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.

Clebsch-Gordon coefficients for adding spin one-half to spin one-half. Only non-zero coefficients are shown.

	\(-1/2, -1/2\)	\(-1/2, 1/2\)	\(1/2,-1/2\)	\(1/2,1/2\)	\(m_{s_1},m_{s_2}\)
[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: \[\chi^{(2)}_{0,0} = \frac{1}{\sqrt{2}}\left(\chi^{(1)}_{-1/2,1/2}-\chi^{(1)}_{1/2,-1/2}\right).\]

Contributors and Attributions

• Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)

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