From 2ac690015a3415fa7b676ea1105f75690c8ff358 Mon Sep 17 00:00:00 2001
From: "T. van der Sar" <t.vandersar@tudelft.nl>
Date: Sun, 22 Mar 2020 21:38:46 +0000
Subject: [PATCH] Update 11_nearly_free_electron_model.md - typo

---
 src/11_nearly_free_electron_model.md | 6 +++---
 1 file changed, 3 insertions(+), 3 deletions(-)

diff --git a/src/11_nearly_free_electron_model.md b/src/11_nearly_free_electron_model.md
index b1626b07..4a460e67 100644
--- a/src/11_nearly_free_electron_model.md
+++ b/src/11_nearly_free_electron_model.md
@@ -113,7 +113,7 @@ $$ E(\delta k) = E_0 \pm \sqrt{v^2\hbar^2\delta k^2 + |W|^2}$$
     that $\psi(x) = u(x) \exp(ikx)$.
 
 
-#### Physical meaning of $W$
+#### Calculating the size of the gaps $W$
 
 To calculate $W=\langle k | V |k' \rangle$, we first express the lattice potential, which is periodic as $V(x)=V(x+a)$, as a Fourier series 
 
@@ -126,9 +126,9 @@ V_n = \frac{1}{a}\int_0^a e^{- i n 2\pi x /a} V(x) dx
 $$
 
 We now calculate the matrix element that couples our basis states $|k\rangle$ and $k'\rangle = |k-2\pi/a '\rangle$
-$$W = \langle k | V | k' \rangle = \frac{1}{a}\int_0^{a} dx \left[e^{i k x}\right]^* V(x) \left[e^{-i k'x}\right] = \frac{1}{a}\int_0^a e^{-2\pi i x /a} V(x) dx = V_1$$
+$$W = \langle k | V | k' \rangle = \frac{1}{a}\int_0^{a} e^{i k x} V(x) e^{-i k'x}  dx = \frac{1}{a}\int_0^a e^{-i 2\pi x /a} V(x) dx = V_1$$
 
-where we have used that $k-k'=2\pi/a$ for the first crossing. We see that the first component of the Fourier-series representation of $V(x)$ determines the strength of the coupling between the two states near the first crossing.
+where we have used that $k'-k =2\pi/a$ because we are analyzing the first crossing. We see that the first component of the Fourier-series representation of $V(x)$ determines the strength of the coupling between the two states near the first crossing.
 
 
 #### Crossings between the higher bands
-- 
GitLab