Solutions to Problem Set 2

Physics 485 - Fall 2014
Solutions to Problem Set 2
Problem 1. Normalization and probability density(Townsend problem 2.20)
(a) (13%) The normalization constant A can be calculated by
∫
∞
1=
−∞
∫
|Ψ|2 dx =
∫
0
−∞
∞
|A|2 e2κx dx +
|A|2 e−2κx dx =
0
∞ ) |A|2
|A|2 ( 2κx 0
e
.
− e−2κx 0 =
−∞
2κ
κ
(1)
Here the parameter κ must be assumed to be real and positive, or the integral will diverge. We can choose
A to be real as well, thus,
A=
(b)
∫
√
κ.
(2)
(12%) The probability of ﬁnding the particle within the range of 1/κ of the origin is
1/κ
(
|Ψ| dx = κ
∫
−1/κ
∫
0
2
2κx
e
1/κ
dx +
−1/κ
0
0
1/2κ )
) 1(
.
= 1 − e−2 = 0.865 .
e−2κx dx = e2κx −1/2κ − e−2κx 0
2
(3)
Here 1/κ is the characteristic length of the delta potential which can be used to describe the size of it. The
”within 1/κ of the origin” can also be understood as the following.
∫
1/2κ
−1/2κ
(
|Ψ|2 dx = κ
∫
∫
0
e2κx dx +
−1/2κ
0
1/2κ
0
1/2κ )
) 1(
.
e−2κx dx = e2κx −1/2κ − e−2κx 0
= 1 − e−1 = 0.632 . (4)
2
The above result shows that the particle in a bound state is mostly conﬁned in the delta potential, but it
still has a small possibility to penetrate outside of the potential. This is diﬀerent from the classic case where
no particle, which has a kinetic energy less than the potential, can move outside the potential well.
Problem 2. Group and phase velocity(Townsend problem 2.26)
(25%) The circular wave pulse propagates at the group velocity, while the surface ripples moves at phase
velocity. It’s important to ﬁnd the relation between the group velocity and the phase velocity for this case.
Firstly, we express the phase velocity of the ripples in terms of the wave number k,
√
vph =
kT
.
ρ
1
(5)
According to the deﬁnition of the phase velocity, we have
√
ω = vph k = k
T
.
ρ
3/2
(6)
From (2.36) in the textbook, the group velocity is then
√
dω
3
vg =
=
dk
2
kT
3
= vph .
ρ
2
(7)
This shows that the group velocity is greater than the phase velocity. Therefore, the surface ripples move
slower than the circular wave pulse. Relative to the latter, they look like move inward through the circular
disturbance.
Problem 3. Expectation values and uncertainties for a particle in
a box(Townsend problem 2.33)
(25%) The expectation value of the position x is given by,
∫
∞
2
⟨x⟩ =
x|Ψ| dx =
L
−∞
∫
L
x sin2
2
0
( πx )
L
dx = .
L
2
(8)
The expectation value of x2 is given by,
∫
∞
2
⟨x ⟩ =
x |Ψ| dx =
L
−∞
2
2
∫
L
x2 sin2
2
0
( πx )
L2 (
3 )
dx =
1− 2 .
L
3
2π
(9)
Thus the standard deviation in position is given by
√
L
△x = ⟨x2 ⟩ − ⟨x⟩2 =
2
√
1
2
− 2.
3 π
(10)
The expectation value of the momentum px is given by,
∫
∞
∗~
2π ~
⟨px ⟩ =
Ψ Ψdx = 2
i
L
i
−∞
∫
L
sin
0
( πx )
( πx )
cos
dx = 0 .
L
L
(11)
The expectation value of p2x is given by,
∫
⟨p2x ⟩
= −~
2
∞
2π 2 ~2
Ψ Ψdx = −
L3
−∞
∗
∫
L
sin2
0
( πx )
~2 π 2
.
dx =
L
L2
(12)
Thus the standard deviation in momentum is given by
△px =
√
π~
.
⟨p2x ⟩ − ⟨px ⟩2 =
L
2
(13)
Therefore the uncertainty relation of position and momentum is given by
(
△x△px = π~
√
1
1 )
− 2 .
12 2π
(14)
Problem 4. Probability density and expectation value of momentum
(a) (9%) The probability density |Ψ(x, t)|2 is,
|Ψ(x, t)|2 = |A|2 e
−(x−p0 t/m)2
σ 2 |β|2
,
(15)
where |β|2 = 1 + ~2 t2 /(m2 σ 4 ). The following ﬁgures plot the density at t = 0 and t = 1 seperately. The
center of the wavepacket is at p0 t/m. The size of the wavepacket can be described by the width of it which
is σ|β|, so the wavepacket spreads as time.
(a) t=0
(b) t=1
Figure 1: Plot the probability density at t=0 and t>0
(b) (8%) Compare the ﬁgures and see that the position of the peak locates at p0 t/m, so as time evolves,
the peak moves at the velocity p0 /m which is the group velocity,.
(c)
(8%) The expectation value of momentum is given by
∫
∞
~ ∂
Ψ (x, t)
⟨p(t)⟩ =
Ψ(x, t)dx = |A|2
i ∂x
−∞
∗
∫
∞
−x′2
(
~x′ −x′2 )
p0 e σ2 |β|2 − 2 e σ2 |β|2 dx′ = p0 .
iσ β
−∞
(16)
For this integral, the ﬁrst term is just the normalization of the wave function, so the integral of this part is
just p0 , while the second term is odd so that the integral of it is zero. This satisﬁes with Ehrenfest’s theorem
since
d⟨p(t)⟩
=0=−
dt
⟨
3
∂V (x)
∂x
⟩
.
(17)

Download Report