Lecture 5

Contains the adjustments to the lecture notes and the excersises.

src/5_atoms_and_lcao.md

@@ -123,13 +123,25 @@ The electrostatic repulsion between them reduces the effective charge of the ato
 ### Two atoms
 
 Consider two atoms next to each other which form a diatomic molecule.
+The total Hamiltonian of the system is
+$$
+H = V₁ + V₂ + K,
+$$
+with $V₁$ the potential due to the first nucleus, $V₂$ due to the second nucleus, and $K$ is the kinetic energy of the electron.
 
 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, such that the shape of the orbitals barely changes due to the presence of the other atom.
+Let's additionally consider the atoms being sufficiently far apart, such that the shape of the orbitals barely changes due to the presence of the other atom.
+
+Let's denote the wave function of an electron bound to the first and second atom $|1⟩$ and $|2⟩$ respectively:
+$$
+\begin{align}
+(V₁ + K)|1⟩ = ɛ_0|1⟩,\\
+(V₂ + K)|2⟩ = ɛ_0|2⟩.
+\end{align}
+$$
 
-Let's denote the wave function of an electron bound on the first and second atom $|1⟩$ and $|2⟩$ respectively.
-We search for a solution in the form:
+Our main idea is to search for a solution in the form:
 $$
 |ψ⟩ = φ_{1}|1⟩ + φ_{2}|2⟩.
 $$
@@ -167,16 +179,28 @@ E \begin{pmatrix} φ_1 \\ φ_2 \end{pmatrix}
 \begin{pmatrix} φ_1 \\ φ_2\end{pmatrix}.
 $$
 
-The eigenvalue problem depends only on two parameters:
-$⟨1|H|1⟩ = ⟨2|H|2⟩ \equiv E_0$ the **onsite energy** and $⟨1|H|2⟩ \equiv -t$ the **hopping integral** (or just **hopping**).
+The eigenvalue problem depends only on two parameters: the **onsite energy** $⟨1|H|1⟩ = ⟨2|H|2⟩ \equiv E_0$ that gives the energy of an electron occupying either of the orbitals, and the **hopping integral** (or just **hopping**) $⟨1|H|2⟩ \equiv -t$ that characterizes the energy associated with the electron moving between the two orbitals.
+
+Let us examine what contitutes the onsite energy and the hopping:
+$$
+E_0 = ⟨1|H|1⟩ = ⟨1|V₁ + V₂ + K|1⟩ = ɛ_0 + ⟨1|V_2|1⟩,
+$$
+where we used that $V₁ + K|1⟩ = ɛ_0|1⟩$.
+In other words the onsite energy is the combination of the energy of the original orbital plus the energy shift $⟨1|V_2|1⟩$ of the electron due to the potential of the neighboring atom.
+Turning to the hopping, we obtain
+$$
+t = -⟨1|H|2⟩ = -⟨1|V₁ + V₂ + K|2⟩ = -⟨1|V₁|2⟩.
+$$
 
-All orbitals $|n⟩$ are purely real because we consider bound states.
+All orbitals $|n⟩$ are purely real because we consider bound state solutions of the Schrödinger equation.
 Hence $t$ is real as well.
 
-We use the previously defined eigenvalue problem to construct the Hamiltonian in matrix form:
-$$ H = \begin{pmatrix} E_0 & -t \\ -t & E_0 \end{pmatrix}$$
+The eigenvalue problem we obtained describes a particle with a discrete $2×2$ Hamiltonian:
+$$
+H = \begin{pmatrix} E_0 & -t \\ -t & E_0 \end{pmatrix}
+$$
 
-Diagonalizing the Hamiltonian yields the following two eigenvalues:
+Diagonalizing this LCAO Hamiltonian yields the following two eigenvalues:
 $$
 E_{±} = E_0 ∓ t.
 $$