# Lecture 2 – Free electron model
_Based on chapters 3 and 4 of the book_
In this lecture we will:
- consider electrons as charged point particles travelling through a solid
- discuss Drude theory
- discuss the Hall experiment
- discuss specific heat of a solid based due to electrons
- introduce mobility, Hall resistance and the Fermi energy
### Drude theory
Ohm's law states that $V=IR=I\rho\frac{l}{A}$. In this lecture we will investigate where this law comes from. We will use the theory developed by Paul Drude in 1900, which is based on three assumptions:
- Electrons have an average scattering time $\tau$.
- At each scattering event an electron returns to momentum ${\bf p}=0$.
- In-between scattering events electrons respond to the Lorentz force ${\bf F}_{\rm L}=-e\left({\bf E}+{\bf v}\times{\bf B}\right)$.
For now we will consider only an electric field (_i.e._ ${\bf B}=0$). What velocity do electrons acquire in-between collisions?
$$
{\bf v}=-\int_0^\tau\frac{e{\bf E}}{m_{\rm e}}{\rm d}t=-\frac{e\tau}{m_{\rm e}}{\bf E}=-\mu{\bf E}
$$
Here we have defined the quantity $\mu\equiv e\tau/m_{\rm e}$, which is the _mobility_. If we have a density $n$ of electrons in our solid, the current density ${\bf j}$ [A/m$^2$] then becomes:
$$
{\bf j}=-en{\bf v}=\frac{n e^2\tau}{m_{\rm e}}{\bf E}=\sigma{\bf E}\ ,\ \ \sigma=\frac{ne^2\tau}{m_{\rm e}}=ne\mu
$$
$\sigma$ is the conductivity, which is the inverse of resistivity: $\rho=\frac{1}{\sigma}$. If we now take $j=\frac{I}{A}$ and $E=\frac{V}{l}$, we retrieve Ohm's Law: $\frac{I}{A}=\frac{V}{\rho l}$.
Scattering is caused by collisions with:
- Phonons: $\tau_{\rm ph}(T)$ ($\tau_{\rm ph}\rightarrow\infty$ as $T\rightarrow 0$)
- Impurities/vacancies: $\tau_0$
Scattering rate $\frac{1}{\tau}$:
$$
\frac{1}{\tau}=\frac{1}{\tau_{\rm ph}(T)}+\frac{1}{\tau_0}\ \Rightarrow\ \rho=\frac{1}{\sigma}=\frac{m}{ne^2}\left( \frac{1}{\tau_{\rm ph}(T)}+\frac{1}{\tau_0} \right)\equiv \rho_{\rm ph}(T)+\rho_0
$$
_Matthiessen's Rule_ (1864). Solid (dashed) curve: $\rho(T)$ for a pure (impure) crystal.
How fast do electrons travel through a copper wire? Let's take $E$ = 1 volt/m, $\tau$ ~ 25 fs (Cu, $T=$ 300 K).
$\rightarrow v=\mu E=\frac{e\tau}{m_{\rm e}}E=\frac{10^{-19}\times 2.5\times 10^{-14}}{10^{-30}}=2.5\times10^{-3}=2.5$ mm/s ! (= 50 $\mu$m @ 50 Hz AC)
### Hall effect
Consider a conductive wire in a magnetic field ${\bf B} \rightarrow$ electrons are deflected in a direction perpendicular to ${\bf B}$ and ${\bf j}$.
${\bf E}_{\rm H}$ = _Hall voltage_, caused by the Lorentz force.
In equilibrium, assuming that the average velocity becomes zero after every collision: $\frac{mv_x}{\tau}=-eE$
The $y$-component of the Lorentz force $-e{\bf v}_x\times{\bf B}$ is being compensated by the Hall voltage ${\bf E}_{\rm H}={\bf v}_x\times{\bf B}=\frac{1}{ne}{\bf j}\times{\bf B}$. The total electric field then becomes
$$
{\bf E}=\left(\frac{1}{ne}{\bf j}\times{\bf B}+\frac{m}{ne^2\tau}{\bf j}\right)
$$
We now introduce the _resistivity matrix_ $\tilde{\rho}$ as ${\bf E}=\tilde{\rho}{\bf j}$, where the diagonal elements are simply $\rho_{xx}=\rho_{yy}=\rho_{zz}=\frac{m}{ne^2\tau}$. The off-diagonal element $\rho_{xy}$ gives us:
$$
\rho_{xy}=\frac{B}{ne}\equiv -R_{\rm H}B
$$
where $R_{\rm H}=-\frac{1}{ne}$ is the _Hall resistance_. So by measuring the Hall resistance, we can obtain $n$, the density of free electrons in a material.
While most materials have $R_{\rm H}>0$, interestingly some materials are found to have $R_{\rm H}<0$. This would imply that the charge carriers either have a positive charge, or a negative mass. We will see later (chapter 17) how to interpret this.
### Sommerfeld theory (free electron model)
Atoms in a metal provide conduction electrons from their outer shells (often s-shells). These can be described as waves in the crystal, analogous to phonons. Hamiltonian of a free electron:
$$
\mathcal{H}=\frac{ {\bf p}^2}{2m}=-\frac{\hbar^2}{2m}\left( \frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2} \right)\ \Rightarrow\ \varepsilon=\frac{\hbar^2}{2m}\left( k_x^2+k_y^2+k_z^2 \right)
$$
Take periodic boundary conditions: $\psi(x,y,z)=\psi(x+l,y+L,z+L)$:
$$
k_x=\frac{2\pi p}{L},\ k_y=\frac{2\pi q}{L},\ k_z=\frac{2\pi r}{L}\ \Rightarrow\ \varepsilon=\frac{2\pi^2\hbar^2}{mL^2}\left( p^2+q^2+r^2 \right)
$$
Comparable to phonons, but: electrons are _fermions_.
- Only 2 (due to spin) allowed per $k$-value
- Fill up from the lowest energy until you run out of electrons
$\rightarrow$ Calculate when you are out of electrons $\rightarrow$ _Fermi energy_.
In order to compute the density of states, we need to perform an integration of k-space. Assuming three dimensions and spherical symmetry (the dispersion in the free electron model is isotropic) we find for the total number of states:
$$
N=2\left(\frac{L}{2\pi}\right)^3\int{\rm d}{\bf k}=2 \left(\frac{L}{2\pi}\right)^34\pi\int k^2{\rm d}k=\frac{V}{\pi^2}\int k^2{\rm d}k ,
$$
where the factor 2 represents spin degeneracy. Using $k=\frac{\sqrt{2m\varepsilon}}{\hbar}$ and ${\rm d}k=\frac{1}{\hbar}\sqrt{\frac{m}{2\varepsilon}}{\rm d}\varepsilon$ we can rewrite this as:
$$
N=\frac{V}{\pi^2}\int\frac{2m\varepsilon}{\hbar^3}\sqrt{\frac{m}{2\varepsilon}}{\rm d}\varepsilon=\frac{Vm^{3/2}}{\pi^2\hbar^3}\int\sqrt{2\varepsilon}\ {\rm d}\varepsilon
$$
So we find for the density of states:
$$
g(\varepsilon)=\frac{ {\rm d}N}{ {\rm d}\varepsilon}=\frac{Vm^{3/2}\sqrt{2\varepsilon}}{\pi^2\hbar^3}\propto\sqrt{\varepsilon}
$$
Similarly,
- For 1D: $g(\varepsilon) = \frac{2 V}{\pi} \frac{ {\rm d}k}{ {\rm d}\varepsilon} \propto 1/\sqrt{\varepsilon}$
- For 2D: $g(\varepsilon) = \frac{k V}{\pi} \frac{ {\rm d}k}{ {\rm d}\varepsilon} \propto \text{constant}$
Total number of electrons:
$$
N=\int_0^{\varepsilon_{\rm F}}g(\varepsilon){\rm d}\varepsilon,
$$
with $\varepsilon_{\rm F}$ the _Fermi energy_ = highest filled energy at $T=0$.
$$
\varepsilon_{\rm F}=\frac{\hbar^2}{2m}\left( 3\pi^2\frac{N}{V} \right)^{2/3}\equiv \frac{\hbar^2 k_{\rm F}^2}{2m},\
k_{\rm F}=\left( 3\pi^2\frac{N}{V} \right)^{1/3}
$$
The quantity $k_{\rm F}=\frac{2\pi}{\lambda_{\rm F}}$ is called the _Fermi wavevector_, where $\lambda_{\rm F}$ is the _Fermi wavelength_, which is typically in the order of the atomic spacing.
For copper, the Fermi energy is ~7 eV $\rightarrow$ thermal energy of the electrons ~70 000 K ! The _Fermi velocity_ $v_{\rm F}=\frac{\hbar k_{\rm F}}{m}\approx$ 1750 km/s $\rightarrow$ electrons run with a significant fraction of the speed of light, only because lower energy states are already filled by other electrons.
The total number of electrons can be expressed as $N=\frac{2}{3}\varepsilon_{\rm F}g(\varepsilon_{\rm F})$.
The bold line represents all filled states at $T=0$. This is called the _Fermi sea_. Conduction takes place only at the _Fermi surface_: everything below $\varepsilon_{\rm F}-\frac{eV}{2}$ is compensated.
Now: **Finite temperature** $\rightarrow$ probability distribution to occupy certain states.
Fermi-Dirac distribution:
$$
f(\varepsilon,T)=\frac{1}{ {\rm e}^{(\varepsilon-\mu)/k_{\rm B}T}+1}
$$
Chemical potential $\mu=\varepsilon_{\rm F}$ if $T=0$. Typically $\varepsilon_{\rm F}/k_{\rm B}$~70 000 K (~7 eV), whereas room temperature is only 300 K (~30 meV) $\rightarrow$ thermal smearing occurs only very close to Fermi surface.
At finite temperature, the total number of electrons $N$ should be:
$$
N=\int_0^\infty f(\varepsilon,T)g(\varepsilon){\rm d}\varepsilon=\int_0^\infty n(\varepsilon,T){\rm d}\varepsilon
$$
We can use this to calculate the electronic contribution to the heat capacity.
Electrons in the top triangle are being excited to the bottom triangle due to temperature increase. Number of excited electrons $\approx\frac{1}{2}g(\varepsilon_{\rm F})k_{\rm B}T=n_{\rm exc}$. Total extra energy $E(T)-E(0)=n_{\rm exc}k_{\rm B}T=\frac{1}{2}g(\varepsilon_{\rm F})k_{\rm B}^2T^2$.
$$
C_{V,e}=\frac{ {\rm d}E}{ {\rm d}T}=g(\varepsilon_{\rm F})k_{\rm B}^2T=\ ...\ =\frac{3}{2}Nk_{\rm B}\frac{T}{T_{\rm F}}\propto T
$$
$T_{\rm F}=\frac{\varepsilon_{\rm F}}{k_{\rm B}}$ is the _Fermi temperature_.
How does $C_{V,e}$ relate to the phonon contribution $C_{V,p}$?
- At room temperature, $C_{V,p}=3Nk_{\rm B}\gg C_{V,e}$
- Near $T=0$, $C_{V,p}\propto T^3$ and $C_{V,e}\propto T$ $\rightarrow$ competition.
New concept: _Fermi surface_ = all points in k-space with $\varepsilon=\varepsilon_{\rm F}$. For free electrons, the Fermi surface is a sphere.
$$
N=2\frac{\frac{4}{3}\pi k_{\rm F}^3}{\left( \frac{2\pi}{L} \right)^3}=\frac{k_{\rm F}^3V}{3\pi^2}\ \Rightarrow\ k_{\rm F}=\left( 3\pi^2\frac{N}{V} \right)^{1/3}
$$
The orange circle represents the Fermi surface at finite current $\rightarrow$ this circle will shift only slightly before the electrons reach terminal velocity $\rightarrow$ all transport takes place near the Fermi surface.
## Useful trick: scaling of $C_V$
Behavior of $C_V$ can be very quickly memorized or understood using the following mnemonic rule
> Particles with energy $E \leq kT$ are thermally excited, and each carries extra energy $kT$.
#### Example 1: electrons
$g(E_F)$ roughly constant ⇒ total energy in the thermal state is $T \times [T\times g(E_F)]$ ⇒ $C_V \propto T$.
#### Example 2: graphene with $E_F=0$ (midterm 2018)
$g(E) \propto E$ ⇒ total energy is $T \times T^2$ ⇒ $C_V \propto T^2$.
#### Example 3: phonons in 3D at low temperatures.
$g(E) \propto E^2$ ⇒ total energy is $T \times T^3$ ⇒ $C_V \propto T^3$.