From 8f1a0ae9b6fe4a75b7ed3b2ed41876427719bbb4 Mon Sep 17 00:00:00 2001 From: "T. van der Sar" <t.vandersar@tudelft.nl> Date: Tue, 5 Feb 2019 13:13:19 +0000 Subject: [PATCH] Update 7_tight_binding.md - polish2 --- src/7_tight_binding.md | 21 ++++++++++----------- 1 file changed, 10 insertions(+), 11 deletions(-) diff --git a/src/7_tight_binding.md b/src/7_tight_binding.md index 4a9ce023..a1f6132c 100644 --- a/src/7_tight_binding.md +++ b/src/7_tight_binding.md @@ -73,9 +73,9 @@ where $x_n=na$ and where we wrote the time-dependent solution of the Schrödinge As usual, periodicity quantizes k-space by requiring $e^{i k L} = 1$ $⇒$ $k = 2\pi p/L$, with $p\in \mathbb{Z}$. -As such, $c_n/A = \exp(2 \pi n p a/L) = \exp(2 \pi n p/N)$, and we see that changing $p→p+N$ corresponds to exactly the same solution, and that we have $N$ different solutions in total. +As such, $u_n/A = \exp(2 \pi n p a/L) = \exp(2 \pi n p/N)$, and we see that changing $p→p+N$ corresponds to exactly the same solution, and that we have $N$ different solutions in total. -We see that solutions with $k$ differing by an integer multiple of $2\pi/a$ are identical by plotting them: +We see that solutions with $k$-values differing by an integer multiple of $2\pi/a$ are identical by plotting them: ```python x = np.linspace(-.2, 2.8, 500) @@ -96,23 +96,23 @@ ax.annotate(s='', xy=(.3, -1.1), xytext=(1.3, -1.1), ax.text(.3 + .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 $c_n$), and therefore we need $N$ normal modes (or eigenstates). +How many different solutions did we expect to find? We have a system with $N$ degrees of freedom (either $u_n$ or $c_n$), and therefore we expect $N$ normal modes (or eigenstates). Because we proposed an ansatz with $N$ different plane-wave solutions, if we find an energy or frequency for each solution, we have fully solved the problem! -## Solution +## Solving the equations of motion ### Phonons First substitute the Ansatz into the equations of motion: -$$ -m \omega^2 c_0 e^{i\omega t - ikx} = \kappa c_0 e^{i\omega t}(-2 e^{-ikx} + e^{-ikx+ika}+ e^{-ikx-ika}),$$ -the exponents and $c_0$ drop out (just like we expected) and we get: +$$ -m \omega^2 A e^{i\omega t - ikx} = \kappa A e^{i\omega t}(-2 e^{-ikx} + e^{-ikx+ika}+ e^{-ikx-ika}),$$ +the exponents and $A$ drop out (just like we expected) and we get: $$ -m \omega^2 = \kappa (-2 + e^{ika}+ e^{-ika})=\kappa [-2 + 2\cos(ka)],$$ or after a further simplification: $$\omega = \sqrt{\frac{2\kappa}{m}}\sqrt{1-\cos(ka)} = 2\sqrt{\frac{\kappa}{m}}|\sin(ka/2)|$$ [here we used that $1-\cos(x)=2\sin^2(x/2)$.] -So we arrive to the dispersion relation +So we arrive at the dispersion relation ```python k = np.linspace(-2*pi, 6*pi, 500) @@ -142,20 +142,20 @@ Before we had $\sum_p → \frac{L}{2\pi}\int_{-\omega_D/v_s}^{\omega_D/v_s}dk$, 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 $\omega \approx \sqrt{\kappa/m} k a$, so we have derived the existence of sound waves. ### Electrons Substitute the Ansatz into the equations of motion: $$ -E c_0 e^{-ikx} = E_0 c_0 e^{-ikx} +t e^{-ikx-ika} + te^{-ikx+ika}, +E e^{-ikx} = E_0 e^{-ikx} +t e^{-ikx-ika} + te^{-ikx+ika}, $$ and after canceling the exponents we immediately get $$ E = E_0 + 2t\cos(ka), $$ -so we come to the dispersion relation: +so we arrive at the dispersion relation: ```python pyplot.figure() @@ -166,7 +166,6 @@ pyplot.xticks([-pi, 0, pi], [r'$-\pi$', 0, r'$\pi$']) pyplot.yticks([-1, 0, 1], ['$E_0+2t$', '$E_0$', '$E_0-2t$']); ``` - Usually electron dispersion has multiple options for $E(k)$, each called a *band*. The complete dispersion relation is also called a *band structure*. The band bottom is at $k=0$ (when $t<0$). There the energy -- GitLab