PX262 - H4 - free electron model (3D)

\begin{matrix} - \frac{ℏ^{2}}{2 m} \nabla^{2} ϕ (\vec{r}) = E ϕ (\vec{r}) \\ ϕ = A \exp (i (k_{x} x + k_{y} y + k_{z} z)) \\ E = \frac{ℏ^{2} k^{2}}{2 m_{e}} = \frac{ℏ^{2}}{2 m_{e}} (k_{x}^{2} + k_{y}^{2} + k_{z}^{2}) \end{matrix}

\begin{matrix} ϕ (0, y, z) = ϕ (L_{x}, y, z) \\ ϕ (x, 0, z) = ϕ (x, L_{y}, z) \\ ϕ (x, y, 0) = ϕ (x, y, L_{z}) \end{matrix}

\begin{matrix} A \exp (i (k_{x} L_{x} + k_{y} y + k_{y} z)) = A \exp (i (k_{y} y + k_{z} z)) \\ ⟹ \exp (i k_{x} L_{x}) = 1 \end{matrix}

\begin{matrix} \exp (i k_{y} L_{y}) = 1 \\ \exp (i k_{z} L_{z}) = 1 \end{matrix}

\vec{k} = (\frac{2 π n_{x}}{L_{x}}, \frac{2 π n_{y}}{L_{y}}, \frac{2 π n_{z}}{L_{z}})

where, $n_{x}, n_{y}, n_{z} \in Z$

each block, contains one allowed $\vec{k}$ vector (each corner shared between eight blocks)

Δ k_{x} Δ k_{y} Δ k_{z} = \frac{8 π^{3}}{L_{x} L_{y} L_{z}} = 8 π^{3} / V

these single electron states can be occupied by two electrons: spins $↑$ and $↓$
ground state: fill all states ( $↑$ and $↓$ ) up to $E = E_{F}$ or $| \vec{k} | \leq k_{F}$

N = 2 \times \frac{Vol. of sphere}{Vol. per k point} = 2 \times \frac{4}{3} π k_{F}^{3} \frac{V}{8 π^{3}} = \frac{V}{3 π^{2}} k_{F}^{3}

where, 2 is for each spin

⟹ k_{F} = {(\frac{3 π^{2} N}{V})}^{1 / 3} = (3 π^{2} ρ_{e})^{1 / 3}

where, $ρ_{e}$ is the electron density

$k_{F}$ is the radius of the fermi sphere, and also the magnitude of the fermi wavevector, ${\vec{k}}_{F}$
fermi wavevector and fermi energy are set by the density, $ρ_{e} :$

E_{F} = \frac{ℏ^{2}}{2 m_{e}} (3 π^{2} ρ_{e})^{2 / 3}

{\vec{v}}_{F} = \vec{\nabla} ω (\vec{k}) = \frac{ℏ \vec{k}}{m_{e}}

and the fermi speed: $ℏ k_{F} / m_{e}$
this is the group velocity of the wave packet with $\vec{k}$ grouped around ${\vec{k}}_{F}$

density of states

for a large $V$ , it is expected that: $Δ k_{x, y, z} \to 0$ , forming a continuum of states in the space of $\vec{k}$ (k-space)
the number of states with an energy below $E = ℏ^{2} k^{2} / 2 m_{e} :$

N (E) = \frac{V k^{3}}{3 π^{2}} = \frac{V}{3 π^{2}} {(\frac{2 m_{e} E}{ℏ^{2}})}^{3 / 2}

n (E) d E = \frac{d N}{d E} d E = \frac{3}{2} \frac{V}{3 π^{2}} {(\frac{2 m_{e}}{ℏ^{2}})}^{3 / 2} E^{1 / 2} d E

ie. the density of states:

n (E) = \frac{V}{2 π^{2}} {(\frac{2 m_{e}}{ℏ^{2}})}^{3 / 2} E^{1 / 2}

N = \int_{0}^{E_{F}} n (E) d E = V α E_{F}^{3 / 2} \frac{2}{3}

E_{T O T} = \int_{0}^{E_{F}} E n (E) d E = V α \int_{0}^{E_{F}} E^{3 / 2} d E = V α \frac{2}{5} E_{F}^{5 / 2}

\frac{E_{T O T}}{N} = \frac{3}{5} E_{F}

	$ρ_{e}$ (m $^{- 3}$ )	$E_{F}$ (eV)	$k_{F}$ (Å $^{- 1}$ )	$v_{F}$ (ms $^{- 1}$ )
Li	$4.7 \times 10^{28}$	$4.74$	$1.12$	$1.29 \times 10^{6}$
Al	$18.1 \times 10^{28}$	$11.7$	$1.75$	$2.05 \times 10^{6}$

the free fermion model is relevant in other areas:
- white dwarfs: $E_{F} \sim 0.3 \times 10^{6}$ eV
- neutron stars: $E_{F} \sim 40 \times 10^{6}$ eV

note: $k_{B} T \leq E_{F} \sim 4.7 eV \to 55 000$ K

f (E, μ, T) = \frac{1}{\exp \frac{E - μ}{k_{B} T} + 1}

where, $μ$ is the chemical potential

\begin{matrix} T = 0 & N = \int_{0}^{E_{F}} n (E) d E & μ = E_{F} \\ T \neq 0 & N = \int_{0}^{E_{F}} f (E, μ, T) n (E) d E & μ = fixed \end{matrix}

E_{T O T} (T) = \int d E n (E) E f (E, μ, T) = E_{T O T} (0) + a T^{2} + \dots

can argue that some electrons in states below $E_{F}$ thermally excited to states above $E_{F}$
$Δ E :$ number of excitations $\times k_{B} T \sim n (E_{F}) k_{B} T \times k_{B} T = a T^{2}$ , where $a \sim n (E_{F}) k_{B}^{2}$
- assuming the variation of $n (E)$ within $k_{B} T$ of $E_{F}$ negligible
the heat capacity:

C_{V} = \frac{d E_{T O T}}{d T} = 2 a T