# Spin 1/2 correlation

by gespex
Tags: 1 or 2, correlation, spin
 P: 57 Hello all, Imagine two spin 1/2 particles that are entangled, going towards two stern-gerlach apparatuses, with some relative angle. Now imagine one stern-gerlach device measures the spin of one of the particles as up. What is the chance that the other stern-gerlach device measures the spin to be down? For 90 degrees it would be 50/50, right? So my guess is $cos^2 ({1 \over 2} \alpha)$. Is that correct? Thanks in advance
P: 110
 Quote by gespex Hello all, Imagine two spin 1/2 particles that are entangled, going towards two stern-gerlach apparatuses, with some relative angle. Now imagine one stern-gerlach device measures the spin of one of the particles as up. What is the chance that the other stern-gerlach device measures the spin to be down? For 90 degrees it would be 50/50, right? So my guess is $cos^2 ({1 \over 2} \alpha)$. Is that correct? Thanks in advance
You can look here , it is the same quastion.
 Sci Advisor PF Gold P: 5,378 Yes, that's correct. I may have steered you wrong at an earlier time.
P: 57
Spin 1/2 correlation

 Quote by sergiokapone You can look here , it is the same quastion.
Thanks for your answer. It doesn't seem to be the same question though - the question of that guy is a lot more advanced than mine. He is asking a question about the implication of the correlations, while I am simply looking for the actual formula for the correlation.

From what I understand from that source, he takes into account two possibilities for the case $\alpha = 90°$: a 50% correlation and a 100% correlation. I thought it was 50%, which would I believe indicate that I was right thinking the correlation is $cos^2({1 \over 2} \alpha$. However, I'm not sure what he meant with the 100% correlation.

Thanks
P: 57
 Quote by DrChinese Yes, that's correct. I may have steered you wrong at an earlier time.
I did ask the question earlier, and you were the one to answer it. But reading up later I assumed you were talking about the correlation as (a - b)/(a + b)? So I did wonder if it was simply a miscommunication, hence me asking the question again.

 Quote by gespex Hello all, For 90 degrees it would be 50/50, right? So my guess is $cos^2 ({1 \over 2} \alpha)$. Is that correct?