# What is the relation between chemical potential and the number of particles?

Chemical potential is defined as the change in energy due to change in the number of particles in a system. Let we have a system which is defined by the following Hamiltonian: $$H = -t \sum_i^L c_i^\dagger c_{i+1} + V\sum_i^L n_i n_{i+1} -\mu \sum_i^L n_i$$ where $$c^\dagger (c)$$ are creation (annihilation) operators, $$n$$ is number operator, $$t$$ is hopping parameter, $$V$$ is nearest-neighbor interaction, $$L$$ is the total number of sites and $$\mu$$ is chemical potential.

What I understand by chemical potential is, if we set $$μ=$$some constant, then no matter how many sites ($$L$$) we add to the system, the number of particles will always be conserved. (Please correct me if I am wrong)

QUESTION:

What is the relation between chemical potential and the number of particles? i.e. if I set $$μ = 10$$ then how many particles are allowed in the system?

• en.wikipedia.org/wiki/… and possible duplicate physics.stackexchange.com/questions/92314/… – N. Steinle Feb 5 at 21:22
• Thermodynamically, $\mu = \partial G / \partial N$ and is the free energy cost of adding another particle. Of you "fix" $\mu$, that says nothing about $N$. – Jon Custer Feb 5 at 21:51
• For an ideal gas of bosons or fermions, the chemical potential determines the mean number of bosons (fermions) at the ground state $N_0=\frac{1}{\exp(-\mu/kT)\mp 1}$ (the minus sign refers to bosons, plus - to fermions . If the particle density is also specified, the chemical potential determines the mean value of the total number of particles $N$. – Aleksey Druggist Feb 6 at 8:43

## 1 Answer

At zero temperature, to find the relation between $$\mu$$ and the particle number you have to know the ground-state energy $$E_N$$ of the system with $$N$$ particles, and then $$\mu= E_{N+1}-E_N$$ Consequently you have to solve your Hubbard model exactly before anything else.

Once you have done this, you can approximate the definition of $$\mu$$ as $$\mu = \frac{\partial E}{\partial N}$$ (where $$E=E(N)=E_N$$) and from this obtain $$N$$ as a function of $$\mu$$ by means of Legendre transformation. Set $$\Phi= E-\mu N$$ Then $$\frac{\partial \Phi}{\partial \mu} = \frac{\partial E}{\partial\mu} -N-\mu \frac{\partial N}{\partial \mu}$$ $$=\frac{\partial E}{\partial N}\frac{\partial N}{\partial\mu } - N- \frac{\partial E}{\partial N}\frac {\partial N}{\partial \mu}$$ $$= -N$$

At finite temperature a thermodynamic system with a fixed chemical potential must be in a grand canonical ensemble and therefore free to exchange particles with a reservoir. Consequently the particle number is not fixed but instead its average $$$$ is determined by $$ = - \frac{\partial \Phi}{\partial \mu}$$ where now $$\Phi \to E-TS-\mu N$$ is the thermodynamic grand potential.

• Thank you. I need $T=0$. Just another quick question, does $\mu$ remain same for all particles? i.e. if $E_2 -E_1 = a$, will $E_{10}-E_9=a$? – Sana Ullah Feb 6 at 11:19
• @Sana Ullah. No. $E_{N+1}-E_N$ will depend on $N$, that is why I said you had to solve your Hubbard model first (and good luck with that! -- it is a famously hard problem!) – mike stone Feb 6 at 14:17
• wow. actually i wanted to fix the chemical potential such that number of particles always remain half of number of sites, for exact diagonalization of the Hubbard model. I need results for only 10 sites. You might want to have a look at my so far progress: physics.stackexchange.com/questions/459178/… . thank you for your time, Sir. – Sana Ullah Feb 6 at 14:31