PHYS 324
Homework 1
Ajay Rawat
Student ID : 1962171
October 12, 2021
1
Question 1
Ψ = A sin(2πBx)e−iCt
1
0 < x < 2B
Ψ∗ = A sin(2πBx)e+iCt
(a) Normalizing the Wave Function
Z
1
2B
|Ψ|2 dx = 1
0
Z
1
2B
ΨΨ ∗ dx = 1
0
Z
1
2B
A2 sin2 (2πBx)eiCt−iCt dx = 1
0
A
2
Z
1
2B
sin2 (2πBx)dx = 1
0
A
2
1
2B
1 − cos(2 × 2πBx)
dx = 1
2
0
1
sin(4πBx)) 2B
2 x
A
−
]
=1
2
8πB
0
Z
1
1 QUESTION 1
2
A2 [
1
sin2π − sin0
−0−(
)] = 1
4B
8πB
1
A2
=1
4B
A2 = 4B
√
A = ±2 B
Since A is a positive, real constant
√
A=2 B
(b) Ψ = 4B 2 sin(2πBx)e−iCt
Z
hxi =
1
2B
Ψ∗ xΨdx
0
Z
=
1
2B
xA2 sin2 (2πBx)dx
0
by parts; u = x; dv = A2 sin2 (2πBx)dx
du = dx; v = 4B( x2 − sin(4πBx)
)
8πB
4Bx2
sin(4πBx)
hxi =
−x
]
2
π
1
2B
Z
− 4B
0
0
2
1
2B
x sin(4πBx)
−
dx
2
8πB
1
x
cos(4πBx)
]
− 0 − (0 − 0) − 4B
+
2B
4
4π 2 B
1
1
1
1
=
−
−[ 2 − 2 ]
2B 4B
4π B 4π B
1
=
−0
4B
1
=
4B
1/2B
=
hxi =
1
4B
0
1 QUESTION 1
3
x
2
1
2B
Z
Ψ∗ x2 Ψdx
=
0
1
2B
Z
x2 A2 sin2 (2πBx)dx
=
0
1
2B
Z
x2 sin2 (2πBx)dx
= 4B
0
u = x2 ; dv = sin2 (2πBx)dx
du = 2xdx; v = ( x2 − sin(4πBx)
)
8πB
x
2
x sin(4πBx)
)]
= 4B x2 ( −
2
8πB
1
2B
− 4B
0
0
= 2B(1/2B)3 − 0 − (0 − 0) − 8B
Z
1
2B
(
0
=
=
=
=
=
3
1
x
−
8B
]
4B 2
6
1
2B
1
2B
Z
Z
1
2B
x sin(4πBx)
( −
)2xdx
2
8πB
x2 x sin(4πBx)
−
)xdx
2
8πB
x sin(4πBx)
xdx
8πB
0
0
1
Z 1
2B cos(4πBx)
1
1
x cos(4πBx) 2B
−
8B
−
8B
2xdx
−
8B
4B 2
48B 3
32π 2 B 2
32π 2 B 2
0
0
1
1
cos(2π)
− 8B 2 − 4
−0
2
4B
6B
32π 2 B 2
1
1
− 2 2
2
12B
8π B
2π 2 − 3
24π 2 B 2
+ 8B
x
2
2π 2 − 3
=
24π 2 B2
1 QUESTION 1
4
p
hx2 i − (hxi)2
r
2π 2 − 3 2
1
= (
) − ( )2
2
2
4B
r 24π B
2
π −6
=
48π 2 B 2
σ(x) =
r
σ(x) =
π2 − 6
48π 2 B2
1 QUESTION 1
5
(c)
p = −ih̄
Z
hpi =
1
2B
Ψ∗ (−ih̄
0
Z
= −ih̄
1
2B
d
Ψ)dx
dx
Ψ∗ (
0
Z
= −ih̄2πB
d
dx
1
2B
d
Ψ)dx
dx
A2 sin(2πBx) cos(2πBx)dx
0
2
Z 1
2B
A
= −ih̄2πB
sin(4πBx)dx
2 0
1
A2
= −ih̄2πB
[− cos(4πBx)]|02B
2
A2
= −ih̄2πB [− cos(2π) + cos 0]
2
A2
= −ih̄2πB [− cos(2π) + cos 0]
2
=0
hpi = 0
1 QUESTION 1
6
p2 = −h̄2
Z
hpi =
1
2B
Ψ∗ (−h̄2
0
2
Z
= −h̄
1
2B
d2
dx2
d2
Ψ)dx
dx2
(A sin(2πBx))(−4π 2 B 2 sin(2πBx))dx
0
2
2 2
Z
1
2B
= 4π B h̄
A2 sin2 (2πBx)dx
0
2
2 2
Z
1
2B
= 4π B h̄
A2 sin2 (2πBx)dx
0
2
2 2
2
2 2
Z
= 4π B h̄
1
2B
|Ψ(x)|2 dx
0
= 4π B h̄
p2 = 4π 2 B2 h̄2
p
hp2 i − (hpi)2
p
= 4π 2 B 2 h̄2 − 0
= 2πBh̄
σ(p) =
σ(p) = 2πBh̄
1 QUESTION 1
7
(d)
r
π2 − 6
2πBh̄
2 2
r 48π B
π2 − 6
h̄
=
12
= 0.57h̄
h̄
≥
2
σ(x)σ(p) =
Thus
σ(x)σ(p) ≥
h̄
2
It is consistent with the Heisenberg uncertainty principle
1 QUESTION 1
8
(e) From the time dependent Schrödinger equation, we know that the En∂
gergy Operator H = ih̄ ∂t
∂
A sin(2πBx)e−iCt
∂t
= ih̄(−iC)A sin(2πBx)e−iCt
HΨ = ih̄
= h̄CA sin(2πBx)e−iCt
Thus Energy of the system =
Z
1
2B
=
Ψ∗ HΨdx
0
Z
1
2B
=
A sin(2πBx)eiCt h̄CA sin(2πBx)e−iCt dx
0
Z
=
1
2B
A sin(2πBx)eiCt h̄CA sin(2πBx)e−iCt dx
0
Z
= h̄C
1
2B
A2 sin2 (2πBx)dx
0
Z
1
2B
= h̄C
|Ψ|2 dx
0
= h̄C
Energy of the system is h̄C
Q2
a)