# Energy of Quasi-particles in a BCS superconductor

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?

• BCS is a theory of solids, you are not speaking of ground state there but of energy relatively to the Fermi level. Jun 30 at 16:47
• 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? Jul 7 at 4:53