Microcanonical Ensemble of a two state system

Question

I have a system consisting of N distinguishable particles. Each particle has two states, one with energy E and the other with energy 0. The number of particles in the state with energy 0 is $n_{0}$ and the number of particles in the second state is $n_{1}$. The number of particles in the different states satisfy $N=n_{0}+n_{1}$. Now I want to compute the number of available microstates in the microcanonical ensemble that are in agreement with the energy constraint $U=n_{1}*E$.

Since the particles are distinguishable I can arrange them in $N!$ ways. And since I have two states the total number of possibilities to arrange this system is $2*N!=2*(n_{0}+n_{1})$. Hence the partition function is:

$$ \Omega(U,N)=2*(n_{0}+n_{1})! $$

Now I want to compute the entropy in the microcanonical ensemble and maximize it with respect to U. I guess the maximum will be at $n_{0}=n_{1}=N/2$ But I don't know how to show this. My feeling somehow tells me I should use Lagrange multipliers and maximize $\Omega$.

Answer 1

Since the particles are distinguishable, the answer is simply the number of possible combinations of $n_1$ elements drawn from a set of $N$ elements:

$$ \Omega (U,N) = \binom{N}{n_1} $$

Where $U = n_1 E$.

You can easily show that the entropy is maximized at $n_1 = N/2$ via the properties of the number of combinations - it is maximized at that number.

Answer 2

I figured my problem now out maybe it is also of interest for others. I have in total N labeled particles and hence I can arrange them in $N!$ ways. Now I distribute the particles in two boxes. The boxes dont't have any drawers or anything and hence in the end I am not able to discriminate between the different orders in which the particles where put into the boxes. In other words my result has to be independent of the drawing process. And hence I have to divide the total number of arrangements $N!$ by the number of particle permutations for every box. This is then just:

$$ W=\frac{N!}{n_{1}!n_{0}!} = { N \choose n_{1}} $$ like already pointed out by Al Nejati. Next the entropy in the microcannonical ensemble is given by (k_{b}=1):

$$ S=ln\left ( \frac{N!}{n_{1}!n_{0}!} \right )=ln(N!)-ln(n_{1}!)-ln(n_{0}!)\approx \\ Nln(N)-N -n_{1}ln(n_{1}) + n_{1} - n_{0}ln(n_{0}) + n_{0}=\\ Nln(N)-n_{1}ln(n_{1})-n_{0}ln(n_{0}) $$

Now I want to maximize the entropy under the constraint U=n_{1}\epsilon. Therefore the function to maximize is:

$$ f(n_{1},N)=Nln(N)-n_{1}ln(n_{1})-n_{0}ln(n_{0}) - \lambda( U - n_{1}\epsilon ) $$ Differentiating with respect to $n_{1}$:

$$ \frac{\partial f}{\partial n_{1}} = ln(n_{1}) + 1 - \lambda\epsilon $$

setting this zero gives the maximum $n_{1}$: $$ n_{1}=e^{-1+\lambda\epsilon} $$