Advanced Solid State Physics

Figure 4: Quantization of k-space

For further analysis the sum is converted to an integral. Before this can be done, thedensity of states
Dmust be evaluated. For an electromagnetic field in a cubic region of lengthLwith periodic boundary
conditions, the longest wavelengthλmax, which fullfills the boundary conditions, isλmax=L. The
next possible wavelengthλmax− 1 isλmax− 1 =L 2 and so on. This gives possible values fork(x,y,z):

k(x,y,z)=

2 π

λ

= 0,±

2 π

L

,±

4 π

L

,...

The volume ink-space occupied by one state in 3D is given by

(

2 π

L

) 3
(see fig. 4). The number of
states in 3D is

L^3 D(k)dk= 2

4 πk^2 dk

(

2 π

L

) 3 =

k^2 L^3

π^2

dk,

where the factor 2 occurs because of the two possible (transverse) polarizations of light (for phonons
there would be 3 : 2 transversal, 1 longitudinal). This gives the density of statesD(k):

D(k) =

k^2

π^2

With the variableωinstead ofk, withω=ck→dω=cdkthis gives:

D(ω) =

ω^2

c^3 π^2

Now theHelmholtz free energy(density)fcan be calculated with eqn. (33):

f=

kBT

L^3

∑

s

ln

(

1 −e−

~ωs

kBT

)

≈kBT

∫∞

0

ω^2

c^3 π^2

ln

(

1 −e−

k~ω

BT

)

dω

Withx=k~BωT and integration by parts:

f=

(kBT)^4

~^3 c^3 π^2

∫∞

0

x^2 ln(1−e−x)dx=−

(kBT)^4

3 ~^3 c^3 π^2

∫∞

0

x^3

ex− 1

dx.