What Is the Probability of Measuring Spin-up for Electron in State Alpha?

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The discussion centers on the probability of measuring the spin state of an electron, specifically when the spin state is defined as alpha=(a,b) and the operator measured is S_y. The correct probability of obtaining the result h/2 is calculated using the expression ()^2, while the incorrect solution conflates measurement with operator application. The confusion arises from a misunderstanding of quantum measurement principles, leading to an erroneous calculation of the probability amplitude.

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the_doors
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hello guys

Suppose an electron is in the spin state alpha=(a,b). if s_y is measured, what is the probability of result h/2 ?


s_y eigenvector for +h/2 is |y+>=1/sqrt(2) (1,i)) so the probability is (<y+|alpha>)^2 . but in the solution i attached , the solution is different ! I'm confused !



thank you for your help
 

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Don't bother, your solution is correct.
 
so the solution is incorrect ?
 
well there's certainly something wrong when you measure a probability in terms of \hbar2...
 
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the_doors said:
so the solution is incorrect ?
Yes, the solution is incorrect. Whoever wrote it made the common mistake of conflating making a measurement with applying the associated operator. In other words, the "solution" calculates ##\langle +_y \lvert \hat{S}_y \rvert \alpha,\beta\rangle## for the probability amplitude while you correctly calculated ##\langle +_y \vert \alpha,\beta\rangle##.
 
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