Question about what a simultaneous measurement of entangled spins means I was working through a problem I found online and ran into something that is confusing me. We have a system of three spin-1/2 particles, in the state 
$$ |\psi\rangle = \frac{1}{\sqrt{2}}(|1/2,1/2,1/2\rangle - |-1/2,-1/2,-1/2\rangle). $$
Now, one can show that this is an eigenstate with eigenvalue $1$ of the operator
$$ M = \sigma_x^1 \sigma_y^2 \sigma_y^3.$$
I.e., the $x$ pauli matrix on the first spin, and the $y$ pauli matrix on the other two spins. Now, the question later asks what the possible measurement outcomes are if an observer simultaneously measures $S_x$ of the first particle and $S_y$ of the other two particles. According to the answer, the possible outcomes are (+--), (-+-), (--+), (+++). Now, I get that the product of these parities is +1, in line with the fact that $|\psi \rangle$ is an eigenstate of $M$ with eigenvalue $1$. But, I guess I'm confused about what a simultaneous measurement actually means here. For example, if we performed the measurements sequentially, then I believe that the $S_x$ measurement would have to yield $-1$ since $|1/2\rangle - |-1/2\rangle$ is an eigenstate of $S_x$ with eigenvalue $-1$. So then what does a "simultaneous measurement" actually mean here? Now sure, if one measured the product operator $S_x^1 S_y^2 S_y^3$ then clearly the only possible outcome would be $(\hbar/2)^3$ but the problem seems to be asking what happens if you measure the individual spins at the same time, not just the single observable $M$. I'm not sure what is actually going on in that case.
 A: You can't factorise out the state of the first qubit a pure eigenstate of $S_x$ as you seem to want to (the state is not a product state). To see the the possible outcomes of the measurement of $S_x$ on the first particle you need to rewrite the parts of the related to that particle in the $x$ basis. I'm going to use $|+\rangle$ and $|-\rangle$ to represent the eigenstates of $S_x$.
First note that:
\begin{align}
   |1/2\rangle &= \frac{1}{\sqrt{2}}\left(|+\rangle + |-\rangle\right)\\
   |-1/2\rangle &= \frac{1}{\sqrt{2}}\left(|+\rangle - |-\rangle\right)\\
\end{align}
So we can write:
\begin{align}
    |\psi\rangle &= \frac{1}{2}\left(\left(|+\rangle + |-\rangle\right)|1/2\rangle|1/2\rangle - \left(|+\rangle - |-\rangle\right)|-1/2\rangle|-1/2\rangle\right)\\
\end{align}
Then combine factors:
\begin{align}
    |\psi\rangle &= \frac{1}{2}\left(|+\rangle\left(|1/2\rangle|1/2\rangle - |-1/2\rangle|-1/2\rangle\right) + |-\rangle\left(|1/2\rangle|1/2\rangle + |-1/2\rangle|-1/2\rangle\right)\right)\\
\end{align}
Your reasoning would have been correct if we could write this in the form $|\psi\rangle = |-\rangle|\text{some state for the other particles}\rangle$ but we can't do that.
The overall aim in doing this algebra is to show that the order you do the measurements in doesn't change the outcomes you can get at all. We can freely say "simultaneous" without worrying about what it means because it doesn't matter what order we do the measurements in! You should definitely go through it and convince yourself of this. All you need to do is factorise the expression into the eigenstates of each operator in turn and use that to work out what the outcomes of each series of measurements should be. It might be less tedious to calculate everything with a two qubit system (and observable) though.
Note that my algebra may be totally wrong (it usually has some errors) but the technique is correct. You factorise the system according to the eigenvectors of the observable you're going to measure first and use that to work out the state after each measurement outcome, then repeat with the new state and the remaining measurements.
