Fermi Liquid Theory

Studying the Fermi liquid theory

Here are the questions that I have, and they are answered by Claude. The answers are not copied and pasted directly from the site. Instead, I included some modifications based on my understanding.

Q1. How did the scattering phase space in the Fermi surface become zero in the second order?

This results from the energy conservation and Pauli’s principle. If the quasiparticle $1$ scatters off $2$ to produce $3$ and $4$, the energy conservation gives $\varepsilon_1 + \varepsilon_2 = \varepsilon_3 + \varepsilon_4$. In addition, Pauli’s principle states that the final states $3$ and $4$ must be empty ($\varepsilon_3, \varepsilon_4 > 0$), while the initial state $2$ must be filled. Since $0 < \varepsilon_3, \varepsilon_4 < \varepsilon$, there are $\varepsilon^2$ possible states. This limits the phase space. Therefore, $1/\tau \propto \varepsilon^2$ and the scattering on the Fermi surface compresses.

Additional question. Is chaotic scattering neglected by this approach? No. The adiabatic continuity suppresses the scattering near the Fermi surface. The chaotic scattering becomes important away from the Fermi surface, which breaks the Fermi liquid theory.

Q2. Why does occupancy parameterize the excitation energy?

The Fermi liquid theory assumes that the low-energy eigenstates are labeled by $\mathbf{p}$ and $\sigma$. Here, the labeling indicates that there is some operator that commutes with the Hamiltonian and forms the compatible set of commuting operators. In addition, labeling by momentum $\mathbf{p}$ and spin $\sigma$ implicitly indicates that there is translational symmetry and $SU(2)$ symmetry. The operator that satisfies these conditions is the density operator: $n^{(0)}_{\mathbf{p}\sigma} = c^{\dagger}_{\mathbf{p}\sigma} c_{\mathbf{p}\sigma}$.

However, if the interaction $V$ turns on, $[n^{(0)}_{\mathbf{p}\sigma}, V] \neq 0$ and the eigenstates generally cannot be labeled by $\mathbf{p}$ and $\sigma$. Here comes the adiabatic connection. It connects the eigenstate of noninteracting Hamiltonian $\ket{\Psi^{(0)}_{n}}$ and that of interacting one:

\[\begin{equation} \ket{\Psi_{n}} = U(0,-\infty)\ket{\Psi^{(0)}\_{n}} \end{equation}\]

If a gap is opened, there is no topological phase transition or spontaneous symmetry breaking during the adiabatic connection, the same labeling $n_{\mathbf{p}\sigma}$ is allowed for interacting system. This labeling, however, becomes significant only when the quasiparticle peaks $Z_{\mathbf{p}}\delta(\omega-\varepsilon^{*}_{\mathbf{p}})$ are well-defined. It alternatively implies that it captures the physics for the low-energy excitations near the Fermi surface. That is why one expands the energy around the ground state density. As a matter of fact, the quasiparticle energy is written as a functional of $\delta n_{\mathbf{p}\sigma} = n_{\mathbf{p}\sigma} - n^{(0)}_{\mathbf{p}\sigma}$:

\[\begin{equation} \mathcal{E} = \mathcal{E}_{0} + \sum_{\mathbf{p}\sigma}(E^{(0)}\_{\mathbf{p}\sigma}-\mu)\delta n_{\mathbf{p}\sigma} + \frac{1}{2}\sum_{\mathbf{p}\sigma,\mathbf{p}'\sigma'}f_{\mathbf{p}\sigma,\mathbf{p}'\sigma'}\delta n_{\mathbf{p}\sigma}\delta n_{\mathbf{p}'\sigma'} \end{equation}\]

where $E^{(0)}_{\mathbf{p}\sigma}-\mu = \delta\mathcal{E}/\delta n_{\mathbf{p}\sigma}$ and $f_{\mathbf{p}\sigma,\mathbf{p}’\sigma’} = \delta^{2} \mathcal{E} / \delta n_{\mathbf{p}\sigma}\delta n_{\mathbf{p}’\sigma’}\vert_{\delta n_{\mathbf{p}’‘\sigma’’} = 0}$. Naturally, the quantity $\delta n_{\mathbf{p}\sigma}$ is nonzero only on the shell of Fermi surface, because the excitations occur on the Fermi surface. In addition, Fermi liquid function $f_{\mathbf{p}\sigma,\mathbf{p}’\sigma’}$ is expanded as Legendre polynomials. This is because of the spherical symmetry implicilty assumed in Fermi liquid theory.

Additional note. The fact that symmetry is lowered does not mean that labeling is entirely impossible. If there is discrete translational symmetry such as a lattice, $\mathbf{p}$ will be within the Brillouin zone, and if spin-orbit coupling is introduced, total angular momentum can be used instead of $\sigma$. However, if symmetry is spontaneously broken (e.g. ferromagnet), $Z_{\mathbf{p}} \sim 0$ (Mott insulator), or there is a topological phase transition (Cooper condensation), the assumption does not work and Fermi liquid theory fails.

RG perspective. In renormalization group language, the second order $\delta n$ remains as marginal. The higher-order $\delta n$ is irrelevant and converges to zero in the low-energy sector (Shankar 1994, Polchinski 1992).

Q3. What is the difference between $f_{\mathbf{p}\sigma,\mathbf{p}’\sigma’}$ and $a_{\mathbf{p}\sigma,\mathbf{p}\sigma’}$?

The fundamental difference lies whether the parameter reflects the response of the Fermi sea. The Fermi liquid parameter $f_{\mathbf{p}\sigma,\mathbf{p}’\sigma’}$ contains the interaction which assumes that the Fermi sea is frozen. However, the scattering amplitude $a_{\mathbf{p}\sigma,\mathbf{p}’\sigma’}$ allows the Fermi sea to respond to the interaction. It is given by the solution of Bethe-Salpeter equation. Physically, $a_{\mathbf{p}\sigma,\mathbf{p}’\sigma’}$ indicates the screened scattering amplitude actually measured by an experiment, while $f_{\mathbf{p}\sigma,\mathbf{p}’\sigma’}$ explains the energy change when a quasiparticle changes the occupancy.

Q4. Is it safe to understand ‘zero sound’ as fast-moving collective motion of electrons?

The zero sound is the collective motion of quasiparticles and is propagated without collisions. Also, it moves ‘fast.’ However, rather than the fast-moving collective motion, its meaning is close to the deformation of the Fermi surface. The zero sound is the “oscillation of the Fermi surface” with a constant group velocity.

Q5. Is the difference between charged and neutral Fermi liquid placed in the presence of nonlocal Coulomb interaction?

Yes. The neutral Fermi liquid (e.g. $^{3}$He) has the finite interaction $V(q)$ at $q\rightarrow 0$. In addition, the Fermi liquid parameter is well-defined. However, the charged Fermi liquid (e.g. the electron system) has the divergent interaction $V(q)$. Silin resolved this by separating the long-range component $e^{2}/(\epsilon_{0}q^{2})$ from the Fermi liquid parameter:

\[\begin{equation} f_{\mathbf{p}\sigma,\mathbf{p}'\sigma'}(\mathbf{q}) = \frac{e^{2}}{\epsilon_{0}q^{2}} + f_{\mathbf{p}\sigma,\mathbf{p}'\sigma'} \end{equation}\]

where the short-range component $f_{\mathbf{p}\sigma,\mathbf{p}’\sigma’}$ is handled by Landau Fermi liquid theory. In addition, $f_{\mathbf{p}\sigma,\mathbf{p}’\sigma’}$ is mediated by “quantum fluctuations” of quasiparticles.

Q6. Why $\chi_{c}(\mathbf{q}) \sim \chi_{n}$ for $q \ll p_{F}$?

In the charged Fermi liquid, $F^{s}_{0}$ is replaced by $F^{s}_{0}(\mathbf{q}) = e^{2}\mathcal{N}^{*}(0) / \epsilon_{0}q^{2} + \bar{F}^{s}_{0}$. Since the compressibility for the neutral system is given by $\chi_{n} = \mathcal{N}^{*}(0)/(1+\bar{F}^{s}_{0})$. The one for the charged Fermi liquid is written as:

\[\begin{equation} \chi_{c}(\mathbf{q}) = \frac{\mathcal{N}^{*}(0)}{1 + (e^{2}\mathcal{N}^{*}(0)/\epsilon_{0}q^{2} + \bar{F}^{s}_{0})} = \frac{\chi_{n}}{1 + \kappa^{2}/q^{2}} \end{equation}\]

where $\kappa^{2} = e^{2}\chi_{n}/\epsilon_{0}$ is the Thomas-Fermi wavenumber. If the Fermi liquid is neutral, $\kappa = 0$ and the compressibility is reduced to the one for the neutral Fermi liquid. In addition, $\chi_{c}(\mathbf{q})$ acts as the neutral one for large $q$; $\chi_{c}(\mathbf{q}) \sim \chi_{n}$ for $q \gg \kappa$.