苏汝铿量子力学课后习题及答案
合集下载
相关主题
- 1、下载文档前请自行甄别文档内容的完整性，平台不提供额外的编辑、内容补充、找答案等附加服务。
- 2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
- 3、如文档侵犯您的权益，请联系客服反馈,我们会尽快为您处理(人工客服工作时间：9:00-18:30)。
2ikA ˜ 2ik−V ˜A V ˜ 2ik−V
(13)
(14)
(15)
= = 3
ik A, ik−mV /¯ h2 2 mV /¯ h A. ik−mV /¯ h2
(16)
So the transmission ratio is
ALL RIGHTS RESERVED, BY SHAO-YU YIN, YI LI, JIA ZHOU NOT FOR DISTRIBUTION
2.2. Discuss the reflection and transmission ratio of a particle with mass m and positive energy E on a Dirac δ barrier, U (x) = V δ (x) (V > 0). How about if it is a δ well (V < 0)? Solution: √ h ¯ 2 d2 − ψ + V δ (x)ψ = Eψ, 2m dx2 d2 ˜ δ (x)ψ = 0. ψ + k2ψ − V 2 dx Integrate in the range (− , ), where is an infinitesimal value, ˜ ψ (0) = 0. ψdx − V
1
07-12-27, 23:59
I.
CHAPTER 1
ALL RIGHTS RESERVED, BY SHAO-YU YIN, YI LI, JIA ZHOU NOT FOR DISTRIBUTION
1.1. Using light or electrons as probe, if we want to observe an erythrocyte (red blood cell), how is the order of amplitude of the minimum energy needed respectively? What if a atom is observed? Solution: The probe must has the wave length less than the scale of the observed object. erythrocyte: 7 ∼ 8µm; atom: ∼ 1˚ A= 0.1nm. wavelength of light: λphoton = wavelength of electron (nonrelativistic): λelectron = wavelength of electron (relativistic): λelectron = For erythrocyte, Ephoton = hc 1.24nm · keV = 7µm 7µm 0.177eV; Eelectron = h2 2me (7µm)2 3.06 ∗ 10−8 eV. (4) h ¯ = p
1 h ¯ω ψ ∗ (x, t) mω 2 x2 ψ (x, t)dx = , 2 4 2
C (p, t) = √
1 2πh ¯
h ψ (x, t)e−ipx/¯ dx =
p2 1 − 2mω − iωt h ¯ 2 , e 1 / 4 (mωπh ¯)
(7) (8)
ALL RIGHTS RESERVED, BY SHAO-YU YIN, YI LI, JIA ZHOU NOT FOR DISTRIBUTION
∞
(26)
(iii). The probability distribution function of momentum is
∞ i 1 1 p·r 2 ¯ C (p) = √ e−r/a0 e− h r dr sin θdθdφ 3 3 ( 2πh ¯) 0 πa0 ∞ i 1 1 −r/a0 − h pr cos θ 2 ¯ √ e r dr sin θdθdφ = e 3 ( 2π h 3 ¯) 0 πa0 ∞ i 2π 1 h ¯ pr cos θ π ¯ √ = e−r/a0 (e− h |0 )rdr 3 ( 2π h 3 ip ¯) 0 πa0 1 2¯ h ∞ −r/a0 2π pr √ e rdr = sin 3 p h ¯ ( 2 π h ¯ ) πa3 0 0 ∞ 1 2¯ h 2π 1 (− + ip )r a0 h ¯ √ = Im [ e rdr] 3 ( 2π h 3 ¯) p πa0 0 2π 1 2¯ h −1 ip −2 ∞ z √ = e zdz ] Im [( + ) 3 p a h ¯ ( 2 π h ¯ ) 0 πa3 0 0
πa0
4
(ii) The expectation value of potential −e2 /r. (iii) The probability distribution function of momentum.
ALL RIGHTS RESERVED, BY SHAO-YU YIN, YI LI, JIA ZHOU NOT FOR DISTRIBUTION
2 2 Ek /c
hc ; E
(1)
h h =√ ; p 2me E
(பைடு நூலகம்)
h ¯ , + 2mEk
(3)
For atom, Ephoton hc 1.24nm · keV = = 0.1nm 0.1nm 12.4keV; Eelectron h2 = 2me (0.1nm)2 150eV. (5)
(nonrelativistic relation is good enough.)
Exercises for the Course of Quantum Mechanics, 2007
ALL RIGHTS RESERVED, BY SHAO-YU YIN, YI LI, JIA ZHOU NOT FOR DISTRIBUTION
Prof.
Ru-Keng Su
Shaoyu Yin Jia Zhou & Yi Li Department of Physics, Fudan University, Shanghai 200433, China
−
(10)
take k =
˜ = 2mV /h 2mE/h ¯, V ¯ 2 , equation becomes (11)
ψ ( ) − ψ (− ) + k 2
(12)
Taking → 0+ , notice that ψ is finite at x = 0, we get ˜ ψ (0). ψ (0+ ) − ψ (0− ) = V Choose the following ansatz Aeikx + Be−ikx , x<0; ψ (x) = A eikx , x>0, the connection condition of ψ at x = 0 yields A+B =A, ˜A , ikA − ikA + ikB = V the results are A = B=
(18)
Notice that the above derivation is independent on the sign of V , so if the particle is reflected by a δ well instead of a δ barrier, the above derivation remains unchanged except for the substitution of V by −V , but the final result for the reflection and transmission ratio are the same since they only depend on V 2 . 2.3. What’s the energy of ground state of a neutron in the gravitation field? Solution: eigenfunction of energy in stationary state: h ¯ 2 d2 − ψ + mn gzψ = Eψ, 2mn dz 2 with boundary condition ψ (z = 0) = 0. Introduce dimensionless parameters ξ = (z − λ)/l, where l3 = The equation becomes h ¯2 , 2m2 ng (20) (19)
II.
CHAPTER 2
2.1. (QM book of Su, Ex.2.7.) A 1-d harmonic oscillator is in the state of ψ (x, t) =
α π 1/2
e−
α2 x2 − iωt 2 2
. (i). The expectation value of potential energy? (ii) The probability
A D= A the reflection ratio is R= B A
2
2
=
k2 1 , 4 = mV 2 2 2 2 k + m V /h ¯ 1 + 2¯ h2 E
(17)
=
m2 V 2 /h ¯4 1 = 1 − D. 4 = 2 2 2 h2 E k + m V /h ¯ 1 + 2¯ 2 mV
λ=
E . mn g
(21)
d2 ψ − ξψ = 0, dξ 2
(22)
with a boundary condition ψ (ξ = − λ ) = 0. This is a 1/3 order Bessel function, whose l solution is the Airy function. Corresponding to the ground state, the first zero appears at ξ1 = −2.338, so Eground = mn gλ1 = 2.338mn g h ¯2 2m2 ng
T =
h ¯ω p2 C (p, t) C (p, t)dp = =− 2m 4
∗
h ¯ 2 d2 ψ (x, t) ψ (x, t)dx. 2m dx2
∗
Or using the Virial theorem (QM book of Su, Chapter 3.8, P117 ), T = 1 dU 1 h ¯ω x = U = E = . 2 dx 2 4 (9)
distribution of the momentum. (iii). The expectation value of kinetic energy? Solution: Compared with the standard harmonic oscillator eigenfunction, it is easy to find that this is the ground state and α = U = mω/h ¯, (6)
Solution: (i). Check that the wave function ψ = √1 3 e−r/a0 is normalized:
πa0 ∞
4π
0
ψ † ψr2 dr =
0
∞
1 = Γ(3) = 1. 2 So
∞
4 −2r/a0 2 4 a0 r dr = 3 ( )3 e 3 a0 a0 2
∞ 0
x2 e−x dx (24)
r = 4π
0
ψ † rψr2 dr =
4 a0 = 3 ( )4 a0 2 (ii). −e2 /r = −e2 4π
4 −2r/a0 3 e r dr a3 0 0 ∞ a0 3a0 x3 e−x dx = Γ(4) = . 4 2 0
∞
(25)
∞ 1 4 −2r/a0 ψ † ψr2 dr = −e2 e rdr r a3 0 0 0 ∞ −e2 e2 4 a0 xe−x dx = Γ(2) = − . = −e2 3 ( )2 a0 2 a0 a0 0
1/3
1.41 ∗ 10−12 eV.
(23)
2.4. (QM book of Su, Ex.2.14.) The state of electron in Hydrogen atom is ψ = √1 3 e−r/a0 , where a0 is the Bohr radius. Try to find: (i) The expectation value of r.
(13)
(14)
(15)
= = 3
ik A, ik−mV /¯ h2 2 mV /¯ h A. ik−mV /¯ h2
(16)
So the transmission ratio is
ALL RIGHTS RESERVED, BY SHAO-YU YIN, YI LI, JIA ZHOU NOT FOR DISTRIBUTION
2.2. Discuss the reflection and transmission ratio of a particle with mass m and positive energy E on a Dirac δ barrier, U (x) = V δ (x) (V > 0). How about if it is a δ well (V < 0)? Solution: √ h ¯ 2 d2 − ψ + V δ (x)ψ = Eψ, 2m dx2 d2 ˜ δ (x)ψ = 0. ψ + k2ψ − V 2 dx Integrate in the range (− , ), where is an infinitesimal value, ˜ ψ (0) = 0. ψdx − V
1
07-12-27, 23:59
I.
CHAPTER 1
ALL RIGHTS RESERVED, BY SHAO-YU YIN, YI LI, JIA ZHOU NOT FOR DISTRIBUTION
1.1. Using light or electrons as probe, if we want to observe an erythrocyte (red blood cell), how is the order of amplitude of the minimum energy needed respectively? What if a atom is observed? Solution: The probe must has the wave length less than the scale of the observed object. erythrocyte: 7 ∼ 8µm; atom: ∼ 1˚ A= 0.1nm. wavelength of light: λphoton = wavelength of electron (nonrelativistic): λelectron = wavelength of electron (relativistic): λelectron = For erythrocyte, Ephoton = hc 1.24nm · keV = 7µm 7µm 0.177eV; Eelectron = h2 2me (7µm)2 3.06 ∗ 10−8 eV. (4) h ¯ = p
1 h ¯ω ψ ∗ (x, t) mω 2 x2 ψ (x, t)dx = , 2 4 2
C (p, t) = √
1 2πh ¯
h ψ (x, t)e−ipx/¯ dx =
p2 1 − 2mω − iωt h ¯ 2 , e 1 / 4 (mωπh ¯)
(7) (8)
ALL RIGHTS RESERVED, BY SHAO-YU YIN, YI LI, JIA ZHOU NOT FOR DISTRIBUTION
∞
(26)
(iii). The probability distribution function of momentum is
∞ i 1 1 p·r 2 ¯ C (p) = √ e−r/a0 e− h r dr sin θdθdφ 3 3 ( 2πh ¯) 0 πa0 ∞ i 1 1 −r/a0 − h pr cos θ 2 ¯ √ e r dr sin θdθdφ = e 3 ( 2π h 3 ¯) 0 πa0 ∞ i 2π 1 h ¯ pr cos θ π ¯ √ = e−r/a0 (e− h |0 )rdr 3 ( 2π h 3 ip ¯) 0 πa0 1 2¯ h ∞ −r/a0 2π pr √ e rdr = sin 3 p h ¯ ( 2 π h ¯ ) πa3 0 0 ∞ 1 2¯ h 2π 1 (− + ip )r a0 h ¯ √ = Im [ e rdr] 3 ( 2π h 3 ¯) p πa0 0 2π 1 2¯ h −1 ip −2 ∞ z √ = e zdz ] Im [( + ) 3 p a h ¯ ( 2 π h ¯ ) 0 πa3 0 0
πa0
4
(ii) The expectation value of potential −e2 /r. (iii) The probability distribution function of momentum.
ALL RIGHTS RESERVED, BY SHAO-YU YIN, YI LI, JIA ZHOU NOT FOR DISTRIBUTION
2 2 Ek /c
hc ; E
(1)
h h =√ ; p 2me E
(பைடு நூலகம்)
h ¯ , + 2mEk
(3)
For atom, Ephoton hc 1.24nm · keV = = 0.1nm 0.1nm 12.4keV; Eelectron h2 = 2me (0.1nm)2 150eV. (5)
(nonrelativistic relation is good enough.)
Exercises for the Course of Quantum Mechanics, 2007
ALL RIGHTS RESERVED, BY SHAO-YU YIN, YI LI, JIA ZHOU NOT FOR DISTRIBUTION
Prof.
Ru-Keng Su
Shaoyu Yin Jia Zhou & Yi Li Department of Physics, Fudan University, Shanghai 200433, China
−
(10)
take k =
˜ = 2mV /h 2mE/h ¯, V ¯ 2 , equation becomes (11)
ψ ( ) − ψ (− ) + k 2
(12)
Taking → 0+ , notice that ψ is finite at x = 0, we get ˜ ψ (0). ψ (0+ ) − ψ (0− ) = V Choose the following ansatz Aeikx + Be−ikx , x<0; ψ (x) = A eikx , x>0, the connection condition of ψ at x = 0 yields A+B =A, ˜A , ikA − ikA + ikB = V the results are A = B=
(18)
Notice that the above derivation is independent on the sign of V , so if the particle is reflected by a δ well instead of a δ barrier, the above derivation remains unchanged except for the substitution of V by −V , but the final result for the reflection and transmission ratio are the same since they only depend on V 2 . 2.3. What’s the energy of ground state of a neutron in the gravitation field? Solution: eigenfunction of energy in stationary state: h ¯ 2 d2 − ψ + mn gzψ = Eψ, 2mn dz 2 with boundary condition ψ (z = 0) = 0. Introduce dimensionless parameters ξ = (z − λ)/l, where l3 = The equation becomes h ¯2 , 2m2 ng (20) (19)
II.
CHAPTER 2
2.1. (QM book of Su, Ex.2.7.) A 1-d harmonic oscillator is in the state of ψ (x, t) =
α π 1/2
e−
α2 x2 − iωt 2 2
. (i). The expectation value of potential energy? (ii) The probability
A D= A the reflection ratio is R= B A
2
2
=
k2 1 , 4 = mV 2 2 2 2 k + m V /h ¯ 1 + 2¯ h2 E
(17)
=
m2 V 2 /h ¯4 1 = 1 − D. 4 = 2 2 2 h2 E k + m V /h ¯ 1 + 2¯ 2 mV
λ=
E . mn g
(21)
d2 ψ − ξψ = 0, dξ 2
(22)
with a boundary condition ψ (ξ = − λ ) = 0. This is a 1/3 order Bessel function, whose l solution is the Airy function. Corresponding to the ground state, the first zero appears at ξ1 = −2.338, so Eground = mn gλ1 = 2.338mn g h ¯2 2m2 ng
T =
h ¯ω p2 C (p, t) C (p, t)dp = =− 2m 4
∗
h ¯ 2 d2 ψ (x, t) ψ (x, t)dx. 2m dx2
∗
Or using the Virial theorem (QM book of Su, Chapter 3.8, P117 ), T = 1 dU 1 h ¯ω x = U = E = . 2 dx 2 4 (9)
distribution of the momentum. (iii). The expectation value of kinetic energy? Solution: Compared with the standard harmonic oscillator eigenfunction, it is easy to find that this is the ground state and α = U = mω/h ¯, (6)
Solution: (i). Check that the wave function ψ = √1 3 e−r/a0 is normalized:
πa0 ∞
4π
0
ψ † ψr2 dr =
0
∞
1 = Γ(3) = 1. 2 So
∞
4 −2r/a0 2 4 a0 r dr = 3 ( )3 e 3 a0 a0 2
∞ 0
x2 e−x dx (24)
r = 4π
0
ψ † rψr2 dr =
4 a0 = 3 ( )4 a0 2 (ii). −e2 /r = −e2 4π
4 −2r/a0 3 e r dr a3 0 0 ∞ a0 3a0 x3 e−x dx = Γ(4) = . 4 2 0
∞
(25)
∞ 1 4 −2r/a0 ψ † ψr2 dr = −e2 e rdr r a3 0 0 0 ∞ −e2 e2 4 a0 xe−x dx = Γ(2) = − . = −e2 3 ( )2 a0 2 a0 a0 0
1/3
1.41 ∗ 10−12 eV.
(23)
2.4. (QM book of Su, Ex.2.14.) The state of electron in Hydrogen atom is ψ = √1 3 e−r/a0 , where a0 is the Bohr radius. Try to find: (i) The expectation value of r.