Landau quasiparticle (${\xi }_p$)

A Landau quasiparticle is a fermionic excitation connected continuously to a Fermi gas particle when interactions preserve the phase. Near an isotropic Fermi surface, its energy relative to the chemical potential is ${\xi }_p={ϵ}_p-\mu \simeq v_F(p-p_F)$, with Fermi momentum $p_F$ and Fermi velocity $v_F$.^[ref]

The Pauli exclusion principle restricts decay. For a three-dimensional normal Fermi liquid with regular short-range interactions, a representative low-energy lifetime model isHere $E_{\ast }$ is an energy scale and $A$ is dimensionless; sharpness requires $\hslash /{\tau }_p\ll |{\xi }_p|$ at zero temperature.^[ref]

For an original example, stipulate $A=1$ and $T=0$. At $|{\xi }_p|/E_{\ast }=0.02$, the decay energy is $0.0004E_{\ast }$, just $2\%$ of the excitation energy; halving $|{\xi }_p|$ reduces that ratio to $1\%$. Approaching the Fermi surface improves the particle description even though the microscopic interaction has not weakened. This quadratic illustration is three-dimensional; low-dimensional corrections require separate analysis.

In a regular three-dimensional Fermi liquid, the Pauli exclusion principle leaves only a small set of final states available to a low-energy Landau quasiparticle. Writing $\xi =ϵ-\mu$, the leading decay rate has the scaleThe coefficient depends on the interaction and band structure.

At zero temperature, an excitation of energy $\xi >0$ creates an additional particle–hole pair. Conservation of energy leaves two small independent energy intervals, producing the quadratic phase-space factor. Its spectral width $\Gamma \sim \hslash /\tau$ then satisfies $\Gamma /|\xi |\to 0$ as $\xi \to 0$. This zero-temperature limit is the sense in which excitations become sharp at the Fermi surface. ^[ref]

At fixed nonzero temperature, the rate instead approaches a finite value proportional to $T^2$ as $\xi \to 0$. One must not infer a vanishing ratio $\Gamma /|\xi |$ in that different limit. A joint low-temperature, low-energy limit compares the width with the relevant excitation scale.

A particle loss rate can include the factor $1-f(\xi )$, whereas a spectral decay width also contains the complementary inverse collision process. Their sum has the regular quadratic low-energy scale ${\xi }^2+(\pi k_{B}T{)}^2$. Particle and hole decay must not be confused with an occupied-state loss probability.

Changing one Landau quasiparticle occupation changes the energies of others. To second order around equilibrium,Here $\mathcal{V}$ is volume, occupations are per momentum and spin state, and $f$ has units of energy times volume.^[ref]

For a toy two-state change at fixed volume, take $\delta n_a=\delta n_b=\eta$, $f_{aa}=f_{bb}=0$, $f_{ab}=f_{ba}=g\mathcal{V}$, and unperturbed energies ${\xi }_a={\xi }_b=e$. The expression gives $2e\eta +g{\eta }^2$. The factor $1/2$ counts the interaction of the same pair once; it does not halve each excitation's energy. This finite-state illustration only explains the bookkeeping of the Landau quasiparticle energy functional.

For a symmetric interaction kernel, $\delta E/\delta n_a={ϵ}_a+V^{-1}{\sum }_b{f}_{ab}\delta n_b$. Smooth Fermi-surface kernels and fixed-density chemical-potential shifts give an $O(T^2)$ thermal correction by the Sommerfeld expansion. The reference dispersion already includes zero-temperature interactions.

For a three-dimensional isotropic unpolarized spin-one-half Fermi liquid, define $f^s=(f_{\uparrow \uparrow }+f_{\uparrow \downarrow })/2$ and $f^a=(f_{\uparrow \uparrow }-f_{\uparrow \downarrow })/2$. Dimensionless Landau parameters use the spin-summed density of states per volume:$P_{\ell }$ is a Legendre polynomial; $s,a$ label spin symmetry. No extra $(2\ell +1)$ multiplies this expansion.^[ref] The superscript $s$ on $F_0^s$ is unrelated to the dimensionless wave speed $s$ below.

In a three-dimensional uniform isotropic continuum obeying the Principle of Galilean relativity, with one bare particle mass $m$, the Fermi-liquid effective mass satisfiesA band effective mass need not satisfy this relation.^[ref]

For the stipulated $m^{\ast }=2m$ in the occupation figure, this relation requires $F_1^s=3$ if those assumptions apply. At fixed $p_F$, $v_F$ is half its free-particle value, so moving by $0.04p_F$ from the Fermi surface costs approximately $0.02p_F^2/m$ instead of $0.04p_F^2/m$. The factor of three belongs to the three-dimensional angular average, not to a universal definition of effective mass.

For an isotropic spinless continuum Fermi liquid with the Principle of Galilean relativity, let $D^{\ast }=A{m}^{\ast }/(2\pi {\hslash }^2)$ be the total density of states and use a Landau quasiparticle energy functional whose double sum has no explicit $1/A$. ThenIndeed, a small boost has radial displacement $p_0\cos \theta$. The one-quasiparticle term costs $N{p}_0^2/(2m^{\ast })$, and the quadratic interaction multiplies that by $D^{\ast }\int f\cos \varphi d\varphi /(2\pi )$, since $\int \cos (\theta +\varphi )\cos \theta d\theta =\pi \cos \varphi$. Matching the Galilean boost energy of an interacting gas proves the formula. With Fourier series $D^{\ast }f=F_0+{\sum }_{\ell \ge 1}{F}_{\ell }\cos (\ell \varphi )$, the angular average is $F_1/2$. If the functional instead contains $1/A$, use its intensive kernel $Af$ and the density of states per area.

The quasiparticle residue is the coherent pole weight of the single-particle Green's function. Its coherent spectral term $Z\delta (E-{\xi }_p)$ contributes $Z\Theta (-{\xi }_p)$ to the zero-temperature momentum occupation. A smooth remainder leaves a jump $Z$ at the Fermi surface; Landau quasiparticle occupations have a unit step.^[ref]

For the local illustrative distribution in the figure, put $x=(p-p_F)/p_F$ and$\Theta$ is the Heaviside step function. Its one-sided limits are $0.825$ and $0.175$, giving $Z=0.65$. The smooth term is stipulated; this local curve does not specify a normalized distribution over all momenta. In particular, $Z$ is neither the fraction of atoms above the Fermi surface nor $m/m^{\ast }$.

The particle-hole continuum is the range of energy and momentum transfers that can excite an occupied fermion state into an unoccupied state, leaving a hole behind. For a free gas at zero temperature, choose $|\mathbf{k}|<k_F$ and $|\mathbf{k}+\mathbf{q}|>k_F$; the excitation energy is ${ϵ}_{\mathbf{k}+\mathbf{q}}-{ϵ}_{\mathbf{k}}$.

Varying the initial momentum produces a continuum at fixed transferred wavevector. For small $q$ in an isotropic gas, its positive frequencies extend approximately up to $v_{F}q$. A collective mode overlapping these excitations can decay by creating particle–hole pairs; a zero sound branch outside that region can avoid this particular damping channel. Interactions change the response and effective parameters, so the free-gas boundary is a useful starting model rather than a universal exact spectrum.