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The carbon atom has its in its final subshell $2p$ simply two electrons. These electrons must be in a state given by:

$$\left|{2,1,m_l,m_s}\right.\rangle$$

Where $m_l=-1, 0, 1$, and $m_s=\dfrac{1}{2}, -\dfrac{1}{2}$. Due to the Pauli exclusion principle, there are only 15 possible ways to arrange these two electrons. However, in one of the arrangements:

$$\left|{2,1,0,m_{s1}}\right.\rangle \otimes \left|{2,1,-1,m_{s2}}\right.\rangle$$

The total orbital angular momentum number $L$ is -1. Is this allowed? Is it simply the total angular momentum $J$ that has to be greater than zero?

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  • $\begingroup$ How is the total orbital angular momentum quantum number -1 there? According to what you've written down, the third index in the ket is $m_l$, not $l$. Can you clarify? $\endgroup$
    – march
    Commented Jan 17, 2023 at 18:38
  • $\begingroup$ Ignoring spin, if you have two states with orbital angular momentum 1, there are nine direct product states. Five linear combinations of these nine direct product states have total orbital angular momentum 2, three have total orbital angular momentum 1, and one has total orbital angular momentum 0. Five plus three plus one equals nine. Surprise surprise. $\endgroup$
    – hft
    Commented Jan 17, 2023 at 22:42

2 Answers 2

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No, $L$ must be greater than $0$.

Note that the quantum number $L$ represents the shape of the orbitals. By definition $L\ge0$. And we usually express the values of $L$ as $0, 1, 2$... With the letters $s, p, d$... Respectively.

However, the magnetic quantum number, $m_l$, can be negative. This indicates the orientation of the orbital, and takes values such that $m_l=(-l,...,0,...,+l)$

In your last ket $L=1$ and $m_l=-1$.

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However, in one of the arrangements:

$$\left|{2,1,0,m_{s1}}\right.\rangle \otimes \left|{2,1,-1,m_{s2}}\right.\rangle$$

The total orbital angular momentum number $L$ is -1.

No, it is not! The total $M_{L}$ is -1. The above state could either have total angular momentum 2 or 1.

When you add two states that both have total orbital momentum 1, you can only get states with total orbital angular momentum 0, 1, and 2.

How do you do it? I will show you, but I will ignore spin.

Also note that, ignoring spin, you have nine direct product states. From these you can form five linear combinations that have total orbital angular momentum 2. You can form three linear combinations that have total orbital angular momentum 1. You can form one linear combination that has total orbital angular momentum 0.

You start from the "stretch state": $$ |n=2,L=1,M_L=1\rangle\otimes|n=2,L=1,M_L=1\rangle\;, $$ which is a state that necessarily has total orbital angular momentum 2 (since otherwise there would be no way for it to have z-angular momentum 2, which it obviously has).

Then apply the total lowering operator $$ L^{\text total}_- = L^{(1)}_-\times 1 + 1 \times L^{(2)}_- $$ to get the other four $L^{\text total}=2$ states.

Then find a state with $M_{L^{\text total}}=1$ that is orthogonal to the state you already have with $M_{L^{\text total}}=1$. This is an $L^{\text total}=1$ state. Now apply the lowering operator to this state twice to generate the other two $L^{\text total}=1$ states.

Then find a state with $M_{L^{\text total}}=0$ that is orthogonal to the other two states you already have with $M_{L^{\text total}}=0$. This is your $L^{\text total}=0$ state.

If you find this method tedious, look into something called "Clebsch-Gordan coefficients."

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