LAST NAME FIRST NAME
Useful constants MKS (SI)
Planck’s constant
Avagadro’s number
h 6.58211889 26( ) 10 16–× eV s=
h 1.054571596 82( ) 10 34–× J s=
e 1.602176462 63( ) 10 19–× C=
m0 9.10938188 72( ) 10 31–× kg=
mn 1.67492716 13( ) 10 27–× kg=
mp 1.67262158 13( ) 10 27–× kg=
kB 1.3806503 24( ) 10 23–× J K 1–=
kB 8.617342 15( ) 10 5–× eV K 1–=
ε0 8.8541878 10 12–× F m 1–=
µ0 4π 10 7–× H m 1–=
c 1 ε0µ0⁄=
aB 4πε0h2
2
PROBLEM 1 A metal ball is buried in an ice cube that is in a bucket of water.
(a) If the ice cube with the metal ball is initially under water, what happens to the water level
when the ice melts?
(b) If the ice cube with the metal ball is initially floating in the water, what happens to the water
level when the ice melts?
(c) Explain how the Earth’s average sea level could have increased by at least 100 m compared
to about 20,000 years ago.
(d) Estimate the thickness and weight per unit area of the ice that melted in (c). You may wish
to use the fact that the density of ice is 920 kg m-3, today the land surface area of the Earth is about
148,300,000 km2 and water area is about 361,800,000 km2.
PROBLEM 2
Sketch and find the volume of the largest and smallest convex plug manufactured from a sphere
of radius r = 1 cm to fit exactly into a circular hole of radius r = 1 cm, an isosceles triangle with
base 2 cm and a height h = 1 cm, and a half circle radius r = 1 cm and base 2 cm.
PROBLEM 3
An initially stationary particle mass m1 is on a frictionless table surface and another particle
mass m2 is positioned vertically below the edge of the table. The distance from the particle mass
m1 to the edge of the table is l. The two particles are connected by a taught, light, inextensible
string of length L > l.
(a) How much time elapses before the particle mass m1 is launched off the edge of the table?
(b) What is the subsequent motion of the particles?
(c) How is your answer for (a) and (b) modified if the string has spring constant κ?
PROBLEM 4
The velocity of water waves in shallow water may be approximated as where g is the
acceleration due to gravity and h is the depth of the water. Sketch the lowest frequency standing
water wave in a 5 m long garden pond that is 0.9 m deep and estimate its frequency.
PROBLEM 5
(a) What is the dispersion relation of a wave whose group velocity is half the phase velocity?
(b) What is the dispersion relation of a wave whose group velocity is twice the phase velocity?
(c) What is the dispersion relation when the group velocity is four times the phase velocity?
PROBLEM 6
A stationary ground-based radar uses a continuous electromagnetic wave at 10 GHz frequency
to measure the speed of a passing airplane moving at a constant altitude and in a straight line at
1000 km hr-1. What is the maximum beat frequency between the out going and reflected radar
beams? Sketch how the beat frequency varies as a function of time. What happens to the beat fre-
quency if the airplane moves in an arc?
v gh=
PROBLEM 7
How would Maxwell’s equations be modified if magnetic charge g (magnetic monopoles) were
discovered? Derive an expression for conservation of magnetic current and write down a general-
ized Lorentz force law that includes magnetic charge. Write Maxwell’s equations with magnetic
charge in terms of a field .
PROBLEM 8
The capacitance of a small metal sphere in air is . A thin dielectric film
with relative permittivity uniformly coats the sphere and the capacitance increases to
. What is the thickness of the dielectric film and what is the single electron charging
energy of the dielectric coated metal sphere?
PROBLEM 9
(a) A diatomic molecule has atoms with mass m1 and m2. An isotopic form of the molecule has
atoms with mass m'1 and m'2. Find the ratio of vibration oscillation frequency ω / ω' of the two
molecules.
(b) What is the ratio of vibrational frequencies for carbon monoxide isotope 12 ( ) and
carbon monoxide isotope 13 ( )?
PROBLEM 10
(a) Find the frequency of oscillation of the particle of mass m illustrated in the Fig. The particle
is only free to move along a line and is attached to a light spring whose other end is fixed at point A
located a distance l perpendicular to the line. A force F0 is required to extend the spring to length l.
(b) Part (a) describes a new type of child’s swing. If the child weighs 20 kg, the length l = 2.5
m, and the force F0 = 450 N, what is the period of oscillation?
G εE i µH+=
εr1 10=
Solution 1
(a) The water level decreases. If ice has volume V then net change in volume of water in the
bucket is .
(b) Again, the mass of the volume of liquid displaced equals the mass of the floating object.
Assuming the metal has a density greater than that of water, the water level decreases when the ice
melts.
(c) If there is just ice floating in the bucket, the water level does not change when the ice melts.
This fact combined with the results from part (a) and (b) allows us to conclude that only the ice
melting over land contributes to increasing the sea level.
(d) Today 71% water, 29% land, ratio is 2.45. The average thickness of ice on land is simply
100 m × 2.45 = 245 m. If one assumes half the land area under ice, then average thickness of ice is
490 m. If this is distributed uniformly from thin to thick ice, then one might expect maximum ice
thicknesses near 1000 m (i.e. ~ 3300 ft high mountains of ice). Ice weighs one metric tone per
cubic meter so the weight per unit area is the thickness in meters multiplied by tones.
Solution 2
We are asked to sketch and find the volume of the largest and smallest convex plug manufac-
tured from a sphere of radius r = 1 cm to fit exactly into a circular hole of radius r = 1 cm, an isos-
celes triangle with base 2 cm and a height h = 1 cm, and a half circle radius r = 1 cm and base 2 cm.
The minimum volume is 1.333 cm3 corresponding to a geometry consisting of triangles placed
on a circular base as shown in the Fig.
To calculate this minimum volume of the plug consider the triangle at position x from the origin
has height k, base length 2l, and area kl.
V V ρice ρwate r⁄ 1–( )=
2 cm
Volume of the plug is
Since r = 1 cm the total volume is exactly 4/3 = 1.333 cm3.
To find the maximum volume of the plug manufactured from a sphere of radius r = 1 cm we
note that the geometry is a half sphere with two slices cut off as shown in the following Fig. The
geometry is found by passing the sphere along the three orthogonal directions x, y, and z and cutting
using a circle, a triangle, and a half circle, and a triangle respectively. As will be shown, the volume
is 1.608 cm3. This is the maximum convex volume of the plug manufactured from a sphere of
radius r = 1 cm.
To calculate the volume we first calculate the volume sliced off the half sphere as illustrated in
following Fig.
x rr
r h
Vol 2 k2 xd 0
r
r
6
so that area of disk radius l at position x is
The volume of the disk is the area multiplied by the disk thickness dx. The total volume is the
integral from to . Remembering to multiply by 2 because there are two slices
gives total volume sliced off
The volume of the plug is just the volume of a half sphere minus .
Since r = 1 cm the total volume is 1.608 cm3. If we lift the restriction that the plug is manufactured from a sphere of radius 1 cm, then calcu-
lating the maximum convex volume turns out to be a bit complicated as may be seen in the follow-
ing Fig. The geometry is found by cutting along the three orthogonal directions x, y, and z using a
circle, a triangle, and a half circle, and a triangle respectively. The volume is 1.659 cm3.
x r
Area π r2 x2–( )=
Voloff 2π r2 x2–( ) xd r 2⁄
r
1
6 2 ----------–
Solution 3
(a) and (b). Force due to gravity acts on particle mass m2. This force is F = gm2 where g is the acceleration due to gravity. The light string transmits the force to the second particle mass m1 that is free to slide horizontally on the table surface. If x is the vertical position of the particle with mass
m2 then and, starting from rest, speed with displacement
. Eventually, particle m1 is launched with velocity v1 from the end of the table, at
which point gravity causes it to accelerate in the x-direction under its own weight. If the distance of
particle mass m1 to the edge of the table is l, then it takes time for m1 to reach
the edge. The launch velocity is . The initial angular momentum of
the system when particle mass m1 is launched from the edge of the table is Lv1m1m2/(m1+m2). (c) If the string has a spring constant then part of the energy of the system can be stored in the
string during its trajectory. For example, this will happen during the initial acceleration phase.
Solution 4
We are given wave velocity, . This is a constant because h and g are fixed. Hence,
group velocity and phase velocity are the same and . For the
.
(a) .
(b) Because the group velocity is twice the phase velocity we have
and so . Integrating both sides of this equation allows one to write
so that the dispersion relation is where A is a constant.
(c) .
Solution 6
For a source moving at velocity v, the Doppler shifted frequency f’ is where
f0 is the original frequency. The ± sign refers to the source moving towards or away from the detec-
tor. Hence, the maximum change in frequency relative to f0 is
. In our case, v = 3.6 × 108 m s-1, c = 3 × 108 m s-1,
and f0 = 1010 Hz giving f = 9.26 kHz. The component of velocity in the direction of the detector is
v × cos(θ) where θ is the angle between the ground and the plane flying at height h. A simple
expression for the time dependence of the angle θ = θ(t) is found knowing that the aircraft is mov-
ing at a constant velocity (and assuming the airplane flies right over the ground station). If the dis-
t2
2
m1 m2+ -------------------=
λ 2 5 m×=
ω Ak0.5=
kd dω 2
ω Ak4=
f′ f 0 1 v c⁄±( )⁄=
f f 0 1 1 1 v c⁄±( )⁄–( ) f± 0 v c v+( )⁄( )= =
8
tance to the plane (range) is l then (assuming we set t = 0 when the plane is
overhead). Then . The Doppler shift as a
function of time is f × cos(θ(t)). If the plane does not fly overhead it is necessary to introduce
another angle φ to describe its position.
If the plane is flying in an arc the Doppler shift can decrease. For example, it will be zero if the
plane is flying in a circle centered on the ground station. Assuming the plane flys at a constant
speed, whatever trajectory the plane describes, it cannot have a Doppler shift that is greater than the
maximum value we calculated f for that velocity.
The reason why a typical radar antenna at an airport is highly directional and rotates is to find
the angular position of the aircraft with respect to the radar detector. Pulsed radar can be used to
find the range (distance) to the aircraft. The trajectory of the aircraft can be obtained by comparing
sequential position updates. Doppler radar is not required.
Solution 7
If magnetic charge g existed Maxwell’s equations would have to include magnetic charge den-
sity and magnetic current density as well as the usual electron charge density and elec-
tron current density . The new equations would be
where and .
Conservation of magnetic current is expressed as
For a particle carrying both electric charge e and magnetic charge g, the Lorentz force is
Reminding ourselves of the SI units used for magnetic intensity H[A m-1], magnetic flux density B[T] where B = µH, electric field strength E[V m-1],the displacement vector field D = εE, permia-
bility µ[N A-2] or [H m-1], permittivity ε[A2 s2 N-1 m-2] or [F m-1], and where c is
the speed of light in free-space. If corresponds to energy density
then it makes sense to write since and have dimensions of energy
density U[J m-3]. Maxwell’s equations in terms of E and H with magnetic charge can be written
l 2 h2 v2t 2+=
θ t( )( )cos vt l⁄ v t h2 v2t2+⁄ 1 h2 v2t2⁄ 1+⁄= = =
ρm Jm ρe
ε0 1 µ0⁄ c2=
G ε 2 --- E i µ
2 --- H+= ε E2 µ H 2
ε 2 --- ∇ E⋅
since
The complex field G can be used to simplify the usual four Maxwell equations to just two equa-
tions.
Solution 8
We are given the information that a small metal sphere has capacitance in air. First, we calculate the radius r0 of the sphere using the formula
It follows that
Now consider the case when the metal sphere radius r0 with charge Q is coated with a dielectric of
relative permittivity to radius r1 and from r1 radius to r2. The voltage drop from r0 through
the dielectrics to a much larger metal sphere of radius r2 is found by integrating
For this particular problem we now take the limit . Capacitance
µ 2 --- H µ
2 --- H∇×+∇× ε
2 ---ε
C0 4πε0r0=
εr1 εr2
r1
and , so that
and we may conclude that the thickness of the dielectric film is . Charging energy . This value is similar to
ambient thermal energy .
Solution 9
(a) The molecule consists of two particles mass and mass with position and
respectively. The center of mass coordinate is and relative position vector is . We assume that
the potential depends only on the difference vector and if the origin is the center of
mass then , so that
Since, for example,
Now, combining center of mass motion and relative motion, the Hamiltonian is the sum of kinetic and potential energy terms
where the total kinetic energy is
or
εr1 10= εr2
1= C0 1.1 10 18– F×= C 2.2 10 18– F×=
r0 10 nm=
r1 1 10–
4 --- r0 22.5 nm= = =
r1 r0– 12.5 nm= E e2 2C⁄ 1.6 10 19– 4.4 10 18–×⁄× 36 meV= = =
kBT 25 meV=
m2
2 m1m2 2
and the reduced mass is
Because the potential energy of the two interacting particles depends only on the separation
between them, . The frequency of oscillation of the molecule is deter-
mined by the potential and given by where κ is the spring constant and m is the
reduced mass . Since the isotope form of the molecule interacts in the same
way, , and the ratio of oscillation frequency is given by
We could get the same result quite simply if we assume motion is restricted to one dimension.
In that case the force on mass m1 and mass m2 is given by their relative displacement multiplied by
the spring constant κ.
giving a characteristic equation
M m 1 m2+=
ω κ m⁄=
κ κ′=
ω′ ω -----
m1m2 m ′1 m ′2+( ) m ′1m ′2 m1 m2+( ) -----------------------------------------=
m1 m2
u v
e iωt–
κu– κ m2ω2–( )+ v 0=
12
and
and, as before, since the isotope form of the molecule interacts in the same way, , and
the ratio of oscillation frequency is given by
(b) To find the ratio of vibrational frequencies we use the result in part (a) and put in the atomic
masses to give . As expected, the lighter molecule
vibrates at a higher frequency (by about 2%) than the molecule.
Solution 10
(a) For small displacement x we assume F0 doesn’t change much so for extension the work
done is . Hence, the potential energy of the spring is equal to the force F0 multiplied by the
extension . For we have , so that the potential energy is
. This is the potential energy of a harmonic oscillator with spring constant
. This oscillator has frequency .
(b) The frequency of oscillation in radians per second is given by
, so the period of oscillation is
, i.e., about 2 seconds.
κ m1m2
m1 m2+ -------------------
ω2 mω2= =
m1m2 m ′1 m ′2+( ) m ′1m ′2 m1 m2+( ) -----------------------------------------=
ω C12
ω C13
--------- 13 16× 12 16+( )× 12 16× 13 16+( )× ----------------------------------------------- 1.02= = C O1612
C O1613
l x l« l l2 x2+ l– x 2 2l⁄= =
V F0x2 2l⁄ κx2 2⁄= =
κ F0 l⁄= ω κ m⁄ F0 ml⁄= =
ω F0 ml⁄ 450 20 2.5×⁄ 9 3 rad s 1–= = = =
τ 1 f⁄ 2π ω⁄ 2.09 s= = =
Applied quantum mechanics 1
LAST NAME FIRST NAME
Useful constants MKS (SI)
Planck’s constant
Avagadro’s number
h 6.58211889 26( ) 10 16–× eV s=
h 1.054571596 82( ) 10 34–× J s=
e 1.602176462 63( ) 10 19–× C=
m0 9.10938188 72( ) 10 31–× kg=
mn 1.67492716 13( ) 10 27–× kg=
mp 1.67262158 13( ) 10 27–× kg=
kB 1.3806503 24( ) 10 23–× J K 1–=
kB 8.617342 15( ) 10 5–× eV K 1–=
ε0 8.8541878 10 12–× F m 1–=
µ0 4π 10 7–× H m 1–=
c 1 ε0µ0⁄=
aB 4πε0h2
2
PROBLEM 1 (a) The Sun has a surface temperature of 5800 K and an average radius 6.96 × 108 m. Assuming
the mean Sun-Mars distance is 2.28 × 1011 m, what is the total radiative power per unit area inci-
dent on the upper Mars atmosphere facing the Sun?
(b) If the surface temperature of the Sun was 6800 K, by how much would the total radiative
power per unit area incident on Mars increase?
PROBLEM 2
If a photon of energy 2 eV is reflected from a metal mirror, how much momentum is
exchanged? Why can the reflection not be modeled as a collision of the photon with a single elec-
tron in the metal?
(a) Draw an energy level diagram for the Li++ ion.
(b) Derive the expression for the energy (in eV) and wavelength (in nm) of emitted light from
transitions between energy levels.
(c) Calculate the three longest wavelengths (in nm) for transitions terminating at .
(d) If the lithium ion were embedded in a dielectric with relative permittivity εr = 10, what
would be the expression for the energy (in eV) and wavelength (in nm) of emitted light from transi-
tions between energy levels.
PROBLEM 4
(a) There is a full symmetry between the position operator and the momentum operator. They
form a conjugate pair. In real-space momentum is a differential operator. Show that in k-space
position is a differential operator, , by evaluating expectation value
in terms of φ(kx) which is the Fourier transform of ψ(x).
(b) The wave function for a particle in real-space is . Usually it is assumed that position
x and time t are continuous and smoothly varying. Given that particle energy is quantized such that
, show that the energy operator for the wave function is .
PROBLEM 5
A simple model of a heterostructure diode predicts that current increases exponentially with
increasing forward voltage bias. Under what conditions will this predicted behavior fail?
PROBLEM 6
Write down the Hamiltonian operator for (a) a one-dimensional simple harmonic oscillator, (b)
a Helium atom, (c) a Hydrogen molecule, (d) a molecule with nn nuclei and ne electrons.
n 2=
ih px∂ ∂
∞
∫=
∂
PROBLEM 7
Calculate the classical velocity of the electron in the n-th orbit of a Li++ ion. If this electron is
described as a wave packet and its position is known to an accuracy of x = 1 pm, calculate the
characteristic time τx for the width of the wave packet to double. Compare τx with the time to
complete one classical orbit.
PROBLEM 8
What is the Bohr radius for an electron with effective electron mass in a
medium with low-frequency relative permittivity corresponding to the conduction
band properties of single crystal InAs?
PROBLEM 9
Because electromagnetic radiation possesses momentum it can exert a force. If completley
absorbed by matter, the absorbed electromagnetic radiation energy per unit time per unit area is a
pressure called radiation pressure.
(a) If the maximum radiative power per unit area incident on the upper Earth atmosphere facing
the Sun is 5.5 kW m -2, what is the corresponding radiation pressure?
(b) Estimate the photon flux needed to create the pressure in (a).
(c) Compare the result in (a) with the pressure due to one atmosphere.
m* 0.021 m0×=
Solution 2
Assume the photon of energy E = 2 eV is incident from free-space at angle θ normal to a flat
metal mirror. The magnitude of photon momentum is where frequency ω is
given by the quanta of photon energy . The momentum change upon reflection is
which, for , gives a value
. The reason why this reflection can
not be modeled as a collision of the photon with a single electron in the metal is that in such a situ-
ation it is not possible to conserve both energy and momentum and have the photon maintain its
wavelength. Collision with an electron would take away energy from the photon and cause a
change in photon wavelength. Since, by definition, reflected light has the same wavelength as the
incident light, photon energy cannot change.
Another way to express this is that the difference in dispersion relation for free electrons and
photons is such that light cannot couple directly to a free electron.
Solution 3
Emission wavelength is , giving , ,
Solution 4
pph hk hω c⁄= =
p 2hω c⁄( ) θcos= θ 0=
2E c⁄ 2 2 1.6 10 19– 3 108×⁄××× 2.1 10 27– kg m s 1–×= =
En Ry Z2
λ5 2, 48.1 nm=
εr 10= En2 En1
∞
∫ 1
∞–
∞
∫ ∞–
∞
∫ ∞–
∞
∫= =
x⟨ ⟩ 1 2π ------ dx dk′φ* k ′( )e i– k ′xx dkφ k( )
k∂ ∂ e ikx
(b) The expectation value of the energy operator is
Hence, in t-space energy is a differential operator, .
Solution 5
Series resistance. Space charging at current density . For bipolar devices with direct a
band gap can have stimulated emission.
Solution 6
We are asked to write down the Hamiltonian operator for (a) a one-dimensional simple har-
monic oscillator, (b) a Helium atom, (c) a Hydrogen molecule, (d) a molecule with nn nuclei and ne
electrons.
(a) The Hamiltonian for a particle mass m restricted to harmonic oscillatory motion of fre-
quency ω in the x-direction is
(b) and (c) may be found from the general solution (d) for a molecule with nn nuclei and ne elec-
trons in which the Hamiltonian is
x⟨ ⟩ 1 2π ------ dx dk′φ* k′( )e i– k′ xx φ k( )e ikx
ix -------
ix -------
2πi -------- dx dk ′φ* k′( )e i– k′ x dk
k∂ ∂ φ k( )
e i kx
∞–
∞
∞
∂ φ k( ) ∞–
∞
∫ 1
∞–
∞
∫ ∞–
∞
∫ ∞–
∞
∫= =
E⟨ ⟩ 1 2π ------ dω d t′φ * t ′( )e iω– t ′hω d tφ t( )
t∂ ∂ e iω t′

∞–
∞
∫ ∞–
∞
∫ ∞–
∞
∫=
E⟨ ⟩ 1 2π ------ dω dt ′φ* t ′( )e iω– t ′hω φ t( )e iω t
iω -------
iω -------
∞–
∞
∫–
∞–
∞
∫ ∞–
∞
∫=
E⟨ ⟩ h– 2πi -------- dω dt ′φ* t ′( )e iω– t ′ dt
t∂ ∂ φ t( )
∞–
∞
∂ φ t( ) ∞–
∞
∂ φ t( ) ∞–
∂ φ t( ) ∞–
x2
2
2m0 --------- ∇2 V r( )+
6
Here, the first term is the contribution to the kinetic energy from the nn nuclei of mass Mj, the sec-
ond term is the kinetic energy contribution from the ne electrons of mass m0, and V(r) is the poten-
tial energy given by
The first term in the potential is the electron-electron coulomb repulsion between the electrons i and
i'. The second term is the nucleus-nucleus coulomb repulsion between the nuclei j and j' with
charges Zj and Z j' respectively. The third term is the coulomb attraction between electron i and
nucleus j.
Solution 7
For n = 1, orbit time ~ 10-17 s, dispersion time ~ 10-20 s.
Solution 8
Solution 9
(a) The maximum radiative power per unit area incident on the upper Earth atmosphere facing
the Sun is Stotal = 5.5 kW m-2. The radiation pressure us just .
(b) From Exercise 1 Chapter 2, so the number of photons per second per
unit area is .
(c) The radiation pressure in (a) is a small value compared to one atmosphere which is about
. Never-the-less, electromagnetic radiation from the Sun can be absorbed, exert a force,
and change the direction of small particles in space. For example, it is responsible for continuously
sweeping dust particles out of the solar system. As another example, uncharged dust particles from
a comet can form a dust tail whose direction and shape is determined by electromagnetic radiation
pressure from the Sun. Interestingly, comets can also shed charged dust particles to form what is
called a gas tail. The gas tail is swept along by a stream of charged particles and magnetic field
lines emitted by the Sun that is called the solar wind. Often a comet will have two separate tails
because the solar wind often does not point radially outward from the Sun.
V r( ) e2
4πε0rjj ′ ------------------
S total c⁄ 1.8 10 5– N m 2–×=
hωaverage 1.92 eV=
S total ehωaverag e⁄ 1.8 1022× s 1– m 2–=
105 N m 2–
Applied quantum mechanics 1
LAST NAME FIRST NAME
Useful constants MKS (SI)
Planck’s constant
Avagadro’s number
h 6.58211889 26( ) 10 16–× eV s=
h 1.054571596 82( ) 10 34–× J s=
e 1.602176462 63( ) 10 19–× C=
m0 9.10938188 72( ) 10 31–× kg=
mn 1.67492716 13( ) 10 27–× kg=
mp 1.67262158 13( ) 10 27–× kg=
kB 1.3806503 24( ) 10 23–× J K 1–=
kB 8.617342 15( ) 10 5–× eV K 1–=
ε0 8.8541878 10 12–× F m 1–=
µ0 4π 10 7–× H m 1–=
c 1 ε0µ0⁄=
aB 4πε0h2
2
PROBLEM 1 Prove that particle flux (current) is zero if the one-dimensional exponential decaying wave func-
tion in tunnel barrier of energy V0 and finite thickness L is , where κ is a real positive number and particle energy .
PROBLEM 2
(a) Use a Taylor expansion to show that the second derivative of a wavefunction sampled
at positions , where j is an integer and h0 is a small fixed increment in distance x, may be
approximated as
(b) By keeping additional terms in the expansion, show that a more accurate approximation of
the second derivative is
PROBLEM 3
Using the method outlined in Exercise 7 of Chapter 3 as a starting point, calculate numerically the first four energy eigenvalues and eigenfunctions for an electron with effective mass
confined to a potential well V(x) = V0 of width L = 10 nm with periodic boundary conditions.
Periodic boundary conditions require that the wave function at position x = 0 is connected (wrapped around) to position x = L. The wave function and its first derivative are continuous and smooth at this connection.
Your solution should include plots of the eigenfunctions and a listing of the computer program you used to calculate the eigenfunctions and eigenvalues.
PROBLEM 4
Using the method outlined in Exercise 7 of Chapter 3 as a starting point, calculate numerically the first six energy eigenvalues and eigenfunctions for an electron with effective mass
confined to a triangular potential well of width L = 20 nm bounded by barriers of infinite energy at x < 0 and x > L. The triangular potential well as a function of distance x is given by V(x) = V0 × x / L where V0 = 1 eV.
Explain the change in shape of the wave function with increasing eigenenergy. Your solution should include plots of the eigenfunctions and a listing of the computer program
you used to calculate the eigenfunctions and eigenvalues.
PROBLEM 5
Calculate the transmission and reflection coefficient for an electron of energy E , moving from left to right, impinging normal to the plane of a semiconductor heterojunction potential barrier of energy , where the effective electron mass on the left-hand side is and the effective electron mass on the right-hand side is .
If the potential barrier energy is V0 = 1.5 eV and the ratio of effective electron mass on either
side of the heterointerface is m1 / m2 = 3, at what particle energy is the transmission coefficient
unity? What is the transmission coefficient in the limit E → ∞ ?
ψ x t,( ) Be κ x– iωt–= E hω V0<=
ψ x( )
x2
2
d
d ψ x j( ) ψ x j 1–( ) 2ψ x j( ) ψ x j 1+( )+–
h0 2
d d ψ x j( )
ψ– x j 2–( ) 16+ ψ x j 1–( ) 30ψ x j( ) 16ψ x j 1+( ) ψ x j 2+( )–+–
12h0 2---------------------------------------------------------------------------------------------------------------------------------------=
SOLUTIONS
Solution 1 To prove that particle flux (current) is zero if the one-dimensional exponential decaying wave
function in tunnel barrier of energy V0 and finite thickness L is , where κ is a real positive number and particle energy , we substitute the wave function into the current operator
Solution 2
(a) Consider the Taylor expansion for the function f(x).
where j is an integer and h0 is a small fixed increment in distance x. Keeping terms to order , we see that the first derivative is
Hence, the first derivative of a wavefunction sampled at positions may be approxi- mated as
To find the second derivative we keep terms to order so that
and
or
so that the second derivative of a wavefunction sampled at positions may be approximated as
This is the three-point approximation to the second derivative accurate to second order, . (b) By keeping five terms in the expansion instead of three one may obtain a more accurate
approximation of the second derivative accurate to fourth order, . To see how, we use the fol-
lowing Taylor expansions with derivatives up to fourth-order.
ψ x t,( ) Be κx– iω t–= E hω V0<=
J i– eh 2m
----------- ψ* x t,( ) xd
d ψ * x t,( )– =
J i– eh 2m
----------- B*e κx– iωt+ κB– e κx– iωt–( ) Be κ x– iωt– κB– *e κx– iω t+( )–( )=
J κieh 2m
----------- B 2 e 2κx– e 2κ x––( ) 0= =
f xj 1+( ) f xj( ) f′ x j( )h0 f′′ x j( )
2! -------------- h0
---------------- h3 0 …+ + + +
n⋅⋅ n ∑= =
h0 ---------------------------------=
ψ x j 1+( ) ψ x j( )– h0
--------------------------------------=
h0 2
f xj 1+( ) f xj( ) f′ x j( )h0 f′′ x j( )
2 -------------- h0
2+ +=
f xj 1–( ) f xj( ) f ′ x j( )h0 f ′′ x j( )
2 -------------- h0
2+–=
f xj 1–( ) f xj 1+( )+ 2f xj( ) f′′ x j( )h0 2+=
f ′′ x j( ) f xj 1–( ) 2 f xj( )– f xj 1+( )+
h0 2
x2
2
d d ψ x j( )
ψ x j 1–( ) 2ψ x j( ) ψ x j 1+( )+–
h0 2
4
To eliminate the terms in the first, third, and fourth derivative we multiply each equation by coeffi- cients a, b, c, d, respectively, to obtain
The coefficients that eliminate terms in the first, third, and fourth derivative must satisfy
The solution to this set of linear equations is
Setting we obtain
so that
so that the second derivative of a wavefunction sampled at positions may be approximated as
Solution 3
To find numerically the first four energy eigenvalues and eigenfunctions for an electron with effective mass confined to a potential well V(x) = V0 of width L = 10 nm with periodic boundary conditions we descretize the wavefunction and potential appearing in the time- independent Schrödinger equation
using a discrete set of equally-spaced points such that position where the index
, and is the interval between adjacent sampling points so one may define
. The region in which we wish to solve the Schrödinger equation is of length
. At each sampling point the wave function has value and the potential is
f xj 2+( ) f xj( ) 2f′ x j( )h0 2f′ ′ x j( )h0 2 4
3 --- f′′′ x j( )h0
3 2 3 --- f ′′′′ x j( )h0
4+ + + +=
f xj 1+( ) f xj( ) f′ x j( )h0 1 2 --- f ′′ x j( )h0
2 1 6 --- f ′′′ x j( )h0
3 1 24 ------ f ′′′′ x j( )h0
4+ + + +=
f xj 1–( ) f xj( ) f′ x j( )h0 1 2 --- f ′′ x j( )h0
2 1 6 --- f ′′′ x j( )h0
3– 1 24 ------ f ′′′′ x j( )h0
4+ +–=
f xj 2–( ) f xj( ) 2f ′ x j( )h0 2f ′′ x j( )h0 2+–
4 3 --- f′ ′ ′ x j( )h0
3 2 3 --- f′′′′ x j( )h0
4+–=
a f xj 2+( ) bf xj 1+( ) c f xj 1–( ) d f xj 2–( )+ + + a b c d+ + +( ) f xj( ) 2a b c– 2d–+( ) f′ x j( )h0
2a b 2 --- c
3 ------ b
6 --- c
6 ---– 4d
3 ------ b
24 ------ c
24 ------ 2d
+ +
+ +
=
0 4a 3 ------ b
a d=
b 16d–=
c 16d–=
a 1=
f xj 2+( ) 16f xj 1+( )– 16f xj 1–( )– f xj 2–( )+ 30– f x j( ) 12f ′′ x j( )h0 2–=
f ′′ x j( ) f xj 2+( ) 16 f xj 1+( ) 30f xj( ) 16 f xj 1–( ) f xj 2–( )–+–+–
12h0 2
x2
2
d d ψ x j( )
ψ– x j 2–( ) 16+ ψ x j 1–( ) 30ψ x j( ) 16ψ x j 1+( ) ψ x j 2+( )–+–
12h0 2---------------------------------------------------------------------------------------------------------------------------------------=
L Nh0= ψ j ψ x j( )=
Applied quantum mechanics 5
. The second derivative of the discretized wave function we use the three-point finite-
difference approximation which gives
Substitution into the Schrödinger equation gives a matrix equation
where the Hamiltonian is a symmetric tri-diagonal matrix. The diagonal matrix elements are
and the adjacent off-diagonal matrix elements are
As usual, the wavefunction must be continuous and smooth. In addition, the periodic boundary
conditions require so that . In this situation the Hamiltonian
becomes a matrix H and the Schrödinger equation is
where I the identity matrix. The matrix elements in the upper right and lower left corners are due to
the periodic boundary conditions. For the case we are interested in, the first four eigenenergies are:
E1 = 0 eV, E2 = 0.215 eV, E3 = 0.215 eV, E4 = 0.860 eV
Note the trivial solution with a constant wavefunction has a zero energy eigenvalue. Eigenenergy
E2 and E3 are degenerate with orthogonal (sine and cosine) wavefunctions.
V j V x j( )=
x2
2
d d ψ x j( )
ψ x j 1–( ) 2ψ x j( ) ψ x j 1+( )+–
h0 2
------------------------------------------------------------------=
Hψ x j( ) u jψ– x j 1–( ) dψ x j( ) u j 1+ ψ x j 1+( )–+ Eψ x j( )= =
d j h2
mh0 2--------- Vj+=
u j h2
N N×
H EI–( )ψ
d1 E–( ) u– 2 0 0 . . . u1– u2– d2 E–( ) u3– 0 . . . .
0 u3– d3 E–( ) u4– . . . .
0 0 u4– d4 E–( ) . . . . . . . . . . . . . . . . . . uN 2–– 0 . . . . . . dN 1– E–( ) uN 1––
u1– . . . . . uN– dN E–( )
Solution 4
The first six eigenenergies are: E1 = 0.259 eV, E2 = 0.453 eV, E3 = 0.612 eV, E4 = 0.752 eV, E5 =
0.884 eV, E6 = 1.02 eV
Solution 5
For a potential step V0, impedance matching occurs when the ratio of effective electron masses is
or
If the potential barrier energy is V0 = 1.5 eV and the ratio of effective electron mass on either
side of the heterointerface is m1 / m2 = 3, then
and the transmission coefficient in the limit E → ∞ is found by noting that
so that
For m1 / m2 = 3 this gives 0.928 transmission and 0.072 reflection. One does not expect a particle with infinite energy to be reflected by a finite potential step!
m2
m1 ------
k2
k1 ----
E ∞→
m1
m2 ------=
LAST NAME FIRST NAME
Useful constants MKS (SI)
Planck’s constant
Avagadro’s number
h 6.58211889 26( ) 10 16–× eV s=
h 1.054571596 82( ) 10 34–× J s=
e 1.602176462 63( ) 10 19–× C=
m0 9.10938188 72( ) 10 31–× kg=
mn 1.67492716 13( ) 10 27–× kg=
mp 1.67262158 13( ) 10 27–× kg=
kB 1.3806503 24( ) 10 23–× J K 1–=
kB 8.617342 15( ) 10 5–× eV K 1–=
ε0 8.8541878 10 12–× F m 1–=
µ0 4π 10 7–× H m 1–=
c 1 ε0µ0⁄=
aB 4πε0h2
2
PROBLEM 1 Write a computer program in Matlab that uses the propagation matrix method to find the trans-
mission resonances of a particle of mass (where is the bare electron mass). (a) Use your computer program to find transmission as a function of energy for a particle mass
m0 through 12 identical one-dimensional potential barriers each of energy 1 eV, width 0.1 nm, sequentially placed every 0.5 nm (so that the potential well between each barrier has width 0.4 nm). What are the allowed (band) and disallowed (band gap) ranges of energy for particle transmission through the structure? How do you expect the velocity of the transmitted particle to vary as a func- tion of energy?
(b) How do these bands compare with the situation in which there are only three barriers, each with 1 eV barrier energy, 0.1 nm barrier width, and 0.4 nm well width?
Your solution should include plots of transmission as a function of energy and a listing of the computer program you used.
PROBLEM 2 Vertical-cavity surface-emitting lasers (VCSELs) operating at wavelength λ = 1300 nm are
needed for local area fiber optic applications. (a) Use the propagation matrix to design a high-reflectivity Bragg mirror for electromagnetic
radiation with center wavelength λ0 = 1300 nm incident normal to the surface of an AlAs / GaAs periodic dielectric layer stack consisting of 25 identical layer-pairs. Each individual dielectric layer has a thickness λ / 4n, with n being the refractive index of the dielectric. Use for the refractive index of AlAs and for GaAs. Calculate and plot optical reflectivity in the wavelength range .
(b) Extend the design of your Bragg reflector to a two-mirror structure similar to that used in the design of a VCSEL. This may be achieved by increasing the number of pairs to 50 and making the thickness of the central GaAs layer one wavelength long. Recalculate and plot the reflectivity over the same wavelength range as in (a). Using high wavelength resolution to find the bandwidth of this optical pass band filter near .
Your results should include a printout of the computer program you used and a computer-gener- ated plot of particle transmission as a function of incident wavelength.
m m0= m0
λ 1300 nm=
Center region
nAlAs
nGaAs
nAlAs
nGaAs
nGaAs
nAlAs
λ0 / nGaAs
PROBLEM 3
Write a computer program in Matlab that uses the propagation matrix method to find the trans- mission resonances of a particle of mass (where is the bare electron mass) for the following one-dimensional potentials.
(a) A uniform electric field falls across the double barrier and single well structure as shown in the Fig. The right-hand edge of the 2.25-nm-thick barrier of energy 1.13 eV is at a potential -0.6 eV below the left-hand edge of the 3-nm-thick barrier of energy 0.5 eV. The well width is 2.7 nm. Comment on the changes in transmission you observe.
(b) Rewrite your program to calculate transmission of a particle as a function of potential drop caused by the application of an electric field across the structure. Calculate the specific case of ini- tial particle energy E = 0.025 eV with the particle incident on the structure from the left-hand side.
PROBLEM 4 Write a computer program to solve the Schrödinger wave equation for the first 17 eigenvalues
of an electron with effective mass confined to the periodic potential sketched in the following Fig. with periodic boundary conditions. Each of the eight quantum wells is of width 6.25 nm. Each quantum well is separated by a potential barrier of thickness 3.75 nm. The barrier potential energy is 0.9 eV. How many energy bandgaps are present in the first 17 eigenvales and what are their values? Plot the highest energy eigenfunction of the first band and the lowest energy eigenfunction of the second band.
10
0.6 eV
4
PROBLEM 5 Use the results of Problem 4 with periodic boundary conditions to approximate a periodic one-
dimensional delta function potential with period 10 nm by considering 8 potential barriers with energy 20 eV and width 0.25 nm. Plot the lowest and highest energy eigenfunction of the first band. Explain the difference in the wave functions you obtain.
Applied quantum mechanics 5
Solution 1
Essentially the same as exercise 3 in chapter 4 (Chap4Exercise3 Matlab file) of the book.
Solution 2
Essentially the same as exercise 4 in chapter 4 (Chap4Exercise4 Matlab file) of the book.
Solution 3
Essentially the same as exercise 7 in chapter 4 (Chap4Exercise7 Matlab file) of the book.
Solution 4
This requires putting in periodic boundary conditions, but is otherwise the same as exercise 9 in
chapter 4 of the book. There are eight quantum wells and so there are eight eigenfunctions associ-
ated with each band. The first seventeen eigenfunctions include two complete bands each contain-
ing eight states. The eigenvalues and band gaps are:
E1 = 0.086418 eV
E2 = 0.086616 eV
E3 = 0.086616 eV
E4 = 0.087097 eV
E5 = 0.087097 eV
E6 = 0.087584 eV
E7 = 0.087584 eV
E8 = 0.087787 eV
E9 = 0.3354 eV
E10 = 0.33669 eV
E11 = 0.33669 eV
E12 = 0.33987 eV
E13 = 0.33987 eV
E14 = 0.34314 eV
E15 = 0.34314 eV
E16 = 0.34453 eV
E17 = 0.69779 eV
Solution 5
Use code from Problem 4. The difference in form of the wavefunction for lowest eigenenergy
value in the band (E1 = 0.049316 eV) and highest eigenenergy value in the band (E8 = 0.053691
eV) is due to value of Bloch wave vector k in the wavefunction ψk(x) = uk(x) exp(ik.x) that
describes a particle in a potential of period L. For E1 the value of k = 0 and for E8 the value of k =
π/L.
0.0
0.9
Distance, x (nm)
Distance, x (nm)
Distance, x (nm)
Distance, x (nm)
LAST NAME FIRST NAME
Useful constants MKS (SI)
Planck’s constant
Avagadro’s number
h 6.58211889 26( ) 10 16–× eV s=
h 1.054571596 82( ) 10 34–× J s=
e 1.602176462 63( ) 10 19–× C=
m0 9.10938188 72( ) 10 31–× kg=
mn 1.67492716 13( ) 10 27–× kg=
mp 1.67262158 13( ) 10 27–× kg=
kB 1.3806503 24( ) 10 23–× J K 1–=
kB 8.617342 15( ) 10 5–× eV K 1–=
ε0 8.8541878 10 12–× F m 1–=
µ0 4π 10 7–× H m 1–=
c 1 ε0µ0⁄=
aB 4πε0h2
2
PROBLEM 1 An electron in an infinite, one-dimensional, rectangular potential well of width L is in the simple superposition state consisting of the ground and third excited state so that
Find expressions for:
(d) The average momentum, .
(e) The current flux, .
PROBLEM 2 (a) Show that the density of states for a free-particle of mass m in two-dimensions is
(b) At low temperature, electrons in two electrodes occupy states up to the Fermi energy, EF. The two electrodes are connected by a two dimensional conductance region. Derive an expression for the conductance of electrons flowing between the two electrodes as a function of applied voltage V, assuming the transmission coefficient through the two-dimensional region is unity. Consider the two limiting cases eV >> EF and eV << EF
PROBLEM 3
Derive expressions for the two-dimensional and one-dimensional density of photon states in a homogeneous dielectric medium characterized by refrac- tive index, nr.
PROBLEM 4 (a) In a particular system the dispersion relation for electrons in one-dimension is
, where t and L are constants and the wave vector in the x-
direction is . This dispersion relation can be derived using a nearest neighbor tight binding model where t is the overlap integral between atomic orbitals. Choosing one hundred equally spaced discrete values of kx, write a computer program
and plot the electron density of states using
and . (b) If one includes next nearest neighbor interactions, the dispersion relation in (a) can, to within a scaling factor, be written . Write a com- puter program to plot the dispersion relation. Then calculate and plot the electron den- sity of states using , , and compare with the result you obtained in (a) including a comparison with the effective electron mass at the band edges.
ψ x t,( ) 1
ψ x t,( ) 2
0 kx π L⁄< <
N E( ) Γ π⁄ E Ek–( )2 Γ 2⁄( )2+
--------------------------------------------- k ∑= Γ t 10⁄=
Γ t 10⁄= t 1–= t ′ 0.2–=
Applied quantum mechanics 3
(c) Write a computer program to plot the electron density of states for a square lattice and cubic lattice both with lattice constant L and for which the dispersion relation is
and
respectively. Use and . (c) The dispersion relation for a two dimensional hexagonal lattice is
where is the overlap integral with the next nearest neighbor. Calculate the electron
density of states using , , and . Compare with the result
you obtain when .
PROBLEM 5 A hydrogen atom is in its ground state with electron wave function
In this expression aB is the Bohr radius and r is a radial coordinate. Use spherical coordinates to find the expectation value of position r and momentum pr for the electron in this state. You should use the fact that in radial coordinates the Her-
mitian momentum operator is .
PROBLEM 6 Using the fact that the Hamiltonian H appearing in the Schrödinger equation
is Hermitian (i.e., ), show that the time dependence of the average
value of the operator is
PROBLEM 7 Show that:
).
(c) The momentum operator acting on the wavefunction is Hermitian.
PROBLEM 8 Classical electromagnetic theory uses real magnetic and electric fields coupled via Maxwell’s equations. The magnetic and electric fields each have physical meaning. Both fields are needed to describe both the instantaneous state and time evolution of the
Ek 2t kxL( ) kyL( )cos+cos( )= Ek 2t kxL( ) kyL( ) kzL( )cos+( )cos+cos( )=
Γ t 10⁄= t 1–=
=
t′ 0=
xd d ψ x( )
4
system. Quantum mechanics uses one complex wave function to describe both the instantaneous state and time evolution of the system. It is also possible to describe quantum mechanics using two coupled real wave functions corresponding to the real and imaginary parts of the complex wave function. However, such an approach is more complicated. In addition, the real and imaginary parts of the wave function have no special physical meaning. (a) If classical electromagnetism were described in terms of a single complex field
show that Maxwell’s equations in free space and in the absence of free charges may written as the complex equations
and
(b) Show that the energy flux density in the electromagnetic field given by the Poynting vector is
(c) If the field G is purely real, what is the value of S?
(d) Show that the electromagnetic energy density is .
G ε0 2⁄ E i µ0 2⁄ H⋅+⋅=
∇ G⋅ 0=
i t∂
SOLUTIONS
Solution 1 (a) Let the potential in the region be zero and infinity elsewhere. The
eigenfunctions are , where ,
, and . Hence,
and the probability distribution of the particle in the superposition state
is
(b) Expectation value of position is
By symmetry, the first two terms in the integrand are zero, leaving
making use of gives
which, after integrating by parts, results in
(c) The momentum probability density is found using the Fourier transform
L 2⁄ x L 2⁄< <–
ψn x t,( ) 2 L⁄ kn x L 2⁄+( )( )e iω nt–
sin= kn nπ L⁄=
ω n hkn 2 2m⁄= n 1 2 3 …, , ,=
ψ 1 x t,( ) 2 L⁄ π L⁄( ) x L 2⁄+( )( )e iω 1 t–
sin 2 L⁄ πx L⁄( )e iω1 t–
( )cos= =
ψ 4 x t,( ) 2 L⁄ 4π L⁄( ) x L 2⁄+( )( )e iω4 t–
sin 2 L⁄ 4πx L⁄( )e iω4 t–
sin= =
ψ x t,( ) 1 2⁄ ψ1 x t,( ) ψ 4 x t,( )+( )=
ψ x t,( ) 2 1 2 --- ψ1 x( ) 2 ψ4 x( ) 2 2 ψ 1 x( ) ψ4 x( ) ω4 ω1–( )t( )cos+ +( )=
ψ x t,( ) 2 1 L --- cos 2
xπ L ------
sin2 4πx
L2-----------=
x t( )⟨ ⟩ ψ x t,( ) 2x xd∫=
x t( )⟨ ⟩ 1 2 --- ψ1 x( ) 2x ψ4 x( ) 2x 2x ψ1 x( ) ψ4 x( ) ω4 ω1–( ) t( )cos+ +( )
L 2⁄–
L 2⁄
L ------
2 x( ) y( )cossin x y+( ) x y–( )sin+sin=
x t( )⟨ ⟩ ω 4 ω1–( ) t( )cos 1 L --- 5πx
L ---------
x t( )⟨ ⟩ ω4 ω 1–( )t( )cos 1 L --- 2L2
25π2 ----------- 2L2
ψ px t,( ) 2
ψ px t,( ) 1
ip xx– xd∫
ipx x– ψ4 x t,( )e
ipx x– +( ) xd
iω 1t– π x L
------ cos e
6
, , .
The terms and by symmetry, leaving
and
iω 1 t π x L
------ cos e
L ---------
L ------
xd L 2⁄–
L ------
L ---------
∫
+ +=
eax bx( )cos xd∫ eax a bx( )cos b bx( )sin+( )
a2 b2+ ---------------------------------------------------------------=
eax bx( )sin xd∫ eax a bx( ) b bx( )cos–sin( )
a2 b2+ ---------------------------------------------------------------=
ip xL 2⁄ e
e i– px L 2⁄
–( )
Lπ e
e ipx L 2⁄
-------- Bsin2
A 2Lπ
2----------------------------------------= C 8– L π h π2 L2px
2–( ) 4π( )2 L2px 2–( )
-----------------------------------------------------------------=
-------- ψ1 * ψ4
-------- 2 L --- 4π
4πx L
--------- coscos x
--------- sin πx
-------- 8π L2------ e
2 ----- 2π
15π ---------+=
i ω4 ω1–( )t– e
i ω4 ω 1–( ) t +( )=
Applied quantum mechanics 7
(e) The current operator is
Solution 2 (a) Density of states of particle mass m in two dimensions is
per spin.
(b) Current is assumed proportional to applied voltage, electron velocity , trans-
mission coefficient , and density of states.
If we assume , then
For the other situation when we use the binomial expansion
so that . Conductance per electron
spin
Solution 3 The two dimensional and one dimensional photon density of states in an isotropic medium of refractive index is
px t( )⟨ ⟩ 16h 15L --------- ω4 ω1–( )t( )sin=
J x t,( ) ieh–
----------- ψ1 * ψ4
∂ ψ1 * ψ4
L ------
x∂
4πx L
–( )
+



=
L ------
x∂
mL2 ------------ 4 π
D2 E( ) m 2πh2------------=
I e v E( )T E( )D2 E( ) Ed EF
EF e V+
I e 2m 2πh2------------ E Ed
EF
EF eV+ ⋅ 2m
EF ------+
3 2⁄
nr
8
Solution 4 (a) We are given the dispersion relation for electrons in one-dimension as
, where and . We are asked to plot the density
of states using and a sum over 100 states.
Notice that the cosine dispersion in one-dimension has a van-Hove singularity the band edge energies of ±2t.
% HW5Exercise4a clear all; clf;
natoms=100; %number of lattice points (atoms) in each direction npoints=100; %number of points in plot emax=6.5; %max energy in units of t emin=-6.5; %min energy in units of t if emax<=emin;error('Error: require emax > emin');end; estep=(emax-emin)/(npoints-1);
dos=zeros(npoints); %set dos vector to zero w=emin:estep:emax; %energy t=-1; %set energy scale - for s-orbital, + for p-orbital gamma = 0.1; %Energy broadening in units of t gamma=gamma*abs(t); %use units of t gamma2=(gamma/2)^2; const=gamma/pi; kx0=pi/(natoms); %kx constant kx=0:kx0:kx0*natoms;
for ix=1:natoms+1 % Nearest neigbor dispersion for the linear lattice
Ek(ix)=2.0*t*cos(kx(ix)); for j=1:npoints
dos(j)=dos(j)+const/((w(j)-Ek(ix))^2.+gamma2); end
end figure(1); plot(w,dos); ttl=['Nearest neighbor tight binding dispersion, \gamma =',num2str(gamma),'t, natoms=',num2str(natoms)]; title(ttl); xlabel('Energy, E (t)'),ylabel('Density of states, N(E)');
D2 ω( ) ω nr
Ek 2t kxL( )cos= t 1–= 0 kx π L⁄< <
N E( ) Γ π⁄ E Ek–( )2 Γ 2⁄( )2+
--------------------------------------------- k ∑= Γ t 10⁄=
Applied quantum mechanics 9
figure(2); plot(Ek); title(ttl); xlabel('Wave vector, k_x ( \Deltax \pi/L)'),ylabel('Energy, E_k(t)');
(b) The one-dimensional density of states increases at high energy band edge and there is a corresponding increase in effective electron mass in this energy range. When we include next nearest neighbors the density of states is no longer symmetric in energy. The next nearest energy interaction pushes eigenvalues to higher energy.
% HW5Exercise4b clear all; clf;
natoms=100; %number of lattice points (atoms) in each direction npoints=100; %number of points in plot emax=6.5; %max energy in units of t emin=-6.5; %min energy in units of t if emax<=emin;error('Error: require emax > emin');end; estep=(emax-emin)/(npoints-1);
dos=zeros(npoints); %set dos vector to zero w=emin:estep:emax; %energy t=-1; %set energy scale - for s-orbital, + for p-orbital tprime=-0.2 %tprime gamma = 0.1; %Energy broadening in units of t gamma=gamma*abs(t); %use units of t gamma2=(gamma/2)^2; const=gamma/pi; kx0=pi/natoms; %kx constant kx=0:kx0:kx0*natoms;
for ix=1:natoms+1 % Nearest neigbor dispersion for the linear lattice
Ekprime(ix)=(2.0*t*cos(kx(ix)))+(2.0*tprime*cos(2*kx(ix))); Ek(ix)=2.0*t*cos(kx(ix));
for j=1:npoints dos(j)=dos(j)+const/((w(j)-Ekprime(ix))^2.+gamma2);
end end figure(1); plot(w,dos); ttl=['Nearest neighbor tight binding dispersion, \gamma =',num2str(gamma),'t, natoms=',num2str(natoms)]; title(ttl); xlabel('Energy, E (t)'),ylabel('Density of states, N(E)'); figure(2); plot(Ek,'b'); hold; plot(Ekprime,'r'); title(ttl); xlabel('Wave vector, k_x ( \Deltax \pi/L)'),ylabel('Energy, E_k(t)'); hold off;
10
(c) For the square lattice with nearest neighbor interactions only, we have
and for the cubic lattice
Notice that the band edge energy for simple cubic lattice of dimension d occurs at ±2td.
(d) For the hexagonal lattice with second nearest neighbor interactions
which is different compared to the result for the hexagonal lattice with only nearest neighbor interactions shown in the following figure.
Applied quantum mechanics 11
where we made use of the standard integral for .
where A is a real number. Because the measurable quantity must be a real num-
ber, and hence
Solution 6 The Hamiltonian H in the Schrödinger equation is Hermitian and so it is its own Hermi- tian adjoint, i.e., and . The time dependence of the expe-
cation value of the operator is
r⟨ ⟩ φ 100 * rφ100d3r
r3 rd 0
pr⟨ ⟩ ih φ100 * 1
0
0
A
t∂ ∂ψ| ⟩+ +=
i h--- ψ⟨ |AH ψ| ⟩ ψ⟨ |
t∂ ∂ A ψ| ⟩+–=
h--- ψ⟨ |HA ψ| ⟩ i h--- ψ⟨ |AH ψ| ⟩ ψ⟨ |
t∂ ∂ A ψ| ⟩+–=
Solution 7
(a) The position operator is Hermitian if it is its own Hermitian adjoint, i.e.,
or . For the operator this is easy to show
(b) To show that the operator is anti-Hermitian we integrate by parts and make use
of the fact that the wavefunction as
and .
(c) Because the operator is anti-Hermitian, it follows that the momentum operator
is Hermitian since .
and
x x† x=
∞
∞
∞
d ψ x( ) xd
xd d ψ* x( )
i t∂
Applied quantum mechanics 13
Another way of looking at this is to recall that
Now, because we consider an electromagnetic wave in free-space, the speed of light
and , and we have
U S c
U S c
14
LAST NAME FIRST NAME
Useful constants MKS (SI)
Planck’s constant
Avagadro’s number
h 6.58211889 26( ) 10 16–× eV s=
h 1.054571596 82( ) 10 34–× J s=
e 1.602176462 63( ) 10 19–× C=
m0 9.10938188 72( ) 10 31–× kg=
mn 1.67492716 13( ) 10 27–× kg=
mp 1.67262158 13( ) 10 27–× kg=
kB 1.3806503 24( ) 10 23–× J K 1–=
kB 8.617342 15( ) 10 5–× eV K 1–=
ε0 8.8541878 10 12–× F m 1–=
µ0 4π 10 7–× H m 1–=
c 1 ε0µ0⁄=
aB 4πε0h2
2
PROBLEM 1 (a) Write down the Hamiltonian for a particle of mass m in a one-dimensional har-
monic oscillator potential in terms of momentum and position x. (b) If one defines new operators
show that the Hamiltonian can be expressed as
(c) Derive the commutation relation . (d) Using your result from (c) to show that the Hamiltonian is
PROBLEM 2 (a) Find the expectation value of position and momentum for the first excited state
for a particle of mass m in a one-dimensional harmonic oscillator potential. (b) Find the value of the product in uncertainty in position x and momentum px
for the first exited state of a particle of mass m in a one-dimensional harmonic oscillator potential.
PROBLEM 3
Often an operator is time-independent but the corresponding numerical value of the
observable A has a spread in values A about an average value and varies with
time because the system is described by a wavefunction which is not an eigen-
state. The change in in time interval t is the slope multiplied by t.
Hence, the exact time t at which the numerical value of the observable A passes through a specific value will actually have a spread in values t such that
(a) Use the generalized uncertainty relation for time independent
operators and to show that .
(b) Show that the spread in photon number and phase for light of frequency ω is
and that for a Poisson distribution of such photons
px
2 ---+
n φ
Applied quantum mechanics 3
PROBLEM 4 A particle of charge e, mass m, and momentum p oscillates in a one-dimensional har-
monic potential and is subject to an oscillating electric field
. (a) Write down the Hamiltonian of the system.
(b) Find .
(c) Find and show that . Under what conditions is the
quantum mechanical result the same Newton’s second law in
which force on a particle is ?
(d) Find .
(e) Use your results in (b) and (c) to find the time dependence of the expectation value
of position . What happens to the maximum value of as a function of time
when and when is close in value to ?
V x( ) mω0 2x2 2⁄=
Ex ω t( )cos
4
SOLUTIONS
Solution 1 (a) The Hamiltonian for a particle of mass m in a one-dimensional harmonic oscillator potential is
where the particle moves in the potential and oscillates at frequency
, where is a force constant. The operator .
(b) The Hamiltonian can be factored
so it makes sense to define new operators
which, can be written in terms of operators and to give
Substituting this into the Hamiltonian gives
(c) To derive the commutation relation we can write out the differ- ential form of the operators and have them act on a dummy wave function. This gives
and
H mω2
b b †
or, in even more compact form,
If we do not want to use differential operators and a dummy wavefunction then we
could write
where we made use of the commutation relation
Solution 2 (a) The expectation value of position and momentum for the first excited state
of a particle of mass m in a one-dimensional harmonic oscillator potential is found using
and
x ihpx
i pxx xpx–( ) i px x[ , ] i i– h( ) 1= = = = =
px x[ , ] i– h=
bn| ⟩ n1 2⁄ n 1–| ⟩=
m n⟨ | ⟩ δmn=
1 2⁄
px⟨ ⟩ 1⟨ | px 1| ⟩ i hmω 2
------------
------------
6
(b) To find the value of the product in uncertainty in position x and momentum px for the first exited state of a particle of mass m in a one-dimensional harmonic oscillator potential we use
and
and since and we will be interested in finding the value of
and . Starting with , we have
and one can see that for the general state one has . Now
turning our attention to we have
and one can see that for the general state one has .
For the particular case we are interested in and the uncertainty product is
For the general state the uncertainty product
Solution 3 (a) We start with the reasonable assumption that the expectation value of an observable
associated with operator evolves smoothly in time such that
which may be written as
In this problem the operator is time-independent so that
since
In addition, the generalized uncertainty relation for operators and is
which may be re-written as
x x2⟨ ⟩ x⟨ ⟩ 2–( ) 1 2/=
px p2 x⟨ ⟩ px⟨ ⟩ 2–( ) 1 2/=
x⟨ ⟩ 0= px⟨ ⟩ 0= x2⟨ ⟩
p2 x⟨ ⟩ x2⟨ ⟩
x2⟨ ⟩ h 2mω ------------
1⟨ | b b †
† bb
† b
2 --- h= =
A
t∂ ∂ A⟨ ⟩ 0=
2 A B A B,[ ]⟨ ⟩≥
Applied quantum mechanics 7
If operator is the Hamiltonian H, then corresponds to so that
Since the A terms on the far left and far right of the equation cancel , we have
It might be tempting to make a comparison between this result and the uncertainty rela-
tion for two non-commuting operators such as momentum and position for which
follows directly from the generalized uncertainty relation. However, such a
comparison would be incorrect since time t is not an operator in quantum mechanics, it is merely a parameter that advances the system in time. Notice that we could not derive
directly from the generalized uncertainty relation because t is not an opera-
tor.
(b) The energy of the n photons oscillating at frequency ω is so that
. The time dependence of the oscillator and its measurable quantities will
be proportional to so that or where is the
spread in values of phase . From (a) we have
and hence
For Poisson statistics so that . This situation corresponds to quasi-classical light in which there are a large number of photons (the classical limit of large numbers of particles, n) but they have a single frequency of oscillation ω and a
phase coherence that increases with increasing n as .
Solution 4
(a) .
(b) Because the position operator does not explicitly depend on time, , we
have
B B E
d A⟨ ⟩ h A t -------= =
Et h 2 ---≥
=
ωt φ–( )cos ω t φ= t φ ω⁄= φ
φ
Et hω nφ ω⁄ hnφ h 2 ---≥= =
nφ 1 2 ---≥
n⟨ ⟩
t∂ ∂ x⟨ ⟩ 0=
h---- x H[ , ]⟨ ⟩= = =
td d x⟨ ⟩ i–
2m -------[ , ] p+⟨ ⟩ p⟨ ⟩
m --------= = =
8
where we made use of the fact that . Notice, that while the position operator
does not explicitly depend on time (i.e., ), the expectation value of the position
operator can explicitly depend on time (i.e., ).
(c) Because we need only consider the commutator of the momentum oper- ator with the potential appearing in the Hamiltonian. Hence,
where we used . The spatial derivative of the expectation value of the
potential is
Hence, in quantum mechanics, . This would be the same as
Newton’s second law for force on a particle if position
and if . The conditions under which the latter is
true is when the potential is slowly varying so that higher order terms in the expansion
of the force may be
neglected so that . This requires that the uncertainty is small.
(d)
(e)
this can be written in the familiar form of an undamped forced oscillator
where . This has an oscillatory solution of the form
which contains two harmonic oscillators, one at the natural frequency and the other
at the frequency of the input force. The constants and are found from the initial
contitions. If the amplitude of the undamped oscillator grows to infinity with the particular solution increasing linearly in time as
x p[ , ] ih=
x x t( )≠
x⟨ ⟩ x⟨ ⟩ t( )=
p2 p[ , ] 0=
mω0 2x2
2 x⟨ ⟩– e Ex ω t( )cos–= = =
p i– h x∂
td d p⟨ ⟩= =
m t 2
d V x⟨ ⟩( )=
F x( ) F x⟨ ⟩( ) x x⟨ ⟩–( )F ′ x⟨ ⟩( ) x x⟨ ⟩–( ) 2F ′′ x⟨ ⟩( ) …+ + +=
F x( ) F x⟨ ⟩( )≅ x x x⟨ ⟩–( )=
td d H⟨ ⟩ i
h--- H H[ , ]⟨ ⟩ t∂
H H[ , ] 0=
Fx e– Ex=
m ω0 2 ω2–( )
which, since , may be written as
The sum frequency is modulated by the difference frequency so
that the amplitude of beats at the difference frequency .
x⟨ ⟩ p t( ) Fx t ω0t( )sin
2mω0 ----------------------------=
2 x( ) y( ) x y–( ) x y+( )cos–cos=sinsin
x⟨ ⟩ t( ) 2Fx
LAST NAME FIRST NAME
Useful constants MKS (SI)
Planck’s constant
Avagadro’s number
h 6.58211889 26( ) 10 16–× eV s=
h 1.054571596 82( ) 10 34–× J s=
e 1.602176462 63( ) 10 19–× C=
m0 9.10938188 72( ) 10 31–× kg=
mn 1.67492716 13( ) 10 27–× kg=
mp 1.67262158 13( ) 10 27–× kg=
kB 1.3806503 24( ) 10 23–× J K 1–=
kB 8.617342 15( ) 10 5–× eV K 1–=
ε0 8.8541878 10 12–× F m 1–=
µ0 4π 10 7–× H m 1–=
c 1 ε0µ0⁄=
aB 4πε0h2
2
PROBLEM 1 (a) Write a computer program to calculate the chemical potential for n non-interact-
ing electrons per unit volume at temperature, T. (b) Calculate the value of the chemical potential for the case when electrons of
effective mass and carrier density are at tem- perature
(c) Repeat (b), only now for the case when electrons have effective electron mass .
(d) Plot the Fermi-Dirac distribution function for the situations described by (b) and (c).
(e) Repeat (b), (c), and (d), only now for the case when temperature .
Your answer should include a print out of your computer program and plots.
PROBLEM 2
(a) Show that
(b) A semiconductor consists of a valance band with electron energy dispersion relation and a conduction band with electron energy dispersion relation such that
, where is a constant such that the conduction band and valence band are separated by an energy band gap, . Show that when particle number is con- served, the chemical potential is in the middle of the band gap with value and is independent of temperature.
m* 0.07 m0×= n 1.5 1018× cm 3–= T 300 K=
m* 0.50 m0×=
T 77 K=
Eg
SOLUTIONS
Solution 1 (a) and (b) use computer program for Fig. 7.5. (c) and (d)
Solution 2
(a)
Because it will be useful for part (b), we modify our notation for the Fermi-Dirac distribution so that
so that
(b) We are given that the semiconductor valance band has an electron dispersion relation and a conduction band with dispersion relation such that
, where is a constant such that the conduction band and valence band are separated by an energy band gap, . This means that the conduction and val- ance bands along with the corresponding density of states will be symmetric around the middle of the band gap energy that occurs at energy .
One way to visualize this is to assume a simple tight-binding band of s-orbitals in one dimension with lattice spacing L for which .
1 1
e µ E–( ) kBT⁄
+ --------------------------------- 1
EVB E k( )= ECB E0 E k( )–= E0
Eg
4
If the density of states for the valence band is , then the total density of states is . To calculate the chemical potential we use
where the Fermi-Dirac distribution is and we note that this has the property
At temperature the integral for the carrier density can be written
Now, because particle number is conserved, we can equate the integral with
the integral. Using our new notation for the Fermi-Dirac distribution, this is written
which requires
This means that the chemical potential is independent of temperature and has a value that is in the middle of the band gap. This is a result of symmetry built into the density of electron and hole states in this exercise. If the density of states is not symmetric (as is usually the case in semiconductors) then the chemical potential is not temperature independent.
π/L Wave vector, k
n D E( )f E( ) Ed E 0=
∞
∞
∫= =
f E µ–( ) 1 f µ E–( )–= T 0 K=
n D1 E( ) Ed E 0=
∞
EF
EF
∫ D1 E( ) f E µ–( ) f E0 E– µ–( )+( ) Ed∫= =
f E µ–( ) f E0 E– µ–( )+ 1=
f E µ–( ) 1 f µ E– 0 E+( )–( )+ 1=
f E µ–( ) f µ E– 0 E+( )– 0=
E– µ µ E– 0 E+ + + 0=
µ E0
LAST NAME FIRST NAME
Useful constants MKS (SI)
Planck’s constant
Avagadro’s number
h 6.58211889 26( ) 10 16–× eV s=
h 1.054571596 82( ) 10 34–× J s=
e 1.602176462 63( ) 10 19–× C=
m0 9.10938188 72( ) 10 31–× kg=
mn 1.67492716 13( ) 10 27–× kg=
mp 1.67262158 13( ) 10 27–× kg=
kB 1.3806503 24( ) 10 23–× J K 1–=
kB 8.617342 15( ) 10 5–× eV K 1–=
ε0 8.8541878 10 12–× F m 1–=
µ0 4π 10 7–× H m 1–=
c 1 ε0µ0⁄=
aB 4πε0h2
PROBLEM 1 In first-order time-dependent perturbation theory a particle initially in eigenstate of
the unperturbed Hamiltonian scatters into state with probability after the perturbation V is applied. (a) Show that if the perturbation is applied at time then the time dependent coef-
ficient is
where the matrix element and is the difference in
eigenenergies of the states and . (b) A particle of mass m is initially in the ground state of a one dimensional harmonic
oscillator. At time t = 0 a perturbation is applied where V0 and τ are constants. Using the result in part (a), calculate the probability of transition to each excited state of the system in the long time limit, .
PROBLEM 2 An electron is in ground state of a one-dimensional rectangular potential well for which
in the range and elsewhere. It is decided to control
the state of the electron by applying a pulse of electric field at
time t = 0, where is a constant, is the unit vector in the x-direction, and is the maximum strength of the applied electric-field. (a) Calculate the probability that the particle will be found in the first excited state
in the long time limit, .
(b) If the electron is in a semiconductor and has an effective mass ,
where is the bare electron mass, and the potential well is of width , cal-
culate the value of for which . Comment on your result.
PROBLEM 3 An electron is initially in the ground state of a one-dimensional rectangular potential well for which in the range and elsewhere. The
ground state energy is and the first excited state energy is . At time t = 0 the sys-
tem is subject to a perturbation . Calculate and then use a computer program to plot the probability of finding the particle in the first excited state as a func- tion of time for . In your plot, normalize time to units of and consider the three
values of , , and , where . Explain your results.
n| ⟩
t ′ 0=
t′ t=
m| ⟩ n| ⟩
t ∞→
E t( ) E0 e t2 τ2⁄– x⋅=
τ x E0
E1 E2
t 0≥ τ
ω21 1 2π⁄= ω21 1= ω21 2π= hω21 E2 E1–=
Applied quantum mechanics 3
Use the Einstein spontaneous emission coefficient to estimate
the numerical value of the spontaneous emission lifetime of the 2p excited state of atomic hydrogen. Use your results to estimate the spontaneous emission lifetime of the 2p transition of atomic He+ ions. Describe the spontaneous emission spectral line- shape you expect to observe. Do you expect the He+ ion 2p spontaneous emission line spectrum to have a larger or smaller full-width at half maximum compared to atomic hydrogen?
PROBLEM 5 A particle in a continuum system described by Hamiltonian is prepared in eigen-
state with eigenvalue . Consider the effect of a perturbation turned on at
time that is harmonic in time such that , where is the spatial part of the potential and ω is the frequency of oscillation. Start by writing down the Schrödinger equation for the complete system including the perturbation and then go on to show that the scattering rate in the static limit ( ) is
, where the matrix element
couples state to state via the static potential , the density of final contin-
uum states is , and ensures energy conservation.
PROBLEM 6 (a) In a uniform dielectric the dielectric function is a constant over space but depends on wave vector so that . Given an impurity potential at position r due to a
charge e at position is , derive an expression for .
(b) Use the expression for and Fermi's Golden rule to evaluate the total elastic scattering rate for an electron of initial energy E(k) due to a single impurity in a dielec- tric with dielectric function . Describe any assumptions you have made. Outline how you might extend your calculations to include elastic scattering from n ionized impurities in a substitutionally doped crystalline semiconductor. (c) What differences in scattering rate do you expect in (b) for the case of randomly positioned impurities and for the case of strongly correlated impurity positions?
A 4ω3e2
n| ⟩ En hωn=
t 0= W x t,( ) V x( ) ωt( )cos= V x( )
ω 0→
1 τn ---- 2π
h------ W mn 2 D E( )δ Em En–( )⋅= W mn m⟨ |W x t,( ) n| ⟩=
n| ⟩ m| ⟩ V x( )
ε ε q( )=
Ri vq r R i–( ) e2– εq r R i– ----------------------= v q( )
v q( )
4
SOLUTIONS Problem 1 (a) Consider a quantum mechanical system described by Hamiltonian and for which we know the solutions to the time-independent Schrödinger equation
and the time-dependent Schrödinger equation
.
At time we apply a time-dependent change in potential W(t) so the new Hamil-
tonian is and the state evolves in time according to
where and are time dependent coefficients.
Substitution of the state in to the time-dependent Schrödinger equation gives
Using the product rule for differentiation ((fg)' = (f 'g + fg')), one may rewrite the left- hand-side as
and remove the term
Multiplying both sides by and using the orthonormal relationship gives
However, since
we may write
If the system is initially in an eigenstate of the Hamiltonian then
and for . There is now only one term on the right-hand-side and we can write
H 0( )
ih t∂
=
ih t∂
n ∑= an t( )
n ∑ H 0( ) W t( )+( ) an t( ) n| ⟩e
iωn t–
iω n t–
n ∑ an t( )H 0( ) n| ⟩e
iωn t–
t∂ ∂
iω nt–
ih td
d am t( ) an t( ) m⟨ |W t( ) n| ⟩e iωm n t
n ∑=
iωm t Wφ n x( )e
iω nt– xd∫ W mne
iωm n t = =
hωmn Em En–= W mn φ m * x( )Wφn x( ) xd∫
ih td
n ∑=
ih td
=
Applied quantum mechanics 5
Integration from the time when the perturbation is applied at gives
(b) The transition probability from state to state is . Using our solution in part (a) we have
In the problem we have initial state and we consider the long time limit,
, this allows us to write
where, for the harmonic oscillator, the non-zero positive integer m multiplied by the frequency is related to the difference in energy eigenvalue by . The time integral is
and the matrix elements are found using , where
is the particle mass. In our case only transitions and are allowed.
where we used , , , and
. Hence,
and
So, in the long time limit, , the maximum probability of the transition taking
place occurs as .
Problem 2 The eigestates of a particle in a one-dimensional rectangular potential well for which
in the range and elsewhere are
where , , and eigenenergies are .
iω mn t ′ t′d
t ′ 0=
t ′ t=
Pnm 1 h2 ----- m⟨ |V x t,( ) n| ⟩e
iω mn t ′ t′d
t ′ 0=
t ′ t=
∫ 2 V0
iωm n 1 τ⁄–( )t ′ t′d
t′ 0=
t′ t=
t′ 0=
e imω 1 τ⁄–( )t ′ t ′d t′ 0=
t′ ∞=
∫ 2
b †
b+( )=
m⟨ |x3 0| ⟩ h 2m*ω⁄( ) 3 2⁄
b †3
6δm 3= 3δm 1=+( )= =
b †
n| ⟩ n 1+( )1 2⁄ n 1+| ⟩= b n| ⟩ n1 2⁄ n 1–| ⟩= b 0| ⟩ 0=
m n⟨ | ⟩ δmn=
t′ 0=
e imω 1 τ⁄–( )t ′ t′d t′ 0=
t′ ∞=
n| ⟩ 2 L --- knxsin=
kn nπ L⁄= n 1 2 3 …, , ,= En h2kn 2 2m⁄ hωn= =
6
The perturbation is turned on at time t = 0, where and
is the maximum strength of the applied electric-field. The probability that the par-
ticle will be found in the first excited state in the long time limit, is
where and
where and we made use of the standard integral
.
(b) If the electron is in a semiconductor and has an effective mass ,
where is the bare electron mass, and the potential well is of width , we
can calculate the value of for which is a maximum.
so that . We now find for which .
This may seem like a large electric field but it corresponds to a voltage drop of only 0.66 V across the potential well of width L = 10 nm.
V x t,( ) ex– E0 e t2 τ2⁄–= τ E0
P12
t ∞→
iω21 t ′ t ′2
2⟨ |x 1| ⟩ 2 L --- x k2x( ) k1x( )sinsin xd
0
L
9hπ2------------ 2
τ2 ------–
9hπ2 ------------
2
t ′ 0=
t′ ∞=
P12 πe2 E0
2
e0.5 2 27 1.05 10 34–×( ) 2
× π3× 32 0.07 9.1× 10 31–×× 1.6 10 19–× 10 24–×× --------------------------------------------------------------------------------------------------------
1.65×= =
Applied quantum mechanics 7
Problem 3 The eigestates of a particle in a one-dimensional rectangular potential well for which
in the range and elsewhere are
where , , and eigenenergies are .
The particle is in the ground state prior to a perturbation being
turned on at time t = 0. The probability that the particle will be found in the first excited state at time t is
where and
Now we notice that is related to via so and
In the limit the probability becomes
In this long time limit and normalizing time to , we see that
and
so, if we want to plot the results on the same graph, it is best to normalize to these long time values.
V x( ) 0= 0 x L< < V x( ) ∞=
n| ⟩ 2 L --- knxsin=
kn nπ L⁄= n 1 2 3 …, , ,= En h2kn 2 2m⁄ hωn= =
V x t,( ) V0x2e t τ⁄–=
P12
2⟨ |x2 1| ⟩ 2 L --- x2 k2x( ) k1x( )sinsin xd
0
L
t ′d t ′ 0=
1– 1 τ⁄ i– ω21
-------------------------------- 2 1 e 2 t τ⁄– 2e t τ⁄– ω21t( )cos–+
ω 21 2 1 τ2⁄+
----------------------------------------------------------------= =
2 1 e 2t τ⁄– 2e t τ⁄– ω21t( )cos–+
ω21 2 1 τ2⁄+
P12 t( ) 16V0
6mω21 ----------------
2 1 e 2 t τ⁄– 2e t τ⁄– ω21t( )cos–+
ω21 2 1 τ2⁄+
V016 6m
V016 6m
8
Problem 4 Differences in energy between eigenfunctions of hydrogen are given by
where Ry = 13.6 eV is the Rydberg energy and ni is the principle quantum number of the state.
For the 2p state E = hω = -3.401 + 13.606 = 10.205 eV corresponding to a wavelength which is λ = 122 nm.
An estimate for the matrix element is , where is the Bohr
radius and we obtain
Hence, our estimate of the spontaneous emission lifetime for hydrogen is .
For the helium ion we note that Z = 2. The Bohr radius scales as . For a hydrogenic atom with ion charge number Z the Bohr radius for the electron is
. The Rydberg energy which determines E scales as Z2 so
. Hence, the linewidth of the He+ ions is
and the lifetime is . The
ψnlm
j⟨ |r k| ⟩ aB∼ aB 4πε0h2
m0e2 -----------------=
τsp 0.89 ns=
2⁄–( )=
A* A Z6 Z2⁄× 16 A× 1.79 1010 s 1–×= = = τsp * 0.056 n s=
Applied quantum mechanics 9
Lorentzian line for helium is 16 times wider than that of the corresponding hydrogen line.
Problem 5 The Hamiltonian of the unperturbed system has solutions to the time-independent Schrödinger equation given by
The time-independent eigenvalues are and the orthonormal eigenfunctions
are . The eigenfunction evolves in time according to
and satisfies
At time we apply a time-dependent change in potential W(x, t) whose effect is to create a new Hamiltonian
and state which evolves in time according to
The wavefunction may be expressed as a sum over the known unperturbed eigenstates
where are time dependent coefficients. It follows that
Multiplying both sides by and using the orthonormal relation