Let’s examine the simplest case concretely: \(j_1=\frac{1}{2}\) and \(j_2=\frac{1}{2}\).
We can form a \(J=1\) state (called a triplet) and
We can form a \(J=0\) state (called a singlet)
Obviously \(j_1=j_2=\frac{1}{2}\)
always. Then \(\left|j,m_j,j_1,j_2\right\rangle=\left|j,m_j\right\rangle\)
can be written as \[\left|j=1,m_j=+1\right\rangle=\left|m_1=+\frac{1}{2},m_2=+\frac{1}{2}\right\rangle=\left|\uparrow\uparrow\right\rangle,\]
since this is the only way to get \(+1\) for \(m_j\) and \[\left|j=1,m_j=-1\right\rangle=\left|m_1=-\frac{1}{2},m_2=-\frac{1}{2}\right\rangle=\left|\downarrow\downarrow\right\rangle,\]
but for \(m_j=0\) we have two
possibilities \[\begin{aligned}
\left|j=1,m_j=0\right\rangle=& \
\alpha\left|m_1=+\frac{1}{2},m_2=-\frac{1}{2}\right\rangle+\beta\left|m_1=-\frac{1}{2},m_2=+\frac{1}{2}\right\rangle\\
=& \
\alpha\left|\uparrow\downarrow\right\rangle+\beta\left|\downarrow\uparrow\right\rangle.
\end{aligned}\] How do we find \(\alpha\) and \(\beta\)?
Answer: Use the total spin lowering operator. We know
\[\begin{aligned}
J^-\left|j=1,m_j=1\right\rangle=& \
\hbar\sqrt{(j+m_j)(j-m_j+1)}\left|j=1,m_j=0\right\rangle \\
=& \ \hbar\sqrt{2\cdot1}\left|j=1,m_j=0\right\rangle \\
=& \ \sqrt{2}\hbar\left|j=1,m_j=0\right\rangle
\end{aligned}\] But \(J^-=J_1^-+J_2^-=S_1^-+S_2^-\) where \(S^-\left|\uparrow\right\rangle=\hbar\left|\downarrow\right\rangle\).
So, \[\begin{aligned}
(S_1^-+S_2^-)\left|\uparrow\uparrow\right\rangle=& \
\hbar\left|\downarrow\uparrow\right\rangle+\hbar\left|\uparrow\downarrow\right\rangle
\\
=& \ \sqrt{2}\hbar\left|j=1,m_j=0\right\rangle
\end{aligned}\] which tells us that \(\alpha=\beta=\frac{1}{\sqrt{2}}\). Hence,
\[\boxed{\left|j=1,m_j=0\right\rangle=\frac{1}{\sqrt{2}}\left|\uparrow\downarrow\right\rangle+\frac{1}{\sqrt{2}}\left|\downarrow\uparrow\right\rangle}\]
How to find \(\left|j=0,m_j=0\right\rangle\)? Well, it
must be orthogonal to \(\left|j=1,m_j=0\right\rangle\), since \[\left\langle
j=1,m_j=0\middle|j=0,m_j=0\right\rangle=0,\] so it must be \[\boxed{\left|j=0,m_j=0\right\rangle=\frac{1}{\sqrt{2}}\left|\uparrow\downarrow\right\rangle-\frac{1}{\sqrt{2}}\left|\downarrow\uparrow\right\rangle}.\]
Note that we have an ambiguity of a \(\pm
1\) for the state.
Here, we consider a general case where \(j_1\) is arbitrary and \(j_2=\frac{1}{2}\). Then, \[j=j_1+\frac{1}{2}\text{ or }j_1-\frac{1}{2}\] Obviously, \[\left|j=j_1+\frac{1}{2},m_j=j_1+\frac{1}{2}\right\rangle=\left|m_1=j_1,m_2=\frac{1}{2}\right\rangle.\] To find \(\left|j=j_1+\frac{1}{2},m_j=j_1-\frac{1}{2}\right\rangle\), use the lowering operator, recalling that \[\begin{aligned} J^-\left|j=j_1+\frac{1}{2},m_j=j_1+\frac{1}{2}\right\rangle=& \ \hbar\sqrt{\left (j_1+\frac{1}{2}+j_1+\frac{1}{2}\right )\left(j_1+\frac{1}{2}-j_1-\frac{1}{2}+1\right )}\left|j=j_1+\frac{1}{2},m_j=j_1-\frac{1}{2}\right\rangle \\ =& \ \hbar\sqrt{2j_1+1}\left|j=j_1+\frac{1}{2},m_j=j_1-\frac{1}{2}\right\rangle. \end{aligned}\] But \(J^-=J_1^-+S_2^-\) so \[J^-\left|j_1,m_2=\frac{1}{2}\right\rangle=\hbar\sqrt{2j_1}\left|m_1=j_1-1,m_2=\frac{1}{2}\right\rangle+\hbar\left|m_1=j_1,m_2=\frac{1}{2}\right\rangle.\] So \[\boxed{\left|j=j_1+\frac{1}{2},m_j=j_1-\frac{1}{2}\right\rangle=\frac{1}{\sqrt{2j_1+1}}\left|m_1=j_1, m_2=-\frac{1}{2}\right\rangle+\sqrt{\frac{2j_1}{2j_1+1}}\left|m_1=j_1-1,m_2=\frac{1}{2}\right\rangle}.\] The state \(\left|j_1=j_1-\frac{1}{2},m_j=j_1-\frac{1}{2}\right\rangle\) is orthogonal to this so \[\boxed{\left|j_1=j_1-\frac{1}{2},m_j=j_1-\frac{1}{2}\right\rangle=\sqrt{\frac{2j_1}{2j_1+1}}\left|m_1=j_1, m_2=-\frac{1}{2}\right\rangle-\frac{1}{\sqrt{2j_1+1}}\left|m_1=j_1-1,m_2=\frac{1}{2}\right\rangle.}\] How do we find lower \(m_j\) values? Use the lowering operator again. \[\begin{aligned} \left|j=j_1+\frac{1}{2},m_j=j_1-\frac{3}{2}\right\rangle=& \ \frac{1}{\hbar\sqrt{2j_1\cdot2}}J^-\left|j=j_1+\frac{1}{2},m_j=j_1-\frac{1}{2}\right\rangle \\ =& \ \frac{1}{\sqrt{2j_1\cdot2(2j_1+1)}}\cdot\sqrt{2j_1\cdot 1}\left|m_1=j_1-1,m_2=-\frac{1}{2}\right\rangle \\ +& \ \frac{1}{\sqrt{2j_1\cdot2(2j_1+1)}}\sqrt{2j_1}\sqrt{(2j_1)2}\left|m_1=j_1-2,m_2=\frac{1}{2}\right\rangle \\ +& \ \frac{1}{\sqrt{2j_1\cdot2(2j_1+1)}}\sqrt{2j_1}\left|m_1=j_1-1,m_2=-\frac{1}{2}\right\rangle. \end{aligned}\] Therefore, \[\boxed{ \left|j=j_1+\frac{1}{2},m_j=j_1-\frac{3}{2}\right\rangle=\sqrt{\frac{2}{2j_1+1}}\left|m_1=j_1-1,m_2=-\frac{1}{2}\right\rangle+\sqrt{\frac{2j_1-1}{2j_1+1}}\left|m_1=j_1-2,m_2=\frac{1}{2}\right\rangle.}\] In general, one finds \[\left|j=j_1+\frac{1}{2},m_j\right\rangle=\sqrt{\frac{j_1-m_j+\frac{1}{2}}{2j_1+1}}\left|m_1=m_j+\frac{1}{2},m_2=-\frac{1}{2}\right\rangle+\sqrt{\frac{j_1+m_j+\frac{1}{2}}{2j_1+1}}\left|m_1=m_j-\frac{1}{2},m_2=\frac{1}{2}\right\rangle.\] Similarly, \[\left|j=j_1-\frac{1}{2},m_j\right\rangle=\sqrt{\frac{j_1+m_j+\frac{1}{2}}{2j_1+1}}\left|m_1=m_j+\frac{1}{2},m_2=-\frac{1}{2}\right\rangle-\sqrt{\frac{j_1-m_j+\frac{1}{2}}{2j_1+1}}\left|m_1=m_j-\frac{1}{2},m_2=\frac{1}{2}\right\rangle.\]
Another way to derive this is as follows. Start with a general expression for the state: \[\left|j,m_j\right\rangle=\alpha\left|m_1=m_j-\frac{1}{2},m_2=\frac{1}{2}\right\rangle+\beta\left|m_1=m_j+\frac{1}{2},m_2=-\frac{1}{2}\right\rangle.\] Using the spherical basis for the dot product, write the total squared angular momentum as \[\begin{aligned} \hat{J}^2=(\hat{J}_1+\hat{J}_2)^2=& \ \hat{J}_1^2+\hat{J}_2^2+2\hat{J}_1\cdot\hat{J}_2 \\ =& \ \hat{J}_1^2+\hat{J}_2^2+2\left(\frac{J_1^+J_2^-+J_1^-J_2^+}{2}\right)+2J_1^zJ_2^z. \end{aligned}\] Then, act \(\hat{J}^2\) on the state and use the known results for the different operators in the uncoupled basis: \[\begin{aligned} \frac{\hat{J}^2}{\hbar^2}\left|j,m_j\right\rangle=& \ \alpha\left|m_1=m_j-\frac{1}{2},m_2=\frac{1}{2}\right\rangle \left(j_1(j_1+1)+\frac{3}{4}+2(m_j-\frac{1}{2})\cdot\frac{1}{2}\right) \\ +& \ \alpha\left|m_1=m_j+\frac{1}{2},m_2=-\frac{1}{2}\right\rangle\left(\sqrt{(j_1-m_j+\frac{1}{2})(j_1+m_j-\frac{1}{2}+1)}\right) \\ +& \ \beta\left|m_1=m_j+\frac{1}{2},m_2=-\frac{1}{2}\right\rangle\left(j_1(j_1+1)+\frac{3}{4}+2(m_j+\frac{1}{2})(-\frac{1}{2})\right) \\ +& \ \beta\left|m_1=m_j-\frac{1}{2},m_2=\frac{1}{2}\right\rangle\left(\sqrt{(j_1+m_j+\frac{1}{2})(j_1-m_j-\frac{1}{2}+1)}\right) \\ =& \ \begin{pmatrix}j_1(j_1+1)+m_j+\frac{1}{4} & \sqrt{(j_1-m_j+\frac{1}{2})(j_1+m_j+\frac{1}{2})} \\ \sqrt{(j_1+m_j+\frac{1}{2})(j_-+m_j+\frac{1}{2})} & j_1(j_1+1)-m_j+\frac{1}{4}\end{pmatrix}\begin{pmatrix}\alpha \\ \beta\end{pmatrix} \end{aligned}\] We want this to be proportional to \(\begin{pmatrix}\alpha \\ \beta\end{pmatrix}\) to be an eigenstate of \(\hat{J}^2\). So, \[\det\begin{pmatrix}j_1(j_1+1)+m_j+\frac{1}{4}-\lambda & \sqrt{(j_1-m_j+\frac{1}{2})(j_1+m_j+\frac{1}{2})} \\ \sqrt{(j_1+m_j+\frac{1}{2})(j_-+m_j+\frac{1}{2})} & j_1(j_1+1)-m_j+\frac{1}{4}-\lambda\end{pmatrix}=0,\] \[\implies \lambda^2+\lambda\left(-2(j_1+\frac{1}{2})^2\right)+\left(j_1+\frac{1}{2}\right)^4-m_j^2-\left(j_1+\frac{1}{2}\right)^2+m_j^2=0.\] Simplifying, we get \[\lambda^2-2\left(j_1+\frac{1}{2}\right)^2\lambda+\left(j_1+\frac{1}{2}\right)^2\left(1+\left(j_1+\frac{1}{2}\right)^2\right)=0.\] Solving for \(\lambda\), we find that \[\lambda=\left(j_1+\frac{1}{2}\right)^2\pm\frac{1}{2}\sqrt{4\left(j_1+\frac{1}{2}\right)^4-4\left(j_1+\frac{1}{2}\right)^4+4\left(j_1+\frac{1}{2}\right)^2}\] \[\implies\left(j_1+\frac{1}{2}\right)\left(j_1+\frac{1}{2}\pm 1\right).\] One root is \(j=j_1+\frac{1}{2}\) and the other is \(j_1-\frac{1}{2}\). Note that we knew this ahead of time, so we did not really need to solve the equation. We now need to find \(\alpha\) and \(\beta\) for each. \(\lambda=(j+1/2)(j_1+1/2\pm 1)\) so, \[\left(\left(j+\frac{1}{2}\right)^2-\left(j+\frac{1}{2}\right)\left(j+\frac{1}{2}\pm1\right)+m_j\right)\alpha+\sqrt{\left(j+\frac{1}{2}\right)^2-m^2}\beta=0\] \[\left(\mp\left(j+\frac{1}{2}\right)+m_j\right)\alpha+\sqrt{\left(j+\frac{1}{2}\right)^2-m^2}\beta=0.\] Therefore, \[\beta=\frac{\pm\left(j+\frac{1}{2}\right)-m_j}{\sqrt{\left(j+\frac{1}{2}\right)^2-m_j^2}}\alpha =\pm\sqrt{\frac{j_1\mp m_j+\frac{1}{2}}{j_1\pm m_j+\frac{1}{2}}}\alpha.\] Now, let \[\begin{aligned} \alpha=& \ C\sqrt{j_1\pm m_j+\frac{1}{2}} \\ \beta=& \ C\sqrt{j_1\mp m_j+\frac{1}{2}} \end{aligned}\] Then, we must demand that \(\alpha^2+\beta^2=1\) \[C^2\left(j_1\pm m_j+\frac{1}{2}+j_1\mp m_j+\frac{1}{2}\right)=1\implies C=\frac{1}{\sqrt{2j_1+1}}.\] Hence, \[\begin{aligned} \left|j_1+\frac{1}{2},m_j\right\rangle=& \ \sqrt{\frac{j_1+m_j+\frac{1}{2}}{2j_1+1}}\left|m_1=m_j-\frac{1}{2},m_2=\frac{1}{2}\right\rangle+\sqrt{\frac{j_1-m_j+\frac{1}{2}}{2j_1+1}}\left|m_1=m_j+\frac{1}{2},m_2=-\frac{1}{2}\right\rangle \\ \left|j_1-\frac{1}{2},m_j\right\rangle=& \ -\sqrt{\frac{j_1-m_j+\frac{1}{2}}{2j_1+1}}\left|m_1=m_j-\frac{1}{2},m_2=\frac{1}{2}\right\rangle+\sqrt{\frac{j_1+m_j+\frac{1}{2}}{2j_1+1}}\left|m_1=m_j+\frac{1}{2},m_2=-\frac{1}{2}\right\rangle. \end{aligned}\]