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

$${\bf S} ={\bf S}_1 + {\bf S}_2$$ (818)

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

$$S_1^{\,2}\, \chi_{s_1,s_2;m_{s_1},m_{s_2}}^{(1)} = s_1\,(s_1+1)\,\hbar^2 \,\chi_{s_1,s_2;m_{s_1},m_{s_2}}^{(1)},$$ (819)

$$S_2^{\,2}\, \chi_{s_1,s_2;m_{s_1},m_{s_2}}^{(1)} = s_2\,(s_2+1)\,\hbar^2 \,\chi_{s_1,s_2;m_{s_1},m_{s_2}}^{(1)},$$ (820)

$$S_{1z}\, \chi_{s_1,s_2;m_{s_1},m_{s_2}}^{(1)} = m_{s_1}\,\hbar \,\chi_{s_1,s_2;m_{s_1},m_{s_2}}^{(1)},$$ (821)

$$S_{2z}\, \chi_{s_1,s_2;m_{s_1},m_{s_2}}^{(1)} = m_{s_2}\,\hbar \,\chi_{s_1,s_2;m_{s_1},m_{s_2}}^{(1)},$$ (822)

$$S_z\, \chi_{s_1,s_2;m_{s_1},m_{s_2}}^{(1)} = m_s\,\hbar \,\chi_{s_1,s_2;m_{s_1},m_{s_2}}^{(1)}.$$ (823)

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

$$S_1^{\,2}\, \chi_{s_1,s_2;s,m_s}^{(2)} = s_1\,(s_1+1)\,\hbar^2 \,\chi_{s_1,s_2;s,m_s}^{(2)},$$ (824)

$$S_2^{\,2}\, \chi_{s_1,s_2;s,m_s}^{(2)} = s_2\,(s_2+1)\,\hbar^2 \,\chi_{s_1,s_2;s,m_s}^{(2)},$$ (825)

$$S^{2}\, \chi_{s_1,s_2;s,m_s}^{(2)} = s\,(s+1)\,\hbar^2 \,\chi_{s_1,s_2;s,m_s}^{(2)},$$ (826)

$$S_z\, \chi_{s_1,s_2;s,m_s}^{(2)} = m_s\,\hbar \,\chi_{s_1,s_2;s,m_s}^{(2)}.$$ (827)

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,

$$m_s = m_{s_1}+ m_{s_2}.$$ (828)

Now, we saw, in the previous subsection, 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.

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

$$\chi^{(2)}_{1,-1} = \chi^{(1)}_{-1/2,-1.2},$$ (829)

$$\chi^{(2)}_{1,0} = \frac{1}{\sqrt{2}}\left(\chi^{(1)}_{-1/2,1/2}+ \chi^{(1)}_{1/2,-1/2}\right),$$ (830)

$$\chi^{(2)}_{1,1} = \chi^{(1)}_{1/2,1/2},$$ (831)

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:

$$\chi^{(2)}_{0,0} = \frac{1}{\sqrt{2}}\left(\chi^{(1)}_{-1/2,1/2}-\chi^{(1)}_{1/2,-1/2}\right).$$ (832)