research35 min studyLesson 554 in Physics

Fermi-liquid quasiparticle residue

Derive a narrow pole from the self-energy and distinguish its residue, lifetime, velocity and Fermi-volume constraint.

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The question for today

What you’ll understand

Extract residue and width from a regular low-energy self-energy.
State when inverse residue is a mass enhancement.
Apply a conventional Luttinger count with its symmetry and phase assumptions.

An interacting metal can behave at low energy as if it contained long-lived fermionic excitations, even though inserting an electron also creates an incoherent cloud of other excitations. The electron Green function then has a narrow pole with residue smaller than one. That residue is an overlap weight, not the fraction of electrons still present. Establishing a Fermi liquid requires a controlled low-energy pole and lifetime behavior, rather than simply fitting a Lorentzian at one temperature.

Expand the inverse Green function near a pole

Write the retarded one-electron Green function using bare dispersion relative to the chemical potential and a complex retarded self-energy. At a Fermi-surface point, its real inverse vanishes at zero energy. If the energy derivative of the real self-energy is finite and the imaginary part is sufficiently small, expand the inverse Green function to first order in energy and displacement from that point. Factoring the energy coefficient defines the quasiparticle residue.

For a stable canonical Fermi-liquid electron pole, the residue lies between zero and one. The rest of the unit spectral area remains in incoherent excitations, not in a missing population. The retarded self-energy has a nonpositive imaginary part on the physical real-frequency axis. Multiplying its negative imaginary component by the residue gives a positive pole half-width. The pole approximation is local in momentum and energy; extrapolating it across all Hubbard bands would violate the structure that the incoherent part restores.

G^R(\mathbf k,E)=\frac1{E-\xi_{\mathbf k}-\Sigma^R(\mathbf k,E)},\qquad Z=\left[1-\partial_E\operatorname{Re}\Sigma^R\right]^{-1},\quad G^R\simeq\frac{Z}{E-\epsilon_{\mathbf k}^*+i\Gamma_{\mathbf k}^*}

E, ξk, Σ and Γ* are energies; Z is dimensionless. Derivatives are evaluated near the quasiparticle pole or Fermi surface, under the assumption that a differentiable narrow-pole expansion exists.↔ Scroll wide equations horizontally.

Check lifetime and velocity separately

In a clean ordinary Fermi liquid, phase space makes the low-energy decay rate vanish faster than excitation energy. A familiar regular three-dimensional form is quadratic in energy and temperature, with material-dependent coefficients. Two-dimensional cases can include logarithmic corrections. What matters for an asymptotically sharp excitation is the ratio of pole width to its energy tending to zero. A broad peak with width comparable to energy is not justified as a long-lived quasiparticle merely because a fitted residue is finite.

The velocity contains both energy and momentum derivatives of self-energy. Only when momentum dependence is negligible, together with the appropriate reference dispersion and Fermi-surface comparison, does a mass enhancement reduce to inverse residue. In general an angularly varying momentum derivative changes the relation. Transport coefficients can require additional vertex and scattering information, so neither residue nor spectral mass alone determines dc conductivity. Calling every small residue a universal mobility reduction would conflate these different measurements.

\Gamma^*=-Z\operatorname{Im}\Sigma^R\ge0,\qquad \mathbf v_F^*=Z\left[\mathbf v_F^0+\frac1\hbar\nabla_{\mathbf k}\operatorname{Re}\Sigma^R\right],\qquad \frac{\Gamma^*}{|E|}\to0\quad(T=0)

The velocity expression uses k as wavevector and E as energy. A momentum-independent self-energy gives vF* = ZvF0 locally; an inverse-Z mass statement also needs a suitable dispersion and fixed Fermi-surface geometry.↔ Scroll wide equations horizontally.

Count the Fermi volume without multiplying by residue

For a translation-invariant conventional spin-degenerate Fermi liquid with conserved charge, an appropriately counted Fermi volume is tied to density modulo filled bands. A small electron residue does not shrink that volume by the same factor. One way to understand the constraint inserts a flux quantum through a periodic system and compares its momentum change with the quasiparticle count. The argument assumes the low-energy state belongs to the stated Fermi-liquid class and uses the correct unit cell.

Broken translation symmetry reconstructs the Brillouin zone, superconductivity changes the conserved-charge setting, and topologically ordered fractionalized phases can modify the conventional counting argument. These are reasons to reexamine assumptions, not permission to declare every broadened experimental contour a violation. One-dimensional Luttinger liquids can retain a density-related momentum count without electron quasiparticles. The two-state lab shows only a normalized pure-state overlap analogy; it has no self-energy, Fermi surface, incoherent continuum or electron lifetime.

\frac{2V_F}{V_{\rm BZ}}=n\pmod{2}

VF is the occupied Fermi volume for one spin species and VBZ the Brillouin-zone volume. This simple form assumes spin degeneracy and a conventional translation-invariant Fermi liquid, with n electrons per primitive cell and filled bands accounted for.↔ Scroll wide equations horizontally.

Work it through

Residue is not the full spectral area

Near a quasiparticle pole let $\partial_E\operatorname{Re}\Sigma=-3$ and $\operatorname{Im}\Sigma=-0.040\,\mathrm{eV}$ . Assume the pole approximation is valid and the self-energy is locally momentum independent. Find Z, its pole half-width and the local velocity ratio.

The residue is $Z=[1-(-3)]^{-1}=1/4$ . Its dimensionless value means the coherent electron pole carries one quarter of the canonical spectral area in the relevant decomposition.
The positive pole half-width is $\Gamma^*=-Z\operatorname{Im}\Sigma=(1/4)(0.040)=0.010\,\mathrm{eV}$ . This is an energy half-width, not a decay time or the full width.
With the stated momentum independence, $v_F^*/v_F^0=Z=1/4$ . The remaining spectral area of three quarters belongs to incoherent weight; it is not evidence that three quarters of the electrons have vanished.

The result

The residue is 0.25, the pole half-width is 10 meV and the local velocity is one quarter of the reference value under the stated locality assumption.

The living notebook

Explore the model

Interactive lab

Interactive lab · quantum

Overlap weight as an explicitly limited analogy

Reset and set BOTH Mixing angle θ = 120° and Relative phase φ = 0°; predict basis-zero probability 1/4. Reset and set θ = 120° and φ = 90°; predict the same probability with a changed Bloch direction. This pure two-state geometry illustrates overlap and phase only. It is not an electron quasiparticle, a many-body spectral decomposition, a Fermi surface or a lifetime calculation.

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

Mixing angle θ65°Relative phase φ30°

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 forAn overlap probability alone does not specify dynamics or a spectral pole.

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Check your understanding

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Read the checkpoint explanation

Residue controls the electron pole weight, while the conventional Luttinger volume counts charge modulo filled bands under its stated assumptions. The two quantities are not proportional.

Your turn

Try it yourself

A translation-invariant one-band spin-degenerate conventional Fermi liquid has $n=0.80$ electrons per primitive cell. Find $V_F/V_{\rm BZ}$ for each spin. If its residue changes from 0.8 to 0.2 without changing density or phase assumptions, what happens to that volume?

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 the density count with a factor of two for spin; residue is not an extra occupation factor in that relation.

Compare with the solution

The one-spin volume fraction is $0.80/2=0.40$ . It remains $0.40$ after the residue change if all the stated assumptions continue to hold. Multiplying the density by Z would confuse a coherent spectral overlap with conserved particle number and would discard the incoherent spectral contribution. If the system instead enters a phase with a different unit cell, broken charge symmetry or topological order, the simple conventional formula must be reconsidered in that new setting.

Keep the idea

Take it with you

A quasiparticle residue measures coherent electron overlap within a valid pole expansion.
Width, velocity and mass involve distinct self-energy information.
A conventional Fermi-volume count constrains density independently of residue.

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