542
10 Quantum Statistics
with 2I a + 1 = 2, 0 (V ) ≡ 2π(2m e /h 2 )
3
2 V , so that 0 (V ))
1
2 represents the
density-of-states for a classical (i.e., spinless) particle.
In the same way, we may write the internal energy U as
U =
∞
0
f (()u(() d ,
(10.4.34)
with u(() given by as
u(() ≡ 4π
2m e
h 2
3
2
V V
3
2 .
(10.4.35)
The integral (10.4.34) for U could be evaluated numerically, but more can be
understood of the behaviour of U if we attempt to evaluate it analytically. To
accomplish this task, we shall integrate by parts, using f (()d as ‘dv’, and u(() as
‘u’ in the integration-by-parts formula
udv = uv −
vdu. Thus, upon carrying
out an integration by parts, we obtain the internal energy as
U =
∞
0
f (()u(() d = −
∞
0
f
(()U (() d ,
(10.4.36)
with U(() given by
U(() ≡
0
u(x) dx =
8π
5
2m e
h 2
3
2
V V
5
2 .
(10.4.37)
Because f (() exhibits a very sharp spike at = μ, most of the contribution to such
integrals comes from the region of around μ, and we may usefully make a Taylor
expansion around this point, and then carry out the integration term-by-term. Thus,
we expand U(() as
U(() = U(μ) + U
(μ)(( − μ) +
1
2!
U
(μ)(( − μ)
2
+ . . . ,
(10.4.38)
which, when substituted into Eq. (10.4.36) gives
U = −
f
(()U (() d
= −
f
(()
U(μ) + U
(μ)(( − μ) +
1
2
U
(μ)(( − μ)
2
+ . . .
d
= −
U(μ)
f
(() dd + U
(μ)
f
(()(( − μ) d