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)


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?

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    $\begingroup$ en.wikipedia.org/wiki/… and possible duplicate physics.stackexchange.com/questions/92314/… $\endgroup$ – N. Steinle Feb 5 '19 at 21:22
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    $\begingroup$ Thermodynamically, $\mu = \partial G / \partial N$ and is the free energy cost of adding another particle. Of you "fix" $\mu$, that says nothing about $N$. $\endgroup$ – Jon Custer Feb 5 '19 at 21:51
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    $\begingroup$ 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 $N$. $\endgroup$ – Aleksey Druggist Feb 6 '19 at 8:43

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 $<N>$ is determined by $$ <N> = - \frac{\partial \Phi}{\partial \mu} $$ where now $$ \Phi \to E-TS-\mu N $$ is the thermodynamic grand potential.

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  • $\begingroup$ 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 $E_{10}-E_9=a$? $\endgroup$ – Sana Ullah Feb 6 '19 at 11:19
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    $\begingroup$ @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!) $\endgroup$ – mike stone Feb 6 '19 at 14:17
  • $\begingroup$ 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. $\endgroup$ – Sana Ullah Feb 6 '19 at 14:31

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