Update 5_atoms_and_lcao_solutions.md - corrected solution last ex

docs/5_atoms_and_lcao_solutions.md

@@ -91,26 +91,28 @@ $$
 
 The eigenstates of the Hamiltonian are given by:
 $$
-    E_{\pm} = E_0\pm\sqrt{t^2+\gamma^2}
+    E_{\mp} = E_0\pm\sqrt{t^2+\gamma^2}
 $$
-Calling the elements of the eigenvector $\alpha$ and $\beta$, we find
+Calling the elements of the eigenvector $\phi_1$ and $\phi_2$, we find
 $$
-\alpha(E_0-\gamma)-\beta t = \alpha E_\pm
+\phi_1(E_0-\gamma)-\phi_2 t = \phi_1 E_\pm
 $$
-From this we find 
+From this we find for the ground state 
 $$
-\beta = -\frac{E_\pm- E_0 + \gamma}{t}\alpha = -\frac{\pm\sqrt{t^2+ \gamma^2} + \gamma }{t}\alpha 
+\phi_2 = -\frac{E_+- E_0 + \gamma}{t}\phi_1 = \frac{\sqrt{t^2+ \gamma^2} - \gamma }{t}\phi_1 
 $$
-Then, using the normalization condition $\alpha^2+\beta^2$=1, we find the normalized eigenfunction.
 
-%The ground state wave function is:
-%$$
-%    |\psi⟩ &= \frac{\gamma+\sqrt{t^2+\gamma^2}}{\sqrt{(\gamma+\sqrt{\gamma^2+t^2})^2+t^2}}|1⟩+\frac{t}{\sqrt{(\gamma+\sqrt{\gamma^2+t^2})^2+t^2}}|2⟩
-%    \end{split}
-%$$
+For simplicity, we now assume that the electric field is small, such that $\gamma/t=\eta \ll 1$. We get
+$$
+\phi_2 \approx (1-\eta)\phi_1 
+$$
+Then, using the normalization condition $\phi_1^2+\phi_2^2$=1, we find 
+$$
+\phi_1\approx\frac{1}{\sqrt{2(1-\eta)}}
+$$
 
 #### Question 4.
-
+We find the polarization using 
 $$
-    P = -\frac{2\gamma^2}{\mathcal{E}}(\frac{1}{\sqrt{\gamma^2+t^2}})
+P=2e\langle\psi|\hat{x}|\psi\rangle = 2e\phi_1^2\langle 1|\hat{x}| 1 \rangle + \phi_1^2\langle 2|\hat{x}| 2 \rangle = ed(\phi_2^2-\phi_1^2) =ed\frac{\eta}{1-\eta}
 $$