# 10.3: 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 ${\bf S} ={\bf S}_1 + {\bf S}_2$ 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 \begin{aligned} S_1^{\,2}\, \chi_{s_1,s_2;m_{s_1},m_{s_2}}^{(1)}&= s_1\,(s_1+1)\,\hbar^{\,2} \,\chi_{s_1,s_2;m_{s_1},m_{s_2}}^{(1)},\\[0.5ex] S_2^{\,2}\, \chi_{s_1,s_2;m_{s_1},m_{s_2}}^{(1)}&=s_2\,(s_2+1)\,\hbar^{\,2} \,\chi_{s_1,s_2;m_{s_1},m_{s_2}}^{(1)},\\[0.5ex] S_{1z}\, \chi_{s_1,s_2;m_{s_1},m_{s_2}}^{(1)}&= m_{s_1}\,\hbar \,\chi_{s_1,s_2;m_{s_1},m_{s_2}}^{(1)},\\[0.5ex] S_{2z}\, \chi_{s_1,s_2;m_{s_1},m_{s_2}}^{(1)}&= m_{s_2}\,\hbar \,\chi_{s_1,s_2;m_{s_1},m_{s_2}}^{(1)},\\[0.5ex] S_z\, \chi_{s_1,s_2;m_{s_1},m_{s_2}}^{(1)}&= m_s\,\hbar \,\chi_{s_1,s_2;m_{s_1},m_{s_2}}^{(1)}.\end{aligned} 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 \begin{aligned} S_1^{\,2}\, \chi_{s_1,s_2;s,m_s}^{(2)}&= s_1\,(s_1+1)\,\hbar^{\,2} \,\chi_{s_1,s_2;s,m_s}^{(2)},\\[0.5ex] S_2^{\,2}\, \chi_{s_1,s_2;s,m_s}^{(2)}&= s_2\,(s_2+1)\,\hbar^{\,2} \,\chi_{s_1,s_2;s,m_s}^{(2)},\\[0.5ex] S^{2}\, \chi_{s_1,s_2;s,m_s}^{(2)}&= s\,(s+1)\,\hbar^{\,2} \,\chi_{s_1,s_2;s,m_s}^{(2)},\\[0.5ex] S_z\, \chi_{s_1,s_2;s,m_s}^{(2)}&=m_s\,\hbar \,\chi_{s_1,s_2;s,m_s}^{(2)}.\end{aligned} Of course, because 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, $m_s = m_{s_1}+ m_{s_2}.$

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.

 $$-1/2, -1/2$$ $$-1/2, 1/2$$ $$1/2,-1/2$$ $$1/2,1/2$$ $$m_{s_1},m_{s_2}$$ [0.5ex] $$1, -1$$ $${\scriptstyle 1}$$ [0.5ex] $$1, 0$$ $${\scriptstyle 1/\sqrt{2}}$$ $${\scriptstyle 1/\sqrt{2}}$$ [0.5ex] $$0, 0$$ $${\scriptstyle 1/\sqrt{2}}$$ $${\scriptstyle -1/\sqrt{2}}$$ [0.5ex] $$1, 1$$ $${\scriptstyle 1}$$ $$s, m_s$$

The Clebsch-Gordon coefficients for adding spin one-half to spin one-half can easily be inferred from Table [t2] (with $$l=1/2$$), and are listed in Table [t4]. It follows from this table that the three triplet states are: \begin{aligned} \chi^{(2)}_{1,-1} &= \chi^{(1)}_{-1/2,-1.2},\\[0.5ex] \chi^{(2)}_{1,0} &= \frac{1}{\sqrt{2}}\left(\chi^{(1)}_{-1/2,1/2}+ \chi^{(1)}_{1/2,-1/2}\right),\\[0.5ex] \chi^{(2)}_{1,1} &= \chi^{(1)}_{1/2,1/2},\end{aligned} where $$\chi^{(2)}_{s,m_s}$$ is shorthand for $$\chi^{(2)}_{s_1,s_2;s,m_s}$$, et cetera. Likewise, the singlet state is written: $\chi^{(2)}_{0,0} = \frac{1}{\sqrt{2}}\left(\chi^{(1)}_{-1/2,1/2}-\chi^{(1)}_{1/2,-1/2}\right).$
