Ciro Santilli @cirosantilli 34

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.

TODO experiment. Likely Zeeman effect.

Fixed total angular momentum.

The direction however is not specified by this number.

To determine the quantum angular momentum, we need the magnetic quantum number, which then selects which orbital exactly we are talking about.

Refinement of the Bohr model that starts to take quantum angular momentum into account in order to explain missing lines that would have been otherwise observed TODO specific example of such line.

They are not observe because they would violate the conservation of angular momentum.

Introduces the azimuthal quantum number and magnetic quantum number.

TODO confirm year and paper, Wikipedia points to: zenodo.org/record/1424309#.yotqe3xmjhe

Quantum numbers appear directly in the Schrödinger equation solution for the hydrogen atom.

However, it very cool that they are actually discovered before the Schrödinger equation, and are present in the Bohr model (principal quantum number) and the Bohr-Sommerfeld model (azimuthal quantum number and magnetic quantum number) of the atom. This must be because they observed direct effects of those numbers in some experiments. TODO which experiments.

E.g. The Quantum Story by Jim Baggott (2011) page 34 mentions:This refers to forbidden mechanism. TODO concrete example, ideally the first one to be noticed. How can you notice this if the energy depends only on the principal quantum number?

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:we have:and so the sum of them has angular momentum $0+1\hslash =1\hslash$. So the value of $m_L$ is 1, we just omit the $\hslash$.

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:and so that corresponds to states on our list: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:

Then for ${}^{3}P$ the choices are: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 $m_S$, we have two electrons with spin up. The angular momentum of each electron is $1/2\hslash$, and so given that we have two, the total is $1\hslash$, so again we omit $\hslash$ and $m_S$ is 1.

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