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The system we consider has constant $N$, $V$ and $T$ (the number of particles, volume and temperature) This is just the thermodynamic variables for the canonical ensemble, why we use fugacity $z$ or chemical potential $\mu$ there?

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In the thermodynamic limit fluctuations in particle number should tend to $0$ and so we should get the same result whether we use the canonical or grand canonical ensemble. If we are ultimately interested in a system with a fixed particle number, we can, if we so choose, set up the problem in the grand canonical ensemble and then invert the equation for $\langle N \rangle$ to find what chemical potential will give us the desired mean particle number.

For systems of identical quantum particles we generally choose to use the grand canonical ensemble because it turns out to be dramatically simpler. In such systems it is convenient to use an occupation number representation for states, rather than listing the state of each individual particle, since it can then be assumed that an appropriate (anti-)symmetrization procedure has been applied and not have to think about it. In the canonical ensemble, however, we now encounter a problem; since the total number of particles is fixed the occupation numbers of different states are not independent, but must satisfy a constraint. This leads to a rather convoluted combinatorics problem to list the set of allowed occupation numbers.

In the grand canonical ensemble we do not encounter this problem, since all values of $N$ are allowed. We simply consider all possible sets of occupation numbers, but since states with very large or very small particle number are exponentially weighted against, they do not contribute. In effect we allow the statistics of the problem to "automatically" find the states with the correct particle number for us.

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If you choose to set $N$, $V$ and $T$, then you are choosing the Canonical Ensemble, so $N$ is fixed. In this approach, the chemical potential $\mu$ is just a parameter that you've to determine by imposing that the total number of particles is $N$.

Viceversa, if you choose the grandcanonical ensemble, you fix $\mu$ and so the number of particles $N$, fluctuates around an average value $\langle N \rangle$. I hope that I've well understood your problem and answered properly.

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    $\begingroup$ Once again, we have a canonical ensemble (N is fixed), .... After several years, I realized that I was completely wrong, of course we do have an ensemble with $<N>$ fixed, and this is equivalent using $\mu$ and grandcanonical ensemble, you were absolutely right $\endgroup$ Jan 20, 2020 at 22:34
  • $\begingroup$ If $N$ is fixed there's no need for $\mu$, i.e. there's no $\mu$, I would say. $\endgroup$
    – lcv
    May 19 at 10:27

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