Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 81 / 4

Let $D_n={\hat{n}}_{n\uparrow }{\hat{n}}_{n\downarrow }$. On a site, empty and doubly occupied states have spin zero, while either singly occupied state has spin $1/2$. ThusThe on-site spin-square normal-ordering identity rewrites the interaction as $3U{\sum }_n{D}_n$. Its contact coefficient is $3U$ in this parameterization, rather than the $U$ convention often used for the Hubbard model.

Using a fermionic coherent-state trace and time slicing, replace each normal-ordered monomial by its Grassmann field symbol. With ${\xi }_{\mathbf{k}}={ϵ}_{\mathbf{k}}-\mu$,The independent fields have antiperiodic thermal boundary conditions. Normal ordering is crucial: the explicit $3\hat{n}/4$ subtraction removes the spin-square contraction, so there is no extra chemical-potential shift from it.

At $U=0$, use fermionic Matsubara frequencies ${\omega }_r=(2r+1)\pi /\beta$. The fermionic Gaussian integral has inverse propagator $-i{\omega }_r+{\xi }_{\mathbf{k}}$. With its thermal normalization, the determinant for one single-particle level is $1+e^{-\beta \xi }$, as also follows from its empty and occupied states. Two spin species therefore giveThe expression labeled Helmholtz free energy in the PDF is the grand potential, since the trace contains $H-\mu \hat{N}$. The thermodynamic Helmholtz free energy at a prescribed electron number $N_e$ is $F={\Omega }_0+\mu N_e$, with $\mu$ fixed by $N_e=2{\sum }_{\mathbf{k}}{n}_F({ϵ}_{\mathbf{k}})$. The Matsubara determinant requires a normalization prescription; a raw infinite sum of logarithms is not an ordinary convergent sum.

Define ${\mathbf{j}}_n={\bar{\psi }}_n\mathbf{\sigma }{\psi }_n$ and $D={\bar{\psi }}_{\uparrow }{\bar{\psi }}_{\downarrow }{\psi }_{\downarrow }{\psi }_{\uparrow }$. The Pauli matrix completeness identity, or direct evaluation of its three components, gives ${\mathbf{j}}^2=-6D$. Consequently the quartic action is $3UD=-U{\mathbf{j}}^2/2$.

The Hubbard–Stratonovich decoupling follows by completing a three-component square at each site and time slice:The constant $C$ is absorbed into the normalized auxiliary-field measure, and $U>0$ ensures convergence of the real Gaussian integral. Thus the spin-vector Hubbard–Stratonovich decoupling yieldsIntegrating out $\mathbf{M}$ reconstructs the normal-ordered interaction exactly. The auxiliary field is periodic in imaginary time, while the fermionic fields remain antiperiodic.

In the homogeneous spin-field sector, choose $\mathbf{M}=M{\hat{\mathbf{e}}}_z$. The fermionic levels become ${ϵ}_{\mathbf{k}}-\mu -\sigma M$, where $\sigma =\pm 1$ is the Pauli eigenvalue. The fermionic one-level thermal determinant can be applied independently to every momentum and spin, while the field cost is $\beta N{M}^2/(2U)$. Hence the homogeneous spin-field effective action isThe remaining homogeneous-field integral is $\mathcal{Z}\simeq \int \mathcal{D}M{e}^{-{\mathcal{S}}_{\mathrm{eff}}}$ in this mean-field restriction. Orientation and fluctuation-independent measure factors do not alter the displayed extensive action or its stationary equation. The two bands are exchange split by $2M$; $M$ has energy units and is proportional to the spin polarization, rather than being an electron number itself.

Differentiate the homogeneous spin-field effective action:At a stationary field, $M=(U/N){\sum }_{\mathbf{k},\sigma }\sigma n_F({ϵ}_{\mathbf{k}}-\sigma M)$. Define the spin-summed density of states per lattice site byThen the requested form isThe factor one half fixes the spin convention. With a single-spin density of states ${\nu }_s=\nu /2$, the prefactor would instead be $U$. In terms of site occupations, the same equation is $M=U(n_{\uparrow }-n_{\downarrow })=2U⟨S^z⟩$. The Fermi distribution includes the chemical potential in its argument definition; it must not be subtracted a second time.

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.