# Lecture 2 : Berry Phase and Chern number¶

## Berry Phase review¶

Assuming a physical system is depended on some parameters $$\mathbf{R}=(R_1,R_2,\cdots,R_N)$$, we have the snapshot Hamiltonian $$H(\mathbf{R})$$, its eigen-values and eigen-states:

$H(\mathbf{R})|n(\mathbf{R})\rangle=E_n(\mathbf{R})|n(\mathbf{R})\rangle$

where $$|n(\mathbf{R})\rangle$$ can have an arbitrary phase prefactor.

The parameters $$\mathbf{R(t)}$$ are slowly changed with time $$t$$, then the adiabatic evolution of time-dependent Schrodinger equation:

$i\frac{d}{dt}|\psi(t)\rangle=H(\mathbf{R(t)})|\psi(t)\rangle$

Take the Ansatz: $$|\psi(t)\rangle=e^{i\gamma_n(t)}e^{-i\int_0^tE_n(\mathbf{R(t')})dt'}|n(\mathbf{R(t)})\rangle$$, we have

$-\left(\frac{d}{dt}\gamma_n\right)|n\rangle+i\left|\frac{d}{dt}n\right\rangle=0$

That gives the Berry phase expression:

$\gamma_n(\mathcal{C})=\int_\mathcal{C}i\langle n(\mathbf{R})|\nabla_{\mathbf{R}}n(\mathbf{R})\rangle d\mathbf{R}$

Define Berry connection:

$\mathbf{A}^{(n)}(\mathbf{R})=i\langle n(\mathbf{R})|\nabla_{\mathbf{R}}n(\mathbf{R})\rangle=-Im\langle n(\mathbf{R})|\nabla_{\mathbf{R}}n(\mathbf{R})\rangle$

Gauge transformation:

$|n(\mathbf{R})\rangle \to e^{i\alpha(\mathbf{R})}|n(\mathbf{R})\rangle$
$\mathbf{A}^{(n)}(\mathbf{R}) \to \mathbf{A}^{(n)}(\mathbf{R})-\nabla_{\mathbf{R}}\alpha(\mathbf{R})$

$$\gamma=\oint\mathbf{A}(\mathbf{R})d\mathbf{R}$$ is gauge invariant.

Gauge and Parallel transportation: recalling the arbitrary phase

$|n(\mathbf{R})\rangle \to e^{i\alpha(\mathbf{R})}|n(\mathbf{R})\rangle$

why shouldn’t we choose one which makes

$\frac{d}{dt}|n\rangle\equiv 0$

from

$-\left(\frac{d}{dt}\gamma_n\right)|n\rangle+i\left|\frac{d}{dt}n\right\rangle=0$

then we have

$\gamma_n=0$

There is no Berry Phase in this frame, which is called inertial frame, the condition $$\frac{d}{dt}|n\rangle\equiv 0$$ is called parallel transportation. All the information resorted to $$|n(\mathbf{R})\rangle$$, similar to a transformation from active frame to passive frame.

## Berry curvature¶

Define the Berry curvature:

$\mathbf{B}(\mathbf{R})=\nabla_{\mathbf{R}}\times \mathbf{A}^{(n)}(\mathbf{R})$

Using Stokes theorem, we have for the Berry Phase:

$\gamma_n(\mathcal{C})=\int_\mathcal{S}\mathbf{B}^{(n)}(\mathbf{R})d\mathcal{S}$

where $$\mathcal{S}$$ is any surface whose boundary is the loop $$\mathcal{C}$$.

Two useful formula:

• $$B_j=\epsilon_{jkl}\partial_kA_l=-Im\epsilon_{jkl}\partial_k\langle n|\partial_ln\rangle=-Im\epsilon_{jkl}\langle\partial_kn|\partial_ln\rangle$$, that is $$\mathbf{B}^{(n)}=-Im\sum\limits_{n'\neq n}\langle\nabla n|n'\rangle\times\langle n'|\nabla n\rangle$$.

• $$\mathbf{B}^{(n)}=-Im\sum\limits_{n'\neq n}\langle\nabla n|n'\rangle\times\langle n'|\nabla n\rangle$$, to calculate $$\langle n'|\nabla n\rangle$$, start from:

$H(\mathbf{R})|n\rangle=E_n|n\rangle$
$\Rightarrow (\nabla H)|n\rangle+H|\nabla n\rangle=(\nabla E_n)|n\rangle+E_n|\nabla n\rangle$
$\Rightarrow \langle n'|\nabla H|n\rangle+\langle n'|H|\nabla n\rangle=E_n\langle n'|\nabla n\rangle$
$\Rightarrow \langle n'|\nabla n\rangle=\frac{\langle n'|\nabla H|n\rangle}{E_n-E_{n'}}$

then we get:

$\mathbf{B}^{(n)}=-Im\sum\limits_{n'\neq n}\langle\nabla n|n'\rangle\times\langle n'|\nabla n\rangle=-Im\sum\limits_{n'\neq n}\frac{\langle n|\nabla H|n'\rangle\times\langle n'|\nabla H|n\rangle}{(E_n-E_{n'})^2}$

which is gauge invariant!

## Berry curvature from perturbation theory¶

We can use time-independent perturbation theory to derive the changes of instant snapshot basis:

$H(\mathbf{R})|n(\mathbf{R})\rangle=E_n(\mathbf{R})|n(\mathbf{R})\rangle$

we have

$|n(\mathbf{R+\Delta R})\rangle=|n(\mathbf{R})\rangle+\sum_{m\neq n}\frac{\langle m|H(\mathbf{R+\Delta R})-H(\mathbf{R})|n\rangle}{E_n-E_m}|m(\mathbf{R})\rangle$

We see that $$\langle n|\Delta n(\mathbf{R})\rangle=0$$, which means we have used parallel transport gauge, more general, we should add a arbitrary phase factor in the above equation for $$|n(\mathbf{R+\Delta R})\rangle$$.

$\nabla_{\mathbf{R}}|n\rangle=\sum_{m\neq n}\frac{\langle m|\nabla_{\mathbf{R}}H|n\rangle}{E_n-E_m}|m\rangle$

From $$\mathbf{B}^{(n)}=-Im\sum\limits_{n'\neq n}\langle\nabla n|n'\rangle\times\langle n'|\nabla n\rangle$$, we also get:

$\mathbf{B}^{(n)}=-Im\sum_{n'\neq n}\frac{\langle n|\nabla H|n'\rangle\times\langle n'|\nabla H|n\rangle}{(E_n-E_{n'})^2}$

Also notice:

$\begin{split}\sum_n\mathbf{B}^{(n)}=&-Im\sum_n\sum_{n'\neq n}\frac{\langle n|\nabla H|n'\rangle\times\langle n'|\nabla H|n\rangle}{(E_n-E_{n'})^2}\\ =&-Im\sum_n\sum_{n'< n}\frac{\langle n|\nabla H|n'\rangle\times\langle n'|\nabla H|n\rangle+\langle n'|\nabla H|n\rangle\times\langle n|\nabla H|n'\rangle}{(E_n-E_{n'})^2}\\ =&0\end{split}$

Which gives:

$\sum_n\gamma_n(\mathcal{C})=\int_{\mathcal{S}}\sum_n\mathbf{B}^{(n)}(\mathbf{R})d\mathcal{S}=0$

## Benchmark: Spin-1/2¶

Gauge!Gauge!Gauge!

2-level Hamiltonian $$H(\mathbf{R})=h_0(\mathbf{R})\sigma_0+\mathbf{h}(\mathbf{R})\cdot\mathbf{\sigma}$$, we can set $$h_0=0$$, because it does not affect the eigenstates, eigen-energy are $$\pm|\mathbf{h}|$$, introduce the unit vector:$$\hat{\mathbf{h}}=\mathbf{h}/|\mathbf{h}|$$, the endpoints of $$\hat{\mathbf{h}}$$ map out the surface of a unit sphere, called the Bloch sphere shows below: