Nr. of micro-states for an Einstein solid

Question

We have $\frac N 3$ atoms in a block made of some material. That means we have $\frac N 3$ 3D oscillators or N-1D oscillators. For one oscillator, its energy is :

$E_n^i= \hbar \omega(n+\frac 1 2)$, n= 0,1,2... and i=1,2...N

Ground state energy of the system would be:

$E_0 = N \frac {\hbar \omega} 2$

Then we define q as:

$q=\frac {E-E_0} {\hbar \omega}$

q counts the amout of energy qunta ($\hbar \omega$) that have been added to the system.

A microstate is chracterized by the set of all the values of $n^i$.

A macrostate is characterized by the value of q.

The multiplicity for a macrostate with an arbitrary q value is (using starts and bars from combinatorics):

$\Omega (q) = \frac {(N-1+q)!}{q!(N-1)!}$

Now my question is about finding the probability of a macrostate. For that I need to do the division between multiplicity of the macrostate with the nr. of microstates. And the problem is that I don't know how to calculate the nr. of microstates for a solid. So how to calculate the nr. of microstates? What is the number of microstates?

This is how I tried to find the nr. of microstates:

If I take into consideration the nr. of microstates (the set of values for the $n^i$'s) corresponding to the q value, then I simply get the multiplicity of the macrostate I am currently observing and the probability will be 1. That is of no interest.

But on the other side q can have infinite values, which means that we have infinite microstates, which means the probability of the above mentioned macrostate, with q-value, is zero. This also makes no sense.

The only thing that I thought, that it can somehow work is if we consider a range of q values, meaning we observe the system (solid) for energy values of the whole system within an interval. That means that different energy values represent different q values, in other words different macrostates. In this case the nr. of microstates is not the same as the multiplicity of the macrosystem, whose probability we are searching and at the same time it's not infinity. But it has a certain value. And the probability is a value between 0 and 1. Which means, there is a probability between 0 and 1 that the system has x value of energy, which is one of the possible values that the system can take, from the total range of energy values.

Is this a way how to find the nr. of microstates for a solid?

Answer 1

Wouldn't it simply be $p(q)=\frac{\Omega(q)}{\sum_s\Omega(s)}$? This would be the probability that the system is in a macroscopic state with the exact value of q. Still, if there are infinite values, the probability doesn't have to be zero! This would be true if you are dealing with probability density functions - this is, if the random variable is continuous-.

You can compute the same using continuous variables: Defining $x=\frac{q}{N}$. Then $x$ characterizes the microstate and $P(x)dx=\frac{\Omega(x)}{\sum_s\Omega(x)g(x)}$ is the probability that the macrostate is in [x,x+dx]. Where $g(x)$ is the density of microstates with a certain value of x.