North-West University
Department of Physics
NPHY221 – Introduction to Quantum Physics
Chapter 5
Oct, 2022
1. In the geometry of Figure 5.1, the wave in region 1 is given by
y1 (x) = C1 sin (2πx/λ1 − ϕ1 ),
where C1 = 11.5, λ1 = 4.97 cm, and ϕ1 = −65.3◦ . In region 2, the wavelength is λ2 = 10.5
cm. The boundary A is located at x = 0, and the boundary B is located at x = L, where
L = 20.0 cm. Find the wave functions in regions 2 and 3.
Let
y2 (x) = C2 sin (2πx/λ2 − ϕ2 )
At x = 0, y1 (x = 0) = y2 (x = 0)
C1 sin ϕ1 = C2 sin ϕ2
At x = 0, y1′ (x = 0) = y2′ (x = 0)
2π
2π
C1 cos ϕ1 =
C2 cos ϕ2
λ1
λ2
ϕ2 = tan−1 (
λ1
4.97
tan ϕ1 ) = tan−1 (
tan − 65.3) = −45.8◦
λ2
10.5
1
C2 = C1
sin ϕ1
sin (−65.3)
= 11.5
= 14.6
sin ϕ2
sin (−45.8)
Thus, y2 (x) = 14.6 sin(2πx/10.5 + 45.8)
For region 3, let
y3 (x) = C3 sin (2πx/λ3 − ϕ3 )
At x = L, y2 (x = L) = y3 (x = L)
2πL
2πL
− ϕ2 = C3 sin
− ϕ3
λ2
λ1
C2 sin
At x = L, y2′ (x = L) = y3′ (x = L)
2πL
2π
2πL
2π
C2 cos
− ϕ2 =
C3 cos
− ϕ3
λ2
λ2
λ1
λ1
ϕ3 = tan−1 (
λ2
10.5
tan ϕ2 ) = tan−1 (
tan − 45.8) = −60.9◦
λ1
4.97
C3 = C2
sin ϕ2
sin (−45.8)
= 7.36
= 14.6
sin ϕ3
sin (−60.9)
Thus, y3 (x) = 7.36 sin (2πx/4.97 − 60.9◦ )
2. A wave has the form y = A cos(2πx/λ + π/3) when x < 0. For x > 0, the wavelength is λ/2.
By applying continuity conditions at x = 0, find the amplitude (in terms of A) and phase of
the wave in the region x > 0. Sketch the wave, showing both x < 0 and x > 0.
Let y1 (x) = A cos (2πx/λ + π/3) for x < 0,
and y2 (x) = A2 cos (2πx/λ2 + ϕ) for x > 0.
The continuity conditions on y and dy/dx at x = 0 give
y1 = y2 → A cos (π/3) = A2 cos ϕ
(1)
dy2
2πA
2πA2
dy1
=
→−
sin π/3 = −
sin ϕ
dx
dx
λ
λ2
(2)
Dividing the (2) by (1), we obtain
tan(π/3) = 2tan ϕ
ie
ϕ = tan−1 [12 tan (π/3)] = 40.91◦
solving the first equation for A2 ,
A2 = A
cos π/3
= 0.661A
cos ϕ
2
3. A particle is trapped in an infinite one-dimensional well of width L. If the particle is in its
ground state, evaluate the probability to find the particle (a) between x = 0 and x = L/3;
(b) between x = L/3 and x = 2L/3; (c) between x = 2L/3 and x = L.
(a) P (0 : L/3)
Z
Z L/3
2
|ψ(x)| dx =
=
L/3
0
0
(b) P (L/3 : 2L/3)
Z
Z 2L/3
|ψ(x)|2 dx =
=
2L/3
L/3
L/3
2
πx
2
sin2
dx =
L
L
π
π/3
Z
2
πx
2
sin2
dx =
L
L
π
0
Z
2
2 u sin 2u π/3
sin2 udu = ( −
)|0 = 0.1955
L
π 4
4
2π/3
π/3
2
2 u sin 2u 2π/3
sin2 udu = ( −
)|π/3 = 0.6090
L
π 4
4
(c) P (2L/3 : L)
Z
L
|ψ(x)|2 dx = .....
=
2L/3
4. The lowest energy of a particle in an infinite one-dimensional well is 5.6 eV. If the width of
the well is doubled, what is its lowest energy?
E1 =
ℏ2
= 5.6 ev, L′ = 2L
8mL2
E1′ =
ℏ2
ℏ2
1 ℏ2
1
1
=
=
= E1 = × 5.6 = 1.4 ev
′2
2
2
8mL
8m(4L )
4 8mL
4
4
5. Show that the wavefunction
ψ(x) = A sin(kx) + B cos (kx)
is a solution to the time-independent Schrödinger equation for a free particle for all values of
the constants A, B.
dψ(x)
= Ak cos (kx) − Bk sin(kx)
dx
d2 ψ(x)
= −Ak 2 sin (kx) − Bk 2 cos(kx) = −k 2 (A sin (kx) + B cos(kx)) = −k 2 ψ
dx2
The time independent Schrodinger equation,
−
ℏ2 d2 ψ
+ V (x)ψ(x) = Eψ
2m dx2
For free particle, V (x) = 0. Thus,
−
ℏ2 d2 ψ
ℏ2
ℏ2 k 2
2
=
−
(−k
ψ)
=
ψ = Eψ
2m dx2
2m
2m
Since we have found that the energy associated with this wavefunction is constant and not
dependent on position,then the wavefunction ψ satisfies the time-independent Schrödinger
ℏ2 k 2
equation with eigenvalue
.
2m
3
6. In a certain region of space, a particle is described by the wave function ψ(x) = Cxe−bx where
C and b are real constants. By substituting into the Schrödinger equation, find the potential
energy in this region and also find the energy of the particle. (Hint: Your solution must give
an energy that is a constant everywhere in this region, independent of x.)
With
ψ(x) = Cxe−bx
dψ
= Ce−bx + Cxbe−bx
dx
d2 ψ
= −2bCe−bx + Cb2 e−bx
dx2
By substituting into the Schrödinger equation we get
−
ℏ
(−2bCe−bx + b2 Cxe−bx ) + U (x)Cxe−bx = ECxe−bx
2m
−
ℏ
(−2b + b2 ) + U (x) = E
2m
ℏb
ℏb2
−
+ U (x)
mx 2m
The cancellation of the two terms depending on x leaves only the remaining term for the
energy:
ℏb2
E=−
2m
E=
4