Understanding summations over microstates of a given function

Question

I am struggling to understand how to sum over microstates in statistical mechanics.

Consider an $N$-spin system where $N \gg 1$ and $\Gamma=\{n_i \}$ for $1 \leq i \leq N$ and each $n_i$ is equal to $0$ or $1$. We want to compute the sum over the microstates $$\sum_{\Gamma}F(\Gamma)=\sum_{\{n_i\}}F(\{n_i\})$$ of a given function $F(\Gamma)$. Evaluate the sum over microstates for $F(\Gamma)=1$, $F(\{n_i\})=a{{\sum_{i=1}^N}n_i}, a>0$.

Question 1) I cannot see why we specify the functions on both sides of the summation, I would have thought specifying only would suffice.

Now it is clear to me that $\sum_{\Gamma}F(\Gamma)=\sum_{\Gamma}1=2^N$, but apparently,

$$\begin{align}\sum_{\{n_i\}}F(\{n_i\}) &= a{{\sum_{i=1}^N}n_i} \end{align}$$

so

$$\sum_{\Gamma} \left( a{\sum_{i=1}^N} n_i \right) = \sum_\Gamma a(n_1+n_2+ \dots n_N)=a\sum_{i=1}^N \sum_{\Gamma}n_i$$

This I am happy with, but apparently

$$\sum_{\Gamma}n_1=\sum_{n_1=0}^1 n_1 \sum_{n_2=0}^1 1 \dots \sum_{n_N=0}^1 1 = 2^{N-1} $$

and so on for $n_2,n_3,\dots$.

Question 2) I cannot see why we have fixed $n_2 \dots n_N$ to be equal to $1$?

Answer 1

The method I will present here is very general and by training this is the first that comes to my mind. There might be a simpler one though.

The idea is to consider another function $G(\Gamma) \equiv e^{-g F(\Gamma)}$ with $g > 0$ and $F(\Gamma) = a \sum_i n_i$.

The sum I want to compute is now $S_g \equiv \sum_{\Gamma} G(\Gamma)$.

This sum is actually quite easy to compute.

One way to do so is to realize that $F(\Gamma)$ will take a value proportional to the number of spins with value 1. Let us call this number $N_+$, we thus have $F(\Gamma) = aN_+$.

We can now split the sum over the configurations $\Gamma$ in the following way:

\begin{eqnarray} \sum_{\Gamma} \rightarrow \sum_{N_+ = 0}^N \sum_{\{\Gamma| F(\Gamma) = aN_+\}} \end{eqnarray}

Now the original sum can be calculated easily:

\begin{eqnarray} S_g = \sum_{N+ = 0}^N \: \frac{N!}{N_+!(N-N_+)!}e^{-gaN_+} = \sum_{N+ = 0}^N \: \frac{N!}{N_+!(N-N_+)!} \left(e^{-ga} \right)^{N_+} \end{eqnarray}

This sum is trivial as to estimate as it is the normal Binomial formula which gives: \begin{eqnarray} S_g = \left(1+ e^{-ga} \right)^N \end{eqnarray}

Now, the trick is to realize that $\left(\frac{\partial S_g}{\partial g}\right)_{g = 0} = - \sum_{\Gamma} F(\Gamma)$

Taking the derivative of the expression we have calculated before gives us then: \begin{eqnarray} \sum_{\Gamma} F(\Gamma) = aN 2^{N-1} \end{eqnarray}

Note that this is consistent with what we could expect from the average value of $N_+$ since the sum you have to calculate is the mean value of $N_+$ up to constant.

Intuitively, since having value 1 is as likely as having value 0, it means that the average value for $N_+$ has to be $N/2$.

This is what we get, by taking our final result and dividing by the first sum (normalization) you calculated, it gives:

\begin{eqnarray} \langle aN_+\rangle \equiv \frac{\sum_{\Gamma} F(\Gamma)}{\sum_{\Gamma} 1} = \frac{aN2^{N-1}}{2^N} = a\frac{N}{2} \end{eqnarray}