Suppose I find a $1/2$ spin particle in the eigenstate of the observable $\hat S_x$ relative to the eigenvalue $\hbar / 2$. I will use the short-hand notation $ \vert \uparrow_x \rangle$.
The goal is to express it in terms of the eigenstates of the observable $\hat S_z$: $\{ \vert \uparrow_z \rangle, \vert \downarrow_z \rangle\}$.
I have to solve the linear system: $$ \left\{ \begin{aligned} &\vert \uparrow_x \rangle = \alpha \vert \uparrow_z \rangle + \beta \vert \downarrow_z \rangle \\ &\vert \downarrow_x \rangle = \alpha' \vert \uparrow_z\rangle + \beta' \vert \downarrow_z \rangle \end{aligned}\right. $$
The condition of normalization $\langle \uparrow_x \vert \uparrow_x \rangle = \langle \downarrow_x \vert \downarrow_x \rangle = 1$ returns $|\alpha|^2 + |\beta|^2 = |\alpha'|^2 + |\beta'|^2 = 1 $, while the ortogonality, $\langle \uparrow_x \vert \downarrow_x \rangle = \langle \downarrow_x \vert \uparrow_x \rangle = 0$ implies $\bar{\alpha}\alpha' + \bar{\beta}\beta' = \bar{\alpha}'\alpha + \bar{\beta}'\beta = 0$.
Now, here comes the problem. Using the identity $[S_+, S_-] = 2 \hbar S_z$ I could write $$ [S_+, S_-]\vert \uparrow_x \rangle = 2 \hbar S_z \vert \uparrow_x \rangle $$ expanding the commutator $$ \begin{aligned} (S_+S_- - \require{cancel}\cancel{S_-S_+}) \vert \uparrow_x \rangle = \hbar ^2 \vert \uparrow_x \rangle \end{aligned} $$ I end up with $$\frac{\hbar}{2}\vert \uparrow_x \rangle = S_x\vert \uparrow_x \rangle = S_z\vert \uparrow_x \rangle $$
But it can't be! I have been checking it for quite a while, but I still don't know what went wrong.
