Now you are in the subtree of Special Topics in Many-Body Theory, Spring 2016 project.

# Quantum impurity models: Anderson and Kondo

In our study of magnetism so far, we concentrated on models like the quantum Heisenberg model where the only low-energy degrees of freedom were local spins. Now we consider the case of local spins interacting with mobile conduction electrons. As we will show, this is an appropriate model for magnetic impurities in a metal (here an impurity is "magnetic" if, in the metallic environment, it has a local electronic magnetic moment) and can be used to explain the puzzling "resistance minimum" seen in some alloys. For example, copper with some iron impurities shows a marked rise in the resistivity as the temperature decreases below a few tens of Kelvins, which is difficult to understand from a weak-interaction picture. A reference for this material is A. C. Hewson, The Kondo Problem to Heavy Fermions (Cambridge U.P.). As there are quite a few equations, I fear that some typos or minus sign errors will invariably creep in, so I encourage readers to consult Hewson for any serious applications of the following.

This lecture starts by considering how to describe impurity scattering in our second-quantization language, then looking at the difference between nonmagnetic and magnetic impurity scattering. Let us start by considering just potential scattering, which you have probably seen before. We assume that we have a Hamiltonian describing Bloch eigenstates of energies $\epsilon_{\bf k}$, plus a localized potential around one point that scatters different eigenstates into each other:
$$$$H = \sum_{{\bf k},\sigma} \epsilon_{\bf k} + c^\dagger_{{\bf k},\sigma} c_{{\bf k},\sigma} + \sum_{{\bf k},{\bf k}^\prime,\sigma} V_{{\bf k},{\bf k}^\prime} c^\dagger_{{\bf k},\sigma} c_{{\bf k}^\prime,\sigma}.$$$$
Here $V_{{\bf k},{\bf k}^\prime} = \langle {\bf k} | V | {\bf k}^\prime \rangle$ is just the matrix element of the potential $V$. Since the above Hamiltonian is quadratic, it can be diagonalized into a new single-particle Hamiltonian with some slightly different eigenstates, which we write as
$$$$H = \sum_{\alpha,\sigma} \epsilon_\alpha c^\dagger_{\alpha,\sigma} c_{\alpha,\sigma}.$$$$
Here the notation $\alpha$ is used because these new states are no longer momentum eigenstates. We would like to start by answering the question of how the density of states is modified from its value without the impurity, $\rho_0(\epsilon) = \sum_{{\bf k}} \delta(\epsilon-\epsilon_{bk})$. The density of states can then be used to express physical quantities like the specific heat and spin susceptibility. We solve this using a Green's function method even though it can also be done directly. Consider an operator $G(t-t^\prime)$ that Hewson calls the resolvent Green's function. We define $G(t-t^\prime)$ as the solution of the equation
$$$$\left(i \hbar {\partial \over \partial t} - H\right)G(t-t^\prime)=\delta(t-t^\prime).$$$$
To be precise we should define boundary conditions on the solution of this equation. The retarded or causal Green's function satisfies $G^-(t)=0$ if $t0$. Their Fourier transforms are
$$$$G^\pm(\epsilon) = \int_{-\infty}^\infty G^\pm(t)e^{i (\epsilon \pm i \eta)/\hbar}\,dt.$$$$
where, as usual, $\eta$ is a positive infinitesimal. Then the Fourier transform of the evolution equation, with a necessary convergence factor, is $G^\pm(\epsilon)$
$$$$(\epsilon \pm i \eta - H) G^\pm(\epsilon) = I.$$$$
Again, this is an operator equation, but formally we can write
$$$$G^\pm(\epsilon)={I \over \epsilon \pm i \eta - H}. \label{gres}$$$$
The matrix element of $G^\pm(\epsilon)$ between two exact eigenstates is then
$$$$\langle \alpha | G^\pm(\epsilon) | \alpha^\prime \rangle = {\delta_{\alpha \alpha^\prime} \over \epsilon \pm i \eta - \epsilon_\alpha}.$$$$
This shows that $G$ is indeed useful to get the density of states:
$$$$\mp {1 \over \pi} Im\,Tr\,G^\pm(\epsilon) = \mp {1 \over \pi} Im \sum_\alpha {1 \over \epsilon \pm i \eta - \epsilon_\alpha} = \sum_\alpha \delta(\epsilon-\epsilon_\alpha) = \rho(\epsilon).$$$$
Recall that the trace is basis-independent, so we could just as well have chosen a different basis in which to evaluate the above.

We now write out a simple perturbation theory to express the full $G$ in terms of $G_0$, defined as the Green's function of the problem without the $V$ terms, i.e., $H=H_0+V$ and $G_0$ is the Green's function of $G_0$. We can write, suppressing the argument $\epsilon$ of all the $G$,
$$$$(\epsilon \pm i \eta - H_0 - V) G^\pm = I$$$$
and therefore
$$\begin{eqnarray} G^\pm &=& {I \over \epsilon \pm i \eta - H_0} {\epsilon \pm i \eta - H_0-V + V \over \epsilon \pm i \eta - H_0-V}\cr &=& {I \over \epsilon \pm i \eta - H_0} + {I \over \epsilon \pm i \eta - H_0} V{I \over \epsilon \pm i \eta - H_0-V} = G^\pm_0 + G^\pm_0 V G^\pm. \end{eqnarray}$$
A compact way to write this is in terms of the $T$-matrix, defined so that
$$$$G^\pm = G^\pm_0 + G^\pm_0 V T(\epsilon \pm i\eta) G^\pm_0.$$$$

The change in the density of states because of the impurity is
$$$$\Delta \rho(\epsilon) = \rho(\epsilon)-\rho_0(\epsilon) = -{1 \over \pi} Im\,Tr\,(G^+(\epsilon)-G^+_0(\epsilon)).$$$$
It turns out that this can be expressed in terms of the generalized phase shift. First recall that trace of log = log of det, so, using (\ref{gres})
$$$$Tr\,G^+(\epsilon)=Tr\,G^+(\epsilon) {\partial (G^+(\epsilon))^{-1} \over \partial \epsilon}= -{\partial \over \partial \epsilon} \log \det G^+(\epsilon).$$$$
Finally
$$$$\Delta \rho(\epsilon) = \rho(\epsilon)-\rho_0(\epsilon) = -{1 \over \pi} Im\,{\partial \over \partial \epsilon}\log\left[\det(G^+(\epsilon) / G_0(\epsilon))\right]= -{1 \over \pi} Im\,{\partial \over \partial \epsilon}\log\left[\det(T(\epsilon\pm i\eta))\right].$$$$

The generalized phase shift is defined as the complex arg of the $T$-matrix:
$$$$\eta(\epsilon) = \arg \det T(\epsilon^+),\quad \epsilon^+ = \epsilon+i\eta.$$$$
Finally we have the important relation
$$$$\Delta \rho(\epsilon) = \frac{1}{\pi} {\partial \eta(\epsilon) \over \partial \epsilon}.$$$$
The connection between more usual definitions of the phase shift and the above definition in terms of the $T$-matrix is worked through in an appendix of Hewson's book. Integrating from zero phase shift at energy $\epsilon=-\infty$ gives for the total charge accumulation near the impurity
$$$$n_{imp} = \int_{-\infty}^{\epsilon_F} \Delta \rho(\epsilon)\,d\epsilon={\eta(\epsilon_F) \over \pi}$$$$
which is known as the Friedel sum rule. Note that, assuming screening, $n_{imp}$ should just balance the total charge on the impurity, so this gives a consistency condition on the integrated phase shift up to the Fermi energy.

What we will do now is compute the above expressions more or less explicitly for the noninteracting "Anderson model" defined below. Our goal is to understand more complex cases than potential scattering: a more realistic model of the impurity takes into account the fact that the impurity has internal states which can be changed by conduction electrons. The noninteracting Anderson model is 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:
$$$$H = \sum_{\sigma} \epsilon_d c^\dagger_{d,\sigma} c_{d,\sigma} + \sum_{{\bf k},\sigma} c^\dagger_{{\bf k},\sigma} c_{{\bf k},\sigma} + \sum_{{\bf k},\sigma} (V_{\bf k} c^\dagger_{d,\sigma} c_{{\bf k},\sigma} + h.c.).$$$$
The full Anderson model is obtained from the above simply by adding an onsite repulsion $U$ if two electrons are on the $d$ orbital:
$$$$H_I = U n_{d\uparrow} n_{d\downarrow}.$$$$
Without the interaction part $H_I$, this model is quadratic and hence solvable. With $H_I$ its behavior is much more interesting and is discussed in the next lecture.

To solve for the Green's function of this model, take matrix elements between the $d$ states of the evolution equation used earlier in this lecture,
$$$$(\epsilon + i \eta - H) G^+(\epsilon) = I,$$$$
to get
$$$$(\epsilon-\epsilon_d) G^+_{d,d}(\epsilon) = 1 + \sum_{\bf k} V_{\bf k}^* G^+_{{\bf k},d} (\epsilon).$$$$
Here $G^+_{d,d}(\epsilon) = \langle d | G^+(\epsilon) | d \rangle$. The same trick but with one $d$ and one $k$ state gives the second coupled equation
$$$$(\epsilon-\epsilon_{\bf k}) G^+_{{\bf k},d}(\epsilon) = V^*_{\bf k} G^+_{d,d}(\epsilon).$$$$
We drop the spin index since for this noninteracting model, spin just creates two copies of the problem. Solving these two equations gives
$$$$G^+_{d,d}(\epsilon) = {1 \over \epsilon + i \eta - \epsilon_d - \sum_{\bf k} {|V_{\bf k}|^2 \over (\epsilon+i\eta-\epsilon_{\bf k})}}.$$$$
Applying the same method to calculate $G^+_{{\bf k},{\bf k}}$, we get
$$$$G^+_{{\bf k},{\bf k}}(\epsilon) = {\delta_{{\bf k},{\bf k}^\prime} \over \epsilon + i \eta - \epsilon_{{\bf k}} } + {V^*_{\bf k} G^+_{d,d}(\epsilon) V_{{\bf k}^\prime} \over (\epsilon + i \eta - \epsilon_{{\bf k}}) (\epsilon + i \eta - \epsilon_{{\bf k}^\prime})}.$$$$
This means that the $T$-matrix, which was previously related to the phase shift, has simple matrix elements
$$$$\langle {\bf k} | T(\epsilon) | {\bf k}^\prime \rangle = V^*_{\bf k} G^+_{d,d}(\epsilon) V_{{\bf k}^\prime}.$$$$

We can calculate the change in the density of states from the above $T$ matrix through the connection to phase shifts mentioned above. Another way is just to calculate the trace of $G^+$,
$$\begin{eqnarray} Tr\,G^+(\epsilon) &=& \sum_{\bf k} G^+_{{\bf k},{\bf k}}(\epsilon) + G^+_{d,d}(\epsilon) \cr &=& \sum_{\bf k} {1 \over \epsilon+i\eta-\epsilon_{\bf k}} + {\partial \over \partial \epsilon} \log \left (\epsilon+i\eta-\epsilon_d-\sum_{\bf k} {|V_{\bf k}|^2 \over \epsilon+i\eta-\epsilon_{\bf k}} \right). \end{eqnarray}$$
We want the imaginary part of this multiplied by $-\pi^{-1}$, or
$$$$\delta \rho(\epsilon) = {1 \over \pi} {\partial \eta (\epsilon) \over \partial \epsilon}$$$$
with the phase shift being equal to the complex angle, which we can find from the arctangent of the ratio of real and imaginary parts of the argument of the logarithm:
$$$$\eta(\epsilon) = {\pi \over 2} - \tan^{-1} \left( {\epsilon_d + \Lambda(\epsilon) - \epsilon \over \Delta(\epsilon)}\right).$$$$
Here the ${\pi \over 2}$ comes from the assumption of no phase shift at $\epsilon=-\infty$ and $\Lambda(\epsilon)$ and $\Delta(\epsilon)$ are the real and imaginary parts of the sum,
$$$$\Lambda(\epsilon) = {\rm P} \sum_{\bf k} {|V_{\bf k}|^2 \over \epsilon-\epsilon_{\bf k}},\quad \Delta(\epsilon) = \pi \sum_{\bf k} |V_{{\bf k}}|^2 \delta(\epsilon-\epsilon_{\bf k}).$$$$
These two terms can be simply interpreted: $\Lambda$ represents a shift in the energy $\epsilon_d$ because of the hybridization $V$, and $\Delta$ is the linewidth induced by the hybridization.

On the last problem set you are asked to calculate these forms exactly for the standard simple model. Assuming that $\epsilon_d$ is well within a flat conduction band of constant density of states $\rho_0$ and ranging over $(\epsilon=-D,\epsilon=D)$, and that $V$ is $k-$independent, then
$$$$\Delta(\epsilon) = \pi \rho_0 \sum_{\bf k} |V|^2 = \Delta$$$$
if $-D < \epsilon < D$, and zero otherwise.

The phase shift for this case is given by
$$$$\eta(\epsilon) = \frac{\pi}{2} - \arctan {{\tilde \epsilon}_d - \epsilon \over \Delta}$$$$
where ${\tilde \epsilon}_d$ is the solution of the self-consistent equation
$$$${\tilde \epsilon}_d = \epsilon_d + \Lambda({\tilde \epsilon}_d).$$$$

Our expectation is that, if the conduction electrons do not modify the state of the impurity, then the earlier potential scattering model is correct. The noninteracting Anderson model describes physics when the conduction electrons can change the occupancy of the spin-up and spin-down impurity states independently. Turning on an interaction energy $H_I$ dramatically changes the physics at low temperature, and explains an experimental puzzle that dates to the 1930s or earlier.

Now we go beyond these two quadratic problems (potential scattering, 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:
$$$$H = \sum_{\sigma} \epsilon_d c^\dagger_{d,\sigma} c_{d,\sigma} + \sum_{{\bf k},\sigma} c^\dagger_{{\bf k},\sigma} c_{{\bf k},\sigma} + \sum_{{\bf k},\sigma} (V_{\bf k} c^\dagger_{d,\sigma} c_{{\bf k},\sigma} + h.c.)+U n_{d\uparrow} n_{d\downarrow}.$$$$
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 $\epsilon_d$, and a doubly occupied state of energy $2 \epsilon_d + U$. We want to consider the situation with $\epsilon_d$ negative and $U$ positive, so that the lowest-energy states are the single-electron states. For obvious reasons the specific choice $\epsilon_d = -U/2$ 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 $\epsilon_d$ 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
$$$$H_{s-d} = \sum_{{\bf k},{\bf k}^\prime} J_{{\bf k},{\bf k}^\prime} (S^+ c^\dagger_{{\bf k},\downarrow} c_{{\bf k}^\prime,\uparrow} + S^- c^\dagger_{{\bf k},\uparrow} c_{{\bf k}^\prime,\downarrow} + S_z (c^\dagger_{{\bf k},\uparrow} c_{{\bf k}^\prime,\uparrow} - c^\dagger_{{\bf k},\downarrow} c_{{\bf k}^\prime,\downarrow}).$$$$
with the spin coupling
$$$$J_{{\bf k},{\bf k}^\prime} = V^*_{\bf k} V_{{\bf k}^\prime} \left[ {1 \over U+\epsilon_d - \epsilon_{\bf k}^\prime} + {1 \over \epsilon_{\bf k} - \epsilon_d}\right].$$$$
The full Hamiltonian consists of $H_{s-d}$ plus the conduction electron Hamiltonian.
(To see the connection 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 $f_E({\bf k})$. Here we integrate over all space, so $f_E({\bf k})$ is the total number of electrons of momentum ${\bf k}$. (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, $\partial _t f_E = 0$, and
$$$${df_E \over dt} = \left( {\partial {\bf k} \over \partial t} \right) \cdot \nabla_k f_E({\bf k}) = C(f_E).$$$$
Here $C(f_E)$ represents collisions between an electron and an impurity ({\it not} between electrons, as in the usual Boltzmann equation). We make the "relaxation-time approximation" that the action of $C$ can be linearized near the Fermi-Dirac equilibrium function $f({\bf k})$:
$$$$C(f_E) = -{f_E({\bf k})-f({\bf k}) \over \hbar \tau_1({\bf k})}. \label{reltime}$$$$
Here $\tau_1({\bf k})$, the relaxation time at momentum ${\bf k}$, is estimated below. The equation of motion is $\hbar \partial_t {\bf k} = -e {\bf E}$, so
$$$$f_E({\bf k}) = -\hbar \tau_1({\bf k}) (-e {\bf E}/\hbar) \cdot \nabla_k f_E({\bf k}) + f({\bf k}) \approx e \tau_1({\bf k}) {\bf E} \cdot \nabla_k f({\bf k})+f({\bf k})$$$$
where in the second step we have assumed a small perturbation to a uniform equilibrium, so that $\nabla_k f_E$ can be replaced by $\nabla_k f$ with an error that is smaller than the retained terms.

The current can be expressed in terms of the average velocity:
$$$${\bf j} = -e \langle {\bf v}_k \rangle = -2 e \int f_E({\bf k}) {\hbar {\bf k} \over m} {{d\bf k} \over (2 \pi)^3}.$$$$
Here the factor of 2 is just for spin. Since $f({\bf k})$ is isotropic it contributes zero current, and assuming the unperturbed system is isotropic $\tau_1({\bf k})$ = $\tau_1(k)$. Also
$$$$\nabla_k f({\bf k}) = {{\bf k} \over m} {\partial f \over \epsilon_k}.$$$$
so
$$$${\bf j} = -2 e^2 \int \tau_1(k) {\bf v}_{\bf k} ({\bf E} \cdot {\bf v}_{\bf k}) {\partial f \over \partial \epsilon_{\bf k}} {d{\bf k} \over (2 \pi)^3}.$$$$
Again, for an isotropic system ${\bf j} || {\bf E}$ and
$$$$\sigma(T) = - {2 e^2 \over 3} \int\,{v_k}^2 \tau_1(k){\partial f \over \partial \epsilon_k} {d {\bf k} \over (2 \pi)^3}.$$$$
At zero temperature we have ${\partial f \over \partial \epsilon_k} = - \delta(\epsilon-\epsilon_k)$, so for free electrons the above is
$$$$\sigma = {n e^2 \tau_1(k_F) \over m}.$$$$

To calculate $\tau_1$, we need to write out the collision term in the linearized approximation. By Fermi's Golden Rule, the rate to scatter from ${\bf k}$ to ${\bf k}^\prime$ is proportional to the squared matrix element $|T_{{\bf k} {\bf k}^\prime}|^2$ times a final density of states, so putting in the numerical factor $2\pi/\hbar$ and writing $c_{imp}$ for the impurity concentration,
$$$$C(f_E) = -{2 \pi c_{imp} \over \hbar} \int \delta(\epsilon_k - \epsilon_{k^\prime}) \left[ |T_{{\bf k} {\bf k}^\prime}|^2 f_E({\bf k}) (1-f_E({\bf k}^\prime))+ |T_{{\bf k}^\prime {\bf k}}|^2 f_E({\bf k}^\prime) (1-f_E({\bf k})) \right] {d {\bf k}^\prime \over (2 \pi)^3}.$$$$
We can simplify this considerably by noting that $T_{{\bf k} {\bf k}^\prime}$ is symmetric under interchange of the two momenta. Then rewriting the above and inserting it into the Boltzmann equation gives
$$$${2 \pi c_{imp} \over \hbar} \int \delta(\epsilon_k - \epsilon_{k^\prime}) |T_{{\bf k} {\bf k}^\prime}|^2 (f_E({\bf k}) - f_E({\bf k}^\prime)) {d {\bf k}^\prime \over (2 \pi)^3} = -{e {\bf E} \over \hbar} \cdot \nabla_k f_E({\bf k}).$$$$
We only need the solution of this to first order in $E$, when it can be assumed that the deviation from equilibrium is small. Substituting in the relaxation-time ansatz gives
$$$${2 \pi c_{imp} \over \hbar} \int \delta(\epsilon_k - \epsilon_{k^\prime}) |T_{{\bf k} {\bf k}^\prime}|^2 e \tau_1(k) {\bf E} \cdot (\nabla_k f({\bf k}) - \nabla_k^\prime f({\bf k}^\prime)) {d {\bf k}^\prime \over (2 \pi)^3} = -{e {\bf E} \over \hbar} \cdot \nabla_k f_E({\bf k}).$$$$
Now use $\nabla_k f = ({\bf k} /m) \partial_\epsilon f$ to get
$$$${2 \pi c_{imp} \over \hbar} \int \delta(\epsilon_k - \epsilon_{k^\prime}) |T_{{\bf k} {\bf k}^\prime}|^2 e \tau_1(k) {\bf E} \cdot {({\bf k} - {\bf k}^\prime) \over m} {\partial f \over \partial \epsilon} {d {\bf k}^\prime \over (2 \pi)^3} = -{e {\bf E} \cdot {\bf k} \over \hbar m} {\partial f \over \partial \epsilon}.$$$$
Now the point of what we've been doing is clear: we can divide by the terms on the right and get
%assume that the $T$ matrix depends only on the angle $\theta$ between ${\bf k}$ and ${\bf k}^\prime$, to get
$$$${1 \over \tau_1(k)} = 2 \pi c_{imp} \int \delta(\epsilon_k - \epsilon_{k^\prime}) |T_{{\bf k} {\bf k}^\prime}|^2 (1 - \cos \theta) {d {\bf k}^\prime \over (2 \pi)^3}.$$$$
and, as we said before, at zero temperature the conductivity is
$$$$\sigma = {n e^2 \tau_1(k_F) \over m}.$$$$
Note that this "transport lifetime" is not just the mean time between collisions; collisions are weighted by $(1-\cos \theta)$ to reflect how much they alter the direction of an electron's motion.

Now we return to the $s-d$ interaction
$$$$H_{s-d} = \sum_{{\bf k},{\bf k}^\prime} J_{{\bf k},{\bf k}^\prime} (S^+ c^\dagger_{{\bf k},\downarrow} c_{{\bf k}^\prime,\uparrow} + S^- c^\dagger_{{\bf k},\uparrow} c_{{\bf k}^\prime,\downarrow} + S_z (c^\dagger_{{\bf k},\uparrow} c_{{\bf k}^\prime,\uparrow} - c^\dagger_{{\bf k},\downarrow} c_{{\bf k}^\prime,\downarrow}).$$$$
and start to calculate perturbatively the resistivity caused by a magnetic impurity. Kondo's result is
$$$$R = {3 \pi m J^2 S(S+1) \over 2 e^2 \hbar \epsilon_F} \left[1 - 4 J \rho_0(\epsilon_F) \log\left({k_B T \over D} \right) \right],$$$$
which diverges at low temperature. We start by calculating the first (non-divergent) term, which corresponds essentially to the Born approximation for scattering.

To lowest order in $J$, we have $T \approx H$ and the following matrix element for spin-preserving scattering:
$$$$\langle {\bf k}, \uparrow | T(\epsilon+i\eta) | {\bf k}^\prime,\uparrow \rangle = J_{{\bf k},{\bf k}^\prime} S_z$$$$
and for spin-flip scattering get, e.g.,
$$$$\langle {\bf k}, \uparrow | T(\epsilon+i\eta) | {\bf k}^\prime,\downarrow \rangle = J_{{\bf k},{\bf k}^\prime} S_-.$$$$
Assuming that $J$ is constant, we get that
$\tau(k_F)^{-1}$ is proportional to $J^2 S(S+1)$, noting that $2 S_z^2 + S^+ S^- + S^- S^+=S(S+1)$.

Above we solved for the density of states in the noninteracting Anderson model and discussed Kondo's perturbative calculation of spin-flip scattering. The $T$ matrix was connected to the conductance via a Boltzmann equation approach, whose predictions can be confirmed by a more detailed calculation using the Kubo formula. In this lecture we will outline Wilson's solution of the Kondo problem using the numerical renormalization group and present a simple Fermi-liquid picture due to Nozieres that explains some of the results given a few assumptions.

First we do a bit more of Kondo's calculation. We previously calculated the $T$ matrix to order $J$, which gave the first term in the resistance formula.
$$$$R = {3 \pi m J^2 S(S+1) \over 2 e^2 \hbar \epsilon_F} \left[1 - 4 J \rho_0(\epsilon_F) \log\left({k_B T \over D} \right) \right].$$$$
To calculate the second term, we would need the $T$ matrix to order $J^2$, which contains several terms. We will just calculate one diverging term to give you the idea.

The important terms turn out to be the terms where the impurity spin is flipped in the intermediate state. An example is the term proportional to
$$$$J^2 \sum_{{\bf k}_1,{\bf k}^\prime_1,{\bf k}_2,{\bf k}^\prime_2} \langle {\bf k}^\prime, \uparrow | S^- c^\dagger_{{\bf k}_1,\uparrow} c_{{\bf k}^\prime_1,\downarrow} (\epsilon + i\eta - H_0)^{-1} S^+ c^\dagger_{{\bf k}_2,\downarrow} c_{{\bf k}^\prime_2,\uparrow} | {\bf k},\uparrow\rangle$$$$
This term will be nonzero if ${\bf k}^\prime = {\bf k}_1$, ${\bf k}^\prime_2={\bf k}$, and ${\bf k}^\prime_1 = {\bf k}_2$, and gives
$$$$J^2 \sum_{{\bf k}_2} S^- S^+ {1 - f(\epsilon_{{\bf k}_2} \over \epsilon + i \eta - \epsilon_{{\bf k}_2}}.$$$$
Here the $1-f$ (where $f$ is just the Fermi distribution) appears because of the requirement that the state ${\bf k}_2$ be initially empty for the above term to be nonzero. After adding together the contributions of many terms, the troublesome parts are proportional to the real part of the function
$$$$g(\epsilon) = \sum_{\bf k} {f(\epsilon_{\bf k}) \over \epsilon_{\bf k} - \epsilon - i \eta}.$$$$
We remark in passing that the reason why magnetic scattering comes out differently from potential scattering is essentially that, if we didn't have the noncommutativity of the $SU(2)$ spin operators, all the divergent terms would just cancel. For details, see Hewson.

How does the function $g(\epsilon)$ behave at low temperature? For the conductivity we are interested in $\epsilon$ near the Fermi level; clearly if the density of states is finite at the Fermi level, then there is a logarithmic divergence at $T=0$; the effect of a nonzero temperature is to cutoff the singularity (giving $\log(D/T)$, where $D$ is the bandwidth). Careful calculation of the prefactor gives the resistivity result quoted above.

Adding the logarithmic divergence found above to the ordinary phonon and nonmagnetic scattering contributions to the resistance gives a reasonable description of the location of the resistance minimum of magnetic alloys. However, we still need to find a valid way of describing the system down to zero temperaure, and also to find a way of calculating other experimental quantities like the spin susceptibility in a magnetic field and the spectral function.

For the most part we will just be describing the phenomena associated with the Kondo effect, but will also show how a Fermi-liquid picture can be used to deduce some of these phenomena from one starting point, the existence of a "Kondo resonance." There is a somewhat intuitive way of understanding what happens at low temperature: the impurity spin binds with the conduction electron spins to form a singlet. We will give a more precise picture of this in a moment; note that the impurity spin must bind to quite a complicated superposition of the impurity electron states in order to explain the nonanalytic behavior of the Kondo temperature.

We start with a set of simplifications of the problem introduced by Wilson. Going back to the $s-d$ model, note that for constant $J$, the impurity spin is only coupled directly to one particular combination of the conduction electron orbitals: we can write the coupling as
$$$$2 {\bf S} \cdot c^\dagger_{0,\sigma} {\bf s}_{\sigma,\sigma^\prime} c_{0,\sigma^\prime}$$$$
where the operator $c_{0,\sigma} \propto \sum_{\bf k} c_{{\bf k},\sigma}$ and ${\bf s}$ is a vector of Pauli matrices. However, the localized orbital filled by $c^\dagger_0$ that we have introduced is not an eigenstate of the conduction electron Hamiltonian $H_c$.

Let us make an orthogonal basis where $H_c$ is tridiagonal. Consider the state $H_c |0\rangle$: this state has in general nonzero projection onto $|0\rangle$, so subtract that out and define the state $|1\rangle = (\gamma_0)^{-1} (H_c |0\rangle - |0\rangle \langle 0 | H_c | 0 \rangle$). The number $\gamma_0$ is chosen to give a normalized state.
Continuing this orthogonalization procedure gives a basis of single-particle states where each state $|n\rangle$ is connected by $H_c$ to $|n-1\rangle$ and $|n+1\rangle$ except for the state $0\rangle$, which is connected to $|1\rangle$ and the impurity spin. The general form is
$$$$|n+1\rangle = {1 \over \gamma_n} \left( H_c |n\rangle - |n\rangle \langle n| H_c |n\rangle - |n-1\rangle \langle n-1 | H_c | n-1\rangle \right).$$$$

We can look at the resulting model as a linear chain with one end coupled to the impurity spin by the coupling $J$, and the other end extending out to $\infty$. The original conduction electron states in the problem will determine the precise energies and couplings between the sites in this new linear chain model, but we can hope that the Kondo effect is a fairly universal phenomenon. Now the conduction electron Hamiltonian looks like
$$$$H_c = \sum_{n,\sigma} \epsilon_n c^\dagger_{n,\sigma} c_{n,\sigma} + \sum_{n,\sigma} (\gamma_n c^\dagger_{n,\sigma} c_{n+1,\sigma} + \gamma^*_n c^\dagger_{n+1,\sigma} c_{n,\sigma}).$$$$
where $\epsilon_n = \langle n | H_c | n \rangle$.
The choice of site energy $\epsilon_n = 0$ and hopping $\gamma_n = D/2$ corresponds to a semicircular band of states, which should be good enough. However, with this choice the numerical procedure described below is not convergent, so what Wilson did is take
$$$$\gamma_n = {D (1 + \Lambda^{-1}) \over 2 \Lambda^{n/2}}$$$$
for some parameter $\Lambda > 1$.
This choice of the $\gamma_i$ corresponds to a discrete rather than continuous spectrum of states, distributed evenly over energy scales (i.e., so that each decade of energy has the same number of states). This corresponds to the idea that a logarithmic divergence means each scale or decade of energies contribute equally in the low-energy limit.

We still have a seemingly difficult many-body problem to solve: find the eigenstates of the coupled impurity spin and linear chain. Now the procedure used by Wilson is to solve the low-energy spectrum of the linear chain by iteratively diagonalizing, keeping the lowest-energy states, then adding another site. This process eventually converges, and one simple qualitative result is that the impurity spin forms a singlet ground state with the conduction electrons. For example, if we start with an odd number of conduction electrons, then at $J=0$ the ground state is four-fold degenerate. Turning on $J$ breaks this degeneracy to give a singlet ground state, and the magnitude of this energy splitting gives an energy scale, which can be used as one definition of the Kondo temperature
$$$$T_K \sim D e^{-1/J \rho_0}.$$$$
Note that this is also the temperature where the two terms in Kondo's resistivity formula become comparable in magnitude.

One of the major accomplishments of the "nanosystem" realizations of the Kondo effect mentioned in the last lecture is that they were able to measure directly the impurity spectral function and observe the so-called Kondo resonance. Going back to the Anderson model, we expect that an STM measurement on top of the impurity, say, would reveal one peak at addition energy $\epsilon_d$ and another peak at energy $\epsilon_d+U$. These peaks would have widths $\Gamma$ which we could estimate using the noninteracting Anderson model studied before.

The Kondo effect results in the creation of another peak in the spectral function, called the Kondo resonance. This resonance is sometimes described as representing the singlet pair of the impurity spin, but this is somewhat imprecise. A detailed calculation of the spectral function using the numerical RG or other methods predicts a resonance centered on the Fermi level, with width of order $T_K$. Although the spectral weight in this resonance can be quite small, it is seen strongly in experiments because its height is large and it moves around with the Fermi level. The Fermi-liquid theory for the strong-coupling fixed point described below is based on these assumptions about the Kondo resonance. Note that, just as with superconductivity, the important Kondo action is on quite low energy scales and involves states close to the Fermi level.

The response of the impurity spin in a magnetic field is modified in quite an interesting way by the Kondo effect. For a decoupled spin the magnetization jumps immediately to its maximum value once a field is applied, but the Kondo ground state "screens" the spin so that only once $g \mu H \sim k T_K$ is the spin significantly aligned with the field. The actual impurity magnetization only approaches its maximum value logarithmically in the ratio $g \mu H / k T_K$. Such logarithmic "slow crossovers" are also seen in the specific heat as a function of temperature.