# How are common extremal eigenvectors of $L^2$, $L_3$, $S^2$, $S_3$ and of $J^2$, $L^2$, $S^2$ related to each other?

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?

Example:

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

• – Dan
Commented Jan 22, 2022 at 12:12
• Answer not clear? Commented Jan 26, 2022 at 23:08

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 !!
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., $$|5/2, 2, 1/2; 5/2\rangle.$$