2

Say we have 3 spin-1/2 particles

In the $|m_1, m_2, m_3 \rangle$ basis we have 8 states where where $m_1,m_2,m_3=\frac{1}{2},-\frac{1}{2}$

$$|\frac{1}{2} \frac{1}{2} \frac{1}{2}\rangle,|\frac{1}{2} \frac{1}{2} -\frac{1}{2}\rangle,|\frac{1}{2} -\frac{1}{2} \frac{1}{2}\rangle,|\frac{1}{2} -\frac{1}{2} -\frac{1}{2}\rangle,$$ $$|-\frac{1}{2} \frac{1}{2} \frac{1}{2}\rangle,|-\frac{1}{2} \frac{1}{2} -\frac{1}{2}\rangle,|-\frac{1}{2} -\frac{1}{2} \frac{1}{2}\rangle,|-\frac{1}{2} -\frac{1}{2} -\frac{1}{2}\rangle,$$

In the $|S, m\rangle$ basis there are only 6 states where $S=\frac{3}{2},\frac{1}{2}$ and $m= \frac{3}{2}, \frac{1}{2}, -\frac{1}{2}, -\frac{3}{2}$ $$|\frac{3}{2} \frac{3}{2}\rangle,|\frac{3}{2} \frac{1}{2}\rangle,|\frac{3}{2} -\frac{1}{2}\rangle, |\frac{3}{2} -\frac{3}{2}\rangle, |\frac{1}{2} \frac{1}{2}\rangle, |\frac{1}{2} -\frac{1}{2}\rangle,$$

Obviously there are 8 states, which I believe is a quartet and two doublets. Where then, have I gone wrong? Are there two states each with $S=\frac{1}{2}$ and $S=-\frac{1}{2}$? Is it incorrect to say that $|S, m\rangle$ is a basis, as $S,m$ do not form a CSCO? Must another quantum number be added to form a basis?

Similarly, if we have 3 spin-1 particles, we get 27 states in the $|m_1,m_2,m_3\rangle$ basis for $m_1,m_2,m_3=1,0,-1$, but only 15 in the $|S,m\rangle$ basis for $S=1,2,3$.

egurns
  • 109
  • 1
    You can do this in two steps. First, you add 2 spins, $S_1$ and $S_2$. Then, you add $S_3$. When you add two spin-1/2 spins you get the spin-$0$ singlet ($S_{12}=0$) and the spin-1 triplet ($S_{12}=1$). Now you add the third spin to each of these. One spin 1/2 and one spin-$0$ ($S_3 = 1/2, S_{12} = 0$) are going to give a spin-1/2 pair while adding spins 1/2 and 1 ($S_3 = 1/2, S_{12} = 1$) will give you two spin 1/2 states and 4 spin 3/2 states. That's the 8 you are looking for. As you can see there are 4 states with $S_{123} = 1/2$, two with $ S_{12} = 0$ and two with $S_{12} = 1$. – secavara Mar 20 '18 at 11:04
  • 1
    Built like this, these 8 states form a basis and are eigenstates of at least $S^2_1,S^2_2,S^2_3,S^2_{12},S^2_{123},S_{z|123}$. The reasoning is similar for the spin-$1$ case. I'll leave that to you to work out, and find that the dimensions of the bases are consistent as well. – secavara Mar 20 '18 at 20:43
  • Related, as well as this one. – Cosmas Zachos Apr 13 '18 at 18:36

0 Answers0