# Sign of the Clebsch-Gordan coefficients

I have been calculating some Clebsch-Gordan coefficients and now I am confused about how to find the sign of the coefficients.

Textbooks usually show the example of the addition of two $\tfrac12$-spin particles and then say that the minus sign there just stems from the orthogonality condition. The problem is how does this work in general? Is there some (easy to apply) rule?

For example: $\left\langle 1, \tfrac12, \!-\!1, +\tfrac12 \:|\: \tfrac12, -\tfrac12\right\rangle=-\sqrt{\tfrac23}$ where I have used the notation $\langle j_1, j_2, m_1, m_2 |J, M \rangle$. Simply using the ladder operators yields $\sqrt{\tfrac23}$, not $-\sqrt{\tfrac23}$.

• No, this CG will vanish, because $m_2 > j_2$. Please correct your question. Commented Aug 19, 2017 at 21:12
• Ups, sorry for those mistakes. I have corrected it, but my question remains. Commented Aug 21, 2017 at 17:43
• You still miss the $\sqrt{.}$. Commented Aug 21, 2017 at 18:08

I don't know how you calculate the coefficients, using a formula or other method. I post here an answer initially as Example therein : Total spin of two spin-$1/2$ particles, but I deleted as so extensive. Ignore any reference to equations not appearing in the Example.

The coefficient you are trying to calculate appears at the end in equation (Ex-26.2) which is repeated here for convenience :

$$\mathbf{\left|\tfrac{1}{2}\;,-\tfrac{1}{2}\right\rangle_{[1]}} =\sqrt{\tfrac{2}{3}}\cdot\left|\tfrac{1}{2},+\tfrac{1}{2}\right\rangle_{\!a}\left|1,\:\!\!-\!1\right\rangle_{\!b}-\sqrt{\tfrac{1}{3}}\cdot\left|\tfrac{1}{2},-\tfrac{1}{2}\right\rangle_{\!a}\left|1,\:0\:\right\rangle_{\!b} \tag{Ex-26.2}$$

This coefficient is $\:+\sqrt{\tfrac{2}{3}}$. But it could be $\:-\sqrt{\tfrac{2}{3}}$ also, since no-one could forbid us to use as basic states of the product system the opposite ones given in equations (Ex-26.2). It's a matter of convention.

You can find calculations of Clebsch-Gordan coefficients using a general formula in Examples therein :How to determine whether an eigenstate of total spin is symmetric or antisymmetric?

Example

$\qquad j_{\alpha}=\frac{1}{2}\:, \quad j_{\beta}=1$

Let the system $\;\alpha\;$ be a particle $\;p_{\alpha}\;$ with spin $\;j_{\alpha}=1/2\;$ and the system $\;\beta\;$ be a particle $\;p_{\beta}\;$ with orbital angular momentum or spin $\;j_{\beta}=1$. Alternatively, the system $\;\beta\;$ may be the same particle $\;p_{\alpha}\;$ with orbital angular momentum $\;j_{\beta}=1$.

So, in system $\;\alpha\;$ $$J^{\boldsymbol{\alpha}}_{1}=\tfrac{1}{2} \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}, \quad J^{\boldsymbol{\alpha}}_{2}=\tfrac{1}{2} \begin{bmatrix} 0 & \!\!\!-i \\ i & 0 \end{bmatrix}, \quad J^{\boldsymbol{\alpha}}_{3}=\tfrac{1}{2} \begin{bmatrix} 1 & 0 \\ 0 & \!\!\!-1 \end{bmatrix} \tag{Ex-01}$$ and $$\left(\mathbf{J}^{\boldsymbol{\alpha}}\right)^{2}=\left(J^{\boldsymbol{\alpha}}_{1}\right)^{2}+\left(J^{\boldsymbol{\alpha}}_{2}\right)^{2}+\left(J^{\boldsymbol{\alpha}}_{3}\right)^{2}= j_{\alpha}\left( j_{\alpha}+1\right)\cdot \mathrm{I}_{\mathbf{a}}=\tfrac{3}{4} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \tag{Ex-02}$$ The basic vectors $\mathbf{a}_{\imath}\: (\imath=1,2)$ are the common eigenvectors of $\left(\mathbf{J}^{\alpha}\right)^{2}$ and $J^{\alpha}_{3}$ : \begin{align} \mathbf{a}_{1} & = \left|j_{\alpha},m^{\alpha}_{1} \right\rangle_{\!a}=\left|\tfrac{1}{2},+\tfrac{1}{2}\right\rangle_{\!a} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}_{\!a} \tag{Ex-03.1}\\ \mathbf{a}_{2} & = \left|j_{\alpha},m^{\alpha}_{2} \right\rangle_{\!a}=\left|\tfrac{1}{2},-\tfrac{1}{2}\right\rangle_{\!a} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}_{\!a} \tag{Ex-03.2} \end{align} A state of system $\alpha$ is represented by a 2-dimensional complex vector $\boldsymbol{\xi}$ \begin{align} \boldsymbol{\xi} & = \xi_{1}\mathbf{a}_{1}\!\!+\!\xi_{2}\mathbf{a}_{2}=\xi_{1}\left|j_{\alpha},m^{\alpha}_{1} \right\rangle_{\!a}\!\!+\!\xi_{2}\left|j_{\alpha},m^{\alpha}_{2} \right\rangle_{\!a} \nonumber\\ & = \xi_{1}\left|\tfrac{1}{2},+\tfrac{1}{2}\right\rangle_{\!a}\!\!+\!\xi_{2}\left|\tfrac{1}{2},-\tfrac{1}{2}\right\rangle_{\!a} = \xi_{1}\!\! \begin{bmatrix} 1 \\ 0 \end{bmatrix}_{\!a} \!\!+\! \xi_{2}\!\! \begin{bmatrix} 0 \\ 1 \end{bmatrix}_{\!a} = \begin{bmatrix} \xi_{1} \\ \xi_{2} \end{bmatrix}_{\!a} \tag{Ex-04} \end{align} in Hilbert space $$\mathsf{H}_{\alpha}\equiv\left\{\boldsymbol{\xi}\in \mathbb{C}^{\boldsymbol{2}}: \boldsymbol{\xi}= \xi_{1}\mathbf{a}_{1}+\xi_{2}\mathbf{a}_{2} \right\} \tag{Ex-05}$$

In system $\;\beta\;$ $$J^{\boldsymbol{\beta}}_{1}=\sqrt{\tfrac{1}{2}} \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}, \quad J^{\boldsymbol{\beta}}_{2}=\sqrt{\tfrac{1}{2}} \begin{bmatrix} 0 & \!\!\!-i & 0 \\ i & 0 & \!\!\!-i \\ 0 & i & 0 \end{bmatrix}, \quad J^{\boldsymbol{\beta}}_{3}= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & \!\!\!-1 \end{bmatrix} \tag{Ex-06}$$ and $$\left(\mathbf{J}^{\boldsymbol{\beta}}\right)^{2}=\left(J^{\boldsymbol{\beta}}_{1}\right)^{2}+\left(J^{\boldsymbol{\beta}}_{2}\right)^{2}+\left(J^{\boldsymbol{\beta}}_{3}\right)^{2}= j_{\beta}\left( j_{\beta}+1\right)\cdot \mathrm{I}_{\mathbf{b}}=2 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \tag{Ex-07}$$ The basic vectors $\mathbf{b}_{\jmath}\: (\jmath=1,2,3)$ are the common eigenvectors of $\left(\mathbf{J}^{\beta}\right)^{2}$ and $J^{\beta}_{3}$ : \begin{align} \mathbf{b}_{1} & = \left|j_{\beta},m^{\beta}_{1} \right\rangle_{\!b}=\left|1,\:\!\!+\!1\right\rangle_{\!b} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}_{\!b} \tag{Ex-08.1}\\ \mathbf{b}_{2} & = \left|j_{\beta},m^{\beta}_{2} \right\rangle_{\!b}=\left|1,\:0\:\right\rangle_{\!b} = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}_{\!b} \tag{Ex-08.2}\\ \mathbf{b}_{3} & = \left|j_{\beta},m^{\beta}_{3} \right\rangle_{\!b}=\left|1,\:\!\!-\!1\right\rangle_{\!b} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}_{\!b} \tag{Ex-08.3} \end{align} A state of system $\beta$ is represented by a 3-dimensional complex vector $\boldsymbol{\eta}$ \begin{align} \boldsymbol{\eta} & =\eta_{1}\mathbf{b}_{1}\!+\!\eta_{2}\mathbf{b}_{2}\!+\!\eta_{3}\mathbf{b}_{3}= \eta_{1}\left|j_{\beta},m^{\beta}_{1}\right\rangle_{\!b}\!+\!\eta_{2} \left|j_{\beta},m^{\beta}_{2} \right\rangle_{\!b}\!+\!\eta_{3}\left|j_{\beta},m^{\beta}_{3} \right\rangle_{\!b} \nonumber\\ & = \eta_{1}\!\left|1,\:\!\!+\!1\right\rangle_{\!b}\!+\!\eta_{2}\!\left|1,\:0\:\right\rangle_{\!b}\!+\!\eta_{3}\!\left|1,\:\!\!-\!1\right\rangle_{\!b} \!=\! \eta_{1}\!\! \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}_{\!b} \!\!\!\!+\!\eta_{2}\!\! \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}_{\!b} \!\!\!\!+\!\eta_{3}\!\! \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}_{\!b} \!=\! \begin{bmatrix} \eta_{1} \\ \eta_{2} \\ \eta_{3} \end{bmatrix}_{\!b} \tag{Ex-09} \end{align} in Hilbert space $$\mathsf{H}_{\beta}\equiv\left\{\boldsymbol{\eta}\in \mathbb{C}^{\boldsymbol{3}}: \boldsymbol{\eta} =\eta_{1}\mathbf{b}_{1}+\eta_{2}\mathbf{b}_{2}+\eta_{3}\mathbf{b}_{3} \right\} \tag{Ex-10}$$

According to the general equations (15) a product state of the composite system is $$\boldsymbol{\chi} = \boldsymbol{\xi} \boldsymbol{\otimes} \boldsymbol{\eta}=\left( \sum_{\imath=1}^{\imath=2}\xi_{\imath}\mathbf{a}_{\imath}\right) \boldsymbol{\otimes}\left( \sum_{\jmath=1}^{\jmath=3}\eta_{\jmath}\mathbf{b}_{\jmath}\right)= \sum_{\imath,\jmath=1,1}^{\imath,\jmath=2,3}\xi_{\imath}\eta_{\jmath}\left( \mathbf{a}_{\imath} \boldsymbol{\otimes }\mathbf{b}_{\jmath}\right) \tag{Ex-11}$$ with matrix representation, in agreement with equation (18) $$\boldsymbol{\chi}= \begin{bmatrix} \begin{array}{c} \chi_{1} \\ \chi_{2} \\ \chi_{3} \\ \chi_{4}\\ \chi_{5}\\ \chi_{6} \end{array} \end{bmatrix}_{\!e}= \begin{bmatrix} \begin{array}{c} \xi_{1}\eta_{1} \\ \xi_{1}\eta_{2} \\ \xi_{1}\eta_{3} \\ \xi_{2}\eta_{1} \\ \xi_{2}\eta_{2} \\ \xi_{2}\eta_{3} \end{array} \end{bmatrix}_{\!e}= \boldsymbol{\xi} \boldsymbol{\otimes} \boldsymbol{\eta} \tag{Ex-12}$$ This representation is relatively to the basis $\:\left\lbrace \mathbf{e}_{k}\right\rbrace \:$ defined according to the general equations (16): \begin{align} \mathbf{e}_{1} & \equiv \mathbf{a}_{1}\boldsymbol{\otimes} \mathbf{b}_{1}=\left|\tfrac{1}{2},+\tfrac{1}{2}\right\rangle_{\!a}\boldsymbol{\otimes}\left|1,\:\!\!+\!1\right\rangle_{\!b}= \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}^{\!\mathsf{T}} \tag{Ex-13.1}\\ \mathbf{e}_{2} & \equiv \mathbf{a}_{1}\boldsymbol{\otimes} \mathbf{b}_{2}=\left|\tfrac{1}{2},+\tfrac{1}{2}\right\rangle_{\!a}\boldsymbol{\otimes}\left|1,\:0\:\right\rangle_{\!b}= \begin{bmatrix} 0 & 1 & 0 & 0 & 0 & 0 \end{bmatrix}^{\!\mathsf{T}} \tag{Ex-13.2}\\ \mathbf{e}_{3} & \equiv \mathbf{a}_{1}\boldsymbol{\otimes} \mathbf{b}_{3}=\left|\tfrac{1}{2},+\tfrac{1}{2}\right\rangle_{\!a}\boldsymbol{\otimes}\left|1,\:\!\!-\!1\right\rangle_{\!b}= \begin{bmatrix} 0 & 0 & 1 & 0 & 0 & 0 \end{bmatrix}^{\!\mathsf{T}} \tag{Ex-13.3}\\ \mathbf{e}_{4} & \equiv \mathbf{a}_{2}\boldsymbol{\otimes} \mathbf{b}_{1}=\left|\tfrac{1}{2},-\tfrac{1}{2}\right\rangle_{\!a}\boldsymbol{\otimes}\left|1,\:\!\!+\!1\right\rangle_{\!b}= \begin{bmatrix} 0 & 0 & 0 & 1 & 0 & 0 \end{bmatrix}^{\!\mathsf{T}} \tag{Ex-13.4}\\ \mathbf{e}_{5} & \equiv \mathbf{a}_{2}\boldsymbol{\otimes} \mathbf{b}_{2}=\left|\tfrac{1}{2},-\tfrac{1}{2}\right\rangle_{\!a}\boldsymbol{\otimes}\left|1,\:0\:\right\rangle_{\!b}= \begin{bmatrix} 0 & 0 & 0 & 0 & 1 & 0 \end{bmatrix}^{\!\mathsf{T}} \tag{Ex-13.5}\\ \mathbf{e}_{6} & \equiv \mathbf{a}_{2}\boldsymbol{\otimes} \mathbf{b}_{3}=\left|\tfrac{1}{2},-\tfrac{1}{2}\right\rangle_{\!a}\boldsymbol{\otimes}\left|1,\:\!\!-\!1\right\rangle_{\!b}= \begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix}^{\!\mathsf{T}} \tag{Ex-13.6} \end{align} where the symbol $\:\mathsf{T}\:$ means the Transpose.

According to the general equation (23) the product space is the 6-dimensional complex Hilbert space $$\mathsf{H}_{f}=\mathsf{H}_{\alpha}\boldsymbol{\otimes}\mathsf{H}_{\beta}\equiv \lbrace \; \boldsymbol{\chi} \; : \;\boldsymbol{\chi}=\sum_{k=1}^{k=6}\chi_{k}\mathbf{e}_{k},\;\chi_{k} \in \mathbb{C} \rbrace \tag{Ex-14}$$ identical to $\mathbb{C}^{6}$.

From equation (69c) with the help of (47), both repeated here for convenience
$$J_{3} = \Bigl(J^{\boldsymbol{\alpha}}_{3}\boldsymbol{\otimes}\mathrm{I}_{\boldsymbol{\beta}}\Bigr)+\Bigl(\mathrm{I}_{\boldsymbol{\alpha}}\boldsymbol{\otimes} J^{\boldsymbol{\beta}}_{3}\Bigr) \tag{69c}$$ $$\mathrm{C}=\mathrm{A}\boldsymbol{\otimes} \mathrm{B} = \begin{bmatrix} a_{11}\mathrm{B} & a_{12}\mathrm{B} & \cdots & a_{1 \rho}\mathrm{B} & \cdots & a_{1r}\mathrm{B} \\ a_{21}\mathrm{B} & a_{22}\mathrm{B} & \cdots & a_{2 \rho}\mathrm{B} & \cdots & a_{2r}\mathrm{B} \\ \vdots & \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{\imath 1}\mathrm{B} & a_{\imath 2}\mathrm{B} & \cdots & a_{\imath \rho}\mathrm{B} & \cdots & a_{\imath r}\mathrm{B}\\ \vdots & \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{r1}\mathrm{B} & a_{r2}\mathrm{B} & \cdots & a_{r \rho}\mathrm{B} & \cdots & a_{rr}\mathrm{B} \end{bmatrix} \tag{47}$$ and the matrix expressions of $\:J^{\boldsymbol{\alpha}}_{3}\:$ and $\:J^{\boldsymbol{\beta}}_{3}\:$ in equations (Ex-01) and (Ex-06) respectively, we have \begin{align} \Bigl(J^{\boldsymbol{\alpha}}_{3}\boldsymbol{\otimes}\mathrm{I}_{\boldsymbol{\beta}}\Bigr) & = \tfrac{1}{2} \begin{bmatrix} 1 & 0 \\ & \\ 0 & \!\!\!-1 \end{bmatrix} \boldsymbol{\otimes} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} \tfrac{1}{2} & 0 & 0 & 0 & 0 & 0\\ 0 & \tfrac{1}{2} & 0 & 0 & 0 & 0\\ 0 & 0 & \tfrac{1}{2} & 0 & 0 & 0\\ 0 & 0 & 0 & \!\!\!-\tfrac{1}{2} & 0 & 0\\ 0 & 0 & 0 & 0 & \!\!\!-\tfrac{1}{2} & 0\\ 0 & 0 & 0 & 0 & 0 & \!\!\!-\tfrac{1}{2} \end{bmatrix} \tag{Ex-15.1}\\ \Bigl(\mathrm{I}_{\boldsymbol{\alpha}}\boldsymbol{\otimes} J^{\boldsymbol{\beta}}_{3}\Bigr)& =\quad \!\! \begin{bmatrix} 1 & 0 \\ & \\ 0 & 1 \end{bmatrix} \boldsymbol{\otimes} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & \!\!\!-1 \end{bmatrix} = \begin{bmatrix} \;1 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & -1 & 0 & 0 & 0\\ 0 & 0 & 0 & \;1 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & -1 \end{bmatrix} \tag{Ex-15.2} \end{align} Adding equations (15.1), (15.2) we find the following diagonal form of $\:J_{\boldsymbol{3}}\:$ $$J_{\boldsymbol{3}} = \begin{bmatrix} \begin{array}{cccccc} \!\!+\frac{3}{2}&0&0&0&0&0\\ 0&\!\!+\frac{1}{2}&0&0&0&0\\ 0&0&\!\!-\frac{1}{2}&0&0&0\\ 0&0&0&\!\!+\frac{1}{2}&0&0\\ 0&0&0&0&\!\!-\frac{1}{2}&0\\ 0&0&0&0&0&\!\!-\frac{3}{2} \end{array} \end{bmatrix} \tag{Ex-16}$$ As expected its diagonal elements, that is its eigenvalues, are all possible sums $\;\left(m^{\boldsymbol{\alpha}}_{\boldsymbol{\imath}}+m^{\boldsymbol{\beta}}_{\boldsymbol{\jmath}}\right) \;$ of the corresponding eigenvalues of its summands. These are the $\;2\cdot3\;$ combinations of \begin{align} m^{\boldsymbol{\alpha}}_{\boldsymbol{\imath}} & = +\tfrac{1}{2},-\tfrac{1}{2} \tag{16.1a}\\ m^{\boldsymbol{\beta}}_{\boldsymbol{\jmath}} & = +1,0,-1 \tag{16.1b} \end{align} From general expression for $\:\mathbf{J}^{\boldsymbol{2}}\:$, equation (80), which we repeat here for convenience $$\mathbf{J}^{\boldsymbol{2}} =\bigl[ j_{\alpha}(j_{\alpha}+1)+ j_{\beta}(j_{\beta}+1) \bigr] \mathrm{I}_{f} +2\sum_{q=1}^{q=3}\Bigl( J^{\boldsymbol{\alpha}}_{\boldsymbol{q}}\boldsymbol{\otimes}J^{\boldsymbol{\beta}}_{\boldsymbol{q}}\Bigr) \tag{80}$$ we have for the first term of the right hand side the following scalar multiple of the $\:6 \times 6\:$ identity matrix $$\bigl[ j_{\alpha}(j_{\alpha}+1)+ j_{\beta}(j_{\beta}+1) \bigr] \mathrm{I}_{f}= \bigl(\tfrac{3}{4}+2 \bigr) \mathrm{I}_{f}= \tfrac{11}{4} \begin{bmatrix} \begin{array}{cccccc} 1&0&0&0&0&0\\ 0&1&0&0&0&0\\ 0&0&1&0&0&0\\ 0&0&0&1&0&0\\ 0&0&0&0&1&0\\ 0&0&0&0&0&1 \end{array} \end{bmatrix} \tag{Ex-17}$$ while using the matrix representations of $\:J^{\boldsymbol{\alpha}}_{\boldsymbol{q}}\:$ and $\:J^{\boldsymbol{\beta}}_{\boldsymbol{q}}\:$ for the three terms in series, equations (Ex-01) and (Ex-06) respectively, we have successively \begin{align} J^{\boldsymbol{\alpha}}_{\boldsymbol{1}}\boldsymbol{\otimes}J^{\boldsymbol{\beta}}_{\boldsymbol{1}} & = \tfrac{1}{2} \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \boldsymbol{\otimes} \sqrt{\tfrac{1}{2}} \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} = \tfrac{1}{2\sqrt{2}} \begin{bmatrix} 0&0&0&0&1&0\\ 0&0&0&1&0&1\\ 0&0&0&0&1&0\\ 0&1&0&0&0&0\\ 1&0&1&0&0&0\\ 0&1&0&0&0&0 \end{bmatrix} \tag{Ex-18.1}\\ J^{\boldsymbol{\alpha}}_{\boldsymbol{2}}\boldsymbol{\otimes}J^{\boldsymbol{\beta}}_{\boldsymbol{2}} & = \tfrac{1}{2} \begin{bmatrix} 0 & \!\!\!-i \\ i & 0 \end{bmatrix} \boldsymbol{\otimes} \sqrt{\tfrac{1}{2}} \begin{bmatrix} 0 & \!\!\!-i & 0 \\ i & 0 & \!\!\!-i \\ 0 & i & 0 \end{bmatrix} =\tfrac{1}{2\sqrt{2}}\!\! \begin{bmatrix} 0&0&0&0&\!\!\!\!-\!1&0\\ 0&0&0&1&0&\!\!\!\!-\!1\\ 0&0&0&0&1&0\\ 0&1&0&0&0&0\\ \!\!-\!1&0&1&0&0&0\\ 0&\!\!\!\!-\!1&0&0&0&0 \end{bmatrix} \tag{Ex-18.2}\\ J^{\boldsymbol{\alpha}}_{\boldsymbol{3}}\boldsymbol{\otimes}J^{\boldsymbol{\beta}}_{\boldsymbol{3}} & = \tfrac{1}{2} \begin{bmatrix} 1 & 0 \\ 0 & \!\!\!-1 \end{bmatrix} \boldsymbol{\otimes} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & \!\!\!-1 \end{bmatrix} =\tfrac{1}{2} \begin{bmatrix} 1&0&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&\!\!\!-\!1&0&0&0\\ 0&0&0&\!\!\!-\!1&0&0\\ 0&0&0&0&0&0\\ 0&0&0&0&0&1 \end{bmatrix} \tag{Ex-18.3} \end{align} so adding equations (18) $$2\sum_{q=1}^{q=3}\Bigl( J^{\boldsymbol{\alpha}}_{\boldsymbol{q}}\boldsymbol{\otimes}J^{\boldsymbol{\beta}}_{\boldsymbol{q}}\Bigr)= \begin{bmatrix} 1&0&0&0&0&0\\ 0&0&0&\sqrt{2}&0&0\\ 0&0&-1&0&\sqrt{2}&0\\ 0&\sqrt{2}&0&-1&0&0\\ 0&0&\sqrt{2}&0&0&0\\ 0&0&0&0&0&1 \end{bmatrix} \tag{Ex-19}$$ while adding equations (17) and (19) we have finally for $\:\mathbf{J}^{\boldsymbol{2}}\:$ $$\mathbf{J}^{\boldsymbol{2}} = \begin{bmatrix} \frac{15}{4}&0&0&0&0&0\\ 0&\frac{11}{4}&0&\sqrt{2}&0&0\\ 0&0&\frac{7}{4}&0&\sqrt{2}&0\\ 0&\sqrt{2}&0&\frac{7}{4}&0&0\\ 0&0&\sqrt{2}&0&\frac{11}{4}&0\\ 0&0&0&0&0&\frac{15}{4} \end{bmatrix} \tag{Ex-20}$$ Now, to find the eigenvalues and eigenvectors of this $\;6\times 6 \;$ symmetric matrix $\:\mathbf{J}^{\boldsymbol{2}}\:$ is not so difficult as it seems from a first glance because :

1. The state $\:\mathbf{e}_{1}\:$ is a common eigenstate of $\:J_{\boldsymbol{3}}\:$ and $\:\mathbf{J}^{\boldsymbol{2}} \:$ of eigenvalue $\widetilde{m}_{1}=+\tfrac{3}{2}$ and $\:\widetilde{\lambda}_{1}=\tfrac{15}{4}=\tfrac{3}{2}\left(\tfrac{3}{2}+1\right)\:$ respectively : \begin{align} J_{\boldsymbol{3}}\mathbf{e}_{1} & = \widetilde{m}_{1}\cdot\mathbf{e}_{1} =\left( +\tfrac{3}{2}\right) \cdot\mathbf{e}_{1} \tag{Ex-21a}\\ \mathbf{J}^{\boldsymbol{2}}\mathbf{e}_{1} & = \widetilde{\lambda}_{1}\cdot\mathbf{e}_{1}=\tfrac{15}{4}\cdot\mathbf{e}_{1}=\tfrac{3}{2}\left(\tfrac{3}{2}+1\right)\cdot\mathbf{e}_{1} \tag{Ex-21b} \end{align}

The operator $\:\mathbf{J}^{\boldsymbol{2}} \:$ leaves invariant the 1-dimensional eigenspace of $\:J_{\boldsymbol{3}}\:$ with eigenvalue $\:\widetilde{m}_{1}=+\tfrac{3}{2}$, that is the subspace spanned by the state $\:\left\lbrace \mathbf{e}_{1}\right\rbrace$.

2. The state $\:\mathbf{e}_{6}\:$ is a common eigenstate of $\:J_{\boldsymbol{3}}\:$ and $\:\mathbf{J}^{\boldsymbol{2}} \:$ of eigenvalue $\widetilde{m}_{6}=-\tfrac{3}{2}$ and $\:\widetilde{\lambda}_{6}=\tfrac{15}{4}=\tfrac{3}{2}\left(\tfrac{3}{2}+1\right)\:$ respectively : \begin{align} J_{\boldsymbol{3}}\mathbf{e}_{6} & = \widetilde{m}_{6}\cdot\mathbf{e}_{6}=\left( -\tfrac{3}{2}\right) \cdot\mathbf{e}_{6} \tag{Ex-22a}\\ \mathbf{J}^{\boldsymbol{2}}\mathbf{e}_{6} & = \widetilde{\lambda}_{6}\cdot\mathbf{e}_{6}=\tfrac{15}{4}\cdot\mathbf{e}_{6}=\tfrac{3}{2}\left(\tfrac{3}{2}+1\right)\cdot\mathbf{e}_{6} \tag{Ex-22b} \end{align}

The operator $\:\mathbf{J}^{\boldsymbol{2}} \:$ leaves invariant the 1-dimensional eigenspace of $\:J_{\boldsymbol{3}}\:$ with eigenvalue $\:\widetilde{m}_{6}=-\tfrac{3}{2}$, that is the subspace spanned by the state $\:\left\lbrace \mathbf{e}_{6}\right\rbrace$.

3.The eigenstates $\:\mathbf{e}_{2}\:$ and $\:\mathbf{e}_{4}\:$ of $\:J_{\boldsymbol{3}}\:$ with eigenvalue $\:+\tfrac{1}{2}\:$ are transformed by $\:\mathbf{J}^{\boldsymbol{2}} \:$ to linear combinations of these same eigenstates \begin{align} \mathbf{J}^{\boldsymbol{2}}\mathbf{e}_{2} & = \tfrac{11}{4}\cdot\mathbf{e}_{2}+\sqrt{2}\cdot\mathbf{e}_{4} \tag{Ex-23a}\\ \mathbf{J}^{\boldsymbol{2}}\mathbf{e}_{4} & = \sqrt{2}\cdot\mathbf{e}_{2}+\tfrac{7}{4}\cdot\mathbf{e}_{4} \tag{Ex-23b} \end{align}

The operator $\:\mathbf{J}^{\boldsymbol{2}} \:$ leaves invariant the 2-dimensional eigenspace of $\:J_{\boldsymbol{3}}\:$ with eigenvalue $\:+\tfrac{1}{2}=\widetilde{m}_{2}=\widetilde{m}_{4}$, that is the subspace spanned by states $\:\left\lbrace \mathbf{e}_{2},\mathbf{e}_{4} \right\rbrace \:$. Its restriction on this subspace is represented by a real symmetric $\:2 \times 2\:$ matrix, so it has in this subspace two real eigenvalues which moreover are positive and different, see in the following.

4.The eigenstates $\:\mathbf{e}_{3}\:$ and $\:\mathbf{e}_{5}\:$ of $\:J_{\boldsymbol{3}}\:$ with eigenvalue $\:-\tfrac{1}{2}\:$ are transformed by $\:\mathbf{J}^{\boldsymbol{2}} \:$ to linear combinations of these same eigenstates \begin{align} \mathbf{J}^{\boldsymbol{2}}\mathbf{e}_{3} & = \tfrac{7}{4}\cdot\mathbf{e}_{3}+\sqrt{2}\cdot\mathbf{e}_{5} \tag{Ex-24a}\\ \mathbf{J}^{\boldsymbol{2}}\mathbf{e}_{5} & = \sqrt{2}\cdot\mathbf{e}_{3}+\tfrac{11}{4}\cdot\mathbf{e}_{5} \tag{Ex-24b}\\ \end{align}

The operator $\:\mathbf{J}^{\boldsymbol{2}} \:$ leaves invariant the 2-dimensional eigenspace of $\:J_{\boldsymbol{3}}\:$ with eigenvalue $\:-\tfrac{1}{2}=\widetilde{m}_{3}=\widetilde{m}_{5}$, that is the subspace spanned by states $\:\left\lbrace \mathbf{e}_{3},\mathbf{e}_{5}\right\rbrace$. Its restriction on this subspace is represented by a real symmetric $\:2 \times 2\:$ matrix, so it has in this subspace two real eigenvalues which moreover are positive and different, see in the following.

$$\widehat{\mathbf{J}}^{\boldsymbol{2}} = \begin{bmatrix} \begin{array}{cc|cccc} \frac{3}{4}& & & & & \\ &\frac{3}{4} & & & &\\ \hline & &\frac{15}{4}& & & \\ & & &\frac{15}{4}& & \\ & & & &\frac{15}{4}& \\ & & & & &\frac{15}{4} \end{array} \end{bmatrix} \tag{Ex-25}$$ $$\widehat{J}_{\boldsymbol{1}} = \begin{bmatrix} \begin{array}{cc|cccc} 0 & \tfrac{1}{2}& & & & \\ \tfrac{1}{2}&0& & & & \\ \hline & &0&\tfrac{\sqrt{3}}{2}&0&0 \\ & &\tfrac{\sqrt{3}}{2}&0&1&0 \\ & &0&1&0&\tfrac{\sqrt{3}}{2} \\ & &0&0&\tfrac{\sqrt{3}}{2}&0 \end{array} \end{bmatrix} \tag{Ex-25.1}$$ $$\widehat{J}_{\boldsymbol{2}} = \begin{bmatrix} \begin{array}{cc|cccc} 0 &\!\!\!-i\tfrac{1}{2}& & & & \\ i\tfrac{1}{2} & 0& & & & \\ \hline & &0&\!\!\!\!-i\tfrac{\sqrt{3}}{2}&0&0\\ & &\!\!\!i\tfrac{\sqrt{3}}{2}&0&-i&0 \\ & &0&i&0&\!\!\!\!-i\tfrac{\sqrt{3}}{2} \\ & &0&0&\!\!\!i\tfrac{\sqrt{3}}{2}&0 \end{array} \end{bmatrix} \tag{Ex-25.2}$$ $$\widehat{J}_{\boldsymbol{3}} = \begin{bmatrix} \begin{array}{cc|cccc} \frac{1}{2}& & & & & \\ &-\frac{1}{2} & & & & \\ \hline & &\frac{3}{2}& & & \\ & & &\frac{1}{2}& & \\ & & & &-\frac{1}{2}& \\ & & & & &-\frac{3}{2} \end{array} \end{bmatrix} \tag{Ex-25.3}$$

Above matrix representations are with respect to the basis $\:\lbrace\mathbf{f}_{k}, k=1,2,3,4,5,6 \rbrace$ (see Table 2). \begin{align} \mathbf{f}_{1} & = \mathbf{\left|\tfrac{1}{2}\;,+\tfrac{1}{2}\right\rangle_{[1]}} =\sqrt{\tfrac{1}{3}}\cdot\left|\tfrac{1}{2},+\tfrac{1}{2}\right\rangle_{\!a}\left|1,\:0\:\right\rangle_{\!b}-\sqrt{\tfrac{2}{3}}\cdot\left|\tfrac{1}{2},-\tfrac{1}{2}\right\rangle_{\!a}\left|1,\:\!\!+\!1\right\rangle_{\!b} \tag{Ex-26.1}\\ \mathbf{f}_{2} & = \mathbf{\left|\tfrac{1}{2}\;,-\tfrac{1}{2}\right\rangle_{[1]}} =\sqrt{\tfrac{2}{3}}\cdot\left|\tfrac{1}{2},+\tfrac{1}{2}\right\rangle_{\!a}\left|1,\:\!\!-\!1\right\rangle_{\!b}-\sqrt{\tfrac{1}{3}}\cdot\left|\tfrac{1}{2},-\tfrac{1}{2}\right\rangle_{\!a}\left|1,\:0\:\right\rangle_{\!b} \tag{Ex-26.2}\\ \mathbf{f}_{3} & = \mathbf{\left|\tfrac{3}{2}\;,+\tfrac{3}{2}\right\rangle_{[2]}} = \left|\tfrac{1}{2},+\tfrac{1}{2}\right\rangle_{\!a}\left|1,\:\!\!+\!1\right\rangle_{\!b} \tag{Ex-26.3}\\ \mathbf{f}_{4} & = \mathbf{\left|\tfrac{3}{2}\;,+\tfrac{1}{2}\right\rangle_{[2]}} =\sqrt{\tfrac{2}{3}}\cdot\left|\tfrac{1}{2},+\tfrac{1}{2}\right\rangle_{\!a}\left|1,\:0\:\right\rangle_{\!b}+\sqrt{\tfrac{1}{3}}\cdot\left|\tfrac{1}{2},-\tfrac{1}{2}\right\rangle_{\!a}\left|1,\:\!\!+\!1\right\rangle_{\!b} \tag{Ex-26.4}\\ \mathbf{f}_{5} & = \mathbf{\left|\tfrac{3}{2}\;,-\tfrac{1}{2}\right\rangle_{[2]}} =\sqrt{\tfrac{1}{3}}\cdot\left|\tfrac{1}{2},+\tfrac{1}{2}\right\rangle_{\!a}\left|1,\:\!\!-\!1\right\rangle_{\!b}+\sqrt{\tfrac{2}{3}}\cdot\left|\tfrac{1}{2},-\tfrac{1}{2}\right\rangle_{\!a}\left|1,\:0\:\right\rangle_{\!b} \tag{Ex-26.5}\\ \mathbf{f}_{6} & = \mathbf{\left|\tfrac{3}{2}\;,-\tfrac{3}{2}\right\rangle_{[2]}} = \left|\tfrac{1}{2},-\tfrac{1}{2}\right\rangle_{\!a}\left|1,\:\!\!-\!1\right\rangle_{\!b} \tag{Ex-26.6} \end{align} Expressions are simplified by omitting the symbol $\;'\boldsymbol{\otimes}'\;$ in the product of states.

Your Clebsch-Gordan doesn't mean anything. First of all, in the $j$ representation, only states with $m=-j,-j+1,\ldots,j$ appear. So if $j_2 = 1/2$, you can only have $m_2 = \pm 1/2$, not $m_2 = 1$. Second, the magnetic quantum number is not conserved: on the one hand, you have $m_1 + m_2 = 3/2$, on the other hand $M = 1/2 \neq 3/2$.

As far as I know, figuring out when a Clebsch-Gordan coefficient vanishes is an open problem in general, see for instance:

J. Raynal, J. Van der Jeugt, S. K. Rao, and V. Rajeswari. "On the zeros of 3j coefficients: polynomial degree versus occurrence order". J. Phys. A, vol. 26, no. 11, pp. 2607–2623, 1993 (also available here).

So I don't think there would be a simple rule for predicting if your coefficient is $>0$ or $<0$ without actually computing it. Note that there are interesting patterns for the signs see: https://mathoverflow.net/questions/222319/branches-of-3j-symbols