Polarization and Thouless Pumping

Bulk polarization in a crystal is not an absolute dipole moment: it is defined modulo the charge transported by shifting one electron through a lattice period. Its change along a gapped path is physical, and the charge transported in a closed adiabatic cycle is an integer Chern number. The gap, cycle orientation, occupied subspace, and unit cell are part of the statement.

Required background. Berry geometry supplies the connection and curvature conventions.

Helpful background. Background responses and invertible phases supplies the effective-response interpretation.

Polarization modulo charge

Consider a one-dimensional insulator with lattice period $a$, cell-periodic occupied states $|u_{nk}\rangle$, and electron charge $-e$. With $k$ oriented from $0$ to $2\pi/a$ and $\mathcal A_k=i\sum_{n\in\mathrm{occ}}\langle u_{nk}|\partial_k u_{nk}\rangle$, the electronic polarization per cell is

$$P_{\rm el}=-\frac{e}{2\pi}\int_{\rm BZ}\mathcal A_k\,dk \pmod e.$$

A periodic frame transformation with determinant winding $w$ changes the integral by $-2\pi w$, hence $P_{\rm el}\mapsto P_{\rm el}+ew$. This is the polarization quantum, not an error. Ionic charges and their chosen cell positions must be added before comparing with a measured total polarization Resta 1994, §§ II–III.

Inversion symmetry quantizes $P$ to $0$ or $e/2$ modulo $e$: inversion sends $P\mapsto-P$, so $2P=0$ modulo $e$. Moving the spatial origin or redefining the unit cell can shift the electronic and ionic pieces; the total symmetry-quantized class remains the invariant after the same convention is used on both sides.

A cyclic Thouless pump

Let $H(k,t)$ be periodic in $t$ with period $T$ and gapped for every $(k,t)$. Differentiating the Berry-phase formula and using $\Omega_{kt}=\partial_k\mathcal A_t-\partial_t\mathcal A_k$ gives

$$Q_{+x}=\Delta P =\frac{e}{2\pi}\int_0^Tdt\int_{\rm BZ}dk\,\Omega_{kt} =eC_{kt}, \qquad C_{kt}\in\mathbb Z.$$

This sign corresponds to the stated $(k,t)$ orientation and positive current toward $+x$. Reversing the cycle reverses $C_{kt}$ and the transported charge. Quantization follows because the occupied states form a bundle over the $(k,t)$ torus Thouless 1983.

For a finite open chain, boundary charge changes as the pump proceeds and edge levels may cross the chemical potential. Those crossings are the boundary realization of bulk transport; the instantaneous boundary charge is termination dependent. In a disordered or interacting ring, a twisted boundary phase replaces $k$, and a unique many-body gap throughout the $(\theta,t)$ torus is the relevant hypothesis.

Where quantization fails

Write the cycle as $H(s)$ with $s=t/T_{\mathrm{cyc}}\in[0,1]$. For instantaneous many-body eigenstates $|m(s)\rangle$, a sufficient finite-system adiabatic condition is

$$\max_{m\ne0,\,s} \frac{\hbar\,|\langle m(s)|\partial_sH(s)|0(s)\rangle|} {T_{\mathrm{cyc}}[E_m(s)-E_0(s)]^2}\ll1,$$

with a nonzero minimum gap $\Delta_{\min}$ throughout the cycle. Thus $T_{\mathrm{cyc}}$ must be long compared with the transition-matrix-element scale, not merely with $\hbar/\Delta_{\min}$, and short compared with heating or decoherence times. A bulk gap closing permits $C$ to change. Partial filling, Landau–Zener excitation, noncyclic driving, or coupling to reservoirs can make transported charge noninteger. Quantization of a change does not make a particular branch of absolute polarization unique.

Exercise

An inversion-symmetric one-dimensional insulator has $P=e/2$ modulo $e$. Show that this is compatible with inversion.

Solution

Inversion requires $P=-P$ modulo $e$, or $2P=ne$. For $P=e/2$, $-e/2=e/2-e$, so the two values differ by one polarization quantum and represent the same bulk class.

References

• Raffaele Resta, “Macroscopic Polarization in Crystalline Dielectrics: The Geometric Phase Approach,” Reviews of Modern Physics 66 (1994) 899–915, doi:10.1103/RevModPhys.66.899.
• David J. Thouless, “Quantization of Particle Transport,” Physical Review B 27 (1983) 6083–6087, doi:10.1103/PhysRevB.27.6083.