# Choosing Clebsch-Gordan coefficients

I have a question about the Clebsch-Gordan coefficients. If I have a quantum state, for example $|1 ,\frac{1}{2} \rangle$ in the base $|m_l,m_s\rangle$ and I want to transform it to the $|j ,m_j\rangle$ base. Does it matter if I look up at a Clebsch-Gordan coefficients table either at $2 \times \frac{1}{2}$ or at $1 \times \frac{1}{2}$? I find my quantum state in both sections but with different coefficients.

• Your question is incomplete in its current form. You need to specify $L$ and $S$ in order to determine which $L$ and $S$ values to use. These depend on $L$ so as a partial answer to your question yes it does matter. Please note that it is "Gordan" with "a" not "Gordon" with "o". – ZeroTheHero Mar 3 '18 at 19:33
• Have you tried comparing the values for various $L$’s? – ZeroTheHero Mar 3 '18 at 23:01

You can't say "the state $|m_\ell\rangle$" without specifying what $\ell$ is. Different $\ell$s will give you different states, and therefore different physical situations. That means that the Clebsch-Gordan* coefficients $$\langle \ell,m_\ell, s,m_s | j,m_j\rangle$$ and $$\langle \ell',m_{\ell'}, s,m_s | j,m_j\rangle$$ will be different if $\ell\neq\ell'$ even if $m_\ell=m_{\ell'}$, because the states $|\ell,m_{\ell}\rangle$ and $|\ell',m_{\ell'}\rangle$ have nothing to do with each other.
• With the data you've posted, it's impossible to tell. You need to know the value of $\ell$ to proceed. – Emilio Pisanty Mar 4 '18 at 15:11