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.
Best Answer
You are mixing properties that are defined for $|\uparrow_z\rangle $ with those defined for $|\uparrow_x\rangle $. The right relations are $$S_z|\uparrow_z\rangle = \frac\hbar2 |\uparrow_z\rangle;$$ $$S_z|\downarrow_z\rangle = -\frac\hbar2 |\downarrow_z\rangle,$$ and thus $$S_z|\uparrow_x\rangle = \alpha\frac\hbar2 |\uparrow_z\rangle-\beta \frac\hbar2 |\downarrow_z\rangle .$$
The other one that is wrong is that you assumed $S_+|\uparrow_x\rangle =0$ which is not, the right condition is $S_+|\uparrow_z\rangle =0$ and $S_+|\downarrow_z\rangle =\hbar|\uparrow_z\rangle$ . Which leads to
$$S_+|\uparrow_x\rangle= \cancel{\alpha S_+|\uparrow_z\rangle} +\beta S_+|\downarrow_z\rangle=\beta\hbar|\uparrow_z\rangle.$$ The rest you can work it out if you work only on the $|\updownarrow_z\rangle$ basis.