Spin 1/2 correlation

  • Thread starter gespex
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  • #1
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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
 

Answers and Replies

  • #2
268
3
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.
 
  • #3
DrChinese
Science Advisor
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Yes, that's correct. I may have steered you wrong at an earlier time.
 
  • #4
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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
 
  • #5
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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.
 
  • #6
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the question of that guy is a lot more advanced than mine.
I am thet guy. :smile:
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.
 
  • #7
268
3
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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