Problem 3.

1 Prove the following theorems:

a) For normalizable solutions, the separation constant E must be real. Hint: Write E (in
Equation 3.6) as E0 + iГ (with E0 and Г real), and show that if Equation 2.20 is to hold for
all t, V must be zero.
b) Ψ can always be taken to be real (unlike Ψ , which is necessarily complex). Note:

This doesn't mean that every solution to the time-independent Schrodinger equation is
real; what it says is that if you've got one that is not, it can always be expressed as a linear
combination of solutions (with the same energy) that are. So in Equation 3.14 you might

as well stick to Ψ 's that are real. Hint: If Ψ ( x ) satisfies the time-independent
Schrodinger equation for a given E, so too does its complex conjugate, and hence also the

real linear combinations ( Ψ +Ψ ¿ ) and i ( Ψ −Ψ

¿
) .

c) If V is an even function [ i. e . , V (−x )=V ( x ) ] , then Ψ ( x) can always be taken to be

either or odd. Hint: if Ψ ( x ) satisfies the time-independent Schrodinger equation for a

given E, so too does Ψ (−x ) , and hence also the even and odd linear combinations

Ψ ( x ) ±Ψ (−x )
Penyelesaian:
∞
2 d ∞ 2
I= ∫ |Ψ| dx ∫ |Ψ| dx=0
dt −∞
a) −∞ not dependent in time
iEt
− ℏ
Ψ ( x, t )=Ψ ( x ) e
Dimana,
E=E 0 +iΓ
i ( E 0+iΓ ) t
− ℏ
Ψ ( x , t )=Ψ ( x ) e
i ( E0 )t
Γ
− t
ℏ
Ψ ( x , t )=Ψ ( x ) e eℏ
Γ iE 0 iE
0
2 2 2ℏt − ℏ t ℏ t
|Ψ ( x, t )| =|Ψ ( x )| e e e
∞ Γ
2 2ℏt
1= ∫ |Ψ ( x )| e dx
−∞
Γ ∞
2ℏt
1=e ∫−∞|Ψ ( x )|2 dx
Suku kedua tidak bergantung terhadap t, jadi jika hasil 1 untuk semua waktu, suku
Γ
2ℏ t
pertama e harus konstan, dan karenanya Γ=0
b) Ψ ( x ) → can always be real
Ψ ( x ) → can always be expressed us a linear combination
2 2
ℏ d Ψ
− +VΨ =EΨ
2 m dx 2
Kompleks conjugate dan V dan E real
ℏ2 d 2 Ψ
− + VΨ ¿=EΨ ¿
2 m dx 2
Ψ ' ( x )=Ψ +Ψ ¿
Ψ ' ( x )=i ( Ψ −Ψ ¿ )
1 ¿ ¿
Ψ ( x ) = [ ( Ψ +Ψ )+i ( Ψ−Ψ ) ]
2
Dimana kombinasi linier adalah 3Ψ =Ψ +Ψ
1 2
2 2
ℏ d Ψ
− +VΨ =EΨ
2 m dx 2
ℏ2 d 2
− ( Ψ +Ψ 2 ) +V ( Ψ 1 +Ψ 2 ) =E ( Ψ 1 +Ψ 2 )
2 m dx 2 1
2 2
ℏ2 d Ψ 1 ℏ2 d Ψ 2
−
2 m dx 2
2
+VΨ 1 + − (
2m dx 2 )
+VΨ 2 =E ( Ψ 1 +Ψ 2 )

ℏ2 d Ψ 3
− +VΨ 3 =EΨ 1 + EΨ 2 =E ( Ψ 1 +Ψ 2 ) =EΨ 3
2 m dx 2
Dengan demikian ( Ψ +Ψ ¿ ) dan i ( Ψ −Ψ ¿ ) yang real. Kesimpulan dari setiap solusi
kompleks, dapat membangun dua solusi real (jika, Ψ adalah real maka yang kedua

1
Ψ= [ ( Ψ + Ψ ¿ ) +i ( Ψ −Ψ ¿ ) ]
akan menjadi nol. Khususnya, karena 2 dapat dinyatakan
sebagai kombinasi linear dari dua solusi real
2 2
ℏ d Ψ (x)
− +V ( x ) Ψ ( x )=E ( x ) Ψ ( x )
c) 2 m dx 2
Jika, V ( x ) =V (−x )
2 2
ℏ d Ψ (−x )
− +V (−x ) Ψ (−x )=E ( x ) Ψ (−x )
2 m dx 2
Ψ ( x ) ∧ Ψ (−x ) adalah solusi
Ψ + ( x )=Ψ ( x ) +Ψ (−x )
Yang genap,
Ψ + (−x ) =Ψ + ( x ) dan Ψ − =Ψ ( x )−Ψ (−x )

Yang ganjil, Ψ − (−x )=Ψ − ( x )

 1
Ψ ( x) = Ψ ( x ) +Ψ − ( x ) )
Tapi, 2( + . Jadi solusi apapun dapat dinyatakan sebagai kombinasi
linear dari solusi genap dan ganjil

Problem 3.3 Show that there is no acceptable solution to the (time-independent) Schrodinger
equation (for the infinite square well) with E = 0 or E < 0. (This is a special case of the
general theorem in Problem 3.2, but this time do it by explicitly solving the Schrodinger
equation and showing that you cannot meet the boundary conditions.)
Penyelesaian:
EΨ = H^ Ψ
ℏ2 d 2 Ψ
EΨ =− +VΨ
2 m dx 2
ℏ2 d2 Ψ
− =( V − E ) Ψ
2 m dx 2
ℏ2 d2 Ψ
=( E−V ) Ψ
2 m dx 2
V =0 ; E=0
ℏ2 d2 Ψ
=0
2 m dx 2
d2 Ψ
=0→
dx 2 konstan atau fungsi linear bukan gelombang
Jika, boundary conditions Ψ ( 0 )=0 ; Ψ ( a )=0
Ψ ( x ) =A +Bx
Ψ ( 0 )= A=0 ⇒Ψ =Bx
Ψ ( a )=Ba=0⇒ B=0
Jadi Ψ =0
Sehingga E=0 tidak diperbolehkan
Jika E<0 dan V=0 maka,
ℏ2 d2 Ψ
=( E−V ) Ψ
2 m dx 2
ℏ2 d2 Ψ
=EΨ
2 m dx 2
Misal,
 Ψ=sin ( kx )
dΨ
=k cos ( kx )
dx
2
d Ψ
2
=−k 2 sin ( kx )
dx
−2 mE
k≡ √ ℏ
Dimana,
kx −kx
Ψ ( x ) =Ae +Be
Jika, boundary conditions Ψ ( 0 )=0 ; Ψ ( a )=0
Ψ ( 0 )= A+B=0 ⇒ B=−A
Ψ= A ( e kx −e−kx )
Ψ ( a )= A ( e ka−e−ka )=0 ⇒ A=0
Ψ=0
Atau,
e ka=e−ka
e 2ka =1
1ka=ln (1 ) =0
k=0
Ψ=0
Sehingga kedua kondisi batas Ψ =0 tidak dinormalisasi

Problem 3.5 Calculate

⟨x⟩ , ⟨x 2⟩ , ⟨ p⟩ , ⟨ p2 ⟩, σ x and σ p , for the nth stationary state of the
infinite square well. Check that the uncertainty principle is satisfied. Which state comes
closest to the uncertainty limit?
Penyelesaian:
2 nπ
Ψ 0=
√ a
sin ( )
a
x


⟨x⟩=∫ x |Ψ|2 dx
a
2 nπ
⟨x ⟩= ∫ x sin2
a0 a
x dx( )
nπ a
y= x ; dx= dy
Misal, a nπ
y : 0→nπ
nπ
2 a a
⟨x ⟩= ∫ y sin2 ( y ) dy
a 0 nπ nπ
 nπ
2a
⟨x ⟩= 2∫
y sin 2 ( y ) dy
( nπ ) 0
y 1
u= y ; du=1 ; dv =sin2 y ; v= − sin2 y
Misal, 2 4
nπ nπ
2a y 1 y 1
⟨x ⟩=
( nπ ) 2 [(2 4)] (
y − sin 2 y −∫ − sin 2 y dy
0 0 2 4
nπ nπ
)
2 a ( nπ )2
⟨x ⟩=
( nπ ) 2 [ ]∫ ∫
2
−
0
y
2
dy +
1
4 0
sin 2 y dy

2 2
2 a ( nπ ) y nπ
⟨x ⟩= [
( nπ )2 2
2
−
4
|0

2
+ −
11
42
cos 2 y|nπ
0 ( )]
2 a ( nπ ) ( nπ ) 1
⟨ x ⟩=
( nπ )2 2[ −
4
− ( 1−1 )
8 ]
2 1
⟨x ⟩=2 a − ( )
4 4
2a
⟨x ⟩=
4
a
⟨x⟩=
2
⟨x 2 ⟩=∫ x2 Ψ Ψ dx
¿

a
2 nπ
⟨x ⟩= ∫ x 2 sin2
2
a 0 a
x dx ( )
nπ a
y= x ; dx= dy
Misal, a nπ
3 nπ
2 a
2
⟨x ⟩=
a nπ ( ) ∫ y sin y dy
0
2 2

y 1
u= y 2 ; du=2 y ; dv=sin2 y ; v= − sin 2 y
Misal, 2 4
nπ
2 a2 2 y 1
2
⟨x ⟩=
( nπ ) 3
y
[ ( nπ y 1
− sin 2 y |0 −2∫ y − sin 2 y dy
2 4
3
0
2 4
nπ
)
nπ
( )]
2 a2 ( nπ ) y2
2
⟨ x ⟩=
( nπ )3 2[ −2∫
0
2
dy+ ∫ y sin 2 y dy
0
]
1
u= y ; du=1 ; dv=sin 2 y ; v=− cos2 y
Misal: 2
3 3 nπ
2
⟨ x ⟩=
( nπ ) 3 2 [
2 a2 ( nπ ) ( nπ ) 1
−
3 2
1
+ − y cos 2 y|nπ
2
1
2 0 (
0 − ∫ cos 2 y dy )]
 2 a3 ( nπ )3 1
⟨x ⟩=2
( nπ ) 6
3 [
− y cos2 y|nπ
4
1 nπ
0 + sin 2 y|0
4 ]
3
2 a3 ( nπ ) nπ
⟨ x 2 ⟩=
( nπ )3 6
− (
4 )
a3 a2
⟨x 2 ⟩= −
3 2 ( nπ )2
d
⟨ p ⟩=m ⟨x ⟩
 dt
⟨x⟩=0
d
⟨ p ⟩=m ⟨0 ⟩
dt
⟨ p⟩=0
ℏ d 2

⟨ p 2 ⟩=∫ Ψ
¿
( )
i dx
Ψ dx

2 d 2Ψ
¿ 2
⟨ p ⟩=∫ Ψ (−ℏ )
dx 2
d2 nπ 2 nπ
dx 2
=−
a ( ) ( )
sin
a
x
a
2 ℏ 2 n2 π 2 nπ
2
⟨ p ⟩=
a a 2 ∫
0
sin 2
a
x ( )
nπ a
u= x ;dx= du
Misal, a nπ
nπ
2 2 ℏ2 n2 π 2 a
⟨ p ⟩= 3 ∫ sin2 du
a nπ 0
n
2 2 n
  sin
2 2
p du
a2 0
nπ
2 ℏ2 nπ u 1
2
⟨ p ⟩= 2
a [ − sin 2 u
2 4 0
]
2 ℏ2 nπ nπ
⟨ p 2 ⟩= 2
a
2 2 2
( ) 2
ℏ n π
⟨ p 2 ⟩=
a2


σ x =√⟨ x 2 ⟩−⟨ x⟩2
a2 a2 2
σ x=
√( −
3 2 ( nπ ) 2
−
a
2 )()
 a2 1 2
σ x=
√ −
4 3 ( nπ )2
a 1 2
( )
σ x=
√
−
2 3 ( nπ )2
σ p = √⟨ p 2 ⟩−⟨ p⟩2
ℏ 2 n2 π 2
σ p=

σ p=
√
ℏ nπ
a 2
−( 0 )
2

a
a 1 2 ℏ nπ

σ x σ p=
√
−
2 3 ( nπ )2 a
2
ℏ ( nπ )
σ x σ p=

Hasil dari
2 3 √ −2
σ x σ p adalah paling kecil untuk n = 1, sehingga
2
σ x σ p=
2 3 √
ℏ ( nπ )
−2
ℏ ℏ
σ x σ p =( 1 .136 ) >
2 2

Problem 3.7 Although the overall phase constant of the wave function is of no physical
significance (it cancels out whenever you calculate a measureable quantity), the relative
phase of the expansion coefficients in Equation 3.14 does matter. For example, suppose we

change the relative phase of Ψ 1 and Ψ 2 in Problem 3.6:

Penyelesaian:

2 nπ
Ψ n=
1
√ a
sin( )
a
x

A=
√2
Dimana,
E1 ∧E2
E2 =ΨE1
E1
ℏ =ω 1
E2
ℏ =ω 2
 1 π −iω1 t 2π
√a ( a
x e e )
iφ −iω1 t
Ψ ( x , t )= (
sin x ) e +sin ( a )
1 π 2π
Ψ ( x , t )= e
√a (
−iω1 t
sin ( x )+sin (
a a ))
x e e
iφ −3 iω1 t

1 π 2π
2
a (
|Ψ ( x , t )| = e e sin ( x )+sin ( x ) e e e
−iωt iωt 2
a a ) 2
e iφ −iφ −3iωt 3iωt
+sin (aπ x) sin (2a π x) e
−iφ 3 iωt
e

π 2π
+sin ( x ) sin ( x ) e e iφ −3 iωt
a a
1
2
(a πa 2a π ) πa 2a π
|Ψ ( x , t )| = sin ( x )+sin ( x ) +sin ( x ) sin ( x ) ( e
2 2 −iφ 3 iωt
e +e iφ e−3iωt )

1 2 π 2 2π π 2π
2
|Ψ ( x , t )| =
a( ( ) ( )) ( ) ( )
sin
a
x +sin
a
x +sin x sin
a a
x (e
i ( 3iωt −φ) iφ −3iωt
+e e )
e iφ +e−iφ=cos φ−i sin φ+cos φ+isin φ=2 cos φ
1 π 2π π 2π
|Ψ ( x , t )|2 =
a ( ( ) ( )) ( ) ( )
sin2 x + sin2
a a
x +sin x sin
a a
x 2 cos ( 3 ωt−φ )

⟨x⟩=∫ x|Ψ ( x,t )|2 dx

a
1 π
⟨x ⟩= ∫ sin2 x dx
a0 a ( )
π a
u= x ; dx = du ; u1 =0 ; u 2=π
Misal, a π
π
a
⟨x 1 ⟩= 2 ∫ y sin 2 y dy
π 0
y 1
u= y ; du=1 ; dv=sin2 y ; v = − sin 2 y
Misal, 2 4
π
a π2
⟨x 1 ⟩= 2 ∫ [
y 1
(
− sin 2 y dy
π 2 0 2 4 ) ]
a π2 π 2 1
⟨x 1 ⟩= 2 [
− + ( cos ( 2 π ) −cos ( 0 ) )
π 2 4 8 ]
a a
⟨x 1 ⟩= ; ⟨ x 2 ⟩=
4 4
 a
2
⟨x 3 ⟩= ∫ sin πa x sin 2a π x cos ( 3 ωt−φ ) dx
a 0 ( ) ( )
π 3π π 3π
⟨x ⟩=
3
2 cos ( 3 ωt−φ )
2a π
a
cos ( 3 ωt−φ ) a 2
a
−
[
x sin ( x ) x sin (

3π
a
a
x ) cos ( x ) cos (
+
(
a2
π
a )
a
−
( 3π
a )
a
2
x)
2
]
⟨x 3 ⟩=
a π2 (
cos ( π ) −cos ( 0 ) − 2 cos ( 3 π )−cos ( 0 )
9π )
a cos ( 3 ωt−φ ) 9 1
⟨x 3 ⟩= 2
π
− +
9 9 ( )
16 a cos ( 3 ωt−φ )
⟨x 3 ⟩=− 2
9π
a a 16 a cos ( 3 ωt−φ )
⟨x ⟩= + + 2
4 4 9π
a 16 a cos ( 3 ωt−φ )
⟨x ⟩= + 2
2 9π
Jika ,
π a 16 a
φ= ⇒ ⟨x ⟩= + 2 sin ( 3 ωt )
2 2 9π
a 16 a
φ=π ⇒ ⟨ x ⟩= − 2 cos (3 ωt )
2 9π

Problem 3.9 Find Ψ ( x,t ) for the initial wave function in Problem 3.8. Evaluate c1, c2, and
c3 numerically, to five decimal places, and comment on these numbers. (cn tells you, roughly

speaking, how much

Ψ n is "contained in" Ψ .) Suppose you measured the energy at time
to > 0, and got the value E3. Knowing that immediate repetition of the measurement must
return the same value, what can you say about the coefficients cn after the measurement?
(This is an example of the "collapse of the wave function", which we discussed briefly in
Chapter 2.)
Penyelesaian:
30
Ψ ( x , 0) = √ x ( a−x )
√a 5
2 nπ
Ψ n=
√ a
a
sin
a( )
x

2 nπ
c n=
a √∫ 0
sin ( x )Ψ ( x , 0 ) dx
a
 b
2 nπ √ 30 ( ax−x )
√ a a (
c n = ∫ sin
a )
x
a
√a 5
2

a [ ( a ) ∫ ( a )]
c =
√60 ax sin nπ x − x sin nπ x 2
n ∫
3
0
nπ ua
u= x ; dx= du ; u1 =0 ; u2 =nπ
Misal, a nπ
x 1
∫ sin ( bx ) dx=− b cos bx + b2 sin bx
∫ x 2 sin ( x )=( 2−u 2) cosu+2 u sin ( u )
a
√ 60
a
x2
c n = 3 a − cos
[[
nπ
x +
a2

2
nπ2
sin
nπ
x −
a3
( a ) ( nπ ) ( a )] ( nπ )
3
[ ( 2−u 2

3
) cos u+2 u sin ( u ) ]
0
π
0
]
a [ ( nπ ]
√60 a a cos ( nπ ) − a [ ( 2−n π ) cos ( nπ )−( 2−0 ) ]
c =
n 3 ) nπ 3 3
2 2

60 a 3 2 a3 2 a3 a3 2 a3
cn= √ 3
a nπ [
cos ( nπ )− 3 3 +
n π ( nπ )3
cos ( nπ ) +
nπ
cos ( nπ ) +
n3 π 3 ]
60
c n = √3 3 2 ( 1−cos ( nπ ) )
n π
2 15
c 1= √ 3 2 ( 1−(−1 ) ) ⇒c 1 =0 , 99925
π
c 2 =0
4 15
c 3 = √ 3 ( 1−(−1 ) ) ⇒ c 3=0 , 05701
27 π

Problem 3.11 Show that the lowering operator cannot generate a state of infinite norm (i.e.,

∫|a− Ψ|2 dx<∞ if Ψ itself is a normalized solution to the Schrodinger equation). What

does this tell you in the case

Ψ=Ψ 0 ? Hint: Use integration by parts to show that
∞ ∞
∫−∞ (a− Ψ ) (a− Ψ ) dx=−∞∫ Ψ ¿ (a+ a− Ψ ) dx
¿

Then invoke the Schrodinger equation (Equation 3.46) to obtain

∞
1
∫−∞ |a−Ψ|2 dx=E− 2 ℏ ω
Where E is the energy of the state Ψ .
Penyelesaian:
 1 ℏ d
a± =
√2 m i dx
±im ωx( )
Ψ 0 =ground state
a− Ψ 0 =0
∫∞|a− Ψ 0|2 dx
∫−∞ |Ψ 0|2 dx=1
∞ ∞
2
∫ |a+ Ψ n| dx=−∞∫ |Ψ n+1|dx=1
−∞
∞ ∞
2
∫ |a− Ψ n| dx=−∞∫ |Ψ n−1|dx=1
−∞
∞ ∞
2
∫−∞ |Ψ n−1| dx=−∞∫ ( a− Ψ n )¿ (a− Ψ n)
∞
1 ℏ d
−∞ √ 2 m ( i dx
∫ − +im ωx ) Ψ n ( a− Ψ n ) dx ¿

∞
1 ℏ dΨ n
∫
−∞ √ 2 m
−
i dx ( +im ωxΨ n ( a− Ψ n ) dx
¿

¿
)
∞
1 ℏ dΨ n
∫
−∞ √ 2 m
− (
i dx − n
¿

( a Ψ ) +im ωxΨ n ( a− Ψ n) dx ¿
)
∞
1 ℏ dΨ n ∞
1
∫
−∞ √ 2 m
−
i dx ( ( a− Ψ n ) + ∫
−∞ √ 2 m
¿

)
im ωxΨ n ( a− Ψ n ) dx ¿

d 1 ℏ dΨ n 1 ℏ ¿

Misal,
u=( a− Ψ n ) ; du= ( a− Ψ n ) ; dv=
dx
−
√2 m i dx
; v=
√2 m i
− Ψn ( ) ¿

1 ∞
1 ℏ dΨ n ∞
1
=
√ 2m 2
ℏ
( ) ∞
− Ψ n ( a− Ψ n )|−∞− ∫ ¿

−∞ √ 2 m
−
i dx
( a− Ψ n ) dx− ∫
mωx 2
−∞ √ 2m
(
im ωxΨ n ( a− Ψ n) dx
¿

) ¿

−i 2
2ℏ
Ψ n = An a n e ⇒ Ψ n ( x ) ¿ e− ±∞ =e−∞ =0
( )

Dimana, + x→±n

∞
1 ℏ dΨ n
=∫
∞
−∞ √ 2 m i dx
( ¿

( a− Ψ n ) +im ωxΨ n ( a− Ψ n ) dx ¿
)
1 ℏ d
∫
−∞ √ 2 m
Ψn ¿
((
−
i dx
+im ωx ( a− Ψ n ) dx ) )
Dimana,
1 ℏ d
a+ =
∞
−
√ 2 m i dx (
+im ωx )
= ∫ Ψ n a+ a− Ψ n dx
¿

−∞
Dimana,
(a a + 12 ℏ ω)Ψ =E Ψ ⇒ a a Ψ =E Ψ − 12 ℏ ωΨ
+ − n n n + − n n n n

∞
1
−∞ 2
¿
(
= ∫ Ψ n E n − ℏ ω Ψ n dx )
Dimana,
En = n+( 12 ) ℏ ω ⇒ E =n ℏ ω+ 12 ℏ ω n

∞
1 1
−∞
∞
2 2 (
= ∫ Ψ n n ℏ ω+ ℏ ω− ℏ ω Ψ n dx
¿
)
1 1
−∞ 2 2 (
= ∫ Ψ n n ℏ ω+ ℏ ω− ℏ ω Ψ n dx
¿

∞
)
1

∞
2 (
= En − ℏ ω )−∞ dx
∫ |Ψ n|2

1
∫−∞ |a− Ψ n|2 dx=En− 2 ℏω
1
∫|Ψ|2 dx=E n− 2 ℏ ω<∞

Problem 3.13 Using the methods and results of this section,

a) Normalize Ψ1 (Equation 3.51) by direct integration. Check your answer against the
general formula (Equation 3.54).

b) Find Ψ 2 , but don't bother to normalize it.

c) Sketch
Ψ 0 , Ψ 1 , and Ψ 2 .

d) Check the orthogonality of

Ψ 0 , Ψ 1 , and Ψ 2 . Note: If you exploit the evenness
and oddness of the functions, there is really only one integral left to evaluate explicitly.
Penyelesaian:
∞ ∞ −mω 2
2 x
a) ∫ |ψ 1| dx=− A 12 2 mω 2 ∫ x 2 e ℏ
dx=1
−∞ −∞

∞ −mω 2
x
2 2 2 ℏ
− A1 2m ω ∫x e dx=1
−∞

∞ −mω 2 −mω 2
x x
2 ℏ 2 ℏ
∫x e dx=z , u=x , dv=e dx
−∞

∞ −mω 2 ∞ −mω 2 −mω 2

x πℏ x x
∫x
−∞
2
e ℏ
dx=z= x 2

√ mω
−2 ∫ x e
−∞
ℏ
dx , u=x , dv=e ℏ
dx
 ∞ −mω 2
x
2

√ πℏ
mω
−2 x
πℏ
−∫ e
mω −∞ [√ ℏ
x

] √
dx = x
2 πℏ
mω [√ √ ]
−2 x
πℏ
mω
−
πℏ
mω
=z

πℏ 2
√ mω
[ x −2 x−2 ]=z
πℏ 2
− A12 2m ω2
√ mω
∞ ∞
[ x −2 x−2 ]−∞=1 , [ x 2−2 x−2 ]−∞=1
mω −1
A 1 2=
√ πℏ 2m ω2
mω i
A1=
√
4

πℏ √ 2m ω2
b) ψ 2= A2 ¿ ¿

c)

d) ∫ ψ 0 ψ 1 dx=∫ ψ 2 ψ 1 dx=0
¿ ¿

−mωx2
∞ 2 mω 2
∫ ψ 2 ψ 0 dx=( A 2 ℏ ω)( A 0 )∫−∞ (
¿ 1−
ℏ )
x e ℏ
dx
−mωx2 −mωx2
= A0 A2 ℏ ω {∫ ∞

−∞
e ℏ 2 mω ∞ 2
dx −ℏ ∫−∞ x e ℏ
dx }
π ℏ 2 mω ℏ πℏ
= A0 A2 ℏ ω {√ mω
−ℏ
2 mω mω
=0
√ }
Problem 3.15 In the ground state of the harmonic oscillator, what is the probability (correct
to three significant digits) of finding the particle outside the classically allowed region? Hint:
Look in a math table under "Normal Distribution" or "Error Function".
Penyelesaian:
Energi untuk kondisi awal yaitu
 1
E0 = ℏ ω
2
Energi klasik untuk getran harmonic adalah
1
E= mω2 x 2
2 0
Atau
1 1
E= ℏ ω= mω2 x 2
2 2 0

2E
x 0=±
ℏ
mω
=±
√mω2
1
√
x 0=±
mω
√ α
α= ℏ
Dimana
1
√ α
Pin = ∫ |Ψ 0|2 dx
− 1
√ α
mω 2
1 x
mω −

Dimana,
Ψ 0=
πℏ ( ) 4 e 2ℏ

mω
1 √ α
−
mω 2
x
Pin =
πℏ( ) 2
∫
1
e 2ℏ
dx

1
−
√ α

α
√ α

√
2
Pin = ∫ e−αx dx
π 1
−
√ α

Misal, √ α x= y
1
1 2
Pin = ∫ e− y dy
√ π −1
1 √π
Pin = [ erf ( 1 )−erf (−1 ) ]
π √4
1
Pin = ( 2−erf ( 1 ) )
2
Pin =0 .8427
Pout =1−Pin
Pout =0 .157

Problem 3.17 A particle in the harmonic oscillator potential has the initial wave function
Ψ ( x , 0 ) =A [ Ψ 0 ( x ) +Ψ 1 ( x ) ]

For some constant A.

a) Normalize Ψ ( x,0 )
2
b) Find Ψ ( x,t ) and |Ψ ( x,t )|
c) Find the expectation value of x as a function of time. Notice that it oscillates sinusoidally.
What is the amplitude of the oscillation? What is its (angular) frequency?

d) Use your result in (c) to determine ⟨p⟩ . Check that Ehrenfest’s theorem holds for this
wave function.

e) Referring to Figure 3.5, sketch the graph of |Ψ| at t = 0, 𝜋/𝜔, 2𝜋/𝜔, 3𝜋/𝜔, and 4𝜋/𝜔.
(Your graphs don't have to be fancy-just a rough picture to show the oscillation.)
Penyelesaian:

a) Normalize Ψ ( x,0 )
Ψ ( x , 0 ) =A [ Ψ 0 ( x ) +Ψ 1 ( x ) ]
1=|A|2∫ ( Ψ 0 +Ψ 1 ) ( Ψ 0 +Ψ 1) dx
¿ ¿

1=|A|2 ∫|Ψ 0|2 dx+∫ Ψ 0 Ψ 1 dx+∫ Ψ 1 Ψ 0 dx +∫|Ψ 1|2 dx

[ ¿ ¿
]
2
1=|A| ( 1+0+0+1 )
=2|A|2
1
A=
√2
−iE 0 t −iE 1 t

b)
Ψ ( x , t )=
1
Ψ e
√2 0
( ℏ
+Ψ1e ℏ
)
1 3
E0 = ℏ ω E1 = ℏ ω
Dimana, 2 dan 2
−i ℏ ωt −3i ℏ ωt
Ψ ( x , t )=
1
(Ψ e
√2 0
2ℏ
+Ψ1e 2ℏ
)
−iωt −3 iωt
Ψ ( x , t )= ( Ψ e )
1 2 2
0 +Ψ 1 e
√2
−iωt
1 2
Ψ ( x , t )=
√2
e (Ψ 0 +Ψ 1 e−iωt )
mω
Jika,
ξ=
√ 1
ℏ
x
−ξ 2
mω
Ψ 0=
πℏ ( ) 4e 2

Dan,
1 −ξ 2
mω 2 1 2 ξe
Ψ 0=
πℏ( ) √2
2

Jadi,
 −iωt 1 −ξ 2
1 mω
Ψ ( x , t )= e
√2
2
( )
πℏ
4e 2
( 1+ √ 2 ξe−iωt )
Apabila,
2
|1+ √ 2ξe−iωt| =( 1+ √2ξe+iωt )( 1+ √ 2ξe−iωt )
2
|1+ √ 2ξe−iωt| =1+√ 2ξ ( e +iωt +e−iωt ) +2ξ 2
2
|1+√ 2ξe−iωt| =1+2ξ 2+2 √ 2ξ cosωt
Jadi,
2
−ξ 2

[ ]
−iωt 1
1 mω 4
Ψ (x,t ) =
2
√2
e 2
πℏ ( )
e 2 ( 1+ √ 2 ξe )
−iωt

1 mω −ξ2
Ψ ( x , t )2=

⟨x⟩=∫ x|Ψ| dx
2 πℏ
2
√
e ( 1+2 ξ2 +2 √ 2 ξ cos ωt )

c)
1 mω ℏ ∞
⟨x ⟩=
√ ∫
2 π ℏ mω −∞
1 ℏ
ξ ( 1+2ξ 2 +2 √ 2 ξ cos ωt )
∞
2
⟨x⟩=
2 πmω
2ℏ
√
2 √ 2cos ωt ∫ ξ 2 e−ξ dξ
√π
−∞

⟨x⟩=
√
πmω
ℏ
cos ωt
2 ( )
⟨x⟩=
√
2mω
cos ( ωt )
ℏ
Dimana, amplitudo
=
√ 2mω dan frekuensi anguler =ω

d ⟨ x⟩
⟨ p⟩=m
d) dt

⟨ p ⟩=m
d (√ 2 mωℏ cos ( ωt ))
dt
ℏ
⟨ p⟩=m
√
ℏ mω
2mω
(−ωsin ωt )

⟨ p⟩=−m
d⟨ p⟩
2
km ω
√
sin ωt

dt
=−
1
2 √
ωcos ( ωt )

V = mω2 x 2
Misal, 2
 dV
=
d ( 12 mω x )
2 2

dx dx
dV
=mω2 x
dx
dV
⟨− ⟩=−mω2 ⟨ x ⟩
dx
dV ℏ
⟨−
dV
dx
⟩=−mω2
ℏ mω
2 mω √
cos ( ωt )

⟨−
dx
ℏ mω
⟩=−
√
2
ωcos ( ωt )
d⟨ p⟩
−
√ 2
ωcos ( ωt )

1
= dt
−iωt
2
Ψ ( x , t )= e
√2
(Ψ 0 +Ψ 1 e−iωt )
e)
Pada t = 0
1
Ψ ( x , t )= ( Ψ +Ψ )
√2 0 1
π
t=
Pada saat, ω
1
Ψ ( x , t )= ( Ψ 0 −Ψ 1 )
√2
2π
t=
Pada saat, ω
1
Ψ ( x , t )= ( Ψ 0 +Ψ 1 )
√2
3π
t=
Pada saat, ω
1
Ψ ( x , t )= ( Ψ 0 −Ψ 1 )
√2
4π
t=
Pada saat, ω
1
Ψ ( x , t )= ( Ψ 0 +Ψ 1 )
√2
Dan seterusnya. Grafiknya adalah
 2π 4π π 3π
t=0 , , t= ,
Untuk , ω ω untuk ω ω

Problem 3.19 Show that the expressions [ Ae ikx+Be−ikx ] , [ C cos kx+ Dsin kx ] ,

[ F cos ( kx+a ) ] and [ Gsin ( kx+β ) ] are equivalent ways of writing the same function of
x, and determine the constants C, D, F, G, α, and β in terms of A and B. (In quantum
mechanics, with V = 0, the exponentials give rise to traveling waves, and are most convenient
in discussing the free particle, whereas sines and cosines correspond to standing waves,
which arise naturally in the case of the infinite square well.) Assume the function is real.
Penyelesaian:
ikx −ikx
Ψ ( x ) =Ae +Be
ikx
Seperti yang kita ketahui bahwa e =cos kx +i sin kx
ikx −ikx
Ae +Be =A ( coskx +i sin kx )+B ( cos kx−isin kx )
( A +B ) coskx+i ( A−B ) sin kx=C cos kx+D sin kx
Dengan,
C=A +B ; D=i ( A−B )
¿
Dimana untuk a fungsi real maka B= A
i ( kx +α ) −i ( kx+α )
F[e +e ]
F cos ( kx+ α )=
2

F cos ( kx+α )= ( F2 e ) e +( F2 e ) e
iα ikx −iα −ikx

ikx −ikx
F cos ( kx+ α )= Ae + Be

Jadi,
( F2 e )= A ; ( F2 e )=B
iα −iα

F 2 F 2
2
|A| +|B| =
2
2
+
2 ( ) ( )
F2 F2
2
|A| +|B| = 2
4
+
4 ( )( )
F2
|A|2 +|B|2 =
2
F2 =2 (| A|2+|B|2 )
F= 2 (|A|2+|B|2)
√
F=2|A|
 A 2iα
=e
B
( A B)
Dimana,
1
α = tan −1
2
[ ]dn

Re (
A
B)

i ( kx+ β ) −i ( kx +β )
G [e +e ]
G sin ( kx+ β ) =
2i
G iβ ikx G −iβ −ikx
G sin ( kx+ β ) = ( ) ( )
2
e e + e
2
e

G G
Jadi,
( 2 ) ( 2 e )=B
e iβ
= A ; −iβ

G 2 G 2
2
|A| +|B| =
2
2
+
2 ( ) ( )
G2 G2
2
|A| +|B| = 2
4
+
4 ( )( )
G2
|A|2 +|B|2 =
2
G2 =2 (| A|2 +|B|2 )
G= 2 (| A|2 +|B|2 )
√
g=2|A|
A 2iβ
=e
B
( A B) − ℏ
Dimana,
1
β= tan −1
2
[ ]dn

Re (
A
B)
2

Problem 3.21 Suppose a free particle, which is initially localized in the range -a < x < a, is
released at time t = 0:

Ψ ( x,0)=¿ { A, if −a<x<a¿¿¿¿
Where A and a are positive real constants.
a) Determine A, by normalizing Ψ

b) Determine φ ( k ) (Equation 2.86)

c) Comment on the behavior of φ ( k ) for very small and very large values of a. How does
this relate to the uncertainty principle?
Penyelesaian:
a) to find the constants A we use the normalizing condition, where:
∞ 2 2 ∞ −2a|x|
1=∫−∞ |ψ(x,0)| dx=|A| ∫−∞ e dx
2 ∞ −2a|x|
but the integrad is an even function over a symmetric region, so write,
1=2|A| −∞ e dx
∫
∞
e−2 a|x|
=2|A|2 [ −2 a ] 0

2
|A|
=
a

And therefore, A= √a

b) to find φ( k) we use equation (2.86)

1 +∞
φ( k)= ∫ψ( x,0)e−ikx dx
√ 2π −∞
A +∞ −a|x| −ikx
= ∫ e e dx
√2 π −∞
A +∞
=
√ π
2 ∫−∞ e−a|x|( cos kx−t sin kx ) dx
+∞ +∞
= √2Aπ ∫−∞ e−a|x|cos kxdx−t 2Aπ ∫−∞ sin kxdx
+∞
= √2Aπ ∫−∞ e−a|x|cos kxdx−0

A +∞ −a|x| eikx +e−ikx

= √2 π ∫−∞ e ( 2 )dx
+∞
= √2Aπ ∫−∞ e−ax ( e ikx +e−ikx ) dx

A +∞ ( ik−a ) x − ( ik+a ) x
= √2 π ∫−∞ ( e +e ) dx
−( ik +a ) x +∞
e
= √2Aπ [ e
( ik−a ) x

ik −a
+
−ik + a ] 0

= √2Aπ −ik−a+ik −a
[ −ik 2 −a2 ]
 = √2Aπ 2a
[ k2 +a2 ]
By substituting the value of A, we have

2a 3 1
φ( k )=
√ [
π 2
k +a
2
]
c) For smaal a the starting wave function (i.e., ψ( x,0) ) will falls off very slowly (seems

1
2 a3

to be flat) and k will dominate the denominator of φ( k) , so

φ( k ) ~
√ π k2 wich is
sharp and narrow, this will mean a well-defined momentum and ill-defined position.

For large a ψ( x,0) will fall off rapidly (which will form a sharp and narrow spike).

2
Meanwhile as a >> k, so
φ( k )= √ πa which is broad and flat, and that would mean a
well-defined position and ill-defined momentum.

Problem 3.23 Evaluate the following integrals:

+1

∫ ( x 3−3 x 2+2 x −1 ) δ ( x+2 ) dx

a) −3
∞
∫ [ cos ( 3 x ) +2 ] δ ( x−π ) dx
b) 0
+1

∫ exp (|x|+3 ) δ ( x−2 ) dx

c) −1

Penyelesaian:
Fungsi delta Dirac ditentukan oleh dua kondisi yaitu.

δ ( x )=¿ {0 , if x≠0¿¿¿¿
Dengan,
+∞
∫ δ ( x ) dx=1
−∞
Untuk x = 0, maka
∞
∫ f ( x ) δ ( x ) dx=f ( 0 )
−∞
Untuk fungsi biasa f(x), maka,
b

∫ f ( x ) δ ( x−k ) dx=¿ {f ( k ) a<k <b ¿ ¿ ¿ ¿ ¿

a
Dengan mensubstitusikan y = x – k dan dx = dy, maka didapatkan
 b b−k

∫ f ( x ) δ ( x−k ) dx= ∫ f ( y+k ) δ ( y ) dx

a a−k

Integral ini adalah f ( y=0 )=f ( k ) untuk batas integral y = 0 adalah –k<0<b-k atau a<k<b
1 1

∫ f ( x ) δ ( x +2 ) dx=∫ ( x3−3 x 2 +2 x−1 ) δ ( x +2 ) dx

a) −3 −3

Dimana, δ ( x +2 ) ⇒ x=−2 atau f (−2 )

1

∫ ( x 3−3 x 2+2 x −1 ) δ ( x+2 ) dx

−3
=(−2 )3 −3 (−2 )2 +2 (−2 ) −1
=8−12−4−1
=−25
∞ ∞
∫ f ( x ) δ ( x−π ) dx=∫ [ cos ( 3 x ) +2 ] δ ( x−π ) dx
b) 0 0

Dimana, δ ( x−π ) ⇒ x=π atau f ( π )

∞
∫ [ cos ( 3 x ) +2 ] δ ( x−π ) dx
0
=[ cos (3 π ) +2 ]
=−1+2
=1
1

∫ e|x|+3 δ ( x−2 ) dx
c) −1

Dimana, δ ( x−2 ) ⇒ x=2 atau f ( 2 )

Karena x = 2 tidak termasuk batas integralnya maka,
1

∫ e|x|+3 δ ( x−2 ) dx=0

−1

Problem 3.25 What is the Fourier transform of 8 (x)? Using Plancherel's theorem, show that
+∞
1 ikx
δ ( x )= ∫ e dk
2π −∞
Comment: This formula gives any respectable mathematician apoplexy. Although the integral
is clearly infinite when x = 0, it doesn't converge (to zero or anything else) when x ≠ 0, since
the integrand oscillates forever. There are ways to patch it up (for instance, you can integrate
from –L to +L, and interpret the integral in Equation 3.126 to mean the average value of the
finite integral, as L ∞). The source of the problem is that the delta function doesn't meet
the requirement (square integrability) for Plancherel's theorem. In spite of this, Equation
3.126 can be extremely useful, if handled with care.
Penyelesaian:
1 ∞
f ( x )= ∫ F ( k ) e ikx dk
√ 2π −∞
1 ∞ 1
F ( k )= ∫ f ( x ) e−ikx dx=
√2π −∞ √ 2π
Apabila, f ( x )=δ ( x )
1 ∞ 1 ikx
f ( x )=δ ( x ) = ∫ e dk
√ 2 π −∞ √2 π
∞
1
f ( x )=δ ( x ) = ∫ eikx dk
2 π −∞

Problem 3.27 Find the transmission coefficient for the potential in Problem 3.26.
Penyelesaian:

ikx −ikx ikx −ikx

Ψ=¿{Ae +Be (a<−a)¿ {Ce +De (−a<x<a) ¿ ¿ impose boundary conditions
−ika ika −ika ika
i. Kesianambungan pada −a : Ae +Be =Ce +De atau βA+ B=βC+D
−2ika
Dengan, β=e
ika −ika ika
ii. Kesianambungan pada +a : Ce +De =Fe atau F=C+βD
2 mα
'
ik ( Ce−ika −Deika ) −ik ( Ae−ika −Be ika )=− 2
( Ae−ika + Beika )
iii. Untuk Ψ pada −a : ℏ
atau
βC−D=β ( γ+1 ) A+B ( γ−1 )
i 2 mα
γ≡
Dengan, ℏ2 k
2 mα
'
ikFe ika−ik ( Ce ika−De −ika ) =− 2
( Feika )
iv. Untuk Ψ pada +a : ℏ atau

C−βD=( 1−γ ) F
Untuk memecahkan nilai C dan D maka,
 Menjumlahkan (ii) dan (iv)
C+βD=F
 C−βD=( γ−1 ) F
+
2C =F + ( 1−γ ) F
2C= (2−γ ) F
 Mengurangi (ii) dan (iv)
C+βD=F
C−βD=( γ−1 ) F
-
2D =F−( 1−γ ) F

2 D= (γ β)F
 Menjumlahkan (i) dan (iii)
βC+ D=βA+B
βC−D=β ( γ+1 ) A+B ( γ−1 )
+
2 βC =βA+B+β ( γ+1 ) A+B ( γ−1 )

2C= ( γ +2 ) A+ (γ β ) B
 Mengurangi (i) dan (iii)
βC+ D=βA+B
βC−D=β ( γ+1 ) A+B ( γ−1 )
-
2D =βA+B−β ( γ+1 ) A−B ( γ −1 )
2 D=−γβ A+ ( 2−γ ) B
Menyamakan dua hasil untuk 2C
2C ii ∧ iv =2 Ci ∧ iii

( 2−γ ) F=( γ + 2 ) A+ ( γ β) B
Menyamakan dua hasil untuk 2D
2 Dii ∧ iv=2 Di ∧ iii

(γ β ) F=−γβ A+( 2−γ ) B

Kalikan dua hasil untuk 2C dengan β ( 2−γ ) dan kalikan dua hasil untuk 2D dengan γ
kemudian kurangi.

β ( 2−γ )2 F= β (−γ 2 ) A+ γ ( 2−γ ) B

 2
γ
( β ) F=−βγ A +γ ( 2−γ ) B 2

-
2
γ
[ β ( 2−γ )2 −
β ] F=β [ 4−γ 2 +γ 2 ] A

γ2
[ β ( 2−γ ) −
2
β ] F=4 βA

F 4
=
A 2
( 2−γ )2− γ
β
i ℏ2 k
g≡ = ; φ=4 ka
Dengan, γ 2 mα
i
γ= , β 2=e−iφ
Jadi, g
2
F 4g
=
A ( 2 g−i )2 +eiφ

( 2 g−i )2 +e iφ=4 g2 −4 ig−1+cos φ+i sin φ

( 2 g−i )2 +e iφ=( 4 g2 −1+cos φ ) +i ( sin φ−4 g )
2
|( 2g−i )2 +e iφ|2 =( 4 g2−1+cosφ ) +i ( sin φ−4 g )2
|( 2g−i )2+e iφ|2=16 g 4+1+cos 2 φ−8g 2−2cos φ+8 g2 cos φ+sin2 φ−8 gsinφ
|( 2g−i )2 +e iφ|2 =16 g 4 +8 g2 +2+( 8 g2 −2 ) cosφ−8 gsinφ
|( 2g−i )2+e iφ|2=8 g 4+4 g2+1+( 8g 2−1 ) cos φ−4 gsin φ
F2
T =| |
A
4 g2 2
T =| |
( 2 g−i )2 +e iφ
8 g4
T=
( 8 g 4 + 4 g2 +1 ) + ( 4 g2 −1 ) cos φ−4 g sin φ
ℏ2 k
g≡ ; φ=4 ka
Dengan, 2 mα
R=1−T
Problem 3.29 Normalize Ψ ( x ) in Equation (3.133) to determine the constants D and F.
Penyelesaian:
∞ ∞
∫−∞ |Ψ| dx=2∫|Ψ|2 dx=1
2

0
a ∞

{
=2 |D| ∫ cos lx dx +|F| ∫ e−2 kx dx
2

0
2 2

0
}
a ∞
=2 |D|{ [ 2 x 1
2 4l 0
2 1 −2 kx
+ sin2 lx +|F| − e
2k ] [ ]}
a
−2kx
a sin 2 la 2e
{ (
=2 |D| +
2 4l
2
+|F|
2k ) }
−ka ka
Tapi, Fe =D cos ( la ) ⇒ F=De cos ( la )
Jadi,
sin ( 2 la ) cos2 la
1=|D|2 a+
2l [ +
k ]
Dimana, k =l tan ( la)
2 sin la cos3 la
1=|D| a+
2l
2
+[l sinla ]
cos la
1=|D|2 a+ [
l sinla
( sin2 la+ cos2 la ) ]
1
1=|D|2 a+
[
l tan la ]
1
1=|D|2 a+
k ( )
1 eka cos la
D= F=
1 1
√ a+
k dan √ a+
k

Problem 3.31 Derive Equations 3.149 and 3.150. Hint: Use Equations (3.147) and (3.148) to
solve for C and D in terms of F:
k k
[ l ] l [ ]
C= sin ( la ) +i cos ( la ) e ika F ; D= cos ( la )−i sin ( la ) e ika F

Plug these back into Equations (3.145) and (3.146). Obtain the transmission coefficient, and
confirm Equation (3.151). Work out the reflection coefficient, and check that T + R = 1.
Penyelesaian:
ika
C sin ( la )+D cos (la )=Fe
 ika
l ( C cos ( la )−D sin (la ) ) =ikFe
ik ika
C cos ( la )−D sin ( la )= Fe
l
ika
Persamaan C sin ( la )+D cos (la )=Fe dikalikan dengan sin ( la ) dan persamaan

ik ika
C cos ( la )−D sin ( la )= Fe
l dikalikan dengan cos ( la ) . Sehingaa,
2 ika
C sin ( la ) +D cos ( la ) sin ( la )=F sin ( la ) e
ik
C cos 2 ( la ) −Dcos ( la ) sin ( la ) = F cos (la ) e ika
l
Dengan menjumahkan persamaan diatas maka didapatkan,
ik
(
C ( sin 2 ( la ) +cos 2 ( la )) =F sin ( la ) +
l )
cos ( la ) e ika

ik
C=F ( sin (la ) + cos ( la ) ) e ika
l
ika
Persamaan C sin ( la )+D cos (la )=Fe dikalikan dengan cos ( la ) dan persamaan

ik ika
C cos ( la )−D sin ( la )= Fe
l dikalikan dengan sin ( la ) . Sehingaa,
2 ika
C sin ( la ) cos ( la ) +D cos ( la )=F cos ( la ) e
ik
C cos ( la ) sin ( la )−D sin2 ( la )= F sin ( la ) e ika
l
Dengan mengurangi persamaan diatas maka didapatkan,
ik
(
D ( sin2 ( la )+cos2 ( la ) )=F cos ( la) −
l )
sin ( la ) eika

ik
D=F (cos ( la )− sin ( la ) ) e ika
l
−ika ika
Substitusikan C dan D ke persamaan Ae +Be =−C sin ( la ) +D cos ( la )
ik ik
(
Ae−ika +Beika=−F sin ( la ) +
l ) ( )
cos (la ) eika sin ( la )+ F cos (la )− sin ( la ) eika cos ( la )
l
ik ik
[
Ae−ika + Beika=F − sin ( la) + ( l ) ( )
cos ( la ) sin (la ) + cos ( la )− sin ( la ) cos ( la ) e ika
l ]
ik ik
[
Ae−ika + Beika=F −sin2 ( la )−
l ]
cos ( la ) sin ( la )+ cos2 ( la )− sin ( la ) cos ( la ) eika
l
 ik
[
Ae−ika + Beika=F cos 2 ( la )−sin2 ( la )−2
l ]
cos ( la ) sin ( la ) eika

ik
[
Ae−ika +Beika=F cos ( 2 la ) −
l ]
sin ( 2 la ) e ika

Substitusikan C dan D ke persamaan [ Ae−ika−Beika ]=l ( C cos ( la )−D sin (la ) )

[ Ae−ika−Beika ]= l (
F sin ( la )+
ik
) ( ik
)
cos ( la ) e ika cos ( la ) +F cos ( la )− sin ( la ) eika sin ( la )
ik l l

[ Ae−ika−Beika ]= l [(
F sin ( la ) +
ik
) ( ik
)
cos ( la ) cos ( la ) + cos ( la )− sin ( la ) sin ( la ) eika
]
ik l l

[ Ae−ika−Beika ]= l [
F cos ( la ) sin ( la )+
ik ik
]
cos 2 ( la ) +cos (la ) sin (la )− sin 2 ( la ) eika
ik l l

[ Ae−ika−Beika ]= l [
F 2cos (la ) sin ( la) +
ik
]
( cos2 (la )−sin2 ( la) ) eika
ik l

[ Ae−ika−Beika ]= l [
F sin ( 2la ) +
ik
cos ( 2 la ) eika
]
ik l
il
[ Ae−ika−Beika ]=F [ ]
cos ( 2la )− sin ( 2 la) e ika
k
Gunakan sifat trigonometri berikut
sin ( A +B )=sin A cos B+sin B cos A
cos ( A+B )=cos A cos B−sin A sin B
Jika, A = B, maka
sin ( 2 A )=2 sin A cos A
2 2
cos ( 2 A )=cos A−sin A
1=cos 2 A +sin 2 A
2 2
1−sin A=cos A
ik
Kemudian kurangi persamaan
[
Ae−ika +Beika=F cos ( 2 la) −
l ]
sin ( 2 la ) e ika
dan persamaan

il
[ Ae−ika−Beika ]=F [ ]
cos ( 2la )− sin ( 2 la) e ika
k
ik il
[
2 Beika =F cos ( 2la )−
l ] [
sin ( 2la ) eika −F cos ( 2la )− sin ( 2 la ) eika
k ]
 ik ik
[
2 Beika =F cos ( 2la )−
l
sin ( 2la )−cos (2 la )− sin ( 2 la ) eika
k ]
2 Beika =F ([ ilk − ikl ) sin ( 2la ) ] e ika

2 Beika
[ kl kl ]
=F i ( − )sin ( 2 la ) e ika

l 2 −k 2
2 Beika =iF [( ) lk
sin (2 la ) eika ]
l2 −k 2
B=iF
2lk( )sin ( 2 la )

ik
Kemudian jumlahkan persamaan
[
Ae−ika +Beika=F cos ( 2 la) −
l ]
sin ( 2 la ) e ika
dan

il
persamaan
[ Ae−ika−Beika ]=F [ ]
cos ( 2la )− sin ( 2 la ) e ika
k

[
2 Ae−ika =F 2 cos ( 2 la )−i ( kl + kl ) sin ( 2 la ) ] e ika

k 2 +l 2
[
2 Ae−ika =F 2 cos ( 2 la )−i ( ) kl
sin (2 la ) eika ]
− 2ika
Ae
F=
k 2 +l2
cos ( 2 la ) −i ( )
2 kl
sin ( 2 la )

−1 A2
T =| |
F
sin (2 la ) 2 2 2
−1
T =|cos ( 2la )−i
2 kl (
( k +l )| )
−1 2 sin 2 (2 la ) 2 2 2
T =cos ( 2la ) + ( k +l )
( 2 kl )2
2 2
Tapi, 1−sin 2la=cos 2la

( k 2 + l 2 )2 ( k 2 + l 2 )2
−1
T =1+sin ( 2la ) + 2
[ ( 2 kl ) 2 ]
−1 =1+
( 2 kl ) 2
sin2 ( 2 la )
 ( k 2 +l 2 ) 2 ( k 2 −l 2 ) 2

[ ( 2 kl )2
−1 =
1
( 2 kl )2
[ ]
k 4
+2 k 2 2
l +l 4
−4 k 2 2
l ] =
1
( 2 kl )2
[ k 4
−2k 2 2
l +l 4
] =
( 2 kl )2

2 mE 2m ( E+V 0 )√
k=√ ; l=
Dimana, ℏ ℏ
2a
2la= ℏ 2 m ( E+V 0 )
√
2
2
√ 2 mE − 2 m ( E+V 0 ) = 1 2 m E−E−V =− 2 mV 0
2 2
k −l = ℏ ( ℏ ℏ2
( ) (√0)
ℏ2
)
2
2mV 0 2m 2
( k 2 −l 2 )2
=
( −
ℏ2 ( ) = V ) =
ℏ2
V 2
0
02

(2 kl )2 2 mE √2 m ( E+V ) 2m 2
4 E ( E+ V ) 2

(2 ℏ ℏ ) ( ℏ )
√ 4 E ( E+V ) 0
2 0
0

V
2a
02
T −1 =1+ sin2 ℏ 2 m ( E +V 0 ) √
4 E ( E+V 0 )
2
B |B|2 |F|2 B
2
R=| | = 2 2 =T | |
A |F| |A| F
V 2
B 2 sin2 ( 2 la ) 2 2 2 0 2a
| |= ( l −k ) = sin 2 ℏ 2m ( E+V 0 ) √
F ( 2 kl ) 2 4 E ( E+V 0 )
V
02 2a
sin 2 ℏ 2 m ( E+V 0 ) √
4 E ( E+V 0 )
R=
V
02 2a
1+ sin2 ℏ 2 m ( E+V 0 ) √
4 E ( E+ V 0 )

Dimana
V
02 2a
θ= sin 2 ℏ 2 m ( E+V 0 ) √
4 E ( E+V 0 )

1 θ 1+θ
T= R= T + R= =1
1+θ ; 1+θ dan 1+θ

Problem 3.33 Consider the step function potential:

V (x )=¿ {0 , if x≤0¿¿¿¿
a) Calculate the reflection coefficient, for the case E < V0 and comment on the answer
b) Calculate the reflection coefficient for the case E < V0
c) For a potential such as this that does not go back to zero to the right of the barrier, the

2 2
transmission coefficient is not simply |F| /|A| , with A the incident amplitude ang F
the transmitted amplitude, because the transmitted wave travels at a different speed. Show
that

E−V 0 |F|2
T=
√
E |A|2
For E > V0. Hint: You can figure it out using Equation (3.81), or more elegantly, but less
informatively from the probability current (Problem 2.9a). What is T for E < V0?
d) For E > V0, calculate the transmission coefficient for the step potential, and check that
T + R =1
Penyelesaian:

2m ( V 0 −E )
a.
Ψ=¿ { Ae ikx +Be−ikx ..... ( x<0 ) ¿ } ¿{}where,l=
Ψ : A +B=F ik
√
ik
2mE
ℏ ;k=
ik
ℏ √
'
Continuity of Ψ :ik ( A−B )=−kF }
A +B=− ( A−B )⇒ A 1+ =−B 1−
k k k ( ) ( )
2
l
B
R=| | =
(21+i
k)
=
1+ ( l / k )
=1
2

A 2 2
l 1+ ( l / k )
(1−i k )
2m ( V 0 −E )
b.
Ψ=¿ { Ae ikx +Be−ikx ..... ( x<0 ) ¿ } ¿{}where,k=
Ψ : A +B=F k k
√ 2mE
ℏ
k
;l= √ℏ
'
Continuity of Ψ :ik ( A−B )=ilF }
A + B= ( A−B )⇒ A 1− =−B 1+
l l l ( ) ( )
2
k
B (21− )
l ( k−l ) 2
( k−l ) 2
2m 2m
R=| | =
A k
= 2
( k +l )
= 2
( k +l ) 2 2 2
, now , k 2 +l 2 =
ℏ
2 [ ( )
E − E +V 0 ] =
ℏ
2
V0
(1+ l)
4
2m ( √ E−√ E−V 0 )
k −l= √ E− E−V so , R=
ℏ [ √ √ 0]
V
02

c. vi vt

vidt vtdt
 2
Pt |F| V t
T= = 2
Pi |A| V i
here Pi is true probability of finding particle in incident beam in

V t √ E−V 0
=
box corresponding to time interval, and likewise for P t. But
Vi √E ( from

T=
√ E−V 0 | F |2
3.81) So √E A alternatively,
iℏ
J i= [ Acikx (−ikA ¿ e−ikc )− A ¿ e−ikc ( ikAe ikx ) ]
2m

J F 2 l F 2 E−V 0
ℏ
m
ℏ
J i= |A|2 k ; J t = |F|2 l; T = t =| | =| |
m Ji A k A E
.QED
√
For E<V 0 ,T =0

k k
F=A + A
( l
−1 )
=A
2
l
=
2k
A
k k k +l
d. For E < V0,
( l +1) ( l +1)
2
F 2l 2k 2
l 4 kl 4 kl ( k−l )2 4 √ E √ E−V 0 ( √ E−V 0 )
T =| | =
A k k +l ( ) = =
k ( k +l )2 ( k 2−l 2 )2
=
V 2
0
2 2
4 kl ( k −l ) 4 kl + k 2 −2 kl+ l 2 k 2 +2 kl+l 2 ( k + l )
T + R= + = = = =1
( k +l )2 ( k +l )2 ( k +l )2 ( k +l )2 ( k + l )2

Problem 3.35 Find the S-matrix for the finite square well (Equation 3.127). Hint: This
requires no new work if you carefully exploit the symmetry of the problem.
Penyelesaian:
F+G=A + B
F−G=( 1+2 iβ ) A−( 1−2 iβ ) B
mα
β=
Dimana, ℏ2 k

2G=−2 iβA+2 ( 1−iβ ) B

1
B= ( iβA +G )
1−iβ
Kalikan persamaan F+G=A + B dengan 1−2iβ dan jumlahkan, maka
 2 (1−iβ ) F−2iβG=2 A
1
F= ( A+iβG )
1−iβ
1 [ iβ 1 ¿ ] ¿
S= ¿ ¿
1−iβ ¿
Untuk potensial yang merata
V (−x ) =V ( x )
Hamburan dari kanan sama dengan hamburan dari kiri, dengan
x ↔−x , A ↔G , B ↔ F
F=S 11 G+S 12 A , B=S21 G+S 22 A

S 11 =S 22 , S 21=S12
−2 ika
e
S 21=
k 2+ l2
( )
cos ( 2 la )−i
2 kl
sin ( 2la )

2 2
l −k
i(
2 kl )
−2 ika
sin ( 2 la ) e
S =
21 2 2
k +l
cos ( 2 la )−i (
2 kl )
sin ( 2la )

1 [ iβ 1 ¿ ] ¿
S= ¿ ¿
1−iβ ¿

l2−k 2
S= 2 2
k +l
e−2ika
¿ i 2 kl sin ( 2 la ) [( ) ]
1¿ ¿ ¿
cos ( 2la )−i
2 kl ( )
sin ( 2la ) ¿

sin ( 2la ) 2 2
S=
cos ( 2la )−i
e
−2ika

sin ( 2la ) 2 2
( k +l )
¿ i
2 kl [
( l −k ) ]
1 ¿ ¿¿
¿
2 kl

Problem 3.37 Find ⟨x⟩ , ⟨ p⟩, ⟨ x 2⟩ , ⟨p2 ⟩, ⟨T ⟩, and V ( x ) for the nth stationary state of the
harmonic oscillator. Check that the uncertainty principle is satisfied. Hint: Express x and
 ( ℏ /i ) ( d /dx ) in terms of ( a + ±a−) , and use Equations 3.52 and 3.53; you may assume
that the states are orthogonal.
Penyelesaian:
Nilai posisi dan momentum operator dalam hal menaikkan dan menurunkan operator
ℏ ℏ mω
x=
√ ( a +a )
2 mω + − dan
p=i
√ 2
( a+ −a− )
Dari persamaan x, maka
ℏ
x 2= ( a + +a− )2
2 mω
ℏ
x 2= a +a a +a
2 mω ( + − )( + − )
ℏ
x 2= ( a a +a a +a a +a a )
2 mω + + − + + − − −
ℏ
x 2= a +a a +a a +a
2 mω ( + 2 − + + − −2 )
Dari persamaan p, maka
ℏ mω
p2 =− ( a+ −a− )2
2
ℏ mω
p2 =− a −a a −a
2 ( + − )( + − )
ℏ mω
p2 =− a a −a a −a a +a a
2 ( + + − + + − − −)
ℏ mω
p2 =− a −a a −a a +a
2 ( +2 − + + − −2)
Dimana a+ adalah menaikkan operator dan a- menurunkan operator
Ketika menaikkan operator yang beroperasi pada kondisi ke-n, maka
a+ Ψ n =√ n+1 Ψ n+1
Ketika menurunkan operator yang beroperasi pada kondisi ke-n, maka
a− Ψ n= √n Ψ n−1
a+ a¿ Ψ n =nΨ n
a− a+ Ψ n =( n+1 ) Ψ n
Kondisi otonormal adalah

∫ Ψ m ( x ) Ψ n ( x )=δmn =1 ,
¿
untuk m = n = 0, dimana m  n
Ketika menaikkan operator dalam kondisi n pada sebuah sistem, itu akan menaiikan sistem

dengan nilai eigen √ n+1 . Ketika menurunkan operator dalam kondisi n pada sebuah

sistem, itu akan menaiikan sistem dengan nilai eigen √n

Jadi,
a− Ψ n= √n Ψ n−1
a+ (√ n Ψ n−1 )= √n √n Ψ n
dan
a+ Ψ n =√ n+1 Ψ n+1
a+ (√ n+1 Ψ n+1 ) =√ n+1 √ n+1 Ψ n =( n+1 ) Ψ n
Harapan nilai pada posisi x adalah

⟨x⟩=∫ Ψ n xΨ n dx ¿

Substitusikan nilai x pada persamaan diatas, maka

ℏ
⟨x⟩=
√ 2mω
∫ Ψ n (a+ +a− ) Ψ n dx
¿

ℏ
⟨x⟩=
√ 2mω
∫ Ψ n (a+ Ψ n ) dx+∫ Ψ n ( a− Ψ n ) dx
¿ ¿

ℏ
⟨x⟩=
√ 2mω
√n+1∫ Ψ n Ψ n+11 dx+ √ n ∫ Ψ n Ψ n−1 dx
¿ ¿

karena

∫ Ψ n Ψ n+11 dx=0
¿
dan ∫ Ψ n Ψ n−1 dx=0
¿

Maka

⟨x⟩=0
Harapan nilai pada momentum p adalah

⟨ p ⟩=∫ Ψ n pΨ n dx
¿

Substitusikan nilai p pada persamaan diatas, maka

ℏ mω
⟨ p ⟩=i
√ 2
∫ Ψ n ( a+−a−) Ψ n dx
¿

ℏ mω
⟨ p ⟩=i
√ 2
∫ Ψ n ( a+ Ψ n ) dx−∫ Ψ n ( a− Ψ n ) dx
¿ ¿
 ℏ mω
⟨ p ⟩=i
√ 2
√ n+1∫ Ψ n Ψ n+11 dx−√ n∫ Ψ n Ψ n−1 dx
¿ ¿

karena

∫ Ψ n Ψ n+11 dx=0
¿
dan ∫ Ψ n Ψ n−1 dx=0
¿

Maka

⟨ p⟩=0
Harapan nilai pada posisi x2 adalah

⟨x 2 ⟩=∫ Ψ n x2 Ψ n dx
¿

Substitusikan nilai x pada persamaan diatas, maka

ℏ
⟨x 2 ⟩= ∫ Ψ n ( a+ +a− ) 2 Ψ n dx
¿
2 mω
ℏ
⟨x 2 ⟩= ∫ Ψ n ( a+ +a−)( a++a−) Ψ n dx
¿
2 mω
ℏ
⟨x 2 ⟩= ∫ Ψ n ( a+2+a− a+ +a+ a−+a−2) Ψ n dx
¿
2 mω
ℏ
⟨x 2 ⟩= [
0+∫ Ψ n ( a+ a− Ψ n ) dx+∫ Ψ n ( a− a+ Ψ n) dx +0
¿ ¿
]
2 mω
ℏ
⟨x 2 ⟩= ∫ Ψ n Ψ n dx+ ( n+1 )∫ Ψ n Ψ n dx
¿ ¿
2 mω
Maka
ℏ
⟨x 2 ⟩= ( 2 n+1 )
2 mω
ℏ 1
⟨ x 2 ⟩=
mω
n+
2 ( )
Dimana,

a+ a− Ψ n =a+ ( √ n ) Ψ n−1=( √ ( n )( n ) ) Ψ n =nΨ n

a+ a− Ψ n =a+ ( √ n+1 ) Ψ n+1= ( √ ( n+1 ) ( n+1 ) ) Ψ n =( n+1 ) Ψ n

a Ψ n =a+ ( √ n+1 ) Ψ n+1 =( √ ( n+1 ) ( n+2 ) ) Ψ n+2
+2

a Ψ n =a+ ( √ n ) Ψ n−1 =( √ ( n ) ( n−1 )) Ψ n−2

−2

Harapan nilai pada momentum p adalah

⟨ p 2 ⟩=∫ Ψ n p2 Ψ n dx ¿

Substitusikan nilai p pada persamaan diatas, maka

 ℏ mω
⟨ p 2 ⟩=− Ψ a −a 2 Ψ dx
2 ∫ n ( + −) n
¿

ℏ mω
⟨ p 2 ⟩=− Ψ a −a a −a Ψ dx
2 ∫ n ( + − )( + − ) n
¿

ℏ mω
⟨ p 2 ⟩=− Ψ a −a a −a a +a Ψ dx
2 ∫ n ( +2 − + + − − 2 ) n
¿

ℏ mω
⟨ p 2 ⟩=− [
0−∫ Ψ n ( a+ a− Ψ n ) dx−∫ Ψ n ( a− a+ Ψ n ) dx +0
¿ ¿
]
2
ℏ mω
⟨ p 2 ⟩=− [
n∫ Ψ n Ψ n dx + ( n+1 )∫ Ψ n Ψ n dx
¿ ¿
]
2
Maka
ℏ mω
⟨ p 2 ⟩= ( 2 n+1 )
2

⟨ p 2 ⟩=ℏ mω n+ ( 12 )
Dimana,

a+ a− Ψ n =a+ ( √ n ) Ψ n−1=( √ ( n )( n ) ) Ψ n =nΨ n

a+ a− Ψ n =a+ ( √ n+1 ) Ψ n+1= ( √ ( n+1 ) ( n+1 ) ) Ψ n =( n+1 ) Ψ n

a Ψ n =a+ ( √ n+1 ) Ψ n+1 =( √ ( n+1 ) ( n+2 ) ) Ψ n+2
+2

a Ψ n =a+ ( √ n ) Ψ n−1 =( √ ( n ) ( n−1 )) Ψ n−2

−2

Harapan nilai dari energi kinetik K adalah

2
⟨p ⟩
⟨K ⟩=
2m
Substitusikan persamaan p2 dari persamaan diatas maka,

⟨ K ⟩=
ℏ mω n+ ( 12 )
2m
1 1
⟨K ⟩= ( ) 2m
n+ ℏ mω
2
1 1
⟨K ⟩= (n+ ) ℏ ω
2 2
Ketidaktentuan posisi adalah

σ x =√⟨ x 2 ⟩−⟨ x⟩2

 1
σ x=
√ ℏ
mω ( )
n+ −0
2

1
σ x=
√ ℏ
mω ( n+ )
2
Ketidaktentuan momentum adalah

σ p = √⟨ p 2 ⟩−⟨ p⟩2
1
√ 2 )−0
(
σ p = ℏ mω n+

1
p
√ 2
σ = ℏ mω ( n+ )

Ketidaktentuan partikel posisi dan momentum adalah

1 1 1
σ x σ p=
√ ℏ
mω( )√ ( ) ( )
n+
2
ℏ mω n+
2
= n+ ℏ≥
2
ℏ
2

Problem 3.39 Solve the time-independent Schrodinger equation for an infinite square well
with a delta-function barrier at the center:

V ( x ) =¿ {αδ ( x ) , for (−a<x<+a) ¿¿¿¿

Treat the even and odd wave functions separately. Don't bother to normalize them. Find the
allowed energies (graphically, if necessary). How do they compare with the corresponding
energies in the absence of the delta function? Comment on the limiting cases α 0 and α ∞.
Penyelesaian:

Problem 3.41 A particle of mass m and kinetic energy E > 0 approaches an abrupt potential
drop V0 (Figure 3.16).
a) What is the probability that it will "reflect" back, if E = V0/3?
b) I drew the figure so as to make you think of a car approaching a cliff, but obviously the
probability of "bouncing back" from the edge of a cliff is far smaller than what you got in
(a)—unless you're Bugs Bunny. Explain why this potential does not correctly represent a
cliff.
Penyelesaian:
 2 mV 0 2 mV 0 8 m V 0 8 mV 0
3 3 3 3
2 2 2 2
−2 2
+ 2
k 1−k 2 k 1 −2 k 1 k 2 +k 2 ℏ ℏ ℏ ℏ2 8 4
a) R= (
k1 + k2 )
= 2
k 1 +2 k 1 k 2 +k 2 2
=
2 mV 0 2m V 0 8 mV 0
=
14 mV 0
= =
14 7
3 3 3 3
2
+2 2
+ 2
ℏ ℏ ℏ ℏ2
b) Potensial tersebut tidak mencerminkan jurang, nilai tegak negatif akan menyebabkan
massa bergerak dengan energi total dari
Problem 3.43 Imagine abead of mass m that slides frictionlessly around a circular wire ring

of circumference a. [This is just like a free particle, except that Ψ ( x ) =Ψ ( x+a ) .] Find the
stationary states (with appropriate normalization) and the corresponding allowed energies.
Note that there are two independent solutions for each energy En—corresponding to
+ −
clockwise and counterclockwise circulation; call them Ψ ( x ) and Ψ ( x ) . How do you
account for this degeneracy, in view of the theorem in Problem 3.42—that is, why does the
theorem fail in this case?
Penyelesaian:

Problem 3.45
a) Show that
1
mω 2 a 2 (
Ψ ( x , t )=
mω
( )
πℏ
4 exp
[ (
−
2ℏ
x + 1+e−2 iωt ) +
2
iℏ t
m
−2 axe−iωt )]

Figure 3.17: The double square well (Problem 3.44).

satisfies the time-dependent Schrodinger equation for the harmonic oscillator potential
(Equation 2.38). Here a is any real constant (with the dimensions of length). This rare
example of an exact closed-form solution to the time-dependent Schrodinger equation
was discovered by Schrodinger himself, in 1926.
 2
b) Find |Ψ ( x,t )| , and describe the motion of the wave packet.

c) Compute ⟨x⟩ and ⟨p⟩ , and check that Ehrenfest's theorem (Equation 2.38) is
satisfied.
Penyelesaian:
mω 2 a2 (
mω 1
[ (
−
2ℏ 2
iℏ t
x + 1+e−2 iωt ) + −2axe−iωt
m )]
a)
Ψ ( x , t )=
πℏ ( ) 4 e

Dimana a konstan.

mω a2
∂Ψ
=− (−2 iωe−2 iωt ) + iℏ −2 ax (−iωt ) e−iωt Ψ
[ ]
∂t 2ℏ 2 m
∂Ψ 1 1
iℏ
∂t 2 [
= − ma2 ω2 e−2 iωt + ℏ ω+max ω 2 e−iωt Ψ
2 ]
mω
∂Ψ
∂x
= −
[( )
2ℏ
( 2 x−2ae−iωt ) Ψ ]
mω
∂Ψ
∂x
= − [( ) (
ℏ
x−ae−iωt ) Ψ ]
2
∂Ψ mω mω ∂Ψ
2
=− ℏ Ψ − ℏ ( x−ae−iωt )
∂x ∂x

∂2 Ψ mω mω mω
∂x 2
=−
ℏ[Ψ−
ℏ
( x−ae−iωt ) −
ℏ
( x−ae−iωt ) Ψ ] [( ) ]
2 2
∂Ψ mω mω
∂x 2 [
= − ℏ + ℏ ( x−ae−iωt ) Ψ ( ) ]
Misalkan,
2 2
[ S ] =− ℏ ∂ Ψ2 + 1 mω2 x 2 Ψ
2m ∂ x 2
Kemudian,
2 2
[ S ] =− ℏ − mω
ℏ +
mω (
ℏ[ ( )
x−ae−iωt
) Ψ +
1 2 2
mω x Ψ ]
2m 2
[ S ] = 1 ℏ ω− 1 mω2 ( x 2 −2 axe−iωt +a2 e−2 iωt ) + 1 mω2 x 2 Ψ
[ ]
2 2 2
1 1
[ ] [
S= ℏ ω+max ω 2 e−iωt − mω2 a 2 e−2iωt Ψ ]
2 2
 [ S ] =i ℏ ∂ Ψ
∂t
Sehingga,
2 2
ℏ ∂ Ψ 1 ∂Ψ
− 2
+ mω2 x 2 Ψ =i ℏ
2m ∂x 2 ∂t
mω 2 a2 ( −2 iωt ) iℏ t
mω [ ( )] e( 2
2 a
− x + 1+e − −2axe−iωt x + ( 1+e−2iωt ) + i ℏ t −2 axe−iωt )
b)
|Ψ| =2
√ πℏ
e
2ℏ 2 m 2 m

mω 2 a2 ( −2 iωt ) iℏ t a2
mω [ (
− x + 1+e )] ( i ℏt
− −2axe−iωt + x2 + (1+e−2 iωt ) + −2 axe−iωt )
|Ψ|2=
√ πℏ
e
2ℏ 2 m 2 m

{ mω 2 2 2
mω − 2 ℏ [ 2 x +a +a cos ( 2ωt ) −4ax cos (ωt ) ] }
|Ψ| =2
πℏ
2
e
√ 2 2
Dengan, a [ 1+cos (2 ωt ) ] =2a cos ( ωt )
Jadi,
mω 2
mω {− ℏ [ x −2ax cos (ωt ) +a cos ( ωt )] }
2 2
2
|Ψ| =
πℏ
e
√ mω
mω {− }
2
ℏ ( x−a cos ( ωt ))
|Ψ|2=
√ πℏ
e
Paket gelombang adalah bentuk tetap Gaussian, dimana pusat berosilasi bolak-balik
dalam bentuk sinusoidal ( cos ( ωt ) ) . Amplitudo osilasi adalah a dan frekuensi sudut
adalah .

c)
⟨x⟩=∫ x|Ψ|2 dx
Dengan, y=x−a cos ( ωt ) dan dx = dy, maka
⟨x⟩=∫ ( y+acos ( ωt ) )|Ψ|2 dy
2
Substitusi dari bagian (b) dengan |Ψ| dan substitusikan dengan y, maka
mω 2
mω
∞
(− ℏ ( y) ) dy
⟨x⟩= ∫
π ℏ −∞
mω
√
( y+a cos ( ωt ) ) e

b=− ℏ
Misal,
∞
π
∫−∞ e−by dy=
2

Kemudian didapat,
√ b

mω πℏ
√
⟨x ⟩=

⟨x⟩=acos ( ωt )
√πℏ
( a cos ( ωt ))
mω
 Dan nilai momentum adalah:
d ⟨ x⟩
⟨ p⟩=m
dt
d⟨a cos ( ωt ) ⟩
⟨ p ⟩=m
dt
⟨ p⟩=−maω sin ( ωt )
dV d ( p )
⟨− ⟩=
Teorema Ehrenfest’s adalah dx dt
d⟨p⟩
=−maω 2 cos ( ωt )
dt
1 dV
V = mω2 x 2 → =mω2 x
2 dx
Maka
dV
⟨− ⟩=−mω2 ⟨ x ⟩
dx
dV
⟨− ⟩=−mω2 a cos ( ωt )
dx
dV d ( p )
⟨− ⟩=
dx dt

Problem 3.47 Consider the moving delta-function well:

V ( x , t ) =−αδ ( x−vt )
Where v is the (constant) velocity of the well.
a) Show that the time-dependent Schrodinger equation admits the exact solution

mα
Ψ ( x, t )= √ ℏ e−mα|x−vt|/ℏ e
2 −i [( E+ ( 1 /2 ) mv 2) t −mvx ] / ℏ

Where E=−ma2 /2 ℏ2 is the bound-state energy of the starfonary delta function.

Hint: Plug it in and check it! Use Problem 3.24b.
b) Find the expectation value of the Hamiltonian in this state, and comment on the result.
Penyelesaian:

mα
Ψ ( x, t )= √ ℏ e−mα|x−vt|/ℏ e
2 −i [( E+ ( 1 /2 ) mv 2) t −mvx ] / ℏ

a)

1
∂Ψ
∂t
mα
[
= − 2 ∂ |x−vt|−i
ℏ ∂t
E+ mv 2
2
ℏ
Ψ
( )
]
Dimana,
∂ |x−vt|=¿ {−v , if x−vt>0 ¿¿¿¿
∂t

θ ( x)=¿ {1 x<0¿¿¿¿
∂ |x−vt|=−v [ 2θ ( x−vt )−1 ]
∂t
∂Ψ
Substitusikan ke persamaan ∂ t

1
∂Ψ mα
[
= 2 v [ 2θ ( x−vt )−1 ]−i
∂t ℏ
Maka,
E+ mv 2
2
ℏ
(
Ψ
)
]
∂Ψ mα 1
iℏ
∂t [
= i ℏ [ 2 θ ( x−vt )−1 ] +E+ mv 2 Ψ
2 ]
∂Ψ mα imv
∂t [
= − 2 ∂ |x −vt|+ ℏ Ψ
ℏ ∂t ]
Dimana,
∂ |x−vt|=¿ { 1 , if x >vt ¿ ¿¿ ¿
∂t
∂Ψ mα imv
∂x [
= i 2 [ 2θ ( x−vt ) −1 ]+ ℏ Ψ
ℏ ]
2
∂2 Ψ mα imv 2 mα
∂x 2
ℏ [ ℏ ∂x ]
= − 2 [ 2θ ( x−vt )−1 ] + ℏ Ψ − 2 ∂ θ ( x−vt ) Ψ
[ ]
Tapi,
∂ θ ( x−vt )=δ ( x−vt )
∂x
Jadi,
2
ℏ2 ∂ 2 Ψ ℏ2 mα imv
−
2m ∂x 2
=−
2m ℏ [ ]
− 2 [ 2θ ( x−vt ) −1 ] + ℏ Ψ −aδ ( x−vt ) Ψ

2
ℏ2 ∂ 2 Ψ ℏ 2 mα imv
−
2m ∂x 2
=−
2m ℏ ( [
− 2 [ 2 θ ( x−vt )−1 ] + ℏ −aδ ( x −vt ) Ψ
] )
 ℏ2 ∂ 2 Ψ ℏ2 m 2 α 2 2 2
2 m v mv mα
−
2m ∂x 2
=−
2m
−
ℏ 4 [[2 θ ( x−vt ) −1 ] −
ℏ 2
−2i ℏ 2 [ 2θ ( x−vt )−1 ] +aδ ( x−vt ) Ψ
ℏ ]
2
Tapi, [ 2θ ( x−vt )−1 ] = 1, maka

ℏ2 ∂ 2 Ψ mα 2 1 2 mv α
−
2 m ∂ x2
= −
[ + mv +i ℏ [ 2θ ( x−vt ) −1 ] +aδ ( x−vt ) Ψ
2 ℏ2 2 ]
ℏ2 ∂ 2 Ψ mα 2 1 2 mv α
−
2 m ∂ x2 2ℏ 2 [
−aδ ( x−vt ) Ψ = − 2 + mv +i ℏ [ 2 θ ( x−vt ) −1 ] Ψ
]
1
E= mv 2
Tapi, 2 , maka
2 2
ℏ ∂ Ψ ∂Ψ
− 2
−aδ ( x−vt ) Ψ =iℏ
2m ∂x ∂t
∞
⟨H ⟩= ∫ Ψ ¿ HΨ dx
b) −∞
∂Ψ mα 1
Dimana,
HΨ =i ℏ
∂t
=i
ℏ[[ 2 ]
2 θ ( x−vt ) −1 ] + E + mv 2 Ψ
∞
mα 1
⟨H ⟩= ∫ i
−∞
∞
[ ℏ [ 2 ]
2 θ ( x−vt )−1 ]+E+ mv 2 Ψ ¿ Ψ dx

mα 1
⟨H ⟩= ∫[
−∞ 2 ]
i ℏ [ 2 θ ( x−vt )−1 ]+E+ mv 2 |Ψ|2 dx
∞

Dari normalisasi
∫−∞ |Ψ|2 dx=1
dan [ 2θ ( x−vt )−1 ] adalah odd function dan integral
odd function dari - sampai  adalah nol, maka
1
⟨H ⟩=E + mv 2
2
Paket gelombang digabung sepanjang (pada kecepatan v) dengan energi total fungsi delta
1 2
mv
yang aka nada dalam fungsi delta tersebut (E), energi kinetic karena gerakannya 2
.

Problem 3.49 The S-matrix tells you the outgoing amplitudes (B and F) in terms of the
incoming amplitudes (A and G):
 ( B ¿) ¿ ¿ ¿
¿
For some purposes it is more convenient to work with the transfer matrix, M, which gives you
the amplitudes to the right of the potential (F and G) in terms of those to the left (A and B):

( F ¿) ¿ ¿ ¿
¿
a) Find the four elements of the M-matrix in terms of the elements of the S-matrix, and vice
versa. Express R1, T1, Rr, and Tr (Equations 3.161 and 3.162) in terms of elements of the
M matrix.
b) Suppose you have a potential consisting of two isolated pieces (Figure 3.19). Show that
the M matrix for the combination is the product of the two M matrices for each section
separately:
M=M 2 M 1
(This obviously generalizes to any number of pieces, and accounts for the usefulness of
the M matrix.)

Figure 3.19: A potential consisting of two isolated pieces (Problem 3.49).

c) Construct the M matrix for scattering from a single delta-function potential at point a:
V ( x ) =−αδ ( x−a )
d) By the method of part (b), find the M-matrix for scattering from the double delta function
V ( x ) =−α [ δ ( x +a )+ δ ( x−a ) ]
What is the transmission coefficient for this potential?