This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: PHYS 507 Answers to Homework VIII (Fall ’05) 1. (a) We have  j 1 j 2  ≤ j ≤ j 1 + j 2 which looks like the triangle inequality, but these are the quantum numbers, not the actual lengths of our vectors. In the actual triangle, the side lengths are a = p j 1 ( j 1 + 1), b = p j 2 ( j 2 + 1) and c = p j ( j + 1) (where I have omitted ¯ h multiplier) which corresponds to the values of p J 2 1 , etc. If θ is the angle between ~ J 1 and ~ J 2 (I will use the internal angle), then we know cos θ = ( a 2 + b 2 c 2 ) / (2 ab ). Using this we can compute all the angles (Note: θ = 180 o corresponds to the case where ~ J 1 and ~ J 2 are exactly parallel and θ = 0 o corresponds to the case they are exactly antiparallel). Our side lengths are a = √ 6, b = √ 2 and c = √ 12 , √ 6 , √ 2. j = 3 , cos θ = 4 2 √ 12 , θ = 125 o , j = 2 , cos θ = 2 2 √ 12 , θ = 73 o , j = 1 , cos θ = 6 2 √ 12 , θ = 30 o . So, maximum j value corresponds to the case where the vectors ~ J 1 and ~ J 2 are more or less parallel. They are not exactly parallel due to the quantum uncertainties in the direction of these vectors. Similarly, minimum j value corresponds to the case where these vectors are more or less antiparallel but again not exactly....
View
Full Document
 Spring '11
 starg
 Vectors, Cos, Eigenvalue, eigenvector and eigenspace, Haplogroup

Click to edit the document details