Nr. of microstates and macrostates for a system

Question

Let's say we have a system S (a quantum gas,either a boson or a fermion-gas), made up by many subsystems, which we will index with $i$. One subsystem is characterized by :

$\bar {\epsilon_i}$ it's average energy value.

$g_i$ nr. of different energy values that a particle located in this subsystem can take.

$n_i$ the number of particles in the subsystem.

Now, if we only observe an arbitrary subsystem with an average energy $\bar {\epsilon_i}$:

A microstate, would be one arrangement of the $n_i$ particles in the $g_i$ energy values. If we for a moment do not concern ourselves with the type of gas (single occupancy or multiple occupancy) and the type of particles (distinguishable or indistinguishable), but we simply say that the number of microstates, the number of possible arrangements of $n_i$ particles in the $g_i$ energy values is $w(i)$.

Now the problem for me is the number of macrostates.

A macrostate of the subsystem can have as it's characteristic the energy value when $n_i$ particles are placed in the $g_i$ energy values. So:

$E_i=\Sigma_{k=1}^g n_i^k\epsilon_i^k$.

I want to know, what is the number of macrostates for the subsystem?

The number of the macrostates should be smaller then the nr. of microstates. For example, we can have x arrangements of the particles, whose totall energy is the same. This is a macrostate with a multiplicity of x. So how do I find the nr. of macrostates?

Answer 1

The terms used here is sort of confusion.Let's first make these straight.

The macroscopic system is specified by some thermo-dynamical parameters, like total number $N$, volume $V$,and average energy $U$. These parameters define a macroscopic state, one macroscopic state.

Then, with these given parameters, what is the number of microscopic states satisfied these macro-parameters? This is knows as the number of microscopic configurations, or the microscopic multiplicity. Precisely, we may call it the microscopic multiplicity of the given macroscopic state.

For your case, we may neglect the index $i$ for simplicity. The constrain conditions for counting the multiplicity is \begin{align*} \sum_k n_k &= n.\,\,\, \text{neglected the index $i$.}\\ \sum n_k \epsilon_k &= E. \end{align*} And the multiplicity is $$ g(n, E) = \frac{n!}{\prod_k n_k!} $$

Then, assume to work in canonical ensemble, we the re-store the index $i$ ($n_i = n$ for canonical ensemble) to sum for partition function: $$ Z_n = \sum_i g_i(n, E_i) e^{-\beta E_i} = \sum_i\frac{n!}{\prod_k n_k!} e^{-\beta E_i} $$

Under grand canonical ensemble ($n\to n_i$, $n_k \to n_{ik}$): $$ Z_n = \sum_i g_i(n_i, E_i) e^{-\beta (E_i - \mu n_i)} = \sum_i \frac{n_i!}{\prod_k n_{ik}!} e^{-\beta (E_i - \mu n_i)} $$