Past exam of the mathematics course of the University of Cambridge / 2017 / ia / Paper 2 / 11F / c

Solution

If exactly

k

spins equal

+ 1

, then

M = (2 k - n) / n

. Therefore its possible values are

E_{n}

, with

k = n (1 + m) /2

for

m \in E_{n}

. Choosing the

k

positive spins gives the binomial coefficient

(k n) = \frac{n !}{[ n ( 1 + m ) /2 ]! [ n ( 1 - m ) /2 ]!} .

Every such configuration has the same Curie–Weiss model weight because

\sum_{i, j} s_{i} s_{j} = (\sum_{i} s_{i})^{2} = n^{2} m^{2}

P (M = m) = \frac{1}{Z _{n, β}} (n ( 1 + m ) /2 n) e^{β n m^{2} /2}, m \in E_{n},

and zero elsewhere. Grouping the original partition function by the same count also gives

Z_{n, β} = \sum_{k = 0}^{n} (k n) exp [\frac{β ( 2 k - n ) ^{2}}{2 n}],

which verifies normalization without counting any configuration twice.

There are no discussions about this article yet.

There are currently no matching articles.