PC 242 Assignment 4: Solutions
1. A wave function can be expressed as a sum over different harmonics as
ψ (x) = 0.7 cos(x) + 0.1cos(3x) +0.2 cos(5x)
(a) What are the constituent wavenumbers and wavelengths ?
(b) What wavelength has the largest amplitude a(k) ?
(a) A real (non-complex) wave function can be written as
ψ (x) = ∑ an cos(kn x)
n
In the above function, there are only 3 terms in the series with
2π
= 2π
k1
2π
k2 = 3 ⇒ λ2 =
= 0.67π
k2
2π
k 3 = 5 ⇒ λ2 =
= 0.4π
k3
The amplitudes are the coefficients an
The first term with λ1 = 2π has the largest value of a1= 0.7
k1 = 1 ⇒ λ1 =
2. A particle is described by the probability density function
−L
#
% AL + 2Ax, for 2 ≤ x ≤ 0
%
L
P(x) = $
AL
−
2Ax,
for0
≤
x
≤
%
2
%
0,
otherwise
&
a) Make a sketch of the probability density function, P(x) as a function of x.
b) If the wave function is real, write down a possible expression for the wave function
ψ(x).
c) Normalize the wave function and determine the constant A in terms of L.
d) What is the probability that the particle lies in the region x < L/6?
L
⇒ P(x) = 0
2
x = 0 ⇒ P(x) = AL
L
x = ⇒ P(x) = 0
2
The equations describe straight lines joining the above points:
x=−
P(x)
AL
-L/2
0
L/2
x
2
(b) Since P(x) = ψ (x) , and the wave function must be real,
−L
$
& ± AL + 2Ax, for 2 ≤ x ≤ 0
&
L
ψ (x) = %
±
AL
−
2Ax,
for0
≤
x
≤
&
2
&
0,
otherwise
'
(c) Normalization: The integral of P(x) over all values of x must be equal to 1.
This integral corresponds to the area under the curve of P(x).
It is easy to see that the area of this triangle is
1
2
L × AL = 1 ⇒ A = 2
2
L
Alternatively, you could have calculated the integral
0
∞
∫
P(x)dx =
∫
L /2
P(x)dx +
− L /2
−∞
= ALx + Ax
2
∫ P(x)dx
0
2 0
+ ALx − Ax 2
− L /2
2
2
L /2
0
2
AL
AL
AL
AL
AL2
2
=
−
+
−
=
=1⇒ A = 2
2
4
2
4
2
L
(d) Probability that the particle lies between in the region x < L/6 is the area under the
curve from x=-L/2 to x=L/6. This is equal to the total area of the full triangle under P(x)
minus the area of the triangle from x=L/6 to x = L/2
Total area under the curve P(x) = 1
Area under P(x) from x=L/6 to x = L/2
1" L L% "
AL % 1 " L % " 2AL % AL2 2
=
= $ − ' × $ AL −
' = $ ' ×$
'=
2# 2 6& #
3 & 2 # 3& # 3 &
9
9
L
2 7
) = 1− =
6
9 9
Thus P(x ≤
Alternatively you could do the integral
L /6
0
L /6
L
P(x ≤ ) = ∫ P(x)dx = ∫ P(x)dx + ∫ P(x)dx
6
−∞
− L /2
0
= ALx + Ax 2
2
2
0
+ ALx − Ax 2
− L /2
2
L /6
0
2
AL
AL
AL
AL 14AL2 7
=
−
+
−
=
=
2
4
6
36
36
9
3. An electron in the n=4 state of a 5nm wide infinite well makes a transition to the
ground state, emitting a photon. What is the wavelength of the photon?
Energy levels in an infinite square well are given by
 2 n 2π 2
En =
, n = 1, 2, 3...
2mL2
A transition from n=4 to n=1 corresponds to an energy of
hc
15 2π 2
ΔE =
= E4 − E1 =
,
λ
2mL2
Setting L=5 nm, we can solve for the wavelength:
2mL2 hc 8mc 2 L2 8 ( 0.511MeV ) (5nm)2
λ=
=
=
= 5.5 µ m
15 2π 2
15hc
15 (1240eV ⋅ nm )
4. An electron is trapped in an infinite well. If the lowest energy transition possible in the
well produces a photon of 450 nm wavelength, what is the width of the well?
Energy levels in an infinite square well are given by
 2 n 2π 2
En =
, n = 1, 2, 3...
2mL2
A transition from n=2 to n=1 corresponds to the lowest possible energy difference
hc
3 2π 2
ΔE =
= E2 − E1 =
,
λ
2mL2
Setting λ=450 nm, we can solve for the width L:
3 2π 2 λ
=
2mhc
L=
3hcλ
= 0.64nm
8mc 2
5. What is the probability that a particle in the n=2 state of an infinite well will be found
in the middle third of the well?
Wave functions in the infinite square well are
2
# nπ x &
φn (x) =
sin %
, n = 1, 2, 3.., 0 ≤ x ≤ L
$ L ('
L
φ (x) = 0, x > L, x < 0
Thus probability distribution functions are
2
" nπ x %
Pn (x) = sin 2 $
, n = 1, 2, 3.., 0 ≤ x ≤ L
# L '&
L
P(x) = 0, x > L, x < 0
Probability of finding a a particle in the n=2 state of an infinite well in the middle third of
the well
P=
2
L
2L/3
2
2 " 2π x %
∫L / 3 sin $# L '& dx = L
x
=
L
2L/3
L/3
2L/3
∫
L/3
1"
" 4π x % %
1 − cos $
dx
$
# L '& '&
2#
1
" 4π x %
−
sin $
# L '&
4π
2L/3
L/3
1 1
" 8π % 1
" 4π %
= −
sin $ ' +
sin $
= 0.19
# 3 & 4π
# 3 '&
3 4π
6. Show that the uncertainty in the momentum of a particle in level n of an infinite well is
given by
nπ
Δp =
L
The uncertainty in momentum is given by
2
p2 − p
Δp =
2
, so we need to first calculate p 2 and p .
By symmetry, the particle is equally likely to travel right as it is to travel left, so the
average momentum p = 0 .
Therefore we only need to calculate p 2 .
2
& 2
nπ x ) &
∂ ) & 2
nπ x )
= ∫ ψ (x)p ψ (x)dx = ∫ (
sin
−i
sin
dx
(
+
L
L +* '
∂x * (' L
L +*
−∞
0'
L
∞
p
2
*
2
2
L
 2 n 2π 2 & 2
nπ x )
 2 n 2π 2
=
sin
dx
=
L2 ∫0 (' L
L +*
L2
Hence Δp =
p2 − p
2
=
n 2π 2  2
nπ 
−0 =
2
L
L
7. An electron is trapped in a finite square well. How far in energy (eV) is it from being
free if the penetration depth into the walls of the well is 1nm?
For a finite well, the energy levels are approximately
n 2π 2  2
En ≈
, n = 1, 2...,
2m(L + 2δ )2
2
δ=

2
( hc ) = 0.038eV
⇒U − E =
=
2
2mδ
8π 2 mc 2δ 2
2m(U − E)
8. A 2kg block oscillates with an amplitude of 10cm on a spring with spring constant
120N/m. What energy level is the block in? What is the energy spacing between
successive energy levels?
The energy levels of a spring are those of a harmonic oscillator:
1
1%
"
Eν = kx 2 = $ ν + ' ω , ν = 0,1, 2....
#
2
2&
⇒ν =
%
% 1 " kmx 2
1 " kx 2
−
1
= $
− 1' = 7 × 10 32
$
'
2 # ω
& 2# 
&
The energy spacing = ω = 
k
= 8.13 × 10 −34 J
m