From 6e4639ce4b54b163a9489cd744badc362e0591b2 Mon Sep 17 00:00:00 2001
From: "T. van der Sar" <t.vandersar@tudelft.nl>
Date: Mon, 7 Mar 2022 10:59:40 +0000
Subject: [PATCH] Update 7_tight_binding_model_solutions.md - plot fix 1st
attempt
---
docs/7_tight_binding_model_solutions.md | 80 ++++++++++++-------------
1 file changed, 37 insertions(+), 43 deletions(-)
diff --git a/docs/7_tight_binding_model_solutions.md b/docs/7_tight_binding_model_solutions.md
index 02ac1980..f7b00b83 100644
--- a/docs/7_tight_binding_model_solutions.md
+++ b/docs/7_tight_binding_model_solutions.md
@@ -36,39 +36,30 @@ $$
m^* = m_e,
$$
-where $m_e$ is the free electron mass. This is expected because the free elctrons are not subject to a potential
+where $m_e$ is the free electron mass. This is expected because the free electrons are not subject to a potential
#### Question 5.
-If the dispersion relation is parabolic, so in the free electron model.
+The effective mass is given by the curvature of the dispersion. Therefore, if the dispersion has a constant curvature, the effective mass is the same for all $k-$values. This is the case for the purely parabolic dispersion $\epsilon(k)\propto^2$ of the free electron model
## Exercise 1: Lattice vibrations
#### Question 1.
The group velocity is given by
-\begin{align}
-v_g(k)&=\frac{\partial \omega(k)}{\partial k}\\
-&= a \sqrt{\frac{\kappa}{m}}\cos(\frac{ka}{2}) \frac{\sin(ka/2)}{|\sin(ka/2)|},
-\end{align}
-which can be written as
$$
-v_g = a \sqrt{\frac{\kappa}{m}}
-\begin{cases}
- &\cos(\frac{ka}{2}), 0<ka<\pi\\
- &-\cos(\frac{ka}{2}), -\pi<ka<0
-\end{cases}
+v_g(k)=\frac{\partial \omega(k)}{\partial k} = a \sqrt{\frac{\kappa}{m}}\cos(\frac{ka}{2}) \text{sign}(k),
$$
-
+where $\text{sign}(k)$ represents the sign of $k$.
#### Question 2.
The density of states is
-\begin{align}
-g(\omega) &= \frac{L}{\pi} \left|\frac{1}{v_g}\right| \\
-&= \frac{L}{a \pi} \sqrt{\frac{m}{\kappa}}\frac{1}{\cos(ka/2)}\\
-&= \frac{L}{a \pi} \sqrt{\frac{m}{\kappa}}\frac{1}{\sqrt{1-\sin^2(ka/2)}}\\
-&=\frac{2L}{a \pi} \frac{1}{\sqrt{4\kappa / m - \omega^2}},
-\end{align}
+$$
+g(\omega) = \frac{L}{\pi} \left|\frac{1}{v_g}\right|
+= \frac{L}{a \pi} \sqrt{\frac{m}{\kappa}}\frac{1}{\cos(ka/2)}
+= \frac{L}{a \pi} \sqrt{\frac{m}{\kappa}}\frac{1}{\sqrt{1-\sin^2(ka/2)}}
+=\frac{2L}{a \pi} \frac{1}{\sqrt{4\kappa / m - \omega^2}},
+$$
where we substituted back the dispersion relation.
#### Question 3.
@@ -109,29 +100,40 @@ Hint: The group velocity is given as $v = \frac{d\omega}{dk}$, draw a coordinate
#### Question 1.
-For the energy we have: $$\langle E \rangle = \int \hbar \omega g(\omega) (n_{BE}(\hbar \omega) + \frac{1}{2})d\omega$$ with $g(\omega)$ being the DOS calculated in exercise 1 and $n_{BE}(\hbar \omega) = \frac{1}{e^{\hbar\omega/k_BT}-1}$.
+For the energy we have:
+$$
+U = \int \hbar \omega g(\omega) (n_{BE}(\hbar \omega) + \frac{1}{2})d\omega
+$$
+with $g(\omega)$ the density of states calculated in exercise 1 and $n_{BE}(\hbar \omega) = \frac{1}{e^{\hbar\omega/k_BT}-1}$ the Bose-Einstein distribution.
#### Question 2.
-
-For the heat capacity we have: $$C = \frac{d \langle E \rangle}{d T} = \int g(\omega) \hbar\omega \frac{d n_{BE}(\hbar \omega)}{d T}d\omega$$
+For the heat capacity we have:
+$$
+C = \frac{d U }{d T} = \int g(\omega) \hbar\omega \frac{d n_{BE}(\hbar \omega)}{d T}d\omega
+$$
## Exercise 3: Next-nearest neighbors chain
#### Question 1.
-
-The Schrödinger equation is given by: $|\Psi\rangle = \sum_n \phi_n |n\rangle$ such that we find $$ E\phi_n = E_0\phi_n - t\phi_{n-1} - t\phi_{n+1} - t'\phi_{n-2} - t'\phi_{n+2}$$
+The Schrödinger equation is $H|\Psi\rangle = E|\Psi\rangle$. The wavefunction is $|\Psi\rangle = \sum_m \phi_m |m\rangle$. By calculating $\langle n |H|\Psi\rangle$, we find
+$$
+E\phi_n = E_0\phi_n - t\phi_{n-1} - t\phi_{n+1} - t'\phi_{n-2} - t'\phi_{n+2}
+$$
#### Question 2.
-
-
-Solving the Schrödinger equation yields dispersion: $$E(k) = E_0 -2t\cos(ka) -2t'\cos(2ka)$$
+We solve the previous equation using the Ansatz $\phi_n=e^{ikna}\phi_0$. Doing so, we find the dispersion relation:
+$$
+E(k) = E_0 - 2t\cos(ka) - 2t'\cos(2ka)
+$$
#### Question 3.
-
-$$m^* = \frac{\hbar^2}{2a^2}\frac{1}{t\cos(ka)+4t'\cos(2ka)}$$
+The effective mass is
+$$
+m^* = \frac{\hbar^2}{2a^2}\frac{1}{t\cos(ka)+4t'\cos(2ka)}
+$$
Plot for t=t':
@@ -152,36 +154,28 @@ pyplot.tight_layout();
#### Question 4.
-Plots for 2t'=t, 4t'=t and 10t'=t:
+Plots for $t=2t'$, $t=4t'$, and $t=10t'$:
```python
def m(k,t):
- return 1/(np.cos(k)+4*t*np.cos(2*k))
+ return 1/(t*np.cos(k)+4*np.cos(2*k))
-k1 = np.linspace(-1.6, -0.83, 300);
-k2 = np.linspace(-0.826, 0.826, 300);
-k3 = np.linspace(0.83, 1.6, 300);
+k1 = np.linspace(-pi, -pi/2-0.01, 300);
+k2 = np.linspace(-pi/2+0.01, pi/2-0.01, 300);
+k3 = np.linspace(pi/2+0.01, pi, 300);
pyplot.plot(k1, m(k1,2),'b');
pyplot.plot(k2, m(k2,2),'b');
pyplot.plot(k3, m(k3,2),'b',label='t=2t\'');
pyplot.xlabel('$k$'); pyplot.ylabel('$m_{eff}(k)$');
-pyplot.xticks([-1.6,0,1.6],[r'$-\pi/a$',0,r'$\pi/a$']);
+pyplot.xticks([-pi,0,pi],[r'$-\pi/a$',0,r'$\pi/a$']);
pyplot.yticks([0],[]);
pyplot.tight_layout();
-k1 = np.linspace(-1.58, -0.81, 300);
-k2 = np.linspace(-0.804, 0.804, 300);
-k3 = np.linspace(0.81, 1.58, 300);
-
pyplot.plot(k1, m(k1,4),'r');
pyplot.plot(k2, m(k2,4),'r');
pyplot.plot(k3, m(k3,4),'r',label='t=4t\'');
-k1 = np.linspace(-1.575, -0.798, 300);
-k2 = np.linspace(-0.790, 0.790, 300);
-k3 = np.linspace(0.798, 1.575, 300);
-
pyplot.plot(k1, m(k1,10),'k');
pyplot.plot(k2, m(k2,10),'k');
pyplot.plot(k3, m(k3,10),'k',label='t=10t\'');
--
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