As always, the best way to get some intuition about an equation is to solve it for some simple cases, so let's give that a try with different fixed potentials.
- www.youtube.com/watch?v=1Z9wo2CzJO8 "Schrodinger equation solved numerically in 3D" by Tetsuya Matsuno. 3D hydrogen atom, code may be hidden in some paper, maybe
- www.youtube.com/playlist?list=PLdCdV2GBGyXM0j66zrpDy2aMXr6cgrBJA "Computational Quantum Mechanics" by Let's Code Physics. Uses a 1D trinket.io.
- www.youtube.com/watch?v=BBt8EugN03Q Simulating Quantum Systems [Split Operator Method] by LeiosOS (2018)
- www.youtube.com/watch?v=86x0_-JGlGQ Simulating the Quantum World on a Classical Computer by Garnet Chan (2016) discusses how modeling only local entanglement can make certain simulations feasible
This is basically how quantum computing was first theorized by Richard Feynman: quantum computers as experiments that are hard to predict outcomes.
TODO answer that: quantumcomputing.stackexchange.com/questions/5005/why-it-is-hard-to-simulate-a-quantum-device-by-a-classical-devices. A good answer would be with a more physical example of quantum entanglement, e.g. on a photonic quantum computer.
Schrödinger equation for a one dimensional particle with . The first step is to calculate the time-independent Schrödinger equation for a free one dimensional particle
Then, for each energy , from the discussion at Section "Solving the Schrodinger equation with the time-independent Schrödinger equation", the solution is: plane wave functions.
The plane wave function appears for example in the solution of the Schrödinger equation for a free one dimensional particle. This makes sense, because when solving with the time-independent Schrödinger equation, we do separation of variable on fixed energy levels explicitly, and the plane wave solutions are exactly fixed energy level ones.
This equation is a subcase of Equation "Schrödinger equation for a one dimensional particle" with .
We get the time-independent Schrödinger equation by substituting this into Equation "time-independent Schrödinger equation for a one dimensional particle":
Now, there are two ways to go about this.
The first is the stupid "here's a guess" + "hey this family of solutions forms a complete bases"! This is exactly how we solved the problem at Section "Solving partial differential equations with the Fourier series", except that now the complete basis are the Hermite functions.
The second is the much celebrated ladder operator method.
A quantum version of the LC circuit!
TODO are there experiments, or just theoretical?
Show up in the solution of the quantum harmonic oscillator after separation of variables leading into the time-independent Schrödinger equation, much like solving partial differential equations with the Fourier series.
Not the same as Hermite polynomials.
www.physics.udel.edu/~jim/PHYS424_17F/Class%20Notes/Class_5.pdf by James MacDonald shows it well.
The operators are a natural guess on the lines of "if p and x were integers".
And then we can prove the ladder properties easily.
The commutator appear in the middle of this analysis.
Is the only atom that has a closed form solution, which allows for very good predictions, and gives awesome intuition about the orbitals in general.
It is arguably the most important solution of the Schrodinger equation.
The explicit solution can be written in terms of spherical harmonics.
In the case of the Schrödinger equation solution for the hydrogen atom, each orbital is one eigenvector of the solution.
Remember from time-independent Schrödinger equation that the final solution is just the weighted sum of the eigenvector decomposition of the initial state, analogously to solving partial differential equations with the Fourier series.
This is the table that you should have in mind to visualize them: en.wikipedia.org/w/index.php?title=Atomic_orbital&oldid=1022865014#Orbitals_table
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:
As the various lines in the spectrum were identified with different quantum jumps between different orbits, it was soon discovered that not all the possible jumps were appearing. Some lines were missing. For some reason certain jumps were forbidden. An elaborate scheme of ‘selection rules’ was established by Bohr and Sommerfeld to account for those jumps that were allowed and those that were forbidden.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?
Determines energy. This comes out directly from the resolution of the Schrödinger equation solution for the hydrogen atom where we have to set some arbitrary values of energy by separation of variables just like we have to set some arbitrary numbers when solving partial differential equations with the Fourier series. We then just happen to see that only certain integer values are possible to satisfy the equations.
Fixed total angular momentum.
The direction however is not specified by this number.
Fixed quantum angular momentum in a given direction.
Can range between .
E.g. consider gallium which is 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p1:
Note that this direction is arbitrary, since for a fixed azimuthal quantum number (and therefore fixed total angular momentum), we can only know one direction for sure. is normally used by convention.
But it is a bit crap that the spin is not given simply as but rather mixes up both the azimuthal quantum number and spin. What is the reason?
- Quantum Mechanics for Engineers by Leon van Dommelen (2011) "5. Multiple-Particle Systems"
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.
- www.youtube.com/watch?v=Og13-bSF9kA&list=PLDfPUNusx1Eo60qx3Od2KLUL4b7VDPo9F "Advanced quantum theory" by Tobias J. Osborne says that the course will essentially cover multi-particle quantum mechanics!
- physics.stackexchange.com/questions/54854/equivalence-between-qft-and-many-particle-qm "Equivalence between QFT and many-particle QM"
- Course: Quantum Many-Body Physics in Condensed Matter by Luis Gregorio Dias (2020) from course: Quantum Many-Body Physics in Condensed Matter by Luis Gregorio Dias (2020) give a good introduction to non-interacting particles
Just ignore the electron electron interactions.
As shown at Schrödinger equation solution for the helium atom, they do repel each other, and that affects their measurable energy.
However, this energy is still lower than going up to the next orbital. TODO numbers.
This changes however at higher orbitals, notably as approximately described by the aufbau principle.
Boring rule that says that less energetic atomic orbitals are filled first.
Much more interesting is actually determining that order, which the Madelung energy ordering rule is a reasonable approximation to.
We will sometimes just write them without superscript, as it saves typing and is useless.
For other atoms with more than one electron, the orbital names are just a very good approximation/perturbation, as we don't have an explicit solution. And the internal electrons do change energy levels.
Looking at the energy level of the Schrödinger equation solution for the hydrogen atom, you would guess that for multi-electron atoms that only the principal quantum number would matter, azimuthal quantum number getting filled randomly.
However, orbitals energies for large atoms don't increase in energy like those of hydrogen due to electron-electron interactions.
This rule is only an approximation, there exist exceptions to the Madelung energy ordering rule.
This notation is so confusing! People often don't manage to explain the intuition behind it, why this is an useful notation. When you see Indian university entry exam level memorization classes about this, it makes you want to cry.
web.chem.ucsb.edu/~devries/chem218/Term%20symbols.pdf puts it well: electron configuration notation is not specific enough, as each such notation e.g. 1s2 2s2 2p2 contains several options of spins and z angular momentum. And those affect energy.
This is why those symbols are often used when talking about energy differences: they specify more precisely which levels you are talking about.
Basically, each term symbol appears to represent a group of possible electron configurations with a given quantum angular momentum.
We first fix the energy level by saying at which orbital each electron can be (hyperfine structure is ignored). It doesn't even have to be the ground state: we can make some electrons excited at will.
The best thing to learn this is likely to draw out all the possible configurations explicitly, and then understand what is the term symbol for each possible configuration, see e.g. term symbols for carbon ground state.
It also confusing how uppercase letters S, P and D are used, when they do not refer to orbitals s, p and d, but rather to states which have the same angular momentum as individual electrons in those states.
It is also very confusing how extremelly close it looks to spectroscopic notation!
The form of the term symbol is:
The can be understood directly as the degeneracy, how many configurations we have in that state.
- chem.libretexts.org/Courses/Pacific_Union_College/Quantum_Chemistry/08%3A_Multielectron_Atoms/8.08%3A_Term_Symbols_Gives_a_Detailed_Description_of_an_Electron_Configuration The PDF origin: web.chem.ucsb.edu/~devries/chem218/Term%20symbols.pdf
- physics.stackexchange.com/questions/8567/how-do-electron-configuration-microstates-map-to-term-symbols How do electron configuration microstates map to term symbols?
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 __
Higher spin multiplicity means lower energy. I.e.: you want to keep all spins pointin in the same direction.
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.
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
For example, on the first line:
m_l +1 0 -1 m_L m_S u_ u_ __ 1 1
and so the sum of them has angular momentum . So the value of is 1, we just omit the .
- one electron with , and so angular momentum
- one electron with , and so angular momentum 0
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.
He then goes for and mentions:
and so that corresponds to states on our list:
- S = 1 so can only be 0
- L = 2 (D) so ranges in -2, -1, 0, 1, 2
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:
ud __ __ 2 0 u_ d_ __ 1 0 u_ __ d_ 0 0 __ u_ d_ -1 0 __ __ ud -2 0
+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:
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.
- S = 2 so is either -1, 0 or 1
- L = 1 (P) so ranges in -1, 0, 1
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.
Can we make any ab initio predictions about it all?
A 2016 paper: aip.scitation.org/doi/abs/10.1063/1.4948309
TODO definition. Appears to be isomers
Molecules that are the same if you just look at "what atom is linked to what atom", they are only different if you consider the relative spacial positions of atoms.
Also known in quantum computing as a qubit :-)