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 Incoming links: Particle-conserving displacement of a Fermi circle

Past exam of the quantum matter course of the University of Oxford / 2023 / 2 / c / Solution 2026-10-25

The shifted Fermi circle satisfies

∣ p - p_{0} \hat{x} ∣ = p_{F}

. Its radial boundary at polar angle

θ

is the positive solution of

p^{2} - 2 p p_{0} cos θ + p_{0}^{2} = p_{F}^{2}

. Thus, for

p_{0} < p_{F}

p_{F} + ν (θ) = p_{0} cos θ + p_{F}^{2} - p_{0}^{2} sin^{2} θ,

(1)

and its leading Taylor expansion is

ν (θ) = p_{0} cos θ + O (p_{0}^{2} / p_{F}) .

(2)

Forward-pointing momenta gain occupied states and backward-pointing momenta lose them. With the paper's explicitly reversed Heaviside step function convention,

Θ^{'} (x) = - δ (x)

; consequently

δ n (p) = ν (θ_{p}) δ (p - p_{F}) + O (p_{0}^{2}),

(3)

with a positive leading sign.

For the later quadratic energy calculation it is useful to distinguish the first-order shape from exact particle number conservation. The particle-conserving displacement of a Fermi circle has

ν (θ) = p_{0} cos θ - \frac{p _{0}^{2}}{2 p _{F}} sin^{2} θ + O (p_{0}^{4} / p_{F}^{3}), δ N = C \int_{0}^{2 π} [p_{F} ν + \frac{ν ^{2}}{2}] d θ = 0.

(4)

The second-order inward correction cancels the apparent area increase one would obtain by treating

ν = p_{0} cos θ

as exact.

Figure 1.
Translation of a Fermi circle and its radial displacement
. Original geometric construction with $p_{0} / p_{F} = 0.25$ . The Fermi sea keeps its area under translation; the first-order radial approximation differs from the exact boundary at second order. **Details**.

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