Density of states of Bogoliubov quasiparticles

Question

For a simple fermionic system the formula for calculating the density of states (DOS) is $N(E) = sum_{n}delta(E-E_{n})$ where ${E_{n}}$ is the set of eigenvalues obtained after diagonalizing the hamiltonian. Now to diagonaloize a hamiltonian with pair correlation terms ($sum_{k}c_{kuparrow}^{dagger}c_{-kdownarrow}^{dagger}$) Bogoliubov transformation ($c_{kuparrow}=u_{k}gamma_{kuparrow}-v_{k}^{ast}gamma_{-kdownarrow}^{dagger}; c_{-kdownarrow}^{dagger}=v_kgamma_{kuparrow}+u_{k}^{ast}gamma_{-kdownarrow}^{dagger}$) is used. Now after diagonalizing we get a set of eigenvalues in the form:${E_n,-E_n}forall n$. Now to find the density of states I found a formula like this: $N(E)=sum_{k}|u_k|^2delta(E-E_k)+|v_k|^2delta(E+E_k)$ where ${E_k}$ is the set of positive eigenvalues only. I don't understand this particular formula for density of states of bogoliubov quaisparticles. If anyone can explain it that would be very helpful.

Victor · Answer

Some information is missing, but I think that maybe if you expand the terms ($sum_{k}c_{kuparrow}^{dagger}c_{-kdownarrow}^{dagger}$) with the Bogoliubov transformation, some ortogonal operators may cancell and so you can separate the hamiltionian in terms of each $gamma$ operator and $|u_k|^2$ and $|v_k|^2$. After that maybe you can separate the whole system into positive and negative energy states.

Answered by Victor on September 23, 2020

Kartik Chhajed · Answer

Write BCS ground state (reference) $|Psi_{BCS}rangle$:
$$|Psi_{BCS}rangle=prodlimits_{bf k}(u_{bf k}+v_{bf k}c_{bf kuparrow}^dagger c_{bf -kdownarrow}^dagger)|0rangle$$
and calculate the number density expectation:
$$langle hat n_{bf k}rangle=langle c^dagger_{bf k}c_{bf k}rangle$$

Density of states of Bogoliubov quasiparticles

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