Lecture 5

Contains the adjustments to the lecture notes and the excersises.

src/5_atoms_and_lcao.md

@@ -11,7 +11,7 @@ pi = np.pi
 ```
 
 # Atoms and bonds
-_(based on chapters 5 and 6.2 of the book)_  
+_(based on chapters 5 and 6.2 of the book)_
 
 !!! success "Expected prior knowledge"
 
@@ -37,7 +37,7 @@ So far we have:
 
 * Introduced the $k$-space (reciprocal space)
 * Postulated the dispersion relation of free electrons and phonons
-* Calculated the heat capacity of free electrons and phonons 
+* Calculated the heat capacity of free electrons and phonons
 
 As a result we:
 
@@ -48,7 +48,7 @@ We made several approximations and postulations through these models.
 However, there are still several mysteries:
 
 * Why is there a phonon cutoff frequency? Why are there no more phonon modes beyond this cutoff frequency?
-* Why don't electrons scatter off from every single atom in the Drude model? 
+* Why don't electrons scatter off from every single atom in the Drude model?
 Atoms are charged and should provide a lot of scattering.
 * Why are some materials not metals? (Think if you know a crystal that isn't a metal)
 
@@ -199,13 +199,13 @@ Because $ψ_- = 0$ between the two atoms, and $ψ_+$ is not, we conclude that $E
 ![](figures/two_atoms.svg)
 
 ### Bonding vs antibonding
-If we decrease the interatomic distance, the two atoms get closer and their atomic orbitals have more overlap. 
+If we decrease the interatomic distance, the two atoms get closer and their atomic orbitals have more overlap.
 The increase in orbital overlap increases the hopping $t$.
 We plot the symmetric and anti-symmetric energies as a function of the inter-atomic distance:
 
 ![](figures/bonding.svg)
 
-When an electron (or two, because there are two states with opposite spin) occupies $|ψ_+⟩$, the atoms attract (or *bond*) because the total energy is lowered. 
+When an electron (or two, because there are two states with opposite spin) occupies $|ψ_+⟩$, the atoms attract (or *bond*) because the total energy is lowered.
 Therefore, if $t$ is positive, $|ψ_+⟩$ is called the **bonding orbital**.
 
 If an electron occupies the $|ψ_{-}⟩$ orbital, the molecular energy increases with decreasing interatomic distance.
@@ -218,7 +218,7 @@ On the other hand, if each atom has 0 or 2 electrons in the outermost shell, the
 ### Summary
 
 * Electrons in atoms occupy shells, with only electrons in the outermost shell (valence electrons) contributing to interatomic interactions.
-* The molecular orbital can be written as a Linear Combination of Atomic Orbitals (LCAO) 
+* The molecular orbital can be written as a Linear Combination of Atomic Orbitals (LCAO)
 * The LCAO method reduces the full Hamiltonian to a finite size problem written in the basis of individual orbitals.
 * If two atoms have one orbital and one electron each, they occupy the bonding orbital.
 
@@ -260,7 +260,7 @@ where $V_0>0$ is the potential strength, $\hat{p}$ the momentum of the electron,
 
     The procedure to find the energy and a wave function of a state bound in a $\delta$-function potential, $V=-V_0\delta(x-x_0)$, is similar to that of a quantum well:
 
-    1. Assume that we have a bound state with energy $E<0$. 
+    1. Assume that we have a bound state with energy $E<0$.
     2. Compute the wave function $\phi$ in different regions of space: namely $x < x_0$ and $x > x_0$.
     3. Apply the boundary conditions at $x = x_0$. The wave function $\phi$ must be continuous, but $d\phi/dx$ is not. Instead due to the presence of the delta-function:
        $$
@@ -271,20 +271,20 @@ where $V_0>0$ is the potential strength, $\hat{p}$ the momentum of the electron,
 
 Let us apply the LCAO model to solve this problem. Consider the trial wave function for the ground state to be a linear combination of two orbitals $|1⟩$ and $|2⟩$:
 $$|ψ⟩ = φ_1|1⟩ + φ_2|2⟩.$$
-The orbitals $|1⟩$ and $|2⟩$ correspond to the wave functions of the electron when there is only a single delta peak present:  
+The orbitals $|1⟩$ and $|2⟩$ correspond to the wave functions of the electron when there is only a single delta peak present:
 
 $$H_1 |1⟩ = \epsilon_1 |1⟩,$$
 $$H_2 |2⟩ = \epsilon_2 |2⟩.$$
 
 We start of by calculating the wavefunction of an electron bound to a single delta-peak.
-To do so, you first need to set up the Schrödinger equation of a single electron bound to a single delta-peak. 
+To do so, you first need to set up the Schrödinger equation of a single electron bound to a single delta-peak.
 You do not have to solve the Schrödinger equation twice—you can use the symmetry of the system to calculate the wavefunction of the other electron bound to the second delta-peak.
 
 1. Find the expressions for the wave functions of the states $|1⟩$ and $|2⟩$: $ψ_1(x)$ and $ψ_2(x)$.
 Also find an expression for their energies $\epsilon_1$ and $\epsilon_2$.
 Remember that you need to normalize the wave functions.
-2. Construct the LCAO Hamiltonian. To simplify the calculations, assume that the orbitals are orthogonal.  
-3. Diagonalize the LCAO Hamiltonian and find an expression for the eigenenergies of the system. 
+2. Construct the LCAO Hamiltonian. To simplify the calculations, assume that the orbitals are orthogonal.
+3. Diagonalize the LCAO Hamiltonian and find an expression for the eigenenergies of the system.
 It was previously mentioned that $V_0>0$.
 Using this, determine which energy corresponds to the bonding energy.
 
@@ -300,7 +300,7 @@ $$
 H_{\textrm{eff}} = \begin{pmatrix}
           E_0&&-t \\
           -t&&E_0
-          \end{pmatrix}.  
+          \end{pmatrix}.
 $$
 
 1. Let us add an electric field $\mathcal{E} \hat{\bf{x}}$ to the system.
@@ -324,4 +324,6 @@ $$
     $$
     Use that ground state you found in 3.2 is a linear superposition of two orthogonal orbitals centered at $-\frac{d}{2}$ and $+\frac{d}{2}$.
 
-[^1]: See the book exercise 6.5 for relaxing the orthogonality assumption.
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+[^1]: See the book exercise 6.5 for relaxing the orthogonality assumption.
+
+* [LCAO]: Linear combination of atomic orbitals
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