2 / 2 2 ) ( x Ae x 1. Let the wave f unction (is constant) Problem: solve the normalization constant A ikx e x ) ( 2. Let the wave f unction Problem: (a) the probability distribution of particle’s position (b) Is the wave function normalized? 3. In the spherical coordinate, the wave functi on can be expressed by ) , , ( r Problem: (a) the probability of finding a pa rticle in (r, r+dr). (b) the probability of finding a partic le in solid angle d(立 立 体 ) of the direction (, ). Exercise ( 第第第 第第第 )
Exercise ( 第二章 波函数 ). 1. Let the wave function. ( is constant). Problem: solve the normalization constant A. 2. Let the wave function. Problem: (a) the probability distribution of particle’s position. (b) Is the wave function normalized?. - PowerPoint PPT Presentation
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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.
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