Addition of two spin 1/2 operators along an arbitrary direction

Question

I'm trying to figure out how the sum of two spin 1/2 operators along an arbitrary direction would work.

These operators are of the form \begin{equation} \textbf{S}_j \cdot \hat{\textbf{n}}_j = \frac{\hbar}{2} \begin{pmatrix} \cos (\theta_j) & e^{-i \varphi_j} \sin (\theta_j)\\ e^{i \varphi_j} \sin (\theta_j) & -\cos (\theta_j) \end{pmatrix}_j \end{equation} where $j = 1, 2$, $\textbf{S}_j = \frac{\hbar}{2} \boldsymbol{\sigma}_j$, and $\hat{\textbf{n}}_j = \sin (\theta_j) \cos (\varphi_j) \hat{\textbf{i}} + \sin (\theta_j) \sin (\varphi_j) \hat{\textbf{j}} + \cos (\theta_j) \hat{\textbf{k}}$. The eigenvalues of these operators are $\pm \frac{\hbar}{2}$ which I'll identify with $\pm$ respectively. Then, the eigenvectors of these eigenvalues are \begin{align*} |\textbf{S}_j \cdot \textbf{n}_j, + \rangle &= \cos \Big( \frac{\theta_j}{2} \Big) |S_z, + \rangle^{(j)} + e^{i \varphi_j} \sin \Big( \frac{\theta_j}{2} \Big) |S_z, - \rangle^{(j)}\\ |\textbf{S}_j \cdot \textbf{n}_j, - \rangle &= \sin \Big( \frac{\theta_j}{2} \Big) |S_z, + \rangle^{(j)} - e^{i \varphi_j} \cos \Big( \frac{\theta_j}{2} \Big) |S_z, - \rangle^{(j)} \end{align*} These are also eigenvectors of $S^2$, both of them with eigenvalue $\frac{3 \hbar^2}{4}$. Let $\textbf{S} \cdot \hat{\textbf{n}} = \textbf{S}_1 \cdot \hat{\textbf{n}}_1 + \textbf{S}_2 \cdot \hat{\textbf{n}}_2$. Since $[\textbf{S}_1 \cdot \hat{\textbf{n}}_1, \textbf{S}_2 \cdot \hat{\textbf{n}}_2] = 0$ then \begin{equation*} (\textbf{S} \cdot \hat{\textbf{n}})^2 = (\textbf{S}_1 \cdot \hat{\textbf{n}}_1)^2 + (\textbf{S}_2 \cdot \hat{\textbf{n}}_2)^2 + 2 (\textbf{S}_1 \cdot \hat{\textbf{n}}_1) \cdot (\textbf{S}_2 \cdot \hat{\textbf{n}}_2) \end{equation*} From our first equation, we can see that $(\textbf{S}_j \cdot \hat{\textbf{n}}_j)^2 = \frac{\hbar^2}{4} \mathbb{1}$, so our last equation becomes \begin{equation*} (\textbf{S} \cdot \hat{\textbf{n}})^2 = \frac{\hbar^2}{2} \mathbb{1} + 2 (\textbf{S}_1 \cdot \hat{\textbf{n}}_1) \cdot (\textbf{S}_2 \cdot \hat{\textbf{n}}_2) \end{equation*} This is the part where I get confused, because how would the term $2 (\textbf{S}_1 \cdot \hat{\textbf{n}}_1) \cdot (\textbf{S}_2 \cdot \hat{\textbf{n}}_2)$ operate, for example, on the state $|\textbf{S}_1 \cdot \textbf{n}_1, + \rangle |\textbf{S}_2 \cdot \textbf{n}_2, - \rangle$? In the usual case where we work with $\textbf{S} = \frac{\hbar}{2} (\sigma_x \hat{\textbf{i}} + \sigma_y \hat{\textbf{j}} + \sigma_z \hat{\textbf{k}})$, said term takes the form \begin{equation*} 2 \textbf{S}_1 \cdot \textbf{S}_2 = S_{+_1} S_{-_2} + S_{-_1} S_{+_2} + 2 S_{z_1} S_{z_2} \end{equation*} Which is useful because we know how these operators work with the states $|S_z, \pm \rangle$; but in the more general case, how would the term $2 (\textbf{S}_1 \cdot \hat{\textbf{n}}_1) \cdot (\textbf{S}_2 \cdot \hat{\textbf{n}}_2)$ expand in order to work with the tensor product between the states $|\textbf{S}_1 \cdot \textbf{n}_1, \pm \rangle$ and $|\textbf{S}_2 \cdot \textbf{n}_2, \pm \rangle$?

Answer 1

Remember that the particles 1 and 2 live in different Hilbert spaces, and the operators associated with each particle (i.e. of measuring the spin of each particle) act only on their respective Hilbert spaces. So you'll have that, e.g.,

$$2(\textbf{S}_1\cdot\hat{\textbf{n}}_1)\cdot(\textbf{S}_2\cdot\hat{\textbf{n}}_2)|\textbf S_1\cdot\textbf n_1,+\rangle|\textbf S_2\cdot\textbf n_2,-\rangle = 2\big[ \big((\textbf{S}_1\cdot\hat{\textbf{n}}_1)|\textbf S_1\cdot\textbf n_1,+\rangle\big) \big((\textbf{S}_2\cdot\hat{\textbf{n}}_2)|\textbf S_1\cdot\textbf n_2,-\rangle\big) \big] = 2\big[\big( \frac\hbar2|\textbf S_1\cdot\textbf n_1,+\rangle \big)\big( {-}\frac\hbar2|\textbf S_2\cdot\textbf n_2,-\rangle \big) \big] = \frac{\hbar^2}{2}|\textbf S_1\cdot\textbf n_1,+\rangle|\textbf S_2\cdot\textbf n_2,-\rangle.$$

To perhaps be even more clear, one could even write the original operator as \begin{equation*} (\textbf{S} \cdot \hat{\textbf{n}})^2 = \big( (\textbf{S}_1 \cdot \hat{\textbf{n}}_1)\otimes\mathbb1_2+\mathbb1_1\otimes(\textbf{S}_2 \cdot \hat{\textbf{n}}_2) \big)^2 = (\textbf{S}_1 \cdot \hat{\textbf{n}}_1)^2\otimes\mathbb1_2 + \mathbb1_1\otimes(\textbf{S}_2 \cdot \hat{\textbf{n}}_2)^2 + 2 (\textbf{S}_1 \cdot \hat{\textbf{n}}_1) \cdot (\textbf{S}_2 \cdot \hat{\textbf{n}}_2). \end{equation*}