Solutions of Time-Independent SchrodSolutions of Time-Independent Schrodinger Equationinger Equation

Solutions of Time-Independent SchrodSolutions of Time-Independent Schrodinger Equationinger Equation

Zero Potential, Step Potential, Barrier Potential, Square Well Potential, Infinite Square Well Potential,

Simple Harmonic Oscillator Potential

Zero Potential, Step Potential, Barrier Potential, Square Well Potential, Infinite Square Well Potential,

Simple Harmonic Oscillator Potential

不含時薛丁格方程式的解不含時薛丁格方程式的解

6-2 The zero potential

Time independent Schrodinger equation is

2 2

2

2

2 2

2

2

H (x) E (x)

d( V) (x) E (x)

2m dx

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

dx2m

(x) (E V) (x) 0

2mE(x) (x) 0

ikx(x) e

22

2mEk k 2mE /

ikx ikx1 2(x) c e c e

k 2mE /

(x) Asin(kx) Bcos(kx)

1 2 1 2A i(c c ) , B c c Let solution is

Eigenfunction of free paritcle

ikx ikx1 2(x) c e c e k 2mE /

E , E 0

iEt / i t(x, t) (x)e (x)e

i(kx t ) i(kx t )1 2c e c e

ikx i(kx t )(x) e , (x, t) e ,wave is traveling in the direction of increasing x.

,wave is traveling in the direction of decreasing x.

ikx i(kx t )(x) e , (x, t) e

Plane wave kx-t = constant, kx+t=constant

dx dt dxk 0 v

dt dt dt k

dx dt dxk 0 v

dt dt dt k

wave velocity

wavefunction of free paritcle

If c1= c2 ， there are two oppositely directed traveling waves that combine to form a standing wave.

Node position:1 t 1

kx t (n ) x (n )2 k 2 k

Consider the wave of free particle traveling in the direction of increasing x.

ikx i(kx t )(x) Ae , (x, t) Ae

Calculate the expectation value of the momentum p*

*

ˆ(x, t)p (x, t)dxp

(x, t) (x, t)dx

i(kx t ) i(kx t )

i(kx t ) i(kx t )

Ae ( i )Ae dxx

Ae Ae dx

2 i(kx t ) i(kx t )

2 i(kx t ) i(kx t )

A e ( k)e dx

A e e dx

k

k 2mE / p 2mE

2 i(kx t ) i(kx t )

2 i(kx t ) i(kx t )

( k)A e e dx

A e e dx

*(x, t) (x, t)dx 1

The wave of free particle traveling in the direction of decreasing x.

ikx i(kx t )(x) Be , (x, t) Be

Calculate the expectation value of the momentum p*

*

ˆ(x, t)p (x, t)dxp

(x, t) (x, t)dx

i(kx t ) i(kx t )

i(kx t ) i(kx t )

Be ( i )Be dxx

Be Be dx

2 i(kx t ) i(kx t )

2 i(kx t ) i(kx t )

B ( k) e e dx

B e e dx

k k 2mE /

p 2mE

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.

x p2

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.

Probability flux（機率通量）2

2

22 * *

( V) (x, t) i (x, t) (1)2m t

( V) (x, t) i (x, t) (2)2m t

*(1) (2) *

* *( ) i[ ( )] 0

t 2m

* * *i

j ( )2m

j 0t

機率密度

機率通量

機率守恆方程

流出＝變化表變化 表流出

* *ij ( )

2m

i(kx t )(x, t) Ae

Consider wavefunction

* *A A 機率密度

機率通量

* *kA A vA A

m

wave velocity

6-3 The step potential (energy less than step height)

0

0 , x 0V(x)

V , x 0

( I ) ( II )

1 12

2 0 22

2m(x) (E 0) (x) 0

2m(x) (V E) (x) 0

1 1

2 2

ik x ik x1 1

k x k x2 2 0

(x) Ae Be ,k 2mE /

(x) Ce De ,k 2m(V E) /

x , (x) 0 C 0 B.C 1

(free particle) Running wave

Exponential decay

1 1

2

ik x ik x1 1

k x2 2 0

(x) Ae Be ,k 2mE /

(x) De ,k 2m(V E) /

Consider continuity of Ψ(x) at x=0 B.C 2

Consider continuity of dΨ(x)/dx at x=0 B.C 3

The wavefunction is iEt /(x, t) (x)e

Reflection coefficient

The combination of an incident and a reflected wave of equal intensities to form a standing wave.

Exponential decayForbidden region

Running wave

Penetration depth

22k x* * * 1 2(x, t) (x, t) D De D D(e )

2 0x 1/ k / 2m(V E)

Penetration depth

Form uncertainty relation

oE E V

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)

( I ) ( II )

0

0 , x 0V(x)

V , x 0

1 12

2 0 22

2m(x) (E 0) (x) 0

2m(x) (E V ) (x) 0

1 1

2 2

ik x ik x1 1

ik x ik x2 2 0

(x) Ae Be ,k 2mE /

(x) Ce De ,k 2m(E V ) /

x 0 , (x) no reflection wave D 0 B.C 1

1 1

2

ik x ik x1 1

ik x2 2 0

(x) Ae Be ,k 2mE /

(x) Ce ,k 2m(E V ) /

Consider continuity of Ψ(x) at x=0 B.C 2

Consider continuity of dΨ(x)/dx at x=0 B.C 3

The wavefunction is iEt /(x, t) (x)e

21 2 1 2

2 21 2 1 2

(k k ) 4k kR T 1

(k k ) (k k )

表粒子數守恆

2

1

jT

j

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。

R+T=1

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

E

E’

( I ) ( II ) ( III )0

0 , x 0

V(x) V , 0 x a

0 , x a

1 12

2 0 22

3 32

2m(x) (E 0) (x) 0 , x 0

2m(x) (V E) (x) 0 , 0 x a

2m(x) (E 0) (x) 0 , x a

0E VCase of

ikx ikx1

x x2 0

ikx ikx3

(x) Ae Be , k 2mE /

(x) Fe Ge , 2m(V E) /

(x) Ce De

x a , (x) no reflection wave D 0 B.C 1

For continuous at x = 0 and x = a for (x) & (x)

1 2

1 2

a a ika2 3

a a ika2 3

F G A B , (0) (0) (1)

ikF G (A B) , (0) (0) (2)

Fe Ge Ce , (a) (a) (3)

ikFe Ge ( )Ce , (a) (a) (4)

ika a ika a

ika a ika a

1 ik ik 1F [(1 )A (1 )B] [( ik)A ( ik)B] , (1) (2) (5)

2 21 ik ik 1

G [(1 )A (1 )B] [( ik)A ( ik)B] , (1) (2) (6)2 2

C ik CF (1 )e ( ik)e , (3) (4) (7)

2 2C ik C

G (1 )e ( ik)e , (3) (2 2

4) (8)

ika a

ika a

( ik)A ( ik)B ( ik)Ce , (5) (7) (9)

( ik)A ( ik)B ( ik)Ce , (6) (8) (10)

2 2 2 a 2 a ika[( ik) ( ik) ]A [( ik) e ( ik) e ]Ce

, (9) ( ik) (10)( ik) (11)

2 2 a a a a ika( 4i k)A [( k )(e e ) 2i k(e e )]Ce a a a a

2 2 ikaA e e e e[( k )( ) 2i k( )]e /( 2i k)

C 2 2

2 2 ika[( k )sinh( a) 2i k cosh( a)]e /( 2i k)

*2 2 2 2 2 2 2 2 2

*

A A[( k ) sinh ( a) 4 k cosh ( a)] /(4 k )

C C

2 2 22 2 2 2

2 2

( k )sinh ( a) cosh ( a) , cosh y 1 sinh y

4 k

2 2 2

22 2

( k )1 sinh ( a)

4 k

* 2 2 22 13

* 2 21

o2 2

o o

v C C ( k )T [1 sinh ( a)]

v A A 4 k

4E(V E)T

4E(V E) V sinh a

穿透係數

For a >> 1

2 a a 2 2 a1 1sinh a (e e ) e

4 4

2 ao2o

16E(V E)e

V

2 21o

o

V sinh aT [1 ]

4E(V E)

2 2 a1o

o

V e[1 ]

16E(V E)

2 2 a1o

o

V e[ ]16E(V E)

0E E V Case of

oi , 2m(E V ) /

2 2o2m(E V ) /

2 2 22 1

2 2

(k )T [1 sinh (i a)]

4 k

22 1o

o

o2

o o

V[1 (i sin a) ]

4E(E V )

4E(E V )

4E(E V ) V sin a