Solutions lecture 7

src/7_tight_binding.md

@@ -58,7 +58,7 @@ For a system of size $L = Na$ with periodic boundary conditions, we also have $u
 
 ### Electrons
 
-The following figure shows the inter-atomic potential with atoms placed at $ka$ with $k \in \z$:
+The following figure shows the inter-atomic potential with atoms placed at $ka$ with $k \in \Z$:
 
 ![](figures/lattice_potential.svg)
 
@@ -81,29 +81,29 @@ and for electrons:
 $$\phi_n = Be^{i E t/\hbar - i k x_n},$$
 where $x_n=na$ and where we wrote the time-dependent solution of the Schrödinger equation to emphasize the similarity between the two systems.
 
-As we have seen before, the periodic boundary conditions (\phi_0=\phi_N) quantize k-space by requiring
+As we have seen before, the periodic boundary conditions ($\phi_0=\phi_N$) quantize k-space by requiring
 
 $$ \phi_0 = e^{ik0 + 2\pi i p} = e^{ikL} = \phi_N \quad ⇒ \quad k = p \frac{2\pi}{L} $$
 
-with $p \in \mathbb{Z}$. As such, $e^{ikx_n} = e^{i p \frac{2\pi}{L} n a} = e^{i \frac{2 \pi n p}{N}}$ and we see that changing $p→p+N$ corresponds to $e^{i \frac{2\pi n(p+N)}{N} = e^{i \frac{2\pi np}{N} + i2\pi n$, which is exactly the same solution. Therefore, we have $N$ different solutions in total. Furthermore, solutions with $k$-values that differ by an integer multiple of $N\frac{2\pi}{L} = \frac{2\pi}{a}$ are identical (see figure).
+with $p \in \mathbb{Z}$. As such, $e^{ikx_n} = e^{i p \frac{2\pi}{L} n a} = e^{i \frac{2 \pi n p}{N}}$ and we see that changing $p→p+N$ corresponds to $ e^{i \frac{2\pi n(p+N)}{N}} = e^{i \frac{2\pi np}{N} + i2\pi n} = e^{i \frac{2\pi np}{N}} $, which is exactly the same solution. Therefore, we have exactly $N$ different solutions in total. Furthermore, solutions with $k$-values that differ by an integer multiple of $N\frac{2\pi}{L} = \frac{2\pi}{a}$ are identical (see figure).
 
 ```python
 x = np.linspace(-.2, 2.8, 500)
 fig, ax = pyplot.subplots()
-ax.plot(x, np.cos(pi*(x - .3)), label=r'$k=\pi/a$')
-ax.plot(x, np.cos(3*pi*(x - .3)), label=r'$k=3\pi/a$')
-ax.plot(x, np.cos(5*pi*(x - .3)), label=r'$k=5\pi/a$')
-sites = np.arange(3) + .3
-ax.scatter(sites, np.cos(pi*(sites - .3)), c='k', s=64, zorder=5)
+ax.plot(x, np.cos(pi*(x)), label=r'$k=\pi/a$')
+ax.plot(x, np.cos(3*pi*(x)), label=r'$k=3\pi/a$')
+ax.plot(x, np.cos(5*pi*(x)), label=r'$k=5\pi/a$')
+sites = np.arange(3)
+ax.scatter(sites, np.cos(pi*(sites)), c='k', s=64, zorder=5)
 ax.set_xlabel('$x$')
 ax.set_ylabel('$u_n$')
 ax.set_xlim((-.1, 3.2))
 ax.set_ylim((-1.3, 1.3))
 ax.legend(loc='lower right')
 draw_classic_axes(ax)
-ax.annotate(s='', xy=(.3, -1.1), xytext=(1.3, -1.1),
+ax.annotate(s='', xy=(0, -1.1), xytext=(1, -1.1),
             arrowprops=dict(arrowstyle='<->', shrinkA=0, shrinkB=0))
-ax.text(.3 + .5, -1.25, '$a$', ha='center');
+ax.text(.5, -1.25, '$a$', ha='center');
 ```
 
 How many different solutions did we expect to find? We have a system with $N$ degrees of freedom (either $u_n$ or $\phi_n$), and therefore we expect $N$ normal modes (or eigenstates).
@@ -114,12 +114,17 @@ Because we proposed an ansatz with $N$ different plane-wave solutions, if we fin
 
 ### Phonons
 
-First substitute the Ansatz into the equations of motion:
-$$ -m \omega^2 A e^{i\omega t - ikx_n} = \kappa A  e^{i\omega t}(-2 e^{-ikx_n} + e^{-ikx_n+ika}+ e^{-ikx_n-ika}),$$
-the exponents and $A$ drop out, and we get:
-$$ -m \omega^2 = \kappa (-2 + e^{ika}+ e^{-ika})=\kappa [-2 + 2\cos(ka)],$$
+We substitute the Ansatz into the equations of motion:
+$$ -m \omega^2 A e^{i\omega t - ikx_n} = \kappa A  e^{i\omega t}(-2 e^{-ikx_n} + e^{-ikx_n+ika}+ e^{-ikx_n-ika})$$
+
+since $Ae^{i\omega t - ikx_n} = 0$ corresponds to the trivial solution, we omit these terms and obtain:
+
+$$ -m \omega^2 = \kappa (-2 + e^{ika}+ e^{-ika})=\kappa [-2 + 2\cos(ka)]$$
+
 or after a further simplification:
-$$\omega = \sqrt{\frac{2\kappa}{m}}\sqrt{1-\cos(ka)} = 2\sqrt{\frac{\kappa}{m}}|\sin(ka/2)|$$,
+
+$$\omega = \sqrt{\frac{2\kappa}{m}}\sqrt{1-\cos(ka)} = 2\sqrt{\frac{\kappa}{m}}|\sin(ka/2)|$$
+
 where we used $1-\cos(x)=2\sin^2(x/2)$.
 
 So we arrive at the dispersion relation
@@ -146,22 +151,22 @@ ax.text(0, -.25, '1st Brillouin zone', ha='center')
 ax.set_ylim(bottom=-.3);
 ```
 
-Here $k_p = 2\pi p / L$ for $0 ≤ p < N$ are *all the normal modes* of the crystal, and therefore we can rederive a better Debye model!
+Here $k_p = 2\pi p / L$ for $0 ≤ p < N$ are *all the normal modes* of the crystal, and therefore we can rederive a better Debye model! (Recall that Debye did not explain \textit{why} we need to include $\omega_D$).
 
 Before we had $\sum_p → \frac{L}{2\pi}\int_{-\omega_D/v_s}^{\omega_D/v_s}dk$, because we introduced the cutoff to fix the problem with the number of modes.
 
 Now $\sum_p → \frac{L}{2\pi}\int_{-\pi/a}^{\pi/a}dk$, the integral is over a finite number of modes because *there is* only a finite number of modes.
 
-**Sound velocity:** at small $k$, $\sin(ka/2)\approx ka/2$, and therefore $\omega \approx \sqrt{\kappa/m} k a$, so we have derived the existence of sound waves.
+**Sound velocity:** at small $k$, $\sin(ka/2)\approx ka/2$, and therefore $\omega \approx \sqrt{\kappa/m} k a = v_sk$, with $v_s$ the velocity of the sound waves so we have derived the existence of sound waves.
 
 ### Electrons
 
 Substitute the Ansatz into the equations of motion:
 
 $$
-E e^{-ikx_n} = E_0 e^{-ikx_n} - t e^{-ikx_n-ika} - te^{-ikx_n+ika},
+E Ae^{iEt/\hbar-ikx_n} = E_0 Ae^{iEt/\hbar-ikx_n} - t Ae^{iEt/\hbar-ikx_n-ika} - t Ae^{iEt/\hbar-ikx_n+ika},
 $$
-and after canceling the exponents we immediately get
+Since we're not interested in $Ae^{iEt/\hbar-ikx_n}=0$ we can omit them and we immediately get:
 $$
 E = E_0 - 2t\cos(ka),
 $$