In this paper we have the following:

The corresponding quasiparticle density is given by the equation $$n_{qp} = 4N_0 \int_\Delta ^\infty dE \frac{E}{\sqrt{E^2 - \Delta^2}} f(E),$$ where $N_0$ is the single-spin density of electron states at the Fermi energy.

Question: why is there a factor of 4 here, rather than a factor of 2? Given that each quasiparticle has two spin states, and the normalized density of states is $\rho(E) = \frac{E}{\sqrt{E^2 - \Delta^2}}$ shouldn't we have $2N_0?$

Reference 21 to the paper, in Appendix A, in fact says:

We define the quasiparticle density $N_{qp}$ as $$N_{qp} \equiv \int_{-\infty}^{+\infty} d\epsilon 2N(0) f(E),$$ where $E=(\epsilon^2 + \Delta^2)^{1/2}$. $N(0)$ is the normal-state single-spin density of states at the Fermi surface. The factor of 2 takes into account the spin degree of freedom for the electrons.

And in Appendix B:

Here the excess quasiparticle density $\delta N_{qp}$ is defined as $$\delta N_{qp} = \int_\Delta ^\infty dE 4N(0) \frac{E}{\sqrt{E^2 - \Delta^2}}[f(E)-f(E,T)].$$ We note that this is not the quasiparticle density difference between the driven state and the equilibrium state, because we have used the steady-state energy gap in Eq. (B4) [above equation] to calculate the quasiparticle density.

  • 1
    $\begingroup$ It looks like the answer might be found in the corresponding references, Refs. (20) and (21). Have you looked at these papers? $\endgroup$
    – march
    Commented Nov 14, 2023 at 18:51
  • $\begingroup$ @march Hi thanks, you are right that I should have checked here initally. Ref (20) seems to not be related. Reference 21 had some suggestive stuff, but it is quite long and dense. I have updated my question. $\endgroup$
    – Jbag1212
    Commented Nov 15, 2023 at 13:47
  • $\begingroup$ Minor comment to the post (v3): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. $\endgroup$
    – Qmechanic
    Commented Nov 15, 2023 at 17:13

1 Answer 1


Additional factor 2 is due to the fact that there are two branches of quasiparticles: with a momentum exceeding the Fermi momentum $p_F$, and with a momentum less than $p_F$. Both of these branches of quasiparticles have a minimum energy $\Delta$ and the same density of states near the minimum. Hence the additional factor 2. This can be seen even for an ideal Fermi gas. The existence of two branches of quasiparticles is also demonstrated by the second formula in the question. Integration in this formula is carried out over $\epsilon$ from $-\infty$ to $\infty$, but the quasiparticle energy $E = \sqrt{\epsilon^2+\Delta^2}$ is an even function of $\epsilon$ and for both signs of $\epsilon$ varies from $\Delta$ to $\infty$ (formally). Thus, the two branches of quasiparticles correspond to the positive and negative sign of $\epsilon$. Here $\epsilon$ is actually the difference between the energy of a free electron and the Fermi energy: $\frac{p^2}{2m} - \varepsilon_F$.

By the way, the first and second formulas in the question actually coincide. Since $E = \sqrt{\epsilon^2+\Delta^2}$ is an even function of $\epsilon$, the second formula can be rewritten as follows $$ N_{qp} = 4N(0)\int_0^\infty\ d\epsilon f(E).\tag{*} $$ Now let's change the integration variable $\epsilon$ to $E$ in the last integral. We have $\epsilon = \sqrt{E^2 - \Delta^2}$ and $d\epsilon = \frac{E\ dE}{\sqrt{E^2-\Delta^2}}$, so we obtain from (*) $$ N_{qp} = 4N(0)\int_{\Delta}^\infty\ dE\ \frac{E}{\sqrt{E^2-\Delta^2}}f(E). $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.