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.

  • $\begingroup$ Doesn't $\delta(E+E_k)$ account for the negative ones? $\endgroup$ – leongz Mar 2 '17 at 8:55
  • $\begingroup$ Yes. Exactly. But I was curious about the coefficients of $\delta(E+E_k)$ and $\delta(E-E_k)$ $\endgroup$ – swagatam Mar 3 '17 at 20:16
  • $\begingroup$ See physics.stackexchange.com/questions/316542/… $\endgroup$ – 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.

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