Semiconductor heterostructures quantum wellsszczytko/LDSN/3_LDSN_2017_QWells.pdf · 2017. 6. 5. · Bandgap engineering 2017-06-05 41 Effects of biaxial strain: decrease of the degeneracy

Semiconductor heterostructures– quantum wells

Faculty of Physics UW

[email protected]

Carriers: Dopants:

holes

electrons

+

-

Acceptors (p-type)

Donors (n-type)

Semiconductors

Be B C N O

Mg Al Si P S

Zn Ga Ge As Se

Cd In Sn Sb Te

II III IV V VI

Group IV: diamond, Si, GeGroup III-V: GaAs, AlAs, InSb, InAs...Group II-VI: ZnSe, CdTe, ZnO, SdS...

Domieszkowanie półprzewodników

2017-06-05 2

Semiconductor heterostructures

2017-06-05 3

Density of statesDensity of states

6/5/2017 4

Born- von Karman boundary conditions

Finite size of the crystal Lx, Ly, Lz

Ψ – the Bloch function

Ψ(x + Lx,y,z) = Ψ(x, y + Ly,z) = Ψ(x, y, z + Lz)

If our crystal has a finite size the set of 𝑘 −vectors is finite (though enormous!). For example: we can assume periodic boundary conditions and then:

kx

ky

Number of states in the

volume 𝑉

Number of states per unit energy 𝜌𝑛𝐷(𝐸) depends on the dimension

2𝜋

𝐿𝑦 2𝜋

𝐿𝑥

𝑘𝑖 = 0,±2𝜋

𝐿𝑖, ±

4𝜋

𝐿𝑖, … , ±

2𝜋𝑛𝑖𝐿𝑖

=2

2𝜋𝐿𝑥

×2𝜋𝐿𝑦

×2𝜋𝐿𝑧

=2𝑉

2𝜋 3

𝑒𝑖𝑘𝑥𝐿𝑥 = 1

𝑒𝑖𝑘𝑦𝐿𝑦 = 1𝑒𝑖𝑘𝑧𝐿𝑧 = 1

Infinite square quantum well

2017-06-05 5

𝜓 𝑥, 𝑡 =2

𝐿sin 𝑘𝑛𝑥 𝑒−𝑖𝜔𝑡

Inside the quantum well:

𝑘𝑛 =𝑛𝜋

𝐿

𝜀𝑛 =ℏ2𝑘𝑛

2

2𝑚=ℏ2𝑛2𝜋2

2𝑚𝐿2


2017-06-05 6

𝜓 𝑥, 𝑡 =2



𝜀𝑛 = 𝐸𝑐 +ℏ2𝑘𝑛

2

2𝑚= 𝐸𝑐 +

ℏ2𝑛2𝜋2

2𝑚0𝑚∗𝐿2

𝑘𝑛 =𝑛𝜋

𝐿

𝐸𝑐

𝜀1 = 𝐸𝑐 +ℏ2𝜋2

2𝑚0𝑚∗𝐿2

𝜀2 = 𝐸𝑐 +2ℏ2𝜋2

𝑚0𝑚∗𝐿2

𝜀3 = 𝐸𝑐 +9ℏ2𝜋2

2𝑚0𝑚∗𝐿2


2017-06-05 7

𝜓 𝑥, 𝑡 =2



𝜀𝑛 = 𝐸𝑣 +ℏ2𝑘𝑛

2

2𝑚= 𝐸𝑣 −

ℏ2𝑛2𝜋2

2𝑚0𝑚∗𝐿2

𝑘𝑛 =𝑛𝜋

𝐿

𝐸𝑣𝜀1 = 𝐸𝑣 −

ℏ2𝜋2

2𝑚0𝑚∗𝐿2

𝜀2 = 𝐸𝑣 −2ℏ2𝜋2

𝑚0𝑚∗𝐿2

𝜀3 = 𝐸𝑣 −9ℏ2𝜋2

2𝑚0𝑚∗𝐿2

Bandgap engineering

2017-06-05 8

How can we change the heterostructure band structure:• selecting a material (eg., GaAs / AlAs)• controlling the composition• controlling the stress

2017-06-05 9

Figure 5. Schematic diagrams depicting the evolution of the conduction band structure in the transverse direction: (a) double heterostructure, (b) separate confinementheterostructure (SCH), (c) graded-index separateconfinement heterostructure (GRIN-SCH), (d) single quantum well heterostructure (QWH), and (e) multiplequantum well (MQW).

The development of the semiconductor laser diode after the first demonstration in 1962 J J Coleman Semicond. Sci. Technol. 27 (2012) 090207

Vapor-phase (VPE, CVD)

wzrost z fazy gazowej dzięki reakcjom chemicznym prekursorów na powierzchni,

często dzielony ze względu na źródłowe gazy na wodorkową VPE i

metalorganiczną VPE (MOCVD); prędkości wzrostu >10 -20 nm/min.

Liquid-phase (LPE)

wzrost z fazy ciekłej na podłożu w temperaturach niższych od temperatury

topnienia hodowanego materiału. Półprzewodnik jest rozpuszczony w cieczy

innego materiału, wzrost w warunkach bliskich równowagi roztworu i depozycji;

prędkości wzrostu 0.1 to 1 μm/min.

Molecular-beam (MBE)

Materiał źródłowy podgrzewany w komórkach produkuje strumień cząsteczek. W

wysokiej próżni (10-8 Pa) cząsteczki docierają do podłoża i osadzają się na nim;

prędkości wzrostu < 1 monowarstwa/s (1 μm/h).

2017-06-05 10

Jak się robi heterostruktury?

Ec

E0

E1D(E)

E

Quantum Well

t

2D

MBE Osadzanie z atomową

precyzją warstw o różnym składzie

lub domieszkowaniu

Ec

Hubert J. Krenner

2017-06-05 11


Wzrost warstw MBE jest monitorowany przez Reflection High Energy Electron

Diffraction (REED). Komputer steruje przesłonami (shutterami) na froncie

podgrzewanych komórek efuzyjnych, co pozwala na precyzyjną kontrolę wzrostu do

poziomu pojedynczej warstwy atomowej.

Wzrost warstw z jamami kwantowymi (quantum wells), kropek kwantowych (quantum

dots) – struktury LD, LED.

2017-06-05 12


komora załadunkowa komora UHV

wzrostu

materiałów III-V

(Ga, Al, In, As, Sb,

N-plazma,

Si lub Te, Be lub Zn,

Mn lub Cr lub Co)

komora UHV

wzrostu

materiałów II-VI

(Zn, Cd, Mg,

S, Se, Te, Mn, Co,

ZnCl2, N-plazma)

komora UHV

przygotowania

podłoży

(odgazowanie

powierzchni)

Urządzenie MBE - do epitaksji z wiązek molekularnych (2 komory wzrostu)

producent SVTA (USA). Zakup przez Wydział Fizyki w r. 2010, program CePT2017-06-05 13


2017-06-05 14


MBE na Wydziale Fizyki UW

2017-06-05 15


MBE na Wydziale Fizyki UW

2017-06-05 16


Reaktor Metal-Organic Chemical Vapour Epitaxy (MOCVD) w Zakładzie Fizyki Ciała

Stałego

Heterostruktury GaInSb, AlGaInAs and AlGaN.

2017-06-05 17


Aixtron CCS 3x2

Bandgap engineering

2017-06-05 18



2017-06-05 19

Investigation of high antimony-content gallium arsenic nitride-gallium arsenic antimonide heterostructures for long wavelength application

Bandgap engineering

2017-06-05

20

Valence band offset (Anderson’s rule)

Valence band offset:

(powinowactwo)

htt

p:/

/en

.wik

iped

ia.o

rg/w

iki/

Het

ero

jun

ctio

n

Bandgap engineering

2017-06-05 21

Valence band offset (Anderson’s rule)

Su-H

uai

Wei

, Co

mp

uta

tio

nal

Mat

eria

ls S

cien

ce, 3

0, 3

37–3

48 (

2004

)

Valence band offset:

(powinowactwo)

htt

p:/

/en

.wik

iped

ia.o

rg/w

iki/

Het

ero

jun

ctio

n

Bandgap engineering

2017-06-05 22

Valence band offset

Wal

uki

ew

icz

Ph

ysic

aB

302

–303

(20

01)

123

–134

Bandgap engineering

2017-06-05 23


Vegard’s law:the empirical heuristic that the lattice parameter of a solid solution of two constituents is approximately equal to a rule of mixtures of the two constituents' (A and B) lattice parameters at the same temperature:

𝑎 = 𝑎𝐴 1 − 𝑥 + 𝑎𝐵𝑥

Bandgap engineering

2017-06-05 24

Vegard’s law:Relationship to band gaps of „binarycompound”:

𝐸 = 𝐸𝐴 1 − 𝑥 + 𝐸𝐵𝑥 − 𝑏𝑥(1 − 𝑥)

b – so-called „bowing” (curvature) of the energy gap

Z D

rid

iet

al.

Sem

ico

nd

. Sci

. Tec

hn

ol.

18N

o9

(Sep

tem

ber

2003

)85

0-85

6


Bandgap engineering

2017-06-05 25

Valence band offsetHow can we change the heterostructure band structure:• selecting a material (eg., GaAs / AlAs)• controlling the composition• controlling the stress


2017-06-05 26

http://beta.globalspec.com/reference/45139/203279/chapter-iii-optical-properties

Thin Solid Films 433 (2003) 22–26

Quaternary compounds


2017-06-05 27

Quinternary barriers push room-temperature operation of GaSb-based type-I lasers further into mid-infrared

http://beta.globalspec.com/reference/45139/203279/chapter-iii-optical-properties

Quinternary compounds

Comments on the conduction band

2017-06-05 28

Depending on the semiconductor the bottom of the conduction band may be constructd from a differnent valleys – the same heterostructure may be a well in a one band (eg. Γ) and a barrierin another (eg. 𝑋).

Comments on the conduction band

2017-06-05 29

Depending on the semiconductor the bottom of the conduction band may be constructd from a differnent valleys – the same heterostructure may be a well in a one band (eg. Γ) and a barrierin another (eg. 𝑋).

GaAs AlAs

Heterostruktury półprzewodnikowe

2017-06-05 32

http://www.springermaterials.com/docs/pdf/10681719_656.html

Bandgap engineering

2017-06-05 33


Bandgap engineering

2017-06-05 34

Saty

amS.

Par

ash

arie

t al

. Ph

ysic

aB

403

(20

08)

3077

–30

88

Elab

syet

al.

Ph

ysic

aB

40

5 (

20

10

) 3

70

9–37

13


Bandgap engineering

2017-06-05 35


Bandgap engineering

2017-06-05 36

Saty

amS.

Par

ash

arie

t al

. Ph

ysic

aB

403

(20

08)

3077

–308

8


Bandgap engineering

2017-06-05 37

Saty

amS.

Par

ash

arie

t al

. Ph

ysic

aB

403

(20

08)

3077

–308

8


Bandgap engineering

2017-06-05 38

Pro

g. Q

uan

t. E

lect

r. V

ol.

21,

No

. 5, p

p. 3

61±

419,

199

8uniaxial


Bandgap engineering

2017-06-05 39

Pro

g. Q

uan

t. E

lect

r. V

ol.

21,

No

. 5, p

p. 3

61±

419,

199

8biaxial


Bandgap engineering

2017-06-05 40

Pro

g. Q

uan

t. E

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r. V

ol.

21,

No

. 5, p

p. 3

61±

419,

199

8

How can we change the heterostructure band structure:• selecting a material (eg., GaAs / AlAs)• controlling the composition• controlling the stress uniaxial biaxial

⊥ allong the strain axis

Bandgap engineering

2017-06-05 41

Effects of biaxial strain: decrease of the degeneracy of the valence band and change of the effective masses in the GaxIn1-xAs / GaxIn1-xAsyyP1-y material system.

S.L. Chuang, Phys. Rev. B 43, p. 9649 (1991). 9, 10

The presence of the well changes the symmetry of the crystal (eg. quantum wells in the direction of [001] corresponds to an uniaxial pressure applied perpendicular to the layer). You have to solve the 𝒌𝒑 perturbation theory (Chemla 1983):

Potencjał harmoniczny

2017-06-05 42

−ℏ2

2𝑚

𝑑2

𝑑𝑧2+1

2𝑚𝜔0

2𝑧2 𝜓 𝑧 = 𝐸𝜓 𝑧 𝜀 =𝐸

ℏ𝜔0𝜉 =

𝑚𝜔0

ℏ𝑧

𝑑

𝑑𝜉− 𝜉

𝑑

𝑑𝜉+ 𝜉 A e

𝜉2

2 = −2𝜀0 + 1 A e𝜉2

2 = 0 ⇒ −2𝜀0 + 1 = 0 ⇒ 𝜀0 =1

2

𝜀𝑛 = 𝑛 +1

2

𝐸𝑛 = ℏ𝜔0 𝑛 +1

2

𝜓𝑛 𝑧 = 𝐴𝑛𝐻𝑛𝑚𝜔0

ℏ𝑧 exp −

𝑚𝜔0

2ℏ𝑧2

𝐻𝑛 - wielomiany Hermite’a

𝐴𝑛 = 2𝑛𝑛!𝜋ℏ

𝑚𝜔

−1/2

Comments on the valence band

2017-06-05 43

𝐸3/2 𝑘 = 𝐴𝑘2 ± 𝐵2𝑘4 + 𝐶2 𝑘𝑥2𝑘𝑦

2 + 𝑘𝑦2𝑘𝑧

2 + 𝑘𝑧2𝑘𝑥

2

From 𝒌𝒑 perturbation theory :010

00100ത1

0ത10

0ത1ത1

01ത1

0ത11

011


2017-06-05 44

𝐸3/2 𝑘 = 𝐴𝑘2 ± 𝐵2𝑘4 + 𝐶2 𝑘𝑥2𝑘𝑦



2 1/2

X.X. Wei,Y. Yu, Y. Bi; Int. J. of Mechanical Sciences Vol. 50, 1499–1509 (2008),

Fig. 7. The 3D constant-energy surfaces of the heavy hole (HH) band, the light hole (LH) band and the split-off (SO) band and their corresponding conductivity masses for unstrained silicon.

From 𝒌𝒑 perturbation theory :


2017-06-05 45

X.X. Wei,Y. Yu, Y. Bi; Int. J. of Mechanical Sciences Vol. 50, 1499–1509 (2008),

Fig. 8. The 3D constant-energy surfaces of the heavy hole (HH) band, the light hole (LH) band and the split-off (SO) band and the corresponding conductivity masses for silicon under the uniaxial compression P ¼ 500kN. .

𝐸3/2 𝑘 = 𝐴𝑘2 ± 𝐵2𝑘4 + 𝐶2 𝑘𝑥2𝑘𝑦



2 1/2



2017-06-05 46

𝐸3/2 𝑘 = 𝐴𝑘2 ± 𝐵2𝑘4 + 𝐶2 𝑘𝑥2𝑘𝑦



2 1/2

𝛾1 = −2𝑚

ℏ2𝐴 𝛾2 = −

𝑚

ℏ2𝐵 𝛾3 =

𝑚

ℏ21

3𝐶2 + 𝐵2

Luttinger dimensionless parameters:

𝐸𝑙,ℎ 𝒌 = −ℏ2

2𝑚𝛾1𝑘

2 ± 4𝛾22𝑘4 + 12 𝛾3

2 − 𝛾22 𝑘𝑥

2𝑘𝑦2 + 𝑘𝑦

2𝑘𝑧2 + 𝑘𝑧2𝑘𝑥2 + heavy holes

− light holes

Expanding around 𝑘 = 0

𝐸ℎℎ 𝒌 = −ℏ2

2𝑚𝛾1 + 2𝛾2 𝒌⊥

2 + 𝛾1 + 2𝛾2 𝑘𝑧2

𝐸𝑙ℎ 𝒌 = −ℏ2

2𝑚𝛾1 − 2𝛾2 𝒌⊥

2 + 𝛾1 − 2𝛾2 𝑘𝑧2



2017-06-05 47


2𝑚𝛾1 + 𝛾2 𝒌⊥

2 + 𝛾1 − 2𝛾2 𝑘𝑧2


2𝑚𝛾1 − 𝛾2 𝒌⊥

2 + 𝛾1 + 2𝛾2 𝑘𝑧2

It follows that the heavy holes have a "light" mass 𝑚

𝛾1+𝛾2in the plane 𝑥 − 𝑦 , while light holes

have heavy mass 𝑚

𝛾1−𝛾2


𝐸

𝐽𝑧 = ±3

2

𝐽𝑧 = ±1

2

𝐸1ℎℎ

𝐸1𝑙ℎ


2017-06-05 48


2𝑚𝛾1 + 𝛾2 𝒌⊥

2 + 𝛾1 − 2𝛾2 𝑘𝑧2


2𝑚𝛾1 − 𝛾2 𝒌⊥

2 + 𝛾1 + 2𝛾2 𝑘𝑧2

𝐸

𝐽𝑧 = ±3

2

𝐽𝑧 = ±1

2

𝐸1ℎℎ

𝐸1𝑙ℎ

𝐸

𝐸1ℎℎ

𝐸1𝑙ℎ

ℏ2

2𝑚𝛾1 − 𝛾2

ℏ2

2𝑚𝛾1 + 𝛾2

𝑘𝑥



2017-06-05 49


2𝑚𝛾1 + 𝛾2 𝒌⊥

2 + 𝛾1 − 2𝛾2 𝑘𝑧2


2𝑚𝛾1 − 𝛾2 𝒌⊥

2 + 𝛾1 + 2𝛾2 𝑘𝑧2

𝐸

𝐽𝑧 = ±3

2

𝐽𝑧 = ±1

2

𝐸1ℎℎ

𝐸1𝑙ℎ

𝐸

𝐸1ℎℎ

𝐸1𝑙ℎ

ℏ2

2𝑚𝛾1 − 𝛾2

ℏ2

2𝑚𝛾1 + 𝛾2

𝑘𝑥

PoLDS 385


2017-06-05 50

Energy bands

Energy bands

2017-06-05 51

Example:

D. Wasik.

Quantum well

2017-06-05 52

http://www.nextnano.de/nextnano3/tutorial/1Dtutorial_AlGaInP_onGaAs.htm

Finite potential well – square well

2017-06-05 53

Simply example has been already calculated


2017-06-05 54

Heterostructures can have different effective masses in different regions:

−ℏ2

2𝑚0𝑚∗

𝑑2

𝑑𝑧2𝜓 𝑧 + V0 z 𝜓 𝑧 = 𝜀𝜓 𝑧

ቤ1

𝑚𝐵

𝑑𝜓

𝑑𝑧z=

a2

= ቤ1

𝑚𝑊

𝑑𝜓

𝑑𝑧z=

a2

Matching conditions at the interface must be modified (equation has to conserve current𝐼𝑊 = 𝐼𝐵 thus 𝑣𝑊 = 𝑣𝐵):

It turns that the conversion 𝑚∗ → 𝑚 𝑧 is not a good solution of the problem (equation ceases to be Hermitean). It has to be done diferently, eg.:

−ℏ2

2𝑚0

𝑑

𝑑𝑧

1

𝑚 𝑧

𝑑

𝑑𝑧𝜓 𝑧 + V0 z 𝜓 𝑧 = 𝜀𝜓 𝑧


2017-06-05 55

Inside the well:

𝜓 𝑧, 𝑡 = 𝐶 ቊcos 𝑘𝑛𝑧

sin 𝑘𝑛𝑧𝑒−𝑖𝜔𝑛𝑡

−𝑎

2< 𝑧 <

𝑎

2

The barrier:

𝜓 𝑧 = 𝐷 exp(±𝜅𝑛𝑧)

ℏ2𝜅2

2𝑚 𝑚𝐵= 𝐸𝐵 − 𝐸𝑛 = 𝐵

𝑘𝑛 =1

ℏ2𝑚𝑚𝑊 𝐸𝑛 − 𝐸𝑊

𝜅𝑛 =1

ℏ2𝑚𝑚𝐵 𝐸𝐵 − 𝐸𝑛

−ℏ2

2𝑚0𝑚𝑊

𝑑2

𝑑𝑧2𝜓 𝑧 = 𝐸𝑛 − 𝐸𝑊 𝜓 𝑧

ቤ1

𝑚𝐵

𝑑𝜓

𝑑𝑧z=

a2

= ቤ1

𝑚𝑊

𝑑𝜓

𝑑𝑧z=

a2

Matching conditions:

𝐶𝑘

𝑚𝑊

−sin 𝑘𝑛𝑎

2

cos 𝑘𝑛𝑎

2

= −𝐷𝜅

𝑚𝐵exp 𝑘𝑛

𝑎

2


2017-06-05 56

We calculate:

𝐶𝑘

𝑚𝑊

−sin 𝑘𝑛𝑎

2

cos 𝑘𝑛𝑎

2

= −𝐷𝜅𝑛𝑚𝐵

exp 𝜅𝑛𝑎

2

ቚ𝜓𝐵 𝑧z=

a2

= ቚ𝜓𝑊 𝑧z=

a2

ቤ1

𝑚𝐵

𝑑𝜓

𝑑𝑧z=

a2

= ቤ1

𝑚𝑊

𝑑𝜓

𝑑𝑧z=

a2

tan 𝑘𝑛𝑎

2

cot 𝑘𝑛𝑎

2

=𝑚𝑊𝜅𝑛𝑚𝐵𝑘𝑛

=𝑚𝑊

𝑚𝐵

2𝑚0𝑚𝑊𝑉0

ℏ2𝑘𝑛2 − 1

Substituting , and (depends only on 𝑚𝑊)𝜃 =𝑘𝑛𝑎

2𝜃02 =

𝑚0𝑚𝑊𝑉0𝑎2

2ℏ2

ቊtan 𝜃−cot 𝜃

=𝑚𝑊

𝑚𝐵

𝜃02

𝜃2− 1


2017-06-05 57

THE SAME mass in the well and in the barrier:


2017-06-05 58

THE DIFFERENT mass in the well and in the barrier:


2017-06-05 59



2017-06-05 60


Harmonic potential

2017-06-05 61

Harmonic potential

2017-06-05 62

𝜀𝑛 = 𝑛 −1

2ℏ𝜔0𝑉 𝑧 =

1

2𝐾𝑧2 =

1

2𝑚𝜔0

2𝑧2−ℏ2

2𝑚

𝑑2

𝑑𝑧2+ 𝑉(𝑧) 𝜓 𝑧 = 𝜀𝜓 𝑧

Harmonic potential

2017-06-05 63

𝜙𝑛+1 𝑧 =1

2𝑛𝑛!

𝑚𝜔0

ℏ𝜋

1/4

exp −𝑚𝜔0𝑧

2

2ℏ𝐻𝑛

𝑚𝜔0

ℏ𝑧

𝐻0 𝑡 = 1𝐻1 𝑡 = 𝑡𝐻2 𝑡 = 4𝑡2 − 2𝐻3 𝑡 = 8𝑡3 − 12𝑡

𝜙1 𝑧 2 =𝑚𝜔0

𝜋ℏexp −

𝑚𝜔0𝑧2

ℏ

The ground state is a Gauss package – does the packet diffuse?

Wavefunction

Hermite polynomials

𝜀𝑛 = 𝑛 −1

2ℏ𝜔0−

ℏ2

2𝑚

𝑑2

𝑑𝑧2+1

2𝑚𝜔0

2𝑧2 𝜓 𝑧 = 𝜀𝜓 𝑧

Harmonic potential

2017-06-05 64

Triangular well

2017-06-05 65

Triangular well

2017-06-05 66

−ℏ2

2𝑚

𝑑2

𝑑𝑧2+ 𝑒𝐹𝑧 𝜓 𝑧 = 𝜀𝜓 𝑧

Triangular well

2017-06-05 67

−ℏ2

2𝑚

𝑑2

𝑑𝑧2+ 𝑒𝐹𝑧 𝜓 𝑧 = 𝜀𝜓 𝑧

𝑑2

𝑑𝑧2𝜓 𝑧 =

2𝑚

ℏ2𝑒𝐹𝑧 − 𝜀 𝜓 𝑧

Transformation:

The equation reduces to Stokes or Airy equation:

𝑑2

𝑑𝑧2𝑓 𝑧 = 𝑧𝑓 𝑧

Its two independent solutions the Airy functions 𝐴𝑖(𝑧) and 𝐵𝑖(𝑧). The solutions of the equation are the zeros of a function 𝐴𝑖(𝑧) (after some rearrangements).

Substituting: 𝑧0 =ℏ2

2𝑚𝑒𝐹

1/3

𝜀0 = 𝑒𝐹𝑧0 =𝑒𝐹ℏ 2

2𝑚

1/3

, ҧ𝑧 =𝑧

𝑧0, ҧ𝜀 =

𝜀

𝜀0

𝑑2

𝑑𝑧2𝜓 𝑧 =

2𝑚

ℏ2𝑒𝐹𝑧 − 𝜀 𝜓 𝑧

Triangular well

2017-06-05 68

𝑉 𝑧 = 𝑒𝐹𝑧 𝜀𝑛 = 𝑐𝑛𝑒𝐹𝑧 2

2𝑚

1/3

WKB approximation

6/5/2017 69

WKB approximation (Wentzel – Krammers – Brillouin) – for slowly changing potential

It is also known as the LG or Liouville–Green method or JWKB and WKBJ, where the "J" stands for Jeffreys or phase integral method or semi-classical approximation.

−ℏ2

2𝑚

𝑑2

𝑑𝑥2+ 𝑉 𝑥 𝜓 𝑥 = 𝐸𝜓 𝑥

What is slowly varing potential? It is for sure 𝑉 𝑥 = 𝑉0 = 𝑐𝑜𝑛𝑠𝑡. The solution for this potential is a plane wave 𝜓 𝑥 = 𝑒𝑖𝑘𝑥 - phase of the wavefunction 𝑘 𝑥 = 𝑘 = 𝑐𝑜𝑛𝑠𝑡 is constant in the

whole space 𝑘2 =2𝑚

ℏ𝐸 − 𝑉0

Let's define 𝑘2 𝑥 =2𝑚

ℏ𝐸 − 𝑉(𝑥) - we want the phase 𝑘 𝑥 to be slowly varying in space, i.e.

𝑑𝑘

𝑑𝑥≪ 𝑘2

(such condition).

We are looking for the solution 𝜓 𝑥 = 𝑒𝑖𝜒 𝑥 where 𝜒 𝑥 is the phase of the wavefunction.

WKB approximation

6/5/2017 70

−ℏ2

2𝑚

𝑑2

𝑑𝑥2+ 𝑉 𝑥 𝜓 𝑥 = 𝐸𝜓 𝑥

Let’s define 𝑘2 𝑥 =2𝑚

ℏ𝐸 − 𝑉(𝑥) - slowly varing in real space 𝑘 𝑥 .

We are looking for the solutuion 𝜓 𝑥 = 𝑒𝑖𝜒 𝑥 where 𝜒 𝑥 its the phase of the wavefunction. Inserting into Schrodinger equation:

𝜒′ 𝑥 2 − 𝑖𝜒′′ 𝑥 =2𝑚

ℏ𝐸 − 𝑉 𝑥 ≡ 𝑘2 𝑥 - this is rigorous.

Zero-order WKB approximation assumes 𝜒′ 𝑥 2 ≫ 𝜒′′ 𝑥 czyli 𝜒′′ 𝑥 ≈ 0

𝜒′ 𝑥 2 = 𝑘2 𝑥 czyli𝜒 𝑥 = ±න

𝑥

𝑘 𝑥′ 𝑑𝑥′

Usually we expand more𝜒′ 𝑥 2 = 𝑘2 𝑥 + 𝑖𝜒′′ 𝑥 = 𝑘2 𝑥 + 𝑖 𝜒′ 𝑥 ′ ≈ 𝑘2 𝑥 ± 𝑖 𝑘 𝑥 ′

Thus:

𝜒′ 𝑥 ≈ ±𝑘 𝑥 1 +𝑖𝑘′ 𝑥

𝑘2 𝑥≈± 𝑘 𝑥 +

𝑖𝑘′ 𝑥

2𝑘 𝑥


WKB approximation

2017-06-05 71

Typically, the WKB method continues into

𝜒′ 𝑥 2 = 𝑘2 𝑥 + 𝑖𝜒′′ 𝑥 = 𝑘2 𝑥 + 𝑖 𝜒′ 𝑥 ′ ≈ 𝑘2 𝑥 ± 𝑖 𝑘 𝑥 ′

Thus:

𝜒′ 𝑥 ≈ ±𝑘 𝑥 1 +𝑖𝑘′ 𝑥

𝑘2 𝑥≈± 𝑘 𝑥 +

𝑖𝑘′ 𝑥

2𝑘 𝑥

therefore:

𝜒 𝑥 = ±න𝑥

𝑘 𝑥′ 𝑑𝑥′ +𝑖

2ln 𝑘 𝑥

We get:

𝜓(𝑥) ≈1

𝑘 𝑥exp ±𝑖 න

𝑥

𝑘 𝑥′ 𝑑𝑥′

The term 1/ 𝑘 𝑥 - the density of probability of fast-moving particles is small for large 𝑘 - OK!


WKB approximation

2017-06-05 72

𝐸

In the case of turning points (ie. the edges of the barriers well) potential changes rapidly compared with the wavelength 𝑘 𝑥 - WKB approximation is not valid (a rigorous approach avoids it by moving into the complex plane). Another way is to note that the potential is linear for a small region around the turning point Δ𝑥𝐿 - and the Airy functions are the exact solutions. Then the solution must match on both regions near 𝑥𝐿.The problem sounds complicated but fortunately the results are simple (we got additional phase).

energia cząstki

𝑉(𝑥)

𝑥

𝑥𝐿

WKB approximation (Wentzel – Krammers – Brillouin) – for slowly varying potential

WKB approximation

2017-06-05 73

𝐸𝜓 𝑥

carrier energy

𝑉(𝑥)

𝑥

𝑥𝐿

𝜓 𝑥 ~2

𝑘 𝑥cos න

𝑥𝐿

𝑥

𝑘 𝑥′ 𝑑𝑥′ −𝜋

4, 𝑥 ≫ 𝑥𝐿

𝜓 𝑥 ~1

𝜅 𝑥exp −න

𝑥𝐿

𝑥

𝜅 𝑥′ 𝑑𝑥′ , 𝑥 ≪ 𝑥𝐿


WKB approximation

6/5/2017 74

𝜓 𝑥 ~2

𝑘 𝑥cos න

𝑥𝐿

𝑥


4, 𝑥 ≫ 𝑥𝐿

Examples:1. „Hard” (infinitely steep) wall – the wave function goes to zero at the boundaries, so an exact number of half-wavelengths must fit between them

න𝑥𝐿

𝑥𝑅

𝑘 𝑥′ 𝑑𝑥′ = 𝑛𝜋

2. „Soft wall” – an allowed state must obey the matching conditions (𝑥 ≫ 𝑥𝐿, above) and at 𝑥𝐿 it

has additional phase −𝜋

4and similarly at 𝑥𝑅 - next −

𝜋

4. Altogether:

න𝑥𝐿

𝑥𝑅

𝑘 𝑥′ 𝑑𝑥′ = 𝑛 −1

2𝜋

𝑘𝑛 =𝑛𝜋

𝐿For : 𝑘 𝑥 = 𝑐𝑜𝑛𝑠𝑡 we have


WKB approximation

2017-06-05 75

𝜓 𝑥 ~2

𝑘 𝑥cos න

𝑥𝐿

𝑥


4, 𝑥 ≫ 𝑥𝐿

3. Triangular well of „hard” and „soft” well:

Eg. Triangular well: 𝑉 𝑥 = 𝑒𝐹𝑥 wtedy 𝑘𝑛 𝑥 =1

ℏ2𝑚 𝐸𝑛 − 𝑉 𝑥 =

1

ℏ2𝑚 𝐸𝑛 − 𝑒𝐹𝑥

න𝑥𝐿

𝑥𝑅

𝑘 𝑥′ 𝑑𝑥′ = 𝑛 −1

4𝜋

න𝑥𝐿

𝑥𝑅

𝑘𝑛 𝑥′ 𝑑𝑥′ = න0

𝐸𝑛/𝑒𝐹 1

ℏ2𝑚 𝐸𝑛 − 𝑒𝐹𝑥′ 𝑑𝑥′ =

2𝑚𝐸𝑛ℏ2

1/2 𝐸𝑛𝑒𝐹

න0

1

1 − 𝑠 𝑑𝑠 =

2𝑚𝐸𝑛ℏ2


න0

1

1 − 𝑠 𝑑𝑠 =2

3

2𝑚𝐸𝑛ℏ2


𝑠 =𝑥

𝑥𝑅

𝐸𝑛 =3

2𝜋 𝑛 −

1

4

2/3𝑒𝐹ℏ 2

2𝑚

1/3


Triangular well

2017-06-05 76

𝐸𝑛 =3

2𝜋 𝑛 −

1

4

2/3𝑒𝐹ℏ 2

2𝑚

1/3


Triangular well

2017-06-05 77

𝐸𝑛 =3

2𝜋 𝑛 −

1

4

2/3𝑒𝐹ℏ 2

2𝑚

1/3

http://www.phys.unsw.edu.au/QED/research/2D_scattering.htm


WKB approximation

2017-06-05 78

𝑘𝑛 𝑥 =1

ℏ2𝑚𝑉 𝑥 =

1

ℏ2𝑚𝑉𝑏 1 −

𝑥

𝑑

2

𝑉 𝑥 = 𝑉𝑏 1 −𝑥

𝑑

2

Low dimensional structures

2017-06-05 79

Full Hamiltonian in our universe has three spatial dimensions 𝑥, 𝑦, 𝑧, 𝑡 = 𝑅, 𝑡

−ℏ2

2𝑚𝛻2 + 𝑉 𝑅 𝜓 𝑅 = 𝐸𝜓 𝑅

For 𝑉 𝑅 = 𝑉(𝑧) we obtain:

−ℏ2

2𝑚

𝜕2

𝜕𝑥2+

𝜕2

𝜕𝑦2+

𝜕2

𝜕𝑧2+ 𝑉 𝑧 𝜓 𝑥, 𝑦, 𝑧 = 𝐸𝜓 𝑥, 𝑦, 𝑧

𝜓 𝑥, 𝑦, 𝑧 = exp 𝑖𝑘𝑥𝑥 exp 𝑖𝑘𝑦𝑦 𝑢(𝑧)

Along directions 𝑥 and 𝑦 mamy ruch swobodny:

We can show (on the blackboard!), that final eigenenergies of the potential 𝑉 𝑧 are:

131

𝜀 = 𝐸 −ℏ2𝑘𝑥

2

2𝑚−ℏ2𝑘𝑦

2

2𝑚−ℏ2

2𝑚

𝑑2

𝑑𝑧2+ 𝑉 𝑧 𝑢 𝑧 = 𝜀𝑢 𝑧


2017-06-05 80

𝐸𝑛 𝑘𝑥, 𝑘𝑦 = 𝜀𝑛 +ℏ2𝑘𝑥

2

2𝑚+ℏ2𝑘𝑦

2

2𝑚

𝜓𝑘𝑥,𝑘𝑦,𝑛 𝑥, 𝑦, 𝑧 = exp 𝑖𝑘𝑥𝑥 exp 𝑖𝑘𝑦𝑦 𝑢𝑛 𝑧 = 𝜓𝒌,𝑛 𝒓, 𝑧 = exp 𝑖𝒌 ∙ 𝒓 𝑢𝑛 𝑧

𝐸𝑛 𝒌 = 𝜀𝑛 +ℏ2𝒌2

2𝑚


2017-06-05 81

𝐸𝑛 𝑘𝑥, 𝑘𝑦 = 𝜀𝑛 +ℏ2𝑘𝑥

2

2𝑚+ℏ2𝑘𝑦

2

2𝑚

𝜓𝑘𝑥,𝑘𝑦,𝑛 𝑥, 𝑦, 𝑧 = exp 𝑖𝑘𝑥𝑥 exp 𝑖𝑘𝑦𝑦 𝑢𝑛 𝑧 = 𝜓𝒌,𝑛 𝒓, 𝑧 = exp 𝑖𝒌 ∙ 𝒓 𝑢𝑛 𝑧

𝐸𝑛 𝒌 = 𝜀𝑛 +ℏ2𝒌2

2𝑚


2017-06-05 82



2017-06-05 83

𝜓 𝑅 = 𝜓𝒌,𝑛 𝒓, 𝑧 = exp 𝑖𝒌 ∙ 𝒓 𝑢𝑛 𝑧

−ℏ2

2𝑚0 𝑚𝑊,𝐵𝛻2 + 𝑉 𝑅 𝜓 𝑅 = 𝐸𝜓 𝑅

Effective mass in the barrier 𝑚𝐵 and in the well 𝑚𝑊

−ℏ2

2𝑚0 𝑚𝑊𝛻2 + 𝐸𝑊 𝜓 𝑅 = 𝐸𝜓 𝑅

For separated wave functions:

−ℏ2

2𝑚0 𝑚𝐵𝛻2 + 𝐸𝐵 𝜓 𝑅 = 𝐸𝜓 𝑅

We got (on the blackboard!):

−ℏ2

2𝑚0 𝑚𝑊

𝑑2

𝑑𝑧2+

ℏ2𝒌2

2𝑚0 𝑚𝑊+ 𝐸𝑊 𝑢𝑛 𝑧 = 𝜀𝑢𝑛 𝑧

−ℏ2

2𝑚0 𝑚𝐵

𝑑2

𝑑𝑧2+

ℏ2𝒌2

2𝑚0 𝑚𝐵+ 𝐸𝐵 𝑢𝑛 𝑧 = 𝜀𝑢𝑛 𝑧


2017-06-05 84

The particle moves in the well which potentiald depends on 𝒌, in fact 𝑘 = 𝒌

−ℏ2

2𝑚0 𝑚𝑊

𝑑2

𝑑𝑧2+

ℏ2𝒌2


−ℏ2

2𝑚0 𝑚𝐵

𝑑2

𝑑𝑧2+

ℏ2𝒌2


𝑉0 𝑘 = 𝐸𝐵 − 𝐸𝑊 +ℏ2𝑘2

2𝑚0

1

𝑚𝐵−

1

𝑚𝑊

The particle gains partially the effective mass of the barrier:

𝐸𝑛 𝑘 = 𝜀𝑛(𝑘) +ℏ2𝑘2

2𝑚0𝑚𝑊≈ 𝜀𝑛(𝑘 = 0) +

ℏ2𝑘2

2𝑚0𝑚𝑒𝑓𝑓

𝑚𝑒𝑓𝑓 ≈ 𝑚𝑊𝑃𝑊 +𝑚𝐵𝑃𝐵

prawdopodobieństwo znalezienia cząstki

Np. w strukturze GaAs-AlGaAs 𝑚𝐵 >𝑚𝑊 więc studnia robi się „płytsza”

energy of the bond state depends on 𝑘


2017-06-05 85

−ℏ2

2𝑚0 𝑚𝑊

𝑑2

𝑑𝑧2+

ℏ2𝒌2


−ℏ2

2𝑚0 𝑚𝐵

𝑑2

𝑑𝑧2+

ℏ2𝒌2


𝑉0 𝑘 = 𝐸𝐵 − 𝐸𝑊 +ℏ2𝑘2

2𝑚0

1

𝑚𝐵−

1

𝑚𝑊

Np. w strukturze GaAs-AlGaAs 𝑚𝐵 >𝑚𝑊 więc studnia robi się „płytsza”

The particle moves in the well which potentiald depends on 𝒌, in fact 𝑘 = 𝒌

Documents

Semiconductor heterostructures quantum wellsszczytko/LDSN/3_LDSN_2017_QWells.pdf · 2017. 6. 5. · Bandgap engineering 2017-06-05 41 Effects of biaxial strain: decrease of the degeneracy