# Glebsch-Gordan coefficient equal to unity (special case)

I'm studying from a quantum mechanics book, and there is a part I don't really understand. First, the Clebsch-Gordan coefficients are $$\langle j_1, j_2; m_1, m_2 \mid j, m \rangle = \langle j, m \mid j_1, j_2; m_1, m_2 \rangle$$ since they are taken real by convention. Now the author of the book says the Clebsch-Gordan coefficients corresponding to the two limiting cases where $m_1 = j_1, m_2 = j_2, j = j_1 + j_2, m = j_1 + j_2$ are equal to one: $$\langle j_1, j_2; j_1, j_2 \mid (j_1 + j_2), (j_1 + j_2) \rangle = 1.$$ He says this can be inferred from the fact that $\mid (j_1 + j_2), (j_1 + j_2) \rangle$ has one element, and from the expression $$\mid j,m \rangle = \sum_{m_1, m_2} \langle j_1, j_2; m_1, m_2 \mid j, m \rangle \mid j_1, j_2; m_1 m_2 \rangle$$ which shows that the bases $\left\{ \mid j_1, j_2; m_1, m_2 \rangle \right\}$ and $\left\{ j, m \right\}$ are connected by a unitary transformation. For the special limiting case described above, this leads then to $$\mid (j_1 + j_2), (j_1 + j_2) \rangle = \langle j_1, j_2; j_1, j_2 \mid (j_1 + j_2), (j_1 + j_2) \rangle \mid j_1, j_2; j_1, j_2 \rangle$$ which is then supposed to show that the coefficients are unity in this case. Still I don't understand this reasoning fully, and I don't see how we can deduce from the expression above that the Clebsch-Gordan coefficient is one $1$ in that case.

• Since both states are normalized, and the coefficient is real, this follows immediately, does't it ? – Adam Mar 30 '16 at 13:00
• Yes, and that shows their norm is unity. But how does it follow from this that the coefficient must equal one? Couldn't they then also be $-1$ ? – Kamil Mar 30 '16 at 16:51

Since there is only one state with projection $m=j=j_1+j_2$, i.e. only one eigenstate of $\hat L_z$ with eigenvalue $j_1+j_2$, it must be that $\vert j_1j_1\rangle\vert j_2j_2\rangle$ is proportional to $\vert jj\rangle$ for $j=j_1+j_2$. For normalized kets this implies $$\vert j_1j_1\rangle\vert j_2j_2\rangle = e^{i\varphi}\vert jj\rangle \, \quad j=j_1+j_2\, ,$$ where $\varphi$ is an arbitrary phase.
There is no mathematical argument to choose $\varphi$. It is convenient to choose the CGs to be real so this limits $e^{i\varphi}=\pm 1$.
Choosing $e^{i\varphi}=+1$ is in line with the most commonly-used phase convention, by Condon and Shortley. Their convention dictates that the seed for the recursion relations for the CG $\langle j_1m_1; j_2m_2\vert j m\rangle$ be constrained to satisfy $$\langle j_1 j_1;j_2 (j-j_1)\vert jj\rangle >0\, .$$ Applied to the specific case where $j=j_1+j_2$, this yields $$\langle j_1 j_1; j_2 j_2\vert jj\rangle >0\, ,$$ which narrows the choice of phase to $+1$.