# Explanation passage in Griffith's - Angular momentum of carbon

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.

• They are angular momenta and have particular "adding rule" (Clebsch-Gordan decomposition of the tensor product space, if that rings a bell). Dec 1, 2017 at 14:57
• Do the Clebsch-Cordan coefficients, e.g. those fancy tables, apply to all kind of angular momentum, e.g. orbital, spin and orbital + spin? Edit: Thought about them being applicable to spin only so far... Dec 1, 2017 at 15:03
• Yes, they are valid for both, "l" is necessarily integer, "s" is necessarily 1/2, m_l can be any integer in the interval [-l,l] and m_s can be either 1/2 or -1/2. Dec 1, 2017 at 15:05
• Remember that angular momentum is a vector so it has magnitude and direction. All the $2p$ have an angular momentum with magnitude $1$, but the $z$ component $L_z$ can be $+1$, $0$ or $-1$. Dec 1, 2017 at 15:36
• The angular momentum addition rules are explained in Section 4.4.3 of your textbook. Nov 15, 2018 at 17:10

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:

$-[{\bf e}^+{\bf e}^- - {\bf e}^0{\bf e}^0 + {\bf e}^-{\bf e}^+] = {\bf I} = {\bf \hat{x}\hat{x}} + {\bf \hat{y}\hat{y}} + {\bf \hat{z}\hat{z}}$

• I just remarked that I'm already stuck with understanding why it is, that the $|1,0\rangle$ state is not allowed in our situation. Doing the math I get $l=l_1 + l_2 = 1+1= 2$ and $m=m_1+m_2=1+0=1$ but according to Clebsch-Gordan tables this state is not available, so whats wrong about it? Dec 2, 2017 at 9:41

Classically two orbital momenta of absolute value $$\hbar$$ can add up to anything between 0 and 2$$\hbar$$. In quantum mechanics orbital angular momentum is quantized to attain only integral multiples of $$\hbar$$.