P460 - square well 1

Finite Square Well Potential• For V=finite “outside” the well. Solutions to S.E. inside the well

the same. Have different outside. The boundary conditions (wavefunction and its derivative continuous) give quantization for E<V 0

• longer wavelength, lower Energy. Finite number of energy levels• Outside:

)(22

)(202

)(

00

2

2

20

2

22

)(

cos,sin

)(

EVmxk

VEmmk

dx

dm

keVE

kVE

andkxkVE

VE

mEk 2

1

P460 - square well 2

Boundary Condition• Want wavefunction and its derivative to be continuous

• often a symmetry such that solution at +a also gives one at -a

• Often can do the ratio (see book) and that can simplify the algebra

x

boundary

x

boundaryx

boundary

x

boundary

boundaryboundary

)(1)(1

)()(

)()(

P460 - square well 3

Finite Square Well Potential

0,0

)(:

)(:

cossin)(:

22

22

2

2

11

FDxasas

GeFexx

DeCexx

xkBxkAxinside

xkxka

xkxka

Equate wave function at boundaries

2/22

2/22

211

211

cossin

cossinakakak

akakak

CeBA

GeBA

And derivative

2/22121

2/22121

211

211

sincos

sincosakakak

akakak

eGkBkAk

eCkBkAk

P460 - square well 4

Finite Square Well Potential

onquantizatikk

xeeBx

wellinxkBxII

onquantizatikk

xeeAx

wellinxkAxI

ak

axkakak

ak

axkakak

221

22/

2

1

221

22/

2

1

1

221

1

221

tan

)cos()(

cos)(:

cot

)sin()(

sin)(:

E+R does algebra. 2 classes. Solve numerically

k1 and k2 both depend on E. Quantization sets allowed energy levels

02 2

222

VEman

n

P460 - square well 5

Finite Square Well PotentialNumber of bound states is finite. Calculate

assuming “infinte” well energies. Get n. Add 1

Electron V=100 eV width=0.2 nm

220

2

2

222 2202

Vma

man

n nVE

)(4

7.1022

2

)14.3()197(

100)2(.51.22

levelsofnumberN

nnmev

eVnmMeV

Deuteron p-n bound state. Binding energy 2.2 MeVradius = 2.1 F (really need 3D S.E………)

stateboundonlyN

nFMeV

MeVFMeV

11

1.022

2

)14.3()197(

2.2)1.2(94022

P460 - square well 6

Finite Square Well PotentialCan do an approximation by guessing at the

penetration distance into the “forbidden” region. Use to estimate wavelength

Electron V=100 eV width=0.2 nm

widerslightlyE

e

amn

n

bx

EVmb

2

222

)2(2

)(21

eVE

nm

nmnmMeV

eVnm

eVMeVeVnm

5.5

02.

2

22

)04.2(.5.2

)197(1

100*5.*2197

P460 - square well 7

Delta Function Potential

-function can be used to describe potential.

• Assume attractive potential V and E<V bound state. Potential has strength /a. Rewrite Schro.Eq

V0

rangeinifdxx

x

01)(

0)0()0(

)()()0()(

..

)0(2

)(

""00)(

22

2

2

xukxuadx

xudES

maxVor

xexceptxV

P460 - square well 8

Delta Function Potential II

• Except for x=0 have exponential solutions.

• Continuity condition at x=0 is (in some sense) on the derivative. See by integrating S.E in small region about x=0

V0

0)(

0)(

xexu

xexukx

kx

quantizeda

k

akeke

ua

xdx

dux

dx

du

xukxuxa

dxdx

xuddx

xukxuxadx

xud

kk

2

0)0()()(

)]()()([)(

)()()()(

)(

22

2

22

2

P460 - square well 9

1D Barriers (Square)

• Start with simplest potential (same as square well)

)(2

0)(2

22

1

0

,0)(

,0)(

0)(

00)(

0)(

00

22

11

VEmEVmmE

xikxik

xkxk

xikxik

kkkwith

VExFeEexor

VExDeCex

xBeAex

xxV

xVxV

V

00

P460 - square well 10

1D Barriers E<V • Solve by having wave function and derivative continuous at x=0

DBA

DBA

Cx

VExDeCex

xBeAex

kik

xx

xkxk

xikxik

1

2

22

11

)0()0(

)0()0(

00)(

,0)(

0)(

solve for A,B. As ||2 gives probability or intensity

|B|2=intensity of plane wave in -x direction

|A|2=intensity of plane wave in +x direction

1Re

)1(2)1(2

2

2

1

22

||

||

1

A

B

kik

kik

flectionR

DBDA

P460 - square well 11

1D Barriers E<V • While R=1 still have non-zero probability to be in region with E<V

cedisnpenetratiod

xeD

k

xk

tan

0||||

2

2

1

222

V

00

A

BD

Electron with barrier V-E= 4 eV. What is approximate

distance it “tunnels” into barrier?

nm

d

eVMeVnmeV

EVmcc

k

1.4*5.*2

*197

)(212

P460 - square well 12

1D Barriers E>V • X<0 region same. X>0 now “plane” wave

)0()(2

2

000

xpVEm

xikxik

k

xDeCe

V

0

A

BD

Will have reflection and transmission at x=0. But

“D” term unphysical and so D=0

ACAB kkk

kkkk

01

1

01

01 2

C

P460 - square well 13

1D Barriers E>V • Calculate Reflection and Transmission probabilities. Note flux is particle/second which is ||2*velocity

1|| 2 dx

T

0

1R

Note R+T=1

1

02

01

21

201

201

2

2

)(

4

)(

)(

||

||

kk

kk

k

kk

kk

A

B

T

R

1

0

kk

vv

E/V 1

“same” if E>V. different k

P460 - square well 14

1D Barriers:Example • 5 MeV neutron strikes a heavy nucleus with V= -50 MeV. What fraction are reflected? Ignore 3D and use simplest step

potential.

0

29.)(

)(

2

5/551

5/551

2

1

1

)(

)(2

01

201

R

REVE

EVE

kk

kk

mEk 2

5

-50 MeV

R

P460 - square well 15

1D Barriers Step E<V • Different if E>V or E<V. We’ll do E<V. again solve by continuity of wavefunction and derivative

0

0)(

)(

0)(

22

11

11

D

axGeFex

axDeCex

xBeAex

xkxkin

xikxik

xikxik

Incoming falling transmitted

No “left” travelling wave

reflected0 a

xa

xa

xx

inin

inin

aa

)()()0()0( ,

)()(,)0()0(

P460 - square well 16

1D Barriers Step E<V • X=0

GkFkBikAik

GFBA

2211

aikakak

aikakak

eikGekFek

CeGeFe122

122

122

X=a

4 equations->A,B,C,F,G (which can be complex)

A related to incident flux and is arbitrary. Physics in:

2

2

2

2

||

||

||

||

Av

Cv

Av

Bv TR Eliminate

F,G,B

)1(161

1

2

222

22

1

)1(16

)(

VE

VEak

ee

eTakif

TVE

VE

akak

P460 - square well 17

Example:1D Barriers Step • 2 eV electron incident on 4 eV barrier with thickness: .1 fm or 1 fm

81.)1(16

6.01

80.)73sinh(.

73.01.

3.7

42

4273.*2

1

)4/21*(4/2*48.

2

2

1*197

2*51.*2)(22

2

73.73.

eTapprox

withorT

akfmfor

fmk

ee

nmeVeVMeV

c

EVmc

With thicker barrier:

6

42

423.7*2

612

2

108.1)1(16

108.17401

3.71

eT

T

akfmfor

approx

exact

P460 - square well 18

Example:1D Barriers Step • Even simpler

For larger V-E

larger k

topbarrierfromfarhowtorelated

wavelengthk

thicknessbarriera

eeOT

EVm

xkak

)(2

2

2

"/1"

)(*)1( 22

More rapid decrease in wave function through

classically forbidden region

P460 - square well 19

Bound States and Tunneling

V

00

• State A will be bound with infinite lifetime. State B is bound but can decay to B->B’+X (unbound) with lifetime which depends on barrier height and thickness.

• Also reaction B’+X->B->A can be analyzed using tunneling

• tunneling is ~probability for wavefunction to be outside well

E=0

B

A

B’

P460 - square well 20

Alpha Decay

• Example: Th90 -> Ra88 + alpha

• Kinetic energy of the alpha = mass difference

• have V(r) be Coulomb repulsion outside of nucleus. But attractive (strong) force inside the nucleus. Model alpha decay as alpha particle “trapped” in nucleus which then tunnels its way through the Coulomb barrier

• super quick - assume square potential

• more accurate - 1/r and integrate

e

KE

KZeKKm

mmZ

r

KzZe

ka

eVEee

squareKVVguess

VKandzr

thicknessbarrieraeT

6.13

)(2)(

)2(''

0

8

2''

4

2

04

22)2(22

0

2

P460 - square well 21

Alpha Decay • Do better. Use tunneling probability for each dx

from square well. Then integrate

• as V(r) is known, integral can be calculated. See E+R and Griffiths 8.2)

MeVeVE

fmr

ZET

emm

em

rZR

KE

99.06.13

24.7

84exp)(

0

40 2

20

0

0

drKrVm

CBAABC

r

ReT

TTTT

''

2 ))((/22

A B C

4Z=large number

P460 - square well 22

Alpha Decay

• T is the transmission probability per “incident” alpha

• f=no. of alphas “striking” the barrier (inside the nucleus) per second = v/2R, If v=0.1c f=1021 Hz

sec105102

10

102.8103.1

99

95.84

710sec/

1

21sec

1339

82848890

yrs

T

ssT

fmfmR

MeVMeVK

PbPoRaTh

rate

rate

Depends strongly on alpha kinetic energy