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 In order to get the probabilities you need to use the Clebsch-Gordan table. From the original equation we know that Y01 is the same as |10> and Y11 is the same as |11>. We also know that electrons have an inherent spin of 1/2, so the chi (+) is |1/2 1/2> and the chi (-) is |1/2 -1/2>. This is evident because chi + is spin up and chi - is spin down and we are working with an electron. So at this point we have: Y01 = |1 0> Y11 = |1 1> Chi (+) = |1/2 1/2> Chi (-) = |1/2 -1/2> So now we plug this back into the first equation and for now we will ignore the radial part (R21). When we do this we get: sqrt(1/3) |10> |1/2 1/2> +sqrt(2/3) |11> |1/2 -1/2> This is where the Clebsch-Godradn table comes into play. We know that the spin of electron is 1/2, and we also know that l is one (we got that because the lower part of the angular equation is on for both Y01, and Y11, and also in the radial equation Rnl) So we go to the 1 X 1/2 clebsch-Gordan table. so |10> |1/2 1/2> = sqrt(2/3) |3/2 1/2> - sqrt(1/3) |1/2 1/2> This one we got from the line highlighted in yellow and |11> |1/2 -1/2> = sqrt(1/3) |3/2 1/2> + sqrt(2/3) |1/2 1/2> This one we got from the line highlighted in green Now we plug this back in and get sqrt(1/3)*(sqrt(2/3) |3/2 1/2> - sqrt(1/3) |1/2 1/2>)+ sqrt(2/3)(sqrt(1/3) |3/2 1/2> + sqrt(2/3) |1/2 1/2>) This equals sqrt(2)/3|3/2 1/2>+sqrt(2)/3|3/2 1/2>+(2/3)|1/2 1/2> - (1/3)|1/2 1/2> Which simplifies to 2*sqrt(2)/3|3/2 1/2>+(1/3)|1/2 1/2> So J2 = (3/2)(3/2+1)hbar2 with a probability of 2*2/9 = (1/2)(1/2+1)hbar2 with a probability of 1/9 And Jz = (1/2)hbar with a probability of 1 Attached Thumbnails