# Atomic physics, determining levels and terms

In atomic physics I understand there a configurations, terms and levels. I think levels for instance appear because of spin-orbit interactions, so that terms are split.

But I'm confused about the correct way to determine levels and terms.

An example is Carbon with configuration $1s^22s^22p^2$, which I think has the ground state terms $^{3}P, ^{1}D, ^{1}S$, which I got by considering that the terms should be anti-symmetric (to satisfy Pauli's exclusion principle I think):

For $^{3}P$, $L = 1$, parity is then $(-1)^{1} = -1$, thus spin part has to be symmetric, i.e. a triplet spin. Is this the correct method of figuring out the terms?

Then when calculating the levels, due to spin-orbit splitting, how do you do that. I know $J = L + S$, such that the values $J$ can take are $|L-S|,|L-S|+1,...,|L+S|+1,|L+S|$ i.e. steps of one.

And then the levels are written in the form: $^{2S+1}L_J$, but in this case $L = 0,1,2$ and $S = 0,1$ as we are considering $2p^2$, where $l = 1, s = 1/2$.

This gives me lots of levels which I don't think all are correct.

I understand Pauli's exclusion principle has to be satisfied for the levels (and terms), but what does that actually mean?

Would be great if anyone could show how to actually do this, as I find it really confusing.

• I don't have time to give a proper answer right now, sorry, but note that Latex does subscripts to the left just fine – zeldredge Apr 21 '15 at 19:06

Ok, so I remember doing this as a student.

take carbon in its ground state with 1s$^2$ 2s$^2$ 2p$^2$.

We only need to worry about the two 2p electrons because the other shells are closed and give net spin and orbital angular momenta of zero.

Now because of Pauli's principle each of the two electrons can have any allowed value of spin, $s$, and orbital, $l$, angular momenta, as long as the values for both $l$ and $s$ are not the same, $viz.$ you can not have $l=+1$ and $s=+1/2$ for both electrons. You can have $l=+1$ and $s=-1/2$ for one electron and $l=+1$ and $s=+1/2$ for the other electron because although the $l$ values are the same the $s$ values are different.

[Please note In fact here where I am writing $s$ and $l$ it should be $m_s$ and $m_l$ which are the projections of $s$ and $l$ with respect to a fixed axis. ]

Now the electrons are indistinguishable, but to try to make things clear I am going to write $l_1$, $s_1$, $l_2$ and $s_2$ for the values of $l$ and $s$ for the two electrons. [again ** it should be $m_s$ and $m_l$ values **]

When we have the $l$ and $s$ values we can add them to get total $L$ and $S$ values for the atom... and then we can worry about getting the $J$ values.

Now for 2p electrons each electron has a spin of 1/2 which can be orientated up or down... or $m_s =+1/2, -1/2$. Each electron also have orbital ang. momentum of 1 which can be oriented as $m_l = +1, 0, -1$ and only these three values.

Now there are 15 different combinations... here goes....

1. $m_l1 = +1, m_s1= +1/2, m_l2 = +1, m_s2= -1/2 \Rightarrow M_L=+2 M_S=0$
2. $m_l1 = +1, m_s1= +1/2, m_l2 = 0, m_s2= +1/2 \Rightarrow M_L=+1 M_S=+1$
3. $m_l1 = +1, m_s1= +1/2, m_l2 = 0, m_s2= -1/2 \Rightarrow M_L=+1 M_S=0$
4. $m_l1 = +1, m_s1= +1/2, m_l2 = -1, m_s2= +1/2 \Rightarrow M_L=0 M_S=+1$
5. $m_l1 = +1, m_s1= +1/2, m_l2 = -1, m_s2= -1/2 \Rightarrow M_L=0 M_S=0$

6. $m_l1 = +1, m_s1= -1/2, m_l2 = 0, m_s2= +1/2 \Rightarrow M_L=+1 M_S=0$

7. $m_l1 = +1, m_s1= -1/2, m_l2 = 0, m_s2= -1/2 \Rightarrow M_L=+1 M_S=-1$
8. $m_l1 = +1, m_s1= -1/2, m_l2 = -1, m_s2= +1/2 \Rightarrow M_L=0 M_S=0$
9. $m_l1 = +1, m_s1= -1/2, m_l2 = -1, m_s2= -1/2 \Rightarrow M_L=0 M_S=-1$

10. $m_l1 = 0, m_s1= +1/2, m_l2 = 0, m_s2= -1/2 \Rightarrow M_L=0 M_S=0$

11. $m_l1 = 0, m_s1= +1/2, m_l2 = -1, m_s2= +1/2 \Rightarrow M_L=-1 M_S=+1$
12. $m_l1 = 0, m_s1= +1/2, m_l2 = -1, m_s2= -1/2 \Rightarrow M_L=-1 M_S=0$

13. $m_l1 = 0, m_s1= -1/2, m_l2 = -1, m_s2= +1/2 \Rightarrow M_L=-1 M_S=0$

14. $m_l1 = 0, m_s1= -1/2, m_l2 = -1, m_s2= -1/2 \Rightarrow M_L=-1 M_S=-1$

15. $m_l1 = -1, m_s1= +1/2, m_l2 = -1, m_s2= -1/2 \Rightarrow M_L=-2 M_S=0$

There are five of these different possibilities that all come from the $L=2, S=0$ atomic state which is $^1D$ -- note that when we look at a projection of $L=2$ then $M_L=+2,+1,0,-1,-2$, but projecting $S=0$ we only have $M_S=0$ --- these five configurations are....

.1. $m_l1 = +1, m_s1= +1/2, m_l2 = +1, m_s2= -1/2 \Rightarrow M_L=+2 M_S=0$

.3. $m_l1 = +1, m_s1= +1/2, m_l2 = 0, m_s2= -1/2 \Rightarrow M_L=+1 M_S=0$

.5. $m_l1 = +1, m_s1= +1/2, m_l2 = -1, m_s2= -1/2 \Rightarrow M_L=0 M_S=0$

.12. $m_l1 = 0, m_s1= +1/2, m_l2 = -1, m_s2= -1/2 \Rightarrow M_L=-1 M_S=0$

.15. $m_l1 = -1, m_s1= +1/2, m_l2 = -1, m_s2= -1/2 \Rightarrow M_L=-2 M_S=0$

There are nine of the configurations that come from the $L=1, S=1$ atomic state which is $^3P$ -- note that when we look at a projection of $L=1$ then $M_L=+1,0,-1$, and projecting $S=1$ we also have $M_S=+1,0,-1$ --- these nine configurations are....

.2. $m_l1 = +1, m_s1= +1/2, m_l2 = 0, m_s2= +1/2 \Rightarrow M_L=+1 M_S=+1$

.6. $m_l1 = +1, m_s1= -1/2, m_l2 = 0, m_s2= +1/2 \Rightarrow M_L=+1 M_S=0$

.7. $m_l1 = +1, m_s1= -1/2, m_l2 = 0, m_s2= -1/2 \Rightarrow M_L=+1 M_S=-1$

.4. $m_l1 = +1, m_s1= +1/2, m_l2 = -1, m_s2= +1/2 \Rightarrow M_L=0 M_S=+1$

.8. $m_l1 = +1, m_s1= -1/2, m_l2 = -1, m_s2= +1/2 \Rightarrow M_L=0 M_S=0$

.9. $m_l1 = +1, m_s1= -1/2, m_l2 = -1, m_s2= -1/2 \Rightarrow M_L=0 M_S=-1$

.11. $m_l1 = 0, m_s1= +1/2, m_l2 = -1, m_s2= +1/2 \Rightarrow M_L=-1 M_S=+1$

.13. $m_l1 = 0, m_s1= -1/2, m_l2 = -1, m_s2= +1/2 \Rightarrow M_L=-1 M_S=0$

.14. $m_l1 = 0, m_s1= -1/2, m_l2 = -1, m_s2= -1/2 \Rightarrow M_L=-1 M_S=-1$

Now there is only one configuration left....

.10. $m_l1 = 0, m_s1= +1/2, m_l2 = 0, m_s2= -1/2 \Rightarrow M_L=0 M_S=0$

this has $M_L=0 M_S=0$ and it comes from $L=0, S=0$, which is $^1S$.

Now this is why for carbon in the ground state of 1s$^2$ 2s$^2$ 2p$^2$ we get the three terms $^1D$, $^3P$ and $^1S$

BUT note that if we had an excited carbon atom with one electron in the 2p and one electron in the 3p and all the other electrons in closed subshells we would get the terms $^1D$, $^3D$, $^1P$, $^3P$, $^1S$ and $^3S$ because, for example, now both electrons could have $m_l = +1, m_s= +1/2$ because the electrons are in different subshells with different quantum number n; n=2 for 2p and n=3 for 3p.

Now for the quantum number $J$. - Above we have assumed 'Russell Saunders' coupling which is reasonable for light atoms -- here all the orbital angular momenta link to give an overall orbital angular momentum of $L$ and the spins combine to give $S$. After we have our $L$ and $S$ values then they need to be combined to give the different overall angular momenta $J$.

For $^1D$ and $^1S$ it is pretty simple because $S$ is zero, which means that for $^1D$ $J=2$ only, and for $^1S$ $J=0$ only. Note that in a magnetic or electric field the $^1D$ $J=2$ state will be split into five sublevels with $M_J=+2,+1,0,-1,-2$.

By contrast the $^3P$ state has three different $J$ values; $J=2,1,0$ depending on how the $L$ and the $S$ combine. We write these as $^3P_2$, $^3P_1$ and $^3P_0$. Note that Note that in a magnetic or electric field the $J=2$ state will be split into five sublevels with $M_J=+2,+1,0,-1,-2$, the $J=1$ state will be split into three sublevels with $M_J=+1,0,-1$ and the $J=0$ state will have a single sublevel with $M_J=0$ - Thus there are a total of 9 different sublevels of the different $J$ states of $^3P$.

• Thanks for taking the time to answer. Though the question is 3 years old, I hope your answer will be useful to others studying atomic physics. – Alexander Mar 10 '18 at 4:53
• @Alexander thanks for marking question as solved - I had no idea it was 3 years old - should have looked at the date. – tom Mar 11 '18 at 0:05