How to choose Clebsch-Gordan coefficients?

I just started learning Clebsch-Gordan coefficients recently. I want to use the expression on Wikipedia (relation to Wigner $$3j$$ symbols):

$$\langle j_{1},m_{1},j_{2},m_{2}|J,M\rangle=(-1)^{-j_{1}+j_{2}-M} \begin{pmatrix} j_{1} & j_{2} & J \\ m_{1} & m_{2} & -M \\ \end{pmatrix}\\$$ For $$M = 5$$, because $$M ∈ \{-J, -J+1, ..., J\}$$, then I choose $$J = 5, 6$$ and calculate $$\bigg\langle\frac{7}{2},\frac{7}{2},\frac{5}{2},\frac{3}{2}\bigg|5,5\bigg\rangle=\frac{\sqrt{\frac{7}{3}}}{2}\\ \bigg\langle\frac{7}{2},\frac{7}{2},\frac{5}{2},\frac{3}{2}\bigg|6,5\bigg\rangle =\frac{\sqrt{\frac{5}{3}}}{2}$$ So I think I could get below relation. $$\bigg|\frac{7}{2},\frac{7}{2},\frac{5}{2},\frac{3}{2}\bigg\rangle=\frac{\sqrt{\frac{7}{3}}}{2}|5,5\rangle\\ \bigg|\frac{7}{2},\frac{7}{2},\frac{5}{2},\frac{3}{2}\bigg\rangle=\frac{\sqrt{\frac{5}{3}}}{2}|6,5\rangle$$ I wonder why there are two expressions. Which one should I choose? I also try other values of $$j_1$$, $$j_2$$, $$m_1$$, $$m_2$$, if the $$(-1)$$ term exists, some results become negative, should I consider?

• Think a bit more about the equations you write after "So I think I could get below relation"... Try simpler examples, suppose |psi> is a combination of |a>, |b> and |c>. Does <psi|a>=2 imply that |psi>=2|a>?
– zakk
Jul 7, 2021 at 12:18

Your error is in going from $$\langle \ell_1 m_1;\ell_2 m_2\vert JM\rangle = a$$ to $$\vert \ell_1 m_1;\ell_2 m_2\rangle = a \vert J M\rangle$$. The Clebsch-Gordan coefficients are change-of-basis coefficients from the uncoupled basis $$\{\vert \ell_1 m_1,\ell_2 m_2\rangle\}$$ to the coupled basis $$\{\vert JM\rangle\}$$. States in the coupled basis are rarely equal to states in the uncoupled basis; in general they will be linear combinations (i.e. sums) of states in the uncoupled basis.
As an analogy, consider a first set of basis vector in the plane $$\{\hat x,\hat y\}$$ and decompose those in a second set $$\{\hat X,\hat Y\}$$ where this last is rotated by $$\pi/4$$ w/r to the first set. Then surely \begin{align} \hat x= \hat X \langle X\vert x\rangle + \hat Y \langle Y\vert x\rangle \end{align} where $$\langle A\vert b\rangle = \vec A\cdot \vec b$$. Here, we have \begin{align} \langle X\vert x\rangle = \langle Y\vert x\rangle =\frac{1}{\sqrt{2}}\, , \tag{1} \end{align} since both sets differ by a $$\pi/4$$ rotation. These overlaps are change-of-basis coefficients, i.e. they play in this example the role of the CG in your original question.
Of course, Eq.(1) does NOT imply \begin{align} \hat X=\frac{1}{\sqrt{2}}\hat x\, ,\quad \hat Y=\frac{1}{\sqrt{2}}\hat x\, , \quad \hat x=\frac{1}{\sqrt{2}}\hat X\, ,\quad \hat x=\frac{1}{\sqrt{2}}\hat Y\, , \end{align} but it does imply that the sum \begin{align} \hat x= \frac{1}{\sqrt{2}}\hat X +\frac{1}{\sqrt{2}}\hat Y \end{align} is an equality.
In your specific case one has (for instance) \begin{align} \big\langle\textstyle{\frac{7}{2}},\frac{7}{2};\frac{5}{2},\frac{3}{2}\big|\,5,5\big\rangle=\frac{\sqrt{\frac{7}{3}}}{2}\, ,\qquad \big\langle\textstyle{\frac{7}{2}},\frac{5}{2},\frac{5}{2};\frac{5}{2}\big|\, 5,5\big\rangle =-\frac{\sqrt{\frac{5}{3}}}{2} \end{align} so that \begin{align} \vert 5,5\rangle = \textstyle\frac{\sqrt{\frac{7}{3}}}{2}\vert \frac{7}{2}, \frac{7}{2};\frac{5}{2},\frac{3}{2}\rangle -\frac{\sqrt{\frac{5}{3}}}{2} \vert \frac{7}{2}, \frac{5}{2};\frac{5}{2},\frac{5}{2}\rangle\, . \end{align}