I am studying Tomonaga-Luttinger Model from Altland and Simon's textbook called Condensed Matter Field Theory. From the derivation, I am stuck with showing that the contribution to the interaction term comes only from $(k, k', q) = (\pm k_F, \pm k_F, 0)$ and $(k, k', q) = (\pm k_F, \mp k_F, 2k_F)$.
To elaborate more on this, the textbook starts from the 1-D jellium model (sorry if the terminology is wrong), namely, $$H=\sum_k a_k^\dagger(\epsilon_k-E_F)a_k+\frac{1}{2L}\sum_{k, k', q\neq0}V_{ee}(q)a_{k+q}^\dagger a_{k'-q}^\dagger a_{k'}a_k \equiv H_0+H_1$$ From the above Hamiltonian, the book first linearizes the first term by expanding in the vicinity of the Fermi energy such that $$H_0=\sum_{s,q}\sigma_sv_Fqa_{sq}^{\dagger}a_{sq}$$ , where s denotes whether $q$ is expanded near $k_F$ or $-k_F$.
Also, noting that $\rho_{sq}=\sum_{k}a^\dagger_{s, k+q}a_{s,k}$, $$H_1 = \frac{1}{2L}\sum_{qs}\left[g_4\rho_{s,q}\rho_{s,-q}+g_2\rho_{\bar{s},q}\rho_{s,-q}\right]$$, where $\bar{s}$ denotes the complement of $s$.
What I do not figure out is the statement of the textbook that only $(k, k', q) = (\pm k_F, \pm k_F, 0)$ and $(k, k', q) = (\pm k_F, \mp k_F, 2k_F)$ contribute to the summation. For, in the course of using the relation $\rho_{sq}=\sum_{k}a^\dagger_{s, k+q}a_{s,k}$, the summation over the entire k and k' seems to have been already conducted. Furthermore, the Fourier transform of 1/r in 1-D is $-2\gamma_E+ln(1/q^2)$, so I understand $q \sim 0$ mostly contributes to the summation but do not figure out how come $q \sim 2 k_F$ also contributes to the summation.
I tried to find any other literature dealing with this issue but in vain. Could anyone please help me understand the abovementioned statement?
Thank you in advance.
