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 Answer


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 \(\omega\).

Answer:
Choosing a local coordinate system, the latter integral becomes

(not complete)

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