I'm currently working through a textbook in statistical mechanics and wanted you to check, wether my attempt on the solution of a problem is right.
The problem is as follows:
Consider a system of $N$ spins where the prohability for each spin to be "spin up" or "spin down" is equal. Let $m$ be the numbers of the spins which are in the state "spin up" Determine the expectation value and the variance of $M=2m-N$ (this is the magnetisation of the system).
My approach is as follows:
i) Clearly the prohability that m particles are in the state "spin up" is $w(N,m)=\binom{N}{m} p^{m}q^{N-m}$ as this is a binomial distribution.
ii) $M=2m-N$ and from this it follows that $w(N,M)=\binom{N}{M/2+N/2}p^{M/2+N/2}q^{N/2-M/2}$
iii) I have to evaluate the sum $E=\sum_{M=-N}^N w(N,M)M$ and $V=\sum_{M=-N}^N w(N,M)M^2$ which gives me $E=0$ and $V=N^2-N$. (using $p=q=1/2$)
Does that make sense to you? Thanks in advance.