Wednesday, July 25, 2012

Some Berry phase computations

 $\newcommand{\bp}{{\boldsymbol{p}}}$ $\newcommand{\bq}{\boldsymbol{q}}$  $\newcommand{\bsi}{\boldsymbol{\sigma}}$ $\newcommand{\pa}{\partial}$ $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\ii}{\boldsymbol{i}}$  $\newcommand{\bsS}{\boldsymbol{S}}$ $\newcommand{\bC}{\mathbb{C}}$ $\newcommand{\lan}{\langle}$ $\newcommand{\ran}{\rangle}$

I'm attending this seminar on topological insulators. I still do not understand the physics, but here is some mathematics.

Consider the Pauli matrices

$$\bsi_x :=\left[\begin{array}{cc} 0 & 1\\
1 & 0
\end{array}
\right],\;\;\bsi_y:=\left[
\begin{array}{cc} 0 & -\ii \\
\ii & 0
\end{array}
\right],\;\;\bsi_z:=\left[
\begin{array}{cc} 1 &0\\
0 & -1
\end{array}
\right]. $$

For $\bp =(x,y,z)\in \bR^3$ we set

$$\bsi(\bp) = x\bsi_x +y\bsi_y+ z\bsi_z=\left[\begin{array}{cc} z & x-\ii y\\
x+\ii y & -z
\end{array}
\right].$$
The family of hermitian  matrices $\bsi(\bp)$ satisfy the  Clifford identities

$$\bsi(\bp)\cdot\bsi(\bq) +\bsi(\bq)\cdot \bsi(\bp) =2\bp\cdot \bq,\;\;\forall \bp,\bq\in\bR^3. $$

In particular

$$\bsi(\bp)^2= |\bp|^2,\;\;\forall\bp\in\bR^3, $$

so that the eigenvalues  of $\bsi(\bp)$ are $\pm |\bp|$.    Consider the unit sphere

$$\bsS^2:=\bigl\lbrace\;\bp\in\bR^3;\;\;|\bp|=1\;\bigr\rbrace. $$

We obtain a family  of  hermitian matrices $\lbrace\bsi(\bp)\rbrace_{\bp\in\bsS^2}$ with eigenvalues $\pm 1$.  Denote by $V_{\bp}$ the $1$-eigenspace      of the matrix $\bsi(\bp)$,  $\bp\in \bsS^2$.   Suppose $\bp=(x,y,z)$ is not the South Pole $P_-=(0,0,-1)$, i.e.,   $z\neq -1$. We set $u=x+\ii y$.  To find a basis  of $V_\bp$ we  need to solve the system

$$ \left[\begin{array}{cc}
 z & \bar{u} \\
u & -z
\end{array}
\right] \cdot \left[
\begin{array}{c}
z_1\\
z_2
\end{array}
\right]=\left[\begin{array}{c}
z_1\\
z_2
\end{array}
\right]. $$

We deduce that

$$z_2= \frac{u}{1+z} z_1.$$

The stereographic projection from the  South Pole  is the map

$$\zeta: \bsS^2\setminus P_-\to\bR^2=\bC$$

that associates to each  point $\bp\in\bsS^2\setminus  P_-$ the  intersection of the line $P_-\bp$ with  the coordinate plane $z=0$. Concretely, if $\bp=(x,y,z)$, then

$$\zeta(\bp)= \frac{u}{1+z}. $$

We deduce that $V_\bp$ is spanned by  the vector

$$\vec{z}(\bp)= (1,\zeta(\bp)). $$

We write $\zeta=\zeta_1+\ii \zeta_2$ so that we can use $(\zeta_1,\zeta_2)$ as coordinates on $\bsS^2\setminus P_-$.  Consider the normalized vector

$$|\Psi_\bp\ran :=\frac{1}{|\vec{z}(\bp)|} \vec{z}(\bp)= \frac{1}{\sqrt{1+|\zeta|^2}}(1,\zeta)$$.


Set

$$ G(\zeta):= \sqrt{1+|\zeta|^2}. $$  Note  that

$$d\left(\frac{1}{G}\right) =-\frac{dG}{G^2}.$$


The Berry connection  $\nabla$ is obtained from the equality

$$\nabla|\Psi_\bp\ran = |\Psi_\bp\ran\lan \Psi_\bp|d\Psi_\bp\ran. $$


Observe that

$$ d|\Psi_\bp\ran = -\frac{dG}{G^2}(1,\zeta)+\frac{1}{G} (0,d\zeta),$$

$$ \lan \Psi_\bp |d\Psi_\bp\ran =-\frac{(1+|\zeta|^2)dG}{G^3}+\frac{1}{G^2}\bar{\zeta} d\zeta $$

$$= -\frac{dG}{G} +\frac{1}{G^2} \bar{\zeta} d\zeta=\underbrace{-d\log G+\frac{1}{1+|\zeta|^2}\bar{\zeta} d\zeta}_{=:\omega}. $$

The  $1$-form associate to the  Berry connection is the above $\omega$. The curvature of the Berry connection is

$$\Omega= d\omega = -\frac{d|\zeta|^2}{(1+|\zeta|^2)^2}\bar{\zeta}d\zeta + \frac{1}{1+|\zeta|^2}d\bar{\zeta}\wedge d\zeta=-\frac{|\zeta|^2}{(1+|\zeta|^2)^2} d\bar{\zeta}\wedge d\zeta +\frac{1}{1+|\zeta|^2} d\bar{\zeta}\wedge d\zeta$$

$$ =\frac{1}{(1+|\zeta|^2)}  d\bar{\zeta}\wedge d\zeta =\frac{2\ii}{(1+|\zeta|^2)^2} d\zeta_1\wedge d\zeta_2. $$

To compute the integral of $\Omega$ we use polar coordinates, $\zeta=r e^{\ii\theta}$ so that

$$d\zeta_1\wedge d\zeta_2 =rdrd\theta$$

and

$$\int_{\bC}\Omega=\ii \int_0^{2\pi}d\theta\int_0^\infty \frac{2rdr}{(1+r^2)^2} dr=\ii\int_0^{2\pi} d\theta\int_0^\infty \frac{d(1+r^2)}{(1+r^2)^2} =2\pi \ii. $$


The Chern form of the Berry connection is $\frac{\ii}{2\pi} \Omega$ and the first Chern number is

$$ \frac{\ii}{2\pi}\int_{\bC} \Omega =-1. $$






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