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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).
Roger V.
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1 Answers1

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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)