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In subsection 5.22 "The periodic table", chapter 5 about identicals particles he states:

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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.

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    $\begingroup$ They are angular momenta and have particular "adding rule" (Clebsch-Gordan decomposition of the tensor product space, if that rings a bell). $\endgroup$
    – DanielC
    Dec 1, 2017 at 14:57
  • $\begingroup$ 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... $\endgroup$ Dec 1, 2017 at 15:03
  • $\begingroup$ 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. $\endgroup$
    – DanielC
    Dec 1, 2017 at 15:05
  • $\begingroup$ 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 $-1$. $\endgroup$ Dec 1, 2017 at 15:36
  • $\begingroup$ The angular momentum addition rules are explained in Section 4.4.3 of your textbook. $\endgroup$ Nov 15, 2018 at 17:10

2 Answers 2

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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:

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

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  • $\begingroup$ 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? $\endgroup$ Dec 2, 2017 at 9:41
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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 $\hbar$.

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