I'm learning about the Berry Phase from the original paper, and from the TIFR Infosys Lectures The Quantum Hall Effect by David Tong (2016).

I have some questions regarding the original derivation of the phase, and an example.

  1. In the derivation, Berry derives the phase as $$\gamma_n(C) = i \oint_C \langle n (\mathbf{R})|\nabla_{\mathbf{R}}n(\mathbf{R})\rangle \cdot d\mathbf{R} $$ He then proceeds to say that the normalization of $|n(\mathbf{R})\rangle$ implies that $\langle n (\mathbf{R})|\nabla_{\mathbf{R}}n(\mathbf{R})\rangle$ is imaginary. Why does this implication arise?

  2. In p. 33 of the notes, the example of a spin-1/2 particle in a magnetic field is used. Here the magnitude of the field remains constant but the direction can be changed. First, the Berry connection is calculated using spherical coordinates as $$\mathcal{F}_{\theta \phi} = -\sin{\theta} $$ And when the coordinates are reverted to Cartesian, it becomes $$\mathcal{F}_{ij} = -\epsilon_{ijk} \frac{B^k}{2|\mathbf{B}|^3} $$ This is called a magnetic monopole in parameter space. However when this quantity is integrated over the sphere of possible configurations (i.e. the sphere of radius $|\mathbf{B}|$ in parameter space), the following result is presented: $$\int \mathcal{F}_{ij}dS^{ij} = 4\pi g$$ where $g$ is the 'charge' of the monopole. How does this result come about?

    One idea I had for this one is that if we consider the Berry connection in Cartesian coordinates are considered in analogy to Coulomb's Law, then the field produced by the monopole should obey a kind of Gauss' Law. And if we take the quantities' analogues as follows: $$\frac{1}{4\pi\epsilon} \rightarrow 1 \text{ and } Q \rightarrow -\frac{1}{2}$$ Then we get the result. However I am not happy/convinced with this argument. Can someone shed some light on this?


2 Answers 2


The argument about the Gauss law is correct. The monopole magnetic field has exactly the same form of the electric field of a point charge, thus the Gauss law can be applied.

However, the integral can be perfectly performed in Cartesian coordinates. In fact, it is a basic exercise in the integration of differential forms. We have:

$$F = -\frac{g}{2} \frac{\epsilon_{ijk}}{|\mathbf{B}|^3}B^k dB^i \wedge dB^j$$ The integration is performed on the surface of the $2$-sphere: ${B^1}^2+{B^2}^2+{B^3}^2 = |\mathbf{B}|^2$

$$\int_{S^2} F = \int_{{B^1}^2+{B^2}^2+{B^3}^2 = |\mathbf{B}|^2} -\frac{g}{2} \frac{\epsilon_{ijk}}{|\mathbf{B}|^3}B^k dB^i \wedge dB^j$$

The integrand has 6 terms corresponding to the different permutations of $i,j,k=1,2,3$. The antisymmetry of the Levi-Civita tensor is countered by the antisymmetry of the wedge products. Thus due to the rotational symmetry of the sphere, all 6 terms have the same contribution and we can integrate one term and multiply by 6:

$$ \begin{align*} \int_{S^2} F &= \int_{{B^1}^2+{B^2}^2+{B^3}^2 = B^2} -6 \times \frac{g}{2} \frac{B^1}{|\mathbf {B}|^3}dB^2 dB^3\\ &= - \frac{3g}{|\mathbf {B}|^3} \int_{-|\mathbf {B}|}^{|\mathbf {B}|} dB^3 \int_{-\sqrt{|\mathbf {B}|^2-{B^3}^2}}^{{\sqrt{|\mathbf {B}|^2-{B^3}^2}}} dB^2 2 \sqrt{|\mathbf {B}|^2-{B^2}^2-{B^3}^2} \end{align*} $$ The integral on the $B^2-B^3$ plane is an integral over the equatorial disc. The factor of 2 in this integral takes into account the integration over the upper and the lower hemisphere. This integral can be performed using the substitution: $$B^2 = \sqrt{|\mathbf {B}|^2-{B^3}^2} \sin\phi$$ The $\phi$ limits will be $\frac{3\pi}{2}$ and $\frac{\pi}{2}$ respectively. Thus: $$ \begin{align*} \int_{S^2} F &= - \frac{3g}{|\mathbf {B}|^3} \int_{-B}^{B} dB^3 (|\mathbf {B}|^2-{B^3}^2)\int_{\frac{3\pi}{2}}^{\frac{\pi}{2}} d\phi 2 \cos^2\phi\\ &= - \frac{3g}{|\mathbf {B}|^3} \int_{-B}^{B} dB^3 (|\mathbf {B}|^2-{B^3}^2)\times -\pi\\ &= \frac{3\pi g}{| \mathbf {B}|^3} \int_{-|\mathbf {B}|}^{|\mathbf {B}|} dB^3 (|\mathbf {B}|^2-{B^3}^2)\\ &= \frac{3\pi g}{| \mathbf {B}|^3} \left (|\mathbf {B}|^2 B^3-\frac{{B^3}^3}{3} \right ) \bigg\rvert_{-|\mathbf {B}|}^{|\mathbf {B}|}\\ &= \frac{3\pi g}{| \mathbf {B}|^3} \frac{4 \pi}{3}|\mathbf {B}|^3 \\ &= 4 \pi g \end{align*} $$

  • 1
    $\begingroup$ And the argument about the imaginary part goes as follow : since $\left\langle n|n\right\rangle =1$ one has $d\left\langle n|n\right\rangle =0\Rightarrow\left\langle dn|n\right\rangle +\left\langle n|dn\right\rangle =\left\langle dn|n\right\rangle +\left\langle dn|n\right\rangle ^{\ast}=0\Rightarrow\Re\left\{ \left\langle dn|n\right\rangle \right\} =0$ so $\left\langle dn|n\right\rangle $ is at best purely imaginary if it doesn't vanish. $\endgroup$
    – FraSchelle
    Dec 18, 2017 at 18:14

For the result $\int F_{ij}dS^{ij}=4\pi g$, just use Gauss's law same as electric field.

For electric field, we have $E=\frac{1} {4\pi \epsilon_0} \frac{q}{r^2}$ and $\oint EdS=\frac{q}{\epsilon_0}=\frac{q}{4\pi \epsilon_0}*4\pi$, where $\frac{q}{4\pi \epsilon_0}$ is the term in E field except $r^2$.

So similarly, we have here $F=\frac{g}{r^2}$, then $\oint FdS=g*4\pi$

This works because $\int \frac{1}{r^2}dS=\int \frac{1}{r^2}r^2sin\theta d\theta d\phi=\int sin\theta d\theta d\phi=\int d\Omega=4\pi$ is the solid angle.


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