From 3e7c6931c9204d8e5292459f506897658714d346 Mon Sep 17 00:00:00 2001
From: BowyLR <b.m.lariviere@student.tudelft.nl>
Date: Sun, 1 Mar 2020 18:06:33 +0000
Subject: [PATCH] Update 5_atoms_and_lcao_solutions.md
---
src/5_atoms_and_lcao_solutions.md | 18 ++++++------------
1 file changed, 6 insertions(+), 12 deletions(-)
diff --git a/src/5_atoms_and_lcao_solutions.md b/src/5_atoms_and_lcao_solutions.md
index 1838bff3..063a1ce4 100644
--- a/src/5_atoms_and_lcao_solutions.md
+++ b/src/5_atoms_and_lcao_solutions.md
@@ -24,18 +24,13 @@
&\sqrt{κ}e^{-κ(x-x_1)}, x>x_1
\end{cases}
$$
-
- Where $κ = \sqrt{\frac{-2mE}{ħ^2}} = \frac{mV_0}{ħ^2}$.
-
- The energy is given by $ϵ_1 = ϵ_2 = -\frac{mV_0^2}{2ħ^2}$
-
- The wave function of a single delta peak is given by
-
- $$
+ Where $κ = \sqrt{\frac{-2mE}{ħ^2}} = \frac{mV_0}{ħ^2}$.
+ The energy is given by $ϵ_1 = ϵ_2 = -\frac{mV_0^2}{2ħ^2}$
+ The wave function of a single delta peak is given by
+ $$
\psi_1(x) = \frac{\sqrt{mV_0}}{ħ}e^{-\frac{mV_0}{ħ^2}|x-x_1|}
- $$
-
- $\psi_2(x)$ can be found by replacing $x_1$ by $x_2$
+ $$
+ $\psi_2(x)$ can be found by replacing $x_1$ by $x_2$
2. $$
H = -\frac{mV_0^2}{ħ^2}\begin{pmatrix}
@@ -49,7 +44,6 @@ $$
3. $$
ϵ_{\pm} = \beta(1/2+\exp(-2\alpha) \pm \exp(-\alpha))
$$
-
Where $\beta = -\frac{mV_0^2}{ħ^2}$ and $α = \frac{mV_0}{ħ^2}|x_2-x_1|$
### Question 3
--
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