# What can you infer from knowing a spin $\dfrac{1}{2}$ particle's $S_z$ is in the $+\dfrac{\hbar}{2}$state?

What can you infer (about the rest of its spin components' states) from knowing a spin $\dfrac{1}{2}$ particle's $S_z$ is in the $+\dfrac{\hbar}{2}$ state?

Or maybe more to point (source of my confusion), what combination of eigenstates can $S_x$, $S_y$, and $S_z$ be in?

I know from Griffiths how to get determine what the possible eigenvalues are, but what combinations can they be in?

If said particle is in spin up state, are all the components have an eigenvalue of positive one half h bar?

If not, how can you determine the sign of the other two if you know one?

$S_x$, $S_y$, and $S_z$ are incompatible observables. That means they do not share a basis of eigenstates. If $S_z$ is $\frac{1}{2}$, then this means it is in an eigenstate of $S_z$, and thus cannot be in an eigenstate of $S_x$ or $S_y$.

As stated, the system is certainly in an eigenstate of $S_z$ and cannot be in an eigenstate of $S_x$ or $S_y$ since these do not commute with $S_z$. Additionally, one can infer using the uncertainty relation $$\Delta S_z\Delta S_x\ge \textstyle\frac{1}{2}\vert\langle S_y\rangle\vert$$ that the average $\langle S_y\rangle=0$ since $\Delta S_z=0$ for an eigenstate of $S_z$. Similarly one can conclude that $\langle S_x\rangle=0$.

Note that since $0$ is not an eigenvalue of either $S_x$ or $S_y$ this confirms that the state is not an eigenstate of either of the above.

Note that the $x$ and $y$ spin operators can be computed form the ladder operators. For example:

$$S_x = \frac{1}{2}(S^+ + S^-)$$

This makes for a fairly easy eigenvalue problem:

$$S_x \psi = m\psi$$

which in the traditional $z$ basis has solutions:

$$\psi_{\pm} = \frac{1}{\sqrt{2}}(\uparrow_z \pm \downarrow_z)$$

where

$$\psi_+ \equiv \uparrow_x$$

is the eigenstate with $\hbar/2$ in the $x$-direction (likewise for $\psi_-$). But you asked about the states in the $x$ and $y$ basis: well, they're just ordered labels, so:

$$\uparrow_z = \frac{1}{\sqrt{2}}(\uparrow_y + \downarrow_y)$$

which is also a equal mix of $\uparrow_x$ and $\downarrow_x$ (but I haven't solved for the phase factor in that basis).