Multichannel and Overscreened Kondo Fixed Points

In a $k$-channel Kondo model, the relation between the number of spin-$1/2$ screening channels and twice the impurity spin classifies the infrared behavior. $k<2S$ leaves a residual moment, $k=2S$ exactly screens it, and $k>2S$ overscreens it and can produce an intermediate-coupling non-Fermi-liquid fixed point. The overscreened point requires channel symmetry and the prescribed bath structure; channel anisotropy and a local field are relevant perturbations that ultimately restore a Fermi-liquid crossover.

Required background. Kondo RG flow supplies exchange scaling and $T_K$.

Helpful background. Affine currents and WZW models supplies the exact boundary-CFT description.

Screening classification

Take

$$H=\sum_{ka\sigma}\varepsilon_kc_{ka\sigma}^\dagger c_{ka\sigma} +J\sum_{a=1}^{k}\mathbf S\mathbin{\cdot}\mathbf s_a(0),$$

where channel $a$ is conserved by scattering. At strong coupling, each channel can contribute a spin-$1/2$ degree of freedom at the boundary.

• If $k<2S$, screening is incomplete and a residual spin $S-k/2$ remains. Ferromagnetic residual coupling produces singular corrections.
• If $k=2S$, the impurity is exactly screened and the ordinary local Fermi liquid is possible.
• If $k>2S$, naive strong coupling has too many screening degrees of freedom and is unstable; the flow ends at an overscreened non-Fermi-liquid fixed point when channel symmetry is exact.

This counting assumes equivalent metallic channels and the same antiferromagnetic coupling. Two physical leads are not necessarily two channels: a basis rotation often leaves only one coupled combination.

Exact fixed-point data

Boundary conformal field theory gives the residual impurity entropy

$$S_{\mathrm{imp}}(0)=\log\left[ \frac{\sin\!\left(\pi(2S+1)/(k+2)\right)} {\sin\!\left(\pi/(k+2)\right)} \right]$$

for the overscreened fixed point. For $S=1/2$, $k=2$, this is $\frac12\log2$. The noninteger boundary degeneracy is not a free localized half-state; it is a universal property of the entangled boundary condition.

Affleck and Ludwig 1991, PRL pp. 161–164 derives the universal noninteger boundary degeneracy.

The two-channel Kondo fixed point has square-root corrections in several observables, such as a $T$-matrix correction proportional to $\sqrt{T/T_K}$, and logarithmic impurity susceptibility and heat-capacity coefficient. Observable exponents and amplitudes depend on which operator couples to the probe. Affleck and Ludwig 1991, §§4–6 derives the boundary entropy and operator spectrum.

Relevant perturbations and crossover

Channel anisotropy $J_1\ne J_2$ is relevant: the more strongly coupled channel eventually screens the impurity, producing a conventional Fermi liquid. A magnetic field is also relevant. Near the two-channel fixed point these perturbations have scaling dimension $1/2$, so the crossover scale behaves parametrically as

$$T^\ast\sim T_K\delta^2,$$

where $\delta$ is the appropriately normalized anisotropy or field. Finite temperature or size above $T^\ast$ can display a broad non-Fermi-liquid crossover even though the asymptotic ground state is ordinary. Experimental claims must demonstrate a scaling collapse and tune the relevant perturbation, not merely fit one fractional power.

Channel realization is often the hardest condition. Charge transfer between purported channels destroys conservation; unequal densities of states generate anisotropy; additional orbital splittings change the impurity representation. The complete device or material symmetry, not the low-energy name alone, decides whether overscreening is possible.

The structure map separates screening count, fixed point, and crossover.

Overscreening is a symmetry-protected boundary fixed point, not a generic multilead effect. Residual entropy and anomalous scaling survive only above crossover scales set by relevant perturbations. Original schematic, not to scale.

See the impurity claim test matrix for fixed-point and crossover tests.

Exercise

Classify three models. Give the screening class for $(S,k)=(1,1)$, $(1,2)$, and $(1/2,2)$.

Solution

For $(1,1)$, $k<2S$ and a residual spin $1/2$ remains: underscreened. For $(1,2)$, $k=2S$: exactly screened. For $(1/2,2)$, $k>2S$: overscreened if the two channels are exactly conserved and symmetric. Breaking that symmetry ultimately produces a one-channel Fermi liquid.

References

• Affleck, I., and Ludwig, A. W. W. (1991). “Critical theory of overscreened Kondo fixed points.” Nuclear Physics B 360, 641–696. doi:10.1016/0550-3213(91)90419-X.
• Affleck, I., and Ludwig, A. W. W. (1991). “Universal noninteger ‘ground-state degeneracy’ in critical quantum systems.” Physical Review Letters 67, 161–164. doi:10.1103/PhysRevLett.67.161.