386 Solutions
10.3: The solution of the Schroedinger equation with a spherically symmetric
potential is
φ
nlm
(r)=R
nl
(r)Y
lm
(θ, ϕ) .
The radial solution for the Coulomb problem is given by
R
nl
(r)=N
n
ρ
l
e(−ρ/2)F (l +1− η;2l +2;ρ)
with the hypergeometric function depending on the variables η (for the energy)
and ρ (for the radial coordinate). For bound states one has E
n
= −R
exc
/n
2
and
ρ =2r/na
B
. Only s-states with l = 0 have a nonvanishing amplitude at r =0,
thus φ
n
(0) = N
n
F (1 − n;2;0) with N
n
=1/na
B
and |φ
n
(0)|
2
=(πa
3
B
n
3
)
−1
.For
the continuum (E =¯hω − E
g
> 0), η =iγ, γ =(R
exc
/(¯hω − E
g
))
1/2
and N =
Γ(1 − iγ)exp(πγ/2) one obtains
|φ(0)|
2
= πγ
exp(πγ)
sinh(πγ)
.
The absorption coefficient is the one of (10.23) for the uncorrelated electron-hole
pairs multiplied by the enhancement factor C(ω).
10.4: Using (10.37) and (10.38) one can write
B
†
νQ
|Ψ
0
= |Ψ
νQ
=
ck
h
,ck
e
k
e
−k
h
=Q
Φ
νQ
(k) c
†
ck
e
c
vk
h
|Ψ
0
.
Thus, in application to the ground state |Ψ
0
, the exciton operator is a linear com-
bination of products c
†
ck
e
c
vk
h
. Let us assume a two-band model, i.e. the excitons
are formed from states with fixed c, v, and consider for simplicity the exciton with
Q = 0 which implies k
e
= k
h
= k . Then, the commutator [B
†
ν,0
,B
ν
′
,0
] is determined
by the commutators
[c
†
vk
c
ck
,c
†
ck
′
c
vk
′
]=c
†
vk
c
ck
c
†
ck
′
c
vk
′
− c
†
ck
′
c
vk
′
c
†
vk
c
ck
= c
†
vk
c
vk
′
δ
k,k
′
− c
†
ck
′
c
ck
δ
k,k
′
= δ
k,k
′
c
†
vk
c
vk
− c
†
ck
c
ck
=(1− n
e
(k) − n
h
(k)) δ
k,k
′
.
Without the electron and hole occupation this relation would lead to [B
ν,0
,B
†
ν
′
,0
]=
δ
ν,ν
′
, classifying excitons as bosons. This relation is valid, ho wever, only if we con-
sider a single exciton. The electron and hole occupations remind of the fact, that
excitons are composed of fermions.
10.5: The zero of (10.42) close to the exciton resonance at ω
ν0
determines the
frequency of the corresponding longitudinal exciton:
ε
1
(ω
L
) ≃ 1+
4e
2
ε
0
m
f
ν0
ω
2
ν0
− ω
2
L
=0,
which for ω
L
≃ ω
ν0
and ∆
LT
=¯h(ω
L
− ω
ν0
) allows to express the LT-splitting in
terms of the exciton oscillator strength:
∆
LT
=
2e
2
¯h
2
ε
0
m
f
ν0
E
ν0
with E
ν0
=¯hω
ν0
.