Baixe Resolução Griffiths-Introduction to Quantum Mechanics 2E solution e outras Provas em PDF para Cultura, somente na Docsity!
Contents
- Preface
- 1 The Wave Function
- 2 Time-Independent Schrödinger Equation
- 3 Formalism
- 4 Quantum Mechanics in Three Dimensions
- 5 Identical Particles
- 6 Time-Independent Perturbation Theory
- 7 The Variational Principle
- 8 The WKB Approximation
- 9 Time-Dependent Perturbation Theory
- 10 The Adiabatic Approximation
- 11 Scattering
- 12 Afterword
- Appendix Linear Algebra
- 2 nd Edition – 1 st Edition Problem Correlation Grid
Preface
These are my own solutions to the problems in Introduction to Quantum Mechanics, 2nd ed. I have made every effort to insure that they are clear and correct, but errors are bound to occur, and for this I apologize in advance. I would like to thank the many people who pointed out mistakes in the solution manual for the first edition, and encourage anyone who finds defects in this one to alert me (griffith@reed.edu). I’ll maintain a list of errata on my web page (http://academic.reed.edu/physics/faculty/griffiths.html), and incorporate corrections in the manual itself from time to time. I also thank my students at Reed and at Smith for many useful suggestions, and above all Neelaksh Sadhoo, who did most of the typesetting.
At the end of the manual there is a grid that correlates the problem numbers in the second edition with those in the first edition.
David Griffiths
© c2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
4 CHAPTER 1. THE WAVE FUNCTION
Problem 1.
(a)
〈x^2 〉 =
∫ (^) h
0
x^2
hx
dx =
h
5 x
5 / 2
h 0
h^2
σ^2 = 〈x^2 〉 − 〈x〉^2 = h^2 5
h 3
h^2 ⇒ σ = 2 h 3
= 0. 2981 h.
(b)
P = 1 −
∫ (^) x+
x−
hx
dx = 1 − 1 2
h
x)
x+
x−
= 1 − √^1
h
x+ − √x−
x+ ≡ 〈x〉 + σ = 0. 3333 h + 0. 2981 h = 0. 6315 h; x− ≡ 〈x〉 − σ = 0. 3333 h − 0. 2981 h = 0. 0352 h.
P = 1 −
Problem 1.
(a)
1 =
−∞
Ae−λ(x−a) 2 dx. Let u ≡ x − a, du = dx, u : −∞ → ∞.
1 = A
−∞
e−λu 2 du = A
π λ
⇒ A =
λ π
(b)
〈x〉 = A
−∞
xe−λ(x−a) 2 dx = A
−∞
(u + a)e−λu 2 du
= A
[∫ ∞
−∞
ue−λu 2 du + a
−∞
e−λu 2 du
]
= A
0 + a
π λ
= a.
〈x^2 〉 = A
−∞
x^2 e−λ(x−a) 2 dx
= A
−∞
u^2 e−λu 2 du + 2a
−∞
ue−λu 2 du + a^2
−∞
e−λu 2 du
= A
[
2 λ
π λ
π λ
]
= a^2 +
2 λ
σ^2 = 〈x^2 〉 − 〈x〉^2 = a^2 +
2 λ − a^2 =
2 λ ; σ =
2 λ
© c2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
CHAPTER 1. THE WAVE FUNCTION 5
(c)
A
a x
ρ(x)
Problem 1.
(a)
|A|^2
a^2
∫ (^) a
0
x^2 dx +
|A|^2
(b − a)^2
∫ (^) b
a
(b − x)^2 dx = |A|^2
a^2
x^3 3
a
0
(b − a)^2
(b − x)^3 3
b
a
= |A|^2
[
a 3
b − a 3
]
= |A|^2
b 3
⇒ A =
b
(b)
a x
A
b
Ψ
(c) At x = a. (d)
P =
∫ (^) a
0
|Ψ|^2 dx =
|A|^2
a^2
∫ (^) a
0
x^2 dx = |A|^2 a 3
a b
P = 1 if b = a, P = 1/2 if b = 2a.
(e)
〈x〉 =
x|Ψ|^2 dx = |A|^2
a^2
∫ (^) a
0
x^3 dx + 1 (b − a)^2
∫ (^) b
a
x(b − x)^2 dx
=^3
b
a^2
x^4 4
a
0
(b − a)^2
b^2 x
2 2 − 2 b x
3 3
4 4
b
a
4 b(b − a)^2
[
a^2 (b − a)^2 + 2b^4 − 8 b^4 /3 + b^4 − 2 a^2 b^2 + 8a^3 b/ 3 − a^4
]
4 b(b − a)^2
b^4 3 − a^2 b^2 +
a^3 b
4(b − a)^2 (b^3 − 3 a^2 b + 2a^3 ) = 2 a + b 4
© c2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
CHAPTER 1. THE WAVE FUNCTION 7
Problem 1.
From Eq. 1.33, d dt〈p 〉= −iℏ
∂t
Ψ∗^ ∂ ∂xΨ
dx. But, noting that (^) ∂x∂t∂^2 Ψ = (^) ∂t∂x∂^2 Ψ and using Eqs. 1.23-1.24:
∂ ∂t
Ψ∗^
∂x
∂t
∂x
+ Ψ∗^
∂x
∂t
[
iℏ 2 m
∂^2 Ψ∗
∂x^2
i ℏ
V Ψ∗
]
∂x
+ Ψ∗^
∂x
[
iℏ 2 m
∂^2 Ψ
∂x^2
i ℏ
V Ψ
]
= iℏ 2 m
[
Ψ∗^ ∂
∂x^3
∂x^2
∂x
]
[
V Ψ∗^ ∂Ψ
∂x
− Ψ∗^ ∂
∂x
(V Ψ)
]
The first term integrates to zero, using integration by parts twice, and the second term can be simplified to V Ψ∗^ ∂ ∂xΨ − Ψ∗V ∂ ∂xΨ − Ψ∗^ ∂V∂x Ψ = −|Ψ|^2 ∂V∂x. So
d〈p〉 dt = −iℏ
i ℏ
−|Ψ|^2
∂V
∂x dx = 〈−
∂V
∂x
〉. QED
Problem 1.
Suppose Ψ satisfies the Schr¨odinger equation without V 0 : iℏ ∂ ∂tΨ = − ℏ 2 2 m
∂^2 Ψ ∂x^2 +^ V^ Ψ. We want to find the solution Ψ 0 with V 0 : iℏ ∂ ∂tΨ^0 = − ℏ 2 2 m ∂^2 Ψ 0 ∂x^2 + (V^ +^ V^0 )Ψ^0. Claim: Ψ 0 = Ψe−iV^0 t/ℏ. Proof: iℏ ∂ ∂tΨ^0 = iℏ ∂ ∂tΨ e−iV^0 t/ℏ^ + iℏΨ
− iV ℏ^0
e−iV^0 t/ℏ^ =
[
2 2 m ∂^2 Ψ ∂x^2 +^ V^ Ψ
]
e−iV^0 t/ℏ^ + V 0 Ψe−iV^0 t/ℏ = − 2 ℏm^2 ∂ (^2) Ψ 0 ∂x^2 + (V^ +^ V^0 )Ψ^0.^ QED This has no effect on the expectation value of a dynamical variable, since the extra phase factor, being inde- pendent of x, cancels out in Eq. 1.36.
Problem 1.
(a)
1 = 2|A|^2
0
e−^2 amx (^2) /ℏ dx = 2|A|^2
√ (^) π (2am/ℏ)
= |A|^2
πℏ 2 am
; A =
2 am πℏ
(b)
∂Ψ ∂t = −iaΨ; ∂Ψ ∂x = − 2 amx ℏ
∂x^2 = − 2 am ℏ
Ψ + x ∂Ψ ∂x
= − 2 am ℏ
1 − 2 amx
2 ℏ
Plug these into the Schr¨odinger equation, iℏ ∂ ∂tΨ = − ℏ 2 2 m
∂^2 Ψ ∂x^2 +^ V^ Ψ:
V Ψ = iℏ(−ia)Ψ +
ℏ^2
2 m
2 am ℏ
2 amx^2 ℏ
[
ℏa − ℏa
2 amx^2 ℏ
)]
Ψ = 2a^2 mx^2 Ψ, so V (x) = 2ma^2 x^2.
© c2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
8 CHAPTER 1. THE WAVE FUNCTION
(c)
〈x〉 =
−∞
x|Ψ|^2 dx = 0. [Odd integrand.]
〈x^2 〉 = 2|A|^2
0
x^2 e−^2 amx (^2) /ℏ dx = 2|A|^2
22 (2am/ℏ)
πℏ 2 am
4 am
〈p〉 = m d〈x〉 dt
〈p^2 〉 =
i
∂x
Ψdx = −ℏ^2
Ψ∗^ ∂
∂x^2 dx
= −ℏ^2
[
2 am ℏ
2 amx^2 ℏ
]
dx = 2amℏ
|Ψ|^2 dx − 2 am ℏ
x^2 |Ψ|^2 dx
= 2amℏ
2 am ℏ 〈x
= 2amℏ
2 am ℏ
4 am
= 2amℏ
= amℏ.
(d)
σ^2 x = 〈x^2 〉 − 〈x〉^2 =
4 am =⇒ σx =
4 am ; σ p^2 = 〈p^2 〉 − 〈p〉^2 = amℏ =⇒ σp =
amℏ.
σxσp =
ℏ 4 am
amℏ = ℏ 2. This is (just barely) consistent with the uncertainty principle.
Problem 1.
From Math Tables: π = 3. 141592653589793238462643 · · ·
(a) P P^ (0) = 0(5) = 3^ / 25 PP^ (1) = 2(6) = 3//^2525 PP^ (2) = 3(7) = 1//^2525 PP^ (3) = 5(8) = 2//^2525 PP^ (4) = 3(9) = 3//^2525
In general, P (j) = N N^ (j ).
(b) Most probable: 3. Median: 13 are ≤ 4, 12 are ≥ 5, so median is 4. Average: 〈j〉 = 251 [0 · 0 + 1 · 2 + 2 · 3 + 3 · 5 + 4 · 3 + 5 · 3 + 6 · 3 + 7 · 1 + 8 · 2 + 9 · 3] = 251 [0 + 2 + 6 + 15 + 12 + 15 + 18 + 7 + 16 + 27] = 11825 = 4.72. (c) 〈j^2 〉 = 251 [0 + 1^2 · 2 + 2^2 · 3 + 3^2 · 5 + 4^2 · 3 + 5^2 · 3 + 6^2 · 3 + 7^2 · 1 + 8^2 · 2 + 9^2 · 3] = 251 [0 + 2 + 12 + 45 + 48 + 75 + 108 + 49 + 128 + 243] = 71025 = 28.4. σ^2 = 〈j^2 〉 − 〈j〉^2 = 28. 4 − 4. 722 = 28. 4 − 22 .2784 = 6.1216; σ =
© c2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
10 CHAPTER 1. THE WAVE FUNCTION
ρ(x)
-2r -r r^ 2r x
∴ ρ(x) =
π√r^2 −x^2 ,^ if^ −^ r < x < r, 0 , otherwise. [Note: We want the magnitude of dx here.]
Total:
∫ (^) r −r 1 π√r^2 −x^2 dx^ =^
2 π
∫ (^) r 0 √^1 r^2 −x^2 dx^ =^
2 π sin
− (^1) x r
∣∣r 0 =^
2 π sin
π ·^ π 2 = 1. (b)
〈x〉 =
π
∫ (^) r
−r
x
r^2 − x^2
dx = 0 [odd integrand, even interval].
〈x^2 〉 =
π
∫ (^) r
0
√^ x^2 r^2 − x^2
dx =
π
[
x 2
r^2 − x^2 + r^2 2 sin
− 1 ( (^) x r
)]∣∣
r
0
π
r^2 2 sin
− (^1) (1) = r^2
σ^2 = 〈x^2 〉 − 〈x〉^2 = r^2 /2 =⇒ σ = r/
To get 〈x〉 and 〈x^2 〉 from Problem 1.11(c), use x = r cos θ, so 〈x〉 = r〈cos θ〉 = 0, 〈x^2 〉 = r^2 〈cos^2 θ〉 = r^2 /2.
Problem 1.
Suppose the eye end lands a distance y up from a line (0 ≤ y < l), and let x be the projection along that same direction (−l ≤ x < l). The needle crosses the line above if y + x ≥ l (i.e. x ≥ l − y), and it crosses the line below if y + x < 0 (i.e. x < −y). So for a given value of y, the probability of crossing (using Problem 1.12) is
P (y) =
∫ (^) −y
−l
ρ(x)dx +
∫ (^) l
l−y
ρ(x)dx =
π
{∫ (^) −y
−l
l^2 − x^2
dx +
∫ (^) l
l−y
l^2 − x^2
dx
=^1
π
sin−^1
( (^) x l
−y −l
( (^) x l
l l−y
=^1
π
[
− sin−^1 (y/l) + 2 sin−^1 (1) − sin−^1 (1 − y/l)
]
sin−^1 (y/l) π −^
sin−^1 (1 − y/l) π. Now, all values of y are equally likely, so ρ(y) = 1/l, and hence the probability of crossing is
P =
πl
∫ (^) l
0
[
π − sin−^1
( (^) y l
− sin−^1
l − y l
)]
dy =
πl
∫ (^) l
0
[
π − 2 sin−^1 (y/l)
]
dy
πl
[
πl − 2
y sin−^1 (y/l) + l
1 − (y/l)^2
l 0
]
πl [l sin−^1 (1) − l] = 1 − 1 +
π
π
© c2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
CHAPTER 1. THE WAVE FUNCTION 11
Problem 1.
(a) Pab(t) =
∫ (^) b a |Ψ(x, t) (^2) dx, so dPab dt =^
∫ (^) b a ∂ ∂t |Ψ| (^2) dx. But (Eq. 1.25):
∂|Ψ|^2 ∂t
∂x
[
iℏ 2 m
Ψ∗^
∂x
∂x
)]
∂t J(x, t).
dPab dt
∫ (^) b
a
∂x J(x, t)dx = − [J(x, t)]|ba = J(a, t) − J(b, t). QED
Probability is dimensionless, so J has the dimensions 1/time, and units seconds−^1.
(b) Here Ψ(x, t) = f (x)e−iat, where f (x) ≡ Ae−amx (^2) /ℏ , so Ψ ∂Ψ ∗ ∂x =^ f e −iat df dx e iat (^) = f df dx , and Ψ∗^ ∂ ∂xΨ = f (^) dxdf too, so J(x, t) = 0.
Problem 1.
(a) Eq. 1.24 now reads ∂Ψ ∗ ∂t =^ −^ iℏ 2 m ∂^2 Ψ∗ ∂x^2 +^ i ℏ V^ ∗Ψ∗, and Eq. 1.25 picks up an extra term:
∂ ∂t |Ψ|
(^2) = · · · + i ℏ |Ψ|
(^2) (V ∗ (^) − V ) = · · · + i ℏ |Ψ|
(^2) (V 0 + iΓ − V 0 + iΓ) = · · · − 2Γ ℏ |Ψ|
and Eq. 1.27 becomes dPdt = − 2Γ ℏ
(^2) dx = − 2Γ ℏ P^.^ QED (b)
dP P =^ −^
ℏ dt^ =⇒^ ln^ P^ =^ −^
ℏ t^ + constant =⇒^ P^ (t) =^ P^ (0)e
−2Γt/ℏ, so τ = ℏ 2Γ.
Problem 1.
Use Eqs. [1.23] and [1.24], and integration by parts:
d dt
−∞
Ψ∗ 1 Ψ 2 dx =
−∞
∂t (Ψ∗ 1 Ψ 2 ) dx =
−∞
∂t
Ψ 2 + Ψ∗ 1 ∂Ψ^2
∂t
dx
−∞
[(
−iℏ 2 m
∂^2 Ψ∗ 1
∂x^2
i ℏ
V Ψ∗ 1
iℏ 2 m
∂^2 Ψ 2
∂x^2
i ℏ
V Ψ 2
)]
dx
iℏ 2 m
−∞
∂^2 Ψ∗ 1
∂x^2
∂^2 Ψ 2
∂x^2
dx
= − iℏ 2 m
[
∂x
∞
−∞
−∞
∂x
∂x dx − Ψ∗ 1 ∂Ψ^2 ∂x
∞
−∞
−∞
∂x
∂x dx
]
= 0. QED
© c2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
CHAPTER 1. THE WAVE FUNCTION 13
(h)
σxσp = √a 7
a
Problem 1.
h √ 3 mkB T
d ⇒ T < h^2 3 mkB d^2
(a) Electrons (m = 9. 1 × 10 −^31 kg):
T < (6.^6 ×^10
3(9. 1 × 10 −^31 )(1. 4 × 10 −^23 )(3 × 10 −^10 )^2
= 1. 3 × 105 K.
Sodium nuclei (m = 23mp = 23(1. 7 × 10 −^27 ) = 3. 9 × 10 −^26 kg):
T < (6.^6 ×^10
3(3. 9 × 10 −^26 )(1. 4 × 10 −^23 )(3 × 10 −^10 )^2
= 3 .0 K.
(b) P V = N kB T ; volume occupied by one molecule (N = 1, V = d^3 ) ⇒ d = (kB T /P )^1 /^3.
T <
h^2 2 mkB
P
kB T
⇒ T 5 /^3 <
h^2 3 m
P 2 /^3
k^5 B/^3
⇒ T <
kB
h^2 3 m
P 2 /^5.
For helium (m = 4mp = 6. 8 × 10 −^27 kg) at 1 atm = 1. 0 × 105 N/m^2 :
T < 1
(1. 4 × 10 −^23 )
(6. 6 × 10 −^34 )^2
3(6. 8 × 10 −^27 )
(1. 0 × 105 )^2 /^5 = 2.8 K.
For hydrogen (m = 2mp = 3. 4 × 10 −^27 kg) with d = 0.01 m:
T <
(6. 6 × 10 −^34 )^2
3(3. 4 × 10 −^27 )(1. 4 × 10 −^23 )(10−^2 )^2 =^3.^1 ×^10
− 14 K.
At 3 K it is definitely in the classical regime.
© c2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
14 CHAPTER 2. THE TIME-INDEPENDENT SCHR ODINGER EQUATION¨
Chapter 2
Time-Independent Schr¨odinger
Equation
Problem 2.
(a) Ψ(x, t) = ψ(x)e−i(E^0 +iΓ)t/ℏ^ = ψ(x)eΓt/ℏe−iE^0 t/ℏ^ =⇒ |Ψ|^2 = |ψ|^2 e2Γt/ℏ. ∫ (^) ∞
−∞
|Ψ(x, t)|^2 dx = e2Γt/ℏ
−∞
|ψ|^2 dx.
The second term is independent of t, so if the product is to be 1 for all time, the first term (e2Γt/ℏ) must also be constant, and hence Γ = 0. QED (b) If ψ satisfies Eq. 2.5, − ℏ 2 2 m ∂^2 ψ dx^2 +^ V ψ^ =^ Eψ, then (taking the complex conjugate and noting that^ V^ and E are real): − ℏ 2 2 m
∂^2 ψ∗ dx^2 +^ V ψ ∗ (^) = Eψ∗, so ψ∗ (^) also satisfies Eq. 2.5. Now, if ψ 1 and ψ 2 satisfy Eq. 2.5, so too does any linear combination of them (ψ 3 ≡ c 1 ψ 1 + c 2 ψ 2 ):
−
ℏ^2
2 m
∂^2 ψ 3 dx^2
ℏ^2
2 m
c 1 ∂^2 ψ 1 dx^2
= c 1
[
2 2 m
d^2 ψ 1 dx^2
]
[
2 2 m
d^2 ψ 2 dx^2
]
= c 1 (Eψ 1 ) + c 2 (Eψ 2 ) = E(c 1 ψ 1 + c 2 ψ 2 ) = Eψ 3. Thus, (ψ + ψ∗) and i(ψ − ψ∗) – both of which are real – satisfy Eq. 2.5. Conclusion: From any complex solution, we can always construct two real solutions (of course, if ψ is already real, the second one will be zero). In particular, since ψ = 12 [(ψ + ψ∗) − i(i(ψ − ψ∗))], ψ can be expressed as a linear combination of two real solutions. QED (c) If ψ(x) satisfies Eq. 2.5, then, changing variables x → −x and noting that ∂^2 /∂(−x)^2 = ∂^2 /∂x^2 ,
−
ℏ^2
2 m
∂^2 ψ(−x) dx^2
so if V (−x) = V (x) then ψ(−x) also satisfies Eq. 2.5. It follows that ψ+(x) ≡ ψ(x) + ψ(−x) (which is even: ψ+(−x) = ψ+(x)) and ψ−(x) ≡ ψ(x) − ψ(−x) (which is odd: ψ−(−x) = −ψ−(x)) both satisfy Eq.
© c2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
16 CHAPTER 2. THE TIME-INDEPENDENT SCHR ODINGER EQUATION¨
〈x^2 〉 =
a
∫ (^) a
0
x^2 sin^2
( (^) nπ a x
dx =
a
( (^) a nπ
) 3 ∫^ nπ 0
y^2 sin^2 y dy
= 2 a^2 (nπ)^3
[
y^3 6
y^3 4
sin 2y − y cos 2y 4
]nπ
0
= 2 a^2 (nπ)^3
[
(nπ)^3 6
nπ cos(2nπ) 4
]
= a^2
[
2(nπ)^2
]
〈p〉 = m d〈x〉 dt =^ 0.^ (Note^ : Eq.^1 .33 is much faster than Eq.^1.^35 .)
〈p^2 〉 =
ψ∗ n
i
d dx
ψn dx = −ℏ^2
ψ∗ n
d^2 ψn dx^2
dx
= (−ℏ^2 )
2 mEn ℏ^2
ψ∗ nψn dx = 2mEn =
nπℏ a
σ^2 x = 〈x^2 〉 − 〈x〉^2 = a^2
2(nπ)^2
a^2 4
(nπ)^2
; σx = a 2
(nπ)^2
σ^2 p = 〈p^2 〉 − 〈p〉^2 =
nπℏ a
; σp = nπℏ a
. ∴ σxσp =
(nπ)^2 3
The product σxσp is smallest for n = 1; in that case, σxσp = ℏ 2
π^2 3 −^ 2 = (1.136)ℏ/^2 >^ ℏ/^2.^
Problem 2.
(a) |Ψ|^2 = Ψ^2 Ψ = |A|^2 (ψ∗ 1 + ψ∗ 2 )(ψ 1 + ψ 2 ) = |A|^2 [ψ 1 ∗ ψ 1 + ψ∗ 1 ψ 2 + ψ 2 ∗ ψ 1 + ψ∗ 2 ψ 2 ].
|Ψ|^2 dx = |A|^2
[|ψ 1 |^2 + ψ∗ 1 ψ 2 + ψ∗ 2 ψ 1 + |ψ 2 |^2 ]dx = 2|A|^2 ⇒ A = 1/
(b)
Ψ(x, t) =
[
ψ 1 e−iE^1 t/ℏ^ + ψ 2 e−iE^2 t/ℏ
]
(but En ℏ = n^2 ω)
= √^1
a
[
sin
( (^) π a x
e−iωt^ + sin
2 π a x
e−i^4 ωt
]
= √^1
a e−iωt
[
sin
( (^) π a x
2 π a x
e−^3 iωt
]
|Ψ(x, t)|^2 =^1 a
[
sin^2
( (^) π a x
( (^) π a x
sin
2 π a x
e−^3 iωt^ + e^3 iωt
2 π a x
)]
a
[
sin^2
( (^) π a x
2 π a x
( (^) π a x
sin
2 π a x
cos(3ωt)
]
©^ c2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
CHAPTER 2. THE TIME-INDEPENDENT SCHR ODINGER EQUATION¨ 17
(c)
〈x〉 =
x|Ψ(x, t)|^2 dx
=^1 a
∫ (^) a
0
x
[
sin^2
( (^) π a x
2 π a x
( (^) π a x
sin
2 π a x
cos(3ωt)
]
dx
∫ (^) a
0
x sin^2
( (^) π a x
dx =
[
x^2 4
x sin
( (^2) π a x
4 π/a
cos
( (^2) π a x
8(π/a)^2
]∣∣
a
0
a^2 4
∫ (^) a
0
x sin^2
2 π a x
dx.
∫ (^) a
0
x sin
( (^) π a x
sin
2 π a x
dx =
∫ (^) a
0
x
[
cos
( (^) π a x
− cos
3 π a x
)]
dx
[
a^2 π^2 cos
( (^) π a x
ax π sin
( (^) π a x
a^2 9 π^2 cos
3 π a x
ax 3 π sin
3 π a x
)]a
0
[
a^2 π^2
cos(π) − cos(0)
a^2 9 π^2
cos(3π) − cos(0)
)]
a^2 π^2
8 a^2 9 π^2
∴ 〈x〉 =
a
[
a^2 4
a^2 4
16 a^2 9 π^2 cos(3ωt)
]
a 2
[
9 π^2 cos(3ωt)
]
Amplitude:
9 π^2
( (^) a 2
= 0.3603(a/2); angular frequency: 3 ω = 3 π^2 ℏ 2 ma^2
(d)
〈p〉 = m d〈x〉 dt =^ m
( (^) a 2
9 π^2
(− 3 ω) sin(3ωt) =
3 a sin(3ωt).
(e) You could get either E 1 = π^2 ℏ^2 / 2 ma^2 or E 2 = 2π^2 ℏ^2 /ma^2 , with equal probability P 1 = P 2 = 1/ 2.
So 〈H〉 = 1 2 (E 1 + E 2 ) =^5 π
4 ma^2 ; it’s the average of E 1 and E 2.
Problem 2.
From Problem 2.5, we see that
Ψ(x, t) = √^1 a e−iωt^
[
sin
( (^) π a x
( (^2) π a x
e−^3 iωteiφ
]
|Ψ(x, t)|^2 = (^1) a
[
sin^2
( (^) π a x
( (^2) π a x
( (^) π a x
sin
( (^2) π a x
cos(3ωt − φ)
]
©^ c2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
CHAPTER 2. THE TIME-INDEPENDENT SCHR ODINGER EQUATION¨ 19
(c)
P 1 = |c 1 |^2 =^16 ·^6 π^4
(d)
〈H〉 =
|cn|^2 En =
π^4
π^2 ℏ^2 2 ma^2
π^2 / 8
48 ℏ^2
π^2 ma^2
π^2 8
6 ℏ^2
ma^2
Problem 2.
(a)
Ψ(x, 0) =
A, 0 < x < a/2; 0 , otherwise.
1 = A^2
∫ (^) a/ 2
0
dx = A^2 (a/2) ⇒ A =
a
(b) From Eq. 2.37,
c 1 = A
a
∫ (^) a/ 2
0
sin
( (^) π a x
dx =
a
[
a π cos
( (^) π a x
)] ∣∣
a/ 2
0
π
[
cos
( (^) π 2
− cos 0
]
π
P 1 = |c 1 |^2 = (2/π)^2 = 0. 4053.
Problem 2.
H^ ˆΨ(x, 0) = − ℏ
2 2 m
∂^2
∂x^2 [Ax(a − x)] = −A
ℏ^2
2 m
∂x (a − 2 x) = A
ℏ^2
m
Ψ(x, 0)∗^ HˆΨ(x, 0) dx = A^2
ℏ^2
m
∫ (^) a
0
x(a − x) dx = A^2
ℏ^2
m
a x^2 2 −^
x^3 3
a
0
= A^2
ℏ^2
m
a^3 2
a^3 3
a^5
ℏ^2
m
a^3 6
5 ℏ^2
ma^2
(same as Example 2.3).
©^ c2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
20 CHAPTER 2. THE TIME-INDEPENDENT SCHR ODINGER EQUATION¨
Problem 2.
(a) Using Eqs. 2.47 and 2.59,
a+ψ 0 = √^1 2 ℏmω
−ℏ d dx
mω πℏ
e−^ mω^2 ℏ^ x 2
2 ℏmω
( (^) mω πℏ
) 1 / 4 [
mω 2 ℏ
2 x + mωx
]
e−^ mω 2 ℏ x 2 =
2 ℏmω
( (^) mω πℏ
2 mωxe−^ mω 2 ℏ x 2 .
(a+)^2 ψ 0 =
2 ℏmω
( (^) mω πℏ
2 mω
d dx
xe−^ mω 2 ℏ x 2
( (^) mω πℏ
) 1 / 4 [
1 − x mω 2 ℏ 2 x
]
e−^ mω 2 ℏ x 2 =
( (^) mω πℏ
) 1 / 4 (^2 mω ℏ x^2 − 1
e−^ mω 2 ℏ x 2 .
Therefore, from Eq. 2.67,
ψ 2 = √^1 2
(a+)^2 ψ 0 = √^1 2
( (^) mω πℏ
) 1 / 4 (^2 mω ℏ x^2 − 1
e−^ mω^2 ℏ^ x 2 .
(b)
(c) Since ψ 0 and ψ 2 are even, whereas ψ 1 is odd,
ψ 0 ∗ ψ 1 dx and
ψ∗ 2 ψ 1 dx vanish automatically. The only one we need to check is
ψ∗ 2 ψ 0 dx: ∫ ψ∗ 2 ψ 0 dx =
mω πℏ
−∞
2 mω ℏ x^2 − 1
e−^ mω ℏ x 2 dx
mω 2 πℏ
−∞
e−^ mω ℏ x 2 dx − 2 mω ℏ
−∞
x^2 e−^ mω ℏ x 2 dx
mω 2 πℏ
(√^
πℏ mω
2 mω ℏ
2 mω
πℏ mω
Problem 2.
(a) Note that ψ 0 is even, and ψ 1 is odd. In either case |ψ|^2 is even, so 〈x〉 =
x|ψ|^2 dx = 0. Therefore 〈p〉 = md〈x〉/dt = 0. (These results hold for any stationary state of the harmonic oscillator.) From Eqs. 2.59 and 2.62, ψ 0 = αe−ξ^2 /^2 , ψ 1 =
2 αξe−ξ^2 /^2. So n = 0:
〈x^2 〉 = α^2
−∞
x^2 e−ξ (^2) / 2 dx = α^2
mω
−∞
ξ^2 e−ξ 2 dξ =
π
mω
π 2
2 mω
©^ c2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
