1549380323-Statistical Mechanics Theory and Molecular Simulation

The ideal fermion gas 425

〈N〉=

4 πg

3

(

m

2 π ̄h^2

) 3 / 2

V εF^3 /^2. (11.5.39)

By a similar procedure, we can obtain an expression for the averageenergy. Recall
that the total energy for a given set of occupation numbers is given by

E{fn}=

∑

m

∑

n

fnmεn. (11.5.40)

Taking the ensemble average of both sides yields

〈H〉=E=

∑

m

∑

n

〈fnm〉εn. (11.5.41)

AtT= 0, this becomes

E=g

∑

n

θ(εF−εn)εn

→g

∫

dnθ(εF−εn)εn

= 4πg

∫∞

0

dn n^2 θ(εF−εn)εn, (11.5.42)

where, as usual, we have replaced the sum by an integral and transformed to spherical
polar coordinates. If the change of variables in Eqn. (11.5.38) is made, we find

E= 4πg

∫∞

0

dεn

1

2

(

mL^2

2 π^2 ̄h^2

) 3 / 2

εn^3 /^2 θ(εF−εn)

= 2πg

(

m

2 π^2 ̄h^2

) 3 / 2

V

∫εF

0

dεnεn^3 /^2

=

4 πg

5

(

m

2 π^2 ̄h^2

) 3 / 2

V εF^5 /^2. (11.5.43)

Combining eqns. (11.5.43) and (11.5.39), the following relation betweenEand〈N〉
can be established:

E=

3

5

〈N〉εF. (11.5.44)

Moreover, sinceεF∼ρ^2 /^3 , we see that the total energy is related to the densityρby

E

V

=CKρ^5 /^3 , (11.5.45)

whereCKis an overall constant,CK= (3 ̄h^2 / 10 m)(6π^2 /g)^2 /^3. Note that if we perform
a spatial integration on both sides of eqn. (11.5.45) over the containing volume, we
obtain the total energy as