\rangle$ are eigenstates of $S^2$ with eigenvalues $s_1(s_1+1)\hbar$ and $s_2(s_2+1)\hbar$, for what values of $s_1$ and $s_2$ can $\langle\psi_1|\sigma_n^x|\psi_2\rangle$ be non-zero?

Small values of $s$ correspond to states where roughly as many spins are pointing in one direction as another, whereas large values of $s$ correspond to states where the spins are aligned in the same direction, so intuitively $\sigma_n^x$ (which flips the $n$th spin in the basis of $\sigma_n^z$ eigenstates) should only connect $s_1$ to $s_2=s_1-1,s_1,s_1+1$. I'm not seeing a quick proof though, and, unlike for $S^z$, it does not seem easy to construct raising and lowering operators for $S^2$.