# Density of states of Bogoliubov quasiparticles

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 ($$\sum_{k}c_{k\uparrow}^{\dagger}c_{-k\downarrow}^{\dagger}$$) Bogoliubov transformation ($$c_{k\uparrow}=u_{k}\gamma_{k\uparrow}-v_{k}^{\ast}\gamma_{-k\downarrow}^{\dagger}; c_{-k\downarrow}^{\dagger}=v_k\gamma_{k\uparrow}+u_{k}^{\ast}\gamma_{-k\downarrow}^{\dagger}$$) 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: $$N(E)=\sum_{k}|u_k|^2\delta(E-E_k)+|v_k|^2\delta(E+E_k)$$ 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.

• Doesn't $\delta(E+E_k)$ account for the negative ones? – leongz Mar 2 '17 at 8:55
• Yes. Exactly. But I was curious about the coefficients of $\delta(E+E_k)$ and $\delta(E-E_k)$ – swagatam Mar 3 '17 at 20:16
• – Everett You Mar 10 '17 at 9:29

Some information is missing, but I think that maybe if you expand the terms ($\sum_{k}c_{k\uparrow}^{\dagger}c_{-k\downarrow}^{\dagger}$) 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.