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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 $S^2$.

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    $\begingroup$ I'm not convinced the intuition about the value of $s$ corresponding to the "alignment" of spins is correct - consider that for 2 particles, the triplet has a state $\lvert \uparrow\downarrow\rangle + \lvert \downarrow\uparrow\rangle$ which has total spin 1, even though the spins of the individual particles are certainly not aligned in any sense. $\endgroup$
    – ACuriousMind
    Jun 11, 2021 at 22:40
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    $\begingroup$ That's the z-basis, but they are aligned in the x basis: $|\rightarrow\rightarrow\rangle-|\leftarrow\leftarrow\rangle$. More generally for $N$ spins, the maximum-spin subspace ($s=N/2$) is exactly the span of all completely aligned states $|\phi\rangle^{\otimes N}$. There are also precise things you can say about how the intuition becomes exact in the classical limit. $\endgroup$ Jun 11, 2021 at 23:03
  • $\begingroup$ Are you fixing $|\psi_1\rangle$ and $|\psi_2\rangle$? Or asking for which pairs of $s_1$ and $s_2$ there can exist states $|\psi_1\rangle$ and $|\psi_2\rangle$ and some Pauli matrix such that the overlap is nonzero? $\endgroup$ Jun 11, 2021 at 23:33
  • $\begingroup$ Side note: when the spins are formed from symmetrized states of $2S$ qubits it's easier to form ladder operators for $S^2$ (use annihilation or creation operators for one of the modes), but that keeps you in the symmetric subspace. $\endgroup$ Jun 11, 2021 at 23:35
  • $\begingroup$ I don't have enough for an answer, but it seems to me that if you start from the totally antisymmetric state ($s=0$ for $N$ even) and flip one spin, you do get a component along the corresponding totally symmetric ($s=N/2$) states. $\endgroup$
    – fqq
    Jun 12, 2021 at 0:17

2 Answers 2

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Let's write the $N$-particle space $H^{\otimes N}$ as $H^{\otimes N-1}\otimes H$ so that we have the basis $$ \lvert \bar{s},m_{\bar{s}}; m_N\rangle$$ where $\bar{s}$ is the total spin for the $N-1$ particles and $m_N$ the $z$-spin of the $N$-th particle (since there can be more than one copy of any given $\bar{s}$ representation, we would also have to have a marker for that duplicity. I'm omitting it here for convenience of notation as I don't think it changes anything about the argument [Edit: confirmed]). Expressing a state of definite total spin $s$ in terms of this is a standard application of Clebsch-Gordan coefficients: $$ \lvert s,m_s\rangle = \sum_{\bar{s}}\sum_{m_{\bar{s}}} \sum_{m_N}C^{sm_s}_{\bar{s}m_{\bar{s}}\frac{1}{2}m_N}\lvert \bar{s},m_{\bar{s}}; m_N\rangle,\tag{1}$$ where the coefficients are only non-zero for $\lvert \bar{s}-\frac{1}{2}\rvert\leq s\leq \bar{s}+\frac{1}{2}$ and $m_s = m_{\bar{s}} + m_N = m_{\bar{s}} \pm\frac{1}{2}$.

The question is now when $\langle s_1, m_{s_1}\lvert \sigma^x_N \lvert s_2,m_{s_2}\rangle$ is non-zero. In the expansion (1), the action of $\sigma^x_N$ is just to flip $m_N$ but doesn't touch the $\bar{s}$ part. Since the $ \lvert \bar{s},m_{\bar{s}}; m_N\rangle$ are an orthonormal basis, that means this overlap can only be non-zero when the two $\lvert s_i,m_{s_i}\rangle$ have at least one non-zero $\bar{s}$ in common. This can only happen when that $\bar{s}$ fulfills both $\lvert \bar{s}-\frac{1}{2}\rvert\leq s_1\leq \bar{s}+\frac{1}{2}$ and $\lvert \bar{s}-\frac{1}{2}\rvert\leq s_2\leq \bar{s}+\frac{1}{2}$. Write $s_2 = s_1 + x$, then $$ \lvert \bar{s}-\frac{1}{2}\rvert \leq s_1\leq \bar{s}+\frac{1}{2} - x,$$ which means $x$ is at most 1, since $\lvert \bar{s}-\frac{1}{2}\rvert$ and $\bar{s} + \frac{1}{2}$ differ at most by 1. This is what we wanted to show.

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  • $\begingroup$ Thanks, you nailed the key idea. I have handled the representation multiplicity issue in my answer here. $\endgroup$ Jun 12, 2021 at 23:42
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It is indeed the case that if the two $\psi_i$ are in respective spin sectors $s_i$ (i.e., that $S^2|\psi_i\rangle = s_i(s_i+1)|\psi_i\rangle$ for $i=1,2$), then $\langle \psi_1 | \sigma_n^x | \psi_2 \rangle$ can be non-zero only if $|s_1-s_2|=0$ or $1$. The key idea is directly from ACuriousMind's answer, which I encourage you to read first. What follows is just a lengthy expansion of that answer to explicitly handle the multiplicity of identical irreducible representations. I expect there is a much more elegant way to do this, so alternative answers are encourage.


For any Hilbert space $\mathcal{H}_{N} = (\mathcal{H}_{1})^{\otimes N} \cong (\mathbb{C}^2)^{\otimes N} \cong \mathbb{C}^{2^N}$ of $N$ spin-1/2 particles, we can use the structure theorem for finite-dimensional von Neumann algebras to decompose it as \begin{align} \mathcal{H}_{N} \cong \bigoplus_{s={0,1/2}}^{N/2}\left(\mathcal{S}^{(N)}_s \otimes \mathcal{E}^{(N)}_s\right) \end{align} where $\mathcal{S}^{(N)}_s \cong {\mathbb{C}}^{2s+1}$ is an irreducible representation of the SU(2) group generated by $\{S^x,S^y,S^z\}$ and is spanned by the $2s+1$ orthonormal vectors $|s,m\rangle$ for $-s\le m \le s$. The direct sum starts at $s=0$ or $s=1/2$, depending on whether $N$ is even or odd, respectively, and the dimension of $\mathcal{E}^{(N)}_s$ — call it $D(N,s)$ — is just the multiplicity associated with how many copies of the spin-$s$ irrep there are. Any state $|\psi\rangle \in \mathcal{H}_{N}$ can therefore be written as \begin{align} |\psi\rangle &= \sum_{s={0,1/2}}^{N/2} \sum_{m=-s}^{s} \sum_{e=1}^{D(N,s)} \alpha(s,m,e) |s:m:e\rangle \end{align} with $\alpha(s,m,e)\in \mathbb{C}$ and where, using an arbitrary orthonormal basis $\{|s;e\rangle\}$ of $\mathcal{E}^{(N)}_{s}$, we have defined the orthonormal basis \begin{align} |s:m:e\rangle := |s,m\rangle\otimes|s;e\rangle \in \mathcal{S}^{(N)}_{s} \otimes \mathcal{E}^{(N)}_{s}. \end{align} For each fixed value of $s$ and $e$, and for $-s\le m \le s$, this spans a distinct spin-$s$ irrep of SU(2) in $\mathcal{H}_N$. Our assumption that the two $|\psi_i\rangle$ live in sectors with a fixed value $s=s_i$ means that, for them, the respective coefficients $\alpha_i(s,m,e)$ vanish except when $s=s_i$, i.e., the sum over $s$ becomes trivial.

We can alternatively decompose any $|\psi\rangle\in\mathcal{H}_N$ with respect to the tensor structure $\mathcal{H}_{N}= \mathcal{H}_{N-1}\otimes \mathcal{H}_{1}$ to get \begin{align} |\psi\rangle &= \sum_{\bar{s}={0,1/2}}^{(N-1)/2} \sum_{\bar{m}=-\bar{s}}^{\bar{s}} \sum_{f=1}^{D(N-1,\bar{s})} \sum_{\hat{m}=-1/2}^{1/2} \beta(\bar{s},\bar{m},f,\hat{m}) |\bar{s}:\bar{m}:f\rangle_{N-1} \otimes |\hat{m}\rangle_1 \end{align} where $\beta(\bar{s},\bar{m},f,\hat{m}) \in \mathbb{C}$. Here we have used the structure theorem on $\mathcal{H}_{N-1}$ but not $\mathcal{H}_{1}$ since it is trivial for the latter. Parameters associated with $\mathcal{H}_{N-1}$ and $\mathcal{H}_{1}$ are indicated with bars ($\bar{s}$ and $\bar{m}$) and hats ($\hat{m}$) respectively. Analogously to before, we have defined the orthonormal basis \begin{align} |\bar{s}:\bar{m}:f\rangle_{N-1} := |\bar{s},\bar{m}\rangle\otimes|\bar{s};f\rangle \in \mathcal{S}^{(N-1)}_{\bar{s}} \otimes \mathcal{E}^{(N-1)}_{\bar{s}}, \end{align} which, for each fixed value of $\bar{s}$ and $f$, and for $-\bar{s}\le \bar{m} \le \bar{s}$, spans a spin-$\bar{s}$ irrep of SU(2) in $\mathcal{H}_{N-1}$. This can be combined with the orthonormal basis $|\hat{m}\rangle \in \mathcal{H}_1$ for $\hat{m} \in\{-1/2,1/2\}$, which (trivially) spans the lone spin-$1/2$ irrep of SU(2) in $\mathcal{H}_{1}$, to form a tensor-product basis for the joint Hilbert space of the pair of spins. In particular, we can perform the Clebsch–Gordan decomposition \begin{align} |s:m:\bar{s}:f\rangle:= \sum_{\bar{m}=-\bar{s}}^{\bar{s}} \sum_{\hat{m}=-1/2}^{1/2} C^{s,m}_{\bar{s},\bar{m};1/2,\hat{m}}|\bar{s}:\bar{m}:f\rangle_{N-1}\otimes|\hat{m}\rangle_1 \end{align} with each allowed fixed choice of $s$, $\bar{s}$, and $f$ spanning a distinct spin-$s$ irrep of SU(2) in $\mathcal{H}_N$ for $-s \le m \le s$.

Recall that in our first decomposition of $|\psi\rangle$ we chose an arbitrary orthonormal basis $\{|s;e\rangle\}$ for $\mathcal{E}^{(N)}_{s}$; all that has happened is that, by fixing a value of $\bar{s}$ and $f$ in the state above, we have effectively picked out one of the basis vectors. In other words, there is a choice of basis for $\mathcal{E}^{(N)}_{s}$ and a function $e(\bar{s},f)$ such that $|s:m:e(\bar{s},f)\rangle=|s:m:\bar{s}:f\rangle$.

We then have \begin{align} \beta(\bar{s},\bar{m},f,\hat{m}) &= \Big[{}_{N-1} \langle\bar{s}:\bar{m}:f|\otimes{}_{1}\langle\hat{m}|\Big]|\psi\rangle\\ &= \sum_{s={0,1/2}}^{N/2} \sum_{m=-s}^{s} \sum_{e=1}^{D(N,s)} \alpha(s,m,e) \Big[{}_{N-1} \langle \bar{s}:\bar{m}:f | \otimes {}_{1}\langle\hat{m}|\Big] |s:m:e\rangle\\ &= \sum_{s={0,1/2}}^{N/2} \sum_{m=-s}^{s} \alpha(s,m,e(\bar{s},f)) C^{s,m}_{\bar{s},\bar{m};1/2,\hat{m}} \end{align} where to get the first (second) equality we used our first (second) decompositions of $|\psi\rangle$. To get the third equality we expanded in terms of Clebsch–Gordan coefficients and used the mapping $e\to e(\bar{s},f)$. Now, following ACuriousMind, we just make the key observation that the Clebsch–Gordan coefficient $C^{s,m}_{\bar{s},\bar{m};1/2,\hat{m}}$ vanishes unless $\bar{s}= s\pm 1/2$. This means that for $|\psi_i\rangle$, for which $\alpha_i(s,m,e)$ vanishes unless $s=s_i$, we can conclude that $\beta_i(\bar{s},\bar{m},f,\hat{m})$ vanishes unless $\bar{s} = s_i \pm 1/2$.

Finally, using the second decomposition for $|\psi_1\rangle$ and $|\psi_2\rangle$, we see that for any local operator on the $N$th qubit, say $\sigma_N^x$, we have \begin{align} \langle\psi_1|\sigma_N^x|\psi_2\rangle &= \sum_{\bar{s}={0,1/2}}^{(N-1)/2} \sum_{\bar{m}=-\bar{s}}^{\bar{s}} \sum_{f=1}^{D(N-1,\bar{s})} \sum_{\hat{m}_1=-1/2}^{1/2} \sum_{\hat{m}_2=-1/2}^{1/2} \beta_1(\bar{s},\bar{m},f,\hat{m}_1)^* \beta_2(\bar{s},\bar{m},f,\hat{m}_2) {}_1 \langle \hat{m}_1|\sigma_N^x |\hat{m}_2\rangle_1 \end{align} where we have used the orthonormality of $|\bar{s}:\bar{m}:f\rangle_{N-1}$. Having observed that $\beta_i(\bar{s},\bar{m},f,\hat{m})$ vanishes unless $\bar{s}=s_i \pm 1/2$, all terms in the above sum vanish except when $s_1 = s_2-1$, $s_2$, or $s_2+1$.

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