In the subject of the addition of the angular momenta how are two common eigenvectors of $L^2$, $L_3$, $S^2$, $S_3$ and of $J^2$, $L^2$, $S^2$ are related to each other?


Suppose that an electron is in a state of orbital angular momentum $l=2$. An orthonormal basis for the states is given by simultaneous eigenstates of $L^2$, $L_3$, $S^2$ and $S_3$ as $|l, m_l; s, m_s\rangle$. Alternatively, we can choose an orthonormal basis as simultaneous eigenstates of $J^2$, $L^2$, $S^2$ and $J_3$ with $J=L+S$ as $|j, l, s; m_j\rangle$

Argue that $|5/2, 2, 1/2; 5/2\rangle = |2, 2; 1/2, 1/2\rangle$:


1 Answer 1


As you probably learned in your course while generating C-G coefficients, the top and bottom states of your ladder, $|m_l|=l, ~ |m_s|=s$, are privileged and unique: They correspond to j=l+s !!

For any other state, there is an ambiguity connecting the sum of L with S, but not here.

You may see this from $\vec J=\vec L + \vec S$, so $$ J^2=(\vec L + \vec S)^2= L^2 +2\vec L \cdot \vec S + S^2\\ =L^2+ S^2 +(2L_3 S_3 + L_+ S_- + L_- S_+). $$

However, for the extremal states such as the r.h.s. one, $$ |2, 2; 1/2, 1/2\rangle, $$ the action of the above operator $J^2$ is unambiguous: This state is annihilated by both $L_+$ and $S_+$, by definition; so, then, by above, the raising-lowering terms vanish, and
$$ J^2~ |2, 2; 1/2, 1/2\rangle\\ = \left (2(2+1)+\frac{1}{2}\left (\frac{1}{2} +1\right )+2(2\times 1/2)\right )|2, 2; 1/2, 1/2\rangle \\ =\frac{5}{2}\left (\frac{5}{2} +1\right ) |2, 2; 1/2, 1/2\rangle, $$ that is, $j=5/2$, unambiguously, thus necessarily identifying this state with the l.h.s., $$|5/2, 2, 1/2; 5/2\rangle.$$


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