# Lecture 1 : 1-d SSH model¶

## The Su-Schrieffer–Heeger (SSH) model¶

The simplest non-trivial topology : 1-d lattice.

Peierls instability makes the atoms dimerize.

Polyacetylene Structure:

## Tight-binding method: first quantization¶

1-d atom chain

Tight-binding method: Single electron total Hamiltonian in atom chain:

$H=\frac{p^2}{2m}+U(x)$

with periodical potential: $$U(x+na)=U(x)$$

Assume single atom potential $$V(x)$$ with Hamiltonian

$H_0=\frac{p^2}{2m}+V(x)$

and well solved eigen-value and eigen-wave-function:(consider only one state)

$H_0\phi(x)=E_0\phi(x)$

Do the combination

$\psi(x)=\sum\limits_n a_n\phi_n$

with $$\phi_n=\phi(x-x_n), x_n=na$$. Define $$\Delta U(x)=U(x)-V(x)$$, and substitute following equation

$H\psi=\left(H_0+\Delta U\right)\psi=E\psi$

we get

$\sum_n \langle \phi_m|\Delta U(x-x_n)|\phi_n\rangle a_n=(E-E_0)a_m$

Define: $$\langle \phi(x-ma+na)|\Delta U(x) |\phi(x)\rangle=-J(x_m-x_n)$$

We get:

$-\sum_n J(x_m-x_n) a_n=(E-E_0) a_m$

Because of the Tranformation symmetry of the Hamiltonian, the resulting wavefunction $$\psi(x)$$ should take Bloch form, which means we should have the solution $$a_m=e^{ikx_m}$$, then, we get

$E-E_0=-\sum_n J(x_n)e^{-ikx_n}$

Consider only the nearest-hopping interaction, define $$J_0=J(0), J=J(\pm a)$$, then we have:

$\begin{split}E-E_0+J_0=&-J(e^{ika}+e^{-ika}) \\ =&-2Jcoska\end{split}$

## Second quantization¶

In the second quantization language, the expectation value of energy becomes a operator, set $$\mathscr{H}=\frac{p^2}{2m}+U(x)$$, we have

$H=\langle \psi|\mathscr{H}|\psi \rangle \Rightarrow \hat{H}=\sum_{m,n}\hat{c}_m^\dagger H_{mn}\hat{c}_n$

with $$\psi \to \hat{\psi}=\sum\limits_n \hat{c}_n \phi_n, H_{mn}=\langle \phi|\mathscr{H}|\phi \rangle$$ $$\phi_n$$ is a orthonormal and complete basis in Hilbert space, like plane-waves $$e^{ikx}$$ or energy eigen-states of $$H_0$$, $$\mathscr{H}$$ is the energy operator in single particle first quantization picture, which can only act on Hilbert space, while the second quantization energy operator $$\hat{H}$$ acts on Fock space. Here, in tight-binding method, $$\phi_n$$ is the wave-function of site $$n$$ of the energy eigen-state $$H_0$$.

Consider only the nearest interaction, we have:

$\hat{H}=\sum_{n=1}^M \hat{c}_n^\dagger \hat{c}_{n+1}t_n+h.c.$

$$h.c.$$ means hermitian conjugation. Rewrite it in matrix form: $$\hat{H}=\sum\limits_{mn}\hat{c}_m \tilde{H}_{mn}c_n$$, we have

$\begin{split}\tilde{H}_{mn}=\begin{pmatrix} 0 & t_1 &0 & \cdots&t_M^*\\ t_1^* &0&t_2&\cdots&0\\ 0&t_2^*&0&\cdots&0\\ \vdots&\vdots&\vdots&\vdots&\vdots\\ t_M&0&0&t_{M-1}^*&0 \end{pmatrix}\end{split}$

In the case when $$t=t_n$$, $$\hat{c}_n$$ satisfy the Bloch condition, we can transform it into momentum space, with $$\hat{c}_n=\frac{1}{\sqrt{M}}\sum\limits_k \hat{c}_k e^{ikx_n}$$,

we can easily get

$\hat{H}=\sum\limits_k \hat{c}_k^\dagger \hat{c}_k(te^{ika}+t^*e^{-ika})=\sum\limits_k \hat{c}_k^\dagger \hat{c}_k E(k)$

which gives us the dispersion relation:

$\begin{split}E(k)=&te^{ika}+t^*e^{-ika}\\ =&2tcoska \qquad \textit{for t is real}\end{split}$

On the other hand, keep in mind that we would get $$c_n=e^{ik(n-1)a}c_1$$, that is

$\begin{split}\hat{H}=\begin{pmatrix}c_1^\dagger&c_2^\dagger&\cdots&c_M^\dagger \end{pmatrix} \begin{pmatrix} 0 & t &0 & \cdots&t^*\\ t^* &0&t&\cdots&0\\ 0&t^*&0&\cdots&0\\ \vdots&\vdots&\vdots&\vdots&\vdots\\ t&0&0&t^*&0 \end{pmatrix} \begin{pmatrix}c_1\\c_2\\\vdots\\c_M\end{pmatrix}\end{split}$

$\begin{split}\Rightarrow &\begin{pmatrix}c_1^\dagger&c_1^\dagger e^{-ika}&\cdots&c_1^\dagger e^{-ik(M-1)a} \end{pmatrix} \begin{pmatrix} 0 & t &0 & \cdots&t^*\\ t^* &0&t&\cdots&0\\ 0&t^*&0&\cdots&0\\ \vdots&\vdots&\vdots&\vdots&\vdots\\ t&0&0&t^*&0 \end{pmatrix} \begin{pmatrix}c_1\\c_1 e^{ika}\\\vdots\\c_1e^{ik(M-1)a}\end{pmatrix}\\ =&\begin{pmatrix}c_1^\dagger&\cdots&c_1^\dagger \end{pmatrix} \begin{pmatrix} te^{ika}+t^*e^{-ika} & 0&0\\ 0&\ddots&0\\ 0&0&te^{ika}+t^*e^{-ika} \end{pmatrix} \begin{pmatrix}c_1\\\vdots\\c_1\end{pmatrix}\end{split}$

which also gives us $$E(k)=te^{ika}+t^*e^{-ika}$$.

More generally, $$t_n$$ can be different from each other, for example, if they are all different up to $$4$$, but have a super-periodicity with $$t_{5}=t_1$$, then there will have 4 sub-bands, in the example we will consider below, we have two $$t$$, $$t_1\neq t_2$$, and we have two sub-bands.

If each atom have a valance electron, then the above mentioned energy band structure $$E(k)=2tcoska$$ is not the stable fundamental mode, it will dimerizes to lower the total energy, that means we’ll get following coupling case:

with the Hamiltonian:

$H=\sum_{n=1}^N(v_n c_{n,1}^\dagger c_{n,2}+w_nc_{n,2}^\dagger c_{n+1,1}+h.c.)$

with $$M=2N$$.

## Review¶

In 1-d atom chain, we have got the Hamiltonian in second-quantization frame as:

$\hat{H}=\sum_{n=1}^M \hat{c}_n^\dagger \hat{c}_{n+1}t_n+h.c.$

$$h.c.$$ means hermitian conjugation. Rewrite it in matrix form: $$\hat{H}=\sum\limits_{mn}\hat{c}_m {H}_{mn}\hat{c}_n$$, we have

$\begin{split}\hat{H}=\begin{pmatrix}c_1^\dagger&c_2^\dagger&\cdots&c_M^\dagger \end{pmatrix}\begin{pmatrix} 0 & t_1 &0 & \cdots&t_M^*\\ t_1^* &0&t_2&\cdots&0\\ 0&t_2^*&0&\cdots&0\\ \vdots&\vdots&\vdots&\vdots&\vdots\\ t_M&0&0&t_{M-1}^*&0 \end{pmatrix}\begin{pmatrix}c_1\\c_2\\\vdots\\c_M\end{pmatrix}\end{split}$

## Energy band and topology¶

$\begin{split}\hat{H}=\begin{pmatrix}c_1^\dagger&c_2^\dagger&\cdots&c_M^\dagger \end{pmatrix}\begin{pmatrix} 0 & t_1 &0 & \cdots&t_M^*\\ t_1^* &0&t_2&\cdots&0\\ 0&t_2^*&0&\cdots&0\\ \vdots&\vdots&\vdots&\vdots&\vdots\\ t_M&0&0&t_{M-1}^*&0 \end{pmatrix}\begin{pmatrix}c_1\\c_2\\\vdots\\c_M\end{pmatrix}\end{split}$

$$t_n$$ can differ from each other.

1. For a open chain with $$M$$ atoms, we have $$t_M=0$$, and this matrix will give us $$M$$ eigen-values and eigen-functions.
2. Possess translational invariance with $$c_{n+1}=c_1e^{ikna}$$, it will be diagonalized with $$H(k)=te^{ika}+t^*e^{-ika}$$.
3. Staggered hopping parameters with $$t_1\neq t_2$$, but have property $$c_{2n+1}=c_1e^{iknb},c_{2n+2}=c_2e^{iknb}$$. We can block the Hamiltonian up in $$2\times 2$$ blocks and also pair up $$c_{2n+1},c_{2n+2}$$.

The Hamiltonian:

$H=\sum_{n=1}^N(v_n c_{n,1}^\dagger c_{n,2}+w_nc_{n,2}^\dagger c_{n+1,1}+h.c.)$

with $$M=2N$$.

For a more beautiful notation, define $$\mathbf{c}_n^\dagger=(c_{n,1}^\dagger,c_{n,2}^\dagger)=(c_{2n-1}^\dagger,c_{2n}^\dagger)$$, then we have

$H=\sum_{m,n=1}^N \mathbf{c}_m^\dagger H_{mn}\mathbf{c}_n$

with

$\begin{split}\mathbf{c}_n^\dagger H_{nn}\mathbf{c}_n=\begin{pmatrix}c_{n,1}^\dagger&c_{n,2}^\dagger\end{pmatrix}\begin{pmatrix}0&v_n\\v_n^*&0\end{pmatrix}\begin{pmatrix}c_{n,1}\\c_{n,2}\end{pmatrix}=\begin{pmatrix}c_{n,1}^\dagger&c_{n,2}^\dagger\end{pmatrix}U_n\begin{pmatrix}c_{n,1}\\c_{n,2}\end{pmatrix}\end{split}$

and

$\begin{split}\mathbf{c}_n^\dagger H_{nn+1}\mathbf{c}_{n+1}=\begin{pmatrix}c_{n,1}^\dagger&c_{n,2}^\dagger\end{pmatrix}\begin{pmatrix}0&0\\w_n&0\end{pmatrix}\begin{pmatrix}c_{n+1,1}\\c_{n+1,2}\end{pmatrix}=\begin{pmatrix}c_{n,1}^\dagger&c_{n,2}^\dagger\end{pmatrix}T_n\begin{pmatrix}c_{n+1,1}\\c_{n+1,2}\end{pmatrix}\end{split}$
$\begin{split}\mathbf{c}_{n+1}^\dagger H_{n+1n}\mathbf{c}_{n}=\begin{pmatrix}c_{n+1,1}^\dagger&c_{n+1,2}^\dagger\end{pmatrix}T_n^\dagger\begin{pmatrix}c_{n,1}\\c_{n,2}\end{pmatrix}\end{split}$

when $$|m-n|>1$$, we have $$H_{mn}=0$$.

For example, for 6 cells (12 sites), we have

$\begin{split}H=\begin{pmatrix}U_1&T_1&0&\cdots&T_6^\dagger\\ T_1^\dagger&U_2&T_2&\cdots&0\\ 0&T_2^\dagger&U_3&\cdots&0\\ \vdots&\vdots&\vdots&\vdots&\vdots\\ T_6&0&0&\cdots&U_6 \end{pmatrix}\end{split}$

$\begin{split}H_{mn}=\begin{pmatrix}U_1&T_1&0&\cdots&T_6^\dagger\\ T_1^\dagger&U_2&T_2&\cdots&0\\ 0&T_2^\dagger&U_3&\cdots&0\\ \vdots&\vdots&\vdots&\vdots&\vdots\\ T_6&0&0&\cdots&U_6 \end{pmatrix}\end{split}$
• Open chain: $$T_6=0$$.

• Closed chain with translational symmetry, $$T_n=T,U_n=U$$, with

$\begin{split}U=\begin{pmatrix}0&v\\v^*&0\end{pmatrix},T=\begin{pmatrix}0&0\\w&0\end{pmatrix}\end{split}$

Using three Pauli matrices

$\begin{split}\sigma_x=\begin{pmatrix}0&1\\1&0\end{pmatrix},\sigma_y=\begin{pmatrix}0&-i\\i&0\end{pmatrix},\sigma_z=\begin{pmatrix}1&0\\0&-1\end{pmatrix}\end{split}$

We get

$U=Re(v)\sigma_x-Im(v)\sigma_y,T=\frac{1}{2}w(\sigma_x-i\sigma_y)$

Using $$\mathbf{c}_n=e^{ik(n-1)b}\mathbf{c}_1$$, we have

$\begin{split}\hat{H}=\begin{pmatrix}\mathbf{c}_1^\dagger&\mathbf{c}_2^\dagger&\cdots&\mathbf{c}_N^\dagger \end{pmatrix} \begin{pmatrix} U & T &0 & \cdots&T^\dagger\\ T^\dagger &U&T&\cdots&0\\ 0&T^\dagger&U&\cdots&0\\ \vdots&\vdots&\vdots&\vdots&\vdots\\ T&0&0&T^\dagger&U \end{pmatrix} \begin{pmatrix}\mathbf{c}_1\\\mathbf{c}_2\\\vdots\\\mathbf{c}_N\end{pmatrix}\end{split}$
$\begin{split}\Rightarrow \begin{pmatrix}\mathbf{c}_1^\dagger&\mathbf{c}_1^\dagger e^{-ikb}&\cdots&\mathbf{c}_1^\dagger e^{-ik(N-1)b} \end{pmatrix} \begin{pmatrix} U & T &0 & \cdots&T^\dagger\\ T^\dagger &U&T&\cdots&0\\ 0&T^\dagger&U&\cdots&0\\ \vdots&\vdots&\vdots&\vdots&\vdots\\ T&0&0&T^\dagger&U \end{pmatrix} \begin{pmatrix}\mathbf{c}_1\\\mathbf{c}_1 e^{ikb}\\\vdots\\\mathbf{c}_1e^{ik(N-1)b}\end{pmatrix}\end{split}$

$\begin{split}=\begin{pmatrix}\mathbf{c}_1^\dagger&\cdots&\mathbf{c}_1^\dagger \end{pmatrix} \begin{pmatrix} U+Te^{ikb}+T^\dagger e^{-ikb} & &\\ &\ddots&\\ &&U+Te^{ikb}+T^\dagger e^{-ikb} \end{pmatrix} \begin{pmatrix}\mathbf{c}_1\\\vdots\\\mathbf{c}_1\end{pmatrix}\end{split}$

which gives us $$H=H(k)\oplus H(k)\cdots \oplus H(k)=\oplus_{n=1}^N H(k)$$ with

$H(k)=U+Te^{ikb}+T^\dagger e^{-ikb}=\mathbf{h}(k)\cdot \mathbf{\sigma}$

with

$\begin{split}h_x(k)=&Re(v)+|w|cos(kb+arg(w))\\ h_y(k)=&-Im(v)+|w|sin(kb+arg(w))\\ h_z(k)=&0\end{split}$

with $$w=|w|e^{i arg(w)}$$.

$H(k)=\mathbf{h}(k)\cdot \mathbf{\sigma}$

with

with eigen-energy

$E(k)=|\mathbf(h)(k)=\pm \sqrt{h_x^2+h_y^2+h_z^2}=\pm \sqrt{|v|^2+|w|^2+2|v||w|cos(kb+arg(v)+arg(w))}$

and eigen-wavefunctions

$\begin{split}|\pm\rangle=\begin{pmatrix}\pm e^{-i\phi(k)}\\ 1\end{pmatrix}\end{split}$

with $$tan\phi=h_y/h_x$$.

For example, set $$arg(v)=arg(w)=0$$, we have

Can not tell the difference $$|v|-|w|=\pm\delta$$.

Energy-band description is not completed, it can give us many information, but not the whole, others are hidden in the wave-function. Alternatively, recalling $$H(k)=\mathbf{h}(k)\cdot \mathbf{\sigma}$$, the Hamiltonian should contain the whole information, but we have only used $$|h|$$, in topological aspect, $$\mathbf{h}(k)$$ will suffices.

Set $$arg(v)=0$$,$$kb=[0,2\pi]$$, we have two cases

• $$|w|<|v|, \mathbf{inter}<\mathbf{intra}$$
• $$|w|>|v|, \mathbf{inter}>\mathbf{intra}$$

two cases

## Winding number¶

$$H(k)=\mathbf{h}(k)\cdot \mathbf{\sigma}$$, $$\mathbf{h}(k)=0$$ is a degenerate point with $$|v|=|w|$$, two bands cross, define $$h(k)=h_x(k)+ih_y(k)$$, we have

$\begin{split}H(k)=\begin{pmatrix}0&h^*(k)\\h(k)&0\end{pmatrix}\end{split}$
$ln(h)=ln(|h|)e^{iarg(h)}=ln(|h|)+iarg(h)$

define

$\nu=\frac{1}{2\pi i}\int_{-\pi}^{\pi}dk\frac{d}{dk}ln(h(k))$

When

• $$|w|>|v|, \nu=1, \mathbf{inter}>\mathbf{intra}$$
• $$|w|<|v|, \nu=0, \mathbf{inter}<\mathbf{intra}$$

A example, $$N=20, M=2N=40, w=1, v=0.5$$, we get eigen-energys:

eigen-energy

Edge-states:

## Chiral symmetry¶

Recalling the Hamiltonian:

$H=\sum_{n=1}^N(v_n c_{n,1}^\dagger c_{n,2}+w_nc_{n,2}^\dagger c_{n+1,1}+h.c.)$

Define projector operators:

$P_A=\sum_n c_{n,1}^\dagger c_{n,1}, P_B=\sum_n c_{n,2}^\dagger c_{n,2}$

and the chiral operator $$\Sigma_z=P_A-P_B$$, The matrix elements of $$\Sigma_z$$ vanish, $$\langle 0|c_r \Sigma c^\dagger_s|0\rangle=0$$ if sites $$r$$ and $$s$$ are in different unit cells.

In first-quantization, we have

$H=\sum_{n=1}^N(v_n |n,1\rangle\langle n,2|+w_n|n,2\rangle\langle n+1,1|+h.c.)$

and

$P_A=\sum_n |n,1\rangle\langle n,1|, P_B=\sum_n |n,2\rangle\langle n,2|$

In matrix form, we have

$\Sigma_z=\sigma_z\oplus\sigma_z\oplus\cdots\oplus\sigma=\oplus_{n=1}^N \sigma_z$

$$\Sigma_z$$ is local, it does not mix site between unit cells, and inherits the algebra from $$\sigma_z$$:

$\Sigma_z^\dagger\Sigma_z=1$
$\Sigma_z^2=1$

Recalling

$\begin{split}\begin{pmatrix} U_1 & T_1 &0 & \cdots&T_N^\dagger\\ T_1^\dagger &U_2&T_2&\cdots&0\\ 0&T_2^\dagger&U_3&\cdots&0\\ \vdots&\vdots&\vdots&\vdots&\vdots\\ T_N&0&0&T_{N-1}^\dagger&U_N \end{pmatrix}\end{split}$

There are no onsite terms in the Hamiltonian, so we have

$\Sigma_z H \Sigma_z=-H$

This is the chiral symmetry.

Note

Actually, here, $$H$$ is defined in momentum space, but $$\Sigma_z$$ in real space, we should write $$\tilde{H}=U^\dagger H U$$ for some unitary matrix $$U$$, but the property survives!

Consequences: For eigenstates $$|\psi_n\rangle$$ of $$H$$, we have

$H|\psi_n\rangle=E_n|\psi_n\rangle$
$H\Sigma_z|\psi_n\rangle=-\Sigma_zH|\psi_n\rangle=-\Sigma_zE_n|\psi_n\rangle=-E_n\Sigma_z|\psi_n\rangle$$If$
• $$E_n\neq 0$$, two orthonormal states $$|\psi_n\rangle, \Sigma_z|\psi_n\rangle$$, which gives

$\begin{split}\begin{pmatrix}\alpha^* &\beta^*\end{pmatrix}\begin{pmatrix}\alpha\\-\beta\end{pmatrix}=0\end{split}$
$\Rightarrow |\alpha|^2=|\beta|^2$
• $$E_n=0$$, we particularly have $$\Sigma_z=\Sigma_zH$$, $$|\psi_n\rangle$$ can be eigen-states of $$H$$ and $$\Sigma$$ simultaneously, $$\Sigma_z|\psi_n\rangle=\pm|\psi_n\rangle$$, gives $$|\psi_n\rangle=\begin{pmatrix}1 \\0\end{pmatrix}$$ or $$|\psi_n\rangle=\begin{pmatrix}0 \\1\end{pmatrix}$$

## Review 2¶

In the last class, we have solved 1-d atom chain with staggered hopping parameters and got the Hamiltonian:

$\begin{split}H(k)=\mathbf{h}(k)\cdot \mathbf{\sigma}=\begin{pmatrix}0&h_x-ih_y\\h_x+ih_y&0\end{pmatrix}=|h(k)|\begin{pmatrix}0&e^{-i\phi(k)}\\e^{i\phi(k)}&0\end{pmatrix}\end{split}$

with $$tan\phi=h_y/h_x$$.

Eigen-values are $$E(k)=\pm|h(k)|$$, with eigen-functions: $$|\pm\rangle=\begin{pmatrix}\pm e^{-i\phi(k)}\\ 1\end{pmatrix}$$.

$$\mathbf{h}(k)$$ depicted as follow:

Topological transition

## Topological transition¶

We have already seen, for $$|w|>|v|$$ or $$|w|<|v|$$, we got different winding number, means there is a topological transition at $$|w|=|v|$$. In the energy-band point of view, it means the gap between two energy bands closes (across each other) and reopens. There are two ways to change the winding number and get a topological transition:

1. Pull the path through the origin in the $$h_x-h_y$$ plane.
2. Lift it out of the plane (breaking the chiral symmetry).

Case 1(a): $$v=0.5, w=0\to 1$$

Case 1(b): $$w=1, v=2.5\to 0$$

Case 2: Sublattice potential $$H_{sublattice}=u\sigma_z$$. Recalling

$H(k)=U+Te^{ikb}+T^\dagger e^{-ikb}$

with $$U=\begin{pmatrix}0&v\\v^*&0\end{pmatrix},U=\begin{pmatrix}0&0\\w&0\end{pmatrix}$$, now $$U$$ changes to $$U=\begin{pmatrix}u&v\\v^*&-u\end{pmatrix}$$, and $$H(k)=\mathbf{h}(k)\cdot\mathbf{\sigma}$$ with $$h_z(k)=u$$.

Set $$w=1, v=1.5-cos\theta, u=sin\theta, \theta=0\to \pi$$,

Case 3: Charge Pump: Using the Hamiltonian $$H=u(t)\sum\limits_{i=1}^N(-1)^{i-1}c_i^{\dagger}c_i+\sum\limits_{i=1}^N[t_0+\delta(t)(-1)^i](c_i^{\dagger}c_{i+1}+h.c.)$$, with $$u(t)=u_0 sin\theta, \delta(t)=\delta_0 cos\theta$$, add time variation with $$\theta=2\pi t/T$$, in previous notation, $$v=t_0-\delta(t), w=t_0+\delta(t)$$. set $$u=sin\theta, v=0.75-0.25cos\theta, w=0.75+0.25cos\theta$$, the vector $$\vec{h(k)}$$ is

The energy band evolution is

energy4

## Winding number v.s. Zak phase¶

From

$\begin{split}H(k)=\begin{pmatrix}0&h^*(k)\\h(k)&0\end{pmatrix}\end{split}$
$ln(h)=ln(|h|)e^{iarg(h)}=ln(|h|)+iarg(h)$

We have defined the winding number

$\nu=\frac{1}{2\pi i}\int_{-\pi}^{\pi}dk\frac{d}{dk}ln(h(k))=\frac{1}{2\pi}arg(h)\left.\right|_{-\pi}^{\pi}=1 \ \textit{or}\ 0$

From the Zak phase definition:

$\gamma=i\oint\langle\psi|\nabla_k|\psi\rangle$

Recalling $$|\psi\rangle=\frac{1}{\sqrt{2}}\begin{pmatrix}\pm e^{-i\phi(k)}\\ 1\end{pmatrix}\Rightarrow\gamma=\frac{1}{{2}}\oint dk\nabla_k\phi=\pm\pi \ \textit{or}\ 0$$.