Solutions to Levi Applied Quantum Mechanics 2nd Ed

Applied Quantum Mechanics

Chapter 1 Problems and Solutions

LAST NAME

FIRST NAME

Useful constants Speed of light in free space Plancks constant

MKS (SI) c = 2.99792458 10 m s8 1

h = 6.58211889 ( 26 ) 10

16

eV s J s C kg kg

h = 1.054571596 ( 82 ) 10 Electron charge Electron mass Neutron mass Proton mass Boltzmann constant e = 1.602176462 ( 63 ) 10 m0 = 9.10938188 ( 72 ) 10 mn = 1.67492716 ( 13 ) 10

34 19

31 27

mp = 1.67262158 ( 13 ) 10 27 kg k B = 1.3806503 ( 24 ) 10 23 J K 1 k B = 8.617342 ( 15 ) 10 5 e V K 1 0 = 8.8541878 10 12 F m 1 0 = 4 10 7 H m 1 c = 1 00 NA = 6.02214199 ( 79 ) 10 a B = 0.52917721 ( 19 ) 102 23 1

Permittivity of free space Permeability of free space Speed of light in free space Avagadros number Bohr radius 4 0 h a B = ---------------m 0e2

mol

10

m

Inverse fine-structure constant 1 = 137.0359976 ( 50 ) 4 0 hc 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 Earths 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 ? 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 Maxwells 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 Maxwells equations with magnetic charge in terms of a field G = PROBLEM 8 The capacitance of a small metal sphere in air is C0 = 1.1 10 2.2 1018 18

E + i H .

F . A thin dielectric film

with relative permittivity r 1 = 10 uniformly coats the sphere and the capacitance increases to 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 / ' of the two molecules. (b) What is the ratio of vibrational frequencies for carbon monoxide isotope 12 ( C O ) and 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 childs 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 A13 16 12 16

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 ( ice 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 x k l l

r r

(r - x)

l = k = ( r + x) ( r x) = r x Volume of the plug is2 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 (r - x) r x r r/21/2 l r x r l (r - x)

l = (r + x ) (r x ) = r x so that area of disk radius l at position x is2 2 2

Area = ( 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 offr

Vol off = 2

r 2

x3 ( r 2 x 2 ) dx = 2 r 2 x --3

r 0

1 1 1 = 2r 3 1 -- ---- ---------- -- 3 2 6 2

3 2 5 Vol off = 2 r - --------- 3 6 2 The volume of the plug is just the volume of a half sphere minus Vol off .

2r 3 2 5 3 1 2 5 3 5 1 Vol = ----------- 2 r - --------- = 2r - - + --------- = 2r ---------- -- - 3 6 2 3 3 6 2 6 2 3 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 .

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 m2 then g m2 gm 2 t dx dx = ------------------ and, starting from rest, speed = ------------------ with displacement 2 m 1 + m2 dt m 1 + m2 dt2

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 = the edge. 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 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, gh k = g h 2 . For the group velocity and phase velocity are the same and = 2 f = f = 10 0.9 10 = 0.3 Hz .

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

Solution 5 (a) = Ak0.5

. d is twice the phase velocity v p = k we have = 2 --dk k k

(b) Because the group velocity v g = and so d dk ------- = 2 ----- . k4

Integrating both sides of this equation allows one to write

ln = 2 ln k + ln A = ln Ak 2 so that the dispersion relation is = Ak 2 where A is a constant. (c) = Ak . 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 frequency8

relative-1

to

f08

is-1

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 , and f0 = 10 10 Hz giving f = 9.26 kHz. The component of velocity in the direction of the detector is v cos() where is the angle between the ground and the plane flying at height h. A simple expression for the time dependence of the angle = (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

tance to the plane (range) is l then l = h + v t overhead). Then cos ( ( t ) ) = v t l = v t another angle to describe its position.2

2

2

2 2 2 2

(assuming we set t = 0 when the plane is h v t + 1 . The Doppler shift as a2 2 2

h +v t = 1

function of time is f cos((t)). If the plane does not fly overhead it is necessary to introduce 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 Maxwells equations would have to include magnetic charge density m and magnetic current density J m as well as the usual electron charge density e and electron current density J e . The new equations would be D = e B = m B E = ------ Jm t H = J e + D ------t where D = E and B = H . Conservation of magnetic current J m = m v is expressed as m + 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 = H, electric field strength E[V m -1 ],the displacement vector field D = E, permia2 bility [N A-2] or [H m -1 ], permittivity [A 2 s2 N-1 m -2] or [F m-1 ], and 0 = 1 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 =

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

density U[J m -3]. Maxwells equations in terms of E and H with magnetic charge can be written e - E = --------2 2 m -- H = ---------2 2 H - E = -- ------- + Jm 2 2 t

8

E -- H = -- ------- + J e 2 2 t It follows that m e G = ---------- + i ---------2 2 and H - E + i -- H = -- ------- 2 2 2 t which can be rewritten as G = iJe Jm 1 dG ---------- + ------------------ G = i dt 2 2 since i dG H 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 E - J m + i -- ------- + i -- J e 2 2 t 2

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 0 r 0 It follows that18 C0 1.1 10 r 0 = ----------- = ----------------------------------------- = 10 n m 12 4 0 4 8.85 10 Now consider the case when the metal sphere radius r0 with charge Q is coated with a dielectric of relative permittivity r 1 to radius r1 and 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 Q Q V = E rd r E r dr = ---------------------- dr ---------------------- dr 2 2 r1 r1 4 0 r 1 r r r 4 0 r 2 r2 2

r0

r1

r0

r1

Q 1 1 1 1 V = ----------- ---------- --------- + ---------- ---------- 4 0 r1 r 0 r 1 r 1 r 2 r 1 r2 r 2 For this particular problem we now take the limit r 2 . Capacitance 4 0 4 0 Q C = -- = ---------------------------------------------------------- = ------------------------------------------------V 1 1 1 ---------- ---------- + ---------- 0 1 1 r2 r1 r r r r - ---------- -- ---------------- r2 r 1 1 0 1 1 r1 r 0 r 1 r 1 r 2

C r2 r C ----------- -- ----------------1 = 4 0 r1 r 0 r 1 r1 r 2


9

C C r2 r ---------- 4 0 = -- ----------------1 -r 1 r1 r r1 r 0 2 r1 r 0 r2 r r0 r2 rr 1 = C ----------------1 -------------------------------- = ----------------1 ------------------------- r2 r1 r C 4 0 r1 r 0 r1 C 0 2 1 ----------- C Putting in the numbers. We have r1 = 10 , r 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, m2 m r 1 = --------- r 2 = ---------2 ( r 1 r ) m1 m1 m2 m r 1 1 + -----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 is2 m1 m 2 2 m 1m 2 2 1 2 1 2 1 m 1 m2 2 2 T = - ( m 1 + m2 )R + --------------------------2 r + --------------------------2 r = - ( m1 + m 2 )R + - -----------------------r 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 = m where 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, = , and the ratio of oscillation frequency is given by ----- = We m1 m 2 ( m1 + m 2 ) ---------------------------------------m1 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 .

m1 u

m2 v

2

m1

du = ( v u) dt 22

dv m2 2 = ( u v ) dt which has solution of the form e ( m 1 )u v = 02 i t

giving

u + ( m2 2 ) v = 0 giving a characteristic equation


11

m1

2

m 22

= ( m 1 2 ) ( m 2 2 ) 2 = 0

m 1 m2 2 - 2 = ------------------ = m m 1 + m 2 and = m1 + m 2 ------------------ = m 1 m2 - m

and, as before, since the isotope form of the molecule interacts in the same way, = , and the ratio of oscillation frequency is given by m1 m 2 ( m1 + m 2 ) ----- = --------------------------------------- m1 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 12C masses to give --------- = 13C 13 16 ( 12 + 16 ) ----------------------------------------------- = 1.02 . As expected, the lighter 12 16 ( 13 + 16 )13 16 12

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 doesnt 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 = x 2 . This is the potential energy of a harmonic oscillator with spring constant2 2

= F0 l . This oscillator has frequency = (b) = The F0 ml = frequency of oscillation 450 20 2.5 =

m = in1

F 0 ml . per the second period of is given by is so oscillation

radians

9 = 3 rad s ,

= 1 f = 2 = 2.09 s , i.e., about 2 seconds.

12



LAST NAME

FIRST NAME


MKS (SI) c = 2.99792458 10 m s8 1

h = 6.58211889 ( 26 ) 10

16

eV s J s C kg kg

h = 1.054571596 ( 82 ) 10 Electron charge Electron mass Neutron mass Proton mass Boltzmann constant e = 1.602176462 ( 63 ) 10 m 0 = 9.10938188 ( 72 ) 10 m n = 1.67492716 ( 13 ) 10

34

19 31 27

m p = 1.67262158 ( 13 ) 10 27 k g k B = 1.3806503 ( 24 ) 10 23 J K 1 k B = 8.617342 ( 15 ) 10 5 eV K 1 0 = 8.8541878 10 12 F m 1 0 = 4 10 7 H m 1 c = 1 00 N A = 6.02214199 ( 79 ) 10 a B = 0.52917721 ( 19 ) 102 23 1


mol

10

m



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 r = 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

x =

dx

*

( x )x ( x )

in terms of (k x) which is the Fourier transform of (x). (b) The wave function for a particle in real-space is ( 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 = h , show that the energy operator for the wave function ( 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 x for the width of the wave packet to double. Compare 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 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 normal to a flat metal mirror. The magnitude of photon momentum is p p h = hk = h c where frequency is given by the quanta of photon energy E = h . p = ( 2h c ) cos which,19 8

The momentum change upon reflection is = 0,1

for27

gives

a

value

2 E c = 2 2 1.6 10

3 10 = 2.1 10

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 Z E n = Ry ----2 n 122.4 For Li++ Z = 3 and so E n = ------------- e V . 2 n 2h c 10.1 Emission wavelength is = -------------------- = ---------------- nm , giving 3, 2 = 72.7 nm , 4, 2 = 53.8 n m , 1 1 En2 En1 --- ---n2 n1 5, 2 = 48.1 n m . m0 Z e 1 For r = 10 we have En 2 En 1 = ---------------------------- -1 ---- , so divide energy differences by -2 2n n1 ( 4 0 r ) h 2 r = 1002 2 4 2

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

x =

dx

*

1 * i kx ikx ( x )x ( x ) = ------ dx dk ( k ) e x d k ( k ) e 2 ikx

1 * ik x e x = ------ dx d k ( k )e x dk ( k ) ------- 2 k ix

4

1 e x = ------ d x d k * ( k )e ik x x ( k ) ------ 2 ix

ikx

e dk k ( k ) ------- ix i kx

1 i * ik x i kx i( k k )x * x = -------- d x dk ( k )e dk k ( k ) e = ------ dxe dk d k ( k ) k ( k ) 2i 2

x = i

d k dk

*

( k )

* * ( k ) ( k k ) = i d k ( k ) ( k ) = ih d k ( k ) ( k ) k k p . p

Hence, in k-space position is a differential operator, ih (b) The expectation value of the energy operator is E =

1 * * i t i t d ( )h ( ) = ------ d d t ( t ) e h d t ( t) e 2 i t

e 1 * i t E = ------ d d t ( t ) e h d t ( t ) -------- 2 t i 1 ei t E = ------ d dt * ( t )e i t h ( t ) ------ 2 i h E = -------- d 2 i

d t t ( t ) ------- i

e i t

*

dt ( t )e*

i t

ih i ( t t ) * i t dt t ( t ) e = ------ d e d td t ( t ) t ( t ) 2

E = ih

dt d t ( t )

* * ( t ) ( t t ) = ih d t ( t ) ( t ) = ih d t ( t ) ( t ) t t t . t

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

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 in the x-direction is h d m 2 H = T + V = ---- 2 + ---------- x ---2m d x 22 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 isj = 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 ) = e22 2

i > i

-----------------4 r

+

0 i i

j > j

j j j ------------------ ---------------4 r 4 r 0 jj i, j

ZZ e

Ze

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 4 0 r h 2 r a * = --------------------- = a B -------- = 0.529 10 10 693 = 36.7 nm B 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 . unit area is S total e h averag e = 1.8 10 22 s 1 m 2 .5

(b) From Exercise 1 Chapter 2, h average = 1.92 eV so the number of photons per second per (c) The radiation pressure in (a) is a small value compared to one atmosphere which is about

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


MKS (SI) c = 2.99792458 10 m s8 1

h = 6.58211889 ( 26 ) 10

16

eV s J s C kg kg


34

19 31 27

m p = 1.67262158 ( 13 ) 10 27 k g k B = 1.3806503 ( 24 ) 10 23 J K 1 k B = 8.617342 ( 15 ) 10 5 eV K 1 0 = 8.8541878 10 12 F m 1 0 = 4 10 7 H m 1 c = 1 00 N A = 6.02214199 ( 79 ) 10 a B = 0.52917721 ( 19 ) 102 23 1


mol

10

m



1

PROBLEM 1 Prove that particle flux (current) is zero if the one-dimensional exponential decaying wave func x i t tion in tunnel barrier of energy V0 and finite thickness L is ( x, t ) = Be , where is a real positive number and particle energy E = h < V0 . PROBLEM 2 (a) Use a Taylor expansion to show that the second derivative of a wavefunction ( 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 ( x j 1 ) 2 ( x j ) + ( x j + 1 ) ( x j ) = ----------------------------------------------------------------2 dx h0 (b) By keeping additional terms in the expansion, show that a more accurate approximation of d2 2

the second derivative is ( x j 2 ) + 16 ( x j 1 ) 30 ( x j ) + 16 ( x j + 1 ) ( x j ) d ( x j ) = --------------------------------------------------------------------------------------------------------------------------------+ 2 -----2 2 dx 12h 02

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 * = 0.07 m0 confined to a potential well V(x) = V0 of width L = 10 nm with periodic boundary e 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 * = 0.07 m0 confined to a triangular potential well of width L = 20 nm bounded by barriers of e 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 x i t function in tunnel barrier of energy V0 and finite thickness L is ( x , t ) = Be , where is a real positive number and particle energy E = h < V 0 , we substitute the wave function into the current operator i eh d d J = ----------- * ( x , t ) ( x, t ) ( x, t ) * ( x , t ) 2m dx dx i eh J = ----------- ( B* e x + i t ( B e x i t ) Be x i t ( B * e x + i t ) ) 2m i eh 2 2x 2 x J = ----------- B ( e e ) = 0 2m Solution 2 (a) Consider the Taylor expansion for the function f(x). f ( x j ) 2 f ( x j ) 3 f ( x j + 1 ) = f ( x j ) + f ( x j ) h 0 + -------------- h 0 + ---------------- h 0 + = 2! 3!

--- f n!n

1

(n )

( xj ) h0n

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 ( x ) sampled at positions x j = j h 0 may be approximated as (x j + 1 ) (x j ) d ( x j ) = ------------------------------------dx h0 To find the second derivative we keep terms to order h 0 so that f ( x j ) 2 f ( x j + 1 ) = f ( x j ) + f ( x j ) h 0 + -------------- h 0 2 and f ( x j ) 2 f ( x j 1 ) = f ( x j ) f ( x j ) h 0 + -------------- 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 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 ( x ) sampled at positions x j = j h 0 may be approximated as ( x j 1 ) 2 ( x j ) + ( x j + 1 ) d ( x j ) = ----------------------------------------------------------------2 dx h2 02 2 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 accurate2

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 + - + -- + 2d f ( x j )h 0 3 6 6 3 2 2 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 + --- + --- + 2d ------ -b- -c- ---- 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 that2

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 2 0 so that the second derivative of a wavefunction ( x ) sampled at positions x j = jh 0 may be approximated as ( x j 2 ) + 16 ( x j 1 ) 30 ( x j ) + 16 ( x j + 1 ) ( x j ) d ( x j ) = --------------------------------------------------------------------------------------------------------------------------------+ 2 -----2 2 dx 12h 02

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 Schrdinger equation h2 d H n ( x ) = ------- 2 + V ( x ) n ( x ) = E n 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 index2

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 Schrdinger equation is of length L = Nh 0 . At each sampling point the wave function has value j = ( 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 ( x j 1 ) 2 ( x j ) + ( x j + 1 ) d ( x j ) = ----------------------------------------------------------------2 2 dx h0 Substitution into the Schrdinger equation gives a matrix equation2

H ( x j ) = u j ( x j 1 ) + d ( x j ) u j + 1 ( x j + 1 ) = E ( x j ) where the Hamiltonian is a symmetric tri-diagonal matrix. The diagonal matrix elements are h d j = --------- + Vj 2 mh 0 and the adjacent off-diagonal matrix elements are2

h u j = -----------2 2mh 0 As usual, the wavefunction must be continuous and smooth. In addition, the periodic boundary2

conditions require ( x ) = ( x + L ) so that 0 ( x 0 ) = N( x N ) . In this situation the Hamiltonian becomes a N N matrix H and the Schrdinger equation is ( d1 E ) u2 u2 ( d2 E ) 0 ( H EI ) = 0 . . . u1 u3 0 . . . . 0 u3 ( d3 E ) u4 . . . . 0 0 u 4 ( d4 E ) . . . . . . . . . . . . . . . . . u1 . . 1 2 3

. . . . 4 = 0 . . . . . uN 2 0 N 2 . ( dN 1 E ) uN 1 N 1 uN ( dN E) N

. .

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 V0 ------ = -------------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 m = -----1 m2

E

so that TransE

=

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


MKS (SI) c = 2.99792458 10 m s8 1

h = 6.58211889 ( 26 ) 10

16

eV s J s C kg kg


34

19 31 27

m p = 1.67262158 ( 13 ) 10 27 k g k B = 1.3806503 ( 24 ) 10 23 J K 1 k B = 8.617342 ( 15 ) 10 5 eV K 1 0 = 8.8541878 10 12 F m 1 0 = 4 10 7 H m 1 c = 1 00 N A = 6.02214199 ( 79 ) 10 23 m o l 1 a B = 0.52917721 ( 19 ) 102 10


m



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 = 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 0 = 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 / 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 > > 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 = 1300 nm .VCSEL structure center wavelength 0 nAlAs nGaAs 25 identical dielectric layer pairs

Bragg mirror center wavelength 0

Center region thickness 0 / n GaAs 25 identical dielectric layer pairs 25 identical dielectric layer pairs n GaAs n AlAs z Incident electromagnetic field, wavelength n GaAs n AlAs Incident electromagnetic field, wavelength

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 0.8 Potential energy, V(x) (eV) 0.6 0.4 0.2 0.0 -0.2 -0.4 -0.6 0 0.6 eV 0.5 eV

3 nm 2.7 nm 2.25 nm

1.13 eV

5 Distance, x (nm)

10

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. PROBLEM 4 Write a computer program to solve the Schrdinger 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.

Potential energy, eV(x) (eV)

3.75 nm 1.0 0.8 0.6 0.4 0.2 0.0 0 20 40 60

6.25 nm

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

0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0 0 10 20 30 40 50 60 70 80

0.08 0.06 Wavefunction, 0.04 0.02 0.00 -0.02 -0.04 -0.06 -0.08 0 10

Potential energy, V(x) (eV)

8

9

20

30

40

50

60

70

80

Distance, x (nm)

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 k (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 = /L.

0.08 20 Potential energy, V(x) (eV) Wavefunction, 16 12 8 4 0 0 10 20 30 40 50 60 70 80 Distance, x (nm) 0.06 0.04 0.02 0.00 -0.02 -0.04 -0.06 -0.08 0 10 20 30 40 50 60 70 8

1

80

Distance, x (nm)

6



LAST NAME

FIRST NAME


MKS (SI) c = 2.99792458 10 m s8 1

h = 6.58211889 ( 26 ) 10 16 eV s h = 1.054571596 ( 82 ) 1034

J s C kg kg kg1

Electron charge Electron mass Neutron mass Proton mass Boltzmann constant

e = 1.602176462 ( 63 ) 10

19 31 27 27

m 0 = 9.10938188 ( 72 ) 10 m n = 1.67492716 ( 13 ) 10 m p = 1.67262158 ( 13 ) 10 k B = 1.3806503 ( 24 ) 10 k B = 8.617342 ( 15 ) 105

23

JK

eV K1

1

Permittivity of free space Permeability of free space Speed of light in free space Avagadros number Bohr radius 4 0 h 2 a B = ---------------m 0e2

0 = 8.8541878 10 0 = 4 10 c = 1 007

12

Fm

H m

1

N A = 6.02214199 ( 79 ) 10 23 m o l 1 a B = 0.52917721 ( 19 ) 1010 m



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

(b) The average particle position, x ( t ) . (c) The momentum probability density, ( p x , t ) (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 2h (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 0 ) .

1d 1d 2 2 p r = ih * - ( r 100 ) 4r d r = ih 100 - ( r 100 ) 4r dr = ihA 100 r dr r dr 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 Schrdinger equation is Hermitian and so it is its own Hermitian adjoint, i.e., H = H and |H = H| . The time dependence of the expe cation value of the operator A is d --- A = | A | + | A | + |A | -- dt t t t d i i --- A = - H |A | - |A H | + | A | dt h h t

d i i --- A = - |HA | - |A H | + | A | dt h h t d i --- A = - [H , A] + A dt h t


11

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

| x =

( x ) ( x ( x ) ) d x =

( x

( x )) ( x )d x =

( x ( x ) )

*

( x ) dx = x |

d is anti-Hermitian we integrate by parts and make use dx of the fact that the wavefunction ( x ) 0 as x (b) To show that the operator | d = dx * * d ( x ) ih d x ( x ) d x = ( x ) ( x ) x= x =

dx

d

*

(x ) (x )d x

|

* d d d = ( x ) ( x ) dx = | d x dx dx

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 0 2 + i H 0 2 ) = 0 0 H 1 1 1 ----= -------------- G = ------------- E + i ------------ H 2 t 00 2 0 2 0 Evaluating real and imaginary parts gives i E G -= i --0 2 t t H t E = 0 and E t (b) We start with H = 0 0 0 i * i -------------- G G = -------------- --0 E i --- H --0 E + i --- H -- -- - 2 2 2 2 00 00 0 00 i i 0 -------------- G * G = -------------- ---- E E + -----0 H H + i ----------0 E H i ---------- H E 2 2 4 4 0 0 00 00 00 i * i -------------- G G = -------------- i -------------- E H + i -------------- E H = E H = S 2 2 0 0 00 (c) If G is purely real then G = G and G G = 0 , so S = 0 .* *

12

0 0 0 * (d) G G = ---- E i ----- H ---- E + i -----0 H 2 2 2 2 0 2 0 2 0 0 1 * G G = ---- E + ----- H i ----------0 E H + i ----------0 H E = - ( E D + B H ) = U 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 0 0 and = 90 , and we have* 2

S 1 G G c * U = ------ = -------------- -------------------- = - G G sin = G c c c 00


13

14



LAST NAME

FIRST NAME


MKS (SI) c = 2.99792458 10 m s8 1

h = 6.58211889 ( 26 ) 10 16 eV s h = 1.054571596 ( 82 ) 1034

J s C kg kg kg1


e = 1.602176462 ( 63 ) 10

19 31 27 27

m 0 = 9.10938188 ( 72 ) 10 m n = 1.67492716 ( 13 ) 10 m p = 1.67262158 ( 13 ) 10 k B = 1.3806503 ( 24 ) 10 k B = 8.617342 ( 15 ) 105

23

JK

eV K1

1


0 = 8.8541878 10 0 = 4 10 c = 1 007

12

Fm

H m

1

N A = 6.02214199 ( 79 ) 10 23 m o l 1 a B = 0.52917721 ( 19 ) 1010 m



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 m b = ------- 2h 1 2

i px x + ------- m

1 2 i px m b = -------- x ------- 2h m show that the Hamiltonian can be expressed as h H = ------- ( b b + b b ) 2 (c) Derive the commutation relation [ b, b ] . (d) Using your result from (c) to show that the Hamiltonian is

1 H = h b b + - 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 Often an operator A is time-independent but the corresponding numerical value of the 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 ( 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 (a) Use the generalized uncertainty relation A B - [ A, B ] for time independent 2 h operators A and B to show that E t -- . 2 (b) Show that the spread in photon number n and phase for light of frequency is 1 n 2 and that for a Poisson distribution of such photons 1 --------------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 ( x ) = m 0 x 2 and is subject to an oscillating electric field

E x cos ( t ) . (a) Write down the Hamiltonian of the system. (b) Find d x . dt (c) Find d d d p and show that p = V ( x ) . Under what conditions is the dt dt dx d d x = V ( x ) the same Newtons second law in 2 dx dt2 2

quantum mechanical result m

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 x ( t ) . What happens to the maximum value of x as a function of time when 0 = and when is close in value to 0 ?


3

SOLUTIONS Solution 1 (a) The Hamiltonian for a particle of mass m in a one-dimensional harmonic oscillator potential is2 p x m 2 H = ------- + ---------- x 2m 2 2

where the particle moves in the potential V ( x ) = x =

2

and oscillates at frequency

m , where is a force constant. The operator p x = i h ----- . x (b) The Hamiltonian can be factored m 2 p x 2 H = ---------- ------------ + x m 22 2 so it makes sense to define new operators i px m 1 2 b = ------- x + ------- 2h m m 1 2 i px b = -------- x ------- 2h m which, can be written in terms of operators x and p x to give h - 1 2 x = ----------- ( b + b ) 2m hm p x = i ----------- 2 -1 2 2

(b b)

Substituting this into the Hamiltonian gives2 2 p x m 2 2 1 hm 2 m h - 2 H = ------- + ---------- x = ------- ----------- ( b b ) + ---------- ----------- ( b + b ) 2m 2 2m 2 2 2 m

h H = ------- ( b b b b + b b + b b ) + 4 h H = ------- ( b b + b b ) 2

h ( b b + b b + b b + b b ) ------- 4

(c) To derive the commutation relation [ b, b ] = b b b b we can write out the differential form of the operators and have them act on a dummy wave function. This gives m h h ( b b ) = ------- x + -------- ----- x -------- ----- 2h m x m x m 2 h h h h ( b b ) = ------- x + -------- + -------- x ----- -------- x ----- ------------ ------- 2h m m x m x m2 2 x 2 and2 2

m h h ( b b ) = ------- x -------- ---- x + -------- ----- 2h m x m x

4

m 2 h h h h ( b b ) = ------- x -------- --------x ----- + -------- x ----- ------------ ------- 2 h- m m x m x m 2 2 x 2 so that the commutation relation simplifies to just2 2

m 2h 2h 2 h ( b b b b ) = ------- -------- -------- x ----- + -------- x ---- = 2h m m x m x or, in even more compact form, [ b, b ] = bb b b = 1 If we do not want to use differential operators and a dummy wavefunction then we could write ih p x ihp x ihp x ihp m m - - - [ b , b ] = b b b b = ------- x + ---------x x --------- ------- x --------- x + --------- - 2h 2h m m m m2 2 2 h px ihp x x ihx p h px ihp x x ihx p m m - 2 [ b , b ] = ------- x 2 + ------------- ------------x + -----------2 ------- x ------------ + -------------x + -----------2 - 2 2 2h 2h m m m m m m 2

im hp x x hx p x [ b , b ] = --------- ----------- ---------- = i ( p x x x p x ) = i [ p x,x ] = i ( i h ) = 1 h - m m since [p x ,x] = i h . (d) The Hamiltonian is h h h H = ------- ( b b + b b ) = ------- ( b b + b b + b b b b ) = ------- ( 1 + 2 b b ) = h b b + 2 2 2 where we made use of the commutation relation [ b, b ] = b b b b = 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 x = ----------- ( b + b ) 2m h m 1 2 p x = i ----------- ( b b ) 2 - and 12 | b n = ( n + 1 ) |n + 1 | b n = n1 2

1 -2

|n 1

m| n = mn so that h 1 2 h 1 2 x = 1| x |1 = ----------- 1|b + b |1 = ----------- ( 1|0 + 2m 2m and hm 1 2 h m 1 2 p x = 1| p x |1 = i ----------- 1| ( b b ) |1 = i ----------- ( 2 1|2 1| 0 ) = 0 2 2 2 1 |2 ) = 0


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 h h 3h 2 x 2 = ----------- 1 | ( b + b ) | 1 = ----------- 1|b b + b b + b b + b b |1 = ---------- 2m 2 m 2 m h 2 and one can see that for the general state |n one has x = ----------- ( 1 + 2 n ) . Now 2m turning our attention to p 2 x we have h m hm 3h m 2 p 2x = ----------- 1 | ( b b ) | 1 = ----------- 1| b b b b + b b + b b | 1 = ------------- 2 - 2 - 22 h m and one can see that for the general state |n one has p 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 associated with operator A evolves smoothly in time such that d A A = ------dt t which may be written as A t = ----------------d A dt In this problem the operator A is time-independent so that d i i A = - [ H ,A] + A = - [ H ,A] h h dt t since A = 0 t

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

If operator B is the Hamiltonian H, then B corresponds to E so that d 2 A E [ A, H ] = h A = h A ------dt t Since the A terms on the far left and far right of the equation cancel, we have h Et -2 It might be tempting to make a comparison between this result and the uncertainty rela tion for two non-commuting operators such as momentum p x and position x 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 Et -- directly from the generalized uncertainty relation because t is not an opera2 tor. (b) The energy of the n photons oscillating at frequency is E = h n + 1 -- so that 2

be proportional to cos ( t ) so that t = or t = where is the spread in values of phase . From (a) we have h Et = h n = h n -2 and hence 1 n 2 For Poisson statistics n = n so that 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 and a phase coherence that increases with increasing n as Solution 4 2 m 0 x p (a) H = T + V = ------- + --------------- + eE x x cos ( t ) . 2 2m (b) Because the position operator does not explicitly depend on time, have d i i i x = - [ H ,x ] + x = - [ H ,x] = ---- [x ,H] h h h dt t Since [x, x] = 0 and [ x, x 2] = 0 , we are left with i i d p p p p x = ---- [x ,------ ] = ---- p [x, ------ ] + [x ,------ ] p = ------- - dt h 2m h 2m 2m m2 2 2

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

n .

= 0 , we x t


7

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

where we used p = i h . The spatial derivative of the expectation value of the x potential is 2m 0 x d d p V ( x ) = ---------------------- + e E x cos ( t ) = 2 dt dx which is Ehrenfests theorem.2

Hence, in quantum mechanics, m

d d x = V ( x ) . This would be the same as dx dt 22

2

d dx Newtons second law for force on a particle F = m 2 = V ( x ) if position dx dt x = x and if d d V ( x ) . The conditions under which the latter is V (x ) = d x dx

true is when the potential is slowly varying so that higher order terms in the expansion 2 of the force F ( x ) = F ( x ) + ( x x )F ( x ) + ( x x ) F ( x ) + may be neglected so that F ( x ) F ( x ) . This requires that the uncertainty x = ( x x ) is small. d i H H H = - [H ,H] + = = eE x x sin ( t ) h t t dt where we used the fact that [H ,H] = 0 . (d)2 m 0 eE x d d p x = -------- = ---------- x -------- cos ( t ) 2 dt m m m dt this can be written in the familiar form of an undamped forced oscillator 2

(e)

d 2 x + m 0 x = Fx cos ( t ) 2 dt where F x = e E x . This has an oscillatory solution of the form m F x cos ( t ) x ( t ) = A cos ( 0 t ) + ------------------------- m ( 2 2) 0 which contains two harmonic oscillators, one at the natural frequency 0 and the other at the frequency of the input force. The constants A and are found from the initial contitions. If 0 = the amplitude of the undamped oscillator grows to infinity with the particular solution increasing linearly in time as

2

8

Fx t sin ( 0 t ) x p ( t ) = --------------------------2m 0 When is close in value to 0 then Fx x ( t ) = ---------------------------( cos ( t ) cos ( 0 t ) ) 2 2 m ( 0 ) which, since 2 sin ( x ) sin ( y ) = cos ( x y ) cos ( x + y ) , may be written as ( 0 + ) ( 0 ) 2 Fx x ( t ) = --------------------------- sin ---------------------t sin --------------------- t 2 2 2 2 m ( 0 ) The sum frequency ( 0 + ) is modulated by the difference frequency ( 0 ) so that the amplitude of x beats at the difference frequency ( 0 ) .


9



LAST NAME

FIRST NAME


MKS (SI) c = 2.99792458 10 m s8 1

h = 6.58211889 ( 26 ) 10 16 eV s h = 1.054571596 ( 82 ) 1034

J s C kg kg kg1


e = 1.602176462 ( 63 ) 10

19 31 27 27

m 0 = 9.10938188 ( 72 ) 10 m n = 1.67492716 ( 13 ) 10 m p = 1.67262158 ( 13 ) 10 k B = 1.3806503 ( 24 ) 10 k B = 8.617342 ( 15 ) 105

23

JK

eV K1

1


0 = 8.8541878 10 0 = 4 10 c = 1 007

12

Fm

H m

1

N A = 6.02214199 ( 79 ) 10 23 m o l 1 a B = 0.52917721 ( 19 ) 1010 m



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 ) k B T ( 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 = 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 2B 1 e +11 1 1 (a) 1 ------------------------------ = ---------------------------------------- = --------------------------------- = -----------------------------( E ) kB T ( E ) kB T ( E ) kB T ( E ) 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

( E ) k T

1 f ( E ) = -----------------------------( E ) kB T e +1 so that f (E ) = 1 f( 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 E VB = E(k) 0 /L D1 (E)

Wave vector, k

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 =

E=0

D ( E )f ( E ) d E =

E=0

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

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

n =

E=0

D 1 (E )d E =

E=0

D 1 ( E ) dE

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 writtenEF

n =

E=0

D 1 ( E ) d E = D 1 ( E ) ( f ( E ) + f ( E0 E ) ) dE

which requires f( E ) + f( E 0 E ) = 1 f( E ) + ( 1 f( E 0 + E ) ) = 1 f( E ) f( E 0 + E ) = 0 E + + E0 + E = 0 E0 = ---2 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


MKS (SI) c = 2.99792458 10 m s8 1

h = 6.58211889 ( 26 ) 10 16 eV s h = 1.054571596 ( 82 ) 1034

J s C kg kg kg1


e = 1.602176462 ( 63 ) 10

19 31 27 27

m 0 = 9.10938188 ( 72 ) 10 m n = 1.67492716 ( 13 ) 10 m p = 1.67262158 ( 13 ) 10 k B = 1.3806503 ( 24 ) 10 k B = 8.617342 ( 15 ) 105

23

JK

eV K1

1


0 = 8.8541878 10 0 = 4 10 c = 1 007

12

Fm

H m

1

N A = 6.02214199 ( 79 ) 10 23 m o l 1 a B = 0.52917721 ( 19 ) 1010 m



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 ) isi mn t 1 a m ( t ) = --- W mn e -d t ih t = 0 t = t

where the matrix element W mn = m | V |n and h 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 oscillator. At time t = 0 a perturbation V ( x, t ) = V 0 x e is applied where V0 and 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 .3 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 et 2 2

x at

time t = 0, where is a constant, x 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 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 and consider the three

values of 21 = 1 2 , 21 = 1 , and 21 = 2 , where h 21 = E 2 E 1 . Explain your results.

2

PROBLEM 42 4 e Use the Einstein spontaneous emission coefficient A = ----------------------- j |r | k to estimate 3 3h c 4 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 = h 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 ( t ) , where V ( x ) is the spatial part of the potential and is the frequency of oscillation. Start by writing down the Schrdinger equation for the complete system including the perturbation and then go on to show that the scattering rate in the static limit ( 0 ) is 1 2 2 ---- = ------ W mn D ( E ) ( Em E n ) , where the matrix element W mn = m | W ( x , t ) | n n h

couples state |n to state |m via the static potential V ( x ) , the density of final continuum states is D ( E ) , and ( 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 = ( q ) . Given an impurity potential at position r due to a e charge e at position Ri is v q ( r R i ) = --------------------- , derive an expression for v ( q ) . q 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 = ( 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?2


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 Schrdinger equation H( 0 ) | n = E n | n and the time-dependent Schrdinger equation i nt in t | ne = 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 | ( t ) evolves in time according to ih | ( t ) = ( H ( 0 ) + W ( t ) ) | ( t ) t

where | ( t ) =

a n ( t ) |n en

i n t

and a n ( t ) are time dependent coefficients.

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

ih

i t i t i t ih a n ( t ) |n e n + a n ( t ) | ne n = ( H ( 0 ) + W ( t ) ) a n ( t ) | ne n t t n n

and remove the term i t ih a n ( t ) | ne n = t n to leave

a n ( t )H ( 0 ) |n en

in t

ih | n en

i n t

a (t) = t n

a n ( t )W ( t ) | nen

i nt

Multiplying both sides by m | and using the orthonormal relationship m|n = mn givesi t d a m ( t ) = a n ( t ) m| W ( t ) | n e m n dt n However, since

ih

m |W ( t ) | n = * ( x ) e m we may write ih d am( t ) = dt

i m t

W n ( x )e

i nt

d x = W mn e

im n t

where h mn = E m En and W mn is defined as the matrix element * ( x ) W n ( x ) dx , m

a n( t ) W mn en

i mn t

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

4

Integration from the time when the perturbation is applied at t = 0 givesi mn t 1 a m ( t ) = --- W mn e -d t ih t = 0 t = t

(b) The transition probability from state | n to state |m is P nm = a m ( t ) . Using our solution in part (a) we have2

1 -P nm = ---2 h

t = t

t = 0

m |V ( x, t ) |n e

i mn t

2

d t

V0 = ----h2

2 t = t

t = 0

m |x 3 | n e

( im n 1 )t

2

d t

In the problem we have initial state |n = 0 and we consider the long time limit, t , this allows us to write V0 2 3 ( im 1 )t d t P nm = ----- m | x |0 e 2 h t = 0 where, for the harmonic oscillator, the non-zero positive integer m multiplied by the frequency is related to the difference in energy eigenvalue by m = ( E m E 0 ) h . The time integral ist = 2 t = 2

t = 0

e ( im 1 )t dt

2

1 = -----------------------------2 2 2 m +13 * 12

and the matrix elements m |x | 0 are found using x = ( h 2m )

( b + b ) , where

m * is the particle mass. In our case only transitions |0 |1 and |0 |3 are allowed. m|x 3 | 0 = ( h 2m * ) where we used m| n = mn . Hence, V2 3 0 P 01 = ----- 1| x |0 2 h and V0 P 03 = ----- 3| x 3 |0 2 h2 t = 2 32

( b + b b + b ) = ( h 2m * )3 2

32

( 6 m = 3 + 3 m = 1 ) b |0 = 0 , and

12 b | n = ( n + 1 ) |n + 1 ,

12 b | n = n | n 1 ,

t = 0 t =

e

( im 1 )t

2

d t

9 V2 h - 3 0 = ------------- --------------------------------- 2m * ( h ) 2 + h 2 2 6V 0 h 3 = ------------- ------------------------------------- 2m * ( 3h ) 2 + h 2 2 2

2

t = 0

e ( im 1 )t d t

2

So, in the long time limit, t , the maximum probability of the transition taking place occurs as . Problem 2 The eigestates of a particle in a one-dimensional rectangular potential well for which V ( x ) = 0 in the range 0 < x < L and V ( x ) = elsewhere are | n = 2 -- sin k n x L2 2

where k n = n L , n = 1, 2 , 3 , , and eigenenergies are E n = h k n 2 m = h n . Applied quantum mechanics 5

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

t

2

2

is turned on at time t = 0, where 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 P12 e E0 2 = ---------------- 2 |x | 1 2 hL 2 2 t t = i 21 t ----2 t = 02

2

e

dt

where h 21 = E 2 E 1 and 2 16 L 2| x |1 = -- x sin ( k 2 x ) sin ( k 1 x ) dx = -----------2 L0 9 so P 12 = e E 02 2

16 L 2 ----------- 9h 2

t t = i 21 t ----2 t = 0

2

2

e

dt

21 2 t 2 t2 21 Let y = - i ---------- so that i 21 t ---- = ------- y and d t = dy . We now can write 2 2 4 2 2 2 2 16 L 2 21 4 2 y P 12 = e E 0 ----------- e e dy 9h 2 t = 0

t =

2

e E0 16L 2 2 2 2 2 21 = ------------------- ----------- e 4 9h 2 2

2

wheret =

21 = 3 h 2mL22

2

and

we

made

use

of

the

standard

integral

t = 0

e

ax

1 - 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 for which P 12 is a maximum. 0 = dP 12, t = 2 2 3 21 d 2 12 . We now find E 0 for which P 12 = 1 .2 2

so that =

2 2 e 2 E0 16L 2 2 1 e 2 E 0 16L 2 2mL- 1 P 12 = ------------------- ----------- ------- e = ------------------- ----------- 2 ------------ e = 1 3 2h 9 h 2 2 9h 2 4 4 21

E0

2

4 9h 3 h 27h - 1 - 1 = ------------- ------------ ------------ e = 2 ------------------ e 2 2 2 16 L 32 meL 3 2mL 2 e2 2 2 2 3 2

E0 =

27h 0.5 2 ------------------- e = 32meL 3 2 3 7 1

27 ( 1.05 10 ) 2 ------------------------------------------------------------------------------------------------------- 1.65 32 0.07 9.1 10 31 1.6 10 19 10 24 3

34 2

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.

6

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 x -n L2 2

where k n = n L , n = 1, 2 , 3 , , and eigenenergies are E n = h k n 2 m = h n . The particle is in the ground state prior to a perturbation V ( x , t ) = V0 x 2 e t 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 P 12 V0 2 = ------- 2| x |1 2 hL 2 2 t = t

t = 0

e

i2 1t t -

2

dt

where h 21 = E 2 E 1 and 2 16 L2 2| x 2 |1 = -- x 2 sin ( k 2 x ) sin ( k 1 x ) dx = -------------L0 9 2 The time integral ist = t

t = 0

e

( 1 i 21 )t

2

dt

21 1 = e -------------------------------1 i 21

( 1 i

)t

2

1 + e 2 t 2e t cos ( 21 t ) = --------------------------------------------------------------2 2 21 + 1

so 16V 0 L 2 2 1 + e 2t 2e t cos ( 21 t ) P 12 ( t ) = ---------------- --------------------------------------------------------------- - 2 2 9h 2 - 21 + 1 Now we notice that 21 is related to L via 21 = 3 2 h 2mL 2 so L 2 = 3 2 h 2 m 21 and 16 V 0 2 1 + e 2 e cos ( 21 t ) P12 ( t ) = --------------- --------------------------------------------------------------- 2 2 6m 21 21 + 1 In the limit t the probability becomes V 0 16 2 1 P 12 ( t ) = --------------- ------------------------- 6m 21 2 + 1 2 21 In this long time limit and normalizing time to , we see that V 0 16 t 1 P 12 - , 21 = ----- = ----------- 38.5 6m 2 V 0 16 t P 12 - , 21 = 1 = ----------- 0.5 6m and V0 16 2 t 4 P 12 - , 21 = 2 = ----------- 6.26 10 6m so, if we want to plot the results on the same graph, it is best to normalize to these long time values.2 2 2t t


7

Problem 4 Differences in energy between eigenfunctions nlm of hydrogen are given by m0 e4 1 1 1 1 E = En 2 En 1 = ------ ----------------------- --- ---- = R y --- ---- 2 n-2 n - 2 ( 4 0 ) 2 h 2 n2 n 2 1 2 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 = h = -3.401 + 13.606 = 10.205 eV corresponding to a wavelength = c2 h E = c 2 which is = 122 nm. 4 0 h 2 An estimate for the matrix element is j |r | k a B , where a B = ---------------- is the Bohr m 0e2 radius and we obtain ( 2 ) e 2 ( 2 ) E e 2 9 1 1 A = ------------- --------------- a B = ----------------------- --------------- a B = 1.12 10 s = ----3 3 3 3 0 h 3 0 h sp h c Hence, our estimate of the spontaneous emission lifetime for hydrogen is sp = 0.89 ns .3 2 3 3 2

For the helium ion we note that Z = 2. The Bohr radius scales as a B 1 Z . For a hydrogenic atom with ion charge number Z the Bohr radius for the electron is a B = 4 0 h m 0 Ze 2 . The Rydberg energy which determines E scales as Z2 so2

E = Z 2 R y ( 1 n 2 1 n 2 ) . 1 2* 6 2

Hence,10

the s1

linewidth

of

the* sp

He +

ions

is

A = A Z Z = 16 A = 1.79 10

and the lifetime is

= 0.056 n s . The

8

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 Schrdinger equation given by H ( 0 ) | n = E n | n The time-independent eigenvalues are E n = h n and the orthonormal eigenfunctions are |n . The eigenfunction | n evolves in time according to | n e = n ( x ) e and satisfies ihi n t i n t

in t i n 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 )

and state | ( x, t ) which

Solutions to Levi Applied Quantum Mechanics 2nd Ed

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