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

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?

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 $S_y\lvert \pm \rangle$.
• why we did not choose ∣±⟩ to be an eigen vector of $S_x$ or $S_y$, I am sorry, but I am still struggling to grasp the answer. – Roshan Shrestha Apr 21 '15 at 14:27
• @RoshanShrestha: You can as well choose it to be an eigenvector of any of those, or an eigenvector of the spin about any other axis. There's nothing distinguishing a special direction. You just need to pick some direction to talk about whether the spin is "up" or "down" compared to that direction, and we conventionally call that direction $z$. – ACuriousMind Apr 21 '15 at 14:31