Statistical Mechanics - Physics at Oregon State University

102 CHAPTER 5. FERMIONS AND BOSONS
Next, we replace the endpoints in all integrals by ±∞, which again introduces
errors of order e −ν . This seems obvious at first, but we need to take some care
since there is a summation in n involved, which easily could give diverging factors.
For large values of ν the additional terms in the integral are approximately
∞
1−ɛ un e −νu du and similar for the other part. Since ɛ is very small, we can replace
this by zero in this integral. We can now easily show that the integral is
proportional to e −ν n!ν−n. Summing this over n, ignoring the small variation
in the binomial factor as a function of n, gives an error term proportional to
1 −ν+ e ν . Hence we get:
4
f 3 (λ) = √ ν
2 3π 5
2
∞
n=0
and using v = νu we finally get:
4
f 3 (λ) = √ ν
2 3π 3
2
∞
ν −n
n=0
3 ∞
2
n −∞
3 ∞
2
n −∞
u n eνu (eνu + 1) 2 du + O(λ−1 ) (5.68)
v n ev (ev + 1) 2 dv + O(λ−1 ) (5.69)
The power series is in terms of inverse powers of ν or log(λ), which are the
slowest terms to go to zero. For large values of λ these are the only important
terms. The ratio of exponents in the integrants is an even function of v, and
hence only even powers of n remain in the expansion, because the integrals are
zero for odd powers.
Therefore, if βµ ≫ 1 it follows that
4
f 3 (λ) ≈
2 3 √ 3
(βµ) 2
π
1 + π2
8 (βµ)−2
(5.70)
Hence in the limit T → 0 we only need to take the first term and we find, with
EF = µ(T = 0):
n
4
= (2S + 1)
nQ(T ) 3 √ π (βEF ) 3
2 (5.71)
and using the definition of the quantum density:
or
2 2π
n
M
which takes the familiar form
3
2
= (2S + 1) 4
3 √ π (EF ) 3
2 (5.72)
√ 2
3 2
3n π 2π
4(2S + 1) M = EF (5.73)
2 6nπ
EF =
2S + 1
2
3 2
2M
(5.74)