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Calculate the probability that a measure on S_y yields h/2

  • Thread starter QFT25
  • Start date
24
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1. Homework Statement
. Suppose an electron is in the spin state (a,B) If sy is measured, what is the probability of the result h/2?

2. Homework Equations
Eigenvectors of the pauli matrix for y are (1,i)/Sqrt[2] (1,-i)/Sqrt[2] and if you are given a wave function of the sort a | +> +b |-> then the probability of getting state | +> is a^2/(a^2+b^2)

3. The Attempt at a Solution

I wrote out (a,B) as a linear combination of the of the two eigenvectors for the pauli matrix and got that the probability of finding the electron with spin h bar/2 to be (|a-ib|^2)/2. I just want to check with all of you if that is right.
 

vela

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Doesn't look right to me. I assume you're just being sloppy and are using b and B to be the same variable.

Please show the calculations you used to arrive at your answer.
 
24
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Certainly (a,B)=(x/Sqrt[2])(1,i)+(y/Sqrt[2])(1,-i). I solved for x and y on Mathematica and got x=(a/Sqrt[2] - iB/Sqrt[2]) and for y= a/Sqrt[2]+iB/Sqrt[2]. I then assuming a^2+B^2=1 I just took the mod square of x and got (|a-i*B|^2)/2 to be my answer. Did I do something wrong?
 

vela

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Nope, my mistake. Your answer is correct.

Because you already worked out ##\lvert +_y \rangle = \frac{1}{\sqrt{2}}(\lvert + \rangle + i\lvert - \rangle)##, an easier way to arrive at the same result is to calculate the amplitude ##\lvert \langle +_y \vert (a,b) \rangle \rvert^2##.
 

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