N 1 ¼ N 0 e
Àħω=k B T ,
ð9:1Þ
where k B is Boltzmann constant and T is absolute temperature. Let N j be the number
of oscillators of the j-th excited state. Then we have
N j ¼ N 0 e
Àjħω=k B T
:
ð9:2Þ
Let N be the total number of oscillators in the system. Then, we get
N ¼ N 0 þ N 0 e
Àħω=k B T
þ Á Á Á þ N 0 e
Àjħω=k B T
þ Á Á Á
¼ N 0
X 1
j¼0
e
Àjħω=k B T
:
ð9:3Þ
Let E be a total energy of the oscillator system in reference to the ground state. That
is, we put a ground-state energy at zero. Then we have
E ¼ 0 Á N 0 þ N 0 ħωe
Àħω=k B T
þ Á Á Á þ N 0 jħωe
Àjħω=k B T
þ Á Á Á
¼ N 0
X 1
j¼0
jħωe
Àjħω=k B T
:
ð9:4Þ
Therefore, an average energy of oscillators E is
E ¼
E
N
¼ ħω
P 1
j¼0 je
Àjħω=k B T
P 1
j¼0 e Àjħω=k B T :
ð9:5Þ
Putting x e
Àħω=k B T [1], we have
E ¼ ħω
P 1
j¼0 jx
j
P 1
j¼0 x j :
ð9:6Þ
Since x < 1, we have
X 1
j¼0
jx
j
¼
X 1
j¼0
jx
jÀ1
Á x ¼
d
dx
X 1
j¼0
x
j
!
x ¼
d
dx
1
1 À x
!
x
¼
x
1 À x
ð
Þ
2
:
ð9:7Þ
X 1
j¼0
x
j
¼
1
1 À x
:
ð9:8Þ
Therefore, we get
340
9 Light Quanta: Radiation and Absorption