Ciro Santilli @cirosantilli 37

Ciro Santilli 37  Updated 2025-07-11  +Created 1970-01-01

This example covered for example at Video 1. "Term Symbols Example 1 by TMP Chem (2015)".

Carbon has electronic structure 1s2 2s2 2p2.

For term symbols we only care about unfilled layers, because in every filled layer the total z angular momentum is 0, as one electron necessarily cancels out each other:

magnetic quantum number varies from -l to +l, each with z angular momentum $- l ℏ$ to $+ l ℏ$ and so each cancels the other out
spin quantum number is either + or minus half, and so each pair of electron cancels the other out

So in this case, we only care about the 2 electrons in 2p2. Let's list out all possible ways in which the 2p2 electrons can be.

There are 3 p orbitals, with three different magnetic quantum numbers, each representing a different possible z quantum angular momentum.

We are going to distribute 2 electrons with 2 different spins across them. All the possible distributions that don't violate the Pauli exclusion principle are:

m_l  +1  0 -1  m_L  m_S
     u_ u_ __    1    1
     u_ __ u_    0    1
     __ u_ u_   -1    1
     d_ d_ __    1   -1
     d_ __ d_    0   -1
     __ d_ d_   -1   -1
     u_ d_ __    1    0
     d_ u_ __    1    0
     u_ __ d_    0    0
     d_ __ u_    0    0
     __ u_ d_   -1    0
     __ d_ u_   -1    0
     ud __ __    2    0
     __ ud __    0    0
     __ __ ud   -2    0

where:

m_l is $m_{l}$ , the magnetic quantum number of each electron. Remember that this gives a orbital (non-spin) quantum angular momentum of $m_{l} ℏ$ to each such electron
m_L with a capital L is the sum of the $m_{l}$ of each electron
m_S with a capital S is the sum of the spin angular momentum of each electron

For example, on the first line:

m_l  +1  0 -1  m_L  m_S
     u_ u_ __    1    1

we have:

one electron with $m_{l} = + 1$ , and so angular momentum $ℏ$
one electron with $m_{l} = + 0$ , and so angular momentum 0

and so the sum of them has angular momentum

0 + 1ℏ = 1ℏ

. So the value of

m_{L}

is 1, we just omit the

ℏ

TODO now I don't understand the logic behind the next steps... I understand how to mechanically do them, but what do they mean? Can you determine the term symbol for individual microstates at all? Or do you have to group them to get the answer? Since there are multiple choices in some steps, it appears that you can't assign a specific term symbol to an individual microstate. And it has something to do with the Slater determinant. The previous lecture mentions it: www.youtube.com/watch?v=7_8n1TS-8Y0 more precisely youtu.be/7_8n1TS-8Y0?t=2268 about carbon.

youtu.be/DAgEmLWpYjs?t=2675 mentions that

^{3} D

is not allowed because it would imply

L = 2, S = 1

, which would be a state uu __ __ which violates the Pauli exclusion principle, and so was not listed on our list of 15 states.

He then goes for

^{1} D

and mentions:

S = 1 so $m_{S}$ can only be 0
L = 2 (D) so $m_{L}$ ranges in -2, -1, 0, 1, 2

and so that corresponds to states on our list:

ud __ __    2    0
u_ d_ __    1    0
u_ __ d_    0    0
__ u_ d_   -1    0
__ __ ud   -2    0

Note that for some we had a two choices, so we just pick any one of them and tick them off off from the table, which now looks like:

 +1  0 -1  m_L  m_S
 u_ u_ __    1    1
 u_ __ u_    0    1
 __ u_ u_   -1    1
 d_ d_ __    1   -1
 d_ __ d_    0   -1
 __ d_ d_   -1   -1
 d_ u_ __    1    0
 d_ __ u_    0    0
 __ d_ u_   -1    0
 __ ud __    0    0

Then for