In subsection 5.22 "The periodic table", chapter 5 about identicals particles he states:
The electrons in the orbitals (1s)&(2s) have no orbital angular momentum ($l=0$), so they shouldn't contribute to the total angular momentum. Meanwhile those two in the orbital (2p) have orbital angular momentum $l=1$. How do you achieve to get a total orbital angular momentum of $0$ or $1$? From my understanding the $l's$ just add up, e.g. $l = l_1 + l_2 = 2$.
Please explain what are the possibles states to attain total orbital angular moment of ${0, 1, 2}$.
I'd would probably be helpful to quickly summarize why and when we can add up those $l's$. Also a link to the actually measured total angular momentum $L^2$, $|L|^2$ would be helpful.
Best Answer
The combination of 2 $l=1$ states can result in total angular momentum of $L=0, 1,$ or $2$. Look up Clebsch-Gordon coefficients to convince your self that in terms of $|m_1, m_2\rangle $, those combinations are:
$L=2$, $M=2$:
$|1,1\rangle$
$L=2$, $M=1$:
$\frac{1}{\sqrt{2}}[|1,0\rangle + |0,1\rangle]$
$L=2$, $M=0$:
$\frac{1}{\sqrt{6}}|1,-1\rangle + \sqrt{\frac{2}{3}}|0,0\rangle +\frac{1}{\sqrt{6}}|-1,1\rangle$
$L=1$, $M=1$:
$\frac{1}{\sqrt{2}}[|1,0\rangle - |0,1\rangle]$
$L=1$, $M=0$:
$\frac{1}{\sqrt{2}}[|1,-1\rangle - |-1,1\rangle]$
$L=0$, $M=0$:
$\frac{1}{\sqrt{3}}|1,-1\rangle - \frac{1}{\sqrt{3}}|0,0\rangle +\frac{1}{\sqrt{3}}|-1,1\rangle$.
You can explicitly verify these by working in the spherical vector bases:
${\bf e}^+ = ({\bf \hat x}+i{\bf \hat y})/(-\sqrt{2})$
${\bf e}^- = ({\bf \hat x}-i{\bf \hat y})/\sqrt{2}$
${\bf e}^0 = {\bf \hat z}$.
Then for instance, the last formula for $L=0, M=0$ becomes:
$-[{\bf e}^+{\bf e}^- - {\bf e}^0{\bf e}^0 + {\bf e}^-{\bf e}^+] = {\bf I} = {\bf \hat{x}\hat{x}} + {\bf \hat{y}\hat{y}} + {\bf \hat{z}\hat{z}}$