Chapter 4: One-Dimensional Potentials

The Potential Step

a) When a neutron with an external kinetic energy K enters a nucleus, it experiences

a potential

b) While a charged particle moves along the axis of two cylindrical electrodes held at

different voltages, its potential energy changes very rapidly when passing from

one to the other. Potential energy function can be approximated by a step potential.

For the step potential, the x-axis breaks up into two regions

0x V

0x 0)x(V

0

Since the potential is time-independent, problem is to solve the time-independent

Schrodinger equation.

There are two cases, 0VE and 0VE

For 0VE :

Schrodinger equations for both regions are

0xfor (x)Eu(x)uVdx

(x)ud

2m-

0xfor (x)Eudx

(x)ud

2m-

IIII02

II

22

I2

I

22

The physically acceptable solutions are

2ikx -ikxI

I I2 2

2iqx iqx0II

II II2 2

d u (x) 2mEu (x) 0 u (x) e e

dx

2m(E-V )d u (x)u (x) 0 u (x) e e

dx

R

T D

where 2

22

1

2mEkk

and

2

022

2

)V-2m(Eqk

At the point x=0, the continuity property of the wave functions requires

I x 0 I x 0

I II0 0

u (x)| u (x)| 1 R T

du (x) du (x)| | (1 R) T

dx dxx x ik iq

E

V11

E

V1-1

qk

q-k R

qk

2k T

0

0

• Reflection probability 2|R| is not zero for 0VE . Only tends to zero in high energy

limit, 0VE (correspondence principle again). Notice that reflection and transmission

probabilities satisfy the relation 2 2| | | | 1R T . This is a simple example of a scattering

problem.

For 0VE :

Schrodinger equations for both regions are

0xfor (x)Eu(x)uVdx

(x)ud

2m-

0xfor (x)Eudx

(x)ud

2m-

IIII02

II

22

I2

I

22

The physically acceptable solutions are

2ikx -ikxI

I I2 2

2q'x0II

II II2 2

d u (x) 2mEu (x) 0 u (x) e e

dx

2m(V )d u (x)u (x) 0 u (x) e

dx

R

EC

-q'xeT

where 2

22

1

2mEkk

and

2

022

2

)E2m(V'qk

At the point x=0, the continuity property of the wave functions requires

I x 0 I x 0

I II0 0

u (x)| u (x)| 1 R T

du (x) du (x)| | (1 R) 'T

dx dxx x ik q

q'ik

iq'-k Rand

q'ik

2k T

x-q'eT illustrates an important difference between classical and quantum physics. While

none of the particles can be found classically in the region x>0, quantum mechanically

there is a nonzero probability density that requires the wave function penetrates this

classically forbidden region

x2q'-

22

2x2q'-22

II e'qk

k4e|T||)x(u|)x(P

The penetration distance xΔ at which the probability density is

E)2m(V22q'

1Δxe|T|

e

1e|T|

)xP(xe

1Δx)xP(x

0

x2q'-2Δx)(x2q'-2

00

00

In classical limit, the term E)2m(V0 is so large compared to 2 , thus xΔ is

immeasurably small.

The probability density 2II |)x(u| for a group wave function incident on the step with

0VE . The group moves up to the step, penetrates slightly into the classically

forbidden region, and then is completely reflected from the step.

• Some tunneling of particles into classically forbidden region even for energies below

step height.

• Tunneling depth depends on energy difference EV0 , E)2m(V22q'

1Δx

0

• But no transmitted particle flux, 100% reflection, like classical case.

The Potential Well

The square well potential is often used in quantum mechanics to represent a situation in

which a particle moves in a restricted region of space under the influence of forces.

III regiona x 0

II regiona xa- V

I region-a x 0

)x(V 0

a) The motion of a neutron in a nucleus can be approximated by assuming that the

particle is in a square well potential with a depth about 50 MeV.

b) A square well potential results from superimposing the potential acting on a

conducting electron in a metal.

The scattering solutions ( E > 0 ):

When E>0 the particle is unconfined and this case corresponds to the scattering problem.

There are unbound states with a continuum range of energies.

2ikx -ikxI

I I2 2

2iqx -iqx0II

II II2 2

2ikx -ikxIII

III III2 2

d u (x) 2mEu (x) 0 u (x) e e

dx

2m(E V )d u (x)u (x) 0 u (x) e e

dx

d u (x) 2mEu (x) 0 u (x) e e

dx

R

A B

T G

At the point ax , the continuity property of the wave functions requires

-ika ika -iqa iqa

I x -a I x -a

-ika ika -iqa iqaI II

u (x)| u (x)| e e e e

du (x) du (x)| | (e e ) ( e e )

dx dxx a x a

R A B

ik R iq A B

At the point ax ,

iqa -iqa ika

II x a III x a

iqa -iqa ikaII III

u (x)| u (x)| e e e

du (x) du (x)| | ( e e ) e

dx dxx a x a

A B T

iq A B ikT

As a result,

2-2ika

22 220

0

2 2-2ika

2 2

2 1e

2 cos 2 ( )sin 21 sin (2 )

4 ( )

( )sin 2e

2 cos 2 ( )sin 2

kqT T

Vkq qa i q k qaqa

E E V

q k qaR i

kq qa i q k qa

Note:

a) If 0Rkq2)kq(VE 220 , no reflection

b) As 0T0k0E 2

c) 1,2,3,...n wherenπqa20R0qa2sin

12

0 En-VE This is called the transmission resonance (Ramsaver-Townsend)

Bound states in a potential well ( E<0 ):

The case E<0 corresponds to a particle which is confined in a bound state. When well is

deep enough, there are bound states with discrete energies.

ax ; eCeC(x)u0(x)u2mE

dx

(x)ud

axa- ;Bsinqx Acosqx(x)u0(x)u|)E|2m(V

dx

(x)ud

-ax ; eCeC(x)u0(x)u|E|2m

dx

(x)ud

αx-'2

αx2IIIIII22

III2

IIII2

0

2

II2

αx-'1

αx1II22

I2

The physically acceptable solutions:

ax ; Te(x)u

axa- ;Bsinqx Acosqx(x)u

-ax ; eC(x)u

αxIII

II

αx1I

At the point ax , the continuity property of the wave functions requires

- a

I x -a I x -a 1

- aI II1

u (x)| u (x)| e cos sin

du (x) du (x)| | e sin cos

dx dxx a x a

C A qa B qa

C qA qa qB qa

At the point ax ,

- a

II x a III x a 2

- aII III2

u (x)| u (x)| cos sin e

du (x) du (x)| | sin cos e

dx dxx a x a

A qa B qa C

qA qa qB qa C

As a result,

qcotgqa α Acosqx (x)u solutions even : 0 Band 0A 1)

qtanqa α Bsinqx (x)u solutions odd : 0 Band 0A 1)0AB

II

II

These transcendental equations cannot be solved directly. They can be solved

numerically on a computer, or graphically.

Discrete spectrum in an attractive potential well:

For the graphical solutions, we have

The even solutions:

tany/yyλαa/y qa tanqa (qa)/V2maαa

qtanqaq/2mV2mE/α E)/-2m(Vq

2220

2

220

220

Note:

a) There is always at least one bound state (for small λ )

b) For large λ there are more bound sates

c) For large λ the intersection points is given approximately 0,1,2,...n , )π2

1(ny

0)1n2(ma8

-V E )2

1(nE)/-2m(V )

2

1(nqay 2

2

22

0222

0

π

ππ

This is familiar the spectrum of even solutions in the infinite box.

The odd solutions:

-cotgy/yyλαa/y cotgqa qa αa 2

Note:

a) The large λ behaviou is 1,2,...n , π ny

0)n2(ma8

-V E nqay 2

2

22

0 π

π

This is familiar the spectrum of odd solutions in the infinite box.

b) There will be an intersection only is

2

2

202

2

πa2mV 0(ππ/2λ

The binding energy of deuteron can be found by using this result.

A second way to calculate the energy eigenvalues in well,

2

2 2 2 20

22

0

2 / 2 a qa

2 ( - ) /

mE mVa R

q m V E

The right side of the equation defines a circle of radius R. The solutions are determined

by the points where the circle intersects the tangent function, tana qa qa and

cota qa qa

Results:

Even solutions at intersections of blue and red curves (always at least one)

Odd solutions at intersections of blue and green curves.

• There is tunneling of particle into classical forbidden region in which EV0 .

• Amount of tunneling depends exponentially on EV0 .

• There are finite number of solutions. The number of bound states depends on the depth

of the well, but there is always at least one (even parity) state.

• Potential is even, so wave functions must be even or odd.

• Limit as 0V , we recover the infinite well solutions.

Example: The quantum well

Quantum well is a “sandwich” made of two different semiconductors in which the

energy of the electrons is different, and whose atomic spacings are so similar that

they can be grown together without an appreciable density of defects:

Now used in many electronic devices (some transistors, diodes, solid-state lasers)

Delta Function Potentials

Single delta function potential

For an attractive form

δ(x)2ma

λV(x)

2

The Schrodinger equation

0u(x) δ(x)a

λu(x)

2mE

dx

u(x)d22

2

The solutions everywhere, except 0x ,

0x ; e

0x ; e )x(u0u(x)

dx

u(x)d

x

x

2

2

2

κ

κ

κ

κκ

The eigenvalue condition:

a2

λκ

The probability of finding the particle in the interval 0xx0

)e1(2

1dx| e |)xx0(P 0x2

0

2x0

κκκ

The probability of finding the particle in the interval 0xx0

)e1(2

1dx| e |)0xx(P 0x2

0

2x0

κκκ

Total probability

0x2e1P

κ

Double delta function potential

a)δ(xa)-δ(x2ma

λV(x)

2

The Schrodinger equation

0u(x) a)δ(xa)-δ(xa

λu(x)

2mE

dx

u(x)d22

2

The solutions everywhere, except x a ,

-ax ; e

axa- ; )xsinh(B)xAcosh(

ax ; e

)x(u0u(x)dx

u(x)d

x

x

2

2

2

κ

κ

κ

κκ

κ

κ

Even solutions

-ax ; e

axa- ; )xAcosh(

ax ; e

)x(u

x

x

κ

κ

κ

κ

κ

1y

ytanh λ

At the intersection point, the eigenvalue condition:

a2a

λκ

λ

Compare with the single delta function potential: the energy for the double potential is

lower than the energy in the single potential

4

1

am2E

am2

22

even

22

λλ

Odd solutions

-ax ; e

axa- ; )xBsinh(

ax ; e

)x(u

x

x

κ

κ

κ

κ

κ

1

1y

ytanh1y

ghycot

λλ

At the intersection point, the eigenvalue condition:

4

1

am2E

a2

22

odd

λλκ

As a result

The odd solutions (if there are a bound state) is less strongly bound than even solution.

|E| |E| EE oddevenoddeven

The Potential Barrier

III regiona x 0

II regiona xa- V

I region-a x 0

)x(V 0

The case 0VE (scattering problems):

2ikx -ikxI

I I2 2

2iqx -iqx0II

II II2 2

2ikx -ikxIII

III III2 2

d u (x) 2mERe : u (x) 0 u (x) e e

dx

2m(E V )d u (x)Re I : u (x) 0 u (x) e e

dx

d u (x) 2mERe : u (x) 0 u (x) e e

dx

gion I R

gion I A B

gion III T G

At the point ax , the continuity property of the wave functions requires

-ika ika -iqa iqa

I x -a I x -a

-ika ika -iqa iqaI II

u (x)| u (x)| e e e e

du (x) du (x)| | (e e ) ( e e )

dx dxx a x a

R A B

ik R iq A B

At the point ax ,

iqa -iqa ika

II x a III x a

iqa -iqa ikaII III

u (x)| u (x)| e e e

du (x) du (x)| | ( e e ) e

dx dxx a x a

A B T

iq A B ikT

As a result,

)VE(E4

)qa2(sinV1

1|T|

0

220

2

The case 0VE :

i) Emission of α particles from radioactive nuclei

ii) Fusion process

iii) Tunnel diode

iv) Cold emission electrons

2ikx -ikxI

I I2 2

2- x x0II

II II2 2

2ikx -ikxIII

III III2 2

d u (x) 2mERe : u (x) 0 u (x) e e

dx

2m(V )d u (x)Re I : u (x) 0 u (x) e e

dx

d u (x) 2mERe : u (x) 0 u (x) e e

dx

gion I R

Egion I A B

gion III T G

At the point ax , the continuity property of the wave functions requires

-ika ika a a

I x -a I x -a

-ika ika a aI II

u (x)| u (x)| e e e e

du (x) du (x)| | (e e ) ( e e )

dx dxx a x a

R A B

ik R A B

At the point ax ,

a a ika

II x a III x a

a a ikaII III

u (x)| u (x)| e e e

du (x) du (x)| | ( e e ) e

dx dxx a x a

A B T

A B ikT

From these four equations

)EV(E4

)a κ2(sinhV1

1

2kκa 2κ)sinhκ(k

2kκ|T|

2κκ)sinhκi(k2κκcosh 2kκ

2kκeT

0

220

2222

22

22

2ika-

This means that there is transmission even though the energy lies below the potential

barrier. This is called tunneling of particles.

Classically we would expect 0|T| 2 (the particle is not permitted to enter the forbidden

region). The quantum wave can penetrate the barrier and give a nonzero probability to

find a particle beyond the barrier.

When 1a κ ,

a2a2a2

02

2

0

e2

1

2

eBeAa2sinh

VE 1a)EV(m2

κκκ

κ

)EV(m2)/a4(

00

a4

222

2

a4

2

222

1κa

2

2

222

2

0eV

E1

V

E16e

κk

4kκ

e4

1

2kκ

κk1

1

2κκsinh2kκ

κk1

1|T|

κ

κ

The fact that 2|T| does not vanish for 0VE is a purely quantum mechanical result.

In general, the barriers that occur in physical phenomena are not square. For an irregular

shaped barrier. WKB method provides one of the most approximations methods.

The transmission probability is an extremely sensitive function of the width of the barrier

and of EV0 .

22

a4

222

22

κk

4kκln2)2κκ)(2a(e

κk

4kκln |T|ln κ

The first term dominates the second one for any reasonable size of aκ .

For a smooth potential barrier

22

2n

n

n0xΔ

n

n

n

barrierspartial

2

barrierpartial

2N

22

21

2

2N

22

21

2

/E)x(Vm2dx2exp |T|

/E)x(Vm2dx2xΔ2lim

xΔ2|T|ln

|T|ln..........|T|ln|T|ln |T|ln

|T|..........|T||T| |T|

κ

κ

Example 1: Cold emission electrons

Electrons in a metal can be removed at a room temperature by application of an external

electric field ε . This phenomena is called as cold emission. The electric field will change

the work function as x eεWW .

22

a

0

22 /mWa3

24exp/xeWm2dx2exp |T| ε

Cold emission has an important application in the scanning tunneling microscope that has

been used for a study of surfaces of metals and some semiconductors. A conducting

probe with a very sharp tip is brought close to a metal. Electrons tunnel through the

empty space to the tip. Tunneling current is so sensitive to the metal (prob) distance

(barrier width) that even individual atoms can be mapped.

Example 2: Emission of α -particle from radioactive nucleus

The transmission probability (the probability of penetrating the barrier) is given by

2

2 1 2

2

1/2 1/2 1/2 1/22 211 2 1 2

2 2

2exp 2 ( ) ( )

22 2 2 cos 1

bG

R

b

R

Z Z emP T e dr V r E E V r b k

b

Z Z e mZ Z em R R RG dr k E k b

r b b b

where 2

Z and ZZZdaughterD

. At low energies, b R .

-1 -1( ) ....2 2

R RCos x x Cos

b b

and

2R

0 b

Classically, the -particle cannot

enter the region 0r>R . Quantum

mechanically, there is a tunneling

through such a barrier

21 2

0 for r R

V(r) Z Z e for r R

rk

22 2α α

2 20 0

m Ze m Zeπ R R R π RG 2 b =2 b 2

2 b b b 2 bπε πε

2 22

α α α 2 20 0 α α

2 2 22 2α α α

2 2 2 200 0 α α 0 0

1 2Ze 1 2ZeE m v b 2

2 bπε πε m v

m Ze m Ze m Ze2Ze π R ZeG 2 2 4 b 4 R

2 b πε vπε πε m v πε πε

Thus the probability of penetrating the barrieras an function of αE becomes

22-G α

1 220 0 α

m ZeZe ZP e exp 8 R exp C C

ε v 4πε E

where C1 and C2 are constants. We then find

α

21

E

ZCClnP

The probability of escape at each collisions is Ge

. Thus the probability of emission per

unit time is G0 e2R

v

and hence the life time (mean life) of the parent nucleus is about

G

0

e v

2Rτ

The emission of 4.2 MeV alpha-particles from HeThU 4

2

234

90

238

92 , the transmission

probability (the probability of penetrating the barrier) is obtained as

2 90 39P T e 10

which is so small and correspondingly

21 -39 1 -18 10

-18 17 10

v1λ P=1.7 10 10 =1.7 10

τ 2R

1/1.7 10 4.1 10 1.3 10

s s

s s y

The experimental half-life is 100.45 10 y that is a remarkable agreement.

Example 3. Fusion occurs by tunneling nuclei through the Coulomb barrier

The Harmonic Oscillator

The classical Hamiltonian is of the form

22

kx2

1

m2

pH

The Schrodinger equation for the harmonic oscillator potential is

0u(y)ydy

u(y)d

Eu(x)u(x)kx2

1

dx

u(x)d

2m-

2

2

2

2

2

22

ε

where αxx mω

y

and ω

2Eε

. The solution of the differential equation gives the

following harmonic oscillator eigenfunctions in terms of Hermite polynomials x) (αHn

and the energy eigenvalues

2 2

1/2

α x /2

n nn

n

αu (x) e H (α x)

π 2 n!

1 E n ω n 0,1,2,....

2

The eigenfunctions (x)un are orthogonal functions.

Energy eigenvalues are quantized.

0E0 is completely quantum mechanical effect.

2 2

1/42

α x /2

0

α u (x) e

π

is a Gauss packet.

n oddfor odd is (x)u

n evenfor even is (x)u(x)u(-1)(-x)u

n

n

nn

n

Important concepts:

The potential step

The potential well problem

The potential barrier

The harmonic oscillator