TODO. Can't find it easily. Anyone?
This is closely linked to the Pauli exclusion principle.
What does a particle even mean, right? Especially in quantum field theory, where two electrons are just vibrations of a single electron field.
Another issue is that if we consider magnetism, things only make sense if we add special relativity, since Maxwell's equations require special relativity, so a non approximate solution for this will necessarily require full quantum electrodynamics.
As mentioned at lecture 1 youtube.com/watch?video=H3AFzbrqH68&t=555, relativistic quantum mechanical theories like the Dirac equation and Klein-Gordon equation make no sense for a "single particle": they must imply that particles can pop in out of existence.
Bibliography:
Hund's rules by Ciro Santilli 40 Updated 2025-07-16
Allow us to determine with good approximation in a multi-electron atom which electron configuration have more energy. It is a bit like the Aufbau principle, but at a finer resolution.
Note that this is not trivial since there is no explicit solution to the Schrödinger equation for multi-electron atoms like there is for hydrogen.
For example, consider carbon which has electron configuration 1s2 2s2 2p2.
If we were to populate the 3 p-orbitals with two electrons with spins either up or down, which has more energy? E.g. of the following two:
m_L -1  0  1
    u_ u_ __
    u_ __ u_
    __ ud __
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:
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:
For example, on the first line:
m_l  +1  0 -1  m_L  m_S
     u_ u_ __    1    1
we have:
and so the sum of them has angular momentum . So the value of is 1, we just omit the .
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 is not allowed because it would imply , 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 and mentions:
  • S = 1 so can only be 0
  • L = 2 (D) so 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 the choices are:
  • S = 2 so is either -1, 0 or 1
  • L = 1 (P) so ranges in -1, 0, 1
so we have 9 possibilities for both together. We again verify that 9 such states are left matching those criteria, and tick them off, and so on.
For the , we have two electrons with spin up. The angular momentum of each electron is , and so given that we have two, the total is , so again we omit and is 1.
Video 1.
Term Symbols Example 1 by TMP Chem (2015)
Source. Carbon atom.
Basically the operators are just analogous to the classical ones e.g. the classical:
becomes:
Besides the angular momentum in each direction, we also have the total angular momentum:
Then you have to understand what each one of those does to the each atomic orbital:
There is an uncertainty principle between the x, y and z angular momentums, we can only measure one of them with certainty at a time. Video 1. "Quantum Mechanics 7a - Angular Momentum I by ViaScience (2013)" justifies this intuitively by mentioning that this is analogous to precession: if you try to measure electrons e.g. with the Zeeman effect the precess on the other directions which you end up modifing.

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