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BCS theory describes a superconducting instability caused by attractive interactions between opposite-momentum fermions near a Fermi surface.

A Fermi liquid is an interacting fermion phase whose low-energy excitations are long-lived quasiparticles near a Fermi surface. Its effective interactions are classified by scaling only the frequency and momentum normal to that surface.

An isotropic spin-polarized metal has a degenerate magnetization direction and a long-wavelength transverse magnon with quadratic dispersion. Nonzero spin density pairs the two transverse broken generators into one type-B Goldstone boson. Exchange-split Fermi surfaces also support fermionic low-energy excitations; the magnon is distinct from longitudinal amplitude motion and the spin-flip continuum.

A Poisson manifold is a smooth manifold with a smooth antisymmetric bivector $\pi =\frac{1}{2}{\pi }^{ij}{\partial }_i\wedge {\partial }_j$ whose function bracket $\{f,g\}={\pi }^{ij}({\partial }_{i}f)({\partial }_{j}g)$ satisfies the Jacobi identity. The bracket is automatically a derivation in each argument. A symplectic manifold has a nondegenerate differential two-form $\omega$ satisfying $d\omega =0$, that is, a closed differential form; nondegeneracy forces its dimension to be even. It determines a nondegenerate Poisson structure. We use matrix convention $\pi =-{\omega }^{-1}$, so that ${\iota }_{X_H}\omega =dH$ corresponds to $X_H(f)=\{f,H\}$.

Examples of genuinely nonsymplectic Poisson geometries include the zero bracket on ${\mathbb{R}}^3$ and the Lie-Poisson bracket $\{x_i,x_j\}={ϵ}_{ijk}{x}_k$ on ${\mathbb{R}}^3$. Their three-dimensional underlying manifold cannot carry a nondegenerate two-form. In the latter case the Poisson rank is two away from the origin and zero at the origin, with spheres and the origin as symplectic leaves. Even in even dimension a given Poisson structure need not come from a symplectic form: $\pi =x{\partial }_x\wedge {\partial }_y$ on ${\mathbb{R}}^2$ satisfies Jacobi but vanishes along $x=0$. The statement concerns this degenerate structure, not a claim that ${\mathbb{R}}^2$ admits no other symplectic structure.

For any candidate bivector, applying the Jacobi identity to coordinate functions gives the necessary conditionIt is sufficient as well: expanding $\{f,\{g,h\}\}$ and its cyclic companions cancels all terms with second derivatives by antisymmetry of $\pi$. What remains is $J^{ijk}({\partial }_{i}f)({\partial }_{j}g)({\partial }_{k}h)$. This is the coordinate Jacobi condition for a Poisson bivector.

For a symplectic form, differentiate $\omega \pi =-I$ to obtain ${\partial }_{\ell }\pi =\pi ({\partial }_{\ell }\omega )\pi$. Substituting this into the cyclic expression givesbecause $d\omega =0$. Thus the inverse symplectic structure always satisfies the Jacobi identity. Conversely, for a nondegenerate Poisson structure, multiplying by three inverse matrices proves closure of its associated two-form.

In momentum space, choose Hamiltonian $H(\mathbf{p})=E(\mathbf{p})$ and the constant Poisson bivectorIts Poisson bracket is $\{f,g\}=e\mathbf{B}\cdot (\nabla f\times \nabla g)$. The Hamiltonian vector field therefore givesexactly the prescribed motion. The sign here uses the signed charge $e$ as printed; an electron's physical charge may be negative. Since the magnetic field is uniform, every derivative of ${\pi }^{ij}$ vanishes, so every term in $J^{ijk}$ is zero. Jacobi holds independently of the dispersion function $E$. This is the momentum-space Poisson structure for a uniform magnetic field.

For $e\mathbf{B}\ne 0$ this bivector has rank two and $C(\mathbf{p})=\mathbf{B}\cdot \mathbf{p}$ is a Casimir function of a Poisson manifold: $\{C,f\}=0$ for every $f$. Its symplectic leaves are the planes $C=c$. For axes with $\mathbf{B}=B_0{\mathbf{e}}_3$, the leaf form is $\omega =(e{B}_0{)}^{-1}d{p}_1\wedge d{p}_2$ in our inverse-sign convention. Also $\dot{E}=\{E,E\}=0$, so trajectories on a Fermi surface lie on its intersections with these planes, typically curves. The energy surface is invariant for this Hamiltonian, but need not be a Poisson submanifold; the bracket has been defined on the whole momentum space. If $e\mathbf{B}=0$, the Poisson structure and motion vanish.

Bloch theorem states that energy eigenfunctions in a Bravais lattice potential may be chosen asThe wavevector is defined modulo a reciprocal lattice vector, so one can restrict it to the first Brillouin zone. This follows because lattice translations commute with the Hamiltonian and have simultaneous eigenvalues $e^{i\mathbf{k}\cdot \mathbf{l}}$.

In one dimension write $V(x)={\sum }_G{V}_G{e}^{iGx}$ with $G=2\pi j/a$ and expand a Bloch state as ${\psi }_k={\sum }_G{c}_G{e}^{i(k+G)x}$. The Schrödinger equation becomesFor zero potential, the parabolas $E_G^0(k)={\hslash }^2(k+G{)}^2/(2m)$ fold into the Brillouin zone. A weak potential shifts a nondegenerate level by $V_0$ at first order, with second-order correction ${\sum }_{G\ne 0}|V_G{|}^2/(E_0^0-E_G^0)$ where the denominators stay away from zero. This expansion fails near crossings, where degenerate perturbation theory is required.

At the first zone boundary, put $G=2\pi /a$ and $k=G/2+q$. The nearly degenerate waves with momenta $\hslash k$ and $\hslash (k-G)$ are coupled by $V_G$. Retaining their two-dimensional subspace givesDiagonalizing producesAt $q=0$ there is a gap $\boxed{\Delta E=2|V_G|}$: the two standing-wave combinations sample the periodic potential differently. Similar Bragg degeneracies occur at $k=j\pi /a$, with gaps controlled to first order by the corresponding Fourier components. If that component vanishes, a gap need not open at first order. Between the gaps the continuous energy eigenvalues form energy bands, periodically represented in the reduced-zone scheme. For the first zone edge, the curvature of the upper and lower branches gives a minimum and a maximum respectively for a sufficiently weak nonzero potential, so the avoided crossing is a genuine local forbidden energy interval.

At zero temperature electrons occupy states up to the Fermi energy. The Fermi surface is the set $E_n(\mathbf{k})=E_F$ separating occupied from unoccupied states within partially filled bands (Fermi points in one dimension). A partially filled band supplies arbitrarily nearby empty states, allowing occupations to respond to a weak electric field and conduct. Completely filled bands give no net current from such a redistribution; if the next empty band is separated by a positive band gap, the system is a band insulator. Band overlap can instead yield a metal even at an integer band filling. At finite temperature an insulator can conduct through thermally excited electrons and holes.

The spin-independent term cancels in the signed sum. The linearized stationary equation becomesThe even powers vanish because the equation is odd under $M\mapsto -M$. Equivalently, the mean-field action has curvatureThus the unpolarized stationary point becomes unstable whenAt low temperature, the derivative of the Fermi distribution samples a narrow window around $\mu$, so ${\chi }_0\simeq \nu (\mu )$ for a smooth density of states. Measuring energy from the Fermi level gives $U_c\nu (0)\simeq 1$, the Stoner criterion in the spin-summed convention fixed above. This identifies local instability of the paramagnetic state; the detailed transition order also depends on higher terms in the action.

In the isotropic ordered phase of itinerant ferromagnetism, continuous spin-rotation symmetry leaves a degenerate direction of magnetization. Its transverse fluctuations produce one gapless, quadratically dispersing mean-field itinerant ferromagnetic Goldstone mode, $E(\mathbf{q})\propto q^2$. The two broken spin generators form a canonical pair, as for the type-B Goldstone boson of the localized ferromagnet. The system also remains metallic, with exchange-split Fermi surfaces and low-energy spin-conserving particle-hole excitations. Longitudinal-amplitude fluctuations and the spin-flip continuum are distinct from the long-wavelength magnon; their energy and damping scales depend on the band structure. Spin anisotropy, if added, could gap the transverse mode, but it is absent from the stated Hamiltonian.

Under Fermi-surface renormalization-group scaling, tangential momentum $k$ is fixed whileIn frequency space the quadratic action is invariant whenequivalently, in the time representation $t\mapsto s^{-1}t$ and $\psi (t,k,\ell )\mapsto s^{-1/2}\psi (t,k,\ell )$. The free Fermi liquid action is therefore marginal.

The deformation $\mu (k){\psi }^{†}\psi$ has no factor of frequency or normal momentum, so it is relevant. It shifts the zero of the quasiparticle energy from $\ell =0$ to $\ell =\mu (k)/v_F(k)$ and hence deforms the Fermi surface.

A generic quartic interaction is irrelevant because momentum conservation fixes a normal component and leaves an additional positive power of $s$. It can be marginal only when all four momenta remain on the Fermi surface while satisfyingFor a smooth generic Fermi surface, the robust possibilities are forward or exchange scattering and the opposite-momentum BCS channel ${\mathbf{k}}_2=-{\mathbf{k}}_1$, ${\mathbf{k}}_4=-{\mathbf{k}}_3$.