Lecture 10

03/23/2023, Th.

Today

Examples of TB models
Boltzmann equation
Semiclassical transport theory

Reading

Girvin & Yang ch. 8

1. Further examples of tight-binding models

1.1 Graphene

The corner of the Brillouin zone:

\begin{matrix} (1) & K = - K^{'} = \frac{b_{1} - b_{2}}{3} \end{matrix}

$K,K'$ by a reciprocal lattice translation.

Image source: https://physics.stackexchange.com/questions/391601/how-is-the-pi-bond-of-carbon-in-graphene-possible

$p_z$ orbital

\begin{matrix} (2) & \hat{H} = - t \sum_{R} \sum_{δ} a_{R}^{+} b_{R + δ} + H.c. \end{matrix}

$\boldsymbol \delta = 0, \boldsymbol a_1, \boldsymbol a_2$ $t$ takes a value between 2.7 and 3 eV for graphene.

Substituting in the Fourier expansion, we find

\begin{matrix} (3) & H = \sum_{k} t_{k} a_{k}^{†} b_{k} + H.c. = \sum_{k} Ψ_{k}^{†} h (k) Ψ_{k} \end{matrix}

where

\begin{matrix} (4) & \begin{matrix} t_{k} = - t (1 + e^{i k \cdot a_{1}} + e^{i k \cdot a_{2}}) \\ h (k) = [\begin{matrix} 0 & t_{k} \\ t_{k}^{*} & 0 \end{matrix}] \end{matrix} \end{matrix}

The dispersion relation is simply

\begin{matrix} (5) & ε_{k} = \pm | t_{k} | . \end{matrix}

$K,K'$ , there are two fold degeneracies

\begin{matrix} (6) & t_{k} = - t (1 + e^{\pm i 2 π / 3} + e^{\mp i 2 π / 3}) = 0 \end{matrix}

$K,K'$ $K$ $\boldsymbol k=K+\boldsymbol q$

\begin{matrix} (7) & t_{q} = \frac{3 t d}{2} (q_{x} + i q_{y}) + O (q^{2}) \end{matrix}

$K$ is

\begin{matrix} (8) & h (q) = ℏ v_{F} (q_{x} σ_{x} + q_{y} σ_{y}) \end{matrix}

$K'$

\begin{matrix} (9) & h (q) = ℏ v_{F} (- q_{x} σ_{x} + q_{y} σ_{y}) \end{matrix}

So we would write

\begin{matrix} (10) & h_{τ} (q) = ℏ v_{F} (τ q_{x} + i q_{y}) \end{matrix}

$\tau=\pm 1$ $K/K'$ $K,K'$ protected by symmetry. The pertinent symmetries here are the time-reversal and inversion, absent of either will lift the degeneracy. But to understand the symmetry protection requires the knowledge of Berry's phase, and cannot be discussed here.

1.2 Polyacetylene and the SSH model

The 1D atomic chain also features a Dirac point at the Brillouin zone boundary. Two ways to lift the degeneracy:

staggered lattice potential: we have discussed.
staggered hopping: Su-Schrieffer-Heeger (SSH) model

\begin{matrix} (11) & \begin{aligned} \hat{H} & = - \sum_{x} (t + δ t) a_{x}^{†} b_{x} + (t - δ t) b_{x}^{†} a_{x + 1} + H.c. \\ = \sum_{k} Ψ_{k}^{†} [\begin{array}{cc} 0 & - t_{+} - t_{-} e^{i k} \\ - t_{+} - t_{-} e^{- i k} & 0 \end{array}] Ψ_{k} . \end{aligned} \end{matrix}

$k=\pi$ $k=\pi+q$

\begin{matrix} (12) & t_{k} = - 2 δ t - i ℏ v_{F} q \end{matrix}

\begin{matrix} (13) & ε_{\pm} (q) = \pm \sqrt{(2 δ t)^{2} + (ℏ v_{F} q)^{2}} \end{matrix}

At half-filling (半满), the change of energy from the distortion is

\begin{matrix} (14) & \begin{aligned} Δ E_{e l} & = \sum_{q} ε_{q}^{-} (δ t) - ε_{q}^{-} (0) \\ = \frac{L}{2 π} \int_{- π / 2}^{π / 2} d q [- \sqrt{{(ℏ u_{F} q)}^{2} + (2 δ t)^{2}} - ℏ u_{F} q] \\ \approx \frac{(2 δ t)^{2} L}{2 π ℏ u_{F}} \log (\frac{2 δ t}{π ℏ v_{F}}) \end{aligned} \end{matrix}

The corresponding lattice energy increase is

\begin{matrix} (15) & Δ E_{latt} = \frac{1}{2} α (δ u)^{2} \cdot L \end{matrix}

where

\begin{matrix} (16) & δ t \sim δ u \end{matrix}

So we find

\begin{matrix} (17) & \frac{Δ E_{tot}}{L} = \frac{1}{2} (α + c \log (\frac{2 δ t}{π ℏ v_{F}})) (δ u)^{2} \end{matrix}

So for sufficiently small distortion, this quantity goes negative. The implication is that the linear monatomic chain with half-filled electron band is always unstable against pairing distortion. This is the Peierl's instability.

2. The Boltzmann equation

\begin{matrix} Semiclassical transport theory = semiclassical eom + Boltzmann equation \end{matrix}

The Boltzmann equation

\begin{matrix} (18) & \frac{d f}{d t} = C [f] \end{matrix}

$f(\lambda,t)$ $C[f]$ $\lambda\rightarrow (\boldsymbol r,\boldsymbol k)$

\begin{matrix} (19) & f = f (r, k, t) \end{matrix}

Consider the change of particle number in a phase space volume element

\begin{matrix} (20) & \begin{aligned} d N_{coll} & = {(\frac{\partial f}{\partial t})}_{coll} Δ t \frac{d^{3} r d^{3} k}{(2 π)^{3}} \\ = (f (r + \dot{r} Δ t, k + \dot{k} Δ t, t + Δ t) - f (r, p, t)) \frac{d^{3} r d^{3} k}{(2 π)^{3}} \end{aligned} \end{matrix}

\begin{matrix} (21) & \frac{\partial f}{\partial t} + \dot{r} \cdot \nabla f + \dot{k} \cdot \nabla_{k} f = C [f] \end{matrix}

Local representative volume in which local equilibrium can be assumed

\begin{matrix} (22) & unit cell ≪ δ V ≪ V \end{matrix}

Locally, the equilibrium is referenced to local temperature, potential

\begin{matrix} (23) & f^{0} (r, k) = \frac{1}{\exp [- β (r) (ε (r, k) - μ (r)] + 1} \end{matrix}

3. The collision integral

We demonstrate the idea by considering the impurity scattering.

$\boldsymbol k$ $\boldsymbol k'$ , from the first Born approximation¹

\begin{matrix} (24) & W_{k^{'} k} = \frac{2 π}{ℏ} | H_{k^{'} k}^{'} |^{2} δ (ε_{k^{'}} - ε_{k}) \end{matrix}

we take an impurity potential of the form

\begin{matrix} (25) & H^{'} = \sum_{r_{i}} U (r - r_{i}) \end{matrix}

Some calculations

\begin{matrix} (26) & \begin{aligned} H_{k^{'} k}^{'} & = ⟨ ψ_{k^{'}} | H^{'} | ψ_{k} ⟩ \\ = \sum_{i} ⟨ ψ_{k^{'}} | u (r - r_{i}) | ψ_{k} ⟩ \\ = \sum_{i} e^{- i (k^{'} - k) \cdot r_{i}} ⟨ ψ_{k^{'}} | U (r) | ψ_{k} ⟩ \\ = \sum_{i} e^{- i (k^{'} - k) \cdot r_{i}} {\tilde{U}}_{k^{'} k} \end{aligned} \end{matrix}

$r_i$ is the lattice point at which an impurity is located

\begin{matrix} (27) & \begin{aligned} W_{k^{'} k} & = \frac{2 π}{ℏ} \overset{―}{{| H_{k^{'} k} |}^{2}} δ (ε_{k^{'} - ε_{k}}) \\ = \frac{2 π}{ℏ} \overset{―}{\sum_{r_{i}, r_{j}} e^{- i (k^{'} - k^{'}) \cdot (r_{i} - r_{j})}} {| {\tilde{U}}_{k^{'} k} |}^{2} δ (ε_{k^{'}} - ε_{k}) \\ = \frac{2 π}{ℏ} N_{i} {| {\tilde{U}}_{k^{'} k} |}^{2} δ (ε_{k^{'}} - ε_{k}) \\ = \frac{2 π}{ℏ} \frac{N_{i}}{V^{2}} {| U_{k^{'} k} |}^{2} δ (ε_{k^{'}} - ε_{k}) \\ = \frac{2 π}{ℏ} \frac{n_{i}}{V} {| U_{k^{'} k} |}^{2} δ (ε_{k^{'}} - ε_{k}) \end{aligned} \end{matrix}

The above refers to a single electron. No information about the distribution is used.
$f$
$\begin{matrix} (28) & \begin{aligned} in & = \int \frac{d^{3} k^{'}}{(2 π)^{3}} W_{k k^{'}} (1 - f_{k}) f_{k^{'}} \\ out & = \int \frac{d^{3} k^{'}}{(2 π)} W_{k^{'} k} f_{k} (1 - f_{k^{'}}) \end{aligned} \end{matrix}$
$f$ $\text d^3k$ is
$\begin{matrix} (29) & \begin{aligned} C [f] & = \int (in - out) d^{3} r \\ = \frac{2 π}{ℏ} n_{i} \int \frac{d^{3} k^{'}}{(2 π)^{3}} {| U_{k^{'} k^{'}} |}^{2} (f_{k^{'}} - f_{k}) δ (ε_{k^{'}} - ε_{k}) . \end{aligned} \end{matrix}$
Further simplification for isotropic system:
$\begin{matrix} (30) & {| U_{k^{'} k} |}^{2} = | U (θ) |^{2} \end{matrix}$ $\begin{matrix} (31) & \begin{aligned} C [f] & = \frac{2 π}{ℏ} n_{i} \int \frac{d Ω^{'}}{4 π} g (ε_{k}) | U (θ^{'}) |^{2} (f_{k^{'}} - f_{k}) \\ \equiv \int W (θ^{'}) (f_{k^{'}} - f_{k}) \frac{d Ω^{'}}{4 π}, \\ W (θ^{'}) & = \frac{2 π}{ℏ} g (ε_{k}) n_{i} | U (θ^{'}) |^{2} . \end{aligned} \end{matrix}$
$W(\theta)$ is ignored right away. But we can be more careful here by including the anisotropic effect in the total collision rate.
$f_k^1=f_k-f_k^0$ $E$ -field is
$\begin{matrix} (32) & f_{k}^{1} = a (ε_{k}) k \cdot E \end{matrix}$ $\begin{matrix} (33) & \begin{aligned} C [f] & = α \int W (θ^{'}) (k^{'} \cdot E - k \cdot E) \frac{d Ω^{'}}{4 π} \\ = α | k | | E | \int W (θ^{'}) ({\hat{k}}^{'} \cdot \hat{E} - \hat{k} \cdot \hat{E}) \frac{d Ω^{'}}{4 π} . \end{aligned} \end{matrix}$
$\hat{\boldsymbol k}=\hat{\boldsymbol z}$
$\begin{matrix} (34) & \begin{aligned} {\hat{k}}^{'} \cdot \hat{E} & = {\hat{k}}_{z}^{'} {\hat{E}}_{z} + \underset{\sim \cos ϕ^{'}}{\underset{⏟}{{\hat{k}}_{⊥}^{'} \cdot {\hat{E}}_{⊥}}} \\ = (\hat{k^{'}} \cdot \hat{k}) (\hat{k} \cdot \hat{E}) + (\dots) \cos ϕ^{'} \end{aligned} \end{matrix}$ $\begin{matrix} (35) & C [f] = \underset{f^{1}}{\underset{⏟}{α | k | | E | \hat{k} \cdot \hat{E}}} \underset{- 1 / τ}{\underset{⏟}{\int W (θ) (\cos θ - 1) \frac{d Ω}{4 π}}} . \end{matrix}$
$\tau$ is the transport relaxation time.¹
$W(\theta)= W_0$ and ignore, that gets you the relaxation time
$\begin{matrix} (36) & \frac{1}{τ_{0}} = W_{0} \end{matrix}$
Now we have the relaxation-time approximation
$\begin{matrix} (37) & C [f] = - \frac{f - f^{0}}{τ} \end{matrix}$
The linearized Boltzmann equation
$\begin{matrix} (38) & \frac{\partial f}{\partial t} + {\dot{r}}_{a} \frac{\partial f}{\partial r_{a}} + {\dot{k}}_{a} \frac{\partial f}{\partial k_{a}} = - \frac{f - f_{0}}{τ} \end{matrix}$

Approximations involved
$U_{k'k}\rightarrow U(\theta)$
advantage: simple enough to get picture and results.

$\partial_t \rightarrow 0$ $T=$ $\mu=$ $f=f(\boldsymbol k)$

\begin{matrix} (39) & {\dot{k}}_{a} \frac{\partial f}{\partial k_{a}} = {\dot{k}}_{a} (\frac{\partial f_{0}}{\partial ε} \frac{\partial ε}{\partial k_{a}} + α E^{a}) = - \frac{1}{τ} (f - f_{0}) \end{matrix}

\begin{matrix} (40) & \begin{aligned} \Rightarrow f - f_{0} & = - τ ℏ {\dot{k}}_{a} (u^{a} \frac{\partial f}{\partial ε} + \frac{α}{ℏ} E^{a}) \\ = - τ q (E_{a} + B_{a b} v^{b}) (v^{a} \frac{\partial f}{\partial ε} + \frac{α}{ℏ} E^{a}) \end{aligned} \end{matrix}

$B=0$

\begin{matrix} (41) & j_{a} = q \int [d k] (v_{a} - Ω_{a b} {\dot{k}}^{b}) (f^{1} + f^{0}) \end{matrix}

So the linear response theory is

\begin{matrix} (42) & j_{a}^{(1)} = \underset{σ_{a b}}{\underset{⏟}{q^{2} \int [d k] [τ (- \frac{\partial f^{0} (ε_{k})}{\partial ε_{k}}) v_{a} v_{b} - f^{0} \frac{1}{ℏ} Ω_{a b}]}} E^{b} \end{matrix}

Isotropic case, the longitudinal conductivity

\begin{matrix} (43) & σ_{a b} = δ_{a b} \frac{q^{2}}{3} (g v^{2} τ)_{F} \end{matrix}

$\tau_0$ $\tau$ is the transport relaxation time, from the Born approximation.¹

The anomalous Hall effect

\begin{matrix} (44) & σ_{a b} = - \frac{q^{2}}{ℏ} \int Ω_{a b} f_{0} [d k] \end{matrix}

1 Landau and Lifshitz. Quantum mechanics - nonrelativistic theory. §126. ↩ ↩ ↩