Solutions of Time-Independent
Schrodinger Equation
不含時薛丁格方程式的解
Zero Potential, Step Potential, Barrier Potential, Square Well
Potential, Infinite Square Well Potential,
Simple Harmonic Oscillator Potential
6-2 The zero potential
Time independent Schrodinger equation is
ˆ  (x)  E (x)
H
2
d2
(
 V) (x)  E (x)
2m dx 2
d2
2m

(x)

(E  V) (x)  0
2
2
dx
2m
(x)  2 (E  V) (x)  0
(x) 
2mE
2
 (x)  0
Let solution is
(x)  e
ikx
k2 
2mE
2
 k   2mE /
(x)  c1eikx  c2eikx
k
2mE /
(x)  Asin(kx)  Bcos(kx)
A  i(c1  c2 )
,
B  c1  c2
Eigenfunction of free paritcle
(x)  c1eikx  c2eikx
k
2mE /
(x, t)  (x)eiEt /  (x)eit
wavefunction of free paritcle
 c1ei(kx t)  c2ei(kx t)
(x)  eikx
E  , E0
, (x, t)  ei(kx t)
,wave is traveling in the direction of increasing x.
(x)  eikx
, (x, t)  ei(kx t)
,wave is traveling in the direction of decreasing x.
Plane wave
k
kx-t = constant, kx+t=constant
dx
dt
dx 
  0 v 

dt
dt
dt k
k
dx
dt
dx

 0 v 

dt
dt
dt
k
wave velocity
If c1= c2 ，there are two oppositely directed traveling waves that combine
to form a standing wave.
1
t
1 

kx


t

(n

)



x


(n

)
Node position:
2
k
2 k
Consider the wave of free particle traveling in the direction of increasing x.
(x)  Aeikx
, (x, t)  Aei(kx t)
Calculate the expectation value of the momentum p

p




 (x, t)pˆ  (x, t)dx

*
 * (x, t)  (x, t)dx








A 2  ei(kx t ) ei(kx t ) dx

 k



A 2  ei(kx t ) ( k)ei(kx t ) dx

)Aei(kx t ) dx
x
Aei(kx t ) Aei(kx t ) dx
Aei(kx t ) (i

( k)A
2



2mE /
e i(kx t ) ei(kx t ) dx
A 2  e i(kx t ) ei(kx t ) dx

p  2mE
k




* (x, t)(x, t)dx  1
The wave of free particle traveling in the direction of decreasing x.
(x)  Beikx
, (x, t)  Bei(kx t)
Calculate the expectation value of the momentum p

p




 (x, t)pˆ  (x, t)dx

*
 * (x, t)  (x, t)dx







)Bei(kx t ) dx
x
Bei(kx t ) Bei(kx t ) dx
Bei(kx t ) (i



B2 ( k)  ei(kx t ) e i(kx t ) dx
B
2




ei(kx t ) ei(kx t ) dx
p   2mE
 k
k
2mE /
x  p 
2
The momentum of the particle is
precisely known
x  
p  0
p k
These wave functions contain
only a single value of the wave
number k.
The probability density * for a group traveling
wave function of a free particle. With increasing time
the group moves in the direction of increasing x, and
also spreads.
Probability flux（機率通量）

(x, t)
2m
t
2
 *
2
*
(
  V) (x, t)  i
 (x, t)
2m
t
(1) *  (2) 
(
2
 2  V)(x, t)  i
(1)
(2)
(*)
i
  [
(*  * )]  0
t
2m
   *

 j  0
t
表變化
表流出
機率密度
i
j
(*  * )
2m
機率守恆方程
流出＝變化
機率通量
Consider wavefunction
(x, t)  Aei(kx t)
機率密度
  *  A*A
機率通量
i
j 
(*  * )
2m
k *

A A  vA*A
m
wave velocity
6-3 The step potential (energy less than step height)
0
V(x)  
V0
, x  0 (free particle) Running wave
, x  0 Exponential decay
( II )
(I)
2m
 

(x)

(E  0)1 (x)  0
2
 1

 (x)  2m (V  E) (x)  0
0
2
2
 2
ik1x
 ik1x


(x)

Ae

Be
, k1  2mE /

1

k2x
k2x

(x)

Ce

De
, k 2  2m(V0  E) /

 2
x   , (x)  0 C  0
B.C 1
ik1x
 ik1x

, k1  2mE /
1 (x)  Ae  Be

 k2x

(x)

De
, k 2  2m(V0  E) /

 2
Consider continuity of Ψ(x) at x=0
Consider continuity of dΨ(x)/dx at x=0
B.C 2
B.C 3
The wavefunction is (x, t)  (x)eiEt /
Reflection coefficient
The combination of an
incident and a reflected
wave of equal intensities
to form a standing wave.
Running wave
Exponential decay
Forbidden region
Penetration depth
* (x, t)(x, t)  D*De2k2x  D*D(e1 )2
Penetration depth
  x  1/ k 2  / 2m(V0  E)
Form uncertainty relation
 E  E Vo
Example 6-1. Estimate the penetration distance x for a very small dust
particle, of radius r=10-6m and density =104kg/m3, moving at the very low
velocity v=10-2m/sec, if the particle impinges on a potential of height equal to
twice its kinetic energy in the region to the left of the step.
Vo-E = K
Example 6-2. A conduction electron moves through a block of Cu at total energy E
under the influence of a potential which, to a good approximation, has a constant value
of zero in the interior of the block and abruptly steps up to the constant value Vo>E
outside the block. The interior value of the potential is essentially constant, at a value
that can be taken as zero, since a conduction electron inside the metal feels little net
Coulomb force exerted by the approximately uniform charge distributions that surround
it. The potential increases very rapidly at the surface of the metal, to its exterior value
Vo, because there the electron feels a strong force exerted by the nonuniform charge
distributions present in that region. This force tends to attract the electron back into the
metal and is, of course, what causes the conduction electron to be bound to the metal.
Because the electron is bound, Vo must be greater than its total energy E. The exterior
value of the potential is constant, if the metal has no total charge, since outside the
metal the electron would feel no force at all. The mass of the electron is m=9×10-31kg.
Measurements of the energy required to permanently remove it form the block. i.e.,
measurements of the work function, show that Vo-E = 4ev. From these data esitmate the
distance x that the electron can penetrate into the classically excluded region outside
the block.
6-4 The step potential (energy greater than step height)
0
V(x)  
V0
,x  0
,x  0
2m
 
 1 (x)  2 (E  0)1 (x)  0

 (x)  2m (E  V ) (x)  0
0
2
2
 2
(I)
( II )

1 (x)  Aeik1x  Beik1x , k1  2mE /


ik 2 x
 ik 2 x

(x)

Ce

De
, k 2  2m(E  V0 ) /

 2
x  0 , (x)
no reflection wave D  0
B.C 1
ik1x
 ik1x

, k1  2mE /
1 (x)  Ae  Be

ik 2 x

(x)

Ce
, k 2  2m(E  V0 ) /

 2
Consider continuity of Ψ(x) at x=0
Consider continuity of dΨ(x)/dx at x=0
B.C 2
B.C 3
The wavefunction is (x, t)  (x)eiEt /
T
j2
j1
(k1  k 2 )2
4k1k 2
R T 

1
2
2
(k1  k 2 ) (k1  k 2 )
表粒子數守恆
R+T=1
當k1、k2互換(入射方向變換)，R與T均不變，表Vo增加與Vo減小的效果是相同的。
所以反射波的產生乃由於V(x)的不連續，與V(x)增加或減小無關。
Example 6-3. When a neutron enters a nucleus, it experiences a potential energy
which drops at the nuclear surface very rapidly from a constant external value V=0
to a constant internal value of about V=-50 Mev. The decrease in the potential is
what makes it possible for a neutron to be bound in a nucleus. Consider a neutron
incident upon a nucleus with an external kinetic energy K=5Mev, which is typical
for a neutron that has just been emitted from a nuclear fission. Estimate the
probability that the neutron will be reflected at the nuclear surface, thereby failing
to enter and have its chance at inducing another nuclear fission.
A neutron of external kinetic energy K incident upon a decreasing potential step
of depth Vo, which approximates the potential it feels upon entering a nucleus. Its
total energy, measured from the bottom of the step potential, is E.
6-5 The barrier potential
0

V(x)  V0
0

,
E’
x0
(I)
, 0xa
,
xa
( II )
( III )
Case of E  V0
2m
 

(x)

(E  0)1 (x)  0 ,
x0
2
 1

2m



(x)

(V0  E) 2 (x)  0 , 0  x  a
 2
2

2m



(x)

(E  0) 3 (x)  0 ,
x a
2
 3

 1 (x)  Aeikx  Be  ikx


x
x

(x)

Fe

Ge
 2
 (x)  Ceikx  De  ikx

 3
x  a , (x)
,
k  2mE /
,
  2m(V0  E) /
no reflection wave D  0 B.C 1
E
For continuous at x = 0 and x = a for (x) & (x)
FG  AB


 F  G  ik (A  B)



a
a
ika
Fe

Ge

Ce


ik
a
a
ika
Fe  Ge  ( )Ce


, 1 (0)   2 (0)
(1)
, 1 (0)  2 (0)
(2)
,  2 (a)   3 (a)
(3)
, 2 (a)  3 (a)
(4)
1
ik
ik
1

F

[(1

)A

(1

)B]

[(  ik)A  (  ik)B]

2


2

G  1 [(1  ik )A  (1  ik )B]  1 [(  ik)A  (  ik)B]

2


2

C
ik ika a C

F  (1  )e

(  ik)eika a

2

2

C
ik ika a C

G  (1  )e

(  ik)eika a

2

2
, (1)  (2)
(5)
, (1)  (2)
(6)
, (3)  (4)
(7)
, (3)  ( 4)
(8)
(  ik)A  (  ik)B  (  ik)Ceika a

ika a
(


ik)A

(


ik)B

(


ik)Ce

, (5)  (7)
(9)
, (6)  (8)
(10)
[(  ik)2  (  ik)2 ]A  [(  ik)2 ea  (  ik)2 ea ]Ceika
, (9)  (  ik)  (10)(  ik)
(11)
(4ik)A  [(2  k 2 )(ea  ea )  2ik(ea  ea )]Ceika
a
a
a
a
A
e

e
e

e
 [( 2  k 2 )(
)  2ik(
)]eika /(2ik)
C
2
2
 [(2  k 2 )sinh(a)  2ik cosh(a)]eika /(2ik)
A*A
2
2 2
2
2 2
2
2 2

[(


k
)
sinh
(

a)

4

k
cosh
(

a)]/(4

k )
*
CC
( 2  k 2 ) 2
2
2
2
2

sinh
(

a)

cosh
(

a)
,
cosh
y

1

sinh
y
2 2
4 k
( 2  k 2 ) 2
2
 1
sinh
(a)
2 2
4 k
穿透係數
v3C*C
( 2  k 2 ) 2
2
1
T

[1

sinh
(

a)]
v1A*A
4 2 k 2
4E(Vo  E)
T
4E(Vo  E)  Vo2 sinh 2 a
For a >> 1
sinh 2 a 
1 a a 2 1 2 a
(e  e )  e
4
4
Vo2 sinh 2 a 1
T  [1 
]
4E(Vo  E)
Vo2e2a
 [1 
]1
16E(Vo  E)
Vo2e2a
[
]1  16E(Vo  E) e2 a
16E(Vo  E)
V2
o
Case of E  E  V0
  i ,   2m(E  Vo ) /
2  2  2m(E  Vo ) /
(k 2  2 )2
2
1
T  [1 
sinh
(i

a)]
42 k 2
Vo2
 [1 
(i sin a) 2 ]1
4E(E  Vo )
4E(E  Vo )

4E(E  Vo )  Vo2 sin a