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Solutions lecture 7

Merged Lars kleyn Winkel requested to merge solutions-lecture-7 into master
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@@ -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 quantize k-space by requiring
As we have seen before, the periodic boundary conditions ($\phi_0=\phi_N$) quantize k-space by requiring
$$e^0 = e^{2i\pi p} = e^{ikL} \quad ⇒ \quad k = p \frac{2\pi}{L} $$
$$ \phi_0 = e^{ik0 + 2\pi i p} = e^{ikL} = \phi_N \quad ⇒ \quad k = p \frac{2\pi}{L} $$
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 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
@@ -146,25 +151,31 @@ 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:
$$
E e^{-ikx_n} = E_0 e^{-ikx_n} - t e^{-ikx_n-ika} - te^{-ikx_n+ika},
E Ae^{iEt/\hbar-ikna} = E_0 Ae^{iEt/\hbar-ikna} - t Ae^{iEt/\hbar-ik(n+1)a} - t Ae^{iEt/\hbar-ik(n-1)a},
$$
and after canceling the exponents we immediately get
Since we're not interested in $Ae^{iEt/\hbar-ikna}=0$ we can omit them and we get:
$$
E = E_0 -te^{-ika} -te^{ika}
$$
$$
E = E_0 - 2t\cos(ka),
$$
so we arrive at the dispersion relation:
```python
@@ -184,7 +195,7 @@ E = E_0 - 2t\cos{ka} \approx E_0 - 2t + t (ka)^2.
$$
Comparing this with $E=(\hbar k)^2/2m$, we see that the dispersion is similar to that of free electrons but with an effective mass given by $m^*=\hbar^2/2ta^2$.
In the following lectures we will see that an electron dispersion usually has multiple options for $E(k)$, each called an energy band. The complete dispersion relation is also called a *band structure*.
Notice that in this particular case we can occupy a continuous band from $E_0+2t$ to $E_0-2t$, in the next lecture we shall see that small deviations in the hopping between two adjacent atoms can change the continuity of the band! The *band structure* of the electrons then consists of two **separate** bands. The complete dispersion relation is called the *band structure*.
### Group velocity, effective mass, density of states
@@ -234,6 +245,8 @@ pyplot.xlabel('$ka$'); pyplot.ylabel('$m_{eff}$')
pyplot.xticks([-pi, 0, pi], [r'$-\pi$', 0, r'$\pi$']);
```
Notice that we can have a negative effective mass, which implies the electrons move opposite to the direction of the force.
### Density of states
The DOS is the number of states per unit energy. In 1D we have
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