Can Fermi liquid be obtained by a canonical transformation? The basic assumption of the Ferm-liquid theory is the one-to-one correspondence between the states of an interacting Fermi gas to those of a gas of non-interacting quasiparticles.  The question is then, whether one can perform a canonical transformation to tranform the interacting Hamiltonian to a non-interacting one.
Remarks:

*

*It is understood that Fermi liquid description is approximate, so I expect some kind of approximatate procedure along the lines of the Schrieffer-Wolff transformation or the the genuine guesses used for spin Hamiltonians (Holstein-Primakoff, Jordan-Wigner, etc.)

*A close relative of the Fermi liquid is Luttinger liquid, which is exactly mapped onto a collection of non-interacting bosons. In this respect, I would like to stress that I am looking for a canonical transformation, similar to canonical bosonization, as described in the reviews of Haldane, Voit or Giamarchi's book (as opposed to more recent popular bosonization via path integrals).

 A: It seems that it can be performed in terms of RPA (random phase approximation). In this set up, one should consider Bose-operators for particle-hole pairs,
$$ c_{p,k}=a^{\dagger}_pa_{p+k},\quad c_{p,k}^{\dagger}=a_{p+k}^{\dagger}a_p,$$
where $|p|<p_F$ and $|p+k|>p_F$. Then particle density operator can be expressed in terms of the introduced Bose-operators,
$$\rho_k=\sum_p(c_{p,k}+c_{-p,-k}^{\dagger}). \tag{*}$$
The exact Hamiltonian looks like
$$H=\sum_{p,\sigma}\zeta(p)a_{p,\sigma}^{\dagger}a_{p,\sigma}+\frac{1}{2}\sum_kV_k\rho_k\rho_{-k}\quad \zeta(p)=\frac{p^2}{2m}-\mu,$$
so in terms of introduced operators $c$ and $c^{\dagger}$ it has "free form". Next step is to consider commutation relations and bla-bla. At the final step it is convenient to rotate operators as
$$\phi_{p,k}\equiv \phi_{-p,-k}^{\dagger}=\frac{1}{\sqrt{2\omega_{p,k}}}(c_{p,k}+c_{-p,-k}^{\dagger}),$$
$$\pi_{p,k}\equiv \pi_{-p,-k}^{\dagger}=\frac{i}{2}\sqrt{2\omega_{p,k}}(c_{p,k}-c_{-p,-k}^{\dagger}),$$
where
$$\omega_{p,k}=\frac{(p+k)^2}{2m}-\frac{p^2}{2m}.$$
In conclusion, the Hamiltonian becomes
$$H_{0}=\frac{1}{2}\sum_{p,k}\left(\pi_{p,k}^{\dagger}\pi_{p,k}+\omega_{p,k}^2\phi_{p,k}^{\dagger}\phi_{p,k}\right),$$
$$H_{\text{int}}=\sum_kV_k\left(\sum_p\sqrt{\omega_{p,k}}\phi^{\dagger}_{p,k}\right)\left(\sum_{p'}\sqrt{\omega_{p',k}}\phi_{p',k}\right).$$
The sum of $H_0$ and $H_{\text{int}}$ can be diagonalized with help of Bogoliubov transformation. Resulting spectrum contains two components: 1) continuous branch $\omega=\omega_{p,k}$ (coincides with non-interacting spectrum), 2) collective branch. The dispersion law for collective branch is given by
$$1=V_k\sum_p\frac{\omega_{p,k}}{\omega^2-\omega_{p,k}^2},$$
which corresponds to plasmon mode.
Where the approximation occurs? In the line $(*)$. The strict denotation is
$$\rho_k = \sum_{p\in R_k}(c_{p,k}+c_{-p,-k}^{\dagger}), \tag{**}$$
where $R_k$ is the sickle-shaped domain that is defined by conditions $|p+k|>p_F$ and $|p|<p_F$. In such region the operator $\rho_k$ is real, which means $\rho_k=\rho^{\dagger}_{-k}$. Such approximation $(**)$ implies that we omit terms $a_p^{\dagger}a_{p+k}$ that raise zero when act on state with single particle-hole pair (and on ground state)
