In BCS theory, we are given the energy of the quasi-particles as $E = \sqrt{\varepsilon^2 + \Delta^2}$ in which $\varepsilon = \frac{\hbar^2k^2}{2m} - \mu$, with $\mu$ being the chemical potential. As such, we find the quasi-particles exist with a minimum energy of $\Delta$.

When we say that the quasi-particles have a minimum energy of $\Delta$, are we saying that the quasi-particles exist at an energy that is at least $\Delta$ above the ground state? So we take the energy of the ground state to be zero? Does this mean that $\mu$ is measured relative to the energy of the ground state? So if $\mu = 5meV$, that means 5meV above the ground state?

  • $\begingroup$ BCS is a theory of solids, you are not speaking of ground state there but of energy relatively to the Fermi level. $\endgroup$
    – Mauricio
    Jun 30 at 16:47
  • $\begingroup$ But BCS theory clearly talks about the existence of a ground state? Usually the ground state is given as $|\Psi_{BCS}\rangle = \prod_k (u_k + v_kc_{k\uparrow}^{\dagger}c_{k\downarrow}^{\dagger})|0\rangle$. While the energy of the ground state is usually not explicitly stated, I would naively think it would lie below the quasi-particle excitations. So do the quasi-particles have an energy that is at least $\Delta$ above the ground state? $\endgroup$ Jul 7 at 4:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.