# Quantum entanglement of spins

Suppose we have a pair of spins in the state $\left|\Psi\right>=\frac{1}{\sqrt2}(\left|\uparrow\downarrow\right>-\left|\downarrow\uparrow\right>)$. We know that the measurements of the spins along the $z$-direction must give opposite results as a spin-up result of the first particle "collapses" the state into $\left|\uparrow\downarrow\right>$ and vice versa. But why can't it collapse into, say, $\frac{1}{\sqrt2}(\left|\uparrow\downarrow\right>+\left|\uparrow\uparrow\right>)$ (which is again an eigenstate of the spin of the first particle along the $z$-direction with the same eigenvalue)?

• "Which is again an eigenstate" of what? – WillO Jul 26 '16 at 14:35
• @WillO See the edited question. – Poon Levi Jul 26 '16 at 15:04
• this is called the preferred basis problem and a sub-problem of the measurement problem. There should be a lot of online resources on the topic. On SE see e.g.: physics.stackexchange.com/questions/65177/… – Wolpertinger Jul 26 '16 at 15:27

The states $\left|\uparrow \downarrow\right>$ and $\left|\downarrow \uparrow\right>$ are in agreement with this correlation. The state $\left|\uparrow \uparrow\right>$ is not, so it is not a possible result.
If you had begun with the unentangled state $$\frac{1}{2}\left(\left|\uparrow \uparrow\right> + \left|\uparrow \downarrow\right> + \left|\downarrow \uparrow\right> + \left|\downarrow \downarrow\right>\right)$$ which is factorable into
$$\frac{1}{2}\left(\left|\uparrow\right>_L + \left|\downarrow\right>_L\right)\left(\left|\uparrow\right>_R + \left|\downarrow\right>_R\right)$$ then the spins would have been independent. Measuring the spin of the left particle would then kill one of the terms for the left particle's spin, but due to the factorable form that does not eliminate any of the right particle's possible states.