### Computing the Berry field strength

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6 months ago by
In "Topological insulators and topological superconductors" by B. Bernevig they derive the Berry curvature and Berry field strength for the Hamiltonian

\begin{align}
H = \vec{d}(\vec{R}) \cdot \sigma
\end{align}

as you can see in the file attached. My question is about equation (2.22) where they write

\begin{align}\label{1}
F_{R_i,R_j} &= F_{\theta\phi} \frac{\partial(\theta,\phi)}{\partial(R_i,R_j)}
\end{align}

with $\frac{\partial(\theta,\phi)}{\partial(R_i,R_j)}$ being "the Jacobian of the transformation from $\vec{R}$ to $(\theta,\phi)$. I do not understand this transformation on different levels;

1. $F_{ij}$ involves two differentiations (one in $\partial_i A_j$ and one in $A_j$ itself), but the Jacobian matrix contains only first derivatives
2. Can you tell me how equation (2.22) is even meant? Is $\frac{\partial(\theta,\phi)}{\partial(R_i,R_j)}$ a matrix or just an entry in that matrix? Why are there two indices in the denominator

I thought of the Jacobian as

\begin{align}
\frac{\partial(\theta,\phi)}{\partial(R_i,R_j)} &= \begin{pmatrix} \frac{\partial\theta}{\partial R_i} &  \frac{\partial \theta}{\partial R_j} \\ \frac{\partial\phi}{\partial R_i} & \frac{\partial \phi}{\partial R_j}\end{pmatrix}
\end{align}

Help is appreciated :)

File attached: Topological-Insulators-and-Topological-Superconductors.pdf (149.23 KB)

1
6 months ago by

It's easier to think of the Berry Phase in a coordinate free manner.

Introduction:

Consider a ground state $|0(t)\rangle$ governed by a Hamiltonian $H(\{\lambda(t)\})$ where $\lambda$ is a set of parameters.
Assuming that the time dependence of $\lambda(t)$ is adiabatic implies that for all $t$, a particle at the ground state $|{0 (t=0)}\rangle$ will remain in the instantaneous ground state $|{0(t)}\rangle$.

Any wave function $|{\Psi(t)}\rangle$ is local gauge invariant $|{\Psi(t)\rangle} = e^{i \gamma(t) - i \int_0^t dt^{\prime} \epsilon(t^{\prime})}|{\Psi(t)}\rangle\text{,}$ where $H |0 (t) \rangle = \epsilon(t) |0 (t) \rangle$ [source]. Inserting the latter into Schrödinger's equation $H(\{\lambda(t)\}) |{\Psi (t)}\rangle = i \partial_t |{\Psi (t)}\rangle \text{,}$
one obtains the Berry phase $\gamma (t)$
$\gamma(t) = \int_0^{t^{\prime}} dt^{\prime} \langle 0(t^{\prime}) | i \partial_{t^{\prime}} | 0(t^{\prime}) \rangle \text{.}$

The Berry phase describes the global phase evolution of a complex vector as it is carried around a path in its vector space [source].

Note that since the instantaneous ground state inherits its time dependence from the parameters $\lambda$ we may write the integral as
$\gamma (t) = \int_c d \lambda \langle 0(\lambda) | i \partial_{\lambda} | 0(\lambda) \rangle$
where $c$ is a line integral in parameter space which starts at $\lambda(t=0)$ and ends at $\lambda(t)$.
It is now clear the geometric origin of the Berry phase and we can elegantly write $\gamma$ in a coordinate free form
$i \int_c \langle 0 | d 0 \rangle$
where $d$ is the exterior derivate.

For a closed path path in parameter space we can make use of Stoke's Theorem:
$\int_{S} d \omega = \int_{\partial S} \phi^* \omega$
where $\phi^*$ is the pull back of the differential form $\omega : dS \rightarrow S$

Thus,
$i \int_c \langle 0 | d 0 \rangle = i \int_{S} d \langle 0 | d 0 \rangle = i \int_{S} \langle d0| \wedge | d 0 \rangle$ since $dd \omega= 0$ for any $k$ differential form $\omega$.