# Why is there a chemical potential term in the Bose-Hubbard Hamiltonian?

When looking at the Bose-Hubbard Hamiltonian $$H_{BH}=-t\sum_{\langle i,j\rangle}a^\dagger_ia_j+\frac{U}{2}\sum_ia^\dagger_ia_i(a_i^\dagger a_i-1)-\mu\sum_ia_i^\dagger a_i,$$ I'm wondering why there is the chemical potential term. In my understanding, the Hamiltonian is supposed to "measure" the energy of the system or more precisely, the eigenvalues are the energies of its eigenstates. While to me, the first two terms seem rather obvious (Energy attached to the "hopping" within the system and on-site interaction due to pseudo contact potential) I don't understand why there is the chemical potential term. Shouldn't this $$\mu=\frac{\partial E}{\partial N}|_{S,V}$$ already be accounted for with the first two terms? And more confusingly,if not, why is the sign negative? Last but not least, it appears that all the particles "carry" the same chemical potential but if we imagine a particle-by-particle filling of the lattice, then obviously this is not the case. So could someone clarify why this term is there and why it looks how it looks?

## 1 Answer

The model assumes your system of $$N$$ particles is connected to a reservoir of infinite particles. The chemical potential $$\mu$$ is the "tilt" between these two bowls of particles, allowing for particles to flow until equilibrium (same number of particles, fixed desired $$N$$) is reached. In the same way a potential difference between the two sides of a charged capacitor exists until both sides have the same total charge. This formalism is known as the grand-canonical ensemble.

$$\mu$$ controls the filling. How many particles there are in your system.
This affects the physics, because even if $$U\rightarrow \infty$$ but you only have $$1$$ particle, nothing interesting will happen. You always need to know what $$J$$ or $$U$$ are compared to $$\mu$$. As an example, if you plot the average number of particles per site in the Mott phase $$\bar n$$, it is a step ladder for integer values of $$\mu / U$$.

Shouldn't this $$\mu=\frac{\partial E}{\partial N}|_{S,V}$$ already be accounted for with the first two terms?

No. The first two terms do not depends on the number of particles. This extra term is telling how much the energy of the system changes by adding a new particle, while the other ones are telling you how much the energy of the system changes as you vary interactions and tunnelling (at fixed particle number).

And more confusingly,if not, why is the sign negative?

The chemical potential of bosons is negative, so really here it should be $$-|\mu|$$. Intuitively, you can justify it with "bosonic enhancement", i.e. bosons like to be in the same state so they lower their energy for that to be more favourable.

Edit: While part of my statement above is true, it is not why there is a $$-\mu$$ term. That also shows up in classical statistical mechanics and is just the result of derivatives and stuff when integrating out the reservoir. See e.g. here.

Last but not least, it appears that all the particles "carry" the same chemical potential but if we imagine a particle-by-particle filling of the lattice, then obviously this is not the case. So could someone clarify why this term is there and why it looks how it looks?

The chemical potential is set by this particle reservoir in the grand-canonical ensemble. It controls how many particles are in the system, it does not care about the system's specifics.

• Thank you very much for the nice explanation! Especially for pointing out that it is not $-\mu$ but $-|\mu|$! I am quite surprised that writing $-\mu$ seems to be a common notation when in fact it needs to be $+\mu$ (At least if we stick to $\mu:=+\frac{\partial E}{\partial N}|_{S,V}$ – Simon Aug 8 '20 at 20:15
• @Simon, Actually I am sorry, I am wrong. The chemical potential is indeed negative for bosons near Tc but the $-\mu$ occurs also in classical statistical physics. I've edited my answer. – SuperCiocia Aug 8 '20 at 21:25