0
$\begingroup$

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)?

$\endgroup$
  • 1
    $\begingroup$ "Which is again an eigenstate" of what? $\endgroup$ – WillO Jul 26 '16 at 14:35
  • $\begingroup$ @WillO See the edited question. $\endgroup$ – Poon Levi Jul 26 '16 at 15:04
  • $\begingroup$ 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/… $\endgroup$ – Wolpertinger Jul 26 '16 at 15:27
2
$\begingroup$

Your initial state is entangled, which means it exhibits a correlation between the spin of the two particles. Each particle is spin up or spin down with equal probability, but the spins are (anti)correlated so that they cannot have the same spin.

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.