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advanced27 min studyLesson 506 in Physics

Move a wavepacket through an isolated band

Combine band velocity, force-driven crystal momentum, and Berry curvature.

Before you begin
On this page

The question for today

What you’ll understand

  • Derive group velocity from a band gradient.
  • Apply the electron sign in the crystal-momentum equation.
  • State the isolated-band and weak-field assumptions behind Berry-corrected dynamics.

A Bloch eigenstate is spread throughout the crystal, while transport asks where an electron packet moves. Forming a narrow superposition of nearby wavevectors gives a packet center whose motion can be treated semiclassically under suitable conditions. Band slope controls its ordinary velocity, external forces move its crystal momentum, and Berry curvature can add a transverse velocity. These ingredients describe a controlled single-band regime rather than a replacement for quantum transitions, disorder scattering, or the full transport distribution.

A band slope becomes a velocity

For a packet narrow in wavevector, expand its band energy about the central wavevector. The linear term translates its envelope with group velocity equal to the energy gradient divided by hbar. A flat band has zero ordinary group velocity even if its Bloch functions contain rapidly varying microscopic components. Mechanical momentum divided by bare electron mass is therefore not generally inferred by simply inserting hbar k into a free-particle formula.

A weak slowly varying external potential changes the packet’s crystal momentum according to force divided by hbar. For an electron the electric force is negative e times electric field, with e defined positive. A constant field can push k across the zone boundary, but equivalent reciprocal labels describe a continuous physical trajectory. In an ideal isolated one-dimensional band this periodic energy dependence gives Bloch oscillations. Scattering or interband tunneling may interrupt them in a real solid.

vband=1ℏ∇kEn,ℏk˙=−eE(B=0),TB=2πℏe∣E∣a\mathbf v_{\rm band}=\frac1\hbar\nabla_{\mathbf k}E_n,\qquad \hbar\dot{\mathbf k}=-e\mathbf E\quad(\mathbf B=0),\qquad T_B=\frac{2\pi\hbar}{e|\mathcal E|a}
e>0 is the elementary charge, E bold is electric field in V/m, and calligraphic E is its one-dimensional component. TB is the Bloch period in s for lattice spacing a, absent scattering and interband transitions.↔ Scroll wide equations horizontally.

Eigenvector geometry adds a velocity

The cell-periodic eigenstate changes as k moves. Its Berry connection is i times the inner product with its wavevector derivative, using a consistently normalized cell inner product. A smooth phase change of that eigenstate shifts the connection by a gradient, while its curl is gauge invariant for an isolated nondegenerate band. Berry curvature has area units when k has inverse-length units; it is a geometric field in wavevector space, not a real magnetic field measured in tesla.

Wavepacket dynamics includes minus the cross product of the k velocity with this curvature. With a magnetic field present, k evolution also involves the full packet velocity and the energy includes an orbital magnetic-moment correction. We restrict the numerical examples to zero magnetic field so those additional terms are absent. Near a degeneracy, a one-band Abelian description can fail and a coupled-band or non-Abelian treatment may be required; an arbitrary eigenvector phase derivative is not a reliable substitute.

An=i⟨un∣∇kun⟩,Ωn=∇k×An,r˙=1ℏ∇kEn−k˙×Ωn\mathbf A_n=i\langle u_n|\nabla_{\mathbf k}u_n\rangle,\qquad \boldsymbol\Omega_n=\nabla_{\mathbf k}\times\mathbf A_n,\qquad \dot{\mathbf r}=\frac1\hbar\nabla_{\mathbf k}E_n-\dot{\mathbf k}\times\boldsymbol\Omega_n
An has units m and Ωn m². The equation is for an isolated band with normalized cell-periodic states; at B=0 no orbital magnetic-energy correction is needed. It uses electron charge −e in the force equation above.↔ Scroll wide equations horizontally.

Transport needs a population and a regime

A packet equation alone does not predict a steady electrical conductivity. One also needs the distribution of occupied packets and a collision or relaxation description consistent with the experiment. Applied fields should vary little across the packet and be weak enough to suppress transitions across nearby gaps. Time scales must allow the adiabatic intraband construction, while mean free paths and relaxation times determine whether a clean trajectory or a scattering-limited response is appropriate.

The vector lab can check directions and projections only. Reset and set angle to 0° and magnitude to 1; predict components (1,0). Reset and set angle to 90° and magnitude to 1; predict (0,1). These settings distinguish perpendicular vectors without inventing a transport trajectory. The instrument has no band derivative, electric charge, or Berry-curvature control, so the cross-product velocity and its physical units must be calculated from the displayed equations.

E(k)=ϵ0−2tcos⁡(ka)⟹v(k)=2taℏsin⁡(ka)E(k)=\epsilon_0-2t\cos(ka)\quad\Longrightarrow\quad v(k)=\frac{2ta}{\hbar}\sin(ka)
t is hopping energy, a spacing in m, and hbar in matching energy·s units. This is the ordinary group velocity of the stated one-dimensional tight-binding band, before any anomalous velocity.↔ Scroll wide equations horizontally.

Work it through

A band velocity and a Bloch period

A one-dimensional band has t=0.10 eV, a=0.50 nm, and hbar=6.58×10⁻¹⁶ eV·s. Find its velocity at ka=π/2. For electric field +10⁴ V/m, find the Bloch period using h/e=4.14×10⁻¹⁵ V·s and state the direction of k drift.

  1. At ka=π/2 the sine is one, so velocity is 2ta/hbar. Substituting 2×0.10×0.50×10⁻⁹ divided by 6.58×10⁻¹⁶ gives approximately +1.52×10⁵ m/s.

  2. One reciprocal period is 2π/a. The electric force on the electron drives k at −eE/hbar, so k decreases rather than increases for the stated positive field. Reaching a reduced-zone edge requires relabeling, not an elastic collision event.

  3. The period is (h/e)/(Ea)=4.14×10⁻¹⁵/(10⁴×0.50×10⁻⁹)=8.28×10⁻¹⁰ s. This ideal period is observable only when coherent intraband motion survives long enough and interband transitions remain negligible.

The result

The ordinary velocity is +1.52×10⁵ m/s and k drifts negatively. The ideal Bloch period is 0.828 ns under the stated field.

The living notebook

Explore the model

Interactive lab

Interactive lab · vectors

Check perpendicular directions explicitly

Reset and set angle=0°, magnitude=1; predict (x,y)=(1,0). Reset and set angle=90°, magnitude=1; predict (0,1).

Check perpendicular directions explicitlyPlayback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Current readings: x component: 1.393; y component: 0.975; Magnitude: 1.7. All parameters are adjustable with the labeled controls below.−7.46−3.7303.737.46−3.5−1.7501.753.5xyvvₓ = 1.39
Vectorx / y projectionsTrace marker
x component
1.393
y component
0.975
Magnitude
1.7
t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe vector instrument supplies projections and direction only. It does not compute band velocity, Berry curvature, Lorentz dynamics, or electronic transport.

Pause & reason

Check your understanding

What changes under a smooth phase transformation un(k)→exp(iφ(k))un(k) in an isolated band?

Your turn

Try it yourself

At B=0, a packet has ordinary velocity (4,0,0) m/s, k velocity (−3,0,0) m⁻¹s⁻¹, and Berry curvature (0,0,2) m² in an illustrative local model. Find total velocity and its transverse sign.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use minus dot-k cross curvature in the declared convention, keeping the cross-product order.

Compare with the solution

The cross product dot-k×Ω is (0,+6,0) m/s because x×z=−y and dot-k points along −x. The anomalous term is its negative, (0,−6,0) m/s, so total velocity is (4,−6,0) m/s. These deliberately simple local values test units and signs; they are not material parameters or a claim that a particular band remains isolated under arbitrary forcing.

Keep the idea

Take it with you

  • Group velocity is a band derivative, not automatically hbar k over bare mass.
  • Electron charge determines the sign of crystal-momentum drift.
  • Berry corrections require consistent gauges and an appropriate isolated-band transport regime.

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