Codex @codex  0

Andre Geim coauthored measurements showing that electric charge carriers in graphene can behave as effectively massless quasiparticles described by the Dirac equation. The 2005 experiments found an unusual sequence of quantum Hall effect plateaus and an oscillation phase associated with Berry phase. Here the Dirac description concerns low-energy quasiparticles in a carbon lattice, rather than free electrons acquiring zero rest mass. The results connect a readily contacted material with concepts usually introduced in relativistic quantum mechanics.^[ref]

A physicist researching condensed matter physics, including graphene.^[ref]

For a normalized, smoothly chosen instantaneous energy eigenstate $|\psi (\mathbf{R})⟩$, the Berry connection has components $A_j=i⟨\psi |{\partial }_{R_j}\psi ⟩$. Its line integral around a closed parameter path $C$ gives the Berry phase in a periodic choice of state vector:Normalization makes the inner product $⟨\psi |{\partial }_{R_j}\psi ⟩$ purely imaginary, so the Berry connection is real. Changing the representative to $e^{i\chi (\mathbf{R})}|\psi ⟩$ changes the Berry connection to $A_j-{\partial }_{R_j}\chi$. A single-valued phase factor along the closed path changes the line integral by an integer multiple of $2\pi$, leaving $e^{i\gamma }$ unchanged. If the chosen vector is not periodic, include the endpoint phase:Here $\psi$ denotes the instantaneous energy eigenstate representative, with its dynamical phase removed. This is freedom to choose a representative of a state ray. ^[ref].

The Berry curvature measures the local twisting of the quantum phase convention of a smoothly varying, nondegenerate energy eigenstate. For the Berry connection $\mathcal{A}={\sum }_j{A}_{j}d{R}_j$, it is the differential formChanging the state representative by $e^{i\chi (\mathbf{R})}$ changes $\mathcal{A}$ by $-d\chi$ but leaves $\mathcal{F}$ invariant. Thus the Berry connection depends on a phase convention, while the Berry curvature has convention-independent local content.

For a surface whose boundary is a closed parameter path, Stokes theorem expresses the Berry phase as the integral of the Berry curvature, provided the chosen patch is smooth on the surface. For a spin one-half aligned with a slowly turning magnetic field, the curvature yields a Berry phase equal to minus half the oriented solid angle for the positive-spin branch. The spectral gap must remain open; a quantum degeneracy can obstruct a single smooth choice over the whole surface.^[ref]

A Berry phase is the geometric phase acquired when a nondegenerate energy eigenstate follows a closed cycle of parameters of a Hamiltonian operator in the ideal limit of the adiabatic theorem. The energy eigenstate must stay separated from other energy eigenstates by a nonzero spectral gap, and the parameters must change sufficiently slowly. Returning the Hamiltonian operator to its starting value returns the state to the same ray, but its quantum phase can remember the journey. The total quantum phase contains both the dynamical phase and the Berry phase.

The beauty of the Berry phase is that, within the adiabatic theorem's limit, it depends on the oriented path of the parameters rather than how long the journey takes. For a spin one-half qubit, the Berry phase of a Bloch-sphere latitude turns that memory into half a solid angle. Berry-phase interference with a reference path converts this geometry into a detection probability.

Michael Berry's 1984 paper formulated the geometric phase for cycles governed by the adiabatic theorem and derived the solid angle result for spin. Its bibliographic details are Proceedings of the Royal Society of London A, volume 392, pages 45–57, DOI 10.1098/rspa.1984.0023. ^[ref]; ^[ref]. The wider concept of geometric phase does not require every cycle to obey the adiabatic theorem: Aharonov and Anandan formulated a geometric phase for general cyclic quantum evolution in 1987. ^[ref].

An isolated qubit that acquires only a global phase has unchanged measurement probabilities. To reveal the Berry phase, use two coherent paths of a Mach-Zehnder interferometer, represented by a path qubit. In one arm the internal spin one-half qubit follows the loop; in the reference arm it stays in its starting energy eigenstate. Assume equal arm amplitudes, identical final internal states, cancelled relative dynamical phases, and calibrated fixed apparatus phases. In the model above, hold the reference Hamiltonian operator at its initial value for the same duration. Both arms then have energy eigenvalue $-\Delta /2$ and acquire the same dynamical phase $\Delta T/(2\hslash )$, so their relative dynamical phase vanishes in the ideal limit of the adiabatic theorem.

The joint quantum state before recombination then factors asThe Berry phase is now a relative quantum phase between the two paths. The calibrated plus output projects onto $|+⟩=(|\mathrm{ref}⟩+|\mathrm{loop}⟩)/\sqrt{2}$. The Born rule givesFor the Berry phase of a Bloch-sphere latitude, the plus-port probabilities at $\theta =0,60^{\circ },90^{\circ },180^{\circ }$ are respectively $1,1/2,0,1$. This ideal Mach-Zehnder interferometer measurement cannot distinguish $\gamma$ from $-\gamma$ or $\gamma +2\pi$ by itself; scanning a known reference phase can resolve the sign. An uncancelled relative dynamical phase would shift the fringe as well.

Choose the computational basis $|0⟩=|{\uparrow }_z⟩$, $|1⟩=|{\downarrow }_z⟩$ for a spin one-half qubit. On the Bloch sphere, let $\mathbf{n}=(\sin \theta \cos \varphi ,\sin \theta \sin \varphi ,\cos \theta )$ and use the normalized quantum stateThis is the $+1$ eigenstate of $\mathbf{n}\cdot \mathbf{\sigma }$, where $\mathbf{\sigma }$ are the Pauli matrices. To make it an instantaneous ground state, take the model Hamiltonian operator $H=-\Delta \mathbf{n}\cdot \mathbf{\sigma }/2$ with constant $\Delta >0$. Its energy eigenvalues are $-\Delta /2$ and $+\Delta /2$, so its spectral gap is $\Delta$ everywhere. The spin points along $\mathbf{n}$; this explicit convention fixes the sign of the Berry phase.

Keep the polar angle $\theta$ fixed and increase the azimuth $\varphi$ from $0$ to $2\pi$, slowly enough for the adiabatic theorem. The state vector written above is periodic in $\varphi$. Directly differentiating givesThus the azimuthal Berry connection is $A_{\varphi }=-{\sin }^2(\theta /2)$ and its line integral isThe north polar cap has oriented solid angle $\Omega ={\int }_0^{2\pi }{\int }_0^{\theta }\sin \vartheta d\vartheta d\varphi =2\pi (1-\cos \theta )$, positive for increasing $\varphi$. Reversing the loop reverses the Berry phase. The opposite spin eigenstate has the opposite Berry phase modulo $2\pi$.

At $\theta =60^{\circ }$, the solid angle is $\pi$ and the Berry phase is $-\pi /2$. At $\theta =90^{\circ }$, the solid angle is $2\pi$ and the Berry phase is $-\pi$. At $\theta =0$, the loop collapses and $\gamma =0$. At $\theta =\pi$, the same expression gives $-2\pi \equiv 0$: the Bloch vector stays at the south pole, while this representative is $e^{i\varphi }|1⟩$. Its winding is only a choice of global phase, so the physically relevant phase factor is again one. The displayed formula is a continuous, unwrapped representative, not a unique real-valued observable.

For an energy eigenstate with instantaneous energy eigenvalue $E(t)$, the dynamical phase isIt comes from the Schrödinger equation and depends on the time spent evolving. In an ideal cycle under the adiabatic theorem, a periodic choice of instantaneous energy eigenstate gives $|\Psi (T)⟩=e^{i(\delta +\gamma )}|\Psi (0)⟩$, with Berry phase $\gamma$.

A geometric phase is a quantum phase associated with the geometry of a path of state rays, after separating the dynamical phase. The Berry phase is its cyclic case governed by the adiabatic theorem; more general cyclic evolution can also have a geometric phase.

Expanding a quantum state in instantaneous energy eigenstates separates dynamical phases from the coupling $A_{mn}$. orthonormality makes $A$ anti-Hermitian. For distinct eigenvalues, differentiating the eigenvalue equation gives $A_{mn}=⟨m|\dot{H}|n⟩/(E_n-E_m)$. Its off-diagonal entries drive nonadiabatic transitions; diagonal entries describe phase changes of the basis and generate the Berry phase.

The conduction and valence dispersions are two cones touching at the Dirac point, with slopes $\pm \hslash v_F$. Here $\mathbf{k}$ is measured relative to one valley centre. Each cone has a density of states proportional to $|E|$, which vanishes at the touching energy. Gate-induced carriers satisfy approximately $n=C_g(V_g-V_D)/e$ per unit area. Negative and positive offsets from $V_D$ produce holes and electrons, respectively. Away from neutrality a simple charge carrier mobility model gives $\sigma \simeq e\mu |n|$, so resistance decreases on either side of a maximum near $V_D$. Disorder, electron–hole puddles, contacts and quantum transport round the peak; the cone dispersion alone fixes neither its height nor its width, nor does it require exact electron–hole symmetry in a real device.

A direct derivation of the magnetic spectrum starts from the single-valley Dirac Hamiltonian operator $H=v_F({\sigma }_x{\pi }_x+{\sigma }_y{\pi }_y)$. Introduce oscillator operators using the magnetic length ${\ell }_B=\sqrt{\hslash /(e|B|)}$. Up to exchanging sublattices or valleys,Squaring gives adjacent oscillator number operators, henceThe positive expression printed in the source describes the electron branch only; the negative-energy partners must also be included. The same square-root dependence follows from quantizing circular orbit area, with the Dirac Berry phase cancelling the usual half-integer offset.

Each spin and valley contributes $Ae|B|/h$ states in area $A$. Including two spins and two valleys gives $N_{\mathrm{LL}}=4Ae|B|/h$, or areal degeneracy $4e|B|/h$. The zero-energy level is half-filled at charge neutrality: the carrier density counts departures from that state.

For $n=10^{16}{\mathrm{m}}^{-2}$, the filling factor and fourfold graphene Hall sequence areIn a plateau, ${\rho }_{xx}\simeq 0$ and $|{\rho }_{xy}|=h/({\nu }_N{e}^2)$. The Hall sign reverses between electron and hole doping; the sketch uses its magnitude to avoid tensor-index sign conventions. Representative plateau-centre fields are $20.68$, $6.893$, $4.136$ and $2.954\mathrm{T}$ for $\nu =2,6,10,14$. Between plateaux, crossing extended states in broadened levels gives longitudinal peaks and Hall steps. Their ideal filling-centre positions have $\nu =4j$, namelyThus peaks near $10.34$, $5.17$ and $3.45\mathrm{T}$ separate the first prominent plateaux as field increases. Many weak oscillations accumulate toward zero field and are damped there. Finite-width plateaux require localized states between extended-state energies; an exactly clean Landau spectrum does not by itself prescribe plateau widths. The illustration uses an explicitly schematic broadening, not inferred experimental widths.

At $6.5\mathrm{T}$, $\nu =6.36$, close to the $\nu =6$ plateau. With $v_F=10^6\mathrm{m}{\mathrm{s}}^{-1}$, $E_1=92.5\mathrm{meV}$ and $E_2=130.8\mathrm{meV}$. The relevant adjacent gap near the chemical potential is therefore $\Delta =E_2-E_1\simeq 38.3\mathrm{meV}$, rather than the gap from the zero level. Resolving levels requires quantum-lifetime broadening $\hslash /{\tau }_q\lesssim \Delta$ and thermal broadening appreciably smaller than $\Delta$, giving the order-of-magnitude limitsThe upper temperature is an energy-scale estimate, not a guaranteed sharp-feature temperature. For oscillation amplitudes the thermal damping factor is $X/\sinh X$, $X=2{\pi }^2{k}_{B}T/(\hslash {\omega }_c^{\ast })$, with $\hslash {\omega }_c^{\ast }=\hslash eB{v}_F^2/E_F\simeq 36.7\mathrm{meV}$ and $E_F=\hslash v_F\sqrt{\pi n}$. Requiring $X\lesssim 1$ gives the more conservative scale $T\lesssim 22\mathrm{K}$ for little thermal damping. Visible weaker features may persist above that scale; their actual cutoff depends on resolution, disorder and temperature-dependent scattering. The transport lifetime and quantum lifetime also need not coincide.

Let $\epsilon =\omega /B\ll 1$. The exact frequency has expansion $\lambda =B-\omega \cos \alpha +O({\omega }^2/B)$. The ratio $(B-\omega \cos \alpha )/\lambda =1+O({\epsilon }^2)$, while the coefficient of the orthogonal state is uniformly $O(\epsilon )$. Hence the supplied solution reduces towith phase error $O({\omega }^{2}t/B)$, which is $O(\epsilon )$ over one cycle. This is the adiabatic approximation: the system remains in the instantaneous positive-energy eigenstate, with energy $E_{+}=\hslash B/2$.

Thus the dynamic phase is $-Bt/2$ and the geometric phase is $-\omega t(1-\cos \alpha )/2$, in the gauge for ${\chi }_{+}$ printed in the PDF. For one cycle $T=2\pi /\omega$, the Berry phase isminus half the solid angle swept by the field direction.

To calculate the requested integral, use $|\psi (\alpha ,\phi )⟩=(\cos (\alpha /2),e^{i\phi }\sin (\alpha /2){)}^T$. It has $i⟨\psi |{\partial }_{\phi }\psi ⟩=-{\sin }^2(\alpha /2)$; its radial and polar-angle connection components vanish in this gauge. Along the circle in parameter space, $\phi$ runs from zero to $2\pi$, so $\Gamma =-{\int }_0^{2\pi }{\sin }^2(\alpha /2)d\phi =-\pi (1-\cos \alpha )$, agreeing with the exact-solution limit.

A ferromagnetic vacuum breaks the special unitary group $SU(2)$ to the subgroup of rotations about its magnetization, so its vacuum manifold is the two-sphere $S^2$. The massless field is therefore a unit vector $\mathbf{n}(x,t)$ with ${\mathbf{n}}^2=1$.

The only rotational scalar linear in $\dot{\mathbf{n}}$ that can be formed directly from $\mathbf{n}$ is $\mathbf{n}\cdot \dot{\mathbf{n}}=\frac{1}{2}{\partial }_t({\mathbf{n}}^2)=0$. A local one-form $\mathbf{A}(\mathbf{n})\cdot \dot{\mathbf{n}}$ can describe the required Berry phase, but no choice of $\mathbf{A}$ is globally nonsingular and strictly rotationally invariant on $S^2$. One therefore needs coordinate patches, an extension, or a redundant spinor parametrization.

Take $\omega >0$ and $T=2\pi /\omega$. A full spin one-half rotation gives $R(T)=-I$, an essential sign. The cyclic phase of a spin in a rotating field has factors and total quantum phasesTo separate the adiabatic dynamical phase, set $B=\sqrt{{\Omega }_0^2+{\Omega }_R^2}>0$ and $\cos \theta ={\Omega }_0/B$. Then $\Omega =B-\omega \cos \theta +O({\omega }^2/B)$. The instantaneous laboratory energies are $\pm \hslash B/2$, so the adiabatic dynamical phases are $\mp BT/2$. Subtracting them from ${\Phi }_{\pm }$ givesThe oriented solid angle $\mathcal{A}$ is swept out by the field direction. These are the Berry phases for the eigenstates parallel and antiparallel to that direction. Retaining the spinor sign makes the quantum phase vanish modulo $2\pi$ when the field does not trace a nontrivial loop. The assumption $\omega \ll B$ suffices for adiabatic following; near degeneracy it must not be silently dropped.^[ref]

An initially occupied, isolated instantaneous energy eigenstate follows its changing eigenspace, acquiring a dynamical phase and generally a Berry phase. For smooth nondegenerate evolution, a useful local smallness condition isA nonvanishing spectral gap and suitably smooth, slow driving throughout the evolution are needed; the local inequality alone is not a universal long-time error bound. Gap closure invalidates the simple nondegenerate approximation. A separated degenerate eigenspace may mix internally even when transitions out of it are suppressed.