Lecture 18

04/24/23, M.

Today

Winding number
Bulk-boundary correspondence
Anomalous Hall effect quantized

1. Winding number

The 1D Hamiltonian matrix

\begin{matrix} (1) & H (k) = h (k) \cdot σ = h_{x} σ_{x} + h_{y} σ_{y} \end{matrix}

$h_z=0$ $\boldsymbol h$ $h(k)\neq 0$ $k$ $h_x$ $h_y$ $\mathbb R^2\backslash \{0\}$ , the punctured plane.

Then we have map

\begin{matrix} (2) & \hat{h} (k) : T^{1} \to S^{1} \end{matrix}

$T^1$ $k$ $S^1$ $h_x$ $h_y$ plane. The map forms a group called the fundamental group of a circle, which is isomorphic (同构) to the integer group (by addition)

\begin{matrix} (3) & π_{1} (S^{1}) ≅ Z \end{matrix}

The circles form a group by path composition, which is part of the elementary discussoin in algebraic topology, which we cannot go into any details here.

$n$ is a topological invariant, called the winding number (绕数).

2. The boundary state

$n=0$ $n=1$ SSH chains

$n=0$ $n=1$ , which is called a soliton (孤子).

$\hat {\boldsymbol h}$ $n=1$ $n=0$ , or vice versa, it is bound to "snag the hole". When this happens, the band gap closes. In other words, there is gauranteed to be some boundary states that have no gap between the two SSH chains.

The winding number is a topological numer classifiying the two phases on the chain. And as the topology changes continuously from one region of the space to another, keeping the symmetry unchanged (chiral symmetry in this case), there is bound to be a place where the gap closes. This is called the principle of bulk-boundary correspondence.

The boundary states arising due to change of topology are then given the name: topologically protected boundary (edge, or surface) states, or zero modes sometimes.

$k=\pi$

\begin{matrix} (4) & \begin{matrix} h (k) ≃ [\begin{array}{cc} 0 & t_{1} - t_{2} + i t_{2} k \\ t_{1} - t_{2} - i t_{2} k & 0 \end{array}] \end{matrix} \end{matrix}

$t_2=1$ $u=t_1/t_2-1$ . Then

\begin{matrix} (5) & \begin{aligned} h (k) & = [\begin{array}{cc} 0 & u + i k \\ u - i k & 0 \end{array}] \\ = [\begin{array}{cc} 0 & u - d/d x \\ u - d/d x & 0 \end{array}] \end{aligned} \end{matrix}

The Schrodinger equation then reads

\begin{matrix} (6) & \begin{matrix} h [\begin{array}{l} ψ \\ ϕ \end{array}] = ε [\begin{array}{l} ψ \\ ϕ \end{array}] \end{matrix} \end{matrix}

So,

\begin{matrix} (7) & \begin{aligned} \Rightarrow & (u + d/d x) ϕ = ε ψ \\ \Rightarrow & (u + d/d x) (u - d/d x) ψ = ε^{2} ψ \\ - ψ^{''} + (u^{2} + u^{'}) ψ = ε^{2} ψ . \end{aligned} \end{matrix}

The zero mode then corresponds to

\begin{matrix} (8) & - ψ_{0}^{''} + (u^{2} + u^{'}) ψ_{0} = 0 \end{matrix}

Consider two regimes:

$|x|\rightarrow\infty$ ,
$\begin{matrix} (9) & - ψ_{0}^{″} + u_{\infty}^{2} ψ_{0} = 0 \end{matrix}$
which means
$\begin{matrix} (10) & ψ_{0} \sim e^{\mp u_{\infty} x} \end{matrix}$
$[-x_0,x_0]$
$\begin{matrix} (11) & \begin{aligned} - ψ_{0}^{''} + ((b x)^{2} - b) ψ_{0} = 0 \\ \Rightarrow ψ_{0} \sim e^{- b x^{2} / 2} \end{aligned} \end{matrix}$

So we have confirmed that there are zero modes localized on the boundary, which is an initial demonstration of the bulk boundary correspondence.

3. 2D Dirac fermion with mass

The Hamiltonian

\begin{matrix} (12) & h (k_{x}, k_{y}) = k_{x} σ_{x} + k_{y} σ_{y} + m σ_{z} \end{matrix}

The Berry flux

\begin{matrix} (13) & \int_{- \infty}^{\infty} d k_{x} \int_{- \infty}^{\infty} d k_{y} Ω_{x k y} = \int_{Σ (m)} d θ d ϕ Ω_{θ ϕ} = π \cdot sign (m) . \end{matrix}

$\pm 2\pi$ . And there is some sort of topology, which looks like wrapping around the sphere once.

Indeed, this corresponds to the so-called Haldane model, first proposed by Duncane Haldane in 1988.

Consider spinless elecrtron on a honeycomb lattice, with nearest-nerighbor (NN最近邻) and next-nearest-neighbor (NNN 次近邻) hoppings

\begin{matrix} (14) & \begin{aligned} H = & \sum_{⟨ R, R^{'} ⟩} (- t) (a_{R}^{+} b_{R}^{'} + H.c .) NN hopping \\ + \underset{H_{A}}{\underset{⏟}{\sum_{⟨ ⟨ R, R^{'} ⟩ ⟩} (- t^{'}) a_{R}^{+} a_{R^{'}}}} + \underset{H_{B}}{\underset{⏟}{\sum_{⟨ ⟨ R, R^{'} ⟩ ⟩} (- t^{'}) b_{R}^{†} b_{R^{'}}}} NNN hopping \end{aligned} \end{matrix}

It should be noticed that the NNN hopping cannot open a band gap, as the Dirac point is protected by inversion and time-reversal symmetries. The NNN hopping breaks neither.

To open a band gap, Haldane introduced a staggered (交错的) magnetic field

Consider the A sites. As the electron moves around a triangle, as shown below, it picks up an Aharonov-Bohm's phase. We can average the phase to the three bonds, because of the 3-fold rotational symmetry

\begin{matrix} (15) & t^{'} \to t^{'} e^{i ϕ} \end{matrix}

Likewise, for the triangle of B sites, the phase picked up has the same magnitude but opposite sign, so we have complex hopping for NNN

\begin{matrix} (16) & t^{'} \to t^{'} e^{- i ϕ} \end{matrix}

Note that the phase would be zero for the hexagon, enclosed by 6 NN hoppings. So the NN hopping remains unchanged.

Then we have

\begin{matrix} (17) & H_{A} = \sum_{⟨ ⟨ R, R^{'} ⟩ ⟩} (- t^{'} e^{- i ϕ}) a_{R}^{+} a_{R^{'}} + m \sum_{R} a_{R}^{+} a_{R}, \end{matrix}

\begin{matrix} (18) & H_{B} = \sum_{⟨ ⟨ R, R^{'} ⟩ ⟩} (- t^{'} e^{+ i ϕ}) b_{R}^{+} b_{R^{'}} - m \sum_{R} b_{R}^{+} b_{R} . \end{matrix}

\begin{matrix} (19) & Denote \vec{k} \cdot {\vec{a}}_{1} = k_{1}, \vec{k} \cdot {\vec{a}}_{2} = k_{2} . \end{matrix}

$\quad \vec{k} \cdot \vec{a}_1=k_1, \quad \vec{k} \cdot \vec{a}_2=k_2$ , and

\begin{matrix} (20) & a_{R} = \frac{1}{\sqrt{N}} \sum_{k} e^{+ i k \cdot R} a_{k} \end{matrix}

we have, for the NNN term on A-sublattice

\begin{matrix} (21) & \begin{aligned} H_{A} & = \sum_{⟨ ⟨ R, R^{'} ⟩ ⟩} (- t^{'}) e^{+ i ϕ (R, R^{'})} \frac{1}{N} \sum_{k k^{'}} e^{- i k \cdot R} e^{i k^{'} \cdot R^{'}} a_{k}^{†} a_{k^{'}} \\ = \sum_{k} a_{k}^{†} a_{k} (- t^{'}) \sum_{⟨ ⟨ R^{'} ⟩ ⟩} e^{+ i ϕ (0, R^{'}) + i k \cdot R^{'}} \end{aligned} \end{matrix}

\begin{matrix} (22) & \begin{aligned} H_{A} = & \sum_{k} a_{k}^{+} a_{k} {- 2 t^{'} [\cos (k_{1} + ϕ) + \cos (k_{2} + ϕ) \\ + \cos (k_{1} + k_{2} - ϕ)] + m} \end{aligned} \end{matrix}

Likewise, for B-sublattice

\begin{matrix} (23) & \begin{aligned} H_{B} = \sum_{k} b_{k}^{†} b_{k} { & - 2 t^{'} [\cos (k_{1} - ϕ) + \cos (k_{2} - ϕ) \\ + \cos (k_{1} + k_{2} + ϕ)] - m} \end{aligned} \end{matrix}

$K\quad (\tau=1)$ $K'\quad (\tau=-1)$

\begin{matrix} (24) & H = \sum_{τ = \pm 1} \sum_{q} Ψ_{τ q}^{†} h_{τ} (q) Ψ_{τ q} \end{matrix}

where

\begin{matrix} (25) & \begin{aligned} h_{τ} (q) = 3 t^{'} \cos ϕ + h \cdot σ \\ h_{x} = τ ℏ v_{F} q_{x}, h_{y} = ℏ v_{F} q_{y} . \\ h_{z} = m + τ 3 \sqrt{3} t^{'} \sin ϕ \end{aligned} \end{matrix}

Non-zero, quantized Berry flux occurs when the two valleys have opposite masses.

This leads to the quantum anomalous Hall effect (QAHE). There is an chiral edge state. Chiral here means the edge state moves one way, not the other. so there is not going to be backscattering. So the edge transport is insensitive to (elastic) scattering, as there can be no back scattering

How to measure the QAHE?

\begin{matrix} (26) & \begin{matrix} σ = [\begin{array}{cc} σ_{bulk} & e^{2} / h \\ - e^{2} / h & σ_{bulk} \end{array}] \Rightarrow ρ = σ^{- 1} = [\begin{array}{cc} ρ_{x x} & ρ_{x y} \\ ρ_{y x} & ρ_{y y} \end{array}] \end{matrix} \end{matrix}

where

\begin{matrix} (27) & \begin{matrix} ρ_{x x} = ρ_{y y} = \frac{σ_{bulk}}{σ_{bulk}^{2} + {(e^{2} / h)}^{2}}, \\ ρ_{x y} = - ρ_{y x} = - \frac{e^{2} / h}{σ_{bulk}^{2} + {(e^{2} / h)}^{2}} \end{matrix} \end{matrix}

source: https://physics.aps.org/articles/v8/41

Another way to understand the chirality. The grand potential is

\begin{matrix} (28) & \begin{matrix} Φ = E - T S - B \cdot M - μ N \end{matrix} \end{matrix}

\begin{matrix} (29) & d Φ = - S d T - p d V - M \cdot d B - N d μ \end{matrix}

So we have the Maxwell relation

\begin{matrix} (30) & {(\frac{\partial N}{\partial B})}_{T, V, μ} = {(\frac{\partial M}{\partial μ})}_{T, V, B} \end{matrix}

\begin{matrix} (31) & {(\frac{\partial n}{\partial B})}_{B = 0} = {(\frac{\partial m}{\partial μ})}_{B = 0} = - \frac{1}{e} σ_{x y} \end{matrix}

We find that in the gap

\begin{matrix} (32) & m = m_{0} - c \frac{e}{h} μ \end{matrix}

So there must be a current circulation that produces the orbital magnetic momentum in the QAHE regime.

Final question, what is the geometric meaning of Chern number in this case?

It is easy to verify that

\begin{matrix} (33) & c = \frac{1}{4 π} \int_{BZ} d k_{x} d k_{y} \hat{h} \cdot \frac{\partial \hat{h}}{\partial k_{x}} \times \frac{\partial \hat{h}}{\partial k_{y}} \end{matrix}

$T^2$ $S^2$ $\mathbb R^3$

\begin{matrix} (34) & \hat{h} (k) : T^{2} \to S^{2} \end{matrix}

which again forms a group by path composition (路径复合, jargon in algebraic topology), that is isomorphic to the integer group. Hence the Chern number.

Two 2D Hamiltonians can have vastly different appearance, but if they have the same Chern number, they are in the same equivalence class and have they same topology.