Question

How to count total spin degeneracies for many spin one half particles?

Answer 1

I can't figure out how many different spin states I can create with a four-electron system. I think I can create a spin-zero state, three spin-one states, and five spin-two states. That gives me nine possible states altogether.

My problem is that I can specify a maximum of eight (complex) numbers to completely describe the spin states of the four electrons. But the nine spin-states I can seemingly create correspond to the spherical harmonic functions and I know for sure that they are linearly independent. It seems very wrong.

This discrepancy in the counting doesn't appear until you get to four electrons. It gets worse as you add more electrons.

Does anyone else have a problem with this?

EDIT: Thanks for the excellent answers, especially from Lagerbeer. It turns out I was even more messed up than I thought I was. I don't run into trouble with fourelectrons....I'm already in trouble with three. I didn't realize it because I was counting 2n electron-parameters instead of n^2, so I had 2, 4, 6, 8... (parameters to describe electron spin) as opposed to 2, 4, 8, 16... as people have pointed out. And this has to relate to (l,m) spin states of 2, 4, 6, 9, 12, 16...

So there's already a problem with three electrons and it boils down to this: you have two different states with z-axis spin 1/2: the are (3/2, 1/2) and (1/2, 1/2). To make them up from electrons you have at your disposal these three states:

{A} = duu

{B} = udu

{C} = uud

Answer 2

¶¶¶¶¶¶Hey mate♂♂♂♂
Using the below mentioned rules or formula:-

◆S¯j=(⨂k=1j−1Ik)⊗(ℏ2σ¯)j⊗(⨂k=j+1NIk)S¯j=(⨂k=1j−1Ik)⊗(ℏ2σ¯)j⊗(⨂k=j+1NIk)

◆J2=∣∣∣∣∑i=1NS¯i∣∣∣∣J2=|∑i=1NS¯i|

◆mn,k = n! (n+1−2k)k! (n+1−k)!.