Problem 1:
The one-dimensional square well
shown in the figure rises to infinity at x = 0 and has range a and depth U1.
(a) Derive the condition for a spinless particle
of mass m to have two and only two bound states in the well.
(b) Sketch the wave function of these two states inside and outside
the well and give their analytic expressions. These expressions may involve undetermined
constants.
Solution:
- Concepts:
Square potentials
- Reasoning:
We want to solve for the bound states in a square well.
- Details of the calculation:
(a)
Let region 1 extend from x = 0 to x = a and region 2 from x = a
to infinity.
For bound states we have E < 0.
Define k2 =
(2m/ħ2)(E + U1), ρ2
=
(2m/ħ2)(-E), and k02 =
(2m/ħ2)U1.
Note: I am assuming that U1 is a positive number denoting the
depth, the potential at the bottom of the well is -U1.
In region 1 we have Φ1(x) =
A sin(kx), since Φ1(0) =
0.
In region 2 we have Φ2(x) = Bexp(-ρx),
since Φ2(∞)
= 0.
At x = a we need hat
Φ(a) and (∂/∂x)Φ(x)|a
are continuous.
A sin(ka) = Bexp(-ρa).
kA cos(ka) = -ρBexp(-ρa).
Therefore cot(ka) = -ρ/k.
1/sin2(ka) = 1 + cot2(ka) = (k2 + ρ2)/k2
= k02/k2.
We can find a graphical solution by plotting |sin(ka)| and k/k0
versus k. The intersections of the two plots in regions where cot(ka)
< 0 gives the values of k for which a solution exist.
It is possible to have no solutions, when π/(2k0a)
> 1.
For only two solution to
exist we need the following condition for the slope 1/k0.
2a/(5π) < 1/k0 < 2a/(3π), 3π/(2a) < k0 <
5π/2a, 3π/(2a) < ((2m/ħ2)U1)1/2
< 5π/2a.
(b) Inside the well Φi(x) =
A sin(kix), outside the well for positive x,
Φi(x) = A sin(kia)exp(ρia)exp(-ρix),
where k is a solution of |sin(kia)| = ki/k0, cot(ka) < 0,
and ρi is
determined once ki is determined.
k1 lies between π/(2a) and π/a, and k2 lies between
3π/(2a) and 2π/a.
Problem 2:
You are given a one dimensional potential barrier of height U
which extends from x = 0 to x = a. A particle of mass m and energy E <
U is incident from the left.
(a) Derive expressions for the transmittance T
and reflectance R for the barrier.
(b) Show that T + R = 1.
Solution:
- Concepts:
This is a "square potential" problem. We solve HΦ(x)
= EΦ(x) in regions where U(x) is constant and apply boundary conditions.
- Reasoning:
We are given a piecewise constant potential. We have a potential that
has a value of U from x = 0 to x =
a and is zero otherwise. We are
asked to find the transmittance T and reflectance R.
- Details of the calculation:
Divide space into 3 regions;
region 1: x < 0, region 2: 0 < x < a, region 3: x > a.
(a) potential barrier, E < U.
The most general solutions in regions 1, 2, and 3 are
Φ1(x) = A1exp(ikx) + A1'exp(-ikx),
Φ2(x) = B2exp(ρx) + B2'exp(-ρx),
Φ3(x)
= A3exp(ikx).
Here k2 = (2m/ħ2)E and ρ2
= (2m/ħ2)(U - E).
The boundary conditions are that
Φ(x) and (∂/∂x)Φ(x)
are continuous at x = 0 and x = a.
x = 0: A1 + A1' = B2 + B2'
, ikA1 - ikA1' = ρB2
- ρB2'.
x = a: B2exp(ρa) + B2'exp(-ρa)
= A3exp(ika),
ρB2exp(ik2a) - ρB2'exp(-ρa)
= ikiA3exp(ika).
We need to solve these equations for A3 and A1' in terms of A1.
(i) Solve for B2 and B2' in terms of A3.
B2 = (½ )exp((ik - ρ)a) (1 + ik/ρ )A3
= C A3.
B2' = (½ )exp((ik + ρ)a) (1 - ik/ρ) A3
= C' A3.
(ii) Now solve for A1 in terms of A3 to find T.
2A1 = B2
+ B2' + (ρ/ik)(B2 - B2') = (C + C' + (ρ/ik) (C
- C'))A3.
A1 = ([(k2 - ρ2)/(2ikρ)]sinhρa
+ coshρa)exp(ika)A3.
T = |A3/A1|2
= 4k2ρ2/[(k2
- ρ2)2sinh2ρa + 4k2ρ2cosh2ρa]
= 4k2ρ2/[(k2
+ ρ2)2sinh2ρa + 4k2ρ2].
[cosh2x - sinh2x = 1.]
T = 4E(U - E)/[U2sinh2[(2m(U
- E)/ħ2)½a] + 4E(U - E)].
(iii)
Now solve for A1' in terms of A1 to find R.
A1 + A1' = B2 + B2' = (C + C')A3.
A1 - A1' = (ρ/(ik))(B2 - B2') =
(ρ/(ik))(C - C')A3.
A1(C + C' - (ρ/(ik)) (C - C')) = A1'(C + C' - (ρ/(ik)) (C
- C')).
C + C' - (ρ/(ik))(C - C') = [[(k2 - ρ2))/(ikρ)]sinhρa
+ coshρa]]exp(ika).
C + C' - (ρ/(ik))(C - C') = [(k2 + ρ2))/(ikρ)]sinhρa
exp(ika).
R = |A1'/A1|2 = (k2+ ρ2)2sinh2ρa
/[(k2
+ ρ2)2sinh2ρa + 4k2ρ2],
or
R = U2sinh2[(2m(U
- E)/ħ2)½a] /[(E + (U - E))2sinh2[(2m(U
- E)/ħ2)½a] + 4E(U - E)].
(b) T + R = 4k2ρ2/[(k2
+ ρ2)2sinh2ρa + 4k2ρ2]
+ (k2+ ρ2)2sinh2ρa /[(k2
+ ρ2)2sinh2ρa + 4k2ρ2]
= 1.
Problem 3:
The wave function ψ(r) of a spinless particle is ψ(r) = Nz2exp(-r2/b2),
where b is a real constant and N is a normalization constant.
(a) If L2 is measured, what results can be obtained and
with what probabilities?
(b) If Lz is measured, what results can be obtained and with
what probabilities?
(c) Is ψ(r) an eigenfunction of L2 or Lz?
Solution:
- Concepts:
The eigenfunctions of the orbital angular momentum operator, the spherical
harmonics
- Reasoning:
The common eigenfunctions of L2 and Lz are the
spherical harmonics. We have to write the given wave functions in terms of the
spherical harmonics.
- Details of the calculation:
ψ(r) = Nz2exp(-r2/b2) =
Nr2exp(-r2/b2)cos2(θ) in spherical coordinates.
We have
Y00 = (4π)-½, Y1±1 =
∓(3/8π)½sinθ
exp(±iφ), Y10 = (3/4π)½cosθ,
Y2±2 = (15/32π)½sin2θ
exp(±i2φ), Y2±1 =
∓(15/8π)½sinθ
cosθ exp(±iφ),
Y20 = (5/16π)½(3cos2θ
- 1).
Therefore cos2θ = [(4π)½/3][2Y20/√5
+ Y00].
Normalize the function of angle on the unit sphere:
Nang2∫dΩ [2Y20/√5
+ Y00] [2Y20/√5
+ Y00]* = 1. Nang = √5/3.
cos2θ = (4π/5)½[2Y20/3 + √5 Y00/3].
ψ(r) = f(r)χ(θ), χ(θ) = 2Y20/3 + √5 Y00/3.
χ(θ) is a normalized function of angle on the unit sphere.
f(r) = N(4π/5)½r2exp(-r2/b2). ∫f(r)2r2dr = 1, since ψ(r)
is normalized
(a) If L2 is measure we can obtain
2(2 + 1)ħ2
with probability 4/9 and 0 with probability 5/9.
(b) If Lz is measured we obtain 0 with probability 1.
(c) ψ(r) is not an eigenfunction of L2.
ψ(r)
is an eigenfunction of Lz.
Problem 4:
A particle moving in one dimension is confined to an
infinite square well of width L. At t = 0 its wave function is piecewise
linear and symmetric about L/2 with a maximum height of A, as shown.
(a) Normalize the wave function and find A.
(b) Write down the normalized eigenfunctions in this potential.
(c) Expand ψ(0) in terms of the eigenfunctions of part (b).
(d) Write an expression for ψ(x,t). How much time T must elapse in order for ψ(x,T) = ψ(x,0)?
(e) Write down an expression for <E>, the expectation value of the particles energy in this state.
∫ x sin(x)dx = sin(x) - x cos(x)
∑n=0, odd∞1/n2 = π2/8
Solution:
- Concepts:
The eigenfunctions of the infinite 1D well, the postulates of quantum mechanics
- Reasoning:
The wave function is not an eigenstate of the Hamiltonian.
We are asked to normalize it and evolve it.
- Details of the calculation:
(a) ψ(x,0) = 2Ax/L, 0 < x < L/2, ψ(x,0) = 2A - 2Ax/L, L/2 < x < L.
∫-∞+∞|ψ(x,0)|2dx = 2∫0L/2(2Ax/L)2dx
= A2L/3 = 1. A = (3/L)½.
(b) ψn(x) = (2/L)½sin(nπx/L), En = n2π2ħ2/(2mL2).
(c) ψ(x,0) = (2/L)½∑an sin(nπx/L), an = (2/L)½∫0L
ψ(x,0) sin(nπx/L) dx.
Symmetry: an = 0
for n = even. For odd n we have
an(L/2)½ = 2∫0L/2(2Ax/L) sin(nπx/L)dx
= (4A/L)(L/nπ)2∫0nπ/2
y sin(y)dy = sin(nπ/2)4AL/(nπ)2.
an = sin(nπ/2)4√6/(nπ)2,
n = 1, 3, 5, ... .
(d) ψ(x,t) = (2/L)½∑an sin(nπx/L) exp(-iEnt/
ħ) = (2/L)½∑an sin(nπx/L) exp(-in2E1t/ħ).
ψ(x,t) = ψ(x,0) for the first time when E1T/ħ
= 2π or when T = 2πħ/E1 = 4mL2/(πħ).
(e) <E> = ∑n=oddn2E1an2 = ∑n=odd48 ħ2/(mL2n2π2)
= 48ħ2/(π2mL2)∑n=odd1/n2
= 48 ħ2/(π2mL2)(π2/8) = 6 ħ2/(π2mL2).
Aside:
How can you find ∑n=odd1/n2 without looking it up in a table?
Explore the Fourier series of a
periodic function of period 2π, for example ψ(x) = 1, 0 < x
< π, ψ(x) = 0, π < x < 2π.
Fourier series:
Assume f(x) is a periodic function of x with fundamental period L.
f(x) = A0/2 +
∑n=1∞ Ancos(n2πx/L)
+ ∑n=1∞ Bnsin(n2πx/L)
with
A0 = (2/L)∫0Lf(x)dx,
Am = (2/L)∫0Lf(x)cos(n2πx/L)dx,
Bm = (2/L)∫0Lf(x)sin(n2πx/L)dx.
∫0L|f(x)|2dx = A02L/4 + ∑n=1∞ An2L/2
+ ∑n=1∞ Bn2L/2
For f(x) = ψ(x), ψ(x) = 1, 0 < x
< π, ψ(x) = 0, π < x < 2π, we have
A0 = (1/π)∫0πdx
= 1,
Am = (1/π)∫0πcos(nx)dx = (1/(πn))∫0nπcos(x')dx'
= 0,
Bm = (1/π)∫0πsin(nx)dx = (1/(πn))∫0nπsin(x')dx'.
Bm = (2/(nπ)) if n = odd, Bm = 0 if n = even.
∫02π|ψ(x)|2dx = π.
But we also have
∫02π|ψ(x)|2dx = π/2 + ∑n=1
odd∞ (4/(πn2).
Therefore ∑n=1 odd∞ (4/(πn2) = π/2.
∑n=0, odd∞1/n2 = π2/8.
Problem 5:
Assume that the
Hamiltonian for a spin ½ particle in a magnetic field B is given by H = g
σ∙B, where
σ = (σx, σy, σz) is a vector for Pauli
matrices and g is a positive real constant with the appropriate units.
The
magnetic field can be written as B = B0(sinθ cosφ, sinθ sinφ,
cosθ), with 0 ≤ θ < π, 0 ≤ φ < 2π.
(a) Find eigenstates and
the corresponding eigenvalues of H.
(b) Let us denote the eigenstate with a negative eigenvalue by |->n.
Calculate A = (Aθ, Aφ), where Aα = in<-|∂/∂α|->n,
for α = θ, φ.
Calculate Ωθ,φ
= ∂Aφ/∂θ - ∂Aθ/∂φ.
(c) Are A and Ωθ,φ
invariant under a local gauge transformation |->n --> eiφ|->n?
Solution:
- Concepts:
The two-dimensional state space of a spin ½ particle, the postulates of quantum
mechanics
- Reasoning:
H = H = g
σ∙B
= g(σxBx + σyBy + σzBz)
To find eigenstates and
the corresponding eigenvalues of H we diagonalize the matrix of H.
- Details of the calculation:
(a) H = gB0(σxsinθcosφ
+ σYsinθsinφ + σzcosθ).
The eigenvalues are E = ±gB0, and the corresponding eigenvectors are
|+>n
= cos(θ/2)exp(-iφ/2)|+> + sin(θ/2)exp(iφ/2)|->,
|->n
= -sin(θ/2)exp(-iφ/2)|+> + cos(θ/2)exp(iφ/2)|->.
(b) ∂/∂θ|n-> =
-½cos(θ/2)exp(-iφ/2)|+>
-½sin(θ/2)exp(iφ/2)|->.
Aθ =
i(½sin(θ/2)cos(θ/2) - ½sin(θ/2) cos(θ/2)) = 0.
∂/∂φ|->n = (i/2)sin(θ/2)exp(-iφ/2)|+> + (i/2)cos(θ/2)exp(iφ/2)|->.
Aφ = ½ sin2(θ/2)
- ½ cos2(θ/2) = -½ cosθ.
Ωθ,φ
= ½sinθ.
(c) eiφ|->n = sin(θ/2)exp(+iφ/2)|+>
+ cos(θ/2)exp(i3φ/2)|->.
Aφ = -½ sin2(θ/2)
- (3/2) cos2(θ/2) = -½ - cos2θ/2.
Ωθ,φ =
½sinθ.
A is not invariant, but Ωθ,φ is
invariant under a gauge transformation |n-> --> eiφ|n->.