From ad91c64e05cb9fd32e0a7374c093d1ee947e6ea3 Mon Sep 17 00:00:00 2001
From: "T. van der Sar" <t.vandersar@tudelft.nl>
Date: Thu, 24 Feb 2022 22:29:05 +0000
Subject: [PATCH] Update 5_atoms_and_lcao_solutions.md - more fixes
---
docs/5_atoms_and_lcao_solutions.md | 14 +++++++-------
1 file changed, 7 insertions(+), 7 deletions(-)
diff --git a/docs/5_atoms_and_lcao_solutions.md b/docs/5_atoms_and_lcao_solutions.md
index 58e39c7d..6cd93be3 100644
--- a/docs/5_atoms_and_lcao_solutions.md
+++ b/docs/5_atoms_and_lcao_solutions.md
@@ -72,19 +72,19 @@ $$
$$
\hat{H} = \begin{pmatrix}
- E_0 & -t\\
- -t & E_0
+ E_0 & -t \\
+ -t & E_0
\end{pmatrix} +e\mathcal{E}\begin{pmatrix}
- \langle 1|\hat{x}|1\rangle & \langle 1|\hat{x}|2\rangle\\
+ \langle 1|\hat{x}|1\rangle & \langle 1|\hat{x}|2\rangle \\
\langle 2|\hat{x}|1\rangle & \langle 2|\hat{x}|2\rangle
\end{pmatrix} = \begin{pmatrix}
- E_0 - \gamma & -t\\
- -t & E_0 + \gamma
+ E_0-\gamma & -t \\
+ -t & E_0+\gamma
\end{pmatrix},
$$
where we defined $\gamma = e d \mathcal{E}/2$ and used
$$
- ⟨1|\hat{x}|1⟩ = -\frac{d}{2}.
+ \langle 1|\hat{x}|1\rangle = -\frac{d}{2}.
$$
#### Question 3.
@@ -114,5 +114,5 @@ $$
#### Question 4.
We find the polarization using
$$
-P=2e\langle\psi|\hat{x}|\psi\rangle = 2e\phi_1^2\langle 1|\hat{x}| 1 \rangle + \phi_1^2\langle 2|\hat{x}| 2 \rangle = ed(\phi_2^2-\phi_1^2) =ed\frac{\eta}{1-\eta}
+P=2e\langle\psi|\hat{x}|\psi\rangle = 2e\left(\phi_1^2\langle 1|\hat{x}| 1 \rangle + \phi_1^2\langle 2|\hat{x}| 2 \rangle \right) = ed(\phi_2^2-\phi_1^2) =ed\frac{\eta}{1-\eta}
$$
--
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