PHYSICS'EXAMINATION'PROBLEMS' SOLUTIONS'AND'HINTS… · physics examination problems solutions and hints for student ... physics examination problems solutions and hints for student

PHYSICS'EXAMINATION'PROBLEMS'SOLUTIONS'AND'HINTS'FOR'STUDENT'SELF5STUDY'

Module'Code' PHY1023'Name'of'Module' Waves'and'Optics'Date'of'Examination' May'2013'

'

'

1.# Bookwork#or#lecture#notes#for#definition#of#variables.#

# For#resonant#angular#frequencies:##ω1#=#1.41#rads#/s#and#ω2#=#0.71#rads#/s.###

Similarly,##f1#=#0.23#Hz#and#f2#=#0.11#Hz.#

Use#boundary#condition#arguments#(x#=#0#at#t#=#0)#to#explain#φ1#and#φ2.#

# A1#=#0.14#m#and#A2#=#0.28#m.#

# Net#forces#:##F1#=#O0.03#N#and#F2#=#O0.03#N.#

# To#prepare#sketching#the#graph,#perform#a#calculation#on#the#total#energy#in#system#1#and#2#respectively:#

# ie.###Total#E1#=#0.002#J#and#total#E2#=#0.004#J.##

#

#

#

#

#

#

#

#

This#figure#shows#the#lines#for#potential#energy#only#(kinetic#energy#lines#not#overlayed#here#to#improve#clarity#of#

figure#for#teaching#purposes).##The#KE#lines#would#mirror#the#form#of#the#PE#lines,#except#they#are#inverted#by#a#

horizontal#reflection#in#the#manner#described#in#lecture#notes.###

#

#

#

Energy#

E2#

E1#

x#A1# A2#

2.# Use#conventional#expressions#for#each#impedance#(from#lecture#notes).####

j"represents#a#phase#term,#indicating#the#phase#relation#between#V#and#I#in#the#impedance#relations.#

At#f#=#100#Hz,##ZC=#159#Ω,##ZL=#6.3#Ω#and#ZR=#100#mΩ,#so#the#capacitor#has#highest#impedance.###

At#f#=#100#kHz,##ZC=#0.16#Ω,##ZL=#6300#Ω#and#ZR=#100#mΩ,#so#the#inductor#has#highest#impedance.###

Sketch#the#standard#graph#for#resonance#of#current#vs#frequency#(see#lecture#notes)#with#appropriate#

explanation#of#filtering#effect#of#L#and#C#components#at#different#frequencies#via#impedance#arguments.#

Resonant#ang.#frequency#=#3162.3#rads#/s,######or###f#=#503.3#Hz.#

Using#conventional#lecture#note#material#(for#parallel#impedances),##

Ztotal =R2 + L

C+ jR ωL − 1

ωC"

#$

%

&'

2R+ j ωL − 1ωC

"

#$

%

&'

#

At#resonance,#Ztotal#=#50#kΩ.##(use#the#simplification#to#the#formula#above#that#happens#at#the#resonance#

condition).#

#

3.# v#represents#the#phase#velocity#of#the#propagating#wave.### v = 1εµ

#

By#differentiating#the#equation#appropriate#(wrt#time#and#wrt#z)#and#then#substituting,#show#that#the#expression#for#phase#velocity#(shown#in#the#previous#line)#arises.#

Refractive#index#=#3.16##(by#substituting#for#ε#and#µ).#

Transmission#(amplitude)#coefficient#=#2z1z1 + z2

#

Substitute#expression#for#T#into#this#equation#and#show#that#it#equals#1.###This#applies#because#it#is#an#expression#related#to#the#conservation#of#energy#(R

2#etc#relates#to#power).#

For#the#numbers#stated#in#the#question,#R#=#0.52.#

#

#

#

#

#

4.# (i)#Use#standard#textbook#definition#and#description#of#obstetric#ultrasonography#(or#similar).#

Required#velocity#of#bat#=#85#m/s.#

#

(ii)##λ#=#0.84#m.####f#=#0.80#Hz.#####vp#=#0.67#m/s.#

Largest#transverse#speed#=#0.1#m/s.#

Tension#=#0.11#N.#

Ratio#AωAω 2 =

1ω= 0.2 .#

#

5.## (i)##Use#standard#graphs#for#ω(k)#and#vg(k)#from#lecture#notes#etc.###Describe#cutOoff#values#and#axes’#

intercepts#etc.#

# CutOoff#freq.#=#45.0#Hz#=#282.8#rads/s.#

# (ii)##Use#standard#lectureOnote#or#bookwork#definitions#of#dispersion,#phase#and#group#velocities.##

Use#standard#math#manipulation#to#produce#the#appropriate#expression.###Follow#this#by#plotting#the#

conventional#“beating”#graph#(featuring#one#high#frequency#and#one#low#frequency#term).###

#

6.# (i)##Use#standard#formula#for#calculation#of#image#distance#u#=#33.3#cm.#

# Magnification#=#O0.33##(i.e.#inverted#and#smaller).##Use#this#to#calculate#actual#size#=#1#cm.#

# Sketch#a#suitable#(typical#from#lecture#notes)#ray#diagram#of#this.#

# (ii)##Define#Huygens’#principle#(lecture#notes#or#textbook).#

# Using#standard#equations#for#position#of#bright#order#and#or#dark#(missing)#order,##then##

slit#spacing#a#=#100#µm.#

slit#width###b#=#25#µm.#

#

#

P.#Vukusic###(2013)#

PHYSICS EXAMINATION PROBLEMS SOLUTIONS AND HINTS FOR STUDENT SELF-STUDY

Module Code PHY1024

Name of module Properties of Matter

Date of examination May 2013

1 (a) 8 × 105 Pa

(b) 71.36 m

(c) The term Nb accounts for the intrinsic volume of the molecules. The parameter b [m3] is the volume of one molecule.

(d) First rearrange the equation in the form ( )

, and then use the hint

provided to expand the right-hand side of this expression, noting that The

result is ( )

( )

2 The time needed to transfer the stated amount of heat is 7.7 s. The increase in the temperature of the Earth will be 57.8 K. In order to obtain an expression for the coefficient of area expansion Ds, make use of the facts that the area A = L2, and the material will expand along the two available directions. This leads to the approximate expression ( ), and therefore Ds = 2D��

3 For three translational degrees of freedom √ .

(a) √

(b) ⟨ ⟩ √

(c) √

(d) In 3D the r.m.s. speed would be 489.9 m s�1.

4 (a)

(b) . Hence at constant temperature ( )

.

Therefore .

(c) 298.5 m s�1 (d) 8 mm (e) First rod: 130 J Second rod 520 J

5 First use the force balance in the vertical direction to obtain the volume of the object, V =

1.335 × 10�4 m3. Then use this value along with the densities provided to calculate the mass of copper in the object, 0.213 kg. (Note that the final answer is very sensitive to the accuracy with which the volume is calculated.)

(a) An order of magnitude estimation would be d ~10�6 m.

(b) If , the volume of the thin membrane will be (approximately) . This can be used to obtain the radius R = 0.106 m. (c) 0.0264 N m�1

6 (a)

( )[ ( )] (b) At the equilibrium separation F = 0, fulfilled for r = r0 (which is independent of D and H). (c) The work that needs to be done against the interatomic forces is

∫ ( ) ∫ ∫ ( )

( )

(d) ( ) ( )

PHYSICS EXAMINATION PROBLEMSSOLUTIONS AND HINTS FOR STUDENT SELF-STUDY

Module Code PHY1026Name of module Mathematics for PhysicistsDate of examination May 2013

1.! (i)! 1, exp 2iπ5

⎛⎝⎜

⎞⎠⎟, exp

4iπ5

⎛⎝⎜

⎞⎠⎟, exp

6iπ5

⎛⎝⎜

⎞⎠⎟, exp

8iπ5

⎛⎝⎜

⎞⎠⎟

! (ii)! (a) 2i (b) 2exp iπ 2( )

! (iii)! ln −i( ) = 3iπ2

or − iπ2

2.! (i)! Using the reciprocal and reciprocity theorems (or otherwise): ∂x∂y⎛

⎝⎜

⎞

⎠⎟z

= x2

1− xy

! (ii)! Using parameterisation, I = 12

3.! (i)! ∇ϕ = x+ y zcos z+ z ycos z− yzsin z( )

! ! Directional gradient = π2 2

! ! Tangent to the surface of constant ϕ means perpendicular to ∇ϕ , so dot product zero.

However there is a mistake in the exam paper, and the vector given is not tangent to the

surface of constant ϕ .

! ! If you wish to practise this type of question, a vector that is tangent to the surface of

constant ϕ at the point given is x + 2πz .

! (ii)! Use divergence theorem to convert to volume integral:

ρ i dSS∫∫ = 2πR2L (twice

volume of cylinder – don’t need to do integral).

4.! ! Upside down parabola with max value of L2 at x = 0. f = 0 at x =±L . f is even.

! ! f x( ) ≈ 2L2

3+ 4L

2

π 2 cosπ xL

⎛⎝⎜

⎞⎠⎟−

L2

π 2 cos2π xL

⎛⎝⎜

⎞⎠⎟+4L2

9π 2 cos3π xL

⎛⎝⎜

⎞⎠⎟

5.! (i)! The differential equation is exact. y = Cln x( )1 2

. Variables separable and integrating

factor methods also work.

! (ii)! Auxiliary equation has equal roots, so system is critically damped:

! ! x = A+ Bt( )exp − t2

⎛⎝⎜

⎞⎠⎟


Module Code PHY2023

Name of module Thermal Physics


1. (i) Isolated system: .0 revesibleS' (ii) Carnot engine cycle in PV variables is clockwise (see lecture notes);

for a heat engine HQW K ; HC TT� 1CarnotK . In TS variables, the diagram is a rectangle with

constS during an adiabatic and constT during an isothermal process. (iii) � � CCHC TTTQW � max . Therefore, the required minimum power is MW1min P .

(iv) � � � � J57.415ln 12 | � RTTnRCS V' ; � � � � J.1285713123

111 |� � eRTeVPU'

2. (i) Bookwork (see notes, lecture 1). (ii) Bookwork. (iii) Use VPSTU ddd � and SVU

VSU

www

ww

w 22

.

(iv) Adiabatic process: � �22112211

23dd0 VPVP

nRVP

nRVPnCTnCUWQ VV � ¸

¹·

¨©§ �� GG .

(v) � �12ln TTCSwater ' ; � � 221 TTTCSres � ' ; 0JK25.74 -1 ! � reswatertotal SSS ''' as

expected from the 2nd Law of thermodynamics.

3. See course notes. Isotropy implies � � � �xx upup � . The number of states with the speed from u to uu d�

is uu d2S . � � � � � � .dexpd/2exp2dd1d22 HHEEHHDDHH � � � � mm

uupm

uuum

Boltzmann factor: � �HE�exp . � �Tk

TkpB

B0

12

21d � ³f

EE

HHHH . Density of states in

2D does not depend on energy. � � � � 37.01expdd)(1

BB

|� ! ³³f

�ftepTkp t

TkHHH .

4. See course notes. ¦ ¸¹

·¨©

§�¸

¹

·¨©

§�

l

lii TkTkp

BB

expexpHH

. ¦ ¸¹

·¨©

§�

l

l

TkZ

B

expH

ensures ¦ i

ip .0

� � � �EE

EHE

HHH

dlnd1

ddZ1exp

dd1exp1

B

ZZZZTkZ

pUr

rr

rr

rrr � � �� ¸

¹

·¨©

§� ¦¦¦ .

78.2531expexpexp 21

B

22

B

11

B

00 |�� ¸

¹

·¨©

§��¸

¹

·¨©

§��¸

¹

·¨©

§� �� ee

Tkg

Tkg

TkgZ HHH

.

� � HHH

H 88.052310exp1 21

B

|u�u� ¸¹

·¨©

§� ��¦ Zee

Tkg

ZU

r

rrr .

� � � � 64.0535311 20 |�� eeeZePP occupiedunoccupied .

5. (i) PVUH � . QSTHPVSTH G �� ddddd (heat flow) for constP . (ii) TSUF � ; extensive; WFTSVPF G� �� dddd for constT .

ZTkF lnB� connects microscopic and macroscopic descriptions of the system, since TV

FP ¸¹·

¨©§ww

� .

(iii) � �� K5277ln1

expexp

0101BB00

B11

0

1 |��

TggEETkTkEgTkEg

pp

.

6. (i) : lnBkS , see course notes. BABABABABA SSkkS � � �� ::::: lnln BB .

252!5!5!10 mp: (same number of heads and tales). .JK1063.7252lnln -123

BB�u| kkS mpmp :

k systems (k=3), n energy levels (n=5): � � 21

)!13(!5)!135(

)!1(!!1

, ��

��

knkn

kn: , 14.07/121/3 | p .

All the energy can be stored in any of three systems; therefore, .JK1052.13ln -123B

�u| kS

.JK1047.26ln6123 -123

B3,2,0�u| � uu kS:

(ii) For 0FB oETk , all the states below FE are filled � � 12.0 F o � EEp .

1 PHY2024

PHY2024

UNIVERSITY OF EXETER

PHYSICS

MAY / JUNE 2013

CONDENSED MATTER I

SOLUTIONS AND HINTS

2 PHY2024

1. (i) The primitive translation vectors of the fcc lattice are a1 = 0.5a(1 1 0),

a2 = 0.5a(0 1 1), a3 = 0.5a(1 0 1), where a is the side of the cubic unit cell.

Calculate the volume of the primitive unit cell. [2]

Calculate each of the three primitive vectors of the reciprocal lattice. [2, 2, 2]

State the definition of the first Brillouin zone. How does its volume relate to that

of the conventional primitive cell of the reciprocal lattice? Calculate the volume of

the first Brillouin zone. [1, 1, 1]

State three different mathematical forms of the condition for wave diffraction in a

crystal, explaining all notations. [2, 2, 2]

HINT: SEE THE LECTURE NOTES

(ii) The geometrical structure factor of a crystal is

� �¦¦

��

�� 1

2

1321

jvzvyvxi

jji

jjjjj efefS SrG

G ,

where j runs over all atoms of the basis and fj is the atomic form factor of the jth

atom of the basis. The expression on the right-hand side is written for a reflection

with indices (v1 v2 v3), corresponding to a reciprocal lattice vector

G = v1b1 + v2b2 + v3b3, where b1, b2 and b3 are the primitive vectors of the reciprocal

lattice and v1 , v2 and v3 are integer numbers.

Given that the basis of the fcc structure has four identical atoms at (0 0 0),

(0 21 2

1 ), ( 21 0 2

1 ) and ( 21 2

1 0), derive the general expression for the structure

factor for a reflection with indices (v1 v2 v3). [4]

HINT: Noting that the atomic form factors of the identical atoms are equal to each

other and therefore e.g. to some value f, obtain the SOLUTION

� � � � � �� 2131321 vvivvivvi eeefS �� SSSG .

3 PHY2024

Hence, show that for a reflection with indices (v1 v2 v3) the structure factor is

(a) equal zero if one and only one of the indices is odd; [1]

(b) equal zero if one and only one of the indices is even; [1]

(c) finite if all of the indices are odd; [1]

(d) finite if all of the indices are even. [1]

HINT: Noting that the complex exponents can only be equal to either 1 or -1 depending

on the integer sum in the parenthesis, analyse each of the given combinations to

show the required properties.

4 PHY2024

2. A diatomic chain consists of atoms of alternating masses M and m, with a nearest

neighbour separation of a/2. The chain supports vibrational waves with dispersion

relation, � �kZ , defined by

� � ¸¸¹

·¨¨©

§¸¹·

¨©§

��r

�

2sin411 2

22 ka

MmmM

mMMmCZ ,

where C is the force constant, ω is the frequency and k is the wave number.

Derive, from the acoustical dispersion branch of the diatomic chain, the dispersion

relation of acoustical waves in a monatomic chain consisting of atoms of mass

µ = 0.5(m + M) with the nearest neighbour separation of a/2. [5]

[Hint: 1 – cos(2x) = 2sin2(x).]

HINT: By a direct substitution of m=M=µ, obtain after simple trigonometrical

manipulations

¸¹·

¨©§

4sin4 22 kaC

PZ ,

And therefore the SOLUTION

¸¹·

¨©§

4sin2 kaC

PZ .

Use the two dispersion relations (one given for the diatomic chain and one derived

for the monatomic chain) to calculate the group velocity of the acoustical modes for

the diatomic and monatomic chains respectively at small values of k>0. [3, 5]

HINT: Using the small angle approximation, obtain for the monatomic chain

kCaP

Z2

| ,


5 PHY2024

PCavgr 2

| .

HINT: Using the small angle approximation and then also 2

11 xx �|� for radicals

(we need to retain only linear terms in the expression for the frequency since the higher order terms will be negligible at small k values), obtain for the diatomic chain

� � kMmCa�

|2

Z ,


� �MmCavgr �

|2 .

Compare the group velocities that you have obtained for the monatomic and

diatomic chains and give a physical interpretation of the results in terms of the

sound velocity in a continuous medium. [3]

HINT: The group velocities are equal if µ=(m+M)/2. To provide the “physical

interpretation”, note that these group velocities are the speeds of sound in the

corresponding continuous materials and analyse what can be concluded about the

average (i.e. “macroscopic”) densities and elastic moduli of the two

“microscopically” different materials.

Assuming that the difference between m and M is small, sketch and label on the

same graph the dispersion curves, � �kZ , for both the monatomic and diatomic

chains, paying particular attention to the wave number axis and clearly labelling the

k values corresponding to the Brillouin zone boundaries. [3, 3]

[Hint: Since the difference between m and M is assumed to be small, you can

assume that the maximum phonon frequencies for the two chains are equal.]

6 PHY2024

Label on the graph the first Brillouin zone for each of the dispersion curves and the

band gap. [1, 1, 1]

7 PHY2024

3. (i) For a non-magnetic dielectric and a non-magnetic metal, sketch the low-temperature

dependence of function C (T) / T, where C (T) is the total heat capacity and T is the

temperature. [2, 2]

From where do the main contributions to the low temperature heat capacity originate

in each case? [2, 2]

Label on your graph the Sommerfeld parameter γ for each curve. [2]

State two of the main reasons why the experimental Sommerfeld parameter values

deviate from those calculated in the free electron model. [2, 2]


(ii) Explain, in words, the concept of the effective mass of an electron in a crystal lattice

structure. [3]


Why is it difficult to use the effective mass concept for acoustical phonons at small

values of k? [2]

HINT: Can one approximate a straight line going through zero by a parabola that is

also going through zero?

8 PHY2024

Consider a one-dimensional electronic energy band described by the expression

¸¹·

¨©§

2sin)( 2

0kaEkH ,

where a is the lattice constant, k is the crystal momentum and E0 is a constant.

Derive mathematically the effective mass of an electronic wave packet in this

energy band and sketch its dependence on the crystal momentum. [4, 2]

SOLUTIONS

)cos(12

02

2*

kaEam !

.

9 PHY2024

4. State Pauli’s exclusion principle. [1]

By considering the volume of a spherical shell in the k-space and the volume

occupied by each electron state, show that the density of states for a gas of free

electrons is given by the expression

212

3

22

22

)( HS

H ¸¹·

¨©§ !

mVD ,

where ε is the electronic energy, V is the volume of the sample and m is the

electronic mass. [6]

Define the Fermi energy for a metal. [2]


Show that for the free electron gas at zero temperature the Fermi energy is given by

� � 322

2

Fermi 32

Nm

SH ! [4]

and the Fermi wave number by

� � 312

Fermi 3 Nk S , [2]

where N is the volume density of electrons.


In the quantum description of the free electron model, only electrons with energy

within kBT (where T is the temperature and kB is the Boltzmann constant) of the

Fermi energy are considered to contribute to the heat transport processes. Justify

this assumption. [3]


10 PHY2024

The probability that an electron state at energy ε is occupied, in an ideal Fermi gas

in thermal equilibrium at temperature T, is given by the Fermi-Dirac distribution

f (ε). For temperature T > 0 K, sketch f (ε), the density of states D (ε) and the

electron distribution function n (ε), all as a function of energy. On the n (ε) plot,

shade the area corresponding to filled electron states. [2, 2, 2, 1]

SOLUTION

11 PHY2024

5. (i) Give the general mathematical form of Bloch waves – the solutions of the

Schrödinger equation for an electron moving in a periodic potential. [2]


Consider an electron moving in a one-dimensional periodic potential with period a.

For a one-dimensional Bloch wave with a wave number of π/2a, sketch the plausible

spatial dependence of the real parts of each of the two factors making up the Bloch

wave, paying particular attention to their periodicity. [2, 2]

Use the graphs that you have drawn to sketch the real part of the resultant Bloch

wave. [2]

SOLUTION:

12 PHY2024

What can be said about the values that the Bloch wave number takes at energies

within electronic band gaps? [1]

SOLUTION: imaginary values.

(ii) The dispersion of an electron in a one-dimensional, piece-wise constant, potential

with a period of a is calculated analytically as

� � ¸¹·

¨©§

¸¹·

¨©§

¸¹

·¨©

§��¸

¹·

¨©§

¸¹·

¨©§

2sin

2sin

21

2cos

2coscos 21

1

2

2

121 akakkk

kkakakka ,

where k1,2 are defined by � � mkU 2/22,12,1 !� H , k is the Bloch wave number and ε

and m are the energy and mass of the electron, respectively.

Show that the band gaps form at energies at which the absolute value of the right

hand side of the equation exceeds unity. [2]

HINT: cos-function cannot return absolute values >1 for real valued arguments.

Demonstrate how the Bloch wave number values corresponding to the Brillouin

zone boundaries can be derived from the equation above. [2]

HINT: The Brillouin zone boundaries should correspond to transition points between

the real and imaginary values of the Bloch wave number.

Show mathematically that the band gaps in the dispersion disappear when the

potential is constant. [4]

HINT: Noting that the constant potential corresponds to k1=k2 in the rhs of the

equation, simply the trigonometric expression obtained in the rhs.

13 PHY2024

Show that in the long wavelength limit, i.e. when ka, k1a, and k2a are all much

smaller than unity, the dispersion relation has the form of that of a free electron. [8]

HINT: Apply the small angle approximation to the functions of ka, k1a, and k2a

retaining terms up to the quadratic order inclusively to obtain k2=0.5(k12 + k2

2).

Then note that k1,2 are defined by � � mkU 2/22,12,1 !� H .

14 PHY2024

6. Describe how the electrical conductivity of metals, insulators and semiconductors

can be qualitatively distinguished by invoking the electron-band theory and the

position of the Fermi energy level. Explain how and why a rise in temperature

affects the electrical conductivity of semiconductors. [8, 2]


The position of the Fermi level (with respect to the top of the valence band) in an

intrinsic semiconductor is given by

¸¹

·¨©

§� *

e

*h

Bg

Fermi ln43

2 mmTk

HH ,

where εg is the electronic band gap width, T is temperature, *hm and *

em are the

effective masses of holes and electrons, respectively, and kB is the Boltzmann

constant.

Describe qualitatively why it depends on the effective masses and temperature in

this way. Explain why the position of the intrinsic Fermi level can be approximated

to being at the centre of the band gap. [7, 2]

HINTS:

1) The numbers of electrons and holes in an intrinsic semiconductor are equal. Hence,

since the densities of electronic and hole states (and therefore also their effective

masses) are generally different, the Fermi level positions itself at different

temperatures so as to counter-balance the difference via different occupancies of the

states.

2) kBT<<εg

15 PHY2024

The concentration of charge carriers in an intrinsic semiconductor is given by

¸¹

·¨©

§�

TkNNn

B

gVC

2i exp

H.

What are the meanings of the terms NC and NV is this expression? [2]


Calculate the concentration of holes in an intrinsic semiconductor at T = 400 K. In

your calculation, assume that the band gap is 0.67 eV, and that

NC(300K) = 1.0 × 1019 cm–3 and NV(300K) = 6.0 × 1018 cm–3 at T = 300 K. [4]

HINTS: 2/3VC , TNNpn ii vv

SOLUTION: pi=2.7*1013 cm-3

1

HINTS AND TIPS PHY2025 JUNE 2013 1. (a) ! ! = !"#$%&!!"!!"#$%&! !"#$!%&' !!"#!

!"#$%&!!"!!"#$%&!!"!! [2]

. !=2 – ‘the number of successful times when a defective screw is drawn’ S=10 – ‘total number of times a different screw can be drawn’ [2] Hence ! ! = !!

!"

B – conditional probability: ‘getting a defective screw if one defective screw already taken out’ ! !|! = !

! !(! !) = ! ! ! !|! = !!

!"!! = 2/90. [3]

(b) Probability of the event occurrence of which depends on the probability of another event. [2] No, because the probability of the second event will depend on the success/failure of the first

event. [2]

(c) ! ! ! = ! ! ! ! [2] (d) x=4 – number of successes, ! = !" – the mean value of the Poisson distribution, where n=13

– number of throws and p=4/52 – the probability of getting an Ace. [2] ! ! = !!

!! !!! =!!!! !!! = 0.0153 [4]

(e) This is an approximation that is not so close to the real value as the formula works best when n tends to infinity and p tends to 0. [1]

2

2.

(a) Fourier sine series ! !! = !"

! where ! = !! ,! = 1,2,3,… [2]

!! = !

! ! ! sin !"! !!!"!!! [3]

(b) !! = !"

! = 2!, ! = 1,2,3,… [2]

[7] (c) Functions !! ! are orthogonal on interval ! ≤ ! ≤ ! if ! ! !! !!

! !! ! = 0 when ! ≠ !. (In this case !(!) = 1.) [2]

(d) !"# 2!!!!!!

!"# 4! !" = 0 can be found via trigonometric identities:

!"# 2!!!!!!

!"# 4! !" = !! cos( 2− 4 !)!"

!!!!!

− !! !"# (2+ 4)! !"

!!!!!

= 0

[4]

3

3. Oscillations of masses are described by the following system: !"#

−=

−+−=

212

1211

22)(23

kxkxxMxxkkxxM

!!!!

[4] assuming a periodic solution in the form iwt

nn exx = where n=1,2. after substitution:

in matrix form: - !!"

#$$%

&!!"

#$$%

&

−

−=!!

"

#$$%

&

2

1

2

12

2225

xx

kkkk

xx

Mω => XAX =2ω where A=- !!"

#$$%

&

−

−

2225

Mk [3]

assuming λω ≡2 and 1≡Mk we get an eigenvalue problem with A= !!

"

#$$%

&

−

−

2225

.

Secular determinant: 0)6)(1(2225

=−−=−−

−−λλ

λ

λ

Eigenvalues: 6,12,1 =λ " Mk /1 =ω Mk /62 =ω [5]

Eigenvectors:

λ = 1: !"

!#

$

=+−

=−

02

02

21

21

xx

xx 2 21 xx = Let x1 = 1 than X1 = !!

"

#$$%

&

21

[1.5]

λ =6 : !"#

=−−

=−−

04202

21

21

xxxx

Let x2 = -1 than X2 = !!"

#$$%

&

−12

[1.5]

Unit eigenvectors: !!"

#$$%

&=

21

21

1U and !!"

#$$%

&

−=

12

21

2U [1]

tt eBeAY 21

12

21 ωω

"#

$%&

'

−+"

#

$%&

'= !

"

#$%

&=!

"

#$%

&

−+!

"

#$%

&=

05

12

21

)0( BAY ! A=1 B=2

[4] the later found from simultaneous equations for A and B.

4

4. (a) !!!!! !, ! = !!!

!!! where !! = !"#$%&#! =10, ! !, ! = ! ! !(!) [2]

After substitution! !!!!!!!!"! =

!!!!!!!! = −!! ! [2]

! ! = ! cos !"#!+ ! sin !"# , ! ! = ! cos !"!+ ! sin !" [2]

From boundary conditions: ! 0 = 0!! ! = 0,!

! ! = 0 = ! sin !" ! !" = !" ! !! = !"/! thus general equation: [2]

!(!, !) = ! sin !"# + ! cos !"#!! sin !"! !! ! ! ! ! ! ! [2] !

(b) [2]

Nodes!at! !! = !!:!none,!at! !! =

!!! :! ! =

!!,!at! !! =

!!! :! ! =

!!,! ! = 2 !!! !

Wave!vectors!accordingly:! !! = !! ,! !! =

!!! ,! !! =

!!! ! ! ! ! [3]

(c) Oscillation frequency ! = !"/2! [2] !! = !!!

!! =!!.! ∙

!"!!! = !15.8!Hz

!! = !!!!! =

!!!.! ∙

!"!! = 31.6 Hz

!! = !!!!! =

!!!.! ∙

!"!! = 47.4 Hz [3]

0 0.5 1 1.5 2 2.5 3 3.5-1

-0.5

0

0.5

1

5

22

2

421

421

0*0

)1(0

43

)(

2

2

ωε

πω

ε

επω

ψψ

ω

ω

m

dxxem

dxxem

dxxUE

xm

xm

!

!

!

!

!

=

=%&

'()

*=

=%&

'()

*=

==

∫

∫

∫

∞+

∞−

−

∞+

∞−

−

+∞

∞−

5. (a) )()())()(( xExxUxL ψψ =+ where !(!) is the Hamiltonian of the ‘exact’ harmonic oscillator [3]

(b) U(x)= 4xε [3] (c) )1(

nn EEE += [3] (d) dxxUE nnn ψψ )(*)1( ∫= [3]

(e) [8]

[2]


Module Code PHY2030Name of module Observing the UniverseDate of examination May/June 2013

1. theta approx lambda / D seeing: bookwork 0.1 arcsec: adaptive optics ; 0.01 arcsec: interferometry (bookwork)

faintest mag approx 26.2 with sky bakckground, approx 22 higher resolution --> smaller aperture --> smaller background count

2. mu * (M1 + M2) = M1 M2

symbols: bookwork compare given equation to r = a(1-e^2) / (1 + e cos (nu) )

form dA = (L/2mu)dt and integrate over 1 period (A = (L/2mu)P) ; sub for L/mu and A as given

3. searching for exoplanets: bookwork (transits)

from plot: fraction of flux eclipsed approx 1 - 0.993 planet radius approx 1.5 RJ

secondary eclipse + phase modulation (bookwork)

orbital period from plot: approx 2.2 days KIII to find a approx 0.038 a.u.

sinusoidal RV curve: circular orbit from plot: K1 approx 200 km/s mass function formula with appropriate approximation: mass approx 2 MJ

know mass and radius: can estimate density (compare to Jupiter) T approx 2300 K --> Hot Jupiter (close orbit, mass similar to Jupiter, high temperature)

4. sketches: saw-tooth for perpendicular, Mexican hat for parallel (bookwork)

relative amplitudes 1:3 for 3:1 mass ratio scaling for parallel Vp / Va = (1+e)/(1-e) = 3

eclipses: bookwork

T1 / T2 = 1.685

F1 / (F1 + F2) = 0.98

5. fluxes/magnitudes, colour index: bookwork

Planck function: bookwork

long wavelength limit: F proportional to T for both filters, ratio is ~constant

asteroid mag approx 6 (limit of visibility)

power in = power out at 1 a.u. (T approx 270 K)

Wien's Law: peak at approx 10 microns

6. distances: bookwork

parallax = 0.25 arcsec; proper motion = 0.4123 arcsec/yr

distance = 4 pc; transverse velocity 8.11 km/s

1 + z = lambda_now / lambda_then = a_now / a_then = T_then / T_now (last step from Wien) recombination (bookwork)

1


Module Code PHY2031

Name of module Lasers, Materials and Nanoscale Probes for Quantum Applications


2. Al mole fraction = 0.30 6. For λ = 600 nm, R = 1.43 × 104 For emission processes to occur at the same rate, temperature T = 3.45 × 104 K

1 PHY3052


Module Code PHY3052

Name of module NUCLEAR & HIGH ENERGY PHYSICS

Date of examination June 2013

1. (i)

!

abbb

bb

q),cos(13)sin(13F(q) 23 � (done by parts)

q = 44 MeV/c

aaab 14822 1023.2

1031*

10582.6 MeV/c) (44q

u uu

�!

Thus for a=6fm: b=1.33 F(q)=0.834

a=6.5fm: b=1.45 F(q)=0.805

a=7fm: b=1.56 F(q)=0.777

a=7.5fm: b=1.67 F(q)=0.748

a ~ 7.4 fm.

Bookwork

(ii) Bookwork

Since = 0.003 is already given, we can work out = 0.041.

Bookwork

Mott

2

expF(q)

OV

OV

dd

dd

2 PHY3052

2. (i) (a) weak, barion numbers conserved and neutrinos involved

` (b) none, leptons numbers not conserved

(c) none, energy not conserve (electron lighter than pion)

(d) EM, photon involved

(e) weak, barion numbers conserved, neutrinos involved

(f) none, charge not conserved

Weak interaction

(ii) Bookwork

(iii)

m = 3e * (0.75)2 / (6x104) * 0.0578

m=2.6 × 10-25 Kg.

3. (i)

Bookwork

B/A

A

~8 MeV/nucleon

4-He 56-Fe

3 PHY3052

233 Th: A=233, Z=90, B = 1780.84 MeV

233 Pa A=233, Z=91, B = 1781.20 MeV

233U A=233, Z=92, B=1780.54 MeV

: Q = B(233Pa) – B(233Th) + (mn – mp – me) c2

= 1781.20-1780.84+0.79 = 1.15 MeV

: Q = B(233U) – B(233Pa) + (mn – mp – me) c2

= 1780.54-1781.20+0.79 = 0.13 MeV

(ii) Bookwork

Distribution given by ~ E2 (Q – E)2. Turning point at E = Q/2:

4. (i) Bookwork

Tauons: EM and weak

Bookwork

(ii)

Energy

Q/2 Q

)(2

)2

()2

(

)211()

211(

11

21

22

3

321

322

22

421

22

422

22

421

22

422

22421

22422

mmp

c

pcm

pcm

cpcmpc

cpcmpc

cpcmpc

cpcmpc

cpcmcpcmE

�

�

��

��

�� '

4 PHY3052

)(sin)

4)(

(sin~

)2

)(2(sin~

)2

)(2(sin~)

2(sin

0

22

12

23

2

21

22

3

2

21

22

3

22

LL

EmmLc

cLmm

Ecc

tmmp

cEt

�

�

�'

!

!

!!

` 03

21

22

21

22

30

4)(

)(4

LcEmm

mmcEL

!

!

�

�

000,10042

2Eccm !

� 23 /109.8 ceV�u

n = 106 /18 * 6×1023 *10 electrons/m3 = 3.33×1029 electrons/m3

Volume = (Collisions/s)/ I n V

r = 584 m.

5) (i) (a) 0,+ (b) 0,+ (c) 5/2, + (d) 1/2, + (e) 7/2, - (f) 7/2, -

Bookwork

Bookwork

Magic numbers: 2,8,20. (a) O168 and (f) Ca47

20 are magic.

(ii) (a) L=1, electric (most intense) L=2, m L=3, e L=4, m L=5, e

(b) L=2, electric (most intense) L=3, m L=4, e L=5, m L=6, e L=7, m

(c) L=1, electric is the only allowed transition

(d) L=2, electric (most intense) L=3, m L=4, e L=5, m

(a) and (c) would have shortest half lives, as they have L=1 transitions.


Module Code PHY3053Name of module General ProblemsDate of examination May 2013

SECTION A

1.! ! Quote (or derive) the energy of the ground state for a particle in a box, or use the uncer-tainty relation to find the momentum and hence energy. Then subtract this energy from 1eV to find the “ionisation” energy, which is 1.005 ×10−19 J. It is an approximation be-cause you are using an infinite well formula for a finite well.

2.! ! How many atoms are there per cubic unit cell? Density is mass * number of atoms / volume of cubic unit cell. Answer: 19.3×103 kg m–3

3.! ! Use conservation of energy to calculate maximum stretched length. This gives the elas-

ticity (in terms of the student’s mass). Then use ω = k m to find the period – the m cancels. Answers: k = 4.91m and Period = 2.84 s

4.! ! Force = Bqv, no need for vectors because all relevant quantities are mutually perpen-dicular. Obtain v from the classical kinetic energy. Answer: 6.14 ×10−15 N

5.! ! 510B+ n→ 5

11B→ 37Li+2

4He . Energy released = 2.793 MeV

6.! ! Use conservation of energy to find kinetic energy at bottom of swing, v = 3.13 m s–1. Equate the net force on the mass at that point with the centripetal force, to find the ten-sion, which is 19.6 N.

7.! ! v = T ρ and , so f ∝ T . Beat frequency is 1.00 Hz.

8.! ! Moment of inertia for rotation about centre of mass is md 2 2 (d the atom separation, m the mass of each atom). TR = 74.7 K.

9.! ! Draw a diagram and balance the forces on one of the rings. Hence obtain an expression for the angle of each side of the rope from the vertical (19.6°). This angle is also re-lated by simple trigonometry to the distance between the rings. Answer: 0.66 m.

10.! ! Measured in the laboratory the mean decay time (= distance/speed, L/v) is the dilated

time. Hence L v = t0 1− v2 c2( )1 2 – solve to find v. Answer: v = 2c 5

11.! ! The log–log plot yields the exponent as its gradient. Experimental value of exponent is n = 1.82

12.! ! Simple radioactive decay, answer: t = 22920 years.

SECTION B

13.! ! (a) AM before = mau, (b) AM after = –mav.! ! Use conservation of AM and conservation of energy (including the energy of the rota-

tion of the rod) to find v in terms of u.! ! Substitute this expression back into the formula for conservation of AM to find ω.

14.!(a)! For the “symmetric” mode, the central mass is fixed so the outer masses behave like

two independent oscillators. Hence ω = k m1 .! (b)! For the antisymmetric mode, one can write equations for the forces on each of the three

masses in terms of their displacements, but the displacements of the two outer atoms are identical, and hence so are the forces on them. This reduces the three equations to the two given in the question.

! (c)! Substitute the solutions given in to the two equations of motion. This gives two equa-tions in the three unknowns C1, C2 and ω (α is π). This enables one to find the ratio of

the amplitudes and ω. Answer: ω =k m1 +m2( )m1m2

⎛⎝⎜

⎞⎠⎟

1 2

15.! ! Use conservation of energy. Gravitational potential energy is GMm R .! ! Sun emits energy equally in all directions. Therefore proportion of total power hitting

comet is (apparent area of comet, πr2 ) / (area of sphere, radius = Sun–comet distance).

! ! Total energy hitting comet is Ldt∫ . Following the hint in the question this can be con-

verted to an integral with respect to distance: −Lr

2

41

2GM

R−3 2 dR∞

R

∫ from which

the answer follows.! ! Energy required to vaporise comet is latent heat * mass. Answer: radius = 3950 m

16.! ! Substituting the ground-state wavefunction back into the Schrödinger equation gives an equation with terms in x2 and x0. The coefficients of these terms should independently be equal on each side of the equation. This gives two conditions from which (a)

λ = 2 mω 0 and (b) E = ω 0 2 .! ! The expectation value of position is zero because the wave function is symmetric about

x = 0.

! ! The new Schrödinger equation is −

2

2md 2ψdx2

+ 12kx2 − Fx⎛

⎝⎜⎞⎠⎟ψ = Eψ .

! ! Substitute the new ground-state eigenfunction in. xF =Fk

and E = 1

2ω 0 −

F2

2k.

1

PHYSICS EXAMINATION PROBLEMS

SOLUTIONS AND HINTS FOR STUDENT SELF-STUDY

Module Code PHY3065

Name of module Quantum optics and photonics

Date of examination May/June 2013

1. Solution for problem plus optional comment/hint as to method if appropriate

2.

3.

4.

5.

1. (i)

ha�k, a†�0k0

i= ��0�kk0 , [a�k, a�0k0

] = 0.

Also acceptable:ha�k, a†�k

i= 0

Photon number operator:

ˆN�k = a†�ka�k.

Action on number states:

ˆN�k|n�ki = n�k|n�ki.

ˆH =

X

�

X

k

~!

ˆN�k +

1

2

�.

The diverging termP

�

Pk ~!/2 in the summation is electromagnetic zero-point energy.

(ii)Expectation value of electric field in state |ni is

hn| Ê(�)|ni = hn|12

�a e�i�

+ a†ei��|ni = hn|1

2

�e�i�

pn|n� 1i+ ei�

pn + 1)|n + 1i

�

= 0.

Similarly, the expectation value of (

Ê(�))

2 in |ni is

hn|( Ê(�))

2|ni =

1

2

✓n +

1

2

◆

The uncertainty �E(�) in the electric field is defined by

(�E(�))

2= hn|( Ê(�))

2|ni � (hn| Ê(�)|ni)2 =

1

2

✓n +

1

2

◆� 0 =

1

2

✓n +

1

2

◆

=) �E(�) =

s1

2

✓n +

1

2

◆.

Rewrite Ê(�) in term of cos� and sin�:

Ê(�) =

ˆX cos�+

ˆY sin�,

2

where the quadrature operators are

ˆX =

1

2

�a + a†

�, ˆY = �1

2

i�a� a†

�.

The expectation values of ˆX and ˆY in |ni are

hn| ˆX|ni = hn|12

�a + a†

�|ni = hn|1

2

�pn|n� 1i+

pn + 1)|n + 1i

�= 0,

hn| ˆY |ni = hn|�i

2

�a� a†

�|ni = hn|�i

2

�pn|n� 1i �

pn + 1)|n + 1i

�= 0.

The expectation values of ˆX2 and ˆY 2 in |ni are similarly

hn| ˆX2|ni =

1

2

✓n +

1

2

◆,

hn| ˆY 2|ni =

1

2

✓n +

1

2

◆.

The uncertainties �X and �Y in the quadratures are defined by

(�X)

2= hn| ˆX2|ni � (hn| ˆX|ni)2 =

1

2

✓n +

1

2

◆

=) �X =

s1

2

✓n +

1

2

◆,

(�Y )

2= hn| ˆY 2|ni � (hn| ˆY |ni)2 =

1

2

✓n +

1

2

◆

=) �X =

s1

2

✓n +

1

2

◆

2. (i)

a4

a3

a2

a1

a3 = Ra1 + T a2, a4 = Ra2 + T a1

ˆN3 =

⇣R⇤a†1 + T ⇤a†2

⌘(Ra1 + T a2) = |R|2a†1a1 +R⇤T a†1a2 + T ⇤Ra†2a1 + |T |2a†2a2,

ˆN4 =

⇣R⇤a†2 + T ⇤a†1

⌘(Ra2 + T a1) = |R|2a†2a2 +R⇤T a†2a1 + T ⇤Ra†1a2 + |T |2a†1a1.

Adding these and using the identities in the question:

ˆN3 +

ˆN4 =

ˆN1 +

ˆN2.

3

This shows the number of photons in the output fields is equal to the number in the input fields.The input state is

|1i1 |0i2 = a†1|0i.

Write a1 in terms of the output operators by inverting the output relations using the identities in the question:

a1 = R⇤a3 + T ⇤a4.

This gives a†1.

|1i1 |0i2 = R|1i3 |0i4 + T |0i3 |1i4.

The right-hand side is the output state, a superposition of one photon in arm 3 and one in arm 4.The probability amplitude for detecting the photon in arm 3 is R; for arm 4 it is T .The probability for detecting the photon in arm 3 is |R|2; for arm 4 it is |T |2.

(ii)The norm of | i is

h | i =

✓1p2

hu|� ip2

hv|◆ ✓

1p2

|ui+

ip2

|vi◆

=

1

2

(hu|ui+ ihu|vi � ihv|ui+ hv|vi)

=

1

2

⇣1 + ie�(v�u)2/2 � ie�(v�u)2/2

+ 1

⌘

= 1,

so | i is normalized.The average photon number is

h |a†a| i =

1

2

�hu|a†a|ui+ ihu|a†a|vi � ihv|a†a|ui+ hv|a†a|vi

�=

1

2

�u2hu|ui+ iuvhu|vi � ivuhv|ui+ v2hv|vi

�

[use hv|ui = e�(v�u)2/2= hu|vi] =

1

2

�u2

+ v2�

For the uncertainty in the photon number, the expectation value of ˆN2 is required. Put ˆN2 into the following form, suitablefor evaluation in a coherent state:

ˆN2= a†aa†a = a†(a†a + 1)a = a†a†aa + a†a.

Then

h | ˆN2| i =

1

2

⇥(u4

+ u2)hu|ui+ i(u2v2

+ uv)hu|vi � i(v2u2+ vu)hv|ui+ (v4

+ v2)hv|vi

⇤

[use hv|ui = e�(v�u)2/2= hu|vi] =

1

2

�u4

+ u2+ v4

+ v2�.

The uncertainty �N in the photon number is defined by

(�N)

2= hn| ˆN2|ni � (hn| ˆN |ni)2

=

1

2

�u4

+ u2+ v4

+ v2�� 1

4

�u2

+ v2�2

=

1

4

⇥(u2 � v2

)

2+ 2(u2

+ v2)

⇤

=) �N =

1

2

p(u2 � v2

)

2+ 2(u2

+ v2).

3.The atom consists of a central nucleus surrounded by electrons. The oscillating electric field in the incident light displaces

the negative from the positive charges, forming an electric dipole. The electric dipole has a potential energy of interaction withthe incident electric field. (The interaction of the oscillating magnetic field with any magnetic dipole moment of the atom isgenerally much weaker than the electric-dipole interaction.)

Hamiltonian of a two-level atom:

ˆHA = ~!1|1ih1|+ ~!2|2ih2|.

4

[or with E1 instead of ~!1, etc.].Electric dipole moment operator:

D =

X

i,j

hi|D|ji|iihj| = h1|D|1i|1ih1|+ h1|D|2i|1ih2|+ h2|D|1i|2ih1|+ h2|D|2i|2ih2|.

[Also acceptable

D = h1|D|2i|1ih2|+ h2|D|1i|2ih1|

because h1|D|1i = h2|D|2i = 0.]The operator a�(k, t) |2ih1| destroys a photon in the mode �k and changes the atom from its ground state |1i to its excited

state |2i (gives zero for atom in state |2i). The operator a†�(k, t) |1ih2| creates a photon in the mode �k and changes the atomfrom its excited state |2i to its ground state |1i (gives zero for atom in state |1i).

A constant absorption or emission rate means the probability to absorb or emit is proportional to time (of interaction).This implies that the “probability” exceeds 1 for long times, which shows that a constant rate is not valid for long times.

The typical lifetime is 1/rate, which is 2.5⇥ 10

�7s.

This time multiplied by the frequency 3⇥ 10

15s

�1 is greater than 10

8. Thus the inequality !0t � 0 is very well satisfied.Integrating

i~ d

dt| (t)i =

ˆV (t)| (t)i,

with respect to time gives

| (t)i = | (0)i � i

~

Z t

0dt0 ˆV (t0)| (t0)i,

where | (0)i it the state vector at t = 0.An approximate solution is obtained by using the first term on the right-hand side (i.e. | (0)i) as an approximate value for

| (t0)i in the second term, i.e.

| (t)i = | (0)i � i

~

Z t

0dt0 ˆV (t0)| (0)i,

The second-order perturbative solution is obtained by using the above first-order solution, which gives

| (t0)i = | (0)i � i

~

Z t0

0dt00 ˆV (t00)| (0)i,

as an approximate solution for | (t0)i in the exact integral equation; this gives

| (t)i = | (0)i � i

~

Z t

0dt0 ˆV (t0)| (0)i+

1

~2

Z t

0dt0

Z t0

0dt00 ˆV (t0) ˆV (t00)| (0)i.

4. (i)In linear optics the polarization (electric dipole moment per unit volume) in a medium due to an applied electric field is taken

to be proportional to the applied field.In nonlinear optics the polarization of the medium depends nonlinearly on the applied electric field; the polarization is gener-

ally taken as a series in the electric field.Nonlinear polarization in a �(3) medium:

PNLi(r, t) = "0X

j,k,l

�(3)ijklEj(r, t)Ek(r, t)El(r, t).

Also acceptable:

PNL(r, t) = "0�(3)

[E(r, t)]3

In the absence of nonlinearity the wave equation the plane-wave solution is

E(z,!) = Aeikz, k =

!

c

p"(!).

5

The slowly varying amplitude approximation takes the solution to be of the form

E(z,!) = E(z,!)eikz,

where the complex amplitude E(z,!) varies slowly with z compared to the oscillating factor eikz; in other words, the amplitudedoes not vary much over the wavelength � = 2⇡/k.

Using E(z,!) = E(z,!)eikz gives

@2zE(z,!) =

⇥@2

z E(z,!)

⇤eikz

+ 2ik⇥@zE(z,!)

⇤eikz � k2E(z,!)eikz,

The slowly varying amplitude approximation implies��@2

z E(z,!)

�� ⌧��k@zE(z,!)

�� ,

so the wave equation becomes

@zE(z,!) =

iµ0!2

2kPNL(z,!)e�ikz

=

iµ0!c

2

p"(!)

PNL(z,!)e�ikz.

Expansion of �(!) to second order:

�(!) = �0 + �1(! � !0) +

�2

2!

(! � !0)2

where �n =

dn�(!)

d!n

��!=!0

.

The phase and group velocities are

vp(!) =

!

�(!)

, vg(!) =

d!

d�=

1

d�(!)d!

.

At ! = !0 these give

vp(!0) =

!

�0, vg(!) =

1

�1.

(ii)Second harmonic generation is production of a beam at twice the frequency of an input beam. In photon terms, two photons

from the input beam are converted to one photon at twice the frequency of the input photon. This process requires a �(2)

nonlinearity.Signal and idler fields are in their vacuum states at z = 0 means the state is |0iS |0iI, defined by

a(0,!)|0iS = 0, a(0,!p � !)|0iI = 0.

The signal field is not in its vacuum state at z > 0 because the annihilation operator at z > 0 does not give zero when acting onthe signal state:

a(z,!)|0iS |0iI = a†(0,!p � !)ei✓psinh [s(!)z] |0iS |0iI = ei✓psinh [s(!)z] |0iS |1iI.

The signal number operator a†(z,!)a(z,!) at z > 0 has an expectation value Sh0| Ih0|a†(z,!)a(z,!)|0iS |0iI in the state|0iS |0iI that is the norm of the state a(z,!)|0iS |0iI above. At z = L this norm is

Sh0| Ih0|a†(L,!)a(L,!)|0iS |0iI = sinh2[s(!)L] Sh0|0iS Ih1|1iI.

Taking the signal and idler number states to be normalized gives an expectation value sinh2[s(!)L].

5.Quantum states of the total system that cannot be written in the product form

| iA |�iB

are called entangled states.

6

The state

a| 1iA |�1iB + b| 1iA |�2iB

is not entangled because it is equal to

| 1iA (a|�1iB + b|�2iB) ,

which is a product state.No-cloning theorem: It is not possible to make a copy of an unknown quantum state.The no-cloning theorem implies that the teleportation of unknown quantum states must require the destruction of the original

state.General qubit:

| i = a|Hi+ b|V i, |a|2 + |b|2 = 1.

EPR state of two photons:

1p2

(|Hi1 |Hi2 + |V i1 |V i2) .

The Schrodinger equation is

i~ d

dt| (t)i =

ˆH| (t)i,

which for time-independent Hamiltonians has the solution

| (t)i = e�iHt/~| (0)i.

The operator e�iHt/~ in this time-evolution equation is unitary⇣e�iHt/~

⌘†e�iHt/~

= eiHt/~e�iHt/~= I

because the Hamiltonian is hermitian (H†= H) and it commutes with itself. Time evolution described by the Schrodinger

equation is thus governed by the unitary operator e�iHt/~ and is accordingly called unitary evolution.The Hadamard gate acting on a qubit gives

UH (a|Hi+ b|V i) =

1p2

(a|Hi+ a|V i+ b|Hi � b|V i)

=

ap2

(|Hi+ |V i) +

bp2

(|Hi � |V i)

It is obvious that U†H = UH , and

UHUH =

1

2

(|HihH|+ |HihV |+ |V ihH|� |V ihV |) (|HihH|+ |HihV |+ |V ihH|� |V ihV |)

=

1

2

(|HihH|+ |HihV |+ |HihH|� |HihV |+ |V ihH|+ |V ihV |� |V ihH|+ |V ihV |)

= |HihH|+ |V ihV |= 1

from the completeness relation. Thus U†HUH = UHU†

H = 1, i.e. UH is unitary.The operator UC must replace |V i1 |Hi2 by |V i1 |V i2 and vice versa, leaving all other basis states unchanged; it is therefore

given by

UC = |Hi1 |Hi2 1hH| 2hH|+ |Hi1 |V i2 1hH| 2hV |+ |V i1 |Hi2 1hV | 2hV |+ |V i1 |V i2 1hV | 2hH|.

The first two terms in UC are equal to their Hermitian conjugates; the Hermitian conjugates of the second two terms are

(|V i1 |Hi2 1hV | 2hV |)† = |V i1 |V i2 1hV | 2hH|, (|V i1 |V i2 1hV | 2hH|)† = |V i1 |Hi2 1hV | 2hV |.

Thus Hermitian conjugation interchanges the second two terms, and we have U†C = UC .

PHYSICS EXAMINATION PROBLEMSSOLUTIONS AND HINTS FOR SELF STUDY

Module code PHY3066Name of module Galaxies and High Energy Astrophysics

Date of examination May/June 2013

1. (i) Redshift 0.20 from z = (�obs��em)/�em [3]. Distance 820 Mpc from cz = H0d. Watchthe units of H0 ' 73km s�1Mpc�1 [3].

(ii) cm wavelengths - bremsstrahlung[2]. mid-IR: thermal dust emission. (These are boththermal processes. You’re told this is a star-forming region, not an active galaxy orsupernova remnant, so relativistic electrons producing synchrotron are unlikely).[2]

(iii) SB galaxy: bar and spiral arms [2]

(iv) fluctuations in the Cosmic Microwave Background [2]

(v) Milky Way from Galactic North Pole [6]:

Note the Galactic coordinate system is centred on the Sun, not the Galactic Centre.OB stars mainly found in spiral arms and inner bulge/ Central Molecular Zone. [2]

Exponential model n(R, z) = n(0, 0) exp(�R/hR) exp(�z/hz) = n(R, 0) exp(�z/hz).Integrate over z to find ⌃R =

R1�1 n(R, 0) exp(�z/hz) dz = 2hzn0(R) where n0(R) =

n(R, 0) is the midplane density at radius R. Note disk has 2 sides hence integral limits±1 and factor 2.[3]

Use ⌃OB = 2hOBn0,OB and similarly for ⌃low to derive ratio. [3]

For OB star formation rate, use NOB = ⌃OB/⌧OB where NOB is the star formation rateper unit area and ⌧OB the OB lifetime, plus the equivalent equation for ⌃low. Answer20 Myr. [3]

For total Galactic OB star formation rate, multiply by the area of the galaxy ⇡R2.Answer (for R = 15 kpc): NOB = 0.35/ yr. This estimate is high, because high massstar formation doesn’t extend out as far as 15 kpc. [3]

1 PHY3066

2. (i) log10 P = 1.5 so MV = �5.5 from graph. [2] mv = Mv + 5 log10(d/10pc) givesd = 10 Mpc.[3]

(ii) Note this question asks for an explanation (‘what is meant by...’) not a definition.Optical depth is a measure of how far light penetrates into or through a medium.Optically thin - transparent, most light passes through. Optically thin - opaque,most light absorbed. [3]

(iii) Upscattered - energies increase. AGN jets, Sunyaev-Zel’dovich e↵ect.[2]

(iv) Supernovae distribute heavy elements, leading to higher metallicity in an enriched ISM;they disrupt molecular clouds and redistribute gas in galactic fountains. [4]

(v) Semimajor axis ⇠ 0.10 arcsec from figure (the semimajor axis is half the long axis).[2] Convert to physical size using r = d✓. [1] To derive the mass, use circular motioneither GM/r2 = r!2 with ! = 2⇡/T (⌘ Kepler’s 3rd law) or GM/r2 = mV 2

circ/r

with Vcirc = 2⇡r/T to find M ' 2.5⇥ 106M�[2]. Assumptions: GC distance 8.5 kpc,orbit in the plane of the sky (not inclined) [2]. Schwarzschild radius R = 2GM/c2 is7.4⇥ 106 km [3].

Another direct observation: radial (Doppler) velocity gives dynamical mass/ furtherconstraints on orbit [3]

Direct observation of Sgr A*: small proper motions indicate dynamically dominantmass. [3]

Mergers transfer gas to inner parts of galaxy increasing the bulge mass. [2] Increasegalaxy mass leading to higher velocity dispersion (virial theorem GM =< V 2 > R).[2]

3. (i) metallicity [1]. Pop I = young, disk, bulge [2]; Pop II = old, halo, bulge [2]

(ii) 2⇥1011 M� from GM = V 2circR with Vcirc = �V/2 – linewidth is twice circular velocity.

[3]

(iii) exp(�⌧) = 0.5, ⌧ = 0.693 [3]

(iv) Dark matter simulations plus analytical equations modelling the behaviour of baryonicmatter. [3]

(v) Virial theorem 2T +W = 0 with T = (1/2)M ⇥3�2 and W = �↵GM2/Re. SubstituteL(M/L) for M . [5]

Luminosity is the surface brightness integrated over the galaxy area L ' ⇡R2eIe. Sub-

stitute in given equation for L to find Re / �2I�1e /(M/L) [3] Substitute (M/L) / L�

to give

Re / �(2/1+2�)I�(�+1)/(1+2�)e [3]

then equate powers with fundamental plane to find � = 0.33[4]

Last part: integrate L = 4⇡R10 2⇡RI(R)dR with substitution x = b(R/Re)1/4, dR =

Re(4/b)(x/b)3 dx, limits unchanged. [5]

2 PHY3066

4. (i) H↵ emission line from gas surrounding young massive stars; dust extinction. [2] Gasand dust are the raw material for star formation.[2].

(ii) I = F/⌦ = L/4⇡A so L = 4⇡ ⇥ ⇡R2 ⇥ I = 8⇥ 109L�. [3]

(iii) Use the dynamical mass V 2circ = GM/R [1] and linewidth �V = 2Vcirc = 120 km s�2.

[2]

(iv) E↵ective potential [4].

(v) AGN components and viewing angles [6,7]

No jet: radio-quiet quasar, Seyfert type I and II.[3]

Maximum radiation in direction of jet (perpendicular to acceleration) [1] Minimum at

✓ ⇠ 1� =q1� v2/c2 = 0.31 rad or 18 degrees [3].

3 PHY3066

5. (i) Any two of: Maser parallax, RR Lyrae distribution or Orbits in the central stellar clusterplus a sentence of explanation. [4]

(ii) rotation curve

[2]

Angular velocity decreases with radius, so inner stars would orbit in less time and windup arm. [2]

(iii) Tully-Fisher L / V 2. Measurement of V gives luminosity; compare to measured flux toobtain distance or measurement of V gives absolute magnitude; compare to apparentmagnitude to obtain distance. [2]

(iv) Set Frad = Fgrav and rearrange to find LEdd = 4⇡GMmpc/�T. [4]

(v) Use ⌧ = ↵L where L is the path length. Absorption coe�cient ↵ / ⌫�2 so more likelyto be optically thick at lower frequency. [3]

Start from equation of RT I = I0 exp(�⌧)+⌃⌫(1� exp(�⌧)). Set background term inI0 to 0. Thermal emission so by Kircho↵’s law ⌃⌫ = B⌫(T, ⌫) the Planck blackbodyfunction. In the low frequency, Rayleigh-Jeans regime so B⌫(T, ⌫) ' 2kT (⌫2/c2). [4]

Optically thick ⌧ � 1, I / ⌫2 at low frequency. Optically thin ⌧ ⌧ 1, (1�exp(�⌧) ' ⌧ ,I / ⌫�0.1 at higher frequencies (but still R-J regime). [4]

[5]

⌧ = 1 constrains temperature T and number of electrons/ions along line of sight viaN2L. [2]

Bremsstrahlung is used to detect ionised Hii regions around massive (OB) stars whichare shortlived so indicate recent star formation. [2]

4 PHY3066

PHY3068: Principles of Theoretical Physics

Hints and tips for the 2012/13 exam

1. The equations of motion are derived in a straightforward way from generalEuler-Lagrange equations of motion. The rotational symmetry is almost obvi-ous: the kinetic energy term is v2, while kinetic energy depends on r2 = q2

1+ q2

2

only. The corresponding conserved quantity is the angular momentum: M =p1q2 − p2q1. The remaining symmetry is due to time uniformity, the conservedquantity is the energy. To derive the expression for the energy, one can eitheruse the general formula, or simply write the Lagrange function as a differenceof kinetic and potential energy: L = T−U . Then the total energy is H = T +U

2. The important facts about Higgs mechanism is that the mass is generated dueto spontaneously broken gauge symmetry. Using Einstein’s notation requiressome care: one has to keep in mind that ∂2µ does not stand for the sum ofthe four squares. Instead, it is an abbreviation for ∂µ∂µ, and raising the indexchanges the signs of spatial coordinates.

The gauge transformation operates additively on χ: χ → χ + f , so that bychoosing f = −χ, one can eliminate the phase of the scalar field. The currentis given by the standard formula: jµ = −∂L/∂Aµ. This gives jµ ∝ −Aµ

The equations for the field A are the standard Maxwell’s equation with thecurrent being proportional to the vector potential. (The equations can be alsoderived from the least action principle via a straightforward calculation.)

3. The fields E and B are given by standard expressions: E = −dA/dt, B =curlA. The wave equation gives the relation for the frequency: ω2 = 2π2c2/b2,where c is the speed of light.

The required components of the stress-energy tensor are σxx, σxy, and σxz. Theformer quantity gives the normal pressure, while the remaining two give shearapplied to the wall. Calculating σxx, one finds that its time average is in factzero.

On a microscopic level, this is due to the compensation between Coulomb force,qE, and Lorentz force, qv × B, which act on surface charges and currents.Another explanation is that the group velocity along the x-axis for the mode inquestion vanishes. Therefore, the photons do not propagate along the x-axis,and thus carry no momentum.

4. The Lagrange function allows an easy mapping to the purely mechanical prob-lem of one-dimensional motion. Indeed, the first term can be interpreted as akinetic energy, while the second term is a position-dependent potential energy.

1

The potential energy is minimal when the “coordinate” φ is a multiple of 2π,which gives equilibrium configurations of the system.

Passing to the imaginary time is straightforward: one has to change either thesign of the potential energy, or the sign of the second derivative in the Newton’ssecond law of motion. Substituting the ansatz for φ(τ), one can find the value

of the parameter b =!

EJ/M .

To calculate the action, one has to substitute this expression into the Lagrangefunction and integrate over time using the equations which are supplied in thepaper. This gives the instanton action S = 8

√MEJ . Note that in this case the

solution is not a bouncing trajectory: it departs from one energy minima φ = 0,and arrives to another one at φ = 2π. Therefore, the action has to be doubledto obtain the exponentially small tunneling rate: Γ ∝ exp (−2S/h).

5. In this confined geometry, the vector potential can be chosen in any form thatgives a correct value of the flux through the ring. For example, taking A = Φ/L,we obtain

"

Adl = Φ, which agrees with the total flux. The correspondingLagrange function is L = mx2/2 + eAx. For a constant vector potential, thesecond term is a total derivative. It is a known fact that such terms giveno contribution to the classical equations of motion. (This can be also easilychecked explicitly.)

The Hamiltonian also takes a standard form, H = (p− eA)2/2m. One can seekthe solutions in the form ψn(x) ∝ e2πinx/L, due to the conservation of angularmomentum. Substituting such wave functions into Schrodinger equation, onefinds the energy levels ϵn ∝ (n−Φ/Φ0)2, where Φ0 = 2πh/e is the flux quantum.

The effect of the vector potential can be interpreted in terms of the phase pickedup by a particle: θ =

#

Adl/h. One can view the ring as a straight line wrappedaround a circle, so that the points x, x±L, x±2L, etc. represent the same pointon the ring. Therefore, the transition amplitude between points x and x′ on thering can be obtained from the transition amplitude between the points x and x′,x and x′ +L, etc. Without a flux, each of these amplitudes is given by Green’sfunction of a free particle. Due to the field, the contribution of a trajectory thatwraps n times around the ring acquires an extra phase, θn = neΦ/h. Amendingthe contributions of the trajectory between x and x′+nL by the factor eiθn , wearrive to the required relation.

2

Hints and Tips

Module Code PHYM001 / PHYM011 (IS)

Name of module Statistical Physics


Question 1

- Use the general definition of the partition function and deduce the consequent probability of a many-body state as well as the free energy F as detailed in the lectures.

- When you identify the many-body states of three electrons in the four single-particle states, pay attention that electrons are fermions and thus obey the Pauli exclusion principle.

- With the energy of the different possible many-body states write down the partition function Z using the definition given above.

- Calculate the average energy E of the system making use of the probability of each state obtained in the first part of the question, with the partition function Z that has just been explicitly calculated.

- Use the conventional expansions of exponentials at KT<<Delta (or KT>>Delta) in order to approximate E at low and high T.

Question 2

- When sketching the dispersion of the two bands pay attention that the two parabolas are shifted rigidly by the same amount in the positive and negative energy direction.

- Calculate the DOS of each subband of 2D massive particles as usual. Pay attention that each subband has a minimum energy related to the corresponding value of lambda. Also pay attention that lambda is a band index, not a spin index. Thus take into account spin degeneracy in each subband.

- When sketching the total DOS take into account the shift between the two bands.- Calculate the occupation N_lambda for each subband as a function of the Fermi energy

EF. Then insert this into the definition of P. Pay attention that, if EF is below the bottom of a given subband, the corresponding occupation vanishes.

- State the general relation between the particle number N and the chemical potential mu at finite T using the concept of DOS and the Fermi-Dirac distribution.

- Discuss the sign and magnitude of μ in the ideal gas approximation. - Use this to expand the Fermi distribution and perform the integral over energy. From the

result, deduce the dependence of μ on T.

Question 3

- When estimating the thermal de Broglie wavelength lambda, pay attention that photons are massless. Given the volume of the cube and the particle number, deduce the typical interparticle distance and, by comparing it with lambda, deduce the crossover temperature between the classical and quantum regimes.

- State the general relation between the particle number N and the chemical potential mu at finite T using the concept of DOS and the Bose-Einstein distribution.

- State the condition on the chemical potential for the onset of BEC.- Calculate the DOS of photons. Take into account the two possible polarisation states.- Calculate the critical temperature for BEC as done in the lectures, using the DOS above.

Notice that in this case, due to the perfect cavity, the number of photons is conserved.- Discuss the Landau criterion of superfluidity as presented in the independent study

package. Apply it to the photonic system with massless dispersion moving at the speed of light.

Question 4

- Sketch the cosine band in 1D. Check limiting values of the energy in the centre and at the edges of the Brillouin zone.

- Define the Fermi energy and chemical potential, and highlight their difference at finite temperatures.

- State the general relation between the particle number N and the chemical potential mu at finite T using the concept of DOS and the Fermi-Dirac distribution.

- Calculate the DOS of this 1D system and sketch it. Include the proper degeneracy factors due to spin and to the k -> -k symmetry. Pay attention that, for energies smaller than the bottom of the band, or higher than the top, the DOS has to vanish. Also pay attention to divergences of the DOS that could emerge.

- Using the given formula and considering T=0, relate the occupation of the band and the value of the Fermi energy EF.

- When calculating the maximum number of particles in the band, send EF to the top of the band and use the given integral. Check your result thinking how many electrons can be accommodated in a linear chain of sites of length L and spacing a.

Question 5

- Describe the assumptions of the Einstein and Debye models as in the lectures. Give the definition of the Debye energy.

- Sketch the dependence of the heat capacity of a metal on temperature, as in the lectures. Do not forget the electronic contribution at low temperatures.

- Use the given dispersion of magnons to estimate the Debye temperature in a crystal with lattice constant d in analogy with that of linearly dispersing phonons.

- Calculate the DOS of the given magnonic branch in 3D as usual. - State the general relation between the total energy E at finite T using the concepts of

DOS for magnons and the Bose-Einstein distribution. Pay attention that magnons have a vanishing chemical potential. Calculate this in the regime of small temperature, i.e. approximate the Debye temperature to be infinite.

- State the definition of the heat capacity as a derivative of E with respect to T, and calculate it from the E above.

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