# Is a quasiparticle just an eigenstate of the Hamiltonian?

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?

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 $$E_0$$.
• Was it intentional that the Hamiltonian in the $\psi$ basis was already diagonal or should $\mathcal{H}_k$ be $\mathcal{H}_{k,l}$? Dec 7, 2020 at 20:30
• $\mathcal{H}_k$ is not necessarily diagonal as I've written it. For instance, one term in the sum over $k$ could look something like $\psi^\dagger_k \mathcal{H}_k\psi_k = \begin{pmatrix}c_k^\dagger & c_k\end{pmatrix}\begin{pmatrix}h_{11}(k) & h_{12}(k) \\ h_{12}^\dagger(k) & h_{22}(k) \end{pmatrix}\begin{pmatrix}c_k \\ c_k^\dagger\end{pmatrix}$