From a8c9a829f2318a06aff721c22f781f9c00cbf196 Mon Sep 17 00:00:00 2001
From: "T. van der Sar" <t.vandersar@tudelft.nl>
Date: Fri, 12 Apr 2024 03:27:04 +0000
Subject: [PATCH] Update 5_atoms_and_lcao.md - changed phi to psi in the
wavefunction of the delta potential
---
docs/5_atoms_and_lcao.md | 6 +++---
1 file changed, 3 insertions(+), 3 deletions(-)
diff --git a/docs/5_atoms_and_lcao.md b/docs/5_atoms_and_lcao.md
index 3e4cc99..59b5d64 100644
--- a/docs/5_atoms_and_lcao.md
+++ b/docs/5_atoms_and_lcao.md
@@ -357,10 +357,10 @@ 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 $E<0$.
- 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 this is not the case for $d\phi/dx$. Instead, due to the presence of the delta-function:
+ 2. Compute the wave function $\psi$ 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 $\psi$ must be continuous, but this is not the case for $d\psi/dx$. Instead, due to the presence of the delta-function:
- $$\left.\frac{d\phi}{dx}\right|_{x_0+\epsilon} - \left.\frac{d\phi}{dx}\right|_{x_0-\epsilon}= -\frac{2mV_0}{\hbar^2}\phi(x_0).$$
+ $$\left.\frac{d\psi}{dx}\right|_{x_0+\epsilon} - \left.\frac{d\psi}{dx}\right|_{x_0-\epsilon}= -\frac{2mV_0}{\hbar^2}\psi(x_0).$$
4. Find at which energy the boundary conditions at $x = x_0$ are satisfied. This is the energy of the bound state.
5. Normalize the wave function.
--
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