2/22

)( xAex 1. Let the wave function ( is constant)

Problem: solve the normalization constant A

ikxex )(2. Let the wave function

Problem: (a) the probability distribution of particle’s position

(b) Is the wave function normalized?

3. In the spherical coordinate, the wave function can be expressed by

),,( r

Problem: (a) the probability of finding a particle in (r, r+dr).

(b) the probability of finding a particle in solid angle d

( 立体角 ) of the direction (, ).

Exercise ( 第二章 波函数 )

2/22

)( xAex 1. Let the wave function ( is constant)

Problem: solve the normalization constant A

Solution:

According to the property of wave function

1)()(*

dxxx

So, we can obtain

122

2

dxeA x

dxe x2Due to So

A

ikxex )(2. Let the wave function

Problem: (a) the probability distribution of particle’s position

(b) Is the wave function normalized?

Solution:

(a) The probability is

dxeedxxx ikxikx)()( *

(b) The wave function is not normalized.

3. In the spherical coordinate, the wave function can be expressed by

),,( r

Problem: (a) the probability of finding a particle in (r, r+dr).

(b) the probability of finding a particle in solid angle d

( 立体角 ) of the direction (, ).

Solution:

(a) the probability of finding a particle in the whole space is

0 0

2

0

2*

*

sin),,(),,(

),,(),,(

ddrdrrr

dVrr

The probability of finding a particle in (r, r+dr) is

0

2

0

2* sin),,(),,() ,( ddrrrdrrr

(b) the probability of finding a particle in solid angle d of the direction (, ) is

0

2* ),,(),,() ,( drrrrd

2. Translate the text contents on the page P68~69.

1. Prove the following formulas:

(1) kijkji pipL ] ,ˆ[

(2) [L, p2]=0

Exercise one （第四章 量子力学中的力学量的算符表

示）

1. Prove the following formulas:

(1) kijkji pipL ] ,ˆ[

(2) [L, p2]=0

Solution:

)(

riprL

)(

)(

)(

xy

yxipypxL

zx

xzipxpzL

yz

zyipzpyL

xyz

zxy

yzx

So

)()(] ,[ yzyyyzxyyxyx zpypppzpypLppLpL

z

zyzyzyyz

pi

ypppypypppyp

)()(] ,[ yzxxyzxxxxxx zpypppzpypLppLpL

0

Hence we can obtain

kijkji pipL ] ,ˆ[

(2) [L, p2]=0

] ,[] ,[ 2222zyx pppLpL

Consider its component x

] ,[] ,[] ,[] ,[] ,[ 2222222zxyxxxzyxxx pLpLpLpppLpL

0

] ,[] ,[] ,[] ,[

] ,[] ,[ 22

yzzyzyyz

zxzzzxyxyyyx

zxyx

ppippippippi

pLpppLpLpppL

pLpL

Using the same deducing, we can obtain the value of other two component y and z is also zero. Therefore, the following relation is true.

[L, p2]=0

consider

n)Hamiltonio is (H ?] ,[ HL

?] ,[ prL

1. Using uncertainty principle, please estimate the ground state energy of helium atom ( 氦原子 ).

2. Prove the following equations

] ,[)( )2(

2 )1(2 pLpLLpi

pipLLp

3. Set the mechanics quantity satisfy the most simple algebra ( 代数 ) equation:

A

0......ˆˆˆ)ˆ( 22

11

nnnn CACACAAf

Where C1, C2, ……, Cn are constant coefficients.

Please prove that the number of engenvalue of is n, and they are the roots of equation f(x) = 0

A

Exercise two （第四章 量子力学中的力学量的算符表

示）

prove , 0]]ˆ ,ˆ[ ,ˆ [ ,0]]ˆ ,ˆ[ ,ˆ [ if 4. BABBAA

]ˆ ,ˆ[2

1ˆˆˆˆ BABABA eee

Imply: set ,prove , then integrate. BABA eeefˆˆˆˆ

)( fBAd

df ,

1. Using uncertainty principle, please estimate the ground state energy of helium atom ( 氦原子 ).

Solution:

2e

-e -e

r1 r2

r12

The whole Hamiltonian of helium atom is

12

2

21

222

21

112)(

2

1

r

e

rrepp

mH

Set the radius which the probability finding atom is maximum is R, ( 最可几半径 )

RrRrr 2

11 ,

111

1221

According to uncertainty principle,

2222

21 / Rpp

So the energy of ground state is

R

e

mRHE

2

2

2

)2

14(

The value R must be satisfy the condition that E is minimum extremum, i.e.

0R

E

So we get

02

2

7

4

7

4a

meR

a0 is Bohr radius

The energy of ground state E is approximate

0

22

4

7

a

eE

] ,[)( )2(

2 )1(2 pLpLLpi

pipLLp

Solution:

yzzyyzzyx LppLLpLppLLp

] ,[] ,[ zyzy pLLp

Using the basic commuting relation

kijkji pipL ] ,ˆ[

(1)

2. Prove the following equations

So we can obtain

]2 pipLLp

(2)

] ,[] ,[] ,[ 222xzxyx pLpLpL

],[],[],[],[ xzxxxzxyxxxy pLpppLpLpppL

x

zyyzyzzy

pLLpi

pLLppLLpi

)(

)(

] ,[)( 2 pLpLLpi So

3. Set the mechanics quantity satisfy the most simple algebra ( 代数 ) equation:

A

0......ˆˆˆ)ˆ( 22

11

nnnn CACACAAf

Where C1, C2, ……, Cn are constant coefficients.

Please prove that the number of engenvalue of is n, and they are the roots of equation f(x) = 0

A

Solution:Set is an arbitrary wavefunction, we can get

0)ˆ( AfIf is the eigenfunction of operator A, so

aA ˆ

Where a is eigenvalue of operator A.

0)......(

)......ˆˆˆ()ˆ(2

21

1

22

11

nnnn

nnnn

CaCaCa

CACACAAf

So

0...... )( 22

11

nnnn CaCaCaaf (1)

The number of roots of equation (1) should be n, i.e, therefore the number of eigenvalue of operator A is n.

If the eigenvalue of A is a1, a2, …, an, the arbitrary eigenfunction of A will satisfy the following equation

0ˆ...ˆˆ21 naAaAaA (2)

Since an arbitrary wavefunction is always described by the linear superposition of eigenfunction of A, equation (2) is true for an arbitrary wavefunction. Therefore

0ˆ...ˆˆ21 naAaAaA

],ˆ ,ˆ[ BAC

CBABA eee 2

1ˆˆˆˆ

Solution:

4. Set , 0]C ,ˆ [ ,0] ,ˆ [ BCA ,0]ˆ ,ˆ[ BA

Prove Glauber formula

B

m

m

Cem

BBA

0

B

!] ,[]e [A,

According to the results of above example, we get

We introduce parameter , and set

BA eefˆˆ

)(

BA eBAed

df

)(

So BA eeffˆˆ

)1( ,1)0(

(2)

(3)

By means of equation (1)

Set ,BB So CC So we obtain

BB CeA, e ][ (1)

)( BAeAe BB

We get

))(()( CBAfCBAeed

df BA

dCBAf

df)(

)(

Integrate

2

2

1)()0(ln)(ln CBAff

2

2

1)(

)0()(

CBA

eff

2

2

1)(

)(

CBA

ef

Hence 2

2

1)( CBABA eee

)(2

1)(

2

BACBA eee

Set 1

We finally get

CBABA eeee 2

1

1. Set a particle is defined in infinite potential well

elsewhere

0 0)(

axxV

Problem: solve its energy eigenvalue, energy eigenfunction and momentum probability distribution.

2. A particle motions in half-infinite potential well V(r)

)(

0 0

)(

12

1

axVV

ax

axV

xV Consider bound state (V2 > E > 0) only

Problem: (1) solve the energy level

(2) If V2=1 eV, V1=1.5 eV, a = 2 nm, the mass of electron is 9.110-31 kg, calculate energy level of electron using the program of Mathematic and draw the sketch of energy level distribution.

Exercise one （第六章 薛定谔方程）

1. Set a particle is defined in infinite potential well

elsewhere

0 0)(

axxV

Problem: (a) solve its energy eigenvalue, energy eigenfunction.

(b) set the particle stays in ground state, solve the

momentum probability distribution.

Solution:

0 a

V

x

Potential along the x-axis

(1) Inside well

)()(2

-2

22

xExxm

Set 2

2 2

mE

k

)()( 22

2

xkxdx

d

The solutions of above equations are

)sin()( kxCx

C is the normalization constant, δ is the phase which must be determined.

0)( and 0)0( a

(2) Outside well

0)( xAccording to boundary condition

(1)

Insert boundary condition into equation (1), we can get

... ,3 ,2 ,1 , and ,0 nnπak

Therefore we can obtain energy eigenfunction and eigenvalue,

22

22

2 ),sin(

2)( n

maEEx

a

n

ax nn

According the normalization condition

dxkxCdVa

)(sin10

22

aC

2

(b) The distribution of momentum probability of a particle staying the ground state

The probability finding a particle in momentum range (p, p+dp) is given by

dppdppp2

)() ,(

Due to the particle staying ground state, the wavefunction is

)0( ),sin(2

)(1 axxaa

x

Using Fourier transformation, we can obtain

dxexp ipx

/

1 )(2

1)(

a ipx dxex

aap

0

/)sin(2

2

1)(

Using the method of integrate step by step, we can get

)/( ),sincos1(.2

.2

1)(

222

pkkaika

ak

a

ap

So

2

cos4

2cos

/4)( 2

22222

22222

2 pa

ap

aka

ak

ap

)(

0 0

)(

12

1

axVV

ax

axV

xV Consider bound state (V2 > E > 0) only

Problem: (1) solve the energy level

(2) If V2=1 eV, V1=1.5 eV, a = 2 nm, the mass of electron is 9.110-31 kg, calculate energy level of electron using the program of Mathematic and draw the sketch of energy level distribution.

Solution:

0 a x

E

V2

V1

2. A particle motions in half-infinite potential well V(r)

In the case of only considering the solution of bound state (V1>V2 > E > 0), and by means of Schordinger equation, we can obtain the wave function in different zone,

axAe

axxkB

xAe

xx

x

,

0 ,)sin(

0 ,

)(2

1

where

/)(2

/2

/)(2

22

11

EVm

mEk

EVm

In the case of x=0, x=a, the wavefunction and its differential are continuous, i.e.

'' )(ln/

is continuous. So we get

2211 /2cot kmVk

2222 /2)cot( kmVkak

Above two equations are equivalent to the following,

12/sin mVk

22/)sin( mVkka

Throw off , and obtain the equation of energy level

... ,2 ,1 ,2

sin2

sin2

1

1

1

n

mV

k

mV

knka

If we can obtain the root kn from above equation, the corresponding energy eigenvalue En is

m

kE n

n 2

22

In general, for every n, we can get a root kn (En ).

In symmetrical potential well, there is at least a solution of bound state, but in non-symmetrical potential well, is there a solution of bound state?

Now we consider the ground state (n=1). For a anti-trigonometric function sin-1x, the independent variable x satisfies

1x

So

/20 i.e. ,12

2

2

mVkmV

k

When the first bound state level presents (n=1),

0/2 2 mVk

121

1

1

2

1 /sin2

sin ,022

sin VVmV

k

mV

k

So the equation of energy level becomes

121

2 /sin2

/2 VVmVa

This is the condition that there exists at least one ground state.

(2) If V2=1 eV, V1=1.5 eV, a = 2 nm, the mass of electron is 9.110-31 kg, calculate energy level of electron using the program of Mathematic and draw the sketch of energy level distribution.

The equation of energy level is

... ,2 ,1 ,2

sin2

sin2

1

1

1

n

mV

k

mV

knka

The approximate solution kn of above equation can be obtained by the graphic method. The solutions is given by the points intersection of linear line y1 and curve y2,

kay 1

... ,2 ,1 ,2

sin2

sin2

1

1

12

n

mV

k

mV

kny

Insert the values given into above equation, get

kkay 91 102

... ,2 ,1 ,1000.1sin1023.1sin 91912 nkkny

Set kx 9102

The above equation becomes

xy 1

... ,2 ,1 ,sin23.1sin 112 nxxny

Using the program of Mathematic, we can obtain the points of intersection of y1 and y2.

In the case of n =1,

xy 1

xxy 112 sin23.1sin

8023.01 x

)eV( 106.1101.92

)105(101

2 1931

286822

x

m

kE n

n

(eV) 00858.0 2x

(meV) 5.58.000858.0 21 E

In the case of n =2,

xy 1

xxy 112 sin23.1sin2

When n =2, 3, …, the points of intersection of y1 and y2 do not exist. So there is only one bound energy level, i.e., ground state,

(meV) 5.51 EE

1. A particle with kinetic energy E motions in the delta potential V(x)

)()( 0 xVxV

Solve the transmission coefficient T

2. A particle motions in the double delta potential V(x)

)]()([)( 0 axaxVxV

(1) Solve the formula of energy level of bound state.

(2) This model can be used to approximately simulate the energy level of hydrogen molecule ion (H2

+). Set a = 10-5 nm, V0 = 0.1 meV, calculate the energy level of electron using Mathematic program.

Exercise two （第六章 薛定谔方程）

1. A particle with kinetic energy E motions in the delta potential V(x)

)()( 0 xVxV

Solve the transmission coefficient T

Solution:

0 x

V(x)

E

The wavefunction satisfies Schordinger equation,

)()()()(2 02

22

xExxVxdx

d

m

At x =0, (x) is continuous, but its differential is not continuous, and satisfies the following equation,

0)]0()0()0([2 0

2

Vm

x 0

/2 ,0)()( 22

2

mEkxkxdx

d

Above equation have two linear independent solutions, i.e.

ikxikx ee ,~

Set the particle moves toward x direction from left side of x = 0, so the incident wavefunction can be expressed by , and the whole wavefunction is given by

ikxe

(1)

a x Se

, xex

ikx

ikxikx

,

0Re)(

Where is reflection term, is transmission term.

According to the continuous condition of (x) at x =0,

SR 1

By means of equation (1),

SmV

RSik2

02)1(

So we obtain

20

20

20 1/ ,1/1

k

imV

k

imVR

k

imVS

ikxRe ikxSe

Therefore transmission coefficient and reflection coefficient are

,2

1/11/12

20

42

20

22

E

mV

k

VmS

,2

1/2 2

20

2

202

E

mV

E

mVR

)]()([)( 0 axaxVxV

(1) Solve the formula of energy level of bound state.

Solution:For the bound state, E < 0.

The wavefunction satisfies Schordinger equation,

)()()()(2 02

22

xExxVxdx

d

m

2. A particle motions in the double delta potential V(x)

(2) This model can be used to approximately simulate the energy level of hydrogen molecule ion (H2

+). Set a = 2 nm, V0 = 0.1 eV.nm, calculate the energy level of electron using Mathematic program.

At x =a and -a, (x) is continuous, but its differential is not continuous, and satisfies the following equation,

)(2

)(2

)0()0(2

0 ab

amV

aa

)(2

)0()0( ab

aa

Where 0

2 / mVb

Set /2mEk

When , Schordinder equation can be expressed by aax ,

0'' 2 k

(1)

(2)

Above equation have two linear independent solutions, i.e. ikxikx ee ,~

Due to V(x)=V(-x), wavefunction has certain parity. In the following, we will discuss it in case of the even parity and the odd parity respectively.

(a) In the case of even parity )()( xx

The solution of wavefunction is given by

axAe

axakxee

-a xAe

x

kx

kxkx

kx

,ch)(2

1

,

)(

By means of the continuous condition of wavefunction at x = a, and equation (1), we can get

kaka

kaka

ee

ee

kb

12

Since wavefunction possesses certain parity, the same result will be given using the continuous condition of wavefunction at x = -a, and equation (2).

(3)

Equation (3) is equivalent to the following equation

kaekb 21 This is the equation of energy level.

(b) In the case of odd parity)()( xx

The solution of wavefunction is given by

axBe

axakxee

-a xBe

x

kx

kxkx

kx

,ch)(2

1

,

)(

By means of the continuous condition of wavefunction at x = a, and equation (1) and (2), we can easily get the equation of energy level

kaekb 21

(2) This model can be used to approximately simulate the energy level of hydrogen molecule ion (H2

+). Set a = 2 nm, V0 = 0.1 eV.nm, calculate the energy level of electron using Mathematic program.

Using the parameter value given, we can get

)nm( 68.0/ 02 mVb

The equation of energy level is

kaekb 21 /2mEk

(a) Even parity

Set kx 910So

xex 4168.0

xey 41 xy 68.0

x =1.475 )(m 10475.110 199 xk

)meV( 742

22

m

kE

In the case of even parity, there is only one bound state energy level.

kaekb 21 /2mEk

(a) odd parity

kx 910

So

xex 4168.0

Set

xey 41

xy 68.0)(m 10465.110 199 xk

)meV( 732

22

m

kE

In the case of odd parity, there may be one bound state energy level.

x =1.465

When b > 2a, there is no solution of bound state.

1. Using the recursion of Hermite polynomials

)(2)(

1

nn nH

d

dH

Prove the following expressions,

)(

2

1)(

2

)(11 x

nx

n

dx

xdnn

n

)()2)(1()()12()()1(2

)(22

2

2

2

xnnxnxnndx

xdnnn

n

And according to these, prove

0 nn pp 2/2/2nnn EmpT

/0 mwhere

Exercise （第七章 谐振子）

The eigenfuction of harmonic oscillator is

)( 22

2

1

xHeN n

x

nn

!2 nN

nn

/0 m

Solution:

Hence

dx

xdHeNx

dx

d nxnn

n )(2/2 22

Using the previous results

)(

2

1)(

2

1)( 11 x

nx

nxx nnn

therefore

)(2

12

)(2

1)(

2

)(

12/

1

11

22

xHeNn

n

xn

xn

dx

xd

nx

n

nnn

)(

2

1)(

2

)(11 x

nx

n

dx

xdnn

n

dx

xdn

dx

xdn

dx

xd nnn )(

2

1)(

2

)( 112

2

So

(1)

Using equation (1), we get

)()2)(1()()12()()1(2

)(22

2

2

2

xnnxnxnndx

xdnnn

n

dx

xdxipp n

nnn

)()(

)(2

1)(

2)( 11 x

nx

nx nnn

Since n(x) is normalized and orthonormality,

0)()()()( 11 xxxx nnnn

0pSo

2

222

22/

dx

d

mmpT n

nnn

Using the above result and the property of normalization and orthonormality of n(x), we get

2/)2/1(2

1)]12(

2[

2

22

nEnnm

T

2. A particle is in the ground state of one-dimension harmonic oscillating potential

211 2

1xkV

Now, k1 abruptly changes to k2, i.e, , and immediately measure the energy of a particle, k1 and k2 are positive real number.

222 2

1xkV

Solve the probability finding that a particle is the ground state of new potential V2.

Solution:The change of wavefunction (x) with t is determined by Schordinger equation,

),(),(

2

),( 2

txVt

tx

mt

txi

When potential V abruptly changes and the change value is finite, how

ever (x) does not change.

Set 0(x) and 0(x) denote the ground state wavefunctions of potential

V1 and V2, respectively. When potential V1 abruptly changes to V2, how

ever wavefunction is 0(x) . The probability finding particle in the state

0(x) is 2

00 )()( xx

The ground state wavefunction of one-dimension harmonic oscillator can be expressed by

)2

1exp( 22

40 x /1 m

)2

1exp( 22

40 x

/2 m

Hence

2/1

22

)(2

1

00

2)()(

22

dxexx

4/1

2

1

2

1

k

k

2/121

4/121

2222

2

00 )/(1

)/(2

/1

/22)()(

kk

kkxx

0 T V E

1. Hydrogen atom stays in ground state, solve the probability finding electron outside the range which classical mechanics is not allowed, i.e,

2. Set potential is

10 ,)(2

22

r

ae

r

erV

a is Bohr radius.

Solve the energy level.

Exercise （第九章 带电粒子在磁场中的运动）

0 T V E

1. Hydrogen atom stays in ground state, solve the probability finding electron outside the range which classical mechanics is not allowed, i.e,

Solution:

When

02

22

2)(

an

eE

r

erV n

nn

We get

022 anrn

When r > rn, this is the range which classical mechanics is not allowed.

For the ground state, the the probability finding electron outside the range which classical mechanics is not allowed is

ddrdrr

sin21000

2

0

*100

1

4

2

22

2

2 0

2

0

2

30

134

sin1

0

0

edyye

ddrdreπa

y

a

a

r

2. Set potential is

10 ,)(2

022

r

ae

r

erV

a0 is Bohr radius.

Solve the energy level.

Solution:

0)(]1

)1()(2

[)(

2

2

22

2

rR

rll

r

eE

m

r

rRl

l

The differential equation about the radial function of hydrogen atom

,2

12

0

2

na

eEn , ..., nlnn rr 21 ,0 ,1

20

222

r

ae

r

e

r

e Now

The differential equation about the radial function becomes

0)(]1

)1()(2

[)(

220

22

22

2

rR

rll

r

ae

r

eE

m

r

rRl

l

Set 222

02

2

1)1(

1)1(

2

rll

rll

r

aem

2

2

0 mea

Insert into above equation, we get

)1(2)1( llll

(1)

Equation (1) becomes

0)(]1

)1()(2

[)(

2

2

22

2

rR

rll

r

eE

m

r

rRl

l

The energy level is

,)(2

12

0

2

na

eEn

, ..., nlnn rr 21 ,0 ,1

2/1

2)12(

81

2

1

2

1

lll

, ..., l 21 ,0