Why does a magnetic impurity in a metal cause the resistivity to increase as temperature decreases, unlike non-magnetic impurities which give a temperature-independent residual resistivity?
AMagnetic impurities attract more electrons at low temperatures
BThe exchange coupling J between the impurity spin and conduction electron spins produces spin-flip scattering. At high T, this scattering is weak (perturbative). As T decreases, higher-order scattering processes (where the electron flips the impurity spin and then flips it back) interfere constructively, producing a logarithmically growing correction: δρ ∝ -J³N(0)² ln(T). This is a many-body resonance that strengthens at low T
CThe magnetic field of the impurity aligns nearby electron spins, creating a local barrier
DMagnetic impurities have larger atomic radii that block electron paths at low temperature
The key is spin-flip scattering involving higher-order (multi-particle) processes. In second-order perturbation theory in J, a conduction electron can virtually flip the impurity spin and then flip it back, and the intermediate state involves the entire Fermi sea (a many-body effect). The resulting scattering amplitude has a ln(T) divergence because the virtual processes involve states at all energies up to the bandwidth, weighted by the Fermi function. This is the Kondo logarithm, and it signals the breakdown of perturbation theory below T_K.
Question 2 Multiple Choice
Wilson's numerical renormalization group (NRG) showed that the Kondo problem flows from a weak-coupling fixed point (free impurity spin) to a strong-coupling fixed point (screened singlet). What is the physical picture at each fixed point?
AAt weak coupling the impurity is superconducting; at strong coupling it is insulating
BAt T >> T_K (weak coupling), the impurity spin is essentially free and the conduction electrons scatter weakly from it — the impurity contributes a Curie-like susceptibility and weak scattering. At T << T_K (strong coupling), the impurity spin is completely screened by a surrounding cloud of conduction electrons forming a many-body singlet, and the impurity site acts as a strong potential scatterer (unitarity limit) with no residual magnetic moment
CBoth fixed points describe free electrons with different effective masses
DThe strong-coupling fixed point has a local magnetic moment
The Kondo crossover from T >> T_K to T << T_K is one of the most beautiful examples of renormalization group flow in physics. At high T, the impurity is a free S = 1/2 moment with a Curie susceptibility χ ∝ 1/T and weak log(T) resistivity corrections. As T decreases through T_K, the effective coupling grows (flows to strong coupling), and the conduction electrons progressively screen the impurity spin. Below T_K, the ground state is a many-body singlet (S_total = 0), the impurity susceptibility saturates (Pauli-like, not Curie), and the resistivity reaches the unitarity limit. Wilson's NRG (Nobel Prize 1982) was the first method to quantitatively describe this crossover.
Question 3 True / False
The Kondo temperature T_K = D exp(-1/JN(0)) has the same non-analytic form as the BCS gap Δ ~ ω_D exp(-1/N(0)V). This is not a coincidence.
TTrue
FFalse
Answer: True
Both expressions reflect a non-perturbative instability of the Fermi sea. In BCS theory, an attractive interaction (however weak) destabilizes the Fermi surface toward Cooper pairing. In the Kondo problem, an antiferromagnetic exchange coupling (however weak) destabilizes the free-spin state toward singlet formation. Both involve logarithmic divergences in perturbation theory that signal the breakdown of the perturbative expansion and the emergence of a new energy scale (Δ or T_K) that is exponentially small in the coupling. The mathematical structure (log divergence → non-perturbative energy scale) is identical and reflects the high density of states at the Fermi level available for forming bound states.
Question 4 Short Answer
Explain why the Kondo effect requires antiferromagnetic exchange coupling (J > 0) and does not occur for ferromagnetic coupling (J < 0).
Think about your answer, then reveal below.
Model answer: For antiferromagnetic coupling (J > 0), the spin-flip scattering processes that produce the Kondo logarithm interfere constructively, making the effective coupling grow at lower energies (asymptotic freedom in reverse — the coupling flows to strong coupling). This produces the Kondo singlet ground state. For ferromagnetic coupling (J < 0), the same processes interfere destructively, and the effective coupling flows to zero at low energies — the impurity spin remains free and the resistivity contribution is simply a constant. The renormalization group beta function has opposite sign for the two cases: β(J) < 0 for AF coupling (relevant, flows to strong coupling) and β(J) > 0 for FM coupling (irrelevant, flows to weak coupling).
This asymmetry is why only AF-coupled magnetic impurities (like Fe, Mn, Cr in copper or gold) show the Kondo effect, while FM-coupled impurities (which are rare in practice) do not. The sign of J depends on the specific hybridization between the impurity d or f orbitals and the conduction band, governed by the Schrieffer-Wolff transformation from the Anderson impurity model.