# Why does the superconductivity hamiltonian have a µ term, while the superfluid does not?

In every discussion of SC and SF that I read (e.g. Simons), the SC Hamiltonian (BCS) has a $\epsilon_k - \mu$ in the kinetic part of the Hamiltonian, while the SF Hamiltonian has just a $\epsilon_k + g$ term, where the $g$ stems from the interaction.

Is there an intuitive reason or is it only tradition?

• It's certainly because the superfluid (SF) is made of bosons, isn't it ? – FraSchelle Apr 18 '15 at 11:39

You are considering a system of fermions (SC hamiltonian) and a system of bosons (weakly interacting BEC). In order for the density to be finite, fermions must have a positive chemical potential $\mu>0$. On the other hand, the chemical potential of a system of bosons is less or equal to the energy of the lowest-energy state. In a non-interacting system this means that $\mu\leq 0$. The condensation occurs exactly when the chemical potential hits the energy of the lowest-energy state from below.
If you consider a Bose-Einstein condensate of weakly interacting particles on the Gross-Pitaevskii equation level (low temperature), you will find that $$\mu = g n,$$ where $g$ is the T-matrix describing the interactions and $n$ is the density of particles. But this is just an artefact of the fact that all the energies are shifted by a mean-field interaction energy, namely, $g n$.