The description of quasiparticles seems to come in two flavors: Completely qualitatively, where it is simply said that different (quasi-)particles interact to "form" a quasiparticle, or quantitatively, but indirectly via characterization of, e.g., the effective mass of interacting electrons, or via association with peaks in spectral functions.

This makes my current understanding of the mathematical definition of a quasiparticle rather unsatisfactory. However, the characterization via peaks in the spectral functions makes me wonder: Is a quasiparticle simply an eigenstate of a (complicated, many-body) Hamiltonian?

I mean this in the following sense: If $|\psi_m\rangle$ and $|\psi_n\rangle$ are eigenstates of $H$, then is a quasiparticle simply the excitation created by the operator $a^{\dagger} = |\psi_m\rangle \langle \psi_n|$ (for appropriately chosen $m,n$)? If not, then what is the relationship between the two?


1 Answer 1


In some contexts yes and in others no.

When we talk about quasiparticles as corresponding to peaks in the one-particle spectral function, the width of the peak tells us the quasiparticle lifetime $\tau$. The very fact we are talking about a finite lifetime means it is not an energy eigenstate (which would of course be a stationary state).

However, sometimes we use the term "quasiparticle" to describe exact excitation states of a mean field many-body Hamiltonian. Such a Hamiltonian might look like $$ H_{MF} = \sum_k \psi_k^\dagger \mathcal{H}_k \psi_k + \text{const.}$$ where $\psi_k$ is a vector of creation/annihilation operators $c_k^\dagger/c_k$ for real electrons and $\mathcal{H}_k$ is a matrix. Through a canonical (unitary or Bogoliubov) transformation of our operators, we can obtain the diagonal form $$ H_{MF} = \sum_kE_k\gamma_k^\dagger \gamma_k +E_0 $$ where the new operators $\gamma_k^\dagger/\gamma$ are linear combinations of the real electron operators $c_k^\dagger/c_k$. Since the transformation is canonical, these new operators behave just like the original $c_k^\dagger/c_k$ and so they can be interpreted as creating/destroying something. We call this thing the quasiparticle. And indeed, from the above diagonal Hamiltonian, we see that these quasiparticles behave like free particles with dispersion $E_k$ and ground state energy $E_0$.

  • $\begingroup$ Was it intentional that the Hamiltonian in the $\psi$ basis was already diagonal or should $\mathcal{H}_k$ be $\mathcal{H}_{k,l}$? $\endgroup$ Dec 7, 2020 at 20:30
  • $\begingroup$ $\mathcal{H}_k$ is not necessarily diagonal as I've written it. For instance, one term in the sum over $k$ could look something like $\psi^\dagger_k \mathcal{H}_k\psi_k = \begin{pmatrix}c_k^\dagger & c_k\end{pmatrix}\begin{pmatrix}h_{11}(k) & h_{12}(k) \\ h_{12}^\dagger(k) & h_{22}(k) \end{pmatrix}\begin{pmatrix}c_k \\ c_k^\dagger\end{pmatrix}$ $\endgroup$
    – jgw
    Dec 8, 2020 at 19:00

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