Past exam of the quantum Condensed Matter Field Theory course of the University of Cambridge / 2013 / Paper 1 / 3

Use $\hslash =k_{\mathrm{B}}=1$ and dimensionless spin operators. At one site, the canonical anticommutation relations imply the on-site spin-square normal-ordering identityIndeed the empty and doubly occupied states have spin zero, while either singly occupied state has ${\mathbf{S}}^2=3/4$. Thus the interaction in normal-ordered form is $3U{\sum }_n{c}_{n\uparrow }^{†}{c}_{n\downarrow }^{†}{c}_{n\downarrow }{c}_{n\uparrow }$. There is no residual $3Un/2$ chemical-potential shift: that contraction has cancelled.

Insert fermionic coherent state resolutions between imaginary-time evolution factors. With ${\xi }_k={ϵ}_k-\mu$ and $D_n={\bar{\psi }}_{n\uparrow }{\bar{\psi }}_{n\downarrow }{\psi }_{n\downarrow }{\psi }_{n\uparrow }$, the grand-canonical coherent-state action isThe independent barred and unbarred Grassmann variables satisfy antiperiodic boundary conditions at $0,\beta$. Adjacent time-slice labels in the normal-ordered symbol fix the continuum prescription and thermal normalization. The interaction coefficient is $3U$, not the $U$ of a differently normalized Hubbard model.

Expand the Grassmann fields in fermionic Matsubara frequencies ${\omega }_l=(2l+1)\pi /\beta$. For $U=0$, the Grassmann Gaussian integral factorizes into thermal fermionic one-level thermal determinants:Each factor counts the empty and occupied states of one spin orbital. Equivalently, the regularized Matsubara sum gives the logarithm of that factor; the raw infinite determinant requires thermal normalization of a fermion determinant. ThereforeThe printed $F$ is the grand potential $\Omega$, because the trace contains $H-\mu \hat{N}$. The Helmholtz free energy at fixed mean particle number is $F_{\mathrm{H}}=\Omega +\mu ⟨\hat{N}⟩$.

Define the Grassmann spin density ${\mathbf{j}}_n={\bar{\psi }}_n\mathbf{\sigma }{\psi }_n$. The Pauli matrices and Grassmann reordering giveFor example $j_z^2=-2D_n$; each of the other two components gives the same result. For $U>0$, complete the square independently at each site and time slice:This normalized spin-vector Hubbard–Stratonovich decoupling yieldsThe auxiliary bosonic quantum field is real and periodic, while the fermionic fields remain antiperiodic. All three vector components are required to reproduce the original $3U$ interaction; replacing this exact transformation by a single-component identity would change its coefficient.

For the uniform mean-field approximation field $\mathbf{M}=M{\hat{\mathbf{e}}}_z$, the inverse fermion propagator is diagonal in momentum, frequency and spin: $-i{\omega }_l+{\xi }_k-\sigma M$. The Grassmann Gaussian integral therefore givesHere $d\mu (M)$ denotes the residual uniform-field measure; integration over its orientation only affects a nonextensive prefactor. HenceThe exchange field separates the two spin bands by $2M$. Integrating out the fermions is exact for a specified auxiliary field; restricting that field to be constant is the mean-field approximation.

Differentiate the uniform effective action:To reproduce the printed coefficient, define the spin-summed density of states per site byThe saddle-point approximation then givesThe uniform spin field and magnetization obey $M=U(n_{\uparrow }-n_{\downarrow })=2U⟨S_z⟩$ for occupations per site. If a one-spin density of states were used instead, the factor $1/2$ would be absent. The spin field is an energy, rather than directly a magnetic moment.

The constant term in the spin sum cancels. Linearizing the saddle point equation givesThe paramagnetic auxiliary-field curvature isConsequently the Stoner criterion for the loss of local paramagnetic stability isThe printed $\nu (0)$ uses energy measured from the Fermi level. For a smooth density of states, $-n_F^{\prime }$ becomes a narrow unit-area peak at $\mu$; a singular density of states or appreciable temperature requires the integral form. The curvature establishes local instability, not the order or location of every possible thermodynamic transition.

With exact spin-rotation symmetry, itinerant ferromagnetism selects a magnetization direction. The two broken transverse generators form one quadratic type-B Goldstone boson, a long-wave ferromagnetic magnon with $E_q\sim D{q}^2$. Longitudinal amplitude motion is not symmetry-protected to be gapless. In a metallic ordered state, particle-hole excitations remain near each surviving exchange-split Fermi surface; these are distinct from the collective magnon. The itinerant spin-flip particle-hole continuum starts at exchange scale $2|M|$ at $q=0$ in this simple mean-field spectrum, while its finite-$q$ boundaries differ. Spin anisotropy, spin-orbit coupling or an external field can gap the transverse mode; none is present in the stated model.