research32 min studyLesson 983 in Physics

Quantum oscillations and Fermi-surface areas

Infer an extremal orbit area while keeping thermal, disorder, spin, and phase ambiguities explicit.

Before you begin

On this page

The question for today

What you’ll understand

Convert an inverse-field period into an extremal momentum-space area.
Separate thermal, Dingle, and spin factors in a conventional oscillation amplitude.
Explain why missing frequencies and fitted phase offsets do not uniquely determine a Fermi surface.

Landau quantization reorganizes electronic states as magnetic field changes. In a conventional metallic regime this produces oscillations in magnetization and often in transport, approximately periodic in inverse field. Their frequency is unusually direct geometric information about an electronic orbit, but their visibility is selective. Heavy masses, disorder, spin cancellation, and limited field range can hide parts of a Fermi surface. The absence of an oscillation is therefore different evidence from the absence of the corresponding electronic states.

An orbit area, not a complete surface

For a closed semiclassical orbit perpendicular to magnetic field, quantized momentum-space areas increase in steps proportional to field. At fixed chemical potential, successive crossings of an extremal Fermi-surface cross section generate an inverse-field period. Extremal means stationary with respect to the momentum component along the field. In three dimensions several neck and belly orbits can contribute; in a quasi-two-dimensional metal the angular dependence needs the actual warped geometry and cannot automatically be replaced by one ideal-cylinder formula.

The Onsager frequency determines area in wave-vector space, not directly a volume or total carrier density. Reconstructing those requires angular scans, multiplicities, spin degeneracy, and an appropriate band geometry. A circular orbit radius follows from its area only after assuming circularity. The ordinary inference also presumes a recognizable semiclassical metallic orbit; field-induced reconstruction, magnetic breakdown, strong chemical-potential oscillation, or unconventional oscillations outside that regime need additional theory rather than a forced interpretation of every peak as a separate pocket.

F=\frac{\hbar A_{\rm ext}}{2\pi e},\qquad \Delta(1/B)=\frac{1}{F},\qquad m_c=\frac{\hbar^2}{2\pi}\left|\frac{\partial A}{\partial E}\right|

F and B are in tesla, Aext is in inverse square metres, e is positive, and the cyclotron mass mc is positive. The derivative is for the relevant orbit; these are conventional semiclassical relations.↔ Scroll wide equations horizontally.

Amplitude contains several filters

A conventional fundamental Lifshitz–Kosevich component contains a thermal factor, a Dingle factor for elastic level broadening, and a spin factor. Thermal broadening averages the oscillatory density of states over the Fermi window. Its temperature dependence can estimate cyclotron mass when the remaining factors and field window are controlled. The Dingle factor instead reflects a quantum lifetime. That lifetime counts broadening from scattering angles differently from the transport lifetime that controls ordinary mobility, so the two should not be identified without a scattering model.

Adding spin-split contributions can reverse the oscillation sign or produce a spin zero. A vanishing fundamental need not imply that the orbit disappeared, and a higher harmonic can remain. The simple spin factor assumes an effective Zeeman splitting with a suitable g factor and comparable amplitudes for the two spin contributions. Dimensional prefactors, harmonics, warping, magnetic breakdown, and field-dependent interactions can modify a full fit. A clean product of three factors is an approximation whose regime must be established.

R_T=\frac{X}{\sinh X},\quad X=\frac{2\pi^2 k_BT m_c}{\hbar eB},\qquad R_D=e^{-\pi m_c/(eB\tau_q)},\qquad R_s=\cos\!\left(\frac{\pi g^*m_c}{2m_e}\right)

These factors describe a conventional fundamental orbit with quantum lifetime tauq. The displayed spin expression uses the cyclotron mass for the chosen orientation and an effective Zeeman g factor; it is not a universal formula for every spin-orbit-coupled band.↔ Scroll wide equations horizontally.

Analyze the window and phase together

A finite inverse-field window limits frequency resolution. Background removal, window choice, nearby frequencies, and beat patterns can move apparent Fourier maxima and bias amplitudes. State the field interval and subtraction procedure, compare several windows, and fit the oscillatory data with a forward model where possible. Because damping varies strongly across the window, assigning a single representative field is itself an approximation. Temperature sweeps should use consistent windows to avoid confusing a changed weighting with a changed mass.

A Landau-index intercept is not automatically a Berry phase. The observable, tensor inversion, three-dimensional curvature phase, Zeeman effects, unresolved frequencies, and indexing convention can all shift it. Cross-check candidate orbits against photoemission and band calculations while allowing that strong field or surface preparation may change the state being probed. The hand calculations below isolate conventional geometric and damping relations; they are hypothetical examples and do not claim an empirical pocket size or a material-specific topological diagnosis.

\delta O(B)=C(B)R_TR_DR_s\cos\!\left(2\pi\frac{F}{B}+\varphi\right)

deltaO is a selected oscillatory observable; C includes its dimensional and geometric prefactor. The total phase phi contains observable and orbit contributions and cannot generally be equated directly with a Berry phase.↔ Scroll wide equations horizontally.

Work it through

An inverse-field period fixes an area

A hypothetical conventional oscillation has period $0.002\,\mathrm T^{-1}$ . Use the rounded conversion $2\pi e/\hbar=9.55\times10^{15}\,\mathrm m^{-2}\mathrm T^{-1}$ . Find its extremal area and, only if circular, its radius.

Invert the period to obtain $F=500\,\mathrm T$ . The frequency is a field scale, although the data are periodic in inverse field.
Multiply by the stated conversion: $A_{\rm ext}=4.775\times10^{18}\,\mathrm m^{-2}=0.04775\,\text{\AA}^{-2}$ . The unit conversion uses one inverse square metre equal to ten to the minus twenty inverse square angstroms.
Under the additional circular assumption, $k_F=\sqrt{A_{\rm ext}/\pi}\simeq0.1233\,\text{\AA}^{-1}$ . A noncircular orbit has the same area without this unique radius.

The result

The measured period implies a 500 tesla frequency and area 0.04775 inverse square angstroms; the conditional circular radius is about 0.1233 inverse angstroms.

The living notebook

Explore the model

Interactive lab

Interactive lab · waves

Cancellation without removing either contribution

For each fresh run Reset and set phase difference and amplitude ratio to $(0,1)$ , then $(180,1)$ . Predict resultant amplitudes 2 and 0. This equal-frequency linear-superposition analogy illustrates cancellation; it does not generate Landau levels, fit a spin factor, scan inverse field, or infer a Berry phase.

SumFirst componentSecond component

Sum at x = 0: 0.866
Resultant amplitude: 1.732
Phase difference: 60°

Phase difference60°Amplitude ratio1

t = 0 s

Two waves share the same frequency and wave number. Playback advances t in sin(2x − t) + a sin(2x − t + phase). 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 forTwo nonzero oscillatory contributions can cancel in the measured channel.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

The spin factor can vanish by phase cancellation. This differs from removing the orbit or all of its harmonics.

Your turn

Try it yourself

For a hypothetical fundamental let $X=\ln2$ , $R_D=1/2$ , and $R_s=1/2$ . Compute $R_TR_DR_s$ . Then identify the fundamental spin factor when $g^*=2$ and $m_c=m_e/2$ within the displayed simple 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

Use sinh(log two)=three quarters. Evaluate the cosine after forming the dimensionless mass ratio.

Compare with the solution

The thermal factor is $4\ln2/3\simeq0.9242$ , so the product is $\ln2/3\simeq0.2310$ . For the second case the spin-factor argument is $\pi/2$ , giving $R_s=0$ . That zero is a cancellation within this assumed Zeeman model, not a zero of the geometric area.

Keep the idea

Take it with you

A conventional frequency measures a selected extremal area rather than a complete carrier census.
Thermal damping, quantum-lifetime damping, and spin cancellation encode different physics.
Frequency visibility and phase inference require field-window and model checks.

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