Download (8.04) Spring 2006 Solutions to Problem Set 2

Massachusetts Institute of Technology
Quantum Mechanics I (8.04) Spring 2006
Solutions to Problem Set 2
Kit Matan
1. Parseval’s theorem (10 points)
Prove the following theorem: A function ψ(x) and its Fourier transform φ(k) have the same
normalization, i.e.
Z
∞
|ψ(x)|2 dx =
−∞
Z
∞
|φ(k)|2 dk
(1)
−∞
§
Z
∞
|ψ(x)|2 dx =
−∞
Z
∞
ψ(x)ψ(x)dx
−∞
Z ∞
!
!
1 Z∞
1 Z∞
0 −ik0 x
0
ikx
√
=
dx √
φ(k )e
dk
φ(k)e dk
−∞
2π −∞
2π −∞
Z ∞
Z ∞ Z ∞
1
0
ix(k−k0 )
e
dx dk 0 dk
=
φ(k )φ(k)
2π −∞
−∞ −∞
Z ∞ Z ∞
1
=
φ(k 0 )φ(k) (2πδ(k − k 0 )) dk 0 dk
2π
−∞ −∞
Z
∞
=
=
φ(k)φ(k)dk
−∞
Z ∞
|φ(k)|2 dk .
(2)
−∞
1
4. The deBroglie wavelength of macroscopic objects. (10 points)
What is the deBroglie wavelength of an automobile (2000 kg) traveling at 25 mile per hour? A
dust particle of radius 1 µm and density 200 kg/m3 being jostled around by air molecules at
room temperature (T = 300 K)? An 87 Rb atom that has been laser cooled to a temperature of
T = 100 µK? Assume that the kinetic energy of the particle is given by 32 kB T .
§ The deBroglie wavelength, λdB , is equal to:
λdB =
h
p
(3)
where h is Planck’s constant and p is a momentum.
For the automobile,
λdB =
6.626 × 10−34 J· s
= 2.96 × 10−38 m .
25 miles/hour · 1609.344 meters/mile
2000 kg ·
3600 s/hour
(4)
For the dust particle at room temperature. First calculate a momentum of the particle.
=⇒ p =
√
p2
2m
= 32 kB T
(5)
3
3mkB T = 3 · 200 kg/m3 43 π · (10−6 ) m3 · 1.38 × 10−23 J/K · 300 K
1/2
= 3.2 × 10−18 kg· m/s .
(6)
(7)
Therefore, we have
λdB =
For the
87
6.626 × 10−34 J·s
= 2.09 × 10−16 m .
3.2 × 10−18 kg·m/s
(8)
Rb atom. Its momentum at T = 100 µK is equal to,
86.909 × 10−3 kg/mole
p =
3mkB T = 3 ·
· 1.38 × 10−23 J/K · 100 × 10−6 K
6.02 × 1023 atoms/mole
= 2.44 × 10−26 kg·m/s .
q
!1/2
(9)
(10)
And, we have
λdB =
6.626 × 10−34 J·s
= 2.72 × 10−8 m .
2.44 × 10−26 kg·m/s
(11)
5. Gaussian wavepacket in free space (30 points)
A free particle of mass m has initial wave function
ψ(x) =
1
−
1/2
(2π)1/4 w0
e
x2
4w2
0
,
where the initial width of the wavepacket w0 is a real, positive constant.
4
(12)
(a) (15 points) Compute the Fourier transform of ψ(x, 0) and show that it is also a Guassian
wavepacket:
2
− x2
1
4w
φ(k) =
e 0.
(13)
1/2
(2π)1/4 k0
Determine the width k0 of the wavepacket in momentum space.
§ Calculate the Fourier transform of ψ(x, 0),
∞
1 Z
φ(k, 0) = √
dxψ(x, 0)e−ikx
2π −∞
∞
2
Z
− x 2 −ikx
1
1
√
dxe 4w0
=
2 1/4
2π (2πw0 ) −∞
(14)
Complete the square,
x2
x
+ ikx =
+ ikw0
2
4w0
2w0
2
+ k 2 w02 .
(15)
Substitute Eq. 15 into Eq. 14, and we obtain,
2
2
2
2
∞
−
e−k w0 Z
1
dxe
φ(k, 0) = √
2 1/4
2π (2πw0 ) −∞
x
+ikw0
2w0
2
e−k w0 q
1
4πw02
= √
2 1/4
(2πw
)
2π
0
=
=
where k0 =
4w02
2π
!1/4
e−k
1
−
1/4
(2πk02 )
e
2 w2
0
k2
4k2
0
,
(16)
1
.
2w0
(b) (15 points)
§ We have
1
∆x∆k = w0 k0 = ,
(17)
2
which is the lower bound of the Heisenberg uncertainty relation ∆x∆k ≥ 1/2 (or equivalently ∆x∆p ≥ h̄/2).
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