1
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ en.wikipedia.org/wiki/… and possible duplicate physics.stackexchange.com/questions/92314/… $\endgroup$ – N. Steinle Feb 5 at 21:22
  • 1
    $\begingroup$ 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 at 21:51
  • 1
    $\begingroup$ 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 at 8:43
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 at 11:19
  • 1
    $\begingroup$ @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 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 at 14:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.