Interaction Term in Tomonaga-Luttinger Model 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.
 A: In the Tomonaga-Luttinger Model we are considering an effective low energy model for a 1D Fermi system. The low energy excitations of a (non interacting) Fermi gas in any dimension take a particle from just below the Fermi surface to just above the Fermi surface. In 1D the Fermi surface consists of two points (the Fermi points) at $\pm k_F$. So there are really two qualitatively different low energy excitations in 1D: Ones that excite a Fermion from just below one Fermi point to just above it ($q\sim 0$) and ones that excite a Fermion from just below one Fermi point to just above the other Fermi point ($q\sim \pm 2k_F$). This is very different from higher dimensions where we can move continuously around the Fermi surface.
Indeed when we linearize the band around the to Fermi points we have already essentially cut the band into two disconnected patches, so have to treat the excitations within one patch differently from excitations between the two patches, as we have thrown away the part of the band (far below the Fermi level) which interpolated between the two.
