In a heavy fermion compound like CeAl₃, the electronic specific heat coefficient γ ≈ 1600 mJ/(mol·K²), compared to ~1 mJ/(mol·K²) for a simple metal like copper. What causes this 1000-fold enhancement?
AThe cerium atoms are much heavier, increasing the electron effective mass through electron-nucleus coupling
BThe hybridization between localized Ce 4f electrons and itinerant conduction electrons, via the Kondo effect at every site, creates a very narrow quasiparticle band at E_F with an enormous density of states g*(E_F) ∝ m*. Since γ = (π²/3)k_B²g*(E_F), the specific heat coefficient is proportional to m* and thus ~1000× enhanced
CThe crystal structure of CeAl₃ creates flat phonon bands
DStrong electron-electron repulsion increases the specific heat
The mechanism is the Kondo lattice effect. Each Ce site has a 4f moment that couples antiferromagnetically to the conduction electrons. At temperatures below the Kondo temperature T_K (typically 1-10 K for these materials), the f-moments are screened and form a coherent, very narrow band of heavy quasiparticles at E_F. The bandwidth is ~k_BT_K ~ 1 meV, compared to ~1 eV for normal conduction bands. The 1000× narrower band means a 1000× higher density of states and effective mass. This is Fermi liquid theory pushed to its extreme: the quasiparticles are 'real' (with well-defined Fermi surfaces observed by dHvA), but enormously massive.
Question 2 Multiple Choice
Heavy fermion systems often exhibit quantum critical points (QCPs) where a magnetic ordering temperature is tuned to zero by pressure or field. What happens to the Fermi liquid description near a QCP?
AFermi liquid theory becomes more accurate near a QCP
BThe quasiparticle effective mass diverges and the quasiparticle lifetime shrinks to zero at the QCP — Fermi liquid theory breaks down. The system enters a non-Fermi-liquid regime with anomalous power laws: resistivity ∝ T (instead of T²), specific heat ∝ T log(T) or T^α with α < 1, and divergent susceptibility. Quantum critical fluctuations on all energy scales replace the well-defined quasiparticle picture
CThe system simply transitions from one Fermi liquid to another
DA QCP has no effect on the heavy fermion character
Quantum critical points are one of the most active topics in condensed matter. At a QCP, the ordering temperature vanishes and quantum fluctuations (not thermal fluctuations) drive the critical behavior. The critical fluctuations extend up to finite temperatures (creating a 'quantum critical fan'), producing non-Fermi-liquid behavior over a wide region of the phase diagram. Heavy fermion systems are ideal platforms for studying QCPs because the energy scales (T_K, magnetic ordering temperatures) are small enough to be tuned by accessible pressures and fields.
Question 3 Short Answer
In the 'Doniach phase diagram' for Kondo lattices, there is a competition between the Kondo effect (which screens local moments) and RKKY interaction (which orders them magnetically). What determines the winner?
Think about your answer, then reveal below.
Model answer: Both the Kondo temperature T_K ∝ exp(-1/JN(0)) and the RKKY ordering temperature T_RKKY ∝ J²N(0) depend on the exchange coupling J between f-electrons and conduction electrons, but with different functional forms. For small J, T_RKKY >> T_K (RKKY wins, magnetic order). For large J, T_K >> T_RKKY (Kondo wins, non-magnetic heavy Fermi liquid). The Doniach phase diagram shows the crossover: at a critical J_c, the two scales are comparable and a quantum phase transition occurs between the magnetically ordered and heavy Fermi liquid ground states. Pressure typically increases J (by increasing hybridization), so applying pressure to a magnetic heavy fermion compound often drives it through the quantum critical point into the non-magnetic heavy fermion state.
The different J-dependences are key: T_K is exponential in 1/J while T_RKKY is polynomial (J²). At small J, the polynomial dominates; at large J, the exponential catches up and surpasses it. This competition is the central organizing principle of heavy fermion physics.
Question 4 Short Answer
Heavy fermion superconductivity (as in CeCu₂Si₂ or UPt₃) is 'unconventional.' What makes it different from BCS superconductivity?
Think about your answer, then reveal below.
Model answer: In conventional BCS superconductors, phonons mediate the pairing and the gap function is isotropic (s-wave, uniform around the Fermi surface). In heavy fermion superconductors, the pairing is believed to be mediated by magnetic fluctuations (spin fluctuations from the nearby magnetic phase), producing anisotropic gap functions with nodes — points or lines on the Fermi surface where the gap vanishes. Evidence includes power-law (not exponential) temperature dependences of specific heat, penetration depth, and NMR relaxation rate. UPt₃ has multiple superconducting phases with different symmetries. The pairing symmetry can be d-wave, p-wave, or more exotic, and determining it is one of the central challenges of heavy fermion research.
The proximity of superconductivity to the quantum critical point in many heavy fermion phase diagrams suggests that quantum critical fluctuations may enhance or mediate the pairing — similar to how antiferromagnetic fluctuations are believed to mediate pairing in cuprate high-T_c superconductors.