1549380323-Statistical Mechanics Theory and Molecular Simulation

The ideal fermion gas 429

Performing the angular integrals, we obtain

ρ 1 (s) =

4 π

V

∫∞

0

dn n^2 θ(εF−εn)

L

2 πins

(

e^2 πins/L−e−^2 πins/L

)

=

4

L^2 s

∫∞

0

dn n θ(εF−εn) sin

(

2 πns

L

)

. (11.5.62)

For the remaining integral overn, transforming fromntoεnis not convenient because
of the sin function in the integrand. However, sincen >0, we recognize that the step
function simply restricts the upper limit of the integral by the condition

2 π^2 ̄h^2

mL^2

n^2 < εF

n <

(

mL^2 εF

2 π^2 ̄h^2

) 1 / 2

≡nF. (11.5.63)

Therefore,

ρ 1 (s) =

4

L^2 s

∫nF

0

dn nsin

(

2 πns

L

)

=

1

π^2 s^3

[

sin

(

2 πnFs

L

)

−

s

lF

cos

(

2 πnFs

L

)]

, (11.5.64)

where

lF=

(

̄h^2

2 mεF

) 1 / 2

. (11.5.65)

Givenρ 1 (s), we can now evaluate the exchange energy. First, we need to transform
from integrations overrandr′to center-of-mass and relative coordinate

R=

1

2

(r+r′), s=r−r′. (11.5.66)

This transformation yields forEx:

Ex=−

1

4

∫

dRds

ρ^21 (s)

s

. (11.5.67)

Integrating overR, transforming thesintegral into spherical polar coordinates, and
performing the angular part of thesintegration gives

Ex=−

V

4

∫

ds

ρ^21 (s)

s

=−πV

∫∞

0

ds sρ^21 (s)