Question

If the angular momentum quantum number is 2, what is the smallest angle L makes with the z-axis?

Answer 1

Since n = 3 (and l isn't 0), l = 1, 2 ... and ml = -l ... l = -2,-1,0,1,2 So then when you draw it.. (I don't seem to be able to get the picture in here] The Lz vector is pointing up along the z-axis and the L vector is some degrees to the right of it. The angle between the z-axis and angular momentum can be written using cosine... c o s ( θ ) = L z L cos(θ)=LzL I understand up to there. But then the next part the professor did is add in numbers and I'm not sure where the numbers came from. He wrote this: c o s ( θ ) = 2 ℏ √ 6 ℏ = 2 √ 6 cos(θ)=2ℏ6ℏ=26 The √ 6 6 comes from L = √ l ( l + 1 ) ℏ L=l(l+1)ℏ using l = 2. But why l = 2 and not l = 1? And I don't know where the 2 on top came from. Then he wrote another one.. c o s ( θ ) = 1 / 2 √ 3 / 4 cos(θ)=1/23/4 I'm assuming this second one is for the electron spin since that is the one we are comparing it to? But.. how do you draw the vectors for electron spin? I'm confused there. So, I'm confused on where the numbers came from and how you take electron spin into account here. If anyone could help, I'd appreciate it!

Reference https://www.physicsforums.com/threads/what-state-has-min-angle-between-angular-momentum-z-axis.456405/