1000 Solved Problems in Modern Physics

80 1 Mathematical Physics

N!→

√

2 πNNNe−N

x!→

√

2 πxxxe−x

(N−x)!→

√

2 π(N−x)(N−x)N−xe−N+x

B(x)→f(x)=

√

N

2 πx(N−x)

pxqN−xNN

xx(N−x)N−x

=

√

N

2 πx(N−x)

(

Np

x

)x(

Nq

(N−x)

)N−x

Letδdenote the deviation ofxfrom the expected valueNp, that is

δ=x−Np

thenN−x=N−Np−δ=Nq−δ

f(x)=

[

2 πNpq

(

1 +

δ

Np

)(

1 −

δ

Nq

)]− 1 / 2

(

1 +

δ

Np

)−(Np+δ)

·

(

1 −

δ

Nq

)−(Nq−δ)

Assume that|δ|Npqso that

∣

∣

∣

∣

δ

Np

∣

∣

∣

∣1 and

∣

∣

∣

∣

δ

Nq

∣

∣

∣

∣^1

The first bracket reduces to (2πNpq)−

(^12)

. Take logeon both sides.

ln

[

f(x)(2πNpq)

12 ]

=−(Np+δ)ln

(

1 +

δ

Np

)

−(Nq−δ)ln

(

1 −

δ

Nq

)

=−(Np+δ)

(

δ

Np

−

δ^2

2 N^2 p^2

+

δ^3

3 N^3 p^3

−···

)

−(Nq−δ)

(

−

δ

Nq

−

δ^2

2 N^2 q^2

−

δ^3

3 N^3 q^3

−···

)

=−

δ^2

2 Npq

−

δ^3 (p^2 −q^2 )

6 N^2 p^2 q^2

Neglect theδ^3 term and puttingσ=

√

Npqandσ=x−Np=x ̄.

f(x)=

1

σ

√

2 π

exp

[

−(x− ̄x)^2

2 σ^2

]

(Normal or Gaussian distribution)