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research33 min studyLesson 546 in Physics

Weyl semimetals and chiral nodes

Treat an isolated three-dimensional band crossing as an oriented curvature source and connect its local charge to lattice pairing and surface arcs.

Before you begin
On this page

The question for today

What you’ll understand

  • Compute the chirality of an anisotropic linear Weyl Hamiltonian.
  • Identify the occupied-band curvature flux through an enclosing sphere.
  • Separate a local node model from lattice constraints, surface projections, and disorder effects.

A Weyl semimetal is gapless, yet a small closed momentum surface around one isolated crossing can be gapped between its two bands. The Berry flux through that surface defines a local topological charge. A linear node's charge depends on the orientation of its velocity map, not on whether all velocities have the same magnitude. A lattice completes the story by requiring compensating charges and by allowing surface states whose constant-energy contours can terminate at projected bulk nodes.

An oriented linear map fixes the node charge

Near a two-band crossing, expand momentum about its node and keep three independent Pauli coefficients linear in the deviations. Their Jacobian maps a small momentum sphere onto an ellipsoid of Hamiltonian directions. Its determinant sign is the chirality. With the positive-i connection used here, the lower eigenstate carries outward curvature flux two pi times that sign; the upper eigenstate carries the opposite flux. Stating the band and convention avoids a common sign ambiguity in the word monopole.

A constant perturbation in any Pauli direction shifts the zero of this invertible map instead of supplying a fourth anticommuting mass matrix. Thus a sufficiently isolated Weyl point cannot generally be gapped by a small uniform two-by-two perturbation; it moves. Opposite charges can annihilate when brought together, and extra bands or translation-breaking couplings can modify the mechanism. This stability statement does not forbid changing the node's energy through an identity term.

H(q)=axqxσx+ayqyσy+azqzσz,χ=sgn⁡(axayaz),12π∮S2Ω−⋅dS=χH(\mathbf q)=a_xq_x\sigma_x+a_yq_y\sigma_y+a_zq_z\sigma_z,\qquad \chi=\operatorname{sgn}(a_xa_ya_z),\qquad \frac{1}{2\pi}\oint_{S^2}\boldsymbol\Omega_-\cdot d\mathbf S=\chi
q is momentum displacement measured as inverse length and ai=ℏvi has units energy×length. All ai are nonzero. The sphere is oriented outward around one isolated node; Ω− is lower-band curvature in the positive-i convention. χ is dimensionless.↔ Scroll wide equations horizontally.

Low-energy state counting depends on all three velocities

Positive-energy surfaces are ellipsoids with radii energy divided by the absolute velocity coefficients. Dividing their momentum volume by the usual three-dimensional state-counting cell gives the number of positive-branch states per real-space volume below that energy. Differentiating produces a density of states proportional to energy squared. The determinant's absolute value controls this magnitude, while its sign controls chirality; they encode different physical properties.

This result assumes a simple untilted cone and an energy window where the linear expansion is accurate. A strong tilt can produce a type-II node with finite electron and hole pockets at the nodal energy, invalidating this vanishing-density formula. Disorder, interactions, and nearby bands can also introduce new low-energy scales. A clean local cone is a controlled effective model, not a complete prediction of a material's carrier density or conductivity.

n+(E)=E36π2∣axayaz∣,g+(E)=dn+dE=E22π2∣axayaz∣(E>0)n_+(E)=\frac{E^3}{6\pi^2|a_xa_ya_z|},\qquad g_+(E)=\frac{dn_+}{dE}=\frac{E^2}{2\pi^2|a_xa_ya_z|}\quad(E>0)
n+ counts positive-band states per real-space volume below E, for one nondegenerate node; g+ has units (energy×volume)⁻¹. This is the untilted three-dimensional linear-cone approximation without an uncounted spin or valley multiplicity.↔ Scroll wide equations horizontally.

A lattice and a surface complete the local picture

An isolated continuum equation can describe one node, but a periodic lattice band structure must compensate its total curvature charge across the Brillouin zone. Appropriate two-dimensional momentum slices between separated opposite nodes carry different Chern numbers, providing an interface-channel interpretation of surface states. At a suitable chemical potential these form Fermi arcs joining projected nodes of compensating total charge rather than standalone closed contours confined entirely to one surface band.

Surface projection matters. Opposite nodes projected to the same point can cancel the relevant surface obstruction, and detailed arc connectivity depends on boundary potentials and other pockets. Translational disorder also spoils a sharply labelled crystal momentum, so a plotted arc is not a universal diagnostic under arbitrary perturbations. The robust starting claim is the quantized flux on a small band-separated enclosing surface and its consistency with the full lattice's node content.

Work it through

Separate chirality from density of states

An untilted node has ax=2, ay=1, az=−1 eV nm. Find χ, lower-band outward Berry flux, and the positive-band density of states at E=0.10 eV. How does that density change at 0.20 eV within the linear regime?

  1. The coefficient product is −2 (eV nm)³, so χ=−1. The lower-band outward flux in the declared convention is −2π, with no factor from a second spin species.

  2. Use the absolute product for state counting: g+(0.10)=0.010/[2π2(2)]=1/(400π2)≈2.53×10−4g_+(0.10)=0.010/[2\pi^2(2)]=1/(400\pi^2)\approx2.53\times10^{-4} eV⁻¹ nm⁻³. A negative chirality does not make the density negative.

  3. Doubling energy multiplies this quadratic density by four, giving approximately 1.01×10−31.01\times10^{-3} eV⁻¹ nm⁻³. The integrated positive-state count instead scales with energy cubed and would increase eightfold.

The result

The node has χ=−1 and lower-band flux −2π; its positive density of states is positive and rises fourfold when energy doubles in the stated linear window.

The living notebook

Explore the model

Interactive lab

Interactive lab · quantum

Opposite directions of a Weyl eigenvector

For the worked node along qz>0, its Pauli coefficient points along −z and its lower eigenstate along +z. Reset and set both controls to θ=0° and φ=0°; predict zero probability 1. For a fresh comparison representing qz<0, Reset and set both controls to θ=180° and φ=0°; predict zero probability 0. This samples two local eigenvector directions only.

Opposite directions of a Weyl eigenvectorA normalized two-level pure state, not an interacting field simulation. Playback adds one radian per second to the relative phase; basis probabilities stay fixed. Current readings: Probability |0⟩: 71.1%; Probability |1⟩: 28.9%; Bloch (x, y, z): 0.78, 0.45, 0.42. All parameters are adjustable with the labeled controls below.xyz|0⟩|1⟩Basis probabilitiesP(0)71.1%P(1)28.9%
Bloch vector / P(0)P(1)Phase latitude
Probability |0⟩
71.1%
Probability |1⟩
28.9%
Bloch (x, y, z)
0.78, 0.45, 0.42
t = 0 s

A normalized two-level pure state, not an interacting field simulation. Playback adds one radian per second to the relative phase; basis probabilities stay fixed. 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 forA node's eigenvectors change direction across momentum space. The pure-qubit lab supplies no momentum surface, Weyl Hamiltonian, density of states, flux integral, or Fermi arc.

Pause & reason

Check your understanding

For an isolated invertible linear two-band Weyl map, what does a sufficiently small constant Pauli-vector perturbation generally do?

Your turn

Try it yourself

For H=2(qxσx+qyσy+qzσz) with coefficients in eV nm, add δσz with δ=0.020 eV. Find the new node momentum, its chirality, and whether a gap opens within this local model.

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

Give me a hint

Solve for the simultaneous zero of the three Pauli coefficients and compare the unchanged Jacobian determinant.

Compare with the solution

The first two coefficients vanish at qx=qy=0, and 2qz+0.020=02q_z+0.020=0 gives qz=−0.010 nm⁻¹. The Jacobian still has three positive diagonal coefficients 2, so χ=+1. The energies meet at zero at this shifted point, and no local gap opens. This conclusion is valid while the shift remains inside the region described by the linear model; a complete lattice could allow eventual collision with an opposite-charge node.

Keep the idea

Take it with you

  • Weyl chirality is an oriented Jacobian sign, while state-counting magnitudes use its absolute value.
  • An enclosing lower-band curvature surface carries quantized flux even though the node itself is gapless.
  • Lattice charge compensation, surface projection, tilt, and disorder qualify local-cone predictions.

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