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Commit 3e7c6931 authored by Bowy La Riviere's avatar Bowy La Riviere
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Update 5_atoms_and_lcao_solutions.md

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...@@ -24,18 +24,13 @@ ...@@ -24,18 +24,13 @@
&\sqrt{κ}e^{-κ(x-x_1)}, x>x_1 &\sqrt{κ}e^{-κ(x-x_1)}, x>x_1
\end{cases} \end{cases}
$$ $$
Where $κ = \sqrt{\frac{-2mE}{ħ^2}} = \frac{mV_0}{ħ^2}$.
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
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_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. $$ 2. $$
H = -\frac{mV_0^2}{ħ^2}\begin{pmatrix} H = -\frac{mV_0^2}{ħ^2}\begin{pmatrix}
...@@ -49,7 +44,6 @@ $$ ...@@ -49,7 +44,6 @@ $$
3. $$ 3. $$
ϵ_{\pm} = \beta(1/2+\exp(-2\alpha) \pm \exp(-\alpha)) ϵ_{\pm} = \beta(1/2+\exp(-2\alpha) \pm \exp(-\alpha))
$$ $$
Where $\beta = -\frac{mV_0^2}{ħ^2}$ and $α = \frac{mV_0}{ħ^2}|x_2-x_1|$ Where $\beta = -\frac{mV_0^2}{ħ^2}$ and $α = \frac{mV_0}{ħ^2}|x_2-x_1|$
### Question 3 ### Question 3
......
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