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Spin 1/2 correlation

by gespex
Tags: 1 or 2, correlation, spin
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gespex
#1
Mar12-12, 08:06 PM
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 [itex]cos^2 ({1 \over 2} \alpha)[/itex]. Is that correct?


Thanks in advance
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sergiokapone
#2
Mar13-12, 01:51 PM
P: 110
Quote Quote by gespex View Post
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 [itex]cos^2 ({1 \over 2} \alpha)[/itex]. Is that correct?


Thanks in advance
You can look here , it is the same quastion.
DrChinese
#3
Mar13-12, 03:14 PM
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Yes, that's correct. I may have steered you wrong at an earlier time.

gespex
#4
Mar13-12, 03:16 PM
P: 57
Spin 1/2 correlation

Quote Quote by sergiokapone View Post
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 [itex]\alpha = 90[/itex]: 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 [itex]cos^2({1 \over 2} \alpha[/itex]. However, I'm not sure what he meant with the 100% correlation.


Thanks
gespex
#5
Mar13-12, 03:21 PM
P: 57
Quote Quote by DrChinese View Post
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.

But thank you again, for your answer now and for your answer last time.
sergiokapone
#6
Mar13-12, 03:30 PM
P: 110
Quote Quote by gespex View Post
the question of that guy is a lot more advanced than mine.
I am thet guy.
As I understood of the answers to my question, the correct answer to your quastion it is
Particle 2 is detected with a 50% probability of having spin −ℏ/2 in the +x direction, and 50% of having spin +ℏ/2 in the -x direction.
sergiokapone
#7
Mar13-12, 03:32 PM
P: 110
Quote Quote by gespex View Post
Hello all,
For 90 degrees it would be 50/50, right? So my guess is [itex]cos^2 ({1 \over 2} \alpha)[/itex]. Is that correct?
Thus, yes, it is correct.


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