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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).
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    $\begingroup$ The Fermi liquid and Luttinger liquid are about as different as can be, actually. In the Fermi liquid, the physically observable excitations share the same discrete quantum numbers as the excitations of the noninteracting theory; they behavior of the interacting theory is a continuous deformation of the noninteracting. However, in a Luttinger liquid, the underlying fundamental particles are fermions, but the observable effects are bosonic, something totally different. $\endgroup$
    – Buzz
    Commented Dec 15, 2021 at 17:15
  • $\begingroup$ @Buzz that's all true. What I mean is that in the Littinger liquid we can (nearly) exactly diagonalize the Hamiltonian, representing it in terms of elementary excitations. The question is whether similar thing can be done in the case of Fermi liquid. $\endgroup$
    – Roger V.
    Commented Dec 15, 2021 at 18:41
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    $\begingroup$ My vague feeling is that it can be related to Bogoliubov transformation. My point is based on superconductivity theory. Bogoliubov transformation is equal to choice of "true vacuum" and consider perturbation near it. It seems that in Fermi liquids the situation is the same: now vacuum related to Fermi surface and you just need to choose correct operators $\endgroup$ Commented Dec 16, 2021 at 8:57

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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)

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