Poor man's scaling in Kondo problem

For the Kondo model: $$H=-t\sum_{i,j}c_i^\dagger c_j+JS\cdot \sigma(0)$$ which only including itinerant electrons with the band-width $ W \in[-D,D]$, and $S$ is the spin operator for local moment of impurity. Next, we can further project out the states with energy higher than $|D'|<|D|$ and obtain the scaling flow for spin exchange coupling $J$:

Coleman, Introduction to Many-Body Physics, Fig. 16.16

which gives us two results:

  • If the spin exchange coupling $J$ is antiferromagnetic(AFM), i.e. $J>0$, $J$ will flow to infinite as temperature decreases, which means the coupling between itinerant electrons and local moment of impurity will be stronger and stronger at lower energy scale.
  • If $J$ is ferromagnetic(FM), i.e. $J<0$, $J$ will flow to zero as temperature decreases, which means the coupling will be weaker and weaker at lower energy scale.


However, from the intuitive picture, if $J$ is FM, the local moment of impurity will "bind" more and more itinerant electrons with the parallel spin, leading to larger and larger "effective local moment", i.e. larger $S$, which equals to larger effective spin exchange $J$. In contrast, if $J$ is AFM, the local moment of impurity will be screened by itinerant electrons with anti-parallel spin, leading to smaller "effective local moment", i.e. smaller $S$, which equals to smaller effective spin exchange $J$.

Such analysis is conflict with the result of poor man's scaling in Kondo problem above, thus I am confused of it.


1 Answer 1


The reasoning about the local moment aligning, if $J$ is FM, and being screened, if $J$ is AFM, is borrowed from the ferromagnetic/antiferromagnetic models, such as Ising model. Kondo model significantly differs from those, since the electrons are mobile and only the local moment is fixed. This means that the local moment constantly flips its spin due to collisions with the incident band electrons of different spin. Thus, the mean magnetization should be zero, regardless of the sign of the coupling. Note that the divergence requiring regularization/renormalization appears in the spin-flip term. Thus, there may be more similarity with xy-model than with the ferromagnets/antiferromagnets - perhaps, somebody could comment on this.

  • $\begingroup$ Thanks for your wonderful answer! I wonder "mean magnetization should be zero", does "magnetization" here means "mean magnetization for impurity"? But from the result of Anderson model, we know that there actually exists non-zero expectation value of magnetization for impurity in this situation. $\endgroup$ Commented Apr 19, 2020 at 8:49
  • $\begingroup$ I mean the magnetization of impurity. You mean the Anderson's first paper with the mean field treatment? It is not applicable there, as far as I know. But I might be wrong about the magnetization, since I was working mainly on transport problems. $\endgroup$
    – Roger V.
    Commented Apr 19, 2020 at 8:58

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