When we are introduced to the density of states in typical band-theory problems we neglect interaction between electrons, and thus we define the density of states of a sigle particle as: $D(E)=2\int_{1st BZ}\delta (E-\epsilon_\mathbf{k})d\mathbf{k}$ since the possible states available to occupy for an electron lie within a band (I assume here there is just one band described by $\epsilon_\mathbf{k}$).
Now, when we switch on the interactions the concept of single-electron density of states seems ill-defined to me, since there is not such a thing as "possible energy of an electron". Take the 1D Hubbard model Hamiltonian within mean-field approximation:
$H=\sum_{k}(\epsilon_{k\uparrow}n_{k\uparrow}+ \epsilon_{k\downarrow}n_{k\downarrow})-U N\langle n_\uparrow \rangle \langle n_\downarrow \rangle$
where $\epsilon_{k\sigma}=-2t\cos k+\langle n_{-\sigma}\rangle U$. $U$ is the on-site Coulomb repulsion and $t$ the hopping term and $\sigma$ the spin (=$\pm$).
In this case the eigenstates are multiparticle states, and thus we can ask only for the energy of the system, or average of one particle. How can I then calculate the DOS in this context? How do I define my bands in first place if I don't know what's the dispersion relation?
Answer
You are right that in general it doesn't make sense to talk about single particle DOS in interacting systems. If the interaction is weak you can treat it as a perturbation, or as an effectively non-interacting system by rescaling some parameters (like in the Fermi liquid theory), and you can still talk about the DOS. In strongly interacting systems you can sometimes identify quasiparticle excitations which are weakly interacting and then talk about the DOS of those quasiparticles.
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