Solutions to Levi Applied Quantum Mechanics 2nd Ed

Applied Quantum Mechanics

Chapter 1 Problems and Solutions

LAST NAME

FIRST NAME

Useful constants

MKS (SI)

Speed of light in free space

c = 2.99792458 × 10 m s

Planck’s constant

h = 6.58211889 ( 26 ) × 10

8

–1

–16

h = 1.054571596 ( 82 ) × 10

eV s

Electron charge

e = 1.602176462 ( 63 ) × 10

Electron mass

– 34 – 19

J s C

m0 = 9.10938188 ( 72 ) × 10

– 31

kg

Neutron mass

mn = 1.67492716 ( 13 ) × 10

– 27

kg

Proton mass

mp = 1.67262158 ( 13 ) × 10 – 27 kg

Boltzmann constant

k B = 1.3806503 ( 24 ) × 10 –23 J K –1 k B = 8.617342 ( 15 ) × 10 –5 e V K– 1

Permittivity of free space

ε 0 = 8.8541878 × 10 – 12 F m –1

Permeability of free space

µ 0 = 4π × 10 – 7 H m – 1


c = 1 ⁄ ε0µ0

Avagadro’s number

NA = 6.02214199 ( 79 ) × 10

Bohr radius

a B = 0.52917721 ( 19 ) ×10

23

–10

mol

–1

m

4πε 0 h a B = ---------------m 0e2

2

Inverse fine-structure constant α –1 = 137.0359976 ( 50 ) 4 πε 0 hc α – 1 = ----------------e2

Applied quantum mechanics

1

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 v =

g h 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 frequency if the airplane moves in an arc? 2

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 generalized Lorentz force law that includes magnetic charge. Write Maxwell’s equations with magnetic charge in terms of a field G =

ε E + i µH .

PROBLEM 8 The capacitance of a small metal sphere in air is C0 = 1.1 × 10

–18

F . A thin dielectric film

with relative permittivity ε r 1 = 10 uniformly coats the sphere and the capacitance increases to 2.2 × 10

–18

F . 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. 12

16

(b) What is the ratio of vibrational frequencies for carbon monoxide isotope 12 ( C O ) and 13

16

carbon monoxide isotope 13 ( C O )? 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? Fixed point A

Spring

Length, l

Mass, m

Displacement, -x


3

SOLUTIONS Solution 1 (a) The water level decreases. If ice has volume V then net change in volume of water in the bucket is ∆ V = V ( ρice ⁄ ρ wate r – 1 ) . (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 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. The minimum volume is 1.333 cm 3 corresponding to a geometry consisting of triangles placed on a circular base as shown in the Fig.

2r = 2 cm

h = 1 cm

r = 1 cm

2 cm

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.

4

h

r

r

r x k l

l (r - x)

l = k = ( r + x) ( r – x) = r – x Volume of the plug is 2

2

2

r

r

2

x3 Vol = 2 ∫ k 2 dx = 2 ∫ ( r 2 – x 2 ) dx = 2 r 2 x – ---3 0 0

r 0

4 = - r3 3

Since r = 1 cm the total volume is exactly 4/3 = 1.333 cm 3 . 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.

2r = 2 cm

h = 1 cm

r = 1 cm

2 cm

To calculate the volume we first calculate the volume sliced off the half sphere as illustrated in following Fig.


5

l l

(r - x) r

(r - x) x

l

r x

r

r

r/21/2

l = (r + x ) (r – x ) = r – x so that area of disk radius l at position x is 2

2

2

Area = π ( r – x ) 2

2

The volume of the disk is the area multiplied by the disk thickness dx. The total volume is the integral from x = r ⁄ 2 to x = r . Remembering to multiply by 2 because there are two slices gives total volume sliced off r

Vol off = 2 π

∫ r⁄ 2

x3 ( r 2 – x 2 ) dx = 2π r 2 x – ---3

r 0

1 1 1 = 2πr 3   1 – --- –  ------- – ----------   3  2 6 2 

3 2 5 Vol off = 2 πr  - – ----------   3 6 2 The volume of the plug is just the volume of a half sphere minus Vol off . 3

2πr 3 2 5 3 1 2 5 3 5 1 Vol = ----------- – 2 πr  -- – ----------  = 2πr  -- – -- + ----------  = 2πr  ---------- – ---      3 3 6 2 3 3 6 2 6 2 3 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 calculating the maximum convex volume turns out to be a bit complicated as may be seen in the following 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 cm 3 .

2r = 2 cm

h = 1 cm z y r = 1 cm x

2 cm

6

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 2

g m2 gm 2 t dx dx = ------------------ and, starting from rest, speed = ------------------- with displacement 2 m 1 + m2 dt m 1 + m2 dt

m2 then

gm 2 t 2 x = --------------------------- . Eventually, particle m1 is launched with velocity v 1 from the end of the table, at 2 ( m 1 + m2 ) 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 t =

2l ( m 1 + m 2 ) ----------------------------- for m1 to reach m 2g

gm2 2l ( m 1 + m 2 ) The launch velocity is v 1 = ----------------------------------------------- . The initial angular momentum of m1 + m2 m 2g the system when particle mass m1 is launched from the edge of the table is Lv 1m1 m2 /(m1 +m2 ). (c) If the string has a spring constant then part of the energy of the system can be stored in the the edge.

string during its trajectory. For example, this will happen during the initial acceleration phase. Solution 4 We are given wave velocity, v =

gh . This is a constant because h and g are fixed. Hence,

group velocity and phase velocity are the same and ω = 2 πf =

gh × k =

g h × 2 π ⁄ λ . For the

lowest frequency standing wave λ = 2 × 5 m and using g ~ 10, then the frequency of the wave is f =

10 × 0.9 ⁄ 10 = 0.3 Hz .

Solution 5 (a) ω = Ak

0.5

.

(b) Because the group velocity v g = and so

dω dk ------- = 2 ------ . ω k

ω dω ∂ω is twice the phase velocity v p = ω ⁄ k we have = 2 ---dk k ∂k

Integrating both sides of this equation allows one to write

ln ω = 2 ln k + ln A = ln Ak 2 so that the dispersion relation is ω = Ak 2 where A is a constant. (c) ω = Ak . 4

Solution 6 For a source moving at velocity v, the Doppler shifted frequency f’ is f′ = f 0 ⁄ ( 1 ± v ⁄ c ) where f0 is the original frequency. The ± sign refers to the source moving towards or away from the detector.

Hence,

the

maximum

change

in

frequency

relative

to

f0

is

∆ f = f 0 ( 1 – 1 ⁄ ( 1 ± v ⁄ c ) ) = ±f 0 ( v ⁄ ( c + v ) ) . In our case, v = 3.6 × 10 m s , c = 3 × 10 m s , 8

-1

8

-1

and f0 = 10 10 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 moving at a constant velocity (and assuming the airplane flies right over the ground station). If the disApplied quantum mechanics

7

2

2

2 2

tance to the plane (range) is l then l = h + v t overhead). Then cos ( θ ( t ) ) = v t ⁄ l = v t ⁄

(assuming we set t = 0 when the plane is

h +v t = 1⁄ 2

2 2

h ⁄ v t + 1 . The Doppler shift as a 2

2 2

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 density ρ m and magnetic current density J m as well as the usual electron charge density ρ e and electron current density J e . The new equations would be ∇ ⋅ D = ρe ∇ ⋅ B = ρm ∂B ∇×E = – ------- – Jm ∂t ∇×H = J e + ∂D ------∂t where D = ε E and B = µ H . Conservation of magnetic current J m = ρ m v is expressed as ∂ρm + ∇ ⋅ Jm = 0 ∂t For a particle carrying both electric charge e and magnetic charge g, the Lorentz force is F = e( E + v × B ) + g( B – v × E ) 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, permia2 bility µ[N A-2] or [H m -1 ], permittivity ε[A 2 s2 N-1 m -2] or [F m-1 ], and ε 0 = 1 ⁄ µ 0 c where c is

the speed of light in free-space. If G G corresponds to energy density U = ( E ⋅ D + B ⋅ H ) ⁄ 2 *

then it makes sense to write G =

ε µ 2 2 -- E + i --- H since ε E and µ H have dimensions of energy 2 2

density U[J m -3]. Maxwell’s equations in terms of E and H with magnetic charge can be written ρe ε -- ∇ ⋅ E = --------2 2ε ρm µ --- ∇ ⋅ H = ---------2 2µ ε ε ∂H - ∇×E = – --- µ ------- + Jm   2 2 ∂t

8

µ µ ∂E --- ∇× H = ---  ε ------- + J e   2 2  ∂t It follows that ρm ρe + i ---------∇ ⋅ G = ---------2µ 2ε and ε µ ε ∂H -- ∇× E + i --- ∇×H = – --- µ ------- – 2 2 2 ∂t which can be rewritten as ∇ ×G =

ε µ ∂E µ - J m + i --- ε ------- + i --- J e 2 2 ∂t 2

iJ Jm 1 dG – ---------- + ---------e---------- ∇ × G = i d t 2ε 2µ εµ since i

dG µ ∂H ε∂E = – --- ------- + i -- ------dt 2 ∂t 2 ∂t 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 C 0 = 1.1 × 10– 18 F in air. First, we calculate the radius r 0 of the sphere using the formula C 0 = 4 πε 0 r 0 It follows that –18 C0 1.1 × 10 = ----------------------------------------= 10 n m r 0 = ----------–12 4πε 0 4π × 8.85 × 10 Now consider the case when the metal sphere radius r0 with charge Q is coated with a dielectric of relative permittivity εr 1 to radius r1 and ε r2 from r 1 radius to r2 . The voltage drop from r 0 through

the dielectrics to a much larger metal sphere of radius r2 is found by integrating r0

r1

r0

r1

Q Q V = – ∫ E rd r – ∫ E r dr = – ∫ ----------------------2 dr – ∫ ----------------------2 dr r1 r1 4πε 0 ε r 1 r r r 4πε 0 ε r 2 r 2

2

Q 1 1 1 1 V = -----------  ---------- – ---------- + ---------- – ----------  4πε 0  ε r1 r 0 ε r 1 r 1 ε r 2 r 1 ε r2 r 2 For this particular problem we now take the limit r 2 → ∞ . Capacitance 4πε 0 4πε 0 Q C = ---- = ---------------------------------------------------------= -----------------------------------------------V 1 1 1  ---------- – ---------- + ---------- – 0   1 1  ε r2 – ε r1  εr r εr r   ---------- – ----  -----------------  ε r2 r 1 1 0 1 1  ε r1 r 0 r 1  ε r 1 ε r  2

C  ε r2 – ε r1  C -  = 4 πε 0 ---------- – ----  ---------------ε r1 r 0 r 1  ε r1 ε r  2


9

C C  ε r 2 – ε r 1 - ---------- – 4πε 0 = ----  ---------------r 1  ε r1 ε r  ε r1 r 0 2 ε r2 – ε r1  ε r1 r 0  ε r2 – ε r1  r0 - --------------------------  -------------------------------=  ---------------r 1 = C  ---------------  ε – r C 4πε ε ε ε ε r1 C 0  r2 0 r1 0  r1 r 2   1 – ---------- C  Putting in the numbers. We have ε r1 = 10 , εr 2 = 1 , C0 = 1.1 × 10 –18 F , C = 2.2 × 10 –18 F , and r 0 = 10 nm , so that r0 1 – 10 9 r 1 =  ---------------  -------------------- = -- r 0 = 22.5 nm 1 4  1 – 10 ------   2 and we may conclude that the thickness of the dielectric film is r 1 – r 0 = 12.5 n m . 2 – 19 – 18 Charging energy ∆E = e ⁄ 2 C = 1.6 × 10 ⁄ 4.4 × 10 = 36 m eV . This value is similar to ambient thermal energy k B T = 25 m eV . Solution 9 (a) The molecule consists of two particles mass m 1 and mass m 2 with position r 1 and r 2 respectively. The center of mass coordinate is R and relative position vector is r . We assume that the potential depends only on the difference vector r = r 2 – r 1 and if the origin is the center of mass then m1 r 1 + m 2 r 2 = 0 , so that m2 r 1 = -----------------------r ( m1 + m 2 ) and m1 r 2 = -----------------------r ( m1 + m 2 ) Since, for example, –m –m r 1 = ---------2 r 2 = ---------2 ( r 1 – r ) m1 m1 m m r 1  1 + ------2  = ------2 r  m 1 m1 r 1 ( m 1 + m2 ) = m 2 r m2 r 1 = ----------------------- r ( m1 + m 2 ) Now, combining center of mass motion and relative motion, the Hamiltonian is the sum of kinetic and potential energy terms m2 m1 1 · · 2 1  · · 2 H = T + V = - m 1  R – -----------------------r + -- m 2 R + ---------------------- r + V ( r ) 2  (m 1 + m 2 )  2  ( m 1 + m2 )  where the total kinetic energy is 2 m1 m 22 · 2 m 1m 2 · 2 1 ·2 1 · 2 1 m 1 m2 · 2 T = -- ( m 1 + m2 )R + -------------------------r + -------------------------r = -- ( m1 + m 2 )R + -- -----------------------r 2 2 2 2 2 ( m1 + m 2 ) ( m1 + m 2 ) ( m 1 + m2 ) or 1 · 2 1 ·2 T = -- M R + -- mr 2 2

10

where M = m 1 + m2 and the reduced mass is m 1 m2 m = ----------------------( m1 + m 2 ) m1 r m2

R

Because the potential energy of the two interacting particles depends only on the separation between them, V = V ( r 2 – r 1 ) = V ( r ) . The frequency of oscillation of the molecule is determined by the potential and given by ω =

κ ⁄ m where κ is the spring constant and m is the

reduced mass 1 ⁄ m = 1 ⁄ m 1 + 1 ⁄ m2 . Since the isotope form of the molecule interacts in the same way, κ = κ′ , and the ratio of oscillation frequency is given by ω′ ----- = ω We

m1 m 2 ( m′1 + m′ 2 ) ----------------------------------------m′1 m′ 2 ( m 1 + m 2 ) 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.

m1 u

κ

m2 v

2

m1

du = κ( v – u) dt 2 2

dv m2 2 = κ ( u – v ) dt which has solution of the form e

–i ωt

giving

( κ – m 1 ω )u – κ v = 0 2

– κu + ( κ – m2 ω 2 ) v = 0 giving a characteristic equation


11

κ – m1 ω

–κ

2

–κ

κ – m 2ω

2

= ( κ – m 1 ω 2 ) (κ – m 2 ω 2 ) – κ 2 = 0

m 1 m2 2 2 κ =  -------------------  ω = mω m1 + m2 and ω =

m1 + m 2 κ  ------------------- = m 1 m2

κ⁄m

and, as before, since the isotope form of the molecule interacts in the same way, κ = κ′ , and the ratio of oscillation frequency is given by m1 m 2 ( m′1 + m′ 2 ) ω′ ----- = ---------------------------------------ω m′1 m′ 2 ( m1 + m 2 ) (b) To find the ratio of vibrational frequencies we use the result in part (a) and put in the atomic ω 12C masses to give --------- = ω 13C

13 × 16 × ( 12 + 16 ) ----------------------------------------------- = 1.02 . As expected, the lighter 12 × 16 × ( 13 + 16 ) 13

12

16

C O molecule

16

vibrates at a higher frequency (by about 2%) than the C O molecule. Solution 10 (a) For small displacement x we assume F0 doesn’t change much so for extension ∆l the work done is F0 ∆ l . Hence, the potential energy of the spring is equal to the force F0 multiplied by the 2 2 2 extension ∆l . For x « l we have ∆l = l + x – l = x ⁄ 2l , so that the potential energy is V = F0 x ⁄ 2l = κx ⁄ 2 . This is the potential energy of a harmonic oscillator with spring constant 2

2

κ = F0 ⁄ l . This oscillator has frequency ω = (b) ω =

The F0 ⁄ ml =

frequency

of

oscillation

450 ⁄ 20 × 2.5 =

κ⁄m = in

12

–1

9 = 3 rad s ,

τ = 1 ⁄ f = 2π ⁄ ω = 2.09 s , i.e., about 2 seconds.

F 0 ⁄ ml .

radians so

per the

second period

of

is

given

oscillation

by is



LAST NAME

FIRST NAME

Useful constants

MKS (SI)


c = 2.99792458 × 10 m s

Planck’s constant

h = 6.58211889 ( 26 ) × 10

8

–1

– 16

h = 1.054571596 ( 82 ) × 10 Electron charge

e = 1.602176462 ( 63 ) × 10

Electron mass

eV s

–34

J s

– 19

C

m 0 = 9.10938188 ( 72 ) × 10

–31

kg

Neutron mass

m n = 1.67492716 ( 13 ) × 10

–27

kg

Proton mass

m p = 1.67262158 ( 13 ) × 10 –27 k g

Boltzmann constant

k B = 1.3806503 ( 24 ) × 10 –23 J K –1 k B = 8.617342 ( 15 ) × 10 – 5 eV K –1


ε 0 = 8.8541878 × 10 – 12 F m –1


µ 0 = 4π × 10 –7 H m –1


c = 1 ⁄ ε0µ0

Avagadro’s number

N A = 6.02214199 ( 79 ) × 10

Bohr radius

a B = 0.52917721 ( 19 ) ×10

23

–10

mol

–1

m

4πε 0 h a B = ---------------m 0e2

2



1

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 incident 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 electron in the metal? PROBLEM 3 Consider a lithium atom (Li) with two electrons missing. (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 n = 2 . (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 transitions 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, ih , by evaluating expectation value ∂px ∞

〈 xˆ 〉 =

∫ dxψ

*

( x )x ψ ( x )

–∞

in terms of φ(k x) which is the Fourier transform of ψ(x). (b) The wave function for a particle in real-space is ψ ( x , t ) . Usually it is assumed that position x and time t are continuous and smoothly varying. Given that particle energy is quantized such that ∂ E = hω , show that the energy operator for the wave function ψ ( x, t ) is ih . ∂t 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.

2

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 ∆τ∆x for the width of the wave packet to double. Compare ∆τ∆x with the time to complete one classical orbit. PROBLEM 8 What is the Bohr radius for an electron with effective electron mass m* = 0.021 × m 0 in a medium with low-frequency relative permittivity ε r 0 = 14.55 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.


3

SOLUTIONS Solution 1 (a) 2.4 kW m -2. (b) 4.5 kW m -2. 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 p p h = hk = hω ⁄ c where frequency ω is given by the quanta of photon energy E = hω . ∆ p = ( 2hω ⁄ c ) cos θ

2 E ⁄ c = 2 × 2 × 1.6 × 10

which, –19

The momentum change upon reflection is θ = 0,

for

⁄ 3 × 10 = 2.1 × 10 8

–27

gives

a

value

–1

k g m s . 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 situation 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 2

Z E n = Ry -----2 n 122.4 For Li++ Z = 3 and so E n = ------------eV . 2 n 2πh c – 10.1 Emission wavelength is λ = -------------------- = ---------------- nm , giving λ 3, 2 = 72.7 nm , λ 4, 2 = 53.8 n m , 1 1 En2 – En1 --- – ----n2 n1 λ 5, 2 = 48.1 n m . 2 4

– m0 Z e  1 1  For ε r = 10 we have En 2 – En 1 = ------------------------------ – ----- , so divide energy differences by 2 ( 4π ε0 ε r ) 2 h  n 2 n 1  ε r = 100 2

Solution 4 (a) The expectation value of the position operator is ∞

〈 xˆ 〉 =

∞

∫ dxψ

–∞

∞

*

∞

∞

1 * –i k′x ikx ( x )x ψ ( x ) = ------ ∫ dx ∫ dk ′φ ( k′ ) e x ∫ d kφ ( k ) e 2π –∞ –∞ –∞ ∞

∞

1 * – ik ′x ∂ e 〈 xˆ 〉 = ------ ∫ dx ∫ d k′φ ( k ′ )e x ∫ dkφ ( k )  -------  2π – ∞ – ∞ ∂ k ix –∞ ikx

4

∞

∞

∞

∞

ikx ∞

1 e 〈 xˆ 〉 = ------ ∫ d x ∫ d k′φ * ( k′ )e – ik′ x x  φ ( k ) ------ 2 π – ∞ –∞ ix ∞

– –∞

∂ e  ∫ dk ∂ k φ ( k )  ------ix  –∞ i kx

∞

∞

∞ ∞

–1 i * – ik′ x ∂ i kx i( k – k′ )x * ∂ 〈 xˆ 〉 = -------- ∫ d x ∫ dk ′φ ( k′ )e ∫ dk  ∂ k φ ( k )  e = -----∫ dxe ∫ ∫ dk d k′φ ( k ′ ) ∂ k φ ( k ) 2πi –∞ –∞ 2π –∞ –∞ – ∞ –∞ ∞ ∞

〈 xˆ 〉 = i ∫

∞

∫ d k dk ′φ

*

( k′ )

–∞–∞

∞

∂ * ∂ * ∂ φ ( k )δ ( k – k ′ ) = i ∫ d kφ ( k′ ) φ ( k ) = ih ∫ d kφ ( k′ ) φ ( k ) ∂k ∂ k ∂ p –∞ –∞

Hence, in k-space position is a differential operator, ih

∂ . ∂p

(b) The expectation value of the energy operator is ∞

〈 Eˆ 〉 =

∞

∞

∞

1 * * –i ωt ′ i ωt ∫ d ωψ ( ω )hω ψ ( ω ) = -----∫ d ω ∫ d t′φ ( t′ ) e h ω ∫ d tφ ( t) e 2π –∞ –∞ –∞ –∞ ∞

∞

∞

∞

∞

∂ e 1 * –i ωt ′ 〈 Eˆ 〉 = ------ ∫ d ω ∫ d t′φ ( t′ ) e h ω ∫ d tφ ( t )  ---------  2 π–∞ –∞ ∂ t iω –∞ 1 ei ωt 〈 Eˆ 〉 = ------ ∫ dω ∫ dt′φ * ( t′ )e –iω t ′ hω  φ ( t ) ------ 2π –∞ –∞ iω ∞

–h 〈 Eˆ 〉 = -------- ∫ dω 2 πi –∞

∞

∫

dt′φ ( t′ )e *

–∞

∫ ∫

∞

∞

– –∞

∂

dt d t′ φ ( t′ ) *

–∞–∞

e iω t

 ∫ d t ∂ t φ ( t )  ------iω 

–∞

∞

∞

∞ ∞

ih i ω( t – t ′ ) * ∂ ∂  i ωt ∫ dt  ∂ t φ ( t )  e = -----∫ d ωe ∫ ∫ d td t′φ ( t′ ) ∂ t φ ( t ) 2 π –∞ –∞ –∞ – ∞ ∞

∞

∞ ∞

〈 Eˆ 〉 = ih

– iω t ′

iω t′

∂ ∂ * ∂ * φ ( t )δ ( t – t′ ) = ih ∫ d tφ ( t′ ) φ ( t ) = ih ∫ d tφ ( t′ ) φ ( t ) ∂ t ∂ t ∂t –∞ –∞

Hence, in t-space energy is a differential operator, ih

∂ . ∂t

Solution 5 Series resistance. Space charging at current density j > nev . 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 harmonic 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 frequency ω in the x-direction is –h d mω 2 H = T + V = -------- 2 + ---------- x 2m d x 2 2

2

2

(b) and (c) may be found from the general solution (d) for a molecule with n n nuclei and ne electrons in which the Hamiltonian is j = nn

H =

–h --------- ∇ 2 + 2M j j=1

∑

2

i = ne

–h --------- ∇ 2 + V ( r ) 2 m0 i=1

∑

2


5

Here, the first term is the contribution to the kinetic energy from the n n nuclei of mass M j, the second term is the kinetic energy contribution from the n e electrons of mass m0 , and V(r) is the potential energy given by V (r ) =

e2

∑ -----------------4πε r

i > i′

0 i i′

+

ZZ e

2

Ze

2

j j′ j – ∑ ---------------∑ -----------------4 πε r 4 πε r

j > j′

0 jj ′

i, j

0 ij

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 4πε 0 ε r h 2 εr a *B = --------------------= a B -------= 0.529 × 10 –10 × 693 = 36.7 nm 2 m 0 m eff e me ff 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 S total ⁄ c = 1.8 × 10 –5 N m– 2 .

(b) From Exercise 1 Chapter 2, hω average = 1.92 eV so the number of photons per second per

unit area is S total ⁄ e hω averag e = 1.8 × 10 22 s –1 m –2 .

(c) The radiation pressure in (a) is a small value compared to one atmosphere which is about 5

10 N m –2 . 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.

6



LAST NAME

FIRST NAME

Useful constants

MKS (SI)


c = 2.99792458 × 10 m s

Planck’s constant

h = 6.58211889 ( 26 ) × 10

8

–1

– 16


e = 1.602176462 ( 63 ) × 10

Electron mass

eV s

–34

J s

– 19

C

m 0 = 9.10938188 ( 72 ) × 10

–31

kg

Neutron mass

m n = 1.67492716 ( 13 ) × 10

–27

kg

Proton mass

m p = 1.67262158 ( 13 ) × 10 –27 k g

Boltzmann constant

k B = 1.3806503 ( 24 ) × 10 –23 J K –1 k B = 8.617342 ( 15 ) × 10 – 5 eV K –1


ε 0 = 8.8541878 × 10 – 12 F m –1


µ 0 = 4π × 10 –7 H m –1


c = 1 ⁄ ε0µ0

Avagadro’s number

N A = 6.02214199 ( 79 ) × 10

Bohr radius

a B = 0.52917721 ( 19 ) ×10

23

–10

mol

–1

m

4πε 0 h a B = ---------------m 0e2

2



1

PROBLEM 1 Prove that particle flux (current) is zero if the one-dimensional exponential decaying wave func– κ x – i ωt tion in tunnel barrier of energy V0 and finite thickness L is ψ ( x, t ) = Be , where κ is a real positive number and particle energy E = h ω < V0 . PROBLEM 2 (a) Use a Taylor expansion to show that the second derivative of a wavefunction ψ ( x ) sampled at positions x j = jh 0 , where j is an integer and h 0 is a small fixed increment in distance x, may be approximated as ψ ( x j – 1 ) – 2ψ ( x j ) + ψ ( x j + 1 ) ψ ( x j ) = ----------------------------------------------------------------2 dx h0 (b) By keeping additional terms in the expansion, show that a more accurate approximation of d

2 2

the second derivative is – ψ ( x j – 2 ) + 16 ψ ( x j – 1 ) – 30ψ ( x j ) + 16ψ ( x j + 1 ) – ψ ( x j + 2 ) d ψ ( x j ) = -------------------------------------------------------------------------------------------------------------------------------------2 2 dx 12h 0 2

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 m *e = 0.07 × m0 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 m *e = 0.07 × m0 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 V 0 , where the effective electron mass on the left-hand side is m 1 and the effective electron mass on the right-hand side is m 2 . 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 → ∞ ?

2

SOLUTIONS Solution 1 To prove that particle flux (current) is zero if the one-dimensional exponential decaying wave – κx – iω t function in tunnel barrier of energy V0 and finite thickness L is ψ ( x , t ) = Be , where κ is a real positive number and particle energy E = hω < V 0 , we substitute the wave function into the current operator – i eh d d J = ----------- ψ * ( x , t ) ψ ( x, t ) – ψ ( x, t ) ψ * ( x , t )   2m  dx dx – i eh J = ----------- ( B* e – κx + i ωt ( – κB e – κx – i ωt ) – Be – κ x – i ωt ( – κB * e – κx + i ω t ) ) 2m κi eh 2 –2κx – 2κ x J = ----------- B ( e –e ) = 0 2m Solution 2 (a) Consider the Taylor expansion for the function f(x). f′′ ( x ) 2 f′′′ ( x ) 3 f ( x j + 1 ) = f ( x j ) + f′ ( x j ) h 0 + --------------j h 0 + ----------------j h 0 + … = 2! 3!

1

-- ⋅ f ∑ -n!

(n )

( xj ) ⋅ h0 n

n

where j is an integer and h0 is a small fixed increment in distance x. Keeping terms to order h 0 , we see that the first derivative is f( x j + 1 ) – f ( x j ) f′ ( x j ) = -------------------------------h0 Hence, the first derivative of a wavefunction ψ ( x ) sampled at positions x j = j h 0 may be approximated as ψ (x j + 1 ) – ψ (x j ) d ψ ( x j ) = ------------------------------------dx h0 2

To find the second derivative we keep terms to order h 0 so that f′′ ( x ) 2 f ( x j + 1 ) = f ( x j ) + f′ ( x j ) h 0 + --------------j h 0 2 and f′′ ( x ) 2 f ( x j – 1 ) = f ( x j ) – f′ ( x j ) h 0 + --------------j h 0 2 Adding these two equations gives f ( x j – 1 ) + f ( x j + 1 ) = 2f ( x j ) + f′′ ( x j ) h 0 or

2

f ( x j – 1 ) – 2 f( x j ) + f ( x j + 1 ) f′′ ( x j ) = ---------------------------------------------------------2 h0 so that the second derivative of a wavefunction ψ ( x ) sampled at positions x j = j h 0 may be approximated as ψ ( x j – 1 ) – 2ψ ( x j ) + ψ ( x j + 1 ) d ψ ( x j ) = ----------------------------------------------------------------2 dx h 20 2

This is the three-point approximation to the second derivative accurate to second order, 0 ( h 0 ) . (b) By keeping five terms in the expansion instead of three one may obtain a more accurate 2

approximation of the second derivative accurate to fourth order, 0 ( h 0 ) . To see how, we use the fol4

lowing Taylor expansions with derivatives up to fourth-order.


3

2 4 4 3 2 f ( x j + 2 ) = f ( x j ) + 2f′ ( x j ) h 0 + 2f′ ′ ( x j ) h 0 + -- f′′′ ( x j ) h 0 + - f′′′′ ( x j ) h 0 3 3 1 1 1 4 3 2 f ( x j + 1 ) = f ( x j ) + f′ ( x j ) h0 + - f′′ ( x j )h 0 + -- f′′′ ( x j )h 0 + ------ f′′′′ ( x j )h 0 24 6 2 1 1 1 4 3 2 f ( x j – 1 ) = f ( x j ) – f′ ( x j ) h 0 + -- f′′ ( x j ) h0 – -- f′′′ ( x j )h 0 + ------ f′′′′ ( x j ) h0 24 6 2 2 4 4 3 2 f ( x j – 2 ) = f ( x j ) – 2f′ ( x j )h 0 + 2f′′ ( x j )h 0 – -- f′ ′ ′ ( x j ) h 0 + -- f′′′′ ( x j ) h 0 3 3 To eliminate the terms in the first, third, and fourth derivative we multiply each equation by coefficients a, b, c, d, respectively, to obtain a f ( x j + 2 ) + bf ( x j + 1 ) + c f ( x j – 1 ) + d f ( x j – 2 ) = ( a + b + c + d ) f ( x j ) + ( 2a + b – c – 2 d ) f′ ( x j ) h 0 + 4 3 2a b c 2d 2 4a b c 4d b c   +  ------ + - – --- – ------ f′′′ ( x j ) h 0 +  ------ + ------ + ------ + ------  f′′′′ ( x j )h 0  3 24 24 3   2a + -2 + -2- + 2d f′′ ( x j )h 0  3 6 6 3  The coefficients that eliminate terms in the first, third, and fourth derivative must satisfy 0 = ( 2 a + b – c – 2d )

4a b c 4d 0 =  ------ + - – --- – ------  3 6 6 3 0 =  2a ------ + ---b--- + ---c--- + 2d ------   3 24 24 3  The solution to this set of linear equations is a = d b = – 16d c = – 16d Setting a = 1 we obtain f ( x j + 2 ) – 16f ( x j + 1 ) – 16f ( x j – 1 ) + f ( x j – 2 ) = –30 f ( x j) – 12f′′ ( x j )h 0 so that 2

– f ( x j + 2 ) + 16 f ( x j + 1 ) – 30f ( x j ) + 16 f ( x j – 1 ) – f ( x j – 2 ) f′′ ( x j ) = -------------------------------------------------------------------------------------------------------------------------12h 20 so that the second derivative of a wavefunction ψ ( x ) sampled at positions x j = jh 0 may be approximated as – ψ ( x j – 2 ) + 16 ψ ( x j – 1 ) – 30ψ ( x j ) + 16ψ ( x j + 1 ) – ψ ( x j + 2 ) d ψ ( x j ) = -------------------------------------------------------------------------------------------------------------------------------------2 2 dx 12h 0 2

Solution 3 To find numerically the first four energy eigenvalues and eigenfunctions for an electron with * effective mass me = 0.07 × m 0 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 timeindependent Schrödinger equation h2 d Hψ n ( x ) =  – ------- 2 + V ( x )  ψ n ( x ) = E n ψ n ( x ) 2m d x using a discrete set of N + 1 equally-spaced points such that position x j = j × h 0 where the index 2

j = 0, 1 , 2 …N , and h 0 is the interval between adjacent sampling points so one may define x j + 1 ≡ x j + h 0 . The region in which we wish to solve the Schrödinger equation is of length L = Nh 0 . At each sampling point the wave function has value ψ j = ψ ( x j ) and the potential is 4

V j = V ( x j ) . The second derivative of the discretized wave function we use the three-point finitedifference approximation which gives ψ ( x j – 1 ) – 2ψ ( x j ) + ψ ( x j + 1 ) d ψ ( x j ) = ----------------------------------------------------------------2 2 dx h0 Substitution into the Schrödinger equation gives a matrix equation 2

Hψ ( x j ) = – u j ψ ( x j – 1 ) + d ψ ( x j ) – u j + 1 ψ ( x j + 1 ) = Eψ ( x j ) where the Hamiltonian is a symmetric tri-diagonal matrix. The diagonal matrix elements are h d j = ---------2 + Vj mh 0 and the adjacent off-diagonal matrix elements are 2

h u j = -------------2 2mh 0 As usual, the wavefunction must be continuous and smooth. In addition, the periodic boundary 2

conditions require ψ ( x ) = ψ ( x + L ) so that ψ 0 ( x 0 ) = ψ N( x N ) . In this situation the Hamiltonian becomes a N × N matrix H and the Schrödinger equation is ( d1 – E ) –u2 –u2 ( d2 – E ) ( H – EI )ψ =

0 –u3

0 0

. . . .

. .

–u1 .

ψ1

. .

.

.

ψ3

0

–u3

( d3 – E )

–u 4

0 . . .

0 . . .

–u4 . . .

( d4 – E ) . . .

–u1

.

.

.

. . . .

ψ2

. . . . ψ4 = 0 . . . . . –uN – 2 0 ψN – 2 . ( dN – 1 – E ) –uN – 1 ψN – 1

. .

– uN

( dN – E)

ψN

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. Applied quantum mechanics

5

Solution 4

The first six eigenenergies are: E1 = 0.259 eV, E2 = 0.453 eV, E3 = 0.612 eV, E4 = 0.752 eV, E 5 = 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 m2 E–V ------ = ---------------0 m1 E or V0 E = ----------------------------1 – ( m2 ⁄ m 1 ) 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 1.5 E = ----------------------- = 2.25 eV 1 – (1 ⁄ 3) and the transmission coefficient in the limit E → ∞ is found by noting that k2 ---k1

E→ ∞

m = ------1 m2

so that Trans

E→ ∞

=

m1 4 ------ ---------------------------2 m 2 m 1 + ------1  m 2

For m 1 / 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!

6



LAST NAME

FIRST NAME

Useful constants

MKS (SI)


c = 2.99792458 × 10 m s

Planck’s constant

h = 6.58211889 ( 26 ) × 10

8

–1

– 16


e = 1.602176462 ( 63 ) × 10

Electron mass

eV s

–34

J s

– 19

C

m 0 = 9.10938188 ( 72 ) × 10

–31

kg

Neutron mass

m n = 1.67492716 ( 13 ) × 10

–27

kg

Proton mass

m p = 1.67262158 ( 13 ) × 10 –27 k g

Boltzmann constant

k B = 1.3806503 ( 24 ) × 10 –23 J K –1 k B = 8.617342 ( 15 ) × 10 – 5 eV K –1


ε 0 = 8.8541878 × 10 – 12 F m –1


µ 0 = 4π × 10 –7 H m –1


c = 1 ⁄ ε0µ0

Avagadro’s number

N A = 6.02214199 ( 79 ) × 10 23 m o l –1

Bohr radius

a B = 0.52917721 ( 19 ) ×10

–10

m

4πε 0 h a B = ---------------m 0e2

2



1

PROBLEM 1 Write a computer program in Matlab that uses the propagation matrix method to find the transmission resonances of a particle of mass m = m 0 (where m 0 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 function 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 n A l A s = 3.0 for the refractive index of AlAs and n GaAs = 3.5 for GaAs. Calculate and plot optical reflectivity in the wavelength range 1200 nm > λ > 1400 nm . (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 λ = 1300 nm . VCSEL structure center wavelength λ0 nAlAs nGaAs

Bragg mirror center wavelength λ 0

25 identical dielectric layer pairs

Center region thickness λ0 / n GaAs 25 identical dielectric layer pairs

25 identical dielectric layer pairs n GaAs n AlAs

z

Incident electromagnetic field, wavelength λ

n GaAs n AlAs Incident electromagnetic field, wavelength λ

Your results should include a printout of the computer program you used and a computer-generated plot of particle transmission as a function of incident wavelength.

2

PROBLEM 3

1.0

3 nm 2.7 nm 2.25 nm

Potential energy, V(x) (eV)

0.8 0.6 0.4 0.5 eV

0.2

1.13 eV

0.0 -0.2

0.6 eV

-0.4 -0.6 0

5

10

Distance, x (nm)

Write a computer program in Matlab that uses the propagation matrix method to find the transmission resonances of a particle of mass m = 0.07 × m 0 (where m 0 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 initial particle energy E = 0.025 eV with the particle incident on the structure from the left-hand side.

Potential energy, eV(x) (eV)

PROBLEM 4 Write a computer program to solve the Schrödinger wave equation for the first 17 eigenvalues * of an electron with effective mass m e = 0.07 × m0 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.

3.75 nm

6.25 nm

1.0 0.8 0.6 0.4 0.2 0.0 0

20

40

60

80

100

Distance, x (nm)


3

PROBLEM 5 Use the results of Problem 4 with periodic boundary conditions to approximate a periodic onedimensional 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.

4

SOLUTIONS 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 associated with each band. The first seventeen eigenfunctions include two complete bands each containing 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 Band gap = E9 - E8 = 0.247613 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 Band gap = E17 - E16 = 0.35326 eV E17 = 0.69779 eV


5

ψ8

ψ9

0.06 Wavefunction, ψ


0.08

0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1

0.04 0.02 0.00 -0.02 -0.04 -0.06 -0.08

0.0 0

10

20

30

40

50

60

70

80

0

10

20

Distance, x (nm)

30

40

50

60

70

80

Distance, x (nm)

Solution 5 Use code from Problem 4. The difference in form of the wavefunction for lowest eigenenergy value in the band (E 1 = 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 E 1 the value of k = 0 and for E8 the value of k = π/L.

0.08

ψ1

0.06

16

Wavefunction, ψ


20

12 8 4

0.04 0.02 0.00 -0.02 -0.04 ψ8

-0.06 -0.08

0 0

10

20

30

40

50

Distance, x (nm)

60

70

80

0

10

20

30

40

50

60

70

80

Distance, x (nm)

6



LAST NAME

FIRST NAME

Useful constants

MKS (SI)


c = 2.99792458 × 10 m s

Planck’s constant

h = 6.58211889 ( 26 ) × 10 – 16 eV s

8

–1


e = 1.602176462 ( 63 ) × 10

Electron mass

–34

J s

– 19

C

m 0 = 9.10938188 ( 72 ) × 10

–31

kg

Neutron mass

m n = 1.67492716 ( 13 ) × 10

–27

kg

Proton mass

m p = 1.67262158 ( 13 ) × 10

–27

kg

Boltzmann constant

k B = 1.3806503 ( 24 ) × 10 k B = 8.617342 ( 15 ) × 10

–5

–23

JK

eV K

–1

–1


ε 0 = 8.8541878 × 10


µ 0 = 4π × 10


c = 1 ⁄ ε0µ0

Avagadro’s number

N A = 6.02214199 ( 79 ) × 10 23 m o l –1

Bohr radius

a B = 0.52917721 ( 19 ) ×10–10 m

–7

H m

– 12

Fm

–1

–1

4πε 0 h 2 a B = ---------------m 0e2 Inverse fine-structure constant α –1 = 137.0359976 ( 50 ) 4 πε 0 hc α – 1 = ----------------e2


1

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 1 ψ ( x, t ) = ------- ( ψ 1 ( x, t ) + ψ 4 ( x, t ) ) 2 Find expressions for: (a) The probability density, ψ ( x , t ) . 2

(b) The average particle position, 〈 x ( t ) 〉 . (c) The momentum probability density, ψ ( p x , t )

2

(d) The average momentum, 〈 px ( t ) 〉 . (e) The current flux, J ( x, t ) . PROBLEM 2 (a) Show that the density of states for a free-particle of mass m in two-dimensions is m D2 ( E ) = ------------2 2πh (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 D 2 ( ω ) and one-dimensional D1 ( ω ) density of photon states in a homogeneous dielectric medium characterized by refractive index, n r. opt

opt

PROBLEM 4 (a) In a particular system the dispersion relation for electrons in one-dimension is Ek = hω k = 2t cos ( k x L ) , where t and L are constants and the wave vector in the xdirection is 0 < k x < π ⁄ L . 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 k x, write a computer program and plot the electron density of states N ( E ) =

Γ ⁄π using Γ = t ⁄ 10 ( E – Ek ) + ( Γ ⁄ 2)

∑ --------------------------------------------2 2 k

and t = – 1 . (b) If one includes next nearest neighbor interactions, the dispersion relation in (a) can, to within a scaling factor, be written E k = 2 t cos ( k x L ) + 2t′ cos ( 2k x L ) . Write a computer program to plot the dispersion relation. Then calculate and plot the electron density of states using Γ = t ⁄ 10 , t = – 1 , t′ = – 0.2 and compare with the result you obtained in (a) including a comparison with the effective electron mass at the band edges.

2

(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 E k = 2t ( cos ( k x L ) + cos ( k y L ) ) and E k = 2t ( cos ( k x L ) + cos ( ( k y L ) + cos ( k z L ) ) ) respectively. Use Γ = t ⁄ 10 and t = – 1 . (c) The dispersion relation for a two dimensional hexagonal lattice is E k = 2t ( 2 cos ( 0.5 × k x L ) cos ( 0.866 × k y L ) + cos ( k x L ) ) + 2t′ ( 2 cos ( 0.866 × k y L ) cos ( 1.5 × k x L ) + cos ( 1.73 × k y L ) ) where t′ is the overlap integral with the next nearest neighbor. Calculate the electron density of states using Γ = t ⁄ 10 , t = – 1 , and t′ = 0.3 . Compare with the result you obtain when t′ = 0 . PROBLEM 5 A hydrogen atom is in its ground state with electron wave function 1⁄2

–r ⁄ a 1 2 ⋅ e B ⋅  ------  φ 100 = ----------3⁄2  4 π aB 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 p r for the electron in this state. You should use the fact that in radial coordinates the Her1∂ mitian momentum operator is pˆ r = – ih -- r . r ∂r

PROBLEM 6 Using the fact that the Hamiltonian H appearing in the Schrödinger equation –i ----H |ψ 〉 = |∂ψ〉 h ∂t is Hermitian (i.e., 〈ψ |Hψ〉 = 〈Hψ |ψ 〉 ), show that the time dependence of the average ˆ is value of the operator A d ˆ i ˆ〉 ˆ ]〉 + 〈 ∂ A --- 〈 A 〉 = -- 〈 [H , A dt h ∂t PROBLEM 7 Show that: (a) The position operator xˆ acting on wavefunction ψ ( x ) is Hermitian (i.e., xˆ † = xˆ ). (b) The operator

d acting on the wavefunction ψ ( x ) is anti-Hermitian (i.e., dx

†

 d  = – d ). dx dx (c) The momentum operator – i h

d acting on the wavefunction ψ ( x ) is Hermitian. dx

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 Applied quantum mechanics

3

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 G = ε 0 ⁄ 2 ⋅ E + i µ 0 ⁄ 2 ⋅ H show that Maxwell’s equations in free space and in the absence of free charges may written as the complex equations ∇⋅G = 0 and ∂G 1 = -------------- ∇ × G ∂t ε0µ 0 (b) Show that the energy flux density in the electromagnetic field given by the Poynting vector is –i * S = E × H = -------------- ( G × G ) ε0µ 0 (c) If the field G is purely real, what is the value of S? i

2

(d) Show that the electromagnetic energy density is U = G .

4

SOLUTIONS Solution 1 (a) Let the potential in the region – L ⁄ 2 < x < L ⁄ 2 be zero and infinity elsewhere. The ψ n ( x, t ) =

eigenfunctions are

2 ⁄ L sin ( k n ( x + L ⁄ 2 ) )e

– iω nt

k n = nπ ⁄ L ,

, where

ω n = hk ⁄ 2m , and n = 1 , 2, 3, … . Hence, 2 n

ψ 1 (x , t ) =

2 ⁄ L sin ( ( π ⁄ L ) ( x + L ⁄ 2 ) ) e

–iω 1 t

=

2 ⁄ L cos ( ( πx ⁄ L ) e

–i ω1 t

–i ω4 t

)

–i ω4 t

ψ 4 ( x , t ) = 2 ⁄ L sin ( ( 4 π ⁄ L ) ( x + L ⁄ 2 ) ) e = 2 ⁄ L sin ( 4πx ⁄ L ) e and the probability distribution of the particle in the superposition state ψ ( x, t ) = is ψ ( x, t )

1 = -- ( ψ 1 ( x ) 2 + ψ 4 ( x ) 2 + 2 ψ 1 ( x ) ψ 4 ( x ) cos ( ( ω 4 – ω 1 )t ) ) 2

2

ψ ( x, t )

1 ⁄ 2 ( ψ 1 ( x, t ) + ψ 4 ( x , t ) )

1 xπ 4 πx πx 4πx h 15π   = --- cos 2 ------  + sin2  ---------  + 2 cos  ------  sin  ---------  cos  ------- -----------t L  L  L  L   2m L 2   L 2

2

h 15 π where we note that ω 4 – ω 1 = ------- ----------. 2m L 2 (b) Expectation value of position is 2

〈 x ( t )〉 =

∫ ψ ( x, t)

2

x dx

L⁄2

1 2 2 〈 x ( t )〉 = -- ∫ ( ψ 1 ( x ) x + ψ 4 ( x ) x + 2x ψ 1 ( x ) ψ 4 ( x ) cos ( ( ω 4 – ω 1 ) t ) ) dx 2 –L ⁄ 2 By symmetry, the first two terms in the integrand are zero, leaving L⁄2

〈 x ( t ) 〉 = cos ( ( ω 4 – ω 1 )t )

2 πx 4 πx --- cos  ------  sin  ---------  x d x L L L –L ⁄ 2

∫

making use of 2 sin ( x ) cos ( y ) = sin ( x + y ) + sin ( x – y ) gives 1 〈 x ( t ) 〉 = cos ( ( ω 4 – ω 1 ) t ) -L

L⁄2

πx  3π x  sin  5-------- + sin  ---------  x dx   L   L  –L ⁄ 2

∫

which, after integrating by parts, results in 2

2

1 2L 2L 〈 x ( t ) 〉 = cos ( ( ω 4 – ω 1 )t ) ---  -----------2 – --------2  x dx L  25π 9 π  32 L 〈 x ( t )〉 = – --------------2 cos ( ( ω 4 – ω 1 ) t ) 225π (c) The momentum probability density ψ ( p x, t ) 2 is found using the Fourier transform L⁄2

– ip x – ip x –i p x 1 1 ψ ( p x, t ) = -------------- ∫ ψ ( x , t )e x d x = -------------- ∫ ( ψ 1 ( x, t )e x + ψ 4 ( x, t ) e x ) dx 2πh 4π h – L ⁄ 2

ψ ( p x, t ) =

– iω t 2 ------------- e 1  4 πhL

L⁄2

L⁄2

4πx –i p x  π x-  e –i px x d x + e –i ω4 t sin  ---------  e x dx  ∫ cos  ----∫   L L  –L ⁄ 2 –L ⁄ 2


5

L⁄2

ψ ( p x, t )

2

 – iω1 t e

ψ ( p x, t )

2

L⁄ 2

iω t iω t 4πx ip x π x ip x 1 = -------------  e 1 ∫ cos  ------  e x dx + e 4 ∫ sin  ---------  e x d x ×      L  L 2πh L –L ⁄ 2 –L ⁄ 2

1 = ------------2πh L

L⁄2

L⁄2

4 πx –i px x   πx- e –ip xx dx + e – iω4 t sin  --------- e d x ∫ cos  ----∫  L L –L ⁄ 2 –L ⁄ 2 L⁄2

∫

–L ⁄ 2 L⁄2 i ( ω4 – ω1 )t

2Re  e

L⁄2

2

π x –i px x cos  ------  e dx +  L πx cos  ------ e L –L ⁄ 2

∫

∫

–L ⁄ 2 –ip x x

2

4πx – ip xx sin  --------- e dx +  L 

4πx i px x sin  --------- e d x L

We make use of the standard indefinite integrals (I. S. Gradshteyn and I. M. Ryzhik, Table of integrals, series, and products, Academic Press, San Diego, 1980, p. 196 (ISBN 0 12 294760 6))

∫e

ax

∫e

ax

e ( a cos ( bx ) + b sin ( bx ) ) cos ( bx ) dx = -------------------------------------------------------------2 2 a +b ax

e ( a sin ( b x ) – b cos ( b x ) -) sin ( bx ) dx = -------------------------------------------------------------2 2 a +b

ψ ( p x, t )

ax

ip L ⁄ 2

2

– ip L ⁄ 2

ip L ⁄ 2

2

–i p L ⁄ 2

2

1 Lπ ( e x + e x ) π( e x – e x ) + = ------------- -----------------------------------------------+ 4L --------------------------------------------------2 2 2 2 2 2 2πh L 16 π – p x L π – px L ip L ⁄ 2

–i p L ⁄ 2

–i p L ⁄ 2

ip L ⁄ 2

 i ( ω4 – ω1 )t Lπ ( e x + e x ) 4 Lπ ( e x – e x )  ------------------------------------------------------------------------------------------------- 2Re  e π 2 – p 2x L 2 16 π 2 – p 2x L 2   ψ (p x, t )

2

p xL p xL  = A cos 2  --------  + Bsin 2  -------+ C sin ( p x L ) sin ( ( ω 4 – ω 1 ) t ) 2 2 

32L π –8L π 2Lπ - , B = ----------------------------------------2 , C = ----------------------------------------------------------------A = ------------------------------. 2 2 2 2 2 2 2 2 2 2 2 2 2 h ( π – L p x ) ( ( 4π ) – L p x ) h( π – L px ) h ( ( 4π ) – L p x ) (d) The expectation value of momentum is 〈 px ( t ) 〉 = –ih ∫ ψ ( x , t ) *

L⁄2

The terms

∫

–L ⁄ 2

ψ4 *

–i h ∂ ψ ( x, t ) dx = -------∂x 2

∂ψ 4 d x = 0 and ∂x

– i h 2 4 π – i( ω4 – ω1 )t 〈 p x ( t )〉 = -------- -- ------ e 2 LL L⁄2

L⁄2

∫

–L ⁄ 2

L⁄2

∫

ψ1

–L ⁄ 2

*

L⁄2

( ψ * + ψ * ) ∂ ( ψ + ψ )  d x 4 4   1 ∂x 1 –L ⁄ 2

∫

∂ψ 1 dx = 0 by symmetry, leaving ∂x

πx 4πx cos  ------ cos  --------- dx +  L  L 

ih 2 π i ( ω4 – ω1 )t  4 πx- sin  πx ----- -- -- e ------ d x ∫ sin  ------- L 2 LL L  –L ⁄ 2 – i h 8π – i( ω 4 – ω1 )t 2L ih 2π i( ω4 – ω1 )t 8L 〈 p x ( t )〉 = -------- ------2 e --------- + ----- ------2 e --------2 L 15 π 2 L 15π i (ω – ω ) t – i h8 – i( ω – ω )t 〈 p x ( t )〉 = ----------- ( e 4 1 + e 4 1 ) 15L and

6

16h 〈 p x ( t )〉 = --------- sin ( ( ω 4 – ω 1 )t ) 15 L (e) The current operator is – ie h * ∂ψ ∂ψ *  J ( x , t ) = -----------  ψ –ψ 2m ∂x ∂x  * * * ∂ ∂ * – ie h J ( x , t ) = -----------  ( ψ 1 + ψ 4 ) ( ψ 1 + ψ 4 ) – ( ψ 1 + ψ 4 ) ( ψ 1 + ψ 4 )  ∂x ∂x 4m – i( ω1 – ω4 )t 4π x i( ω1 – ω4 )t – i eh πx ∂ –e )+ J ( x, t ) = -----------  cos  ------  sin  ---------  ( e L 2L m L ∂x i( ω 1 – ω4 )t 4πx ∂ πx –i ( ω1 – ω4 )t sin  --------- cos  ------ ( e –e ) L ∂x L

4π x 4π x ∂ πx –e h πx ∂ J ( x, t ) = ---------  cos  ------  sin  ---------  – sin  ---------  cos  ------  sin ( ( ω 4 – ω 1 ) t )         L L ∂x L  Lm L ∂x –πe h x x x x J ( x, t ) = -----------4 cos  π ---  cos 4 π ---  – sin  4π --  sin  π ---   sin ( ( ω 4 – ω 1 )t ) 2        L L L L mL Solution 2 (a) Density of states of particle mass m in two dimensions is m D 2 ( E ) = ------------2 per spin. 2π h (b) Current is assumed proportional to applied voltage, electron velocity v ( E ) , transmission coefficient T ( E ) , and density of states. EF + e V

I = e

∫

E F + eV

v ( E )T ( E )D 2 ( E ) d E = e

EF

∫ EF

2mE  m -  --h-- --------- ---------- m h 2 -  T ( E )  2 πh 2  dE

If we assume T ( E ) = T ( E F ) = 1 , then 2m I = e ------------2 2πh

EF + e V

∫ EF

2m 2 3 ⁄ 2 EF + eV 2m 3 ⁄ 2 eV 3 ⁄ 2 E d E = ------------2 ⋅ - [ E ] EF = ------------2 E F  1 + ------  – 1  EF 2π h 3 3πh

When E F « eV , conductance per electron spin I 2m G = --- =  ------------2 V  3 πh  V For the other situation

when

EF » e V

we

use

the

binomial

expansion

3 eV n eV 3 ⁄ 2 ( 1 + x ) = 1 + nx + … so that 1 + ------  = 1 + -- ------ . Conductance per electron EF 2 EF spin 2m I G = --- = ------------2 E F V 2πh Solution 3 The two dimensional and one dimensional photon density of states in an isotropic medium of refractive index n r is Applied quantum mechanics

7

ωn D2 ( ω ) = --------2-r πc

2

nr D1 ( ω ) = -----πc Solution 4 (a) We are given the dispersion relation for electrons in one-dimension as Ek = 2t cos ( k x L ) , where t = – 1 and 0 < k x < π ⁄ L . We are asked to plot the density Γ ⁄π using Γ = t ⁄ 10 and a sum over 100 states. k ( E – E k ) + (Γ ⁄ 2 ) Notice that the cosine dispersion in one-dimension has a van-Hove singularity the band edge energies of ±2t. of states N ( E ) =

∑ --------------------------------------------2 2

% 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); w=emin:estep:emax; t=-1; gamma = 0.1; gamma=gamma*abs(t); gamma2=(gamma/2)^2; const=gamma/pi; kx0=pi/(natoms); kx=0:kx0:kx0*natoms;

%set dos vector to zero %energy %set energy scale - for s-orbital, + for p-orbital %Energy broadening in units of t %use units of t

%kx constant

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)');

8

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); w=emin:estep:emax; t=-1; tprime=-0.2 gamma = 0.1; gamma=gamma*abs(t); gamma2=(gamma/2)^2; const=gamma/pi; kx0=pi/natoms; kx=0:kx0:kx0*natoms;

%set dos vector to zero %energy %set energy scale - for s-orbital, + for p-orbital %tprime %Energy broadening in units of t %use units of t

%kx constant

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;


9

(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.

10

Solution 5 ∞

〈 r〉 =

∞

3 ∫ φ 100 rφ 100 d r = *

0

∞

* 3 4 – 2r ⁄ aB 3 e r dr ∫ φ 100 φ 100 4 πr dr = ----3∫ aB 0 0

Let x = 2 r ⁄ a B , so that ∞

a a B 3! 3 3 –x 〈 r〉 = -----B ∫ x e dx = ---------= - aB 40 4 2 ∞

n –µ x

where we made use of the standard integral ∫ x e

dx = n! µ

–n –1

for Re ( µ > 0 ) .

0 ∞

∞

1d 1d 2 2 〈 p r〉 = – ih ∫ φ *100 -- ( r φ 100 ) 4πr d r = – ih ∫ φ 100 -- ( rφ 100 ) 4πr dr = – ihA r r d r d r 0 0 where A is a real number. Because the measurable quantity 〈 p r〉 must be a real number, A = 0 and hence 〈 p r〉 = 0 Solution 6 The Hamiltonian H in the Schrödinger equation is Hermitian and so it is its own Hermitian adjoint, i.e., H = H† and 〈ψ |Hψ 〉 = 〈 Hψ|ψ 〉 . The time dependence of the expeˆ is cation value of the operator A d ˆ ˆ |ψ 〉 + 〈ψ | ∂ A ˆ |ψ〉 + 〈ψ |A ˆ |∂ψ 〉 ----- 〈 A 〉 = 〈 ∂ψ| A dt ∂t ∂t ∂t d ˆ i ˆ |ψ 〉 – --i 〈ψ |A ˆ H |ψ 〉 + 〈 ψ| ∂ A ˆ |ψ 〉 --- 〈 A 〉 = -- 〈Hψ |A dt h h ∂t

d ˆ i ˆ |ψ 〉 – -i- 〈ψ |A ˆ H |ψ 〉 + 〈 ψ| ∂ A ˆ |ψ 〉 --- 〈 A 〉 = -- 〈ψ |HA dt h h ∂t d ˆ i ˆ ]〉 + 〈 ∂ A ˆ〉 --- 〈 A 〉 = -- 〈 [H , A dt h ∂t


11

Solution 7 (a) The position operator xˆ is Hermitian if it is its own Hermitian adjoint, i.e., xˆ † = xˆ * or 〈ψ |x | φ〉 = 〈φ |x |ψ〉 . For the operator xˆ this is easy to show ∞

〈ψ| xψ 〉 =

∞

∫ψ

*

( x ) ( xψ ( x ) ) d x =

–∞

∫ ( xψ

∞ *

( x )) ψ( x )d x =

–∞

∫ ( xψ ( x ) )

*

ψ ( x ) dx = 〈x ψ|ψ 〉

–∞

d is anti-Hermitian we integrate by parts and make use dx of the fact that the wavefunction ψ ( x ) → 0 as x → ±∞ (b) To show that the operator

d ψ〉 = dx

〈 ψ|

∞

* *  d  ∫ ψ ( x )  ih d x ψ ( x )  d x = ψ ( x )ψ ( x )

–∞

x=∞ x = –∞

∞

–

d

∫ dx ψ

*

–∞

(x )ψ (x )d x 

∞

〈 ψ|

* d d d ψ 〉 = – ∫  ψ ( x ) ψ ( x ) dx = – 〈 ψ| ψ〉 dx d x d x –∞ †

d d and   = – . d x dx (c) Because the operator – ih

d is anti-Hermitian, it follows that the momentum operator dx †

d d d d is Hermitian since  – i h  = ih –  = – i h .  d x  dx dx dx

Solution 8 (a) ∇ ⋅ G = ∇ ⋅ ( E ε 0 ⁄ 2 + i H µ 0 ⁄ 2 ) = 0 µ 0 ∂H 1 1 1 ----= -------------- ∇× G = ------------- ∇×E + i ------------ ∇×H 2 ∂t ε0µ0 2µ 0 2ε 0 Evaluating real and imaginary parts gives i

ε ∂E ∂G – = i ----0 2 ∂t ∂t

∇× E = – µ 0

∂H ∂t

and ∂E ∂t (b) We start with ∇× H = ε 0

ε µ ε µ –i * –i -------------- G × G = --------------  ----0 E – i -----0 H  ×  ----0 E + i -----0 H     2 2 2 2  ε0µ0 ε0µ0 µ ε0µ0 ε0µ0 –i –i ε -------------- G * × G = --------------  ----0 E × E + -----0 H × H + i ---------E × H – i ---------H × E   2 4 4 ε0µ 0 ε0µ0 2 ε0µ0 ε0µ0 –i * –i -------------- G × G = --------------  i -------------E × H + i -------------E × H = E × H = S   2 2 ε0µ 0 ε0µ0 (c) If G is purely real then G = G and G × G = 0 , so S = 0 . *

12

*

µ ε µ ε * (d) G ⋅ G =  ----0 E – i -----0 H  ⋅  ----0 E + i -----0 H  2 2 2 2 ε0 2 µ0 2 ε0µ0 ε0µ 0 1 * H ⋅ E = -- ( E ⋅ D + B ⋅ H ) = U E ⋅ H + i ---------G ⋅ G = ---- E + ----- H – i ---------2 2 2 2 2 Another way of looking at this is to recall that S U = -----c Now, because we consider an electromagnetic wave in free-space, the speed of light c = 1⁄

ε 0 µ 0 and θ = 90° , and we have

S 1 G ×G c * U = ------ = -------------- -------------------- = -- G G sin θ = G c c c ε0µ0 *


2

13

14



LAST NAME

FIRST NAME

Useful constants

MKS (SI)


c = 2.99792458 × 10 m s

Planck’s constant

h = 6.58211889 ( 26 ) × 10 – 16 eV s

8

–1


e = 1.602176462 ( 63 ) × 10

Electron mass

–34

J s

– 19

C

m 0 = 9.10938188 ( 72 ) × 10

–31

kg

Neutron mass

m n = 1.67492716 ( 13 ) × 10

–27

kg

Proton mass

m p = 1.67262158 ( 13 ) × 10

–27

kg

Boltzmann constant

k B = 1.3806503 ( 24 ) × 10 k B = 8.617342 ( 15 ) × 10

–5

–23

JK

eV K

–1

–1


ε 0 = 8.8541878 × 10


µ 0 = 4π × 10


c = 1 ⁄ ε0µ0

Avagadro’s number

N A = 6.02214199 ( 79 ) × 10 23 m o l –1

Bohr radius

a B = 0.52917721 ( 19 ) ×10–10 m

–7

H m

– 12

Fm

–1

–1



1

PROBLEM 1 (a) Write down the Hamiltonian for a particle of mass m in a one-dimensional harmonic oscillator potential in terms of momentum p x and position x. (b) If one defines new operators mω bˆ =  --------  2h 

1⁄ 2

i pˆ x   xˆ + ------ mω

1⁄ 2 i pˆ † mω bˆ =  --------  xˆ – -------x-   2h   mω  show that the Hamiltonian can be expressed as hω † † H = ------- ( bˆ bˆ + bˆ bˆ ) 2 † (c) Derive the commutation relation [ bˆ , bˆ ] . (d) Using your result from (c) to show that the Hamiltonian is † 1 H = hω  bˆ bˆ + --  2

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 ˆ is time-independent but the corresponding numerical value of the Often an operator A observable A has a spread in values ∆A about an average value 〈 A ( t ) 〉 and varies with time because the system is described by a wavefunction ψ ( x , t ) which is not an eigend 〈 A ( t ) 〉 multiplied by ∆t. dt 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 state. The change in 〈 A ( t ) 〉 in time interval ∆t is the slope

∆t = ∆ A ⁄

d 〈 A〉 dt

i ˆ , Bˆ ]〉 for time independent (a) Use the generalized uncertainty relation ∆A ∆B ≥ -- 〈 [ A 2 ˆ and Bˆ to show that ∆E∆ t ≥ h --- . operators A 2 (b) Show that the spread in photon number ∆n and phase ∆φ for light of frequency ω is 1 ∆n∆φ ≥ -2 and that for a Poisson distribution of such photons 1 ∆φ ≥ ---------------2 〈 n〉 2

PROBLEM 4 A particle of charge e, mass m, and momentum p oscillates in a one-dimensional har2 2 monic potential V ( xˆ ) = mω 0 xˆ ⁄ 2 and is subject to an oscillating electric field

E x cos ( ω t ) . (a) Write down the Hamiltonian of the system. (b) Find d 〈 xˆ 〉 . dt (c) Find

d ˆ d ˆ d 〈 p〉 and show that 〈 p〉 = – 〈 V ( xˆ ) 〉 . Under what conditions is the dt dt dx 2

quantum mechanical result m

d d 〈 xˆ 〉 = – 〈 V ( xˆ )〉 the same Newton’s second law in 2 d x dt 2

which force on a particle is F = m

dx = – d V ( x) ? dx d t2

(d) Find d 〈 H〉 . dt (e) Use your results in (b) and (c) to find the time dependence of the expectation value of position 〈 xˆ 〉 ( t ) . What happens to the maximum value of 〈 xˆ 〉 as a function of time when ω 0 = ω and when ω is close in value to ω 0 ?


3

SOLUTIONS Solution 1 (a) The Hamiltonian for a particle of mass m in a one-dimensional harmonic oscillator potential is 2

mω 2 2 pˆ H = -------x + ---------- xˆ 2m 2 where the particle moves in the potential V ( x ) = κx

2

and oscillates at frequency

∂ κ ⁄ m , where κ is a force constant. The operator pˆ x = – i h ----- . ∂x (b) The Hamiltonian can be factored ω =

2

mω 2 pˆ x 2 + xˆ  H = ----------  -----------2 m 2ω2 so it makes sense to define new operators i pˆ mω 1 ⁄ 2 bˆ =  --------  xˆ + -------x-  2h mω † mω 1 ⁄ 2 i pˆ bˆ =  --------  xˆ – -------x-  mω 2h

which, can be written in terms of operators xˆ and pˆ x to give † h 1⁄2 xˆ =  ------------  ( bˆ + bˆ ) 2mω

hmω pˆ x = i  ------------ 2

1⁄ 2

† ( bˆ – bˆ )

Substituting this into the Hamiltonian gives 2 2 2 2 2 pˆ † † mω 2 –1 hmω mω h H = -------x + ---------- xˆ = -------  ------------  ( bˆ – bˆ ) + ----------  ------------ ( bˆ + bˆ ) 2m 2 2m 2 2 2 mω † † † † hω H =  ------- ( – bˆ bˆ – bˆ bˆ + bˆ bˆ + bˆ bˆ ) + 4

ω ˆ ˆ ˆ † ˆ † ˆ ˆ † ˆ † ˆ h ( bb + b b + bb + b b )  ------4

† † hω H = ------- ( bˆ bˆ + bˆ bˆ ) 2 † † † (c) To derive the commutation relation [ bˆ , bˆ ] = bˆ bˆ – bˆ bˆ we can write out the differential form of the operators and have them act on a dummy wave function. This gives † mω h ∂ h ∂ ( bˆ bˆ ) ψ =  --------  x + -------- -----  x – -------- -----  ψ  2h   mω ∂ x  mω ∂x  † mω 2 h h ∂ h ∂ h ------∂ - ( bˆ bˆ ) ψ =  --------  x + -------- + -------- x ----- – -------- x ----- – -----------ψ 2h mω mω ∂ x mω ∂x m2 ω 2 ∂x 2 and 2

† mω h ∂ h ∂ ( bˆ bˆ ) ψ =  --------  x – -------- -----   x + -------- -----  ψ  2h   mω ∂ x  mω ∂x 

4

2

† mω 2 h h ∂ h ∂ h ∂  ( bˆ bˆ ) ψ =  --------  x – -------- – --------x ----- + -------- x ----- – ------------------- ψ 2h mω mω ∂x mω ∂x m 2 ω 2 ∂x 2 so that the commutation relation simplifies to just 2

2

† † mω 2h 2h ∂ 2h ∂ ( bˆ bˆ – bˆ bˆ ) ψ =  --------   -------- – -------- x ----- + -------- x -----  ψ = ψ 2h mω mω ∂x mω ∂ x or, in even more compact form, † † † [ bˆ , bˆ ] = bˆ bˆ – bˆ bˆ = 1 If we do not want to use differential operators and a dummy wavefunction then we

could write ih pˆ ihpˆ ihpˆ ihpˆ mω † † † mω [ bˆ , bˆ ] = bˆ bˆ – bˆ bˆ =  --------  xˆ + ----------x  xˆ – ----------x  –  --------   xˆ – ----------x   xˆ + ----------x          2h   mω  mω mω 2h mω 2 2 2 † ihpˆ x xˆ ihxˆ pˆ x h pˆ x   mω  2 ihpˆ x xˆ ihxˆ pˆ x h pˆ x  mω – + + ----------------------– -----------------x ˆ – ------------- + ----------[ bˆ , bˆ ] =  --------  xˆ 2 + ------------ 2h   mω mω mω mω m 2 ω 2 m 2 ω 2  2h   2

† imω hpˆ x xˆ hxˆ pˆ x – ----------- = i ( pˆ x xˆ – xˆ pˆ x ) = i [ pˆ x,xˆ ] = i ( – i h ) = 1 [ bˆ , bˆ ] =  ----------   ----------h mω mω

since [pˆ x ,xˆ ] = – i h . (d) The Hamiltonian is † † † † † † † † hω hω hω H = ------- ( bˆ bˆ + bˆ bˆ ) = ------- ( bˆ bˆ + bˆ bˆ + bˆ bˆ – bˆ bˆ ) = ------- ( 1 + 2 bˆ bˆ ) = hω  bˆ bˆ +  2 2 2

1 --2

† † † where we made use of the commutation relation [ bˆ , bˆ ] = bˆ bˆ – bˆ bˆ = 1

Solution 2 (a) The expectation value of position 〈 x〉 and momentum 〈 p 〉 for the first excited state | n = 1〉 of a particle of mass m in a one-dimensional harmonic oscillator potential is found using † h 1⁄2 xˆ =  ------------ ( bˆ + bˆ ) 2mω

h mω 1 ⁄ 2 † pˆ x = i  ------------ ( bˆ – bˆ ) 2 and † 1⁄2 | bˆ n〉 = ( n + 1 ) |n + 1 〉 | bˆ n〉 = n

1⁄ 2

|n – 1〉

〈 m| n 〉 = δ mn so that † h 1⁄2 h 1⁄ 2 〈 x〉 = 〈1| x |1 〉 =  ------------  〈 1|bˆ + bˆ |1〉 =  ------------ ( 〈 1|0 〉 +  2mω   2mω

2 〈1 |2〉 ) = 0

and † hmω 1 ⁄ 2 h mω 1 ⁄ 2 〈 pˆ x〉 = 〈1| pˆ x |1 〉 = i  ------------ 〈1| ( bˆ – bˆ ) |1〉 = i  ------------  ( 2 〈1|2〉 – 〈1| 0〉 ) = 0  2   2 


5

(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 ∆x = ( 〈 x 2〉 – 〈 x〉 2 ) 1 / 2 and ∆p x = ( 〈 p 2 x〉 – 〈 px 〉 2 ) 1 / 2 and since 〈 x〉 = 0 and 〈 p x 〉 = 0 we will be interested in finding the value of 〈 x 2〉 and 〈 p 2 x〉 . Starting with 〈 x 2〉 , we have † 2 † † † † h h 3h 〈 xˆ 2〉 =  ------------  〈1 | ( bˆ + bˆ ) | 1〉 =  ------------ 〈 1|bˆ bˆ + bˆ bˆ + bˆ bˆ + bˆ bˆ |1 〉 = ----------- 2mω   2 mω 2 mω

h 2 and one can see that for the general state |n〉 one has 〈 x 〉 =  ------------  ( 1 + 2 n ) . Now  2mω  turning our attention to 〈 p 2 x〉 we have 2 † † † † † h mω hmω 3h mω 〈 p 2x 〉 =  ------------ 〈1 | ( bˆ – bˆ ) | 1〉 =  ------------ 〈 1| – bˆ bˆ – bˆ bˆ + bˆ bˆ + bˆ bˆ | 1〉 = -------------- 2  2 2 2 h mω and one can see that for the general state |n 〉 one has 〈 pˆ x 〉 =  ------------  ( 1 + 2n ) . 2

For the particular case we are interested in | n = 1〉 and the uncertainty product is 3 ∆x ∆p x = ( 〈 x 2〉 〈 p 2x 〉 ) 1 ⁄ 2 = - h 2 h For the general state | n〉 the uncertainty product ∆x ∆p x = -- ( 1 + 2 n ) 2 Solution 3 (a) We start with the reasonable assumption that the expectation value of an observable ˆ evolves smoothly in time such that associated with operator A d ˆ ∆A 〈 A〉 = ------dt ∆t which may be written as ∆A ∆t = -----------------d ˆ 〈 A〉 dt ˆ is time-independent so that In this problem the operator A d ˆ i ˆ ]〉 + 〈 ∂ Aˆ 〉 = -i- 〈 [ H ,A ˆ ]〉 〈 A〉 = -- 〈 [ H ,A h h dt ∂t since 〈

∂ ˆ A〉 = 0 ∂t

ˆ and Bˆ is In addition, the generalized uncertainty relation for operators A i ˆ , Bˆ 〉 ] ∆A ∆ B ≥ -- [ 〈 A 2 which may be re-written as ˆ , Bˆ ] 〉 2 ∆A ∆B ≥ 〈 [ A 6

If operator Bˆ is the Hamiltonian H, then ∆B corresponds to ∆ E so that A ˆ , H ]〉 = h d 〈 A ˆ〉 = h ∆ 2 ∆A ∆E ≥ 〈 [ A ------dt ∆t Since the ∆A terms on the far left and far right of the equation cancel, we have h ∆ E∆t ≥ --2 It might be tempting to make a comparison between this result and the uncertainty relation for two non-commuting operators such as momentum pˆ x and position xˆ for which h ∆ p x ∆x ≥ --- follows directly from the generalized uncertainty relation. However, such a 2 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 h ∆ E∆t ≥ --- directly from the generalized uncertainty relation because t is not an opera2 tor. (b) The energy of the n photons oscillating at frequency ω is E = h ω  n + 

1 --- so that 2

∆ E = hω ∆n . The time dependence of the oscillator and its measurable quantities will

be proportional to cos ( ωt – φ ) so that ω ∆t = ∆ φ or ∆t = ∆φ ⁄ ω where ∆φ is the spread in values of phase φ . From (a) we have h ∆ E∆t = hω∆ n∆φ ⁄ ω = h ∆n∆φ ≥ --2 and hence 1 ∆ n∆φ ≥ -2 For Poisson statistics ∆n = 〈 n〉 so that ∆ φ ≥ 1 ⁄ 4 〈 n 〉 . 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

〈 n〉 .

Solution 4 2 2

2 mω 0 xˆ pˆ + eE x xˆ cos ( ω t ) . (a) H = T + V = ------- + --------------2 2m

(b) Because the position operator does not explicitly depend on time, 〈

∂ ˆx〉 = 0 , we ∂t

have d i ∂ i –i 〈 xˆ 〉 = - 〈 [ H ,xˆ ]〉 + 〈 xˆ 〉 = -- 〈 [ H ,xˆ ] 〉 = ---- 〈 [xˆ ,H]〉 h h h dt ∂t Since [xˆ , xˆ] = 0 and [ xˆ , xˆ 2] = 0 , we are left with –i –i d pˆ pˆ pˆ 〈 pˆ 〉 〈 xˆ 〉 = ---- 〈 [xˆ ,------- ]〉 = ---- 〈 pˆ [xˆ , ------- ] + [xˆ ,------- ] pˆ 〉 = -------dt h 2m h 2m 2m m 2


7

where we made use of the fact that [ xˆ ,pˆ ] = ih . Notice, that while the position operator does not explicitly depend on time (i.e., xˆ ≠ xˆ ( t ) ), the expectation value of the position operator can explicitly depend on time (i.e., 〈 xˆ 〉 = 〈 xˆ 〉 ( t ) ). 2 (c) Because [ pˆ ,pˆ ] = 0 we need only consider the commutator of the momentum operator with the potential appearing in the Hamiltonian. Hence, 2 2 mω 0 xˆ –i 2 d 〈 pˆ 〉 = – i + eE x xˆ cos ( ω t ) ] 〉 = – mω 0 〈 xˆ 〉 – e Ex cos ( ω t ) ---- 〈 [pˆ ,H]〉 = ---- 〈 [ pˆ , --------------dt 2 h h

∂ where we used pˆ = – i h . The spatial derivative of the expectation value of the ∂x potential is 2mω 0 〈 xˆ 〉 d ˆ d 〈 p〉 V ( xˆ )〉 = ---------------------- + e E x cos ( ωt ) = 2 dt dx which is Ehrenfest’s theorem. 2

–〈

2

Hence, in quantum mechanics, m

d d ˆ 〈 x 〉 = – 〈 V ( xˆ ) 〉 . This would be the same as dx dt 2 2

d dx Newton’s second law for force on a particle F = m 2 = – V ( x ) if position dx dt x = 〈 xˆ 〉 and if 〈

d d V ( 〈 xˆ 〉 ) . The conditions under which the latter is V ( xˆ ) 〉 = d 〈 xˆ 〉 dx

true is when the potential is slowly varying so that higher order terms in the expansion 2 of the force F ( xˆ ) = F ( 〈 xˆ 〉 ) + ( xˆ – 〈 xˆ 〉 )F′ ( 〈 xˆ 〉 ) + ( xˆ – 〈 xˆ 〉 ) F′′ ( 〈 xˆ 〉 ) + … may be

neglected so that F ( xˆ ) ≅ F ( 〈 xˆ 〉 ) . This requires that the uncertainty ∆x = ( xˆ – 〈 xˆ 〉 ) is small. d i ∂H ∂H 〈 H〉 = -- 〈 [H ,H] 〉 + 〈 〉 = 〈 〉 = – eE x ω 〈 xˆ 〉 sin ( ωt ) h ∂t ∂t dt where we used the fact that [H ,H] = 0 . (d)

2

2 mω eE d ˆ d 〈 pˆ 〉 〈 x〉 = -------- = – ----------0 〈 xˆ 〉 – --------x cos ( ω t ) 2 d t m m m dt this can be written in the familiar form of an undamped forced oscillator

(e)

2

d ˆ 2 〈 x〉 + mω 0 〈 xˆ 〉 = Fx cos ( ω t ) 2 dt where F x = – e E x . This has an oscillatory solution of the form m

F x cos ( ωt ) 〈 xˆ 〉 ( t ) = A cos ( ω 0 t – δ ) + -------------------------m ( ω 20 – ω 2 ) which contains two harmonic oscillators, one at the natural frequency ω 0 and the other at the frequency ω of the input force. The constants A and δ are found from the initial contitions. If ω 0 = ω the amplitude of the undamped oscillator grows to infinity with the particular solution increasing linearly in time as

8

Fx t sin ( ω 0 t ) 〈 xˆ 〉 p ( t ) = --------------------------2mω 0 When ω is close in value to ω 0 then Fx ( cos ( ωt ) – cos ( ω 0 t ) ) 〈 xˆ 〉 ( t ) = --------------------------2 2 m ( ω0 – ω ) which, since 2 sin ( x ) sin ( y ) = cos ( x – y ) – cos ( x + y ) , may be written as ( ω0 + ω ) (ω 0 – ω ) 2 Fx sin  ---------------------t  sin  --------------------- t 〈 xˆ 〉 ( t ) = --------------------------2 2 2 2 m ( ω0 – ω ) The sum frequency ( ω 0 + ω ) is modulated by the difference frequency ( ω 0 – ω ) so that the amplitude of 〈 xˆ 〉 beats at the difference frequency ( ω 0 – ω ) .


9



LAST NAME

FIRST NAME

Useful constants

MKS (SI)


c = 2.99792458 × 10 m s

Planck’s constant

h = 6.58211889 ( 26 ) × 10 – 16 eV s

8

–1


e = 1.602176462 ( 63 ) × 10

Electron mass

–34

J s

– 19

C

m 0 = 9.10938188 ( 72 ) × 10

–31

kg

Neutron mass

m n = 1.67492716 ( 13 ) × 10

–27

kg

Proton mass

m p = 1.67262158 ( 13 ) × 10

–27

kg

Boltzmann constant

k B = 1.3806503 ( 24 ) × 10 k B = 8.617342 ( 15 ) × 10

–5

–23

JK

eV K

–1

–1


ε 0 = 8.8541878 × 10


µ 0 = 4π × 10


c = 1 ⁄ ε0µ0

Avagadro’s number

N A = 6.02214199 ( 79 ) × 10 23 m o l –1

Bohr radius

a B = 0.52917721 ( 19 ) ×10–10 m

–7

H m

– 12

Fm

–1

–1



1

PROBLEM 1 (a) Write a computer program to calculate the chemical potential for n non-interacting electrons per unit volume at temperature, T. (b) Calculate the value of the chemical potential for the case when electrons of * 18 –3 effective mass m = 0.07 × m0 and carrier density n = 1.5 × 10 c m are at temperature T = 300 K (c) Repeat (b), only now for the case when electrons have effective electron mass * m = 0.50 × m 0 . (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 T = 77 K . Your answer should include a print out of your computer program and plots. PROBLEM 2 1 1 (a) Show that ------------------------------= 1 – -----------------------------( E – µ) ⁄ k B T (µ – E ) ⁄ kB T e +1 e +1 (b) A semiconductor consists of a valance band with electron energy dispersion relation EVB = E ( k ) and a conduction band with electron energy dispersion relation such that EC B = E 0 – E ( k ) , where E0 is a constant such that the conduction band and valence band are separated by an energy band gap, E g . Show that when particle number is conserved, the chemical potential is in the middle of the band gap with value µ = E 0 ⁄ 2 and is independent of temperature.

2

SOLUTIONS Solution 1 (a) and (b) use computer program for Fig. 7.5. (c) and (d)

Solution 2 (µ – E ) ⁄ k T

B 1 e +1–1 1 1 (a) 1 – -----------------------------= ---------------------------------------= --------------------------------= -----------------------------(µ – E ) ⁄ kB T (µ – E ) ⁄ kB T –( µ – E ) ⁄ kB T ( E – µ) ⁄ k B T e +1 e +1 1+e e +1 Because it will be useful for part (b), we modify our notation for the Fermi-Dirac distribution so that

1 f ( E – µ ) = -----------------------------( E – µ ) ⁄ kB T e +1 so that f (E – µ ) = 1 – f( µ – E ) (b) We are given that the semiconductor valance band has an electron dispersion relation E VB = E ( k ) and a conduction band with dispersion relation such that E C B = E 0 – E ( k ) , where E0 is a constant such that the conduction band and valence band are separated by an energy band gap, E g . This means that the conduction and valance bands along with the corresponding density of states will be symmetric around the middle of the band gap energy that occurs at energy E = E 0 ⁄ 2 . 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 E ( k ) = – 2t cos ( k L ) .


3

E(k)

E(k) D1 (E0 - E) E CB = E0 - E(k) Eg D1 (E)

E VB = E(k) 0

Wave vector, k

π/L

Density of states, D(E)

If the density of states for the valence band is D1 ( E ) , then the total density of states is D ( E ) = D 1 ( E ) + D 1 ( E 0 – E ) . To calculate the chemical potential we use ∞

n =

∫

∞

D ( E )f ( E ) d E =

E=0

∫

( D1 ( E ) + D1 ( E 0 – E ) )f ( E ) d E

E=0

1 where the Fermi-Dirac distribution is f ( E ) = ------------------------------ and we note that this has ( E – µ ) ⁄ k BT e +1 the property f ( E – µ ) = 1 – f ( µ – E ) At temperature T = 0 K the integral for the carrier density can be written ∞

n =

∫ E=0

EF

D 1 (E )d E =

∫

D 1 ( E ) dE

E=0

Now, because particle number is conserved, we can equate the T = 0 K integral with the T ≠ 0 K integral. Using our new notation for the Fermi-Dirac distribution, this is written EF

n =

∫

D 1 ( E ) d E = ∫ D 1 ( E ) ( f ( E – µ ) + f ( E0 – E – µ ) ) dE

E=0

which requires f( E – µ ) + f( E 0 – 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 – E0 + E = 0 E µ = ----02 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.

4



LAST NAME

FIRST NAME

Useful constants

MKS (SI)


c = 2.99792458 × 10 m s

Planck’s constant

h = 6.58211889 ( 26 ) × 10 – 16 eV s

8

–1


e = 1.602176462 ( 63 ) × 10

Electron mass

–34

J s

– 19

C

m 0 = 9.10938188 ( 72 ) × 10

–31

kg

Neutron mass

m n = 1.67492716 ( 13 ) × 10

–27

kg

Proton mass

m p = 1.67262158 ( 13 ) × 10

–27

kg

Boltzmann constant

k B = 1.3806503 ( 24 ) × 10 k B = 8.617342 ( 15 ) × 10

–5

–23

JK

eV K

–1

–1


ε 0 = 8.8541878 × 10


µ 0 = 4π × 10


c = 1 ⁄ ε0µ0

Avagadro’s number

N A = 6.02214199 ( 79 ) × 10 23 m o l –1

Bohr radius

a B = 0.52917721 ( 19 ) ×10–10 m

–7

H m

– 12

Fm

–1

–1



1

PROBLEM 1 In first-order time-dependent perturbation theory a particle initially in eigenstate | n〉 of the unperturbed Hamiltonian scatters into state | m〉 with probability a m ( t ) 2 after the perturbation V is applied. (a) Show that if the perturbation is applied at time t = 0 then the time dependent coefficient a m ( t ) is t′ = t

iω mn t ′ 1 a m ( t ) = ----- ∫ W mn e d t′ ih t ′ = 0

where the matrix element W mn = 〈m | V |n〉 and hω mn = E m – En is the difference in eigenenergies of the states |m〉 and | n〉 . (b) A particle of mass m is initially in the ground state of a one dimensional harmonic 3 –t ⁄ τ

oscillator. At time t = 0 a perturbation V ( x, t ) = V 0 x e 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, t → ∞ .

PROBLEM 2 An electron is in ground state of a one-dimensional rectangular potential well for which V ( x ) = 0 in the range 0 < x < L and V ( x ) = ∞ elsewhere. It is decided to control the state of the electron by applying a pulse of electric field E ( t ) = E 0 e

2

–t ⁄ τ

2

⋅ xˆ at

time t = 0, where τ is a constant, xˆ is the unit vector in the x-direction, and E 0 is the maximum strength of the applied electric-field. (a) Calculate the probability P 12 that the particle will be found in the first excited state in the long time limit, t → ∞ . (b) If the electron is in a semiconductor and has an effective mass m* = 0.07 × m 0 , where m 0 is the bare electron mass, and the potential well is of width L = 10 n m , calculate the value of E 0 for which P 12 = 1 . Comment on your result.

PROBLEM 3 An electron is initially in the ground state of a one-dimensional rectangular potential well for which V ( x ) = 0 in the range 0 < x < L and V ( x ) = ∞ elsewhere. The ground state energy is E1 and the first excited state energy is E 2 . At time t = 0 the sys2 –t ⁄ τ tem is subject to a perturbation V ( x , t ) = V0 x e . Calculate and then use a computer program to plot the probability of finding the particle in the first excited state as a function of time for t ≥ 0 . In your plot, normalize time to units of τ and consider the three

values of ω 21 = 1 ⁄ 2π , ω 21 = 1 , and ω 21 = 2 π , where h ω 21 = E 2 – E 1 . Explain your results.

2

PROBLEM 4 2 4ω e Use the Einstein spontaneous emission coefficient A = ----------------------〈j |r | k〉 to estimate 3 3h c 4πε 0 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 lineshape 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? 3 2

PROBLEM 5 A particle in a continuum system described by Hamiltonian H( 0 ) is prepared in eigenstate |n 〉 with eigenvalue E n = hω n . Consider the effect of a perturbation turned on at time t = 0 that is harmonic in time such that W ( x, t ) = V ( x ) cos ( ωt ) , where V ( x ) 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 ( ω → 0 ) is 1 2π 2 ---- = ------ W mn ⋅ D ( E ) δ ( Em – E n ) , where the matrix element W mn = 〈m | W ( x , t ) | n〉 τn h

couples state |n〉 to state |m〉 via the static potential V ( x ) , the density of final continuum states is D ( E ) , and δ ( E m – En ) 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 ε = ε ( q ) . Given an impurity potential at position r due to a 2

–e charge e at position Ri is v q ( r – R i ) = ---------------------- , derive an expression for v ( q ) . εq r – R i (b) Use the expression for v ( q ) 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 dielectric with dielectric function ε = ε ( q ) . 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?


3

SOLUTIONS Problem 1 (a) Consider a quantum mechanical system described by Hamiltonian H( 0 ) and for which we know the solutions to the time-independent Schrödinger equation H( 0 ) | n〉 = E n | n〉 and the time-dependent Schrödinger equation – iω nt – iωn t ∂ | n〉e = H ( 0 ) |n 〉e . ∂t At time t = 0 we apply a time-dependent change in potential W(t) so the new Hamil-

ih

tonian is H = H( 0 ) + W ( t ) and the state |ψ ( t ) 〉 evolves in time according to ih

∂ | ψ ( t )〉 = ( H ( 0 ) + W ( t ) ) | ψ ( t )〉 ∂t

where | ψ ( t )〉 =

∑ a n ( t ) |n〉 e

–i ωn t

and a n ( t ) are time dependent coefficients.

n

Substitution of the state |ψ ( t )〉 in to the time-dependent Schrödinger equation gives – iω t – iω t d a n ( t ) | n〉e n = ( H ( 0) + W ( t ) ) ∑ a n ( t ) |n 〉e n ∑ dt n n Using the product rule for differentiation ((fg)' = (f 'g + fg')), one may rewrite the lefthand-side as

ih

– iω t – iω t – iω t ∂ ∂ ih ∑   a n ( t ) |n 〉e n + a n ( t )  | n〉e n   = ( H ( 0 ) + W ( t ) ) ∑ a n ( t ) | n〉e n       ∂t ∂t n n

and remove the term – iω t ∂ ih ∑ a n ( t ) | n〉e n = ∂ t n to leave

ih ∑ | n〉 e n

–iω n t

∂ a (t) = ∂t n

∑ a n ( t )H ( 0 ) |n 〉e

– iωn t

n

∑ a n ( t )W ( t ) | n〉e

– iω nt

n

Multiplying both sides by 〈m | and using the orthonormal relationship 〈 m|n〉 = δ mn gives iω t d a m ( t ) = ∑ a n ( t ) 〈m| W ( t ) | n〉 e m n dt n However, since

ih

〈m |W ( t ) | n〉 = ∫ φ *m ( x ) e

i ωm t

W φ n ( x )e

– iω nt

d x = W mn e

iωm n t

where hω mn = E m – En and W mn is defined as the matrix element ∫ φ *m ( x ) W φ n ( x ) dx , we may write ih

d am( t ) = dt

∑ a n( t ) W mn e

i ωmn t

n

If the system is initially in an eigenstate | n〉 of the Hamiltonian H ( 0) then a n ( t = – ∞ ) = 1 and a m ( t = – ∞ ) = 0 for m ≠ n . There is now only one term on the right-hand-side and we can write ih

4

i ωm nt d a m ( t ) = a n ( t ) W mn e dt

Integration from the time when the perturbation is applied at t = 0 gives t′ = t

iω mn t ′ 1 a m ( t ) = ----- ∫ W mn e d t′ ih t ′ = 0

(b) The transition probability from state | n〉 to state |m〉 is P nm = a m ( t ) . Using our solution in part (a) we have 2

1 P nm = -----2 h

t′ = t

∫

〈m |V ( x, t ) |n〉 e

iω mn t ′

2 t′ = t

2

V = -----20 h

d t′

t′ = 0

∫

〈m |x 3 | n〉 e

( iωm n –1 ⁄ τ )t ′

2

d t′

t′ = 0

In the problem we have initial state |n = 0〉 and we consider the long time limit, t → ∞ , this allows us to write t′ = ∞

2

2

V0 2 3 ( imω –1 ⁄ τ )t ′ e d t′ P nm = -----2 〈m | x |0〉 ∫ h t′ = 0 where, for the harmonic oscillator, the non-zero positive integer m multiplied by the frequency ω is related to the difference in energy eigenvalue by mω = ( E m – E 0 ) ⁄ h . The time integral is t′ = ∞

∫

e ( imω –1 ⁄ τ )t ′ dt′

2

1 = -----------------------------2 2 2 m ω +1⁄τ

t′ = 0

and the matrix elements 〈m |x | 0〉 are found using x = ( h ⁄ 2m ω ) 3

*

1⁄2

† ( bˆ + bˆ ) , where

m * is the particle mass. In our case only transitions |0 〉 → |1 〉 and |0〉 → |3〉 are allowed. 〈 m|x 3 | 0〉 = ( h ⁄ 2m * ω )

3⁄2

( bˆ + bˆ bˆ + bˆ ) = ( h ⁄ 2m * ω ) †3

†2

†

† 1⁄2 bˆ | n〉 = ( n + 1 ) |n + 1〉 ,

where we used

3⁄2

( 6δ m = 3 + 3δ m = 1 )

1⁄2 bˆ | n〉 = n | n – 1〉 ,

bˆ |0 〉 = 0 , and

〈 m| n 〉 = δ mn . Hence, V2 3 P 01 = -----20 〈1| x |0〉 h and 2

V P 03 = -----20 〈3| x 3 |0〉 h

t′ = ∞ 2

∫

e

( imω –1 ⁄ τ )t ′

2

9 V 20 h -  3  ---------------------------------2 =  ------------2 2 * 2m ω   ( hω ) + h ⁄ τ 

2

6V 0 h 3  =  ------------- ------------------------------------ 2m * ω   ( 3h ω ) 2 + h 2 ⁄ τ 2 

d t′

t′ = 0

t′ = ∞ 2

∫

e ( imω –1 ⁄ τ )t ′ d t′

t′ = 0

2

So, in the long time limit, t → ∞ , the maximum probability of the transition taking place occurs as τ → ∞ . Problem 2 The eigestates of a particle in a one-dimensional rectangular potential well for which V ( x ) = 0 in the range 0 < x < L and V ( x ) = ∞ elsewhere are | n〉 =

2 -- sin k n x L

where k n = n π ⁄ L , n = 1, 2 , 3 , … , and eigenenergies are E n = h k n ⁄ 2 m = hω n . 2 2


5

The perturbation V ( x , t ) = –e x E 0 e

2

2

–t ⁄ τ

is turned on at time t = 0, where τ and E 0

is the maximum strength of the applied electric-field. The probability P 12 that the particle will be found in the first excited state in the long time limit, t → ∞ is 2

2

2

P12

e E0 2 = ---------------〈2 |x | 1〉 2 h

t′ t ′ = ∞ i ω21 t ′ – -----2τ

∫

e

2

dt′

t′ = 0

where h ω 21 = E 2 – E 1 and L

2 – 16 L 〈2| x |1 〉 = -- ∫ x sin ( k 2 x ) sin ( k 1 x ) dx = -----------2 L0 9π so 2

2

2

P 12 = e E 0

2

16 L  2  ---------- 9hπ 2 

t ′t ′ = ∞ i ω21 t ′ – ----2 τ

∫

e

dt′

t′ = 0

ω 21 τ ω 221 t 2 t2 Let y = - – i ---------- so that iω 21 t – ----2 = – ------- – y and d t = τ dy . We now can write τ 2 4 τ t′ = ∞

2 2 2 2 16 L 2 –ω 21 τ ⁄ 4 2 –y P 12 = e E 0  ------------2  e e dy ∫ 9hπ t′ = 0

ω 21 = 3 π h ⁄ 2mL 2

where t′ = ∞

∫

e

–ax

t′ = 0

2

2

and

2

we

2

π e E0 16L 2 –ω221 τ 2 ⁄ 2 2 = -------------------  ------------2  e τ π 4 9hπ 2

made

use

of

the

standard

integral

1 π dx = -- -- . 2 a

* (b) If the electron is in a semiconductor and has an effective mass m = 0.07 × m 0 ,

where m0 is the bare electron mass, and the potential well is of width L = 10 n m , we can calculate the value of τ for which P 12 is a maximum. dP 12, t → ∞ = 2τ – ω 221 τ 3 dτ

0 =

so that τ =

2 ⁄ ω 12 . We now find E 0 for which P 12 = 1 . 2

2

πe 2 E0  16L  2 2π – 1 πe 2 E 0  16L  2  2mL 2 2 – 1 ------------2 ------------------------- ------------2 2π -----------P 12 = ------------------= e  3 π2h  e = 1  9 hπ  ω 2  9h π  4 4 21 E0

2

4  9hπ 3π h  1 27h π -  1 ------------ -------------2 e = 2  -----------------= ------------e 3 2 2  16 L   2mL 32 meL  2π e 2

27h π 0.5 2  -------------------3  e =  32meL  2

E0 =

2

3

2

2

3

2

– 34 2

27 × ( 1.05 × 10 ) × π -  × 1.65 2  ------------------------------------------------------------------------------------------------------ 32 × 0.07 × 9.1 × 10 – 31 × 1.6 × 10 –19 × 10 –24  3

E0 = 6.6 × 10 V m 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. 7

6

–1

Problem 3 The eigestates of a particle in a one-dimensional rectangular potential well for which V ( x ) = 0 in the range 0 < x < L and V ( x ) = ∞ elsewhere are | n〉 =

2 -- sin k n x L

where k n = n π ⁄ L , n = 1, 2 , 3 , … , and eigenenergies are E n = h k n ⁄ 2 m = hω n . 2 2

The particle is in the ground state prior to a perturbation

V ( x , t ) = V0 x 2 e – t ⁄ τ

being

turned on at time t = 0. The probability P 12 that the particle will be found in the first excited state at time t is 2

P 12

V0 2 = ------〈2| x |1〉 2 h

2

t′ = t

∫

e

iω2 1t ′ – t′ -τ

2

dt′

t′ = 0

where h ω 21 = E 2 – E 1 and L

2 – 16 L2 〈 2| x 2 |1〉 = --- ∫ x 2 sin ( k 2 x ) sin ( k 1 x ) dx = -------------L0 9π 2 The time integral is t′ = t

∫

t′ = 0

e

– ( 1 ⁄ τ –i ω21 )t ′

2

dt′

– ( 1 ⁄ τ – iω

)t

21 –1 = e-------------------------------1 ⁄ τ – i ω 21

2

1 + e –2 t ⁄ τ – 2e –t ⁄ τ cos ( ω 21 t ) = --------------------------------------------------------------2 2 ω 21 + 1 ⁄ τ

so 16V 0 L 2 2  1 + e – 2t ⁄ τ – 2e – t ⁄ τ cos ( ω 21 t )  P 12 ( t ) =  ---------------- --------------------------------------------------------------- 2 2 2 9h π   ω 21 + 1 ⁄ τ Now we notice that ω 21 is related to L via ω 21 = 3 π2 h ⁄ 2mL 2 so L 2 = 3π 2 h ⁄ 2 mω 21 and –2t ⁄ τ

–t ⁄ τ

16 V 0  2  1 + e – 2 e cos ( ω 21 t )  P12 ( t ) =  --------------- --------------------------------------------------------------2 2   6mω 21   ω 21 + 1 ⁄ τ In the limit t → ∞ the probability becomes V 0 16  2  1 P 12 ( t → ∞ ) =  --------------- --------------------------  6mω 21   ω 2 + 1 ⁄ τ 2  21 In this long time limit and normalizing time to τ , we see that 2

V 0 16  t 1 - 38.5 P 12  -- → ∞ , ω 21 = ------  =  ----------τ 2π 6m  2

V 0 16 t P 12  -- → ∞ , ω 21 = 1  =  ------------  0.5 τ 6m and V0 16 2 t –4 P 12  -- → ∞ , ω 21 = 2 π =  ------------  6.26 × 10 τ 6m so, if we want to plot the results on the same graph, it is best to normalize to these long time values.


7

Problem 4 Differences in energy between eigenfunctions ψ nlm of hydrogen are given by m e4 1 1 1  -1-- – ---- = R y  -----2 – -----2  ∆E = En 2 – En 1 = ------0 ----------------------n n  2 ( 4 πε 0 ) 2 h 2  n21 n 22 1 2 where R y = 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 λ = c2π h ⁄ ∆E = c 2π ⁄ ω which is λ = 122 nm. 4 πε 0 h 2 - is the Bohr An estimate for the matrix element is 〈j |r | k〉 ∼ a B , where a B = ---------------m 0e2 radius and we obtain ( 2π ) --------------e 2 ( 2π ) ∆ E --------------e 2 9 –1 1 A = ------------⋅ ⋅ a = 1.12 × 10 s = -----⋅ ⋅ a = ----------------------3 3 3 3πε 0 h B 3 πε 0 h B τsp h c λ Hence, our estimate of the spontaneous emission lifetime for hydrogen is τ sp = 0.89 ns . 3

2

3

2

3

For the helium ion we note that Z = 2. The Bohr radius scales as a B α 1 ⁄ Z . For a hydrogenic atom with ion charge number Z the Bohr radius for the electron is a B = 4πε 0 h ⁄ m 0 Ze 2 . The Rydberg energy which determines ∆E scales as Z2 so 2

∆E = Z 2 R y ( 1 ⁄ n 21 – 1 ⁄ n 22 ) .

Hence,

A = A × Z ⁄ Z = 16 × A = 1.79 × 10 *

8

6

2

the 10

s

–1

linewidth

of

the

and the lifetime is τ

* sp

He +

ions

is

= 0.056 n s . The

Lorentzian line for helium is 16 times wider than that of the corresponding hydrogen line.

Problem 5 The Hamiltonian H ( 0 ) of the unperturbed system has solutions to the time-independent Schrödinger equation given by H ( 0 ) | n〉 = E n | n〉 The time-independent eigenvalues are E n = hω n and the orthonormal eigenfunctions are |n 〉 . The eigenfunction | n〉 evolves in time according to –i ωn t

| n〉 e = φn ( x ) e and satisfies

–i ωn t

– iωn t –i ωn t ∂ |n 〉e = H ( 0 ) |n 〉e ∂t At time t = 0 we apply a time-dependent change in potential W(x, t) whose effect is to create a new Hamiltonian H = H( 0 ) + W ( x, t )

ih

and state | ψ ( x, t ) 〉 which evolves in time according to ∂ |ψ ( x, t ) 〉 = ( H ( 0) + W ( x, t ) ) |ψ ( x , t )〉 ∂t The wavefunction |ψ ( x , t )〉 may be expressed as a sum over the known unperturbed eigenstates ih

| ψ ( x, t ) 〉 =

∑ a n ( t ) |n〉 e

–i ωn t

n

where a n ( t ) are time dependent coefficients. It follows that ih

– iω t –i ω t d a n ( t ) | n〉e n = ( H ( 0) + W ( x, t ) ) ∑ a n ( t ) |n〉 e n ∑ dt n n

–i ω t – iω t –iω t ∂ ∂ ih ∑   a n ( t )  |n 〉e n + a n ( t )  | n〉e n  = ( H( 0 ) + W ( x, t ) ) ∑ a n ( t ) | n〉e n       ∂ t ∂ t n n

ih ∑ | n〉e n

– iω n t

∂ a (t) = ∂t n

∑ a n ( t )W ( x, t) |n 〉e

–i ω n t

n

Multiplying both sides by 〈 m| and using the orthonormal relationship 〈 m|n〉 = δ mn gives iω t d a m ( t ) = ∑ a n ( t ) 〈m |W ( x, t ) |n 〉e m n dt n If the perturbation is harmonic and of the form W ( x, t ) = V ( x ) cos ( ωt )

ih

where the spatial part of the potential is given by V ( x ) . The matrix elements W mn in the representation of the unperturbed system described by Hamiltonian H ( 0 ) are just W mn e

i ( ωm – ωn )t

= 〈m | V ( x ) |n〉 e

i( ω m – ωn )t


i (ω

+ ω)t

i( ω

– ω )t

e mn + e mn cos ( ω t ) = 〈m |V ( x ) |n〉 -----------------------------------------------2

9

where ω mn = ω m – ω n . Assuming an initial condition such that a n ( t ≤ 0 ) = 1 for only one eigenvalue and a m ( t ≤ 0 ) = 0 for m ≠ n , then we may write t′ = t

1 a m ( t ) = ----- ∫ W mn dt′ ih t ′ = 0 Performing the integration gives i (ω

– ω )t

i( ω

– ω )t

mn mn – 〈m |V ( x ) |n〉 – 1 + e---------------------------– 1-  a m ( t ) = -----------------------------  e---------------------------- ω mn + ω 2h ω mn – ω 

Taking the static limit ω → 0 i ωm n t

– 〈m |V ( x ) |n〉 e –1 a m ( t ) = -----------------------------  --------------------   ω mn  h i ωm n t ⁄ 2

–i ωmn t ⁄ 2

– 〈m |V ( x ) |n〉 i ωm n t ⁄ 2  e –e ----------------------------------------  a m ( t ) = ----------------------------- e   h ω mn – 2 〈 m |V ( x ) |n 〉 iω t ⁄ 2 sin ( ω mn t ⁄ 2 )  a m ( t ) = -------------------------------- e m n  -----------------------------  h ω mn Assuming independent scattering channels, the probability of a transition out of state |n〉 into any state |m〉 is the sum Pn ( t ) =

∑ am ( t )

2

m

If there are D ( E ) = dN ⁄ dE states in the energy interval dE = h dω′ , then the sum can be written as an integral sin2 ( ( ω′ – ω n ) t ⁄ 2 ) 4 2 dN Pn ( t ) = -----2 〈 m |V ( x ) |n 〉 ⋅ ------- ⋅ ∫ ---------------------------------------------⋅ hdω ′ 2 dE h ( ω′ – ω n )

To perform the integral we change variables so that x = ( ω ′ – ω n ) ⁄ 2 and then take the limit t → ∞ . This gives sin2 ( tx ) -----------------πt x 2

= δ (x ) t→∞

We now note that dE ⁄ d x = 2h since E = h 2x . Hence, the integral can be written

∫

2 2 sin 2 ( ( ω – ω n ) t ⁄ 2 ) dE ( tx ) ⋅ tπ hπt sin ( tx ) -------------------------------------------------- dx = 2h ∫ sin ---------------------------dx = -------- ∫ -----------------dx 2 2 2 dx 2 (ω – ω n ) πt4 x πtx

so that in the limit t → ∞ sin ( ( ω – ω n ) t ⁄ 2 ) dE hπt -------------------------------------------------- dx = -------- ∫ δ ( x ) dx 2 d x 2 (ω – ω n ) 2

∫

One may now write the probability of a transition out of state | n〉 into any state |m〉 as 4 hπt 2π 2 2 Pn ( t ) = -----2 〈 m |V ( x ) |n 〉 D ( E ) -------- = ------ W mn D ( E )t 2 h h Recognizing dP n ( t ) ⁄ dt as the inverse probability lifetime τ n of the state |n 〉 , we can write Fermi’s golden rule

10

1 2π ---- = ------ 〈 m |V ( x ) |n 〉 2 D ( E ) δ ( ω m – ω n ) h τn The delta function is included to ensure energy conservation. Because we took the limit ω → 0 the final state must be the same as the initial state energy. If we did not take the limit ω → 0 so that ω ≠ 0 then the final state energy could be ± hω . The importance of what we have shown is that Fermi’s golden rule can also be applied to the static limit of a harmonic perturbation. The significance of the result is that we can use time dependent perturbation theory to describe scattering from static potentials such as electrons scattering elastically from ionized impurities in a semiconductor.

Problem 6 (a) and (b) see text. (c) Here it is important to recognize that there are different limiting cases to consider: (i) Perfectly ordered impurities in a periodic lattice, (ii) clusters of impurity scattering sites and (iii) anticlustering of scattering sites.


11

12


Final

LAST NAME

FIRST NAME

SI-MKS 8


c = 2.99792458 × 10 m s

Planck’s constant

= = 6.58211889 × 10

– 16

= = 1.054571596 × 10 Electron charge Electron mass Neutron mass Proton mass

e = 1.602176462 × 10

–1

eV s

– 34

– 19

Js

C

m 0 = 9.10938188 × 10

– 31

kg

m n = 1.67492716 × 10

– 27

kg

m p = 1.67262158 × 10

– 27

kg

– 23

JK

–1

eV K

–1

Boltzmann

k B = 1.3806503 × 10

constant

k B = 8.617342 × 10


ε 0 = 8.8541878 × 10


µ 0 = 4π × 10


c = 1 ⁄ ε0 µ0

Avagadro’s number

N A = 6.02214199 × 10

Bohr radius

a B = 0.52917721 ×10 m

–7

–5

Hm

– 12

Fm

–1

–1

23

mol

–1

– 10

2

4πε 0 = a B = ---------------m0e 2

Inverse fine-structure constant α –1 = 137.0359976 4πε 0 =c α –1 = ----------------e2


1

PROBLEM 1 The time-independent Schrödinger equation for Hamiltonian H 0 has known orthonormal eigenfunctions |n〉 such that H 0 |n〉 = E n |n〉 where E n = =ω n . In the presence of a time-dependent change in potential W(x, t) applied at time t > 0 the system evolves in time according to ∂ i= |ψ ( x, t )〉 = ( H 0 + W ( x, t ) ) |ψ ( x, t )〉 = H |ψ ( x, t )〉 ∂t where |ψ ( x, t )〉 =

∑ a n ( t ) |n〉e

– iω n t

n

and a n ( t ) are time-dependent coefficients. (a) Show that for time t > 0 iω t d i= ----- a m ( t ) = ∑ a n ( t )W mn e mn dt n where =ω mn = =ω m – =ω n and W mn = 〈m|W ( x, t ) |n〉 . (30%) (b) In first-order time-dependent perturbation theory W(x, t) is assumed small and a n ( t > 0 ) is approximated by its initial value a n ( t = 0 ) . Show that for time t > 0 the final states after scattering may be found using 1 a m ( t ) = ----i=

t=∞

∫ W mn e

iω mn t′

dt′

(30%)

t=0

(c) If P k → j ( t ) is the probability of a transition from |k〉 at time t′ = 0 to a state |j〉 at time t′ = t and P j → k ( t ) is the probability that the same Hamiltonian induces a transition from state |j〉 to state |k〉 , use first-order time-dependent perturbation theory to show P k → j ( t ) = P j → k ( t ) . (40%)

PROBLEM 2 (a) Show that the root-mean-square deviations in the measured value of the observable associated ˆ for a system in state ψ n is ∆A = ( 〈 A 2〉 – 〈 A〉 2 ) 1 ⁄ 2 . (40%) with operator A (b)

Show that the position operator, xˆ = ( = ⁄ 2 mω )

pˆ = – i ( =mω ⁄ 2 )

1⁄2

1⁄2

†

( bˆ + bˆ ) and momentum operator

† ( bˆ – bˆ ) , for the n-th state of a one-dimensional harmonic oscillator with

= 1 energy eigenvalue E n = =ω  n + --- satisfies the relation ∆x∆p = --- ( 1 + 2n ) . (40%)  2 2 (c) Calculate the room temperature value of ∆x for a Ge atom (atomic mass 72.6) vibrating as a harmonic oscillator in one-dimension at a frequency of 9.0 THz. Compare the value you obtain with the lattice constant of a germanium crystal ( 5.64613 × 10

– 10

m ). (20%)

2

PROBLEM 3 An electron is initially in the ground state of a one-dimensional rectangular potential well for which V ( x ) = 0 in the range 0 < x < L and V ( x ) = ∞ elsewhere. The ground state energy is E 1 , the first excited state energy is E 2 , and E 21 = E 2 – E 1 . At time t ≥ 0 the system is subject to a pertur–t ⁄ τ

bation V ( x, t ) = – e E 0 xe where τ is a constant and E 0 is the maximum strength of an electric field applied in the x-direction. (a) Show that the probability P 12 that the particle will be found in the first excited state as a function of time for t ≥ 0 is – 2t ⁄ τ

–t ⁄ τ

16e E 0 L 2  1 + e – 2e cos ( ω 21 t ) - --------------------------------------------------------------P 12 ( t ) =  --------------------2 2   9π 2   E 21 + ( = ⁄ τ )

(50%) *

(b) The electron in part (a) has an effective mass m = 0.07 × m 0 , where m 0 is the bare electron mass, the potential well is of width L = 10 nm , and the time constant τ = 1 ps . Calculate the value of E 0 for which P 12 = 1 in the long time limit t → ∞ . Comment on your result. (50%)

PROBLEM 4 (a) What determines the selection rules for optical transitions at frequency ω between states |k〉 and |j〉 ? (20%) (b) Show the inverse of the Einstein spontaneous emission coefficient for a transition from the first excited state to the ground state 3

3πε 0 =c 1 --- = -----------------------------2 = τ 2 3 A e ω 〈j|r |k〉 can be rewritten for light emission of wavelength λ from electronic transitions in a harmonic oscillator potential as 1 2 --- = 45 × λ ( µm ) = τ ( ns ) A where wavelength is measured in µm and time τ is measured in ns. You may wish to use the fact † 1⁄2 that the position operator for a one-dimensional harmonic oscillator is xˆ = ( = ⁄ 2 m 0 ω ) ( bˆ + bˆ ) . (50%) (c) Calculate the numerical value of the spontaneous emission lifetime and the spectral linewidth for an electron mass m0 making a transition from the first excited-state to the ground-state of a harmonic oscillator potential characterized by force constant κ = 3.59 × 10


–3

–2

kg s . (30%)

3

4

Solutions to Levi Applied Quantum Mechanics 2nd Ed

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