# Uncertainty principle and entanglement state

Operators corresponding to spin along $x, y$ and $z$ axes are $S_x, S_y, S_z$. Eigenstates of $S_z$ are $\left|+z\right>$ and $\left|-z\right>$. Eigenstates of $S_x$ are $\left|+x\right>$ and $\left|-x\right>$. We consider 2 particles with the following wave function

$$\left|\psi\right> = \frac{\left|+z\right>_1\left|-z\right>_2 - \left|-z\right>_1\left|+z\right>_2}{\sqrt{2}} = - \frac{\left|+x\right>_1\left|-x\right>_2 - \left|-x\right>_1\left|+x\right>_2}{\sqrt{2}}$$

I.e. if we measure the spin of the first particle along $z$ and got $s_z = -z$ then the second particle will have the spin $s_z = +z$. The same result for $x$ axe: if we got for the first particle $s_x = +x$ spin then the second one will have $s_x = -x$ spin.

The article says that

It remains only to show that $S_x$ and $S_z$ cannot simultaneously possess definite values in quantum mechanics. One may show in a straightforward manner that no possible vector can be an eigenvector of both matrices. More generally, one may use the fact that the operators do not commute.

The operators $S_x$ and $S_z$ really don't commute: $$\left[S_x,S_z\right] = -i \hbar S_y$$

Uncertainty principle says that $$\Delta s_x \Delta s_z \ge \frac{\hbar}{2}\left| \left<S_y\right>\right|$$

I assume that $\left<S_y\right> = \left<\psi\right|S_y\left|\psi\right>$, but the problem is that $\left<\psi\right|S_y\left|\psi\right> = 0$ and as result $$\Delta s_x \Delta s_z \ge 0.$$ Thus potentially no problem to measure spin along $x$ and $z$ axes in the same time i. e. uncertainty principle does not apply any limitation on this.

Could anybody point me what's wrong there?

1. You are correct that two operators not commuting does not show that they have no simultaneous eigenvectors. In fact, all it shows is that there is no basis of simultaneous eigenvectors, but in general, the existence of a single or a few such vectors is not forbidden.

2. It is nevertheless true that there are no simultaneous eigenstates of the angular momentum operators $S_x$ and $S_z$. Here's why:

Without loss of generality we assume that we are in an irreducible representation of the rotation group with spin $s$, so it has dimension $2s+1$ and is spanned by eigenvectors $\lvert -s\rangle,\dots,\lvert s\rangle$ of $S_z$ with eigenvalues from $-s$ to $s$ in integer increments. Since all their eigenvalues are different, these are the only eigenvectors of $S_z$ on this space, so any simultaneous eigenvector of $S_x$ and $S_z$ must be one of these. Now we recall that the combinations $S_\pm = S_x \pm \mathrm{i} S_y$ are raising and lowering operators for the $S_z$-eigenstates, i.e. $$S_+\lvert j\rangle = \lvert j + 1 \rangle \quad \land \quad S_-\lvert j\rangle = \lvert j - 1 \rangle,$$ where the states on the r.h.s. is zero in case $j+1 > s$ or $j-1 < -s$. We have that $S_x = \frac{1}{2}(S_+ + S_-)$ and therefore $$S_x \lvert j\rangle = \frac{1}{2}(S_+\lvert j\rangle + S_-\lvert j\rangle) =\frac{1}{2}(\lvert j+ 1\rangle + \lvert j-1\rangle)\neq \lvert j\rangle,$$ so none of the $S_z$ eigenstates is an $S_x$ eigenstate.