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An explanation of the kondo model, a many-body physics model used to describe the behavior of magnetic impurities in metals. The calculation of the resistance minimum for magnetic impurities, where the spin-up and spin-down degrees of freedom are strongly coupled. It includes the derivation of the s-d model and the spin coupling, as well as a discussion on how to calculate the conductivity using the boltzmann equation.
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We continue our study of localized impurity models. The last lecture discussed potential scat- tering, where the impurity contains no degrees of freedom, and the noninteracting Anderson model, where the spin-up and spin-down degrees of freedom are independent. In this lecture we reproduce Kondo’s original calculation of the resistance minimum for the case of magnetic impurities, where the spin-up and spin-down degrees of freedom are strongly coupled.
We start with the full Anderson model, a simplified model for a single localized orbital (labeled d since it is typically a d or f orbital of a transition metal) coupled to conduction electrons:
H =
∑ σ
dc† d,σcd,σ +
∑
k,σ
c† k,σck,σ +
∑
k,σ
(Vkc† d,σck,σ + h.c.) + U nd↑nd↓. (1)
Consider the model with no coupling between impurity and conduction states (V = 0). Then the impurity part contains four states: an empty state of energy 0, two states with a single electron (spin-up or spin-down) of energy d, and a doubly occupied state of energy 2d + U. We want to consider the situation with d negative and U positive, so that the lowest-energy states are the single-electron states. For obvious reasons the specific choice d = −U/2 is known as the symmetric Anderson model.
When we turn on V , we can use second-order perturbation theory to calculate an approximate Hamiltonian for the low-energy states of the system with one electron on the impurity. If U is large and positive and d is large and negative compared to the bandwidth, we expect to have singly occupied impurity states. The term “mixed-valence” is used to describe the situation where the zero- and one-electron states are nearly degenerate. Here we will focus on the situation with single occupancy, so the Kondo or s − d model defined below is valid.
This is nearly the same calculation we did in deriving the antiferromagnetic limit of the half- filled Hubbard model previously, so I won’t write out the details. Again the result is an effective spin model (known as the s − d model) whose interaction term is
Hs−d =
∑
k,k′
Jk,k′ (S+c† k,↓ck′,↑ + S−c† k,↑ck′,↓ + Sz (c† k,↑ck′,↑ − c† k,↓ck′,↓). (2)
with the spin coupling
Jk,k′ = V (^) k∗ Vk′
[ 1 U + d − ′ k
k − d
]
. (3)
The full Hamiltonian consists of Hs−d plus the conduction electron Hamiltonian. (To see the con- nection of this to what we had before for the Hubbard model, recall that before in the intermediate state one site became empty and one became doubly occupied, so the energy denominator was exactly U .)
The next challenge is to figure out a way to calculate the conductivity, given a Hamiltonian like the above. The most correct but also most time-consuming way is to work from the Kubo formula for linear-response conductivity, which expresses the conductivity in terms of the current- current correlation function. Then the current-current correlation function can be evaluated using diagrammatic perturbation theory. Here we will use a quicker and perhaps more physical method
where the approximations are a bit less clear, based on the Boltzmann equation for the evolution of the electronic distribution function fE (k). Here we integrate over all space, so fE (k) is the total number of electrons of momentum k. (We also assume that the reader is at least somewhat familiar with Boltzmann-type equations; more information is in any kinetic theory textbook.)
For a steady-state distribution, ∂tfE = 0, and
dfE dt
( ∂k ∂t
) · ∇kfE (k) = C(fE ). (4)
Here C(fE ) represents collisions between an electron and an impurity (not between electrons, as in the usual Boltzmann equation). We make the “relaxation-time approximation” that the action of C can be linearized near the Fermi-Dirac equilibrium function f (k):
C(fE ) = −
fE (k) − f (k) ¯hτ 1 (k).^ (5)
Here τ 1 (k), the relaxation time at momentum k, is estimated below. The equation of motion is ¯h∂tk = −eE, so
fE (k) = −¯hτ 1 (k)(−eE/¯h) · ∇kfE (k) + f (k) ≈ eτ 1 (k)E · ∇kf (k) + f (k) (6)
where in the second step we have assumed a small perturbation to a uniform equilibrium, so that ∇kfE can be replaced by ∇kf with an error that is smaller than the retained terms.
The current can be expressed in terms of the average velocity:
j = −e〈vk〉 = − 2 e
∫ fE (k)
¯hk m
dk (2π)^3
Here the factor of 2 is just for spin. Since f (k) is isotropic it contributes zero current, and assuming the unperturbed system is isotropic τ 1 (k) = τ 1 (k). Also
∇kf (k) = k m
∂f k
so
j = − 2 e^2
∫ τ 1 (k)vk(E · vk) ∂f ∂k
dk (2π)^3
Again, for an isotropic system j||E and
σ(T ) = − 2 e^2 3
∫ vk^2 τ 1 (k) ∂f ∂k
dk (2π)^3
At zero temperature we have (^) ∂∂fk = −δ( − k), so for free electrons the above is
σ = ne^2 τ 1 (kF ) m
To calculate τ 1 , we need to write out the collision term in the linearized approximation. By Fermi’s Golden Rule, the rate to scatter from k to k′^ is proportional to the squared matrix element