80
1 Mathematical Physics
N ! →
√
2π N N
N e
−N
x! →
√
2π x x
x e
−x
(N − x)! →
2π(N − x)(N − x)
N −x e
−N +x
B(x) → f (x) =
N
2π x(N − x)
p
x q
N −x N
N
x x (N − x) N −x
=
N
2π x(N − x)
N p
x
x
Nq
(N − x)
N −x
Let δ denote the deviation of x from the expected value Np, that is
δ = x − N p
then N − x = N − N p − δ = Nq − δ
f (x) =
2π N pq
1 +
δ
N p
1 −
δ
Nq
−1/2
1 +
δ
N p
−(N p+δ)
·
1 −
δ
Nq
−(Nq−δ)
Assume that |δ| | N pq so that
δ
N p
1 and
δ
Nq
1
The first bracket reduces to (2π N pq)
−
1
2 . Take log e on both sides.
ln
f (x)(2π N pq)
1
2
= −(N p + δ) ln
1+
δ
N p
−(Nq − δ) ln
1−
δ
Nq
= − (N p + δ)
δ
N p
−
δ
2
2N 2 p 2 +
δ
3
3N 3 p 3 − · · ·
− (Nq − δ)
−
δ
Nq
−
δ
2
2N 2 q 2 −
δ
3
3N 3 q 3 − · · ·
= −
δ
2
2N pq
−
δ
3 ( p
2
− q
2 )
6N 2 p 2 q 2
Neglect the δ
3 term and putting σ =
√
N pq and σ = x − N p = ¯
x.
f (x) =
1
σ
√
2π
exp
−(x − ¯
x)
2
2σ 2
(Normal or Gaussian distribution)