for helium it becomes more complex: $\psi({\mathbf r})\rightarrow \psi({\mathbf {r_1, r_2}})$, so
In the case of helium, the Hamiltonian becomes more complex: not only is there a Coulomb attraction between the electrons and the nuclei, but there is also a Coulomb repulsion between the two electrons.
which means we need to find eigenvalues and eigenvectors of a 6-dimensional differential equation!
which means we need to find eigenvalues and eigenvectors of a 6-dimensional partial differential equation!
"Mundane" copper has 29 electrons, so to find the electronic spectrum of Copper we would need to solve an 87-dimensional Schrödinger equation, and there is no way in the world we can do so.
"Mundane" copper has 29 electrons, so to find the electronic spectrum of Copper we would need to solve an 87-dimensional Schrödinger equation, and there is no way in the world we can do so currently.
This exponential growth in complexity with the number of interacting quantum particles is why *many-body quantum physics* is very much an open area in solid state physics.
This growth in complexity with the number of interacting quantum particles is why *many-body quantum physics* is very much an open area in solid state physics and quantum chemistry.
However we need to focus on what is possible to do, and apply heuristic rules based on the accumulated knowledge of how atoms work (hence we will need a bit of chemistry).
However for us the complexity of solving the problem is the reason why we need to accept empirical observations as *chemical laws* even though they work with limited precision, and are "merely" consequences of the Schrödinger equation.
The complexity of solving the problem is the reason why we need to accept empirical observations as *chemical laws* even though they work with limited precision, and are "merely" consequences of the Schrödinger equation.
### Quantum numbers and shell filling
Single electron states have 4 quantum numbers: $|n, l, l_z, s\rangle$
The electrons in a hydrogen orbital can be described by the folowing four quantum numbers: $|n, l, l_z, m_s\rangle$
Quantum numbers:
* $n=1,2,\ldots$ is the azimuthal (principal) quantum number
* $l=0, 1, \ldots, n-1$ is the angular momentum (also known as $s, p, d, f$ orbitals)
* $l_z=-l, -l+1\ldots, l$ is the $z$-component of angular momentum
* $s$ is the spin
* $m_s$ is the $z$-compoment of the spin
It turns out that electrons in all atoms approximately occupy very similar orbitals, only the energies are very different due to the Coulomb interaction.
In the figure below, the orbitals for the first three azimuthal numbers are shown.
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It turns out that electrons in other atoms approximately occupy very similar orbitals, only the energies are very different due to the Coulomb interaction.
Therefore, we can use our knowledge of the hydrogen orbitals to describe other atoms.
When we consider an atom with multiple electrons, we need to determine which orbitals are filled and which are not.
The order of orbital filling is set by several rules:
* Aufbau principle: *first fill a complete shell (all electrons with the same $n, l$) before going to the next one*
* Madelung's rule: *first occupy the shells with the lowest $l+n$, and of those with equal $l+n$ those with smaller $n$*
*_Aufbau principle_: electrons first fill a complete shell (all electrons with the same $n$) before going to the next one
*_Madelung's rule_: electrons first occupy the shells with the lowest $n+l$, and of those with equal $n+l$ those with smaller $n$
Therefore shell-filling order is 1s, 2s, 2p, 3s, 3p, 4s, 3d, ...
If we take the Aufbau principle and Madelung's rule into account, the shell filling order is 1s, 2s, 2p, 3s, 3p, 4s, 3d, ...
Note that there are exceptions to these rules.
Especially for very large atoms.
The electrons in the outermost shell are the only ones participating in chemical reactions and electric conduction.
The rest provides a negatively charged cloud that reduces the attraction to the atomic nucleus.
The valence electrons (those in the outermost shell) are the only ones participating in chemical reactions and electric conduction.
From the valence electrons point of view the bound electrons provide a negatively charged cloud that reduces the effective charge of the atomic nucleus.
## Covalent bonds and linear combination of atomic orbitals
*Here we present the material in the order different from the book because the covalent bonds are the most important for the exercises*
### Two atoms
Consider two atoms next to each other.
Consider two atoms next to each other which form a diatomic molecule.
Since different orbitals of an atom are separated in energy, we consider one orbital per atom (even though this is often a bad starting point and it should only work for $s$-orbitals).
Let's imagine that the atoms are sufficiently far apart, so that the shape of the orbitals or the energy of the orbitals doesn't change much.
Let's imagine that the atoms are sufficiently far apart, such that the shape of the orbitals barely changes due to the presence of the other atom.
If $|1⟩$ is the wave function of an electron bound to the first atom, and $|2⟩$ is the wave function of the electron near the second atom, we will search for a solution in form: $|ψ⟩ = ϕ_{1}|1⟩ + ϕ_{2}|2⟩$, or in other words as a *Linear Combination of Atomic Orbitals (LCAO)*.
If $|1⟩$ is the wave function of an electron bound to the first atom, and $|2⟩$ is the wave function of the electron bound to the second atom, we will search for a solution in the form of:
$$
\|\psi⟩ = \varphi_{1}|1⟩ + \varphi_{2}|2⟩
$$
where $\varphi_{1}$ and $\varphi_{2}$ are the probability amplitudes of the respective orbitals.
The orbital $|\psi⟩$ is called a _molecular orbital_ because it describes the entire orbital of the diatomic molecule.
The molecular orbital is created as a *Linear Combination of Atomic Orbitals (LCAO)*.
For simplicity let's assume now that $⟨1|2⟩=0$, so that $ψ$ is normalized whenever $|ϕ_1|^2 + |ϕ_2|^2 = 1$.
For simplicity, we assume that the atomic orbitals are orthogonal (i.e. $⟨1|2⟩=0$), such that $|\psi>$ is normalized whenever $|\varphi_1|^2 + |\varphi_2|^2 = 1$.
(See the book exercise 6.5 for relaxing this assumption)
Acting with the Hamiltonian on the LCAO wave function we get an eigenvalue problem:
If we apply the Hamiltonian to the molecular orbital $|\psi>$, we obtain
$$
\begin{align}
H|\psi> &= E |\psi>\\
&= \varphi_{1}H|1⟩ + \varphi_{2}H|2⟩\\
\end{align}
$$
Taking the left inner product with either $\langle 1|$, we obtain
