# Given $N$ spins, which eigenspaces of total angular momentum $S^2$ are linked by flipping a single spin?

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 $$S^2$$.

• 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. Jun 11, 2021 at 22:40
• 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. Jun 11, 2021 at 23:03
• 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? Jun 11, 2021 at 23:33
• 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. Jun 11, 2021 at 23:35
• 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.
– fqq
Jun 12, 2021 at 0:17

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.

• Thanks, you nailed the key idea. I have handled the representation multiplicity issue in my answer here. Jun 12, 2021 at 23:42

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 $$s_2+1$$.