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.

  • $\begingroup$ Since both states are normalized, and the coefficient is real, this follows immediately, does't it ? $\endgroup$
    – Adam
    Commented Mar 30, 2016 at 13:00
  • $\begingroup$ 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 $-1$ ? $\endgroup$
    – Kamil
    Commented Mar 30, 2016 at 16:51

1 Answer 1


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 $+1$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.