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

Use a real, even interaction $V_{-\mathbf{k}}=V_{\mathbf{k}}$. Let $\mathcal{V}=L^3$ and $N_{\mathrm{ex}}={\sum }_{\mathbf{k}\ne 0}{b}_{\mathbf{k}}^{†}{b}_{\mathbf{k}}$. In the Bogoliubov theory expansion, keep terms with zero or two noncondensed operators. The zero-momentum term gives $V_0{N}_0(N_0-1)/(2\mathcal{V})$. The terms with two noncondensed operators giveHere $V_0$ is the direct condensate interaction and $V_{\mathbf{k}}$ is its exchange contribution. At fixed total particle number, $N_0=N-N_{\mathrm{ex}}$, soThe second term cancels the direct normal term. Adding the kinetic energy yields the number-conserving Bogoliubov quadratic HamiltonianThe constant is the condensate mean-field theory interaction energy; the normal term describes moving a particle out of the condensate, while the anomalous term creates or annihilates an opposite-momentum pair by exchanging two particles with the condensate. Total particle number remains fixed in the underlying theory; the condensate amplitudes in this notation are treated as classical amplitudes.

For condensate fluctuation power counting, put $\delta =N_{\mathrm{ex}}/N\ll 1$. Formal interaction-amplitude counting gives $H_2-E_{\mathrm{mf}}\sim E_{\mathrm{mf}}\delta$, cubic terms $\sim E_{\mathrm{mf}}{\delta }^{3/2}$ and quartic terms $\sim E_{\mathrm{mf}}{\delta }^2$; these are not uniform bounds on individual modes. Thus the latter are down by $\sqrt{\delta }$ and $\delta$ relative to the quadratic interaction terms. The $N_0-1$ correction is relative $1/N$. Homogeneity eliminates terms linear in nonzero-momentum operators. Small depletion and weak interactions, not low temperature alone, justify the truncation. ^[ref]

For each opposite-momentum pair use the Bogoliubov transformationEquivalently $u_k=\cosh {\theta }_k$, $v_k=\sinh {\theta }_k$, with $\tanh (2{\theta }_k)=B_k/A_k$. The identity $u_k^2-v_k^2=1$ preserves the canonical commutation relation. Vanishing anomalous terms require $B_k(u_k^2+v_k^2)-2A_k{u}_k{v}_k=0$, and hence the Bogoliubov quasiparticle dispersion isSubstitution and normal ordering giveThis is the requested form after putting each $E_k/2$ inside the oscillator term. By momentum-pair counting in bosonic diagonalization, all sums cover the full nonzero-momentum set; no further factor of two is needed. The bosonic Bogoliubov zero-point shift combines the two vacuum sums before taking a continuum limit.

If $V_k\to V_0>0$ smoothly, the low-energy excitations are gapless phonons:For contact repulsion, the healing length $\xi =\hslash /\sqrt{2mn{V}_0}$ separates this regime from the particle-like regime $E_k\simeq {ϵ}_k+n{V}_0$. Stability requires ${ϵ}_k+2n{V}_k>0$ for every nonzero mode. A general momentum-dependent interaction need not share the contact model's entire curve. An unregularized contact model also makes the vacuum-energy shift ultraviolet divergent; a physical finite-range interaction or scattering-length renormalization is needed for that absolute ground-state energy.

In the Bogoliubov quasiparticle vacuum, ${\alpha }_{\mathbf{k}}|0_{\alpha }⟩=0$, but the inverse Bogoliubov transformation gives $⟨b_{\mathbf{k}}^{†}{b}_{\mathbf{k}}⟩=v_k^2$. Consequently the quantum depletion of a Bose condensate isThis expression assumes zero temperature. At nonzero temperature, thermal Bogoliubov quasiparticles add to the depletion:Thus the printed formula is the quantum contribution, or a low-temperature approximation when the thermal contribution is negligible. It is nonnegative for a stable quadratic Hamiltonian and vanishes in the noninteracting zero-temperature limit.

A periodic harmonic chain with $N$ masses $m_0$, spacing $a$ and nearest-neighbour spring constant $K$ hasHamilton's equations give $m_0{\ddot{u}}_j=K(u_{j+1}+u_{j-1}-2u_j)$. The discrete Fourier transform $u_j=N^{-1/2}{\sum }_k{Q}_k{e}^{ikja}$, $p_j=N^{-1/2}{\sum }_k{P}_k{e}^{ikja}$, with $k=2\pi l/(Na)$ in one Brillouin zone, diagonalizes the quadratic form:The harmonic-chain normal-mode reality condition means $Q_k^{†}=Q_{-k}$, not that $Q_k$ and $Q_{-k}$ are independent real coordinates. For canonical quantization, impose $[u_j,p_l]=i\hslash {\delta }_{jl}$, equivalently $[Q_k,P_{k^{\prime }}]=i\hslash {\delta }_{k,-k^{\prime }}$. For $k\ne 0$ setThen the quantum harmonic chain is a set of independent phonon oscillators:The centre-of-mass zero mode of a harmonic chain is the free coordinate $X=N^{-1}{\sum }_j{u}_j$, with $[X,P_{\mathrm{tot}}]=i\hslash$; it cannot be normalized by dividing by $\sqrt{{\omega }_0}$. Retaining it giveswith the corresponding momentum expansion and uniform term $P_{\mathrm{tot}}/N$. The nonzero modes alone give the site commutator $i\hslash ({\delta }_{jl}-1/N)$; the uniform term supplies the missing $1/N$. Each oscillator occupation is a nonnegative integer, and adding one phonon costs $\hslash {\omega }_k$ and crystal momentum $\hslash k$ modulo a reciprocal-lattice vector. ^[ref]

In the continuum limit of a harmonic chain, $\rho =m_0/a$, $Y=Ka$ andThe wave equation and linear low-$k$ dispersion relation follow. The lattice gives an ultraviolet cutoff; the acoustic approximation does not reproduce the zone-edge flattening.

With the centre of mass fixed, the oscillator partition function and Bose-Einstein distribution areFor a long chain, above its finite-size level spacing and below the zone-edge scale, integrating the two acoustic propagation directions gives $C/(Na)=\pi k_{\mathrm{B}}^{2}T/(3\hslash c)$. At high temperature each retained oscillator instead contributes $k_{\mathrm{B}}$ to the heat capacity.

For $H={\hat{p}}^2/(2m)+V(\hat{x})$, divide elapsed time $t$ into $J$ slices $\Delta t=t/J$ and apply the Trotter product formula. Inserting complete position and momentum bases gives the regulated phase-space path integralThe endpoint values are $x_0=x_i$, $x_J=x_f$. Each Gaussian momentum integration in a phase-space path integral can be performed explicitly, leaving the configuration-space path integralIts measure is the time-sliced limit of $(m/(2\pi i\hslash \Delta t){)}^{J/2}{\prod }_{j=1}^{J-1}d{x}_j$, with the usual infinitesimal convergence prescription. The integral sums amplitudes, not positive classical probabilities. Operator ordering remains part of the time-slice definition. Expanding a short-time kernel recovers the Schrödinger equation; composing kernels reproduces the time-evolution operator.

In the classical limit, rapid phase cancellation suppresses paths away from stationary action. Varying $S$ with fixed endpoints gives $m\ddot{x}=-V^{\prime }(x)$. The semiclassical result sums the classical-path phases multiplied by fluctuation functional determinants; for a quadratic Hamiltonian the fluctuation integral is Gaussian and this expansion is exact.

For quantum statistical mechanics, continue $t=-i\tau$ by a Wick rotation. The Euclidean action and thermal trace giveThe trace closes position-space paths. In a many-particle trace one also sums over endpoint permutations with the appropriate exchange signs. In a coherent state field formulation, bosonic fields are periodic and fermionic coherent states instead give antiperiodic Grassmann fields, hence bosonic and fermionic Matsubara frequencies $2\pi n/(\beta \hslash )$ and $(2n+1)\pi /(\beta \hslash )$. ^[ref]

The quantum-to-classical imaginary-time mapping views a local $d$-dimensional quantum problem as a classical-looking field theory in $d+1$ dimensions, with imaginary-time extent $\beta \hslash$. A genuine classical probability interpretation additionally requires a nonnegative weight; fermion signs and Berry terms can obstruct it.

For a smooth potential in the high-temperature or small-$\hslash$ limit, imaginary-time paths fluctuate only over the thermal de Broglie wavelength ${\lambda }_T=\sqrt{2\pi \beta {\hslash }^2/m}$. If this is short compared with the potential's variation scale, the diagonal free kernel givesThis classical phase-space limit of the thermal kernel is the classical phase space partition function for one degree of freedom. A dilute identical-particle gas also needs its $1/N!$ factor; when exchange is appreciable, classical statistics fails. As a check, a single quantum harmonic oscillator gives $Z=[2\sinh (\beta \hslash \omega /2){]}^{-1}\to 1/(\beta \hslash \omega )$, precisely the classical phase-space integral.

For conventional superconductivity, an attractive interaction pairs opposite-spin fermions. A complex pairing Hubbard–Stratonovich transformation introduces a Cooper pair field $\Delta$. The fermionic fields then appear quadratically, giving the pair-field effective action of a superconductorThe trace convention counts each independent spin-pair block once. The uniform saddle is the BCS gap equation; the fermionic excitations have $E_k=\sqrt{{\xi }_k^2+|\Delta {|}^2}$ in the fully gapped case. A slow-field expansion near the transition yields Ginzburg-Landau theory of superconductivity. Normalize its pair field $\Psi$ so $|\Psi {|}^2$ is an effective pair density, with pair mass $m_{\ast }$ and signed charge $q=-2e$:For $a<0$, the uniform minimum has ${\rho }_0=|\Psi {|}^2=-a/b$, with arbitrary constant phase. Writing $\Psi =(\sqrt{{\rho }_0}+h)e^{i\theta }$ separates the amplitude mode $h$ from the phase. The local potential has curvature term $2|a|h^2$ and coherence scale $\xi =\hslash /\sqrt{2m_{\ast }|a|}$. A neutral condensate has a gapless phase Goldstone mode when suitable dynamics is included.

The charged theory is invariant underThe gauge-invariant condensate momentum is $\hslash \nabla \theta -q\mathbf{A}$. Only $\hslash \nabla \theta -q\mathbf{A}$ and $\hslash \dot{\theta }+q\varphi$ enter physical phase dynamics. At nearly constant amplitude, varying $F$ givesFor a smooth phase, taking the curl and using Ampère's law gives ${\nabla }^2\mathbf{B}=\mathbf{B}/{\lambda }^2$: the Meissner effect screens a static bulk magnetic field over the London penetration depth $\lambda$. In a gauge with $\theta =0$, the phase stiffness is a quadratic term in $\mathbf{A}$, the superconducting version of the Higgs mechanism.

To identify the dynamical mode, a simple lossless long-wave superconducting phase-density action isHere ${\chi }_p$ is the pair compressibility coefficient. Eliminating the longitudinal electric potential givesThus the neutral sound mode becomes a gapped longitudinal plasma oscillation; transverse electromagnetic waves acquire ${\omega }^2=c_{\mathrm{vac}}^2{k}^2+c_{\mathrm{vac}}^2/{\lambda }^2$ in this simple bulk model. This redistribution of the phase degree of freedom into electromagnetic collective modes is the Anderson-Higgs mechanism in a superconductor. Local gauge symmetry is a redundancy, not an independently observable symmetry whose breaking creates another massless particle. ^[ref]

The static functional alone does not determine a frequency or damping rate. The displayed mode dispersions require the added dynamical assumptions; real screening, normal carriers, boundaries and dissipation modify them. The amplitude fluctuation is distinct from the phase mode and is not removed by the gauge choice.

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.