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Two Spin One-Half Particles

Consider a system consisting of two spin one-half particles. Suppose that the system does not possess any orbital angular momentum. Let ${\bf S}_1$ and ${\bf S}_2$ be the spin angular momentum operators of the first and second particles, respectively, and let
\begin{displaymath}
{\bf S} ={\bf S}_1 + {\bf S}_2
\end{displaymath} (839)

be the total spin angular momentum operator. By analogy with the previous analysis, we conclude that it is possible to simultaneously measure either $S_1^{ 2}$, $S_2^{ 2}$, $S^2$, and $S_z$, or $S_1^{ 2}$, $S_2^{ 2}$, $S_{1z}$, $S_{2z}$, and $S_z$. Let the quantum numbers associated with measurements of $S_1^{ 2}$, $S_{1z}$, $S_2^{ 2}$, $S_{2z}$, $S^2$, and $S_z$ be $s_1$, $m_{s_1}$, $s_2$, $m_{s_2}$, $s$, and $m_s$, respectively. In other words, if the spinor $\chi_{s_1,s_2;m_{s_1},m_{s_2}}^{(1)}$ is a simultaneous eigenstate of $S_1^{ 2}$, $S_2^{ 2}$, $S_{1z}$, and $S_{2z}$, then
$\displaystyle S_1^{ 2}  \chi_{s_1,s_2;m_{s_1},m_{s_2}}^{(1)}$ $\textstyle =$ $\displaystyle s_1 (s_1+1) \hbar^2
 \chi_{s_1,s_2;m_{s_1},m_{s_2}}^{(1)},$ (840)
$\displaystyle S_2^{ 2}  \chi_{s_1,s_2;m_{s_1},m_{s_2}}^{(1)}$ $\textstyle =$ $\displaystyle s_2 (s_2+1) \hbar^2
 \chi_{s_1,s_2;m_{s_1},m_{s_2}}^{(1)},$ (841)
$\displaystyle S_{1z}  \chi_{s_1,s_2;m_{s_1},m_{s_2}}^{(1)}$ $\textstyle =$ $\displaystyle m_{s_1} \hbar
 \chi_{s_1,s_2;m_{s_1},m_{s_2}}^{(1)},$ (842)
$\displaystyle S_{2z}  \chi_{s_1,s_2;m_{s_1},m_{s_2}}^{(1)}$ $\textstyle =$ $\displaystyle m_{s_2} \hbar
 \chi_{s_1,s_2;m_{s_1},m_{s_2}}^{(1)},$ (843)
$\displaystyle S_z  \chi_{s_1,s_2;m_{s_1},m_{s_2}}^{(1)}$ $\textstyle =$ $\displaystyle m_s \hbar
 \chi_{s_1,s_2;m_{s_1},m_{s_2}}^{(1)}.$ (844)

Likewise, if the spinor $\chi_{s_1,s_2;s,m_s}^{(2)}$ is a simultaneous eigenstate of $S_1^{ 2}$, $S_2^{ 2}$, $S^2$, and $S_z$, then
$\displaystyle S_1^{ 2}  \chi_{s_1,s_2;s,m_s}^{(2)}$ $\textstyle =$ $\displaystyle s_1 (s_1+1) \hbar^2
 \chi_{s_1,s_2;s,m_s}^{(2)},$ (845)
$\displaystyle S_2^{ 2}  \chi_{s_1,s_2;s,m_s}^{(2)}$ $\textstyle =$ $\displaystyle s_2 (s_2+1) \hbar^2
 \chi_{s_1,s_2;s,m_s}^{(2)},$ (846)
$\displaystyle S^{2}  \chi_{s_1,s_2;s,m_s}^{(2)}$ $\textstyle =$ $\displaystyle s (s+1) \hbar^2
 \chi_{s_1,s_2;s,m_s}^{(2)},$ (847)
$\displaystyle S_z  \chi_{s_1,s_2;s,m_s}^{(2)}$ $\textstyle =$ $\displaystyle m_s \hbar
 \chi_{s_1,s_2;s,m_s}^{(2)}.$ (848)

Of course, since both particles have spin one-half, $s_1=s_2=1/2$, and $s_{1z}, s_{2z}=\pm 1/2$. Furthermore, by analogy with previous analysis,
\begin{displaymath}
m_s = m_{s_1}+ m_{s_2}.
\end{displaymath} (849)

Now, we saw, in the previous section, that when spin $l$ is added to spin one-half then the possible values of the total angular momentum quantum number are $j=l\pm 1/2$. By analogy, when spin one-half is added to spin one-half then the possible values of the total spin quantum number are $s=1/2\pm 1/2$. In other words, when two spin one-half particles are combined, we either obtain a state with overall spin $s=1$, or a state with overall spin $s=0$. To be more exact, there are three possible $s=1$ states (corresponding to $m_s=-1$, 0, 1), and one possible $s=0$ state (corresponding to $m_s=0$). The three $s=1$ states are generally known as the triplet states, whereas the $s=0$ state is known as the singlet state.


Table 4: Clebsch-Gordon coefficients for adding spin one-half to spin one-half. Only non-zero coefficients are shown.
  $-1/2, -1/2$ $-1/2, 1/2$ $1/2, -1/2$ $1/2, 1/2$ $m_{s_1},m_{s_2}$
$1, -1$ ${\scriptstyle 1}$        
$1,0$   ${\scriptstyle 1/\sqrt{2}}$ ${\scriptstyle 1/\sqrt{2}}$    
$0, 0$   ${\scriptstyle 1/\sqrt{2}}$ ${\scriptstyle -1/\sqrt{2}}$    
$1, 1$       ${\scriptstyle 1}$  
$s, m_s$          


The Clebsch-Gordon coefficients for adding spin one-half to spin one-half can easily be inferred from Table 2 (with $l=1/2$), and are listed in Table 4. It follows from this table that the three triplet states are:

$\displaystyle \chi^{(2)}_{1,-1}$ $\textstyle =$ $\displaystyle \chi^{(1)}_{-1/2,-1.2},$ (850)
$\displaystyle \chi^{(2)}_{1,0}$ $\textstyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left(\chi^{(1)}_{-1/2,1/2}+ \chi^{(1)}_{1/2,-1/2}\right),$ (851)
$\displaystyle \chi^{(2)}_{1,1}$ $\textstyle =$ $\displaystyle \chi^{(1)}_{1/2,1/2},$ (852)

where $\chi^{(2)}_{s,m_s}$ is shorthand for $\chi^{(2)}_{s_1,s_2;s,m_s}$, etc. Likewise, the singlet state is written:
\begin{displaymath}
\chi^{(2)}_{0,0} = \frac{1}{\sqrt{2}}\left(\chi^{(1)}_{-1/2,1/2}-\chi^{(1)}_{1/2,-1/2}\right).
\end{displaymath} (853)



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Next: Exercises Up: Addition of Angular Momentum Previous: Angular Momentum in the
Richard Fitzpatrick 2010-07-20