From 2e73cfe1d6f2992fac3be2c60f078e9efea6154a Mon Sep 17 00:00:00 2001
From: Bowy La Riviere <b.m.lariviere@student.tudelft.nl>
Date: Fri, 26 Feb 2021 10:25:16 +0000
Subject: [PATCH] updates error of solutions last exercise
---
src/5_atoms_and_lcao_solutions.md | 12 +++++++++---
1 file changed, 9 insertions(+), 3 deletions(-)
diff --git a/src/5_atoms_and_lcao_solutions.md b/src/5_atoms_and_lcao_solutions.md
index 063a1ce4..b442f0f3 100644
--- a/src/5_atoms_and_lcao_solutions.md
+++ b/src/5_atoms_and_lcao_solutions.md
@@ -56,12 +56,18 @@ $$
2.
$$
- H_{eff} = \begin{pmatrix}
+ \hat{H} = \begin{pmatrix}
+ E_0 & -t\\
+ -t & E_0
+ \end{pmatrix} +\begin{pmatrix}
+ ⟨1|ex\mathcal{E}|1⟩ & ⟨2|ex\mathcal{E}|1⟩\\
+ ⟨1|ex\mathcal{E}|2⟩ & ⟨2|ex\mathcal{E}|2⟩
+ \end{pmatrix} = \begin{pmatrix}
E_0 - \gamma & -t\\
-t & E_0 + \gamma
- \end{pmatrix}
+ \end{pmatrix},
$$
-Where $\gamma = e d \mathcal{E}/2$ and where we have used that $$⟨1|H_{eff}|1⟩ = -e d \mathcal{E}/2⟨1|1⟩ = e d \mathcal{E}/2$$
+where $\gamma = e d \mathcal{E}/2$ and have used $$⟨1|ex\mathcal{E}|1⟩ = -e d \mathcal{E}/2⟨1|1⟩ = -e d \mathcal{E}/2$$
3.
--
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