Solve N Ground State in (2s+1)(L)(J) Format

[Domnu]

Homework Statement

Find the ground state of nitrogen in the form (2s+1)(L)(J), where s represents the total spin, L represents the total orbital angular momentum (s,p,d,f), and J represents the total angular momentum.

Attempt at solution
Well, the electron configuration is [He](2s)^2 (2p)^3. Now, we know that everything until the (2p)^2 contributes 0 total angular momentum. We have three electrons of orbital angular momentum 1 (p = 1). These can form orbitals or s, p, d, f (0, 1, 2, 3). Now, the spins of the three electrons are 1/2, so they can either form spins 1/2 or 3/2. By Hund's first, they will take spin 3/2 (if possible). Now, from this, we have the possible electron configurations to be

4S 3/2
4P 1/2, 4P 3/2, 4P 5/2
4D 1/2, 4D 3/2, 4D 5/2, 4D 7/2
4F 3/2, 4F 5/2, 4F 7/2, 4F 9/2

By Hund's third rule, since the atom is less than half filled, we know that the only possible configurations left are

4S 3/2
4P 1/2
4D 1/2
4F 3/2

Now, for Hund's second rule, 4F 3/2 should be the correct one; but this isn't the case; the total spin in this case is 3/2, and by the clebsch gordan table, we have:

|3/2 3/2> = |1 1> |1/2 1/2> = (|1/2 1/2> |1/2 1/2>) |1/2 1/2>,

which is symmetric. So we need the orbital angular momentum part of our configuration to be antisymmetric since electrons are fermions. Now, we look at F: we have 3 particles whose orbital angular momenta are 1 and we need to try to form 3. We can do this as follows:

|3 3> = |2 2> |1 1> = |1 1> |1 1> |1 1>

which is symmetric. So, 4F 3/2 doesn't work because the entire thing is symmetric. Now, we go on to the D levels: I need to have 3 particles of orbital angular momentum 1 have total orbital angular momentum 2. The answer says that this is also symmetric; how is this so?

 

[Domnu]

Any ideas? I know that it has something to do with the Clebsch Gordan coefficients, but I can't figure it out =/

 

[yaychemistry]

Hi, it's been a little while since I've dealt with Term symbols and the CG coefficients, but here goes (e.g. I may be totally wrong, but at least this might get you thinking about the right answer):

The term symbol we're shooting for the [tex]^4\textrm{D}_{1/2}[/tex], so for our spin we need the coupled angular momentum to be |3/2 3/2>, which you've already done. For our D, the ORBITAL angular momentum needs to be |2 ?> where the z-projection of the orbital angular momentum is the '?'. We also know that our TOTAL angular momentum is 1/2. So working backwards, we need to mix the orbital angular momentum |2 ?> with the spin |3/2 3/2> to get a final TOTAL angular momentum 1/2 (with a multiplicity of 2 b/c of two possible z-projections of the total a.m.). So from that, you might be able to figure out what the |2 ?> is (or what combination of |2 ?> you need) to create the correct D state.

This table might help (http://en.wikipedia.org/wiki/Table_of_Clebsch-Gordan_coefficients#j1.3D2.2C_j2.3D3.2F2)

Again, I apologize if that's totally wrong... it's been a while. But I hope it at least helps.

 

[Domnu]

Thanks very much for responding

Yes, this is more or less the idea I used while trying to show that the possible D states are all symmetric: we pretty much need to show that |2 ?> is a symmetric function when expanded in |1 ?>|1 ?>|1 ?>. My first question is... how do you expand it like this? Do we start off with say..

|2 2> = |1 1> |1 1>

and expand one of the |1 1>'s in states of other |1 ?>|1 ?> s, substitute and move on? Even if I try this, I don't get anything symmetric. This looks correct, but I'm probably missing something crucial. Could someone help?

 

[yaychemistry]

Right, so its harder to decouple angular momenta than the couple them together and it involves use of Wigner 3-j symbol's, or at least very careful application of the CG coefficients and I can't remember it right now.

Instead let's take a physical approach:
We know that the TOTAL angular momentum is either |1/2 1/2> or |1/2 -1/2>.
Let's decompose the |1/2 1/2>:
[tex]\left|1/2, 1/2\left(\textrm{total}\right)\rangle = \sum_{m1, m2}a_{m1,m2}\left|2, m1\left(\textrm{orbital}\right)\rangle\left|3/2, m2\left(\textrm{spin}\right)\rangle[/tex].
If you check the clebsch-gordan coefficients then ONE of the terms in the sum will look like |2 2(orbital)>|3/2 -3/2(spin)>. For the spin part, we know that means all electrons are spin down.
We can decompose the |2 2> part, and although we are unsure of the exact summation, we know were will get terms that look like permutations of |1 1>|1 1>|1 0>.
What does that tell us? All of our electrons are in p-orbitals, but one has a z-projection of 0, and two have a z-projection of 1. That suggests we have two electrons paired in the p+1 orbital and one in the p0 orbital. However, from before we know that, at least in this term, they are all spin down... thus two electrons are in p+1, spin down states, which violates the Pauli principle.

While this is somewhat of a hand-wavy argument I think it works. We haven't shown explicitly that the wave function isn't symmetric, we've at least shown it to violate the Pauli exclusion principle.

Hope this helps. I'll try to think about the problem more and remember how to decouple three angular momenta.