Berry Connection as a Mathematical Object

The Berry connection is a locally defined $U(1)$ connection on the complex line bundle formed by a smoothly varying nondegenerate eigenspace.

In a local normalized eigenvector gauge, it is the one-form

$$A_n = i\langle n(R)\rvert d n(R)\rangle.$$

We will derive its gauge transformation and curvature, then calculate the spin-$1/2$ example. For the physical conditions under which this geometry produces an adiabatic phase, continue to Berry Phase.

Helpful background. U(1) Bundles and Quantum Phase explains local phase choices; Exterior Derivative supplies the operation $F=dA$.

An isolated eigenline over parameter space

Let $M$ be a parameter space and let $H(R)$ be a family of Hamiltonians depending smoothly on $R\in M$. Suppose that on the region under discussion there is a nondegenerate eigenvalue $E_n(R)$ separated from the rest of the spectrum:

$$H(R)\lvert n(R)\rangle = E_n(R)\lvert n(R)\rangle.$$

At each $R$, the physical eigenspace is the complex line

$$L_R = \{\lambda\lvert n(R)\rangle:\lambda\in\mathbb C\}.$$

The collection of lines $L_R$ is the Berry line bundle over $M$. A smooth choice of normalized eigenvector $\lvert n(R)\rangle$ is a local frame or local gauge for that line bundle.

The assumption of a separated nondegenerate eigenvalue matters. If the eigenvalue crosses another level, the line bundle can fail to be smooth there. If the eigenspace is degenerate, the appropriate object is a higher-rank vector bundle with a nonabelian Berry connection, not the $U(1)$ connection treated here.

Local Connection One-Form

In a chosen local normalized eigenvector gauge, define

$$A_n = i\langle n\rvert d n\rangle.$$

Here $d$ is the exterior derivative on parameter space. In local coordinates $R^i$,

$$A_n = A_i^{(n)}\,dR^i, \qquad A_i^{(n)} = i\langle n(R)\rvert \partial_i n(R)\rangle.$$

The normalization condition

$$\langle n\rvert n\rangle=1$$

implies

$$d\langle n\rvert n\rangle = \langle dn\rvert n\rangle+\langle n\rvert dn\rangle = 0.$$

Therefore $\langle n\rvert dn\rangle$ is imaginary:

$$\langle dn\rvert n\rangle = \langle n\rvert dn\rangle^* = -\langle n\rvert dn\rangle.$$

Multiplication by $i$ makes $A_n$ a real one-form. This is why it can be integrated as a phase connection.

Gauge Transformation

The local eigenvector is not unique. A different local gauge is

$$\lvert n'(R)\rangle = e^{i\chi(R)}\lvert n(R)\rangle.$$

Then

$$d\lvert n'\rangle = e^{i\chi} \left( i\,d\chi\,\lvert n\rangle+d\lvert n\rangle \right),$$

and hence

$$\begin{aligned} A_n' &= i\langle n'\rvert d n'\rangle\\ &= i \left( i\,d\chi+\langle n\rvert dn\rangle \right)\\ &= A_n-d\chi. \end{aligned}$$

This is exactly the abelian gauge transformation law for a $U(1)$ connection in the convention used on U(1) Bundles and Quantum Phase.

The connection one-form is therefore not gauge invariant. It is local data. Its transformation law is what allows different local phase conventions to describe the same geometry.

Parallel-Transport Gauge

Let $R(t)$ be a path in $M$. Pulling the Berry connection back to the path gives

$$A_t(t) = i\langle n(R(t))\rvert \frac{d}{dt}n(R(t))\rangle.$$

A local parallel-transport gauge along the path sets

$$A_t(t)=0.$$

Under $\lvert n\rangle\mapsto e^{i\chi(t)}\lvert n\rangle$ along the path,

$$A_t\mapsto A_t-\frac{d\chi}{dt}.$$

On an open interval, one can choose

$$\chi(t) = \int_0^t A_\tau(\tau)\,d\tau$$

to make $A_t'=0$. Around a closed loop, the same construction can return the vector representative with a different endpoint phase. That endpoint mismatch is holonomy; see Parallel Transport and Holonomy.

Curvature

The Berry curvature is the curvature two-form of the Berry connection:

$$F_n = dA_n.$$

In coordinates,

$$F_n = \frac12 F_{ij}^{(n)}\,dR^i\wedge dR^j,$$

where

$$F_{ij}^{(n)} = \partial_i A_j^{(n)} - \partial_j A_i^{(n)}.$$

Using $A_i^{(n)}=i\langle n\rvert\partial_i n\rangle$, one obtains

$$F_{ij}^{(n)} = i \left( \langle \partial_i n\rvert \partial_j n\rangle - \langle \partial_j n\rvert \partial_i n\rangle \right).$$

Under $A_n\mapsto A_n-d\chi$,

$$F_n\mapsto d(A_n-d\chi)=dA_n,$$

because $d^2=0$. Thus the curvature is gauge invariant in the abelian case.

Projector Formula

The connection requires a local phase choice, but the curvature can be written directly from the eigenprojector

$$P_n(R) = \lvert n(R)\rangle\langle n(R)\rvert.$$

For a nondegenerate eigenline,

$$F_{ij}^{(n)} = i\,\operatorname{Tr} \left( P_n \left[ \partial_i P_n, \partial_j P_n \right] \right).$$

This formula is useful because $P_n$ is phase independent:

$$\lvert n\rangle\mapsto e^{i\chi}\lvert n\rangle \quad \Longrightarrow \quad P_n\mapsto P_n.$$

The projector expression is also the form that generalizes most cleanly to band theory and Chern-number calculations, where globally smooth eigenvectors may not exist.

Expression from Hamiltonian Derivatives

For a finite-dimensional smooth Hermitian matrix $H(R)$ with nondegenerate $E_n$, differentiating the eigenvalue equation and projecting onto another eigenstate $\lvert m\rangle$ gives, for $m\ne n$,

$$\langle m\rvert \partial_i n\rangle = \frac{ \langle m\rvert \partial_i H\rvert n\rangle }{ E_n-E_m }.$$

Substituting into the curvature formula yields

$$F_{ij}^{(n)} = i \sum_{m\ne n} \frac{ \langle n\rvert \partial_i H\rvert m\rangle \langle m\rvert \partial_j H\rvert n\rangle - \langle n\rvert \partial_j H\rvert m\rangle \langle m\rvert \partial_i H\rvert n\rangle }{ (E_n-E_m)^2 }.$$

For an infinite-dimensional Hamiltonian this sum requires an appropriate spectral resolution and domain and convergence assumptions; a continuous part of the spectrum contributes an integral. The finite-matrix formula above avoids those extra issues.

This expression shows why Berry curvature is sensitive to nearby levels. Small gaps can strongly enhance the curvature, and degeneracies are precisely where the nondegenerate line-bundle description breaks down.

Holonomy and Flux

In a normalized eigenvector frame that is smooth and periodic around a closed loop $C$, the $U(1)$ holonomy is

$$\exp \left( i\oint_C A_n \right).$$

A smooth frame along the interval $0\leq t\leq T$ need not return to the same vector even though the eigenline closes. In that case include the endpoint overlap:

$$e^{i\gamma_n[C]}=\langle n(0)|n(T)\rangle \exp\left(i\int_0^T A_t\,dt\right).$$

The overlap has unit modulus because the endpoints represent the same normalized eigenline. Under $|n(t)\rangle\mapsto e^{i\chi(t)}|n(t)\rangle$, it acquires $e^{i[\chi(T)-\chi(0)]}$, while the exponential acquires the inverse factor. This explains how a parallel-transport frame with $A_t=0$ can still have a nonzero geometric phase.

If $C=\partial\Sigma$ and a smooth gauge is available on the surface $\Sigma$, then Stokes theorem gives

$$\oint_C A_n = \int_\Sigma F_n.$$

This is a mathematical statement about a connection and its curvature. In an adiabatic quantum evolution, the same holonomy becomes the Berry phase factor after the dynamical phase has been separated. The physical conditions for that statement are handled on Berry Phase.

Spin-1/2 Local Gauge Example

For the normalized spinor

$$\lvert +;\theta,\phi\rangle = \begin{pmatrix} \cos(\theta/2)\\ e^{i\phi}\sin(\theta/2) \end{pmatrix},$$

which is a $+1$ eigenvector of $\hat{\mathbf n}\cdot\boldsymbol\sigma$ and is smooth on the sphere except at the south pole, the Berry connection is

$$\begin{aligned} A_+ &= i\langle +;\theta,\phi\rvert d \lvert +;\theta,\phi\rangle\\ &= -\sin^2(\theta/2)\,d\phi\\ &= -\frac{1-\cos\theta}{2}\,d\phi. \end{aligned}$$

Its curvature is

$$F_+ = dA_+ = -\frac12\sin\theta\,d\theta\wedge d\phi.$$

For a surface $\Sigma$ on the parameter sphere,

$$\int_\Sigma F_+ = -\frac{\Omega(\Sigma)}{2},$$

where $\Omega(\Sigma)$ is the oriented solid angle. This recovers the mathematical core of the spin-$1/2$ solid-angle result. The sign depends on eigenstate and Hamiltonian conventions, so physics pages must state the convention before quoting the phase.

Local Versus Global Data

The Berry connection is usually written in a local eigenvector gauge. If the Berry line bundle is topologically nontrivial, one may need multiple patches, with transition functions

$$\lvert n_\beta\rangle = e^{i\chi_{\beta\alpha}}\lvert n_\alpha\rangle$$

on overlaps. The local connections obey

$$A_\beta = A_\alpha-d\chi_{\beta\alpha}.$$

The curvature pieces agree on overlaps and define a global two-form. Holonomy around loops can be computed by patching local connection integrals with transition-function contributions when one gauge does not cover the whole loop or spanning surface.

This is the practical reason the bundle language matters: a local formula can be correct while no single local formula covers the whole parameter space.

Common Mistakes

• Treating $A_n$ as gauge invariant instead of recognizing it as a local connection one-form.
• Forgetting that $F_n=dA_n$ is gauge invariant only after using the abelian gauge law and $d^2=0$.
• Writing a global eigenvector gauge across a degeneracy or coordinate singularity.
• Applying Stokes theorem through a region where the eigenstate gauge is not defined.
• Confusing the mathematical Berry connection with the full physical Berry phase, which also requires an adiabatic evolution and separation from the dynamical phase.
• Dropping the nondegeneracy assumption when using the $U(1)$ formula.
• Missing sign changes caused by different conventions for the Hamiltonian, eigenstate label, or definition of $A_n$.

Exercises

1. Take a constant eigenline represented by $|n(\phi)\rangle=e^{i\phi/2}|n_0\rangle$, with $0\leq\phi\leq2\pi$. Compute the connection integral and endpoint overlap. Is the holonomy $-1$ or $+1$?

Solution

The connection is $A=-d\phi/2$, so $\int A=-\pi$. But $|n(2\pi)\rangle=-|n(0)\rangle$, giving endpoint overlap $-1$. Therefore the holonomy is $(-1)e^{-i\pi}=+1$, as expected for a constant line. Using only the integral would mistake a nonperiodic basis for physical transport.

2. For the displayed $+1$ spinor, integrate $F_+=-\sin\theta\,d\theta\wedge d\phi/2$ over the north polar cap $0\leq\theta\leq\theta_0$. What phase is obtained when $\theta_0=\pi/2$? What is the Chern number on the whole sphere?

Solution

The cap flux is $-\pi(1-\cos\theta_0)=-\Omega/2$. At the equator this is $-\pi$, so the phase factor is $-1$. On the whole oriented sphere the flux is $-2\pi$, hence $C_1=-1$. The north-patch spinor cannot be used smoothly at the south pole, but its curvature extends there; the nonzero Chern number rules out a single global smooth eigenvector.

3. Two levels cross inside a group of occupied bands, but a nonzero gap still separates the whole occupied subspace from unoccupied levels. Which object can remain smooth: each individual eigenline or the total occupied projector?

Solution

The individual eigenlines need not remain smooth or separately defined at the crossing. The projector onto the entire separated occupied subspace can remain smooth. Its connection is matrix valued; the trace of its curvature gives the first Chern number of the occupied bundle. The scalar nondegenerate formula with $(E_n-E_m)^{-2}$ must not be applied to a crossing individual band.

References

• M. V. Berry, “Quantal phase factors accompanying adiabatic changes”, Proceedings of the Royal Society A 392, 45-57, 1984.
• M. Nakahara, Geometry, Topology and Physics, 2nd ed., CRC Press, 2003.
• B. Simon, “Holonomy, the quantum adiabatic theorem, and Berry’s phase”, Physical Review Letters 51, 2167-2170, 1983.