Effective $T$ matrix in Kondo Hamiltonian Consider the Kondo Hamiltonian
$$H=\sum \epsilon_k c^\dagger_{k\sigma} c_{k\sigma} + J^z S^z \sum c^\dagger_{k'\alpha} \sigma_{\alpha\beta}^z c_{k\beta} + J^{\pm} \sum \left( S^+ c^\dagger_{k',-} c_{k,+} + S^- c^\dagger_{k',+} c_{k,-}\right).$$
The below is the part of the article https://www.cond-mat.de/events/correl15/manuscripts/nevidomskyy.pdf.



Eq.(8) is the familiar definition of $T$ matrix. In Eq.(9), it seems that $\Delta \hat T$ is defined as
$$\Delta \hat T=\hat V \frac{1}{w-\hat H_0} \hat V.$$
My question is on Eq.(10). How one can come to this equation? Also, I think that $\hat H_0$ in Eq.(10) is a typo by compared to Eq.(11). The result about $\hat V$ that I can obtain is as follows:
$$\langle k'| \Delta \hat T | k \rangle= \sum_q \langle k|V|q\rangle \langle q \left| \frac{1}{w-\hat H_0} \right| q\rangle \langle q|V|k\rangle  $$
where we ignore spin. Since $\langle q \left| \frac{1}{w-\hat H_0} \right| q\rangle= 1/(w-\epsilon_q)$, there is no way of having $1/(w-\epsilon_q+\epsilon_k)$ in the denominator! Also I am mysterious about the summation range of $q$ in Eq.(10).
 A: 1) I agree that the $\hat{H}_0$ in Eq.10 seems like a typo. I don't see a reason it should be there.
2) The operation of $\hat{V}$ on the conduction electrons is to remove an excitation with energy $\epsilon_k$ and scatter it into an excitation with energy $\epsilon_q$. Therefore, the total change in energy is $\epsilon_q-\epsilon_k$ and this is the term that will appear in the denominator. I think that when you wrote $\langle q | V | k \rangle$ you thought of a single-body setup, where here we have to deal with a lot of particles. So the states are a many-body particles and we have to count holes as negative energies when operating with $\hat{H}_0$.
You plug in the identity operator to get $|q\rangle \langle q|$ but these are not single-body states $|q\rangle \neq c^{\dagger}_q |{\rm vac}\rangle$. Rather, the identity is comprised of all many-body states $I = \sum_N\sum_{\{n_k\}}|k_1,\ldots,k_N\rangle \langle k_1,\ldots,k_N| $ where you sum over number of particles $N$ and different occupation of states $\{n_k\}$ and cover all possible combinations. Now, naturally, the only states that will contribute are those with the same number of particles as the initial state. So if we start with a state $|k_1,\ldots,k_N\rangle$ and operate with $V$ we will get to $|q,k_1,\ldots,\displaystyle{\not} k_j,\ldots,k_N\rangle$. That is your $|q\rangle$, and when $\hat{H}_0$ acts on it its eigenvalue (when we set the groundstate energy to zero) is $\epsilon_q-\epsilon_k$.
3) The range of summation of $q$ is the crux of Anderson's RG. We integrate out the high-momenta (~energy) excitations, and incorporate them into an effective Hamiltonian that acts on low-energy (~momenta) degrees of freedom. By examining how this effective low-energy behaves as we integrate out more and more of the high-energy excitations we can deduce how the setup will behave at low temperatures (that is - determine what will be the dominant interactions and how will they scale with lowering the temperature). To do that, we enforce a cutoff $\Lambda$ and then sum over small range of high-momenta excitations all $\Lambda-\delta \Lambda < q < \Lambda$, with the result being a Hamiltonian with a slightly lower cutoff $\Lambda-\delta\Lambda$, and then repeat the process again and again.
