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In the study of interacting fermion systems, the Fermi surface is often defined by the presence of a sharp discontinuity in the momentum distribution function n(k) at zero temperature. For example, in Luttinger's work [Phys. Rev. 119, 1153 – Published 15 August, 1960], this discontinuity is central to the formal definition of the Fermi surface.

I noticed that the book A Guide to Feynman Diagrams in the Many-body Problem discusses the distribution function n(k) in interacting Fermi systems and mentions the existence of this discontinuity. However, the book does not explain why the discontinuity is necessary for defining the Fermi surface. Here, the discontinuity at the Fermi surface is equal to the quasiparticle weight ZZ.

Why is this sharp discontinuity essential for defining the Fermi surface, especially in interacting systems? Is it possible to characterize the Fermi surface without such a discontinuity, and if not, what physical property or principle enforces its necessity?

I would greatly appreciate any insights, detailed explanations to the underlying principles, or references to relevant literature.

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    $\begingroup$ This is of fundamental importance to the description of materials in general and is an unavoidable milestone in the understanding of statistical thermodynamics and condensed matter physics. You should just consult standard textbooks on the topic. $\endgroup$ Commented Dec 9 at 14:27
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    $\begingroup$ On the subject of Fermi Liquid and Landau quasiparticles (to which the figure refers): physics.stackexchange.com/a/832007/247642, physics.stackexchange.com/a/833396/247642 $\endgroup$
    – Roger V.
    Commented Dec 11 at 20:17
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    $\begingroup$ IMHO, very roughly discontinuity means that at $k_{\mu}$ Fermionic system multiple phases co-exists in thermodynamic equilibrium (or all states are occupied uniformly by fermions),- by analogy like triple point discontinuity in $P(T)$ diagram shows that all three phases - liquid, solid and gas exist in thermodynamic equilibrium. Also concept of Degenerate Fermi gas may help. $\endgroup$ Commented Dec 11 at 21:32
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    $\begingroup$ Also check Fermi-Dirac distribution. $\endgroup$ Commented Dec 11 at 21:34

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In a non-interacting Fermi gas at zero temperature ($T=0$), the momentum distribution function, denoted by $n(\mathbf{k})$, is given by the Heaviside step function: $n(\mathbf{k}) = \theta(k_F - |\mathbf{k}|)$, which equals 1 for $|\mathbf{k}| < k_F$ and 0 for $|\mathbf{k}| > k_F$, where $k_F$ is the Fermi momentum and $\theta(x)$ is the Heaviside step function. This sharp discontinuity at $|\mathbf{k}| = k_F$ arises directly from the Pauli exclusion principle, which dictates that at $T=0$, all single-particle states with momenta $|\mathbf{k}| < k_F$ are occupied while all states with $|\mathbf{k}| > k_F$ are unoccupied.

When interactions are introduced, Landau's Fermi liquid theory describes the low-energy excitations as quasiparticles – long-lived, fermionic excitations with renormalized properties. The momentum distribution function $n(\mathbf{k})$ deviates from the ideal step function; however, a crucial feature persists: a finite discontinuity at $|\mathbf{k}|=k_F$, quantified by the quasiparticle residue, $Z$. This discontinuity is defined as $\Delta n(k_F) = n(k_F^-) - n(k_F^+) = \lim_{\epsilon \to 0} \left[ n(k_F - \epsilon) - n(k_F + \epsilon) \right] = Z$, where $n(k_F^-)$ and $n(k_F^+)$ denote the limits of $n(\mathbf{k})$ as $|\mathbf{k}|$ approaches $k_F$ from below and above, respectively. The quasiparticle residue $Z$, where $0 < Z \leq 1$, is a measure of the overlap between the quasiparticle state and the bare fermion state, and can be expressed in terms of the Green's function $G(k,\omega)$ through the relation $$Z = \left. \frac{\partial}{\partial \omega} G(k_F,\omega)^{-1}\right|_{\omega=0}^{-1}$$.

The persistence of this discontinuity is fundamental. The Fermi surface is defined as the locus of points in k-space where this discontinuity in $n(\mathbf{k})$ occurs, specifically at $|\mathbf{k}|=k_F$. Luttinger's theorem states that the volume enclosed by the Fermi surface, $V_{FS}$, (related to $k_F$ through $V_{FS} \propto k_F^d$, where d is the spatial dimension) is invariant under interactions, provided the system remains in the Fermi liquid phase. This can be formulated in terms of the electron density $n$ as $$\sum_{k,\sigma} n(k, \sigma) = n$$, and $V_{FS}$ can be related to this quantity. This invariance is topologically protected and intimately connected to the conservation of particle number. The volume enclosed by the Fermi surface can also be expressed through the Green's function in terms of a topological invariant: $$V_{FS} = \frac{1}{2}\int \frac{d^dk}{(2\pi)^d} \operatorname{tr} [G(\mathbf{k}, \omega\rightarrow 0^+) - G(\mathbf{k}, \omega\rightarrow 0^-)]$$. The finite quasiparticle residue $Z$ is crucial for the existence of well-defined quasiparticles with long, but finite, lifetimes ($\tau \propto \epsilon^{-2}$, where $\epsilon$ is the energy above the Fermi level) near the Fermi surface ($|\mathbf{k}|\approx k_F$). The presence of $\Delta n(k_F)$ signifies that the interacting system is adiabatically connected to the non-interacting Fermi gas, as long as $Z \neq 0$. The mathematical foundation for these concepts is the single-particle Green's function, $G(\mathbf{k}, \omega)$, which can be expressed as $$G(\mathbf{k}, \omega) = \frac{Z}{\omega - \epsilon_k + i\delta} + G_{inc}(\mathbf{k}, \omega)$$, where $\epsilon_k$ is the quasiparticle dispersion and $G_{inc}$ is an incoherent part. Luttinger's theorem can be derived by analyzing the pole structure of $G(\mathbf{k}, \omega)$ as $\omega \rightarrow 0$, specifically showing that the volume enclosed by the singularity in $G(\mathbf{k}, \omega)$ does not change with interactions. The Green's function near the Fermi surface also has a direct relation to the momentum distribution function: $$n(\mathbf{k}) = -i\int_{-\infty}^{0} \frac{d\omega}{2\pi} G(\mathbf{k},\omega)$$.

In systems where the discontinuity vanishes ($Z \rightarrow 0$), such as one-dimensional Luttinger liquids or certain strongly correlated materials, we are no longer in the Fermi liquid regime. In these non-Fermi liquids, $n(\mathbf{k})$ becomes a smooth function with no discontinuity at $|\mathbf{k}| = k_F$. In this case, defining the Fermi surface can become more abstract, and may require other approaches based on the single-particle Green's function or spectral functions, like the spectral function $$A(k,\omega) = -1/\pi Im[G(k,\omega)]$$, and, for example, the location of the maximum in $A(k, \omega=0)$.

In multi-band systems, each band crossing the Fermi level can give rise to distinct Fermi surfaces, each characterized by its discontinuity $\Delta n(\mathbf{k}_F)$ associated with the specific band's Fermi momentum. This is particularly relevant in strongly correlated systems where Fermi surface reconstructions occur due to complex electronic correlations. The experimental observation of this discontinuity is typically achieved using Angle-Resolved Photoemission Spectroscopy (ARPES), which measures the spectral function which provides information on $n(\mathbf{k})$, allowing for the determination of the existence or absence of the discontinuity, therefore, providing crucial experimental evidence for Fermi or non-Fermi liquid behaviors. Near quantum critical points (QCPs), strong fluctuations and diverging effective masses can suppress $Z$ significantly, leading to non-Fermi liquid behavior where $Z \rightarrow 0$ and the quasiparticle description breaks down.

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  • $\begingroup$ Thank you for the detailed answer! I’ve always found it somewhat intriguing that the definition of the Fermi surface can be twofold. On one hand, there’s the definition based on the presence of a sharp discontinuity in the momentum distribution function n(k) at zero temperature. On the other hand, there’s the definition involving the imaginary part of the spectral function, as you described. $\endgroup$
    – John
    Commented Dec 12 at 15:13
  • $\begingroup$ For myself, I understand it as follows: the first definition applies within the framework of Landau Fermi liquid theory at zero temperature, where the sharp discontinuity in n(k) is a hallmark feature. The second definition, based on the spectral function, is more relevant for realistic calculations, such as DFT+DMFT, and experimental methods like ARPES. $\endgroup$
    – John
    Commented Dec 12 at 15:13
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Why is a sharp discontinuity necessary to define the Fermi surface in interacting fermion systems?

Mathematically, if the Fermi surface were not sharp, it wouldn't be a surface. In other words, Fermi surface is an $(d-1)$-dimensional manifold in $d$-dimensional space of states, separating the filled states from the empty ones (at zero temperature.)

Fermi surface implicitly underlies our conventional understanding of material properties. Its absence is a manifestation of unusual properties.
The Fermi surface arises as a result of Fermi-Dirac statistics - at zero temperature all the states below this surface are fill, while all the states above it are empty. The excitations arise from transfer of particles from below Fermi surface to the vacancies above it.

Such behavior had important implications, some of which are:

  • The excitations are fermionic, i.e., similar to classical particles - this ensures success of simple descriptions of material behavior, such as, e.g., Drude model (see Why do Drude/Sommerfeld models even work?). The success of Fermi liquid theory (for which the figure in the OP is) is in showing that excitations in many interacting electron systems are still particle-like, even with renormalized parameters. This is notably not the case in one dimension - Luttinger liquid has bosonic excitations and very different properties.
  • There's no gap. The excitations can be created by moving particles with infinitesimally small energy from just below the Fermi surface to just above it. This implications for how conductance, specific heat and other properties behave like. An immediate example of a gapped material is superconductors, where inexistence of small energy excitations results in lossless current, and other peculiar behaviors.

Remark:
The important reason why the zero-temperature theory is so successful is that the values of Fermi energy in many materials are very high, ranging from tens to hundreds of Kelvin (in temperature units, where the Fermi energy is called Fermi temperature.) E.g., for iron we have $E_F=11.1$eV$=13\cdot 10^4$K. Thus, broadening of Fermi surface of the order of ambient temperature $T=300$K is often negligeable.

This is not the case in doped semiconductors used in industrial applications, where the Fermi energy is shallow. However, in semiconductor nanostructures, typically studied at Kelvin or milli-Kelvin temperatures, the metallic behavior is routinely present.

Related:
If a non-interacting particle behaves like an undamped wave, can an interacting particle behave like a damped wave?
Particle/hole excitations have finite lifetime

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  • $\begingroup$ In reality, due to thermal broadening, the sharp step in the momentum distribution function becomes smeared at finite temperatures, and the discontinuity at the Fermi surface is no longer present. However, despite this thermal effect, we can still experimentally observe the Fermi surface. Techniques such as ARPES allow us to reconstruct the Fermi surface by identifying features like the spectral weight maxima, which remain well-defined and trace the underlying structure of the Fermi surface even at finite temperatures. $\endgroup$
    – John
    Commented Dec 12 at 15:18
  • $\begingroup$ Therefore, it seems to me that sharp-discontinuity-definition is not directly applicable to real materials. $\endgroup$
    – John
    Commented Dec 12 at 15:18
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    $\begingroup$ @John I was careful in my answer to stress that discontinuity characterizes zero temperature case. Yet, this works with high degree of precision at typical temperature of interest, since Fermi energy of a typical metal corresponds to a temperature of a few thousand Kelvin, so thermal broadening inn ambient conditions does not change the qualitative picture. $\endgroup$
    – Roger V.
    Commented Dec 12 at 15:37

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