The experiment
Enerrgy
Conduction band
Ph t hν
Photon
h
Valance band
(occupied states)
Antonio Polimeni, dipartimento di Fisica, Sapienza Università di Roma
Experimental determination of the band gap:
the GaAs case
Ga
As
Optical absorption measurements
bulk GaAs
E (k)
CB
transmitted
light
white light
k
spectral
analysis
VB
Kph<<2π/a
T
IT ~ I0 e (-α L)
energy
Optical absorption measurements
r
r 2
2π ⎛ eF ⎞
dk
α ( hω ) =
⎜
⎟ ∑ ∫ 3 M v , c ( k ) δ [ E c ( k ) − E v ( k ) − hω ] ≈
h ⎝ mω ⎠ c ,v BZ 4π
2
CB
≈ A M v , c ρ c , v ( hω )
2
E (k)
k
For 3D (bulk) material
ρ c ,v (hω ) = B hω − E gap
GaAs
21 K
90 K
294 K
VB
186 K
Photon energy (eV)
α (hω ) = C hω − E gap
Optical emission
r
r
i , f = u c ,v ( r ) f c ,v ( r )
E (k)
CB
[
i
f
VB
]
Li → f (hω ) ∝ Pif f (Ei ) 1 − f (E f ) → α (hω ) e −(hω ) (k BT )
k
2π
Pi , f =
i Vint f
h
Vint =
r
eF
(εˆ ⋅ p )
2mω
2
δ ( Ei − E f − hω )
Density of states
Calculating quantities weighted over the electronic levels
r
r r
V
Q = 2∑
Q(k ) = 2 3 ∑
Q(k ) ∆k
r
r
8π k
k
r
r
r
Q
dk
f
for
∆k → 0 (V → ∞) q = lim
li
= ∫ 3 Q(k )
V →∞ V
4π
r
1
d −1
−
d k
d −1
2
∝
k
dk ∝ E 2 E 2 dE
d
(2π )
d
If
r h 2k 2
E (k ) =
2m
g (E) =
m
hπ
2
2
2mE
h2
in 3D
Energy
λdeBroglie
h2
= 2π
≈ 10 nm
*
2m E
g ( En ) =
GaAs
ϑ (E )
*
m
n
2
hπ
in 2D
E0 E1 E2
E
Energy
1
Lz
1
g (E) =
hπ
m*
2E
in 1D
Ly
E
E0 E1 E2 Energy
E
ρ(E)
g ( E ) = ∑ 2 δ ( E − En )
n
in 0D
E0 E1
E2 Energy
Quantum heterostructures
Lz
h ≈ ∆z ⋅ ∆p z ; ∆z = Lz
2
Eloc
∆p z
h2
≈
=
*
2
2m
2 m * Lz
∆ pz
A semiconductor quantum heterostructure can be obtained by embedding a smaller
gap material in a larger gap matrix
≈1
1 µm
10 nm
1nm
Excitonic effects
CB
VB
E
e-
e-
h+
h+
r
1 µe 4
E = E gap − 2 2 2 ; n = 1, 2, 3, ...
n 2ε h
without
ith t excitonic
it i effects
ff t
Excitonic effects
⎡ p 2 e2 ⎤ r
r
−
ϕ
(
r
)
=
η
ϕ
(
r
)
⎢
⎥
⎣ 2µ ε r ⎦
r
ground state ϕ (r ) =
a0ex
For GaAs
[
Bound states for η < 0
1
]
12
π (a0ex )3
h 2ε
= 2
µe
R0ex
h 2ε
a = 2 ≈ 15 nm
µe
ex
0
e
( − r / a0ex )
µe 4
= 2 2
2ε h
µe 4
R = 2 2 ≈ 4.2 meV
2ε h
ex
0
Photoluminescence (PL) spectroscopy
CB
eCB
CB
Ed
h
hν
laser
Ea
luminescence
VB
h+
Laser
VB
VB
EXCITON:
IMPURITY
e‐h
LEVELS
system
F1 M1
Sample
p
F3 F2
L1
Monochromator
M2
L3
L2
Cryostat
Acquisition Detector
and Control System
Exciton diamagnetic shift
Using
ε h2
length unit
a0 =
µ e2
µ e4
Ry = 2 2 energy unit
2ε h
B=
2 µ Ry
magnetic field unit
he
the Hamiltonian is
⎡ 2
∂ 2 γ2 2
Hψ ( r ) = ⎢ − ∇ − i γ
− +
x + y2
∂φ r 4
⎣
(
)
½ hω c ε 2 h 3 B
γ=
= 2 3
Ry
me
where
Ry = 13.6
⎤
⎥ ψ ( r ) = Eψ ( r )
⎦
m
ε
2
eV
hωc heB 0.58 ×10 B
eV
=
=
m
2
2m
−4
m = 0.054 m0
ε = 12
γ=
½ hω c
= 0.21 B
Ry
Hydrogen atom vs exciton in semiconductors
Strength of magnetic field:
• Magnetic field
H
BH Ry
1
= exc =
Bexc Ry 6⋅
6 10− 4
⇒ hωc=(heB)/m
If Bexc ~ 40 T ⇒ BH ~ 2000 T !
Magneti superconduttori
“Economici” e compatti
Campi stabili e uniformi
Lunga durata
Campo massimo 20 T
~ 103 spire/cm
Lfilo~10 km
B=12 T
I=120 A
Se tutto Cu (R=230 Ω)
a 100 A, P~2 MW
Magneti superconduttori
NbTi
NbSn
~ 1 mm Ø (Cu)
µm Ø ((Nb/Ti))
~ 20 µ
La tecnica “cable-in-conduit conductor” permette di:
- aumentare la superficie efficace a contatto con il liquido criogenico;
- rimuovere velocemente il calore generato durante il funzionamento e portare la corrente
che scorre nel superconduttore in caso di quench (ρCu=1.8⋅10
1.8 10-8 Ωm, ρNb=15⋅10
15 10-8 Ωm);
- la matrice di rame conferisce stabilita’ meccanica (NbTi e’ fragile)
Magneti resistivi
Ag-coated Cu
<1 mm-thick
aste p
per il
posizionamento
scorre acqua (250 l/s; 30 atm)
deionizzata (non conduce) per
raffreddare il magnete
Magneti resistivi
Per completare un passo dell’elica sono necessari
~10 plate intervallati da un disco isolante.
Il disco di isolante serve a far compiere alla
corrente un cammino elicoidale e per evitare
corto-circuiti lungo l’asse della bobina.
Circa 1000 dischi di rame compongono un
solenoide.
Magneti resistivi
Le bobine sono percorse dalla
corrente in serie a partire da
quella piu’ interna.
~3 m
- Alimentati da ~20 MW a 700 V (~3⋅104 A dc)
- Raffreddati da 104 l/min a 30 atm di acqua deionizzata (isolante)
a ~5 °C. L’acqua esce a ~40 °C. Si utilizza una riserva di 16 ⋅106 l.
costo: 1500 euro/ora
High-field limit
Ec (kz,nc)
1⎞
h 2 k z2
⎛
Ec (k z , nc ) = Ec 0 + ⎜ nc + hωce +
2⎠
2me
⎝
h 2 k z2
1⎞
⎛
Ev (k z , nv ) = Ev 0 + ⎜ nv + ⎟hωch +
2⎠
2mh
⎝
k zc = k zv
nc = nv
Ev (kz,,nv)
1 ⎞⎛ 1
1 ⎞
1 ⎞ heB
B
⎛
⎛
⎟⎟ ehB = E (0) + ⎜ n + ⎟
E n (B ) = E (0) + ⎜ n + ⎟⎜⎜
+
2 ⎠⎝ me mh ⎠
2⎠ µ
⎝
⎝
(elettrone-lacuna liberi)
r
r
i εˆ ⋅ p f = uc εˆ ⋅ p uv
3r
* r
* r
d
r
f
(
r
)
f
c
v (r )
∫
−ξ c2 2
−ξ v2 2
3r
* r
* r
3 r − i (k xc ⋅ x + k zc ⋅ z ) i (k xv ⋅ x + k zv ⋅ z )
(
)
d
r
f
(
r
)
f
(
r
)
=
d
r
e
e
e
H
ξ
e
H n ,v (ξ c )
c
v
n ,c
c
∫
∫
c
v
dxdz
(
⋅
⋅
⋅
)
→
k
=
k
x
x
∫
∫ dy(⋅ ⋅ ⋅) → n
c
= nv
k zc = k zv
12 T
E(k)
0T
(e,C)
C
k
hν ( e , C ) ( B ) = E
0
CBM
eh
−E +
B − ∆E C ( B )
2me
0
C
Quantum heterostructures
Lz
h ≈ ∆z ⋅ ∆p z ; ∆z = Lz
2
electroon energy
Eloc
∆p z
h2
≈
=
*
2
2m
2 m * Lz
∆pz
A semiconductor quantum heterostructure can be obtained by embedding a smaller
gap material in a larger gap matrix
In--plane defect engineering
In
x=0.92%
µ-PL
10 µm
5 µm
1 µm
2 µm
20 µm
GaAsN
GaAsN:H ≡ GaAs
Ga(AsN)
GaAs
GaAsN
G A N epilayer
il
M. Felici et al., Adv. Mater. 18, 1993 (2006).
0.5 µm
Spatially-controlled
modulation of carrier
potential
In--plane defect engineering
In
hydrogen
GaAsN
GaAs
In--plane defect engineering
In
e beam lithograpy at IFN
e-beam
IFN-CNR
CNR (A.
(A Gerardino and M
M. Francardi)
200 nm dots
courtesy of L. Felisari (TASC)
Toward 0D0D-nanostructures
Near-surface GaAsN/GaAs quantum well
[N]=0.9% well width L=6 nm
surface distance ss=30
30 nm
dot diameter d=80 nm
GaAs
Ti dot
100 nm
GaAsN
Toward 0D0D-nanostructures
0.9%
0.4%
20 nm
0%
Low-field limit
R. Trotta et al., Advanced Materials 23, 2706 (2011)
PL Intensity (arb. units)
d=100 nm
1.390
Energ
gy (eV)
T 5K
T=5
1.389
1.388
1.387
1.386
0
5
10
15
B (T)
20
25
30
d 80 nm
d=80
B=30 T
B=0 T
1.386 1.388 1.390
Energy (eV)
R. Trotta et al., Advanced Materials 23, 2706 (2011)
2
-3
β80 nm=2.2×10 meV/T
2
2
∆EZ (m
meV)
∆Ed (m
meV)
3
d = 80 nm
1
0
gexc=1.04
1
d = 80 nm
0
0
5 10 15 20 25 30
∆E d =
8µ exc
µ exc = 0.095 m0
10 15 20 25 30
B (T)
B (T)
e2
5
0 x2 + y2 0 B2
∆EZ = g exc µ B B
d exc = 2a B = 2 0 x 2 + y 2 0 = 4.3 nm
Intermediate-field limit
µexc=0.053 m0
9
⎛
⎞
(
)
m
µ
0
∆E d = 13.6 ⋅ 10 3 ⎜⎜
⋅ ∑ Ai
2
⎝ ε r ⎠ i =1
⎡
⎤
⎛
⎞
ε
r
⎢4.26 ⋅ 10 −6 ⋅ ⎜⎜
B⎥
⎢⎣
⎝ (µ m0 ) ⎠ ⎥⎦
2
i
1 ⎞ heB
⎛
En (B ) = E (0) + ⎜ n + ⎟
2 ⎠ µexc
⎝
µexc=0.07 m0
Exciton mass overestimation