Past exam of the quantum matter course of the University of Oxford / 2023 / 2 / e

Use the total density of states $D^{\ast }=A{m}^{\ast }/(2\pi {\hslash }^2)$ and $C=A/(2\pi \hslash {)}^2$ from part (b). In the paper's double-sum convention, $f(\varphi )$ has units of energy and scales as $A^{-1}$. Define its first angular moment byThis convention matters: the Landau quasiparticle energy functional written with an explicit $1/A$ instead uses the intensive kernel $\tilde{f}=Af$.

At zero temperature, the displayed functional is the change in $E-\mu N$; a true Galilean boost preserves $N$, so $\Delta \mathcal{F}=\Delta E$. For a radial displacement $\nu (\theta )$, the first term in the Landau quasiparticle energy functional becomesThe oriented radial integral also works for negative $\nu$: removing occupied levels below the chemical potential costs positive energy. Substituting $\nu =p_0\cos \theta +O(p_0^2/p_F)$ and using $N=C\pi p_F^2$ givesThe exact second-order correction to the boundary is unnecessary here because the unperturbed excitation energy vanishes on the Fermi surface; it changes this term only beyond quadratic order. It must nevertheless be understood when identifying $\Delta \mathcal{F}$ with a fixed-particle number energy change.

For the quadratic interaction term, the first-order occupation change alone suffices. The radial Dirac delta function integrals givePut $\varphi =\theta -{\theta }^{\prime }$. The remaining angular integral usesso only the first cosine component of the interaction survives. Since $D^{\ast }=2\pi C{m}^{\ast }$,Adding both contributions and matching the bare-mass result of part (a) yields the two-dimensional Landau effective-mass relation:Equivalently, using the density of states per unit area $g^{\ast }=D^{\ast }/A$,If dimensionless Landau parameters are defined through the Fourier series $D^{\ast }f(\varphi )=F_0+{\sum }_{\ell \ge 1}{F}_{\ell }\cos (\ell \varphi )$, then $D^{\ast }{\overline{f}}_1=F_1/2$ andDefining the cosine expansion with an extra factor of two would instead absorb that factor into $F_1$; the angular-integral formula above fixes the convention unambiguously. The physical content is that interaction changes accompanying a moving Fermi sea compensate the renormalized quasiparticle velocity, so the total boost energy is governed by the bare mass. For an interaction independent of angle, ${\overline{f}}_1=0$ and this consistency condition requires $m^{\ast }=m$.