Download Optical Transitions in Semiconductors

Material Physics: Physics of Nanostructured Materials
Exercise 6: Solutions
Optical Transitions in Semiconductors
1) Interband Transitions in 3D Systems
a) In silicon, an electron from the top of the valence band is taken to the bottom of the
conduction band by photon absorption. The conduction band minima for silicon are at a kvalue of 2π
a (0.85,0,0) as shown in figure (a). Calculate the change in the electron momentum.
Can this momentum difference be provided by a photon?
Solution
The top of the valence band has a k-value of 0. The change in the momentum is
~∆k = ~
2π
(0.85) ≈ 1.037 · 10−24 kg · m · s−1
a
with a300K
= 5.43 · 10−10 m.
Si
A photon which has an energy equal to the silicon bandgap can only provide a momentum
of
2π
≈ 6.251 · 10−28 kg · m · s−1
λ
The λ for silicon bandgap is 1.06 µm and thus the photon momentum is about a factor of
1660 too small to balance the momentum needed for the momentum conservation. The lattice
vibrations produced by thermal vibration are needed for the process.
~kph = ~
b) The figure above shows the absorption coefficients of a direct and an indirect semiconductor as a function of wavelength. Which curve belongs to the direct and which one
belongs to the indirect semiconductor? Explain your answer. What is the band gap of both
semiconductor materials? Which material could it be?
Solution
Curve (a) represents an indirect band gap semiconductor whereas curve (b) represents a
direct band gap semiconductor. This can be seen from the slope of the curves. As λ is
decreased, which means that the photon energy becomes larger, the absorption of semiconductor (a) increases slowly, i.e. no efficient absorption. Whereas for semiconductor (b) as
soon as the photon energy is high enough also the absorption is high. For photon wavelengths
below 0.5µm curve (a) has a strongly increasing slope meaning that the photon energies are
sufficient for the direct transition.
The absorption starts at:
(a): λa = 1.1 µm ⇒ Eg,a = h ·
(b): λb = 0.85 µm ⇒ Eg,a = h ·
c
≈ 1.13 eV ⇒ Si
λ
c
≈ 1.46 eV ⇒ GaAs
λ
Material Physics: Physics of Nanostructured Materials
Exercise 6: Solutions
2) Absorption in a Quantum Well
The absorption of electromagnetic waves between two states i and f is associated with the
dipole matrix element:
α ∝ |hj|~ep̂|ii|2
The wave function of a state in the quantum well is given by
Ψnmk (~r) = Ω1/2 · χm (~r) · unk (~r)
Here, u is the periodic contribution of the Bloch wave (characterized by the band index n
and the wave vector ~k ) and χ is the envelope function, given by the superposition of a plane
wave in the plane of the well and a bound state in z-direction
~
χm (~r) = A1/2 · eik||~r|| · φm (z)
φ is characterized by the subband-index m.
a) General matrix element
• Show that hnj mj kj |~ep̂|ni mi ki i for two states i and j in the quantum well results in
hj|~ep̂|ii = huj |~ep̂|ui ihχj |χi i + huj |ui ihχj |~ep̂|χi i
• Which parts contribute to interband transitions and which to inter-subband transitions?
Solution
Z
Ψ∗j p̂Ψi d3 r
hnj mj kj |~ep̂|ni mi ki i = ~e
Ω
Z
= ~e
Ω
χ∗mj u∗nj kj p̂χmi uni ki d3 r
Considering that u varies typically much faster than χ, we can show that:
Z
Z
Z
Z
Z
∗ ∗
3
∗
3
∗
3
∗
3
~e χj uj p̂χi ui d r = ~eΩ
uj p̂ui d r ·
χj χi d r + ~eΩ
uj ui d r ·
χ∗j p̂χi d3 r
Ω
Ω0
Ω
Ω0
Here, Ω0 is the volume of the unit cell.
→ hj|~ep̂|ii = huj |~ep̂|ui ihχj |χi i + huj |ui ihχj |~ep̂|χi i
|
{z
} |
{z
}
interband
i=j→inter−subband
Ω
b) Selection rules for interband-transitions
• Discuss the interband matrix element of the envelope function with respect to the kvector and subband index m, respectively.
• Discuss the interband-matrix element of the periodic part of the Bloch wave.
Solution
Discussion of hχj |χi i:
1
hχj |χi i =
A
k-vector:
Z
0
i(~k|| −~k|| )~
r||
Due to the oscillating term e
Z
Z
dz
d2 rφ∗j (z)e
0
i(~k|| −~k|| )~
r||
φi (z)
0
the integral over d2 r is non-zero only for ~k|| = ~k|| :
2
...d r =
(
A
0
0
if ~k|| = ~k||
0
if ~k|| 6= ~k||
The in-plane wave vector remains conserved; only vertical transitions are allowed.
∆m:
Assumption: the QW in the conduction band (CB) as well as the QW in the valence band
(VB) are symmetric (in z-direction) and infinitely deep:
• Solutions φm are equal for CB and VB
• Solutions φm are either even or uneven
R
The integral φ∗mj (z)φmi (z)dz is only deviating from 0, if both wave functions are even, i.e.
the parity remains conserved.
→ Selection rule: ∆m = 0
Discussion of huj |~ep̂|ui i (periodic part of Bloch waves):
The Bloch waves of the VB’s (heavy holes, light holes, so-holes) have p-type wavefunctions
with contributions of px , py and pz . In contrast to the valence band, the CB is s-like. The
operator p̂ changes the parity of the Bloch wave and the interband transition VB → CB is
allowed (hCB|~ep̂|VBi =
6 0).
Material Physics: Physics of Nanostructured Materials
Exercise 6: Solutions
c) Inter-subband absorption in a quantum well
• Consider the matrix element for inter-subband transitions for different propagation directions of the light wave:
1. perpendicular to the QW, e.g. ~e = (1,0,0)
2. in the plane of the QW, e.g. ~e = (0,0,1)
3. show that for propagation in the plane of the QW only ”‘vertical”’ transitions are
allowed, i.e. optical transitions for which the wave vector k is conserved
• Using the transition matrix element, one should explain how the symmetry of the QW
determines the selection rules for optical transitions. Why do we observe in the spectrum
the transition |1i → |2i , but not |1i → |3i?
Solution
• inter-subband transitions huj |ui ihχj |~ep̂|χi i for different propagation directions:
(because of parity reasons: huj |ui i = δui uj )
∂
1. polarization in x-direction: ~e = (1,0,0) → ~ep̂ = −i~ ∂x
, analog for y-polarization
→ apply to wavefunction of QW: ~ep̂χ(~r) = ~kx χ(~r) since the momentum operator
applies only to plane waves in χ(~r)
→ no light absorption, because matrix element is zero: hχj |~ep̂|χi i = ~kx hχj |χi i = 0
∂
2. polarization in z-direction: ~e = (0,0,1) → ~ep̂ = −i~ ∂z
R i(~k0 −~k )
R
∂
→ matrix element: hχj |~ep̂|χi i = −i~ A1 e || || d2 r φ∗j (z) ∂z
φi (z)dz
3. due to oscillating term: e
0 −~
i(~k||
k|| )
the integral results in
R
...d2 r = δ~k0 ~k
|| ||
→ in-plane wave vector remains conserved, only vertical transitions are allowed
• symmetry of hχj |~ep̂|χi i:
→ symmetry of QW V (−z) = V (z) describes symmetry of wave functions
→ solution of Schrdinger equation leads to even or uneven wave functions
∂
→ apply the momentum operator p̂z = i~ ∂z
changes the parity of the wave function
R ∗
∂
→ matrix element hφj |p̂z |φi i = −i~ φj (z) ∂z φi (z)dz 6= 0 if one function is even and
the other function is uneven
→ this restriction corresponds to a selection rule, absorption from ground state (n=1)
can only result in even states n=2,4,...