Quantum Spin – 3 Particles Spin Using Clebsch-Gordan Coefficients

Question

I know that there are a lot of question similar to mine on this website and I have checked them all out and still can't seem to to figure this out…

If I'm adding together let's say the spins of three particles with spin 1/2 a system consisting of a neutron a proton and a neutron so : $|npn\rangle $. If I first couple together the spins of a neutron and the proton I get: $s_z =0 $ and $ s=0,1$. The remainging neutron has $s_z =-1/2 $ and $s=1/2$. If I then check the Clebsch-Gordan coefficients table I can get $|\frac{1}{2} -\frac{1}{2}\rangle$ or $\sqrt{\frac{2}{3}} |\frac{3}{2} -\frac{1}{2}\rangle+\sqrt\frac{1}{3}|\frac{1}{2} -\frac{1}{2}\rangle $ depending on whether I couple the first two spins to $s=0$ or $s=1$ ?

Are both states possible ? Is this the right way to do this ?

Answer 1

First let's add the first two spins $\mathbf{s}_1+\mathbf{s}_2=\mathbf{s}_{12}$, this can occur in two ways: $s_{12}=0$ and $s_{12}=1$. The general rule of momentum coupling is $$ |s_1s_2:s_{12}m_{12}\rangle_{12}=\sum_{m_1m_2}( s_1m_1s_2m_2|s_{12}m_{12})|s_1m_1\rangle_1|s_2m_2\rangle_2. $$ In the case of two spin-one-half particles it leads to familiar singlet and triplet states: $$ {\textstyle\left|\frac12\frac12:00\right\rangle_{12}}=\frac{|\uparrow\rangle_1|\downarrow\rangle_2-|\downarrow\rangle_1|\uparrow\rangle_2}{\sqrt2}, $$ $$ {\textstyle\left|\frac12\frac12:11\right\rangle_{12}}=|\uparrow\rangle_1|\uparrow\rangle_2,\quad {\textstyle\left|\frac12\frac12:10\right\rangle_{12}}=\frac{|\uparrow\rangle_1|\downarrow\rangle_2+|\downarrow\rangle_1|\uparrow\rangle_2}{\sqrt2},\quad{\textstyle\left|\frac12\frac12:1-1\right\rangle_{12}}=|\downarrow\rangle_1|\downarrow\rangle_2 $$ (in the following we will omit $\frac12\frac12:$ in these state vectors).

Now let's add the third spin $s_3=\frac12$ to $s_{12}=0,1$, the resulting total spin is $s$ and its projection is $m$. It occurs according to the rule $$ |s_1s_2(s_{12})s_3:sm\rangle=\sum_{m_{12}m_3}(s_{12}m_{12}s_3m_3|sm)|s_{12}m_{12}\rangle_{12}|s_3m_3\rangle_3. $$

The possible cases are:

A) $s_{12}=0$, $s=\frac12$, $m=\pm\frac12$: $$ \textstyle\left|\frac12\frac12(0)\frac12:\frac12\frac12\right\rangle=\left|00\right\rangle_{12}\left|\uparrow\right\rangle_3. $$ $$ \textstyle\left|\frac12\frac12(0)\frac12:\frac12-\frac12\right\rangle=\left|00\right\rangle_{12}\left|\downarrow\right\rangle_3. $$

B) $s_{12}=1$, $s=\frac12$, $m=\pm\frac12$: $$ {\textstyle\left|\frac12\frac12(1)\frac12:\frac12\frac12\right\rangle}=\sqrt{\frac23}{\textstyle\left|11\right\rangle_{12}\left|\downarrow\right\rangle_3}-\sqrt{\frac13}{\textstyle\left|10\right\rangle_{12}\left|\uparrow\right\rangle_3}, $$ $$ {\textstyle\left|\frac12\frac12(1)\frac12:\frac12-\frac12\right\rangle}= \sqrt{\frac13}{\textstyle\left|10\right\rangle_{12}\left|\downarrow\right\rangle_3}-\sqrt{\frac23}{\textstyle\left|1-1\right\rangle_{12}\left|\uparrow\right\rangle_3}, $$

C) $s_{12}=1$, $s=\frac32$, $m=\pm\frac12,\pm\frac32$: $$ {\textstyle\left|\frac12\frac12(1)\frac12:\frac32\frac32\right\rangle}= {\textstyle\left|11\right\rangle_{12}\left|\uparrow\right\rangle_3}, $$ $$ {\textstyle\left|\frac12\frac12(1)\frac12:\frac32\frac12\right\rangle}= \sqrt{\frac13}{\textstyle\left|11\right\rangle_{12}\left|\downarrow\right\rangle_3}+\sqrt{\frac23}{\textstyle\left|10\right\rangle_{12}\left|\uparrow\right\rangle_3}, $$ $$ {\textstyle\left|\frac12\frac12(1)\frac12:\frac32-\frac12\right\rangle}= \sqrt{\frac23}{\textstyle\left|10\right\rangle_{12}\left|\downarrow\right\rangle_3}+\sqrt{\frac13}{\textstyle\left|1-1\right\rangle_{12}\left|\uparrow\right\rangle_3}, $$ $$ {\textstyle\left|\frac12\frac12(1)\frac12:\frac32-\frac32\right\rangle}= {\textstyle\left|1-1\right\rangle_{12}\left|\downarrow\right\rangle_3}. $$