Why do $S_x$ and $S_y$ flip up/down spin states but $S_z$ does not?

Question

By using the notation $S\lvert s,m_s\rangle$, such that $\bigl\lvert\frac{1}{2},\frac{1}{2}\bigr\rangle=\lvert+\rangle$ and $\bigl\lvert\frac{1}{2},-\frac{1}{2}\bigr\rangle=\lvert-\rangle$ we can obtain following the eigenvalue equation for the components of the spin vector: $$\begin{align} S_x\lvert\pm\rangle&=\frac{\hbar}{2}\lvert\mp\rangle \\ S_y\lvert\pm\rangle&=\pm i\frac{\hbar}{2}\lvert\mp\rangle \\ S_z\lvert\pm\rangle&=\pm\frac{\hbar}{2}\lvert\pm\rangle \end{align}$$ However, I am struggling to interpret the result. If we consider the $x$ and $y$ component of spin, then there seems to be flipping of the state function, but it's not the same for the $z$-component. Why is that?

Answer 1

You chose the $\lvert \pm \rangle$ to be an eigenvector of $S_z$ with eigenvalue $\pm\frac{1}{2}$ - that's what the $m_s$ is: The eigenvalue of the state w.r.t. the $z$-spin.

Since $S_x$ and $S_y$ do not commute with $S_z$, $\lvert \pm \rangle$ is not an eigenvector of them, hence the state cannot stay the same after they are applied to it.

That the spin is precisely flipped comes from the form of the commutation relations $$ [S_i,S_j] = \mathrm{i}\epsilon_{ijk}S_k$$ and you can see that the spin is flipped by evaluating $S_z$ on $S_x\lvert \pm \rangle$ and $S_y\lvert \pm \rangle$.