Introduction to Quantum Mechanics, 3rd ed, by
Griffiths, Schroeter. Chapter 1-12
SOLUTION MANUAL
,2
Contents
1 The Wave Function 4
2 The Time-Independent Schrödinger Equation 16
3 Formalism 78
4 Quantum Mechanics in Three Dimensions 109
5 Identical Particles 168
6 Symmetries and Conservation Laws 197
7 Time-Independent Perturbation Theory 235
8 The Variational Principle 301
9 The WKB Approximation 333
10 Scattering 354
11 Quantum Dynamics 372
12 Afterword 420
A Linear Algebra 427
,4 CHAPTER 1. THE WAVE FUNCTION
Chapter 1
The Wave Function
Problem 1.1
(a)
h ji2 = 212 = 441.
1 X 1
h j2 i = j2 N ( j) = (142) + (15 2) + 3(162) + 2(222) + 2(242) + 5(252)
N 14
1 6434
= (196 + 225 + 768 + 968 + 1152 + 3125) = 459.571.
= 14
14
j ∆j = j − hji
14 14 − 21 = −7
15 15 − 21 = −6
(b) 16 16 − 21 = −5
22 22 − 21 = 1
24 24 − 21 = 3
25 25 − 21 = 4
1 X 1
σ2 = (∆j)2 N ( j) = (−7)2 + (−6)2 + (−5)2 · 3 + (1)2 · 2 + (3)2 · 2 + (4)2 · 5
N 14
1 260
= (49 + 36 + 75 + 2 + 18 + 80) = = 18.571.
14 14
√
σ = 18.571 = 4.309.
(c
)
h j2i − hji2 = 459.571 − 441 = 18.571. [Agrees with (b).]
,CHAPTER 1. THE WAVE FUNCTION 5
Problem 1.2
(a) 1 2
dx = √ x5/2 h 2
2 Z h
= h .
hx i = x2 √1
0 2 2 h 5
0
hx
5
h 2 2h
= 4 h2 ⇒ σ =
2
σ 2 = hx2i − hxi2 = − h √ = 0. 2981h.
5 3 45 3 5
(b
) Z x+ 1 1 √ x+ 1 √
P =1− √ dx = 1 − √ (2 x) =1−√ √ − x− .
2 hx x+
x− 2 h x− h
x+ ≡ hxi + σ = 0.3333h + 0.2981h = 0.6315h; x− ≡ hxi − σ = 0.3333h − 0.2981h = 0.0352h.
√ √
P =1− 0.6315 0.0352 = 0.393.
+
Problem 1.3
(a)
Z ∞ 2
1 = Ae−λ(x −a) dx. Let u ≡ x − a, du = dx, u : −∞ → ∞.
−∞
Z∞ r r
2 π λ
1= e−λu du = A
λ ⇒ A= .
A −∞ π
(b
)
2
Z ∞ Z ∞
− λ ( x− a )
hx i = A xe dx = A (u + a)e−λu du
2
−∞ −∞
Z ∞ Z ∞ r
ue−λu du + a e−λu du = A π = a.
2 2
=A 0+a
−∞ −∞ λ
Z ∞
h x2 i = A x2e−λ (x −a) dx
2
−∞
Z ∞ Z ∞ Z ∞
=A u2e−λu du + 2a
2
ue−λu du + a2
2
e−λu 2du
−∞ −∞ −∞
2 1
=A + 0 + a2 = a + .
2λ λ λ 2λ
1 1 1
σ 2 = hx2i − hxi2 = a2 + − a2 = ; σ= .
√
2λ 2λ 2λ
,6 CHAPTER 1. THE WAVE FUNCTION
(c
x)
)
A
a x
Problem 1.4
(a)
2Z a b
( a b
)
2 |A| 2 Z 1 x3 3
|A| x dx + (b 2
x) dx = A 2 + (b − x)
1= 2 1
− −
a2 (b − a) a | | 3 (b − a)2 3
0 a2 0 a
b−a b r
a+ 3
= |A|2 = |A|2
3 3 3 ⇒ A= b
.
(b)
A
a b x
(c) At x = a.
(d)
Z a 2 Z a P =1 if b = a, X
2 |A| 2 2a a
P = |Ψ| dx = x dx = |A| = .
0 a2 0 3 b P = 1/2 if b = 2a. X
(e
)
Z Z 1 Z b
hx i = 1 a x3 dx +
2
x|Ψ| dx = |A| 2 x(b − x)2dx
( a2 0 (b − a)2 a
)
=3 1 x2
a x3 x4 b
1
b2 — 2b
x4
+ (b − a)2 2 3 + 4
b a2 4 0 a
3
= a2 (b − a)2 + 2b4 − 8 b4 /3 + b4 − 2a2 b2 + 8a3 b/3 − a4
4b(b − a)2
3 b4 2 3 1 2a + b
= − a2b2 + ab = (b3 − 3a2b + 2a3) = .
4b(b − a)2 3 3 4(b − a)2 4
,CHAPTER 1. THE WAVE FUNCTION 7
Problem 1.5
(a)
Z Z ∞ e−2 λx
∞ 2
−2 λx √
e dx = 2|A| 2 |A| A = λ.
= λ ;
2 2
1 = |Ψ| dx = 2|A|
0 2λ
− 0
(b)
Z Z ∞
hx i = xe−2λ|x |dx = [Odd integrand.]
x|Ψ|2dx = |A|2 0.
−∞
Z ∞ (2λ)3 1
2
hx i = 2|A| 2 x2e−2λx dx = 2λ =2 .
0 2λ2
(c
) √
1 1 √
2λ
2
= 0.2431λ.
2 2
σ = hx i − hxi = 2
. |Ψ(±σ )| = |A| e2 2 −2 λσ
= λe −2 λ/ = λe−
2λ2 ; σ= √
2λ
.24
x
Probability outside:
∞
Z ∞ Z ∞ e−2 λx √
2
e−2λx dx = 2λ e− = 0.2431.
2 |Ψ|2dx = 2|A|2 −2λ =e −2 λσ
=
o σ
σ
Problem 1.6
For integration by parts, the differentiation has to be with respect to the integration variable – in this case the
differentiation is with respect to t, but the integration variable is x. It’s true that
∂ ∂x ∂ ∂
(x Ψ 2 ) = Ψ 2+x Ψ2=x Ψ 2,
∂t | | ∂t | | ∂t | | ∂t |
|
but this does not allow us to perform the integration:
Zbx Z b ∂
∂ |Ψ|2dx = (x|Ψ|2)dx 6= (x|Ψ|2) b.
a ∂t a ∂t a
,8 CHAPTER 1. THE WAVE FUNCTION
Problem 1.7
dhpi R ∂
From Eq. 1.33, = −i~ Ψ∗ ∂Ψ dx. But, noting that ∂ Ψ 2
= ∂ 2Ψ and using Eqs. 1.23-1.24:
dt ∂t ∂x ∂x∂t ∂t∂ x
∂ ∂Ψ ∂Ψ∗ ∂ Ψ ∂ ∂Ψ i~ ∂ 2 Ψ ∗ i ∂Ψ ∂ i~ ∂ 2 Ψ i
Ψ∗ = + Ψ∗ = − + V Ψ∗ + Ψ∗ − VΨ
∂t ∂x ∂t ∂x ∂x ∂t 2m ∂x2 ~ ∂x ∂x 2m ∂x2 ~
i~ ∗ ∂ 3 Ψ ∂2 Ψ ∗ ∂ Ψ i ∂Ψ ∂
= Ψ − + V Ψ∗ − Ψ∗ (V Ψ)
2m ∂x3 ∂x2 ∂x ~ ∂x ∂x
The∗∂first term integrates
∗ ∂Ψ ∗ ∂V
to zero, using
2 ∂V
integration by parts twice, and the second term can be simplified to
VΨ Ψ
∂x − Ψ V ∂x − Ψ ∂x Ψ = −|Ψ| ∂x . So
Z ∂V ∂V
dhpi i −|Ψ|2 dx = h− i. QED
= −i~
dt ~ ∂x ∂x
Problem 1.8 ∂2 Ψ
+ V Ψ. We want to find the solution
Suppose Ψ satisfies the Schrödinger equation without V0 : i~ ∂Ψ = − ~2
2 ∂t 2 m ∂x 2
Ψ0 with V0: i~ ∂Ψ0 = − ~2 ∂ Ψ0 + (V + V )Ψ .
∂t 2 m ∂x2 0 0
Claim: Ψ0 = Ψe−iV0t/~ . h i
Proof: i~ ∂Ψ0 = i~ ∂Ψ e−iV0 t/~ + i~Ψ − iV0 e−iV0 t/~ = − ~ ∂ Ψ
2
+2 V Ψ e−iV0 t/~ + V Ψe−iV0 t/~
∂t ∂t2 ~ 2 m ∂x2 0
~ ∂ 2Ψ0
= − 2m ∂x2 + (V + V 0 )Ψ0 . 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 1.9
(a)
Z r r
∞
2 1 π π~
1 = 2|A|2 e−2amx /~dx = 2|A|2 = |A|2 ;
A= 2am 1/4
.
0 2 (2am/~) 2am π~
(b
)
∂Ψ ∂Ψ 2amx ∂2 Ψ 2am ∂Ψ 2am 2amx2
Ψ+x
= −ia Ψ; =− Ψ; ∂x = −
2
∂x =− 1− Ψ.
∂t ∂x ~ ~ ~ ~
Plug these into the Schrödinger equation, i~ ∂Ψ = − ~ 2 ∂ 2Ψ
+ V Ψ:
∂t 2 m ∂x 2
~2 2am 2amx2
V Ψ = i~(−ia)Ψ + − 1− Ψ
2m ~ ~
2amx2
= ~a − ~a Ψ = 2a2mx2Ψ, so V (x) = 2ma 2x 2.
1−
~
,CHAPTER 1. THE WAVE FUNCTION 9
(c)
Z ∞
hx i = 0. [Odd integrand.]
x|Ψ|2dx =
−∞
r
Z ∞
2
π~ ~
1
2
hx i = 2|A| 2
x e
2 −2amx /~ 2
dx = 2|A| 22(2am/~) = 4am .
0 2am
dhxi
0.
hpi = m =
dt
Z 2 Z
~ ∂
2
2 2 ∂Ψ
hp i = Ψ ∗ Ψdx = −~ Ψ∗ dx
i ∂x ∂x2
Z Z Z
2 2am 2amx2 2am
= −~ Ψ∗ − Ψ dx = 2am~ 2
|Ψ| dx − x2|Ψ|2dx
~ 1− ~
~
2am 2am ~
2
= 2am~ = 2am~ = 2am~
1− hx i 1− 4am 1 = am~.
~ ~ 2
(d)
~ r
2 2 2 ~ √
σ = hx i − hxi ⇒ x
2 2
σp = hp i − hpi
2
= am~ =⇒ σp = am~.
x
= = σ = 4am ;
4am
q
~
√
σx σ p = 4am
am~ = ~2 . This is (just barely) consistent with the uncertainty principle.
Problem 1.10
From Math Tables: π = 3.141592653589793238462643 · · ·
P (0) = 0 P (1) = 2/25 P (2) = 3/25 P (3) = 5/25 P (4) = 3/25
(a)
P (5) = 3/25 P (6) = 3/25 P (7) = 1/25 P (8) = 2/25 P (9) = 3/25
N (j)
In general, P (j) = N
.
(b) Most probable: 3. Median: 13 are ≤ 4, 12 are ≥ 5, so median is 4.
Average: hji = 125[0 · 0 + 1 · 2 + 2 · 3 + 3 · 5 + 4 · 3 + 5 · 3 + 6 · 3 + 7 · 1 + 8 · 2 + 9 · 3]
118
1 [0 + 2 + 6 + 15 + 12 + 15 + 18 + 7 + 16 + 27] = 4.72.
= 25 25
=
1
(c) h j2i = [0
25
+ 12 · 2 + 22 · 3 + 32 · 5 + 42 · 3 + 52 · 3 + 62 · 3 + 72 · 1 + 82 · 2 + 92 · 3]
710
1 [0 + 2 + 12 + 45 + 48 + 75 + 108 + 49 + 128 + 243] = = 28.4.
= 25 25
√
σ 2 = h j2i − hji2 = 28.4 − 4.722 = 28.4 − 22.2784 = 6.1216; σ = 6.1216 = 2.474.
,10 CHAPTER 1. THE WAVE FUNCTION
Problem
1.11
(a) 1 r
mv2 + V = E 2
2 → v(x) = (E − V (x)) .
m
(b r
Z Z b
) b 1 dx = m 1
p dx.
T= a
q
E − 12kx2 k a (2E/k) — x2
m2
p
r vZ=b 0 ⇒ E = V = 1 kb
Turning points: 2
2 r⇒ b = 2E/k ; ba = −
r b.
m 1 m x m
T=2 √ dx = 2 sin−1 =2 sin−1(1)
k 0 b2 − x2 k b 0 k
r r
=2 m π = π m.
k 2 k
1 1
ρ(x ) = q = √ . x)
pm 2
E−1 π b2 − x2
π
2 kx
21
k Zb
m
Z b 2 1
ρ(x) dx = √ dx = 2 π = 1. X
a π 0 b2 − x2 π 2
-b b x
(c hxi = 0.
)
1 Z x Z b
h x2i =
b
√
2 2 x2
dx = √ dx
π −b b2 − x2 π 0 b2 − x2
b2 x b b2 b2 π b2
2 x b2 − x2 +
p E
sin−1 = sin−1(1) = = = .
= − b π π2
π 2 2 0 2 k
p p
σx = x 2 i − hxi2 = h x2 i = √ =b r
h E.
2 k
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Griffiths, Schroeter. Chapter 1-12
SOLUTION MANUAL
,2
Contents
1 The Wave Function 4
2 The Time-Independent Schrödinger Equation 16
3 Formalism 78
4 Quantum Mechanics in Three Dimensions 109
5 Identical Particles 168
6 Symmetries and Conservation Laws 197
7 Time-Independent Perturbation Theory 235
8 The Variational Principle 301
9 The WKB Approximation 333
10 Scattering 354
11 Quantum Dynamics 372
12 Afterword 420
A Linear Algebra 427
,4 CHAPTER 1. THE WAVE FUNCTION
Chapter 1
The Wave Function
Problem 1.1
(a)
h ji2 = 212 = 441.
1 X 1
h j2 i = j2 N ( j) = (142) + (15 2) + 3(162) + 2(222) + 2(242) + 5(252)
N 14
1 6434
= (196 + 225 + 768 + 968 + 1152 + 3125) = 459.571.
= 14
14
j ∆j = j − hji
14 14 − 21 = −7
15 15 − 21 = −6
(b) 16 16 − 21 = −5
22 22 − 21 = 1
24 24 − 21 = 3
25 25 − 21 = 4
1 X 1
σ2 = (∆j)2 N ( j) = (−7)2 + (−6)2 + (−5)2 · 3 + (1)2 · 2 + (3)2 · 2 + (4)2 · 5
N 14
1 260
= (49 + 36 + 75 + 2 + 18 + 80) = = 18.571.
14 14
√
σ = 18.571 = 4.309.
(c
)
h j2i − hji2 = 459.571 − 441 = 18.571. [Agrees with (b).]
,CHAPTER 1. THE WAVE FUNCTION 5
Problem 1.2
(a) 1 2
dx = √ x5/2 h 2
2 Z h
= h .
hx i = x2 √1
0 2 2 h 5
0
hx
5
h 2 2h
= 4 h2 ⇒ σ =
2
σ 2 = hx2i − hxi2 = − h √ = 0. 2981h.
5 3 45 3 5
(b
) Z x+ 1 1 √ x+ 1 √
P =1− √ dx = 1 − √ (2 x) =1−√ √ − x− .
2 hx x+
x− 2 h x− h
x+ ≡ hxi + σ = 0.3333h + 0.2981h = 0.6315h; x− ≡ hxi − σ = 0.3333h − 0.2981h = 0.0352h.
√ √
P =1− 0.6315 0.0352 = 0.393.
+
Problem 1.3
(a)
Z ∞ 2
1 = Ae−λ(x −a) dx. Let u ≡ x − a, du = dx, u : −∞ → ∞.
−∞
Z∞ r r
2 π λ
1= e−λu du = A
λ ⇒ A= .
A −∞ π
(b
)
2
Z ∞ Z ∞
− λ ( x− a )
hx i = A xe dx = A (u + a)e−λu du
2
−∞ −∞
Z ∞ Z ∞ r
ue−λu du + a e−λu du = A π = a.
2 2
=A 0+a
−∞ −∞ λ
Z ∞
h x2 i = A x2e−λ (x −a) dx
2
−∞
Z ∞ Z ∞ Z ∞
=A u2e−λu du + 2a
2
ue−λu du + a2
2
e−λu 2du
−∞ −∞ −∞
2 1
=A + 0 + a2 = a + .
2λ λ λ 2λ
1 1 1
σ 2 = hx2i − hxi2 = a2 + − a2 = ; σ= .
√
2λ 2λ 2λ
,6 CHAPTER 1. THE WAVE FUNCTION
(c
x)
)
A
a x
Problem 1.4
(a)
2Z a b
( a b
)
2 |A| 2 Z 1 x3 3
|A| x dx + (b 2
x) dx = A 2 + (b − x)
1= 2 1
− −
a2 (b − a) a | | 3 (b − a)2 3
0 a2 0 a
b−a b r
a+ 3
= |A|2 = |A|2
3 3 3 ⇒ A= b
.
(b)
A
a b x
(c) At x = a.
(d)
Z a 2 Z a P =1 if b = a, X
2 |A| 2 2a a
P = |Ψ| dx = x dx = |A| = .
0 a2 0 3 b P = 1/2 if b = 2a. X
(e
)
Z Z 1 Z b
hx i = 1 a x3 dx +
2
x|Ψ| dx = |A| 2 x(b − x)2dx
( a2 0 (b − a)2 a
)
=3 1 x2
a x3 x4 b
1
b2 — 2b
x4
+ (b − a)2 2 3 + 4
b a2 4 0 a
3
= a2 (b − a)2 + 2b4 − 8 b4 /3 + b4 − 2a2 b2 + 8a3 b/3 − a4
4b(b − a)2
3 b4 2 3 1 2a + b
= − a2b2 + ab = (b3 − 3a2b + 2a3) = .
4b(b − a)2 3 3 4(b − a)2 4
,CHAPTER 1. THE WAVE FUNCTION 7
Problem 1.5
(a)
Z Z ∞ e−2 λx
∞ 2
−2 λx √
e dx = 2|A| 2 |A| A = λ.
= λ ;
2 2
1 = |Ψ| dx = 2|A|
0 2λ
− 0
(b)
Z Z ∞
hx i = xe−2λ|x |dx = [Odd integrand.]
x|Ψ|2dx = |A|2 0.
−∞
Z ∞ (2λ)3 1
2
hx i = 2|A| 2 x2e−2λx dx = 2λ =2 .
0 2λ2
(c
) √
1 1 √
2λ
2
= 0.2431λ.
2 2
σ = hx i − hxi = 2
. |Ψ(±σ )| = |A| e2 2 −2 λσ
= λe −2 λ/ = λe−
2λ2 ; σ= √
2λ
.24
x
Probability outside:
∞
Z ∞ Z ∞ e−2 λx √
2
e−2λx dx = 2λ e− = 0.2431.
2 |Ψ|2dx = 2|A|2 −2λ =e −2 λσ
=
o σ
σ
Problem 1.6
For integration by parts, the differentiation has to be with respect to the integration variable – in this case the
differentiation is with respect to t, but the integration variable is x. It’s true that
∂ ∂x ∂ ∂
(x Ψ 2 ) = Ψ 2+x Ψ2=x Ψ 2,
∂t | | ∂t | | ∂t | | ∂t |
|
but this does not allow us to perform the integration:
Zbx Z b ∂
∂ |Ψ|2dx = (x|Ψ|2)dx 6= (x|Ψ|2) b.
a ∂t a ∂t a
,8 CHAPTER 1. THE WAVE FUNCTION
Problem 1.7
dhpi R ∂
From Eq. 1.33, = −i~ Ψ∗ ∂Ψ dx. But, noting that ∂ Ψ 2
= ∂ 2Ψ and using Eqs. 1.23-1.24:
dt ∂t ∂x ∂x∂t ∂t∂ x
∂ ∂Ψ ∂Ψ∗ ∂ Ψ ∂ ∂Ψ i~ ∂ 2 Ψ ∗ i ∂Ψ ∂ i~ ∂ 2 Ψ i
Ψ∗ = + Ψ∗ = − + V Ψ∗ + Ψ∗ − VΨ
∂t ∂x ∂t ∂x ∂x ∂t 2m ∂x2 ~ ∂x ∂x 2m ∂x2 ~
i~ ∗ ∂ 3 Ψ ∂2 Ψ ∗ ∂ Ψ i ∂Ψ ∂
= Ψ − + V Ψ∗ − Ψ∗ (V Ψ)
2m ∂x3 ∂x2 ∂x ~ ∂x ∂x
The∗∂first term integrates
∗ ∂Ψ ∗ ∂V
to zero, using
2 ∂V
integration by parts twice, and the second term can be simplified to
VΨ Ψ
∂x − Ψ V ∂x − Ψ ∂x Ψ = −|Ψ| ∂x . So
Z ∂V ∂V
dhpi i −|Ψ|2 dx = h− i. QED
= −i~
dt ~ ∂x ∂x
Problem 1.8 ∂2 Ψ
+ V Ψ. We want to find the solution
Suppose Ψ satisfies the Schrödinger equation without V0 : i~ ∂Ψ = − ~2
2 ∂t 2 m ∂x 2
Ψ0 with V0: i~ ∂Ψ0 = − ~2 ∂ Ψ0 + (V + V )Ψ .
∂t 2 m ∂x2 0 0
Claim: Ψ0 = Ψe−iV0t/~ . h i
Proof: i~ ∂Ψ0 = i~ ∂Ψ e−iV0 t/~ + i~Ψ − iV0 e−iV0 t/~ = − ~ ∂ Ψ
2
+2 V Ψ e−iV0 t/~ + V Ψe−iV0 t/~
∂t ∂t2 ~ 2 m ∂x2 0
~ ∂ 2Ψ0
= − 2m ∂x2 + (V + V 0 )Ψ0 . 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 1.9
(a)
Z r r
∞
2 1 π π~
1 = 2|A|2 e−2amx /~dx = 2|A|2 = |A|2 ;
A= 2am 1/4
.
0 2 (2am/~) 2am π~
(b
)
∂Ψ ∂Ψ 2amx ∂2 Ψ 2am ∂Ψ 2am 2amx2
Ψ+x
= −ia Ψ; =− Ψ; ∂x = −
2
∂x =− 1− Ψ.
∂t ∂x ~ ~ ~ ~
Plug these into the Schrödinger equation, i~ ∂Ψ = − ~ 2 ∂ 2Ψ
+ V Ψ:
∂t 2 m ∂x 2
~2 2am 2amx2
V Ψ = i~(−ia)Ψ + − 1− Ψ
2m ~ ~
2amx2
= ~a − ~a Ψ = 2a2mx2Ψ, so V (x) = 2ma 2x 2.
1−
~
,CHAPTER 1. THE WAVE FUNCTION 9
(c)
Z ∞
hx i = 0. [Odd integrand.]
x|Ψ|2dx =
−∞
r
Z ∞
2
π~ ~
1
2
hx i = 2|A| 2
x e
2 −2amx /~ 2
dx = 2|A| 22(2am/~) = 4am .
0 2am
dhxi
0.
hpi = m =
dt
Z 2 Z
~ ∂
2
2 2 ∂Ψ
hp i = Ψ ∗ Ψdx = −~ Ψ∗ dx
i ∂x ∂x2
Z Z Z
2 2am 2amx2 2am
= −~ Ψ∗ − Ψ dx = 2am~ 2
|Ψ| dx − x2|Ψ|2dx
~ 1− ~
~
2am 2am ~
2
= 2am~ = 2am~ = 2am~
1− hx i 1− 4am 1 = am~.
~ ~ 2
(d)
~ r
2 2 2 ~ √
σ = hx i − hxi ⇒ x
2 2
σp = hp i − hpi
2
= am~ =⇒ σp = am~.
x
= = σ = 4am ;
4am
q
~
√
σx σ p = 4am
am~ = ~2 . This is (just barely) consistent with the uncertainty principle.
Problem 1.10
From Math Tables: π = 3.141592653589793238462643 · · ·
P (0) = 0 P (1) = 2/25 P (2) = 3/25 P (3) = 5/25 P (4) = 3/25
(a)
P (5) = 3/25 P (6) = 3/25 P (7) = 1/25 P (8) = 2/25 P (9) = 3/25
N (j)
In general, P (j) = N
.
(b) Most probable: 3. Median: 13 are ≤ 4, 12 are ≥ 5, so median is 4.
Average: hji = 125[0 · 0 + 1 · 2 + 2 · 3 + 3 · 5 + 4 · 3 + 5 · 3 + 6 · 3 + 7 · 1 + 8 · 2 + 9 · 3]
118
1 [0 + 2 + 6 + 15 + 12 + 15 + 18 + 7 + 16 + 27] = 4.72.
= 25 25
=
1
(c) h j2i = [0
25
+ 12 · 2 + 22 · 3 + 32 · 5 + 42 · 3 + 52 · 3 + 62 · 3 + 72 · 1 + 82 · 2 + 92 · 3]
710
1 [0 + 2 + 12 + 45 + 48 + 75 + 108 + 49 + 128 + 243] = = 28.4.
= 25 25
√
σ 2 = h j2i − hji2 = 28.4 − 4.722 = 28.4 − 22.2784 = 6.1216; σ = 6.1216 = 2.474.
,10 CHAPTER 1. THE WAVE FUNCTION
Problem
1.11
(a) 1 r
mv2 + V = E 2
2 → v(x) = (E − V (x)) .
m
(b r
Z Z b
) b 1 dx = m 1
p dx.
T= a
q
E − 12kx2 k a (2E/k) — x2
m2
p
r vZ=b 0 ⇒ E = V = 1 kb
Turning points: 2
2 r⇒ b = 2E/k ; ba = −
r b.
m 1 m x m
T=2 √ dx = 2 sin−1 =2 sin−1(1)
k 0 b2 − x2 k b 0 k
r r
=2 m π = π m.
k 2 k
1 1
ρ(x ) = q = √ . x)
pm 2
E−1 π b2 − x2
π
2 kx
21
k Zb
m
Z b 2 1
ρ(x) dx = √ dx = 2 π = 1. X
a π 0 b2 − x2 π 2
-b b x
(c hxi = 0.
)
1 Z x Z b
h x2i =
b
√
2 2 x2
dx = √ dx
π −b b2 − x2 π 0 b2 − x2
b2 x b b2 b2 π b2
2 x b2 − x2 +
p E
sin−1 = sin−1(1) = = = .
= − b π π2
π 2 2 0 2 k
p p
σx = x 2 i − hxi2 = h x2 i = √ =b r
h E.
2 k
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