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Oct 30, 2020 at 18:59 comment added astronautgravity My pleasure. Glad I could help! Good luck with physics going forward.
Oct 30, 2020 at 18:54 vote accept R. M.
Oct 30, 2020 at 18:52 comment added R. M. I see, in this case it wasn't necessary since the vectors $|\textbf{S}_j \cdot \hat{\textbf{n}}_j, \pm \rangle$ are already eigenvectors of the operators $\textbf{S}_j \cdot \hat{\textbf{n}}_j$. Thank you for both your answer and replies.
Oct 30, 2020 at 18:41 comment added astronautgravity In general $\textbf S|+\rangle$ is non-trivial because $|+\rangle$ is an eigenvector of $S_z$ not of $\textbf S$. So if you want to understand $\textbf{S}|+\rangle$, a good approach is to write $S_x$ and $S_y$ in terms of operators whose effect on $|+\rangle$ you know.
Oct 30, 2020 at 18:30 comment added R. M. I see, when we're working with Pauli matrices, in theory the coupling term (I think that's what it's called) is $2 \textbf{S}_1 \cdot \textbf{S}_2 |+ \rangle |- \rangle = 2 (\textbf{S}_1 |+ \rangle)(\textbf{S}_2 |- \rangle)$, it's just that in this case $S_x$ and $S_y$ are expressed in terms of the ladder operators. Why wasn't such substitution necessary in the general case?
Oct 30, 2020 at 18:21 history answered astronautgravity CC BY-SA 4.0