SOLUTIONS AND HINTS FOR EXAMINATION...

1

SOLUTIONS AND HINTS FOR EXAMINATION QUESTIONS

Module code PHY1024

Name of module Properties of Matter

Year of examination 2014-15

1. (a, b) Refer to Lecture-07 (Elasticity), Slides 6–9. (c) pfloor = psurface + ρgΔh ≈1.03×103 kg m−3 × 9.81m s−2 ×11×103 m =111MPa or 1.11 kbar

(d) ΔV V0 =−ΔpB

= −111MPa140 GPa

×100% = −0.079%

U = − pdVV0

Vf∫ = pV0Bdp

p0

pf⌠⌡⎮ = V0

B12 p

2⎡⎣ ⎤⎦p0

p0+ρgΔh = V02B

p0 + ρgΔh( )2 − p02⎡⎣

⎤⎦

= MCu2BρCu

ρgΔh( )2 + 2p0ρgΔh⎡⎣

⎤⎦ ≈

1.4 kg × 111MPa( )22 ×140 GPa ×8960kgm−3 = 6.88 Nm = 6.9 J

The volume of efficient of expansion is three times the linear coefficient

ΔV = 6.9 J × 3×16.5×10−6 K−1

385Jkg−1K−1 ×1.4kg× 1.4kg8960kgm−3 = 9.9 ×10

−11m3 = 0.099mm3

2. (i) (a)FvdW = −α r7 , Frep = β r13

(b) Add the forces and integrate to find the potential, e.g.:

− dUdr

= Fnet = − αr7

+ βr13

=> U r( ) = − Ar6

+ Br12

(c) Find the force by differentiation:

F r( ) = − dVdr

= −V0ddr

ar"

#$%

&'2−2ar

(

)**

+

,--= −2aV0

r − a( )r3

or 2a2V0r3

−2aV0r2

(d) Equilibrium occurs when: F r( ) = 0 => − 2aV0r − a( )r3

= 0 => r = a

(e) At equilibrium: Veq =V a( ) =V0aa

⎛⎝⎜

⎞⎠⎟2− 2aa

⎡

⎣⎢⎢

⎤

⎦⎥⎥= −V0

2. (ii) For the sketch, refer to Lecture-05 (Maxwell-Boltzmann Distribution) Slide-18 (Effect of

Temperature).

P v1 < v < v2( ) = 4π m2πkBT

⎛⎝⎜

⎞⎠⎟

3 2

exp − mv2

2kBT⎛

⎝⎜⎞

⎠⎟v1

v2⌠

⌡⎮ v2 dv.

2


Module code PHY1024



3. (i) Self-study pack 1.

(ii) Refer to Self-Study Pack 1 - Latent Heat and Phase Changes. Heat lost by water is gained by ice so when TF is the final temperature:

ΔQ =M ice TMP −Tice( )cice +M iceL +M ice TF −TMP( )cwater = −M toniccwater TF −Ttonic( )

M ice = 36.7×10−3 kg = 37g

(iii) Refer to Self-Study Pack 2 - Dimensional Analysis t∝Mαρβκ χcδ

s = kg[ ]α kgm−3"#

$%βWm−1K−1"#

$%χJK−1 kg−1"#

$%δ

s = kg[ ]α kgm−3"#

$%βkgm2 s−2 s−1m−1K−1"#

$%χkgm2 s−2 K−1 kg−1"#

$%δ

result depends on ‘s’ but not on ‘K’ so δ = −χ and χ = −1 leaving

s = kg⎡⎣ ⎤⎦

α kgm−3⎡⎣⎢

⎤⎦⎥β

sm kg−1⎡⎣⎢

⎤⎦⎥

Result is independent of ‘m’, so β = 1

3 and s = kg⎡⎣ ⎤⎦

α kg1/3 m−1⎡⎣⎢

⎤⎦⎥

sm kg−1⎡⎣⎢

⎤⎦⎥

hence α = 2

3

tduck =75g50g!

"#

$

%&2/3

× 4min ~ 5.2min

(iv) Refer to Self-Study Pack 3 - Thermal Expansion. Definitions of linear expansion and Young’s modulus give

ΔL =αL0ΔT and ΔL = FL0AY

F =αAYΔT = 22×10−6 K−1×69GPa×380mm2 ×13K = 7.5kN

+

–V

J1 J2

Isothermal�connector block

3


Module code PHY1024



4. (i) Refer to Lecture-16 Slides-04 and -07 for definitions.

(ii) The nth level has energy

En = n + 12( )! Λ

mr

where n is a natural number, Λ is the ‘force constant’ and mr is the reduced mass.

Refer to Lecture-14 Slide-21 for sketch of energy levels.

(a) mr =m1m2m1 +m2

= 1.67×10−27 kg × 58.07×10−27 kg

1.67×10−27 kg + 58.07×10−27 kg=1.623×10−27 kg

(b)

E0 = 2.974 ×10−20 J E1 = 8.922 ×10

−20 J E2 =1.487×10−19 J

0.72×10−18 J = n+ 12( )×5.948×10−20 J ⇒ n = 0.72×10−18 J5.948×10−20 J

−12

n =11.6

Therefore the molecule would have dissociated when n = 12.

a b

(a)

(b)

(c)

PHYSICS EXAMINATION PROBLEMSSOLUTIONS AND HINTS FOR STUDENT SELF-STUDY

Module Code PHY1026Name of module Mathematics for PhysicistsDate of examination May 2014

1.! (i)! 83 = 2, 2exp i2π3

⎛⎝⎜

⎞⎠⎟ , 2exp i4π

3⎛⎝⎜

⎞⎠⎟

!

SOLUTION TO DEGREE EXAMINATION QUESTIONfor Coordinator

Name of setter Usher Paper Question

Name of module Mathematics for Physicists

Year of examina-tion

2015 PHY1026 1

Initials of checker

! (ii)! z = 12

−1+ i 3( )

! (iii)! exp iz( ) = exp −3 3( ) = 0.00554 3sf( )⎡⎣ ⎤⎦

! ! Start by writing z in the form a + ib, then take the exponential and find its modulus.

2.! (i)! dfdt

= ∂ f∂x y

dxdt

+ ∂ f∂y x

dydt

.

! ! By both methods, dfdt

= 2 sin 4t( ) + 2t 3( )

! (ii)! Parameterise, using ds = dxdt

⎛⎝⎜

⎞⎠⎟2

+ dydt

⎛⎝⎜

⎞⎠⎟2

dt , I = 1/4.

3.! (i)! ∇ϕ = x ∂ϕ∂x y,z

+ y ∂ϕ∂y z,x

+ z ∂ϕ∂z x,y

! ! A = –10

! ! Unit vector required is in direction of ∇ϕ . It is 173x + 2y − 6z( )

! (ii)! Conservative force: work done when point of contact of force moves from A to B is

independent of path.

! ! F i dr

C!∫ = 0

! ! Stokes’ theorem – see Section 3 lecture notes ~page 63, and do not forget to specify

that C encloses S.

! ! By Stokes’ theorem,

∇× F( ) i dSS∫∫ = 0 , but in fact the curl of the vector given is zero,

so no integral has to be performed – the small number of marks for this part of the

questions hints at this.

4.! ! Fourier series definition – see page 19 of lecture notes for Section 4. All four equations

required.

!


Name of setter Usher Paper Question

Name of module Mathematics for Physicists

Year of examina-tion

2015 PHY1026 4

Initials of checker

! ! f x( ) ≈ L2− 4Lπ 2 cos

π xL

⎛⎝⎜

⎞⎠⎟ −

4L9π 2 cos

3π xL

⎛⎝⎜

⎞⎠⎟

5.! ! d 2ydx2

= dpdx

! ! dpdx

+ 4 p = 4x

! ! p = x − 14+C exp −4x( )

! ! y = x2

2− x4− C4exp −4x( ) + D

! ! y = x2

2− x4− 116exp −4x( ) + 1

16

PHYSICS EXAMINATION PROBLEMS SOLUTIONS AND HINTS FOR STUDENT SELF-STUDY

Module Code PHY2023

Name of module Thermal Physics

Date of examination May 2015

1. (i) K91

2Mgun

2bullet

�

KK'

cum

T . Momentum conservation .22

;bullet

2

gun

2

bulletgun mp

Mpppp ��

(ii) (a) kJ.10-1

exhausttaken

KQ

Q (b) s.5.0Power

takencycle

Qt

K

(iii) W.02.0PowerPower

Powercoldhot

coldlostCarnot

ref

lostreq

�u

TTT

K (iv) .JK5.9 1-

SunEarthtotal �

TQ

TQS'

2. � �^ `

.exp B

ZTk

p ii

H� Explain also the case of degenerate states. � �¦¦

f

f

� ¸

¹

·¨©

§�

00 Bexpexp

ii

i

iTk

Z EHH

.

� � � �^ `.

ln1exp1exp

1

00 EEEEH

EHHHww

� ww

� w�w

� � ¦¦f

f

ZZZZZ i

i

iii For the considered system:

� � � � � � � � � � � � .36.091exp95

41exp43

1exp;6.1291exp541exp31exp HHH �|¿¾½

¯® ��

� |�� Z

Z

Probability that the ground state is not occupied� �

|� Z

p 1exp1-1 0 0.78.

3. (i) .PVUH � For an ideal monoatomic gas: RTnU23

and RTnPV : TRnH25

� .

QHPVSTPVVPUH G �� ddddddd for constP .

SP P

HVSHTPVSTH ¸

¹·

¨©§ww

¸¹·

¨©§ww

�� andddd . Since PS S

VPT

SPH

PSH

¸¹·

¨©§ww

¸¹·

¨©§ww

�ww

w

www 22

.

(ii) The number of states in the interval is pp d4 2S , Boltzmann’s factor is .21

,exp3

BB¸¹

·¨©

§ ¸

¹

·¨©

§�

TkcA

Tkcp

� � ¸¹

·¨©

§� �v

TkBfp

B

2 expHHHH ,

� �3B2

1

TkB . Thus, TkBmp 2 H , TkB3 H , TNkCV B3 .

4. (i) : lnBkS , see course notes. This is also valid out of equilibrium. BB 320ln kkS | .

(ii) The total number of Fermions82

8d

24 FF

F

F

nhppLkLNk

k

� ¸¹·

¨©§ ³

�!SS

, where LNn .

8FFnh

pEE DDD � p . .88 FF

hLE

NhE

nDD

�

For -16 ms10 D , mm1.0 L and eV1.0F E : 41093.1 u|N .

EXAM SOLUTIONS AND HINTS 1 PHY2024

PHY2024

UNIVERSITY OF EXETER

PHYSICS

MAY 2015

CONDENSED MATTER I

SOLUTIONS AND HINTS


1. (i) The diagram below shows a section of a crystal plane within the unit cell.

xy

z

What are the Miller indices of the crystal plane? [2]

The majority answered this question correctly.

Answer: (2 0 1).

Consider a crystal structure with a simple cubic primitive cell with lattice constant a.

Using the Miller indices of the plane, calculate the distances between the lattice

planes that give the three smallest Bragg angles. [8]

Here, the main problem appeared to be in being unable to visualise the cubic unit

cell and its cross-sections by the various possible crystal planes. It also proved

difficult for some to deduce from the Bragg law that the Bragg angles scale

approximately inversely with the inter-planar distance.

Answers:

h = 1, k = l =0 (optionally: and similar planes obtained by cyclic index

permutations) → d = a

h = k = 1, l =0 (optionally: and similar planes obtained by cyclic index

permutations) → d = a / sqrt(2)

h = k = l = 1 → d = a / sqrt(3)

Sketch the Wigner-Seitz cell of a rectangular reciprocal lattice. [4]


This one was done generally well, provided the student could recall what is meant

by the Wigner-Seitz unit cell.

Answer:

Indicate on your sketch two examples of the incident x-ray wave vectors that lead to

diffraction peaks. [2]

Here, it was not enough to sketch some arrows. Instead, one should have recalled

that the X-ray wave vectors giving rise to diffraction peaks needed to form a

specific geometrical relationship with the reciprocal lattice, as shown on the

sketch above.

Answer: The vectors k1 and k2 (ending on the BZ boundaries) on the sketch

above.

(ii) Define the lattice energy of a crystal structure with ionic bonding. [2]

The lattice energy, U lattice, of a crystal with N ions is written as

� �¦z

N

jiijij RUU

21

lattice ,

where U ij is the energy of interaction between ions i and j located at distance Rij.

Explain the origin of the factor of 1/2 in the equation. [2]

Some recalled it, some did not.


Answers:

Lattice energy – energy that must be added to an ionic crystal to separate its

components into free ions at rest and with the same electronic configuration.

The factor of 1/2 takes into account that the interaction of each pair of ions is

counted twice in the sum.

(iii) What is the total number of possible two-dimensional Bravais lattices? [1]


Answer: 5

The figure shows (a) oblique and (b) rhombic (centred rectangular) Bravais lattices

together with their primitive lattice vectors. The two lattice types share the same

conditions that 21 aa z and qz 90M . Derive the additional condition that the

rhombic lattice must satisfy to differ from the oblique lattice. [4]

(a) (b)

Here, the key was to appreciate that the conventional unit cell in (b) contains one

point in the centre of the rectangle and then to express this fact mathematically

(e.g. 22

221 aaa � ). I have counted three different correct ways to do this in

your answers, each credited with full marks.


2. (i) Explain what is meant by a hole in the electronic theory of solids. [2]

Some recalled it, some did not. I awarded full marks irrespective of whether the

definition was provided in the real or reciprocal space.

Answer: e.g. Hole is a vacant electronic state in a valence band.

Define the effective mass in terms of the dispersion curvature and explain the

physical significance of the negative value of the effective mass of an electron near

the Brillouin zone boundary. [4]

The majority defined the effective mass by a formula (which was acceptable,

earning a mark). However, it was also required that you pointed to the Bragg

scattering and to the fact that, depending on the sign of the curvature (and

therefore effective mass), the group velocity might actually decrease as the linear

momentum and energy increase.

(ii) 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.

State a general expression for the group velocity. [1]


Answer: � �k

kvww

Z

group .

Explain the physical meaning of the group velocity vector. [2]


Some recalled it, some did not. Answers "direction of energy propagation" and

"velocity of a wave packet" were both correct, yielding full marks.

Derive from the above equation a mathematical expression for the group velocity of

acoustic phonons in the chain as a function of their wave number. [5]

The calculation was not the simplest one, and so, relatively few succeeded here.

The statement that acoustic branch corresponds to the minus sign was worth one

point as such! So, many got this point, while some who have done the math right

lost the point for not making the choice between the plus and minus signs.

Answer: � � � ��

� ��

¸¹·

¨©§

��

�

2sin41

sin2

22

kaMm

mMk

kaMm

aCkvgroupZ

Your expression should suggest that the group velocity should be zero for wave

numbers close to the centre of the Brillouin zone. Explain why it is, in fact, finite

(non-zero)? [2]

The limit of small k values needed to be considered here. However, only few

reached this stage anyway.

Answer: While the sine function in the nominator tends to zero at k ≈ 0, the same

is true for the frequency in the denominator. As result, in the limit of k ≈ 0, the

ratio produces a finite value

(iii) The terms Bloch waves and Bloch oscillations refer to distinctly different

phenomena. Describe qualitatively the physical meaning of each term. [2, 2]


Answers:


Bloch waves are solutions of the Schrödinger (or more generally, wave) equation

for a periodic potential (periodically modulated medium).

Bloch oscillations refer instead to the oscillatory motion of an electron in a

periodic potential caused by the Bragg scattering.

Derive a mathematical expression for the frequency of Bloch oscillations in terms of

the electron charge, e, the magnitude of the applied electric field, E, and the lattice

constant, a. [5]

Some recalled the derivation from the lecture slides; some did not; some attempted

to derive it "from scratch" (clearly, not remembering the topic), occasionally

earning a few points.

HINT: See the derivation in the lecture slides.


3. (i) State the three main assumptions of Drude's free electron model. [3]

The majority answered this question correctly.

HINT: See the lecture slides.

Taking the electron spin into account, use Sommerfeld's model to derive an

expression for the density of states for a two-dimensional free electron gas confined

within area A. [6]

Hence, derive an expression for the Fermi energy, assuming that N is the number of

electrons. [2]

Pleasantly, these two questions were dealt with reasonably well by more than half

of the class, although some marks were still lost here and there.

Answers:

� � 222 242

!! SS

SAm

kmkAED u

AmNEF

2!S

(ii) Sketch the IV-curve of a pn-junction, clearly labelling the voltages corresponding to

the forward and reverse bias. [2]


current J

Vbias

forward biasreverse

bias


Sketch how the depletion layer in a pn-junction differs under forward and reverse

bias conditions. [3, 3]

Some recalled it, some did not. Of note is that this question generated two

versions of graphical answers. I considered both to be correct, provided the

details were correct and the main physical message was conveyed properly.

Answer:

pn-junction under forward bias

p-typeEc

Ev

-e(Vdiffusion – Vbias)

n-type

μtoteVbias + -

forward bias

np

n

- +reverse bias

ppn-junction under reverse bias

p-type

-e(Vdiffusion + Vbias)

n-type

-eVbias

Ec

Ev

μtot

Hence describe how forward and reverse bias affect the electrical resistance of the

pn-junction. [2]

The majority answered this correctly, i.e. that the resistance decreases in the

forward bias and increases in the reverse bias.

(iii) Explain electron-hole recombination in a semiconductor. How does the

recombination rate depend on the concentrations of electrons and holes in the

semiconductor? [2, 2]



Answers:

The electron-hole recombination is a process whereby an electron occupies the

empty state represented by a hole.

The recombination rate scales with the product of the electron and hole

concentrations.


4. (i) Consider an insulating crystalline material that has a simple cubic lattice with lattice

constant a and a basis of one atom with mass m. The sound velocity in the material

is vs.

For this material, give the mathematical definitions of

(a) the Debye wave number, k Debye, [2]

(b) the Debye frequency, ω Debye, [2]

(c) the Debye temperature, T Debye, [2]

identifying all variables.

The majority answered this correctly. The most frustrating (at least, for me)

pitfall was in giving correct answers for the first two questions and then using the

free electron dispersion to answer the third…

Answers:

The heat capacity of this material is strongly suppressed at low temperatures

(T << T Debye). Explain why. [4]

This question demonstrated that the topic of the phonon energy quantisation

escaped the attention of the majority. Answers based on pointing out the steep

mathematical dependence yielded a point. The main pitfall was to confuse the

heat capacity and thermal conductivity.

Answer: The phonon heat capacity is strongly suppressed at low temperatures due

to the quantisation of the energy at each vibrational state. As a result of the


quantisation, the lowest energy that a state can have as a result of thermal

excitation is ħω, and so, the probability of getting excited is strongly reduced at

temperatures for which kT < ħω.

How does the low temperature heat capacity scale with the Debye temperature? [1]


Answer: 3

1

DebyeV T

C v

Consider another insulator that is identical to the one above with the exception of its

sound velocity, which is twice as high, i.e. 2vs. Calculate the ratio of the phonon

contributions to the heat capacity in the two materials and hence the ratio of their

thermal conductivities at low temperatures (T << T Debye). [5]

About a quarter took the solution as far as find the ratio of the heat capacities,

while the rest could not reach even that stage.

Answer: 2

(ii) Sketch and explain the temperature dependence of the chemical potential for an n-

type semiconductor, clearly labelling the valence and conduction band edges and the

donor level. [5, 4]


Answers:

At sufficiently low temperature, only impurity carriers of one type are efficiently

(yet only partly) excited into one of the bands (the closest of the valence or

conduction bands), and so, the chemical potential is located near the impurity

levels. At sufficiently high temperatures, the new carriers are excited by creation


of an electron-hole pair via excitation of an electron from the valence into

conduction band, and so, the chemical potential relocates to the middle of the

band gap.

T

μ ECond

EVal

Egap

Eimpurity(donor)Egap/2


Module Code PHY2030Name of module Observing the UniverseDate of examination May 2015

1 (i) 300, 30 (remember that magnitudes are logarithmic; convert to fluxes first) 3600 photelectrons total in 2 minutes; noise on this is sqrt(3600)=60, so require 600 signal photoelectron for S/N=10. Magnitude equivalent is approx 26. Distance=175kpc. Interstellar extinction, other sources of noise in detector would prevent this.

Adaptive optics (bookwork).

(ii) two independent observations means sqrt(2) milliarcseconds uncertainty. 5 x sqrt(2) milliarcsec for accuracy stated. (distance = 140pc approx)

need to separate proper motion from parallax.

reference field has its own parallax (reduces measured parallax according to relative parallax formula). measured distance is approx 163 pc

GAIA uses absolute parallaxes (bookwork).

2. symbols: bookwork. L/mu = sqrt ( a(1-e)^2 G(M1+M2) )

integrate over one period (dA = (L/2mu) dt leads to Area = (L/2mu)P). Sub for L/2mu.

Use KIII as instructed to find free-fall time (express M/R^3 that results in terms of density) Free-fall time is approx 2.4Myr. 10^9 solar masses in this time gives star formation rate of approx 400 solar masses per year, much higher than observed (factor of approx 100)

Answer could include some or all of: natural inefficiency, feedback preventing further star formation, support mechanisms (turbulence, magnetic fields)

3. Semi-amplitude 325 km/s (measure it from graph!)

symmetry suggests that star is in elliptical orbit, receding at periastron and approaching at apastron, projected orbital velocities subject only sin(i) projection factor at these points. VMAX approx 450 km/s, VMIN approx 200 km/s (magnitudes) VMAX / VMIN = (1 + e) / (1 - e) , rearrange for e approx 0.38

Make appropriate approximation to mass function equation. m sin(i) approx 7.7 Jupiter masses. Use KIII given period and mass to find semimajor axis (approx 0.48 a.u.)

Sketching transits (bookwork). Be careful: primary and secondary eclipse will not be half a period apart. Note that a body is receding if the radial velocity is positive. When it returns to zero velocity after reaching its maximum positive velocity, it has reached is most distant point (and obviously will be further away than the other body).

Eclipse depth. Find radius of planet knowing its mass and density. Find fractional area blocked given radius of star. (eclipse depth approx 0.988 i.e. 0.012 blocked)

Find temperature of planet assuming power in = power out. Use Wien's Law (approx 540K, approx 5.4 microns) 4. Write down the energy per unit mass of blob on the surface, if it is at distance R from the centre and has a velocity v, and set it to zero. Rearrange and express the resulting M/R^3 factor in terms of the density. Identify this (zero energy) condition as the critical case (any slower, the blob will fall back, any faster then it has a higher expansion velocity than it needs). Define H0=V/R and relate this to the density in the critical case (rho_crit). Define Omega = rho/rho_crit

Ignore rest of Universe provided it is homogeneous, isotropic. Less restrictively, must be spherically symmetric about us. But now impose condition of no special place.

R(t) = Chi * a(t) V = dR/dt = Chi * da/dt V/R = H(t) = (da/dt) / a

integrate H dt = (1/a) da from t0 to t1 with H=constant H(t1-t0) = ln (a(t1)/a(t0)) a(t1) = a(t0) e^(H(t1-t0))

1


Module Code PHY2031

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


2. Photoelectron spectroscopy: v = 1.9 × 107 ms-1 Auger electron spectroscopy: EKE = 9.1 eV


Module Code PHY3052

Name of module NUCLEAR & HIGH ENERGY PHYSICS

Date of examination June 2015

1. (i)

Weak, Weak, Weak, Strong (or EM)

g

w −

s u u

u u

w −

(ii) Bookwork

Rcharge ~ 4.09 fm

Rmass ~ 4.59 fm

~9 MeV/nucleon

2. Bookwork

Mass = 105546. - 94.433 Z + 0.984357 Z^2 MeV/c^2

(or equivalent analytical expression)

40 42 44 46 48 50 52 54

103290

103300

103310

103320

103330

103340

Turning point at Z = 47.97 (48)

No products: Cd nucleus cannot beta decay, since it is the lightest nucleus in the series.

V(r)

r

(b) neutron

(a) proton

(ii) Bookwork (self study)

ψ2

2= sin2( L

L0)

( LL0)2 ≈ 0.02

L0 =L0.02

=7300.02

= 5162km

m2c2 ≈

4!EcL0

=4(6.582x10−22 )(15, 000)(3x108)

(5.162x106 )m2 = 0.000000048MeV / c

2 = 0.048eV / c2 = 8.52x10−38Kg

3. (i) Bookwork

(a) Total charge = Z e =

Solving for charge density:

(b)

Then simplify.

(c) Calculating q ( ) and substituting into given equation for form factor

equation gives 0.849.

= (0.849)^2 * 0.01 = 0.0072 barns/radian

Mott

2

expF(q)

λσ

λσ

dd

dd

=

(d)

(ii) (a) vµ + p → µ − + p + π+ (b) π + → µ+ + vµ

(c) Σ– → Λ0 + e− + ev (d) µ + → e+ + ev + vµ

(e) τ − → π– + π0 + vτ (f) e− + p → n + ev

(g) n → e− + p + ev (h) ev + p → n + e+

Feynman diagram: bookwork

4. (i) Bookwork (self study) Distribution given by ~ E2 (Q – E)2. Turning point at E = Q/2: Bookwork (self study)

Energy

Q/2 Q

20 degrees: exp = 0.0072 barns/rad, Mott = 0.01 barns/rad

B(14-N) – B(14-O) = Q - (mp – mn – me) c2 = 1.84 + 1.81 = 3.65 MeV 14-O 14-N* 14-N

Bookwork (self study)

(ii) (a) EM (b) Weak (c) Strong, EM (d) Strong (e) Weak

(iii) (a) L=4, magnetic L=5, electric L=6, magnetic

(b) L=2, electric L=3, magnetic (c) L=2, magnetic L=3, electric L=4 magnetic L=5, electric L=6, magnetic (d) L=1, electric L=2, magnetic

Decay (d) will have shortest half-life, since it is the only L=1 decay

The Pu nucleus will have the shortest half life since it is the largest nucleus, and half life

is inversely proportional to radius.

β

γ

3.65 MeV

2.31 MeV


Module Code PHY3053Name of module General ProblemsDate of examination May 2015

SECTION A

1.! ! Quote (or derive) the energies for a particle in a box. Hence calculate the transition en-

ergy. E = hc λ . Answer: d = 3hλ8mc

⎛⎝⎜

⎞⎠⎟1 2

2.! ! Density = mass/volume. How much mass is there in the volume of a conventional cu-bic unit cell? Final answer: density = 2180 kg m–3

3.! ! Use conservation of energy and work in the centre of mass frame. Initially the two spacecraft only have gravitational potential energy. When they collide they have a dif-ferent PE and also some KE. Noting that the speed of each spacecraft in the CM frame is half the relative speed: Answer: relative speed is1.55 ×10−4 m s–1.

4.! ! v = fλ and λ = 2L for the fundamental mode. Answer: 329 Hz. v = T ρ so

ρ = 2.73×10−4 kg m–1.

5.! ! Obtain release speed from circular motion. Release angle for maximum range is 45 de-grees. Then use constant acceleration formulae. Maximum range is 258 m.

6.! ! Reaction is 12H + 1

2H→ 23He + 0

1n +Q . Energy released is difference in rest mass ener-

gies, 5.12 ×10−13 J.

7.! ! Time dilation ′t = γ t0 . Answer: v c = 0.958 .

8.! (a)! Plot ln(P) vs ln(V), gradient is −γ .! (b)! Values of V should be chosen such that ln(V) is evenly spaced. Either calculate this or

choose values that approximately achieve it, e.g. 0.01, 0.02, 0.05, 0.1, 0.2, 0.5, 1.! (c)! A constant offset in V will cause the graph no longer to be straight.

9.! ! For small EK the force is EM (Coulomb repulsion), for large EK it is strong nuclear. Obtain threshold by working out coulomb energy at a distance equal to the nuclear ra-dius. EK ≈ 8 ×10−13 J (or equivalent in MeV).

10.! ! Torque = force × perpendicular distance from force to centre of rotation( )∑ . Answer:

7.07 ×10−24 N m.

11.! ! !Q = −KA dT

dx, and !Q is the same everywhere along the composite bar. Answer:

T = 20.7 °C.

12.! ! Use constant acceleration formulae to find the acceleration, then Newton 2 to find the total force (which includes a gravitational component). Answer: frictional force = 34.9 N.

SECTION B

13.! ! vn =v0Rn 2 =

v010n

. tn =2v0 cosφgRn 2 , hn =

v02 cos2φ2gRn , h1 = 948 m, t1 = 1.04 ×10

4 s,

h2 = 9.48 m, t2 = 1.04 ×103 s. Total horizontal distance = v0

2 sin2φg

11−1 R

= 676 m.

!

14.!(a)! Steady state →∂c ∂t = 0 . Integrate the Laplacian twice and apply the boundary condi-

tions: c r( ) = c0 1−Rr

⎛⎝⎜

⎞⎠⎟ .

! (b)! R = 15 μm.! (c)! J = −4πDRc0 , and rate at which molecules are metabolised is –J. Answer

2.04 ×1016 molecules per hour.

15.! ! v = GMr

! ! Force must be attractive, so opposite charges.

! ! v = Q1Q2

4πε0mr⎛⎝⎜

⎞⎠⎟

1 2

= 0.23 m s–1.

! ! For slow evolution of orbit use conservation of angular momentum r1v1 = r2v2 . r2 =r12

,

i.e. it is decreased.! ! For rapid change, apply conservation of energy and conservation of angular momentum

at the two extremes of the eccentric orbit. rmin =r13

.

16.!(a)! V0 x( ) = kx2 = 12mω 2x2 so that ω = k

m.

! (b)! − !

2

2m∂2ψ∂x2

+ 12mω 2x2ψ = Eψ

! ! En = n + 1

2⎛⎝⎜

⎞⎠⎟ !ω

! (c)! Displacement of equilibrium position is x0 =qEk

= qEmω 2

⎛⎝⎜

⎞⎠⎟ .

! (d)! V1 x( ) = −qEx

! (e)! After completing the square, the new potential energy can be written

! V x( ) = 12mω 2 x − qE

mω 2⎛⎝⎜

⎞⎠⎟2

− q2E2

2mω 2

! ! This can be interpreted as a harmonic oscillator with the same frequency as before,

oscillating about a new equilibrium position, and with a constant potential energy

superimposed.

! En = − q2E2

2mω 2 + n + 12

⎛⎝⎜

⎞⎠⎟ !ω

!

! (f)! Natural frequency is unchanged – one might be able to see this intuitively without

doing the maths in (e).

! (c)!

!


Name of setter Mariani Paper Question Name of module General ProblemsYear of exam 2014 PHY3053 16 contInitials of checker AU

V = 8W3π

1


Module Code PHY3065Name of module Quantum optics and photonicsDate of examination MAY/JUNE 2015

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

2.

3.

4.

5.

1. (i)

⇢ =

1

8

|0ih0| + 1

2

|1ih1| + 3

8

|2ih2|

Matrix elements of ⇢:

hm|⇢|ni =

1

8

�m0�n0 +

1

2

�m1�n1 +

1

8

�m2�n2.

Average number of photons detected is the expectation value of the number operator

ˆN = a†a in the state ⇢, which is

Tr(⇢ ˆN) =

X

n

hn|⇢ ˆN |ni =

X

n

hn|✓

1

8

|0ih0| ˆN +

1

2

|1ih1| ˆN +

3

8

|2ih2| ˆN

◆|ni =

1

2

+

3

8

2 =

5

4

.

(Alternatively, take average directly from the ensemble of number states: hNi =

180 +

121 +

382 =

54 ).

Expectation value of

ˆN2is

Tr(⇢ ˆN2) =

X

n

hn|✓

1

8

|0ih0| ˆN2+

1

2

|1ih1| ˆN2+

3

8

|2ih2| ˆN2

◆|ni =

1

2

+

3

8

2

2= 2.

(Alternatively, take average directly from the ensemble of number states: hN2i =

180

2+

121

2+

382

2= 2). Thus uncertainty �N

is given by

(�N)

2= 2 �

✓5

4

◆2

=

32 � 25

16

=

7

16

=) �N =

p7

4

.

(ii)Coherent state |↵i is defined by a|↵i = ↵|↵i.Expectation value of electric field in state |↵i is

h↵| Ê(�)|↵i = h↵|12

�a e�i�

+ a†ei��|↵i =

1

2

�↵ e�i�

+ ↵⇤ei��.

Take ↵ = |↵|ei✓:

h↵| Ê(�)|↵i =

1

2

|↵|⇣e�i(��✓)

+ ei(��✓)⌘

= |↵| cos(�� ✓).

The expectation value of (

Ê(�))

2in |↵i is

h↵|( Ê(�))

2|↵i = h↵|14

�a e�i�

+ a†ei��

a e�i�

+ a†ei��|↵i =

1

4

h↵|�aa e�2i�

+ aa†+ a†a + a†a†e2i�

�|↵i

=

1

4

h↵|⇣|↵|2e�2i(��✓)

+ 1 + 2a†a + |↵|2e2i(��✓)⌘|↵i =

1

4

|↵|2⇣e�2i(��✓)

+ 2 + e2i(��✓)⌘

+

1

4

=

1

2

|↵|2 cos[2(�� ✓)] +

1

2

|↵|2 +

1

4

= |↵|2 cos

2(�� ✓) +

1

4

.

Thus the uncertainty �E(�) is given by

(�E(�))

2= |↵|2 cos

2(�� ✓) +

1

4

� [|↵| cos(�� ✓)]2 =

1

4

.

2

DE=1

2

p2 p 2 p

c-q

1

-1

2

-2

Field

Required figure is a rough version of the following:

Inserting the over-completeness relation:

|ni =

1

⇡

Zd2↵ |↵ih↵|ni =

1

⇡

Zd2↵ exp

��|↵|2/2

�(↵⇤

)

n

pn!

|↵i

This expresses the number state as a superposition of coherent states.

(iii)Let

Â and

ˆB be arbitrary hermitian operators.

(

Â ˆB)

†=

ˆB†Â†

=

ˆB Â.

Thus (

Â ˆB)

†is not equal to

Â ˆB unless

Â ˆB =

ˆB Â.

2. (i)EPR state:

1p2

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

Basis vector |�i is a superposition

|�i = a|Hi + b|V i

h+|�i =

1

2

a +

p3

2

b = 0 =) a = �p

3 b

h�|�i = aa⇤+ bb⇤ = 1 =) |b|2 (3 + 1) = 1 =) |b| =

1

2

=) |�i = �p

3

2

|Hi +

1

2

|V i dropping an arbitrary overall phase factor.

Inverting the transformation:

hH|+i =

1

2

, hH|�i = �p

3

2

=) |Hi =

1

2

|+i �p

3

2

|�i

hV |+i =

p3

2

, hV |�i =

1

2

=) |V i =

p3

2

|+i +

1

2

|�i

3

The EPR above, in the +/� basis, is

1p2

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

=

1p2

" 1

2

|+i1 �p

3

2

|�i1

! 1

2

|+i2 �p

3

2

|�i2

!+

p3

2

|+i1 +

1

2

|�i1

! p3

2

|+i2 +

1

2

|�i2

!#

=

1p2

"1

4

|+i1 |+i2 �p

3

4

|+i1 |�i2 �p

3

4

|�i1 |+i2 +

3

4

|�i1 |�i2

+

3

4

|+i1 |+i2 +

p3

4

|+i1 |�i2 +

p3

4

|�i1 |+i2 +

1

4

|�i1 |�i2

#

=

1p2

(|+i1 |+i2 + |�i1 |�i2) .

The state is thus also an EPR state in the +/� basis. A measurement in the +/� basis yields + polarization with probability

1/2, or � polarization with probability 1/2.

(ii)Integrating i~ d| (t)i/dt =

ˆV (t)| (t)i gives

| (t)i = | (0)i � i

~

Zt

0dt0 ˆV (t0)| (t0)i.

To first order replace | (t0)i in integral with | (0)i:

| (t)i = | (0)i � i

~

Zt

0dt0 ˆV (t0)| (0)i.

| (0)i = |0, ei, where the first entry is the number state of the light mode.

| (t)i = |0, ei � i

~

Zt

0dt0 D

ha|eihg|ei(!0�!)t0

+ a†|gihe|e�i(!0�!)t0i|0, ei

[a|0, ei = 0] = |0, ei � i

~

Zt

0dt0 D

he�i(!0�!)t0 |1, gi

i= |0, ei � i

~De�i(!0�!)t � 1

�i(!0 � !)

|1, gi

= |0, ei +

D(e�i(!0�!)t � 1)

~(!0 � !)

|1, gi

Probability amplitude for atom to be in state |gi is

D(e�i(!0�!)t�1)~(!0�!) , so probability is

��D(e�i(!0�!)t � 1)

~(!0 � !)

��2

=

D2(e�i(!0�!)t � 1)(ei(!0�!)t � 1)

~2(!0 � !)

2=

D2(2 � e�i(!0�!)t � ei(!0�!)t

)

~2(!0 � !)

2

=

2D2(1 � cos[(!0 � !)t])

~2(!0 � !)

2=

4D2sin

2[(!0 � !)t/2]

~2(!0 � !)

2

For ! ! !0, probability is

D2t2

~2.

Probability cannot exceed 1, so this result cannot be valid for times t such that D2t2/(2~2) ⇡ 1, or t ⇡

p2~/D.

3. (i) a3 = Reikz1Ra1 + T eikz2T a1 + Reikz1T a2 + T eikz2Ra2

=

�eikz1R2

+ eikz2T 2�a1 +

�eikz1 + eikz2

�RT a2.

a4 = T eikz1Ra1 + Reikz2T a1 + T eikz1T a2 + Reikz2Ra2

=

�eikz2R2

+ eikz1T 2�a2 +

�eikz1 + eikz2

�RT a1.

4

a4

a3

a2

a1

z1

z2

From the sketch and the input/output relations, a beam in arm 1 completely exits at right angles in arm 3 if eikz1 + eikz2 = 0,

which is also the condition for a beam in arm 2 to exit completely at right angles in arm 4.

eikz1 + eikz2 = eikz1⇣1 + eik(z2�z1)

⌘= 0 =) eik(z2�z1)

= �1 =) k(z2 � z1) = (2n + 1)⇡, n = 0,±1,±2, . . .

=) z2 � z1 = (2n + 1)

⇡

k= (n + 1/2)�, n = 0,±1,±2, . . .

(ii)

[

ˆX, ˆY ] =

i

4

[a + a†,�a + a†] =

i

4

(1 + 1) =

i

2

[

ˆX✓

, ˆY✓

] =

i

4

[aei✓ + a†e�i✓,�aei✓ + a†e�i✓

] =

i

4

(1 + 1) =

i

2

a =

ˆX + i ˆY , a†=

ˆX � i ˆY

=) ˆX✓

=

1

2

h(

ˆX + i ˆY )ei✓ + (

ˆX � i ˆY )e�i✓

i=

ˆX cos ✓ � ˆY sin ✓

ˆY✓

= � i

2

h(

ˆX + i ˆY )ei✓ � (

ˆX � i ˆY )e�i✓

i=

ˆX sin ✓ +

ˆY cos ✓

Expectation values of quadratures:

h | ˆX| i =

1

4

(h2| + h4|)�a + a†�

(|2i + |4i) = 0, h | ˆY | i =

i

4

�h2| + h4|)(�a + a†�

(|2i + |4i) = 0.

Expectation values of squares of quadratures:

h | ˆX2| i =

1

8

(h4| + h6|)⇥a2

+ aa†+ a†a + (a†

)

2⇤(|4i + |6i)

=

1

8

⇥h4|(aa†

+ a†a)|4i + h4|a2|6i + h6|(aa†+ a†a)|6i + h6|(a†

)

2|4i⇤

=

1

8

h2(4) + 1 +

p(6)(5) + 2(6) + 1 +

p(6)(5)

i=

1

8

⇣22 + 2

p30

⌘

h | ˆY 2| i = �1

8

(h4| + h6|)⇥a2 � aa† � a†a + (a†

)

2⇤(|4i + |6i)

= �1

8

h�(2(4) + 1) +

p(6)(5) � (2(6) + 1) +

p(6)(5)

i=

1

8

⇣22 � 2

p30

⌘

5

Uncertainties in quadratures:

�X =

1

2

q11 +

p30, �Y =

1

2

q11 �

p30

�X > 1/2 so X is not squeezed. Y is squeezed if

�Y <1

2

=) (�Y )

2 <1

4

=) 11 �p

30 < 1 =) 10 <p

30 =) 100 < 30

which is not true so there is no quadrature squeezing in this state.

4. (i)In linear optics the polarization (electric dipole moment per unit volume) 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.

PNLi(r, t) = "0X

j,k,l

�(3)ijkl

Ej

(r, t)Ek

(r, t)El

(r, t)

Also acceptable:

PNL(r, t) = "0�(3)

[E(r, t)]2 E(r, t)

The linear polarization is PL = "0�(1)E, so

D = "0E + P = "0E + "0�(1)E + "0�

(3)E2E

Defining the effective permittivity by D = "0"e↵E we get

"e↵ = 1 + �(1)+ �(3)E2.

The effective refractive index is

ne↵ =

p"e↵ =

p1 + �(1)

+ �(3)E2=

qn2

0 + �(3)E2= n0

s

1 +

�(3)E2

n0⇡ n0 +

�(3)E2

2n0,

for �(3) ⌧ �(1).

(ii)Entanglement occurs when the quantum state of a system containing subsystems A and B cannot be written as a product

| iA

|�iB

of quantum states of the subsystems.

Completeness relation:

|HihH| + |V ihV | = I.

General operator

ˆO:

ˆO = a|HihH| + b|HihV | + c|V ihH| + d|V ihV |.

�x

= |HihV | + |V ihH|, �z

= |HihH|� |V ihV |,1p2

(�x

+ �z

) =

1p2

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

1p2

(I � i�y

) =

1p2

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

6

(�x

+ �z

)/p

2 on photon 1 in state gives

1p2

(|Hi1 1hH| + |Hi1 1hV | + |V i1 1hH|� |V i1 1hV |) 1

2

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

=

1

2

p2

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

=

1p2

(|Hi1|Hi2 � |Hi1|V i2) =

1p2

|Hi1 (|Hi2 � |V i2)

(I � i�y

)/p

2 on photon 1 in this state gives

1

2

(|Hi2 2hH| + |V i2 2hV |� |Hi2 2hV | + |V i2 2hH|) |Hi1 (|Hi2 � |V i2)

=

1

2

|Hi1 (|Hi2 + |V i2 � |V i2 + |Hi2) = |Hi1|Hi2

The final state, |Hi1|Hi2, is unentangled and is obtained from two unitary operators acting separately on each photon; the

initial state is thus given by the inverse operators acting on the final state:

1p2

(�x

+ �z

)

† |Hi1�

1p2

(I � i�y

)

† |Hi2�

which is a product state so initial state is also not entangled.

Alternatively one can explicitly write the initial state as the product state

1

2

(|Hi1 + |V i1) (|Hi2 � |V i2) .

PHY3068: Principles of Theoretical Physics

Hints and tips for May 2015 exam

1. Least action principle states that the actual trajectory of a mechanical system isa stationary point of the action functional. This stationary point can be foundby computing the variation of the action and setting it to zero to the first orderin variation. The Nother theorem states that any continuous symmetry of amechanical system is associated with a conserved quantity. Time uniformity(t → t + δt) results in energy conservation. For nonrelativistic systems, L =T − U , while H = T + U .

The similarity between the action and the free energy is obvious, the effectivemass being m = α, and the potential energy U = −β

4 (M2 − M2

0 )2 (note the

sign!). The differential equation is M ′′ = − dUdM = βM(M2 − M2

0 ), where theprime denotes the derivative with respect to x. The integration of this equationis facilitated by employing the “energy” conservation: H = α

2M′2 − β

4 (M2 −

M20 )

2. Far away (x = ±∞), H = 0; since H is x-independent, we find M ′ =

±!

β2α(M

2 −M20 ). Thus, the solution to the equation is given by

"

dM

M2 −M20

= ±

#

β

2αx .

There are several ways to compute the integral: one can expand it into partialfractions, or to use hyperbolic trigonometric substitution: M = M0 tanh u.Either way, the result is

"

dM

M2 −M20

=1

2M0log

M0 −M0

M0 +M= −

1

M0Arth

M

M0.

Therefore, the magnetisation distribution is M(x) = ±M0 tanh$

M0

!

β2αx

%

.

2. Spontaneous symmetry breaking occurs when the Lagrangian possesses certainsymmetry, but its individual minima (vacua) break the symmetry; this is pos-sible if the lowest-energy states are degenerate. Goldstone theorem states thatunder these conditions the excitation spectrum must include a massless (orgapless) mode. The most important difference between Goldstone bosons andHiggs bosons is that Higgs bosons occur in systems including gauge fields, i.e.,in systems of charged particles (or fields).

The Lagrangian is symmetric if b′ = b. This can be shown by several methods.First, one can notice that for b′ = b the potential energy is a function of φ2

1+φ22,

which is invariant under rotations. One can also analyse the behaviour of Lunder infinitesimal rotation: δφ1 = ϵφ2, δφ2 = −ϵδφ1, and show that the change

1

in L is proportional to (b−b′)ϵφ1φ2(φ21−φ2)2; this quantity vanishes if b = b′. Yet

another approach is to represent the quartic terms through the use of identititiesφ41 + φ4

2 =12

&

(φ21 + φ2

2)2 + (φ2

1 − φ22)

2'

φ21φ

22 =

14

&

(φ21 + φ2

2)2 − (φ2

1 − φ22)

2'

. One

can then notice that the combination (φ21−φ2

2) is not invariant under rotations,while φ2

1 + φ22 is.

After setting b = b′, we recover the standard Mexican-hat potential. Sponta-neous symmetry breaking occurs for a > 0, and φ2

1 + φ22 =

ab .

The Lagrangian can be rewritten in terms of φ and φ∗ as L = 12 |∂µφ|

2+ a2 |φ|

2−b4 |φ|

4. This is, essentially, the action of scalar electrodynamics, and rotationalsymmetry is nothing but the phase symmetry of this action. The interactionwith the gauge field can be incorporated through the covariant derivative ∂µφ →Dµφ ≡ ∂µφ+

iehAµφ. In a spontaneously broken state, one can replace φ by the

condensate value φ0, so that the mass of the vector field is M2A = e2φ2

0. Themass vanishes in the e = 0 limit: the vector field decouples from the condensate,and its longitudinal component, Aµ ∼ kµ becomes the new Goldstone mode. )

3. The rate of quantum tunneling is proportional to the exponential of imaginary-time action of the bouncing trajectory: Γ ∼ exp

(

− Sh

)

. The bouncing trajectorystarts at t = −∞ at the initial state of tunneling and returns there at t = ∞,reaching the exit point at some intermediate time.

The zero-point motion energy is the ground level of the quantum harmonic os-cillator, E0 =

12 hω. One can show that this energy is small compared to 1

2mω2x20

if x0 ≫!

hmω . This allows one to use E = 0 approximation in the analysis of

quantum tunneling. For E = 0, the turning points are x = 0 and x = ±x0.Thus, the bouncing trajectory should start at x = 0 at τ = −∞, should reachthe point x = x0, at some intermediate time (say, at τ = 0), and return to x = 0at τ = +∞. The imaginary-time action is

S ="

*

m

2x2 +

mω2

2x2

+

dτ ;

the corresponding quation of motion is x = ω2x. One can show that the equa-tion is solved by the trajectory of the form x(τ) = Ae−α|τ |, if α = ω. Thetrajectory reaches the exit point x = ±x0 if A = ±x0. The two terms in theaction give equal contributions, so that S = mωx2

0 (do not forget to integrateover positive and negative times), and the tunneling rate Γ is proportional to

the exponential: Γ ∼ exp(

−mω2x20

h

)

.

4. In Feynman’s approach, the contribution of an individual trajectory to the totaltransition amplitude is given by a complex exponential of the classical action

2

for that trajectory, so that the phase is α = S[r(t)]h . For a classical particle in

an external magnetic field the action takes the form

S[r(t)] ="

dt

*

mr2

2+ eA(r)r

+

.

The contribution of the second term to the phase is eh

,

Adr, as rdt = dr.Applying Stokes’s theorem is trivial, and curlA = B by the definition of A.

For a solenoid, the integral, after rewriting it in polar coordinates, simplifies tothe form

, Φ2πdθ; this gives the flux Φ for an arbitrary loop encircling the origin

once. This also implies there is no flux through any loop not encircling theorigin; hence there is no magnetic field outside the tube. The quantum phasefor a full counter-clockwise loop is eΦ

h . For the half-loop shown in Figure 1, thephase is one half of this value. If Φ = Φ0 =

he , one finds the quantum phase to

be φ1/2 = eh2h = π. Picking up a phase of π is equivalent to changing the sign

of the wave functions. This shows that the Aharonov-Bohm phase mimicks theeffects of Fermi statistics: Ψ(1, 2) = −Ψ(2, 1).

3

PHYM003


PHYSICS

May/June 2015

CONDENSED MATTER II

Hints and Tips

Question 1

a) In geometry, a square is a regular quadrilateral, which means that it has four equal sides and four equal angles (90-degree angles, or right angles). The simplest primate unit cell is a square

b) Write the coordinates of two sides of a quadrilateral with common origin

The Bloch theorem describes the physically acceptable solutions for the Schrodinger equation in a periodic potential.

The energy band width for the given energy dispersion corresponds to the difference between the maximum and minimum energy values that charge carriers can take in the band.

The expression for the effective mass tensor can be found in the lecture slides.

Calculate the number of states with wave vectors having a magnitude smaller than or equal to k for the two-dimensional lattice. Consider the density of states in k space and in energy.

Question 2

a-e) The expressions for each quantity can be found in the lecture slides.

The material is transparent for frequencies higher than the plasma frequency.

In the photoelectron spectroscopy one has to consider the kinetic energy of the particles.

The phase coherent transport in a hollow cylinder is the Sharvin & Sharvin experiment covered in the lecture slides.

Question 3

In the collisionless limit the scattering time is infinite and the longitudinal resistance goes to zero. The off diagonal terms are due to the Hall effect.

Resistivity is the inverse of conductivity. Calculate the inverse of a 3x3 matrix.

In the two band limit, the equivalent circuit is the parallel of two resistors. Hence the magneto-conductivity will simply be the sum of the individual magneto-conductivities of each band.

Question 4

Draw a thin slab and draw arrows in the plane of the slab. Consider if the magnetic field lines penetrate in the slab.

In the lecture slides there is a step by step guide on how to calculate the differential equation governing the magnetic flux density in a superconductor using the second London equation.

Solve the differential equation of the magnetic flux for a thin slab of thickness 2L. Remember to impose the right boundary conditions, such as the magnetic field flux is B outside the slab.

This is a type II superconductor since when a perpendicular magnetic field flux is applied to the plane of the slab, it is energetically more favourable for the system to let the field flux pass through vortices than to expel it.

PHYSICS HINTS AND TIPS

Module Code PHYM006

Name of module Relativity and Cosmology


Question 1 The first part is a definition of an inertial frame, which is given several times in the lectures. The question on four velocity and four momentum is related to material in lectures 4 and 5. Now we are asked to use the invariant quantity given to get the relativistic energy. This is given in the lectures but I’ll summarise here: ηµνU

µUν = c2 Now$multiply$though$by$m2 $to$get$$

m2ηµνUµUν =ηµν p

µ pν = m2c2 $$Now$we$know$

ηµν pµ pν = E2 / c2 − px

2 − py2 − pz

2 = E2 / c2 − p2 = m2c2 $So$finally$ E2 = p2c2 +m2c4 and$

E = p2c2 +m2c4 $$Now$we$move$to$the$“unseen”$question$about$the$electron$and$massive$particle$colliding.$$This$is$directly$related$to$the$questions$you’ve$just$done.$Think$first$about$the$energies$of$the$particles.$Before$the$collision$the$electron$has$energy$Ee = mec

2 $$since$it$is$at$rest.$The$massive$particle$has$energy$EM = γ Mc2 .$After$the$collision$the$electron$gains$energy$T $$and$the$massive$particle$loses$the$same$amount.$$So$we$can$then$use$the$equation$above,$rearranged$for$momentum$to$get$the$momenta$after$the$collision:$

pe =(T +mec

2 )2

c2−me

2c2⎡

⎣⎢

⎤

⎦⎥

1/2

$and$

pM = (γ Mc2 −T )2

c2−M 2c2⎡

⎣⎢

⎤

⎦⎥

1/2

$$

Now$the$momentum$of$the$massive$particle$before$the$collision$is$ pM' = [γ 2M 2c2 −M 2c2 ]1/2 $$and$the$

collision$is$headEon.$So$we$set$the$momentum$before$the$collision$to$the$momentum$after$the$collision:$ pM

' = pe + pM $.$You$then$need$to$do$some$algebra$to$get$an$expression$for$T:$

T = 2meM2 (γ 2 −1)c2

me2 +M 2 + 2γ meM

$

But$noting$that$me << M $$and$γ 2 −1= γ 2v2

c2$$we$get$T = 2meγ

2v2 $$as$required.$The$last$part$of$the$

question$is$straight$from$the$lectures$and$is$absolutely$basic$knowledge$that$you$must$learn.$$ $

Question)2)$Remember$that$a$proper$time$interval$is$the$elapsed$time$between$two$events$as$measured$by$a$clock$that$passes$through$both$events.$The$second$part$of$the$question$is$to$reproduce$the$Schwarzschild$metric,$direct$from$the$lectures.$The$last$stable$orbit$derivation$is$given$in$lecture$$16.$It$is$the$last$part$of$the$question$that$is$“unseen”.$The$four$velocity$in$Schwarzschild$coordinates$is$

[U µ ]= c dtdτ, drdτ, dθdτ, dφdτ

⎛⎝⎜

⎞⎠⎟ $

Now$we$have$the$invariant$quantity$$ gµνU

µUν = c2 Now$we$can$write$

c2 = c2 1− 2GMc2r

⎛⎝⎜

⎞⎠⎟

dtdτ

⎛⎝⎜

⎞⎠⎟2

− 1− 2GMc2r

⎛⎝⎜

⎞⎠⎟−1 dr

dτ⎛⎝⎜

⎞⎠⎟2

− r2 dθdτ

⎛⎝⎜

⎞⎠⎟2

− r2 sin2θ dφdτ

⎛⎝⎜

⎞⎠⎟2

$$$

$Noting$that$for$a$circular$orbit$in$the$π / 2 $plane$we$have$ sin2θ = 1$,$ dθ / dτ = 0 $and$dr / dτ = 0 $$we$get$the$expression$in$the$question.$Now$for$a$circular$orbit$we$can$write$

dφdt

⎛⎝⎜

⎞⎠⎟2

= GMr3

= Ω2 $$

which$is$Kepler’s$third$law$(this$was$part$of$the$Problem$Sheet$4).$Noting$that$$dφdτ

= dφdt

dtdτ

$$we$get$

c2 = c2 1− 2GMc2r

⎛⎝⎜

⎞⎠⎟

dtdτ

⎛⎝⎜

⎞⎠⎟2

− r2Ω2 dtdτ

⎛⎝⎜

⎞⎠⎟2

= c2 1− 3GMc2r

⎛⎝⎜

⎞⎠⎟

dtdτ

⎛⎝⎜

⎞⎠⎟2

$

Now$we$just$use$ r = 7GMc2

$$and$solve$for$dt / dτ $$which$tells$us$that$P∞ / Pship = 7 / 4 $.$

$ $

Question)3)$This$question$is$a$collation$of$questions$asked$in$the$Problem$Sheets.$$For$completeness$the$nonEzero$Christoffel$symbols$are$ Γθφ

φ = Γφθφ = cotθ

Γφφθ = −sinθ cosθ $$

The$geodesic$equations$are$therefore$d 2θdτ 2

− sinθ cosθ dφdτ

⎛⎝⎜

⎞⎠⎟2

= 0 $$

and$d 2φdτ 2

+ 2cotθ dθdτ

dφdτ

= 0 $ $

and$by$choosing$θ = π / 2 $as$we$are$free$to$do$we$can$show$that$φ = τ / R (a$great$circle)$is$the$solution$to$both$geodesic$equations.$Now$use$the$equation$for$the$Ricci$tensor$components$from$the$Christoffel$symbols$given$as$a$hint$in$the$question$to$get R00 = 1 $$and$ R11 = sin

2θ $.$Now$calculate$the$Ricci$tensor:$

R = gµνRµν =1R2.1+ 1

R2 sin2θsin2θ = 2

R2

$Thus$demonstrating$that$the$2Esphere$has$constant$positive$curvature.$$ $

Question)4))Remember$that$the$Cosmological$Principle$is$that$when$viewed$on$sufficiently$large$scales$the$Universe$is$both$homogeneous$and$isotropic.$The$answers$to$the$first$we$parts$of$this$question$are$from$Lectures$18,$19$and$20.$$Remember$the$EinstenEde$Sitter$model$is$flat$( k = 0 $)$and$has$no$radiation$or$dark$energy$so$the$first$Friedmann$equation$gives$us$

1RdRdt

⎡⎣⎢

⎤⎦⎥

2

= 8πG3

ρm,0

(R / R0 )3 $$

so$

dRdt

= 8πG3

ρm,0R03/2

R1/2

So$$

dRR1/2

∝ dt⎯→⎯ R ∝ t 2/3

To$get$the$fluid$equation$you$should$differentiate$the$first$Friedmann$equation$with$respect$to$time$and$then$substitute$the$second$Friedmann$equation$and$then$the$first$Friedmann$equation$into$the$result.$A$bit$of$simple$algebra$will$then$lead$you$to$the$fluid$equation.$(Check$Lambourne$if$you$don’t$manage$it).$$For$the$final$“unseen”$question$we$start$with$the$fluid$equation$and$substitute$for$ p $$with$the$equation$of$state$given,$which$should$give$you$$

dρdt

+ 3γρR

dRdt

= 0 Now$set$this$up$in$a$similar$form$

1ρdρdt

= − 3γRdRdt

Which$you$can$use$to$show$ ρ ∝ R−3γ $$Now$we$use$the$first$Friemann$equation$to$give$

1RdRdt

⎡⎣⎢

⎤⎦⎥

2

= 8πG3

ρ0RR0

⎛⎝⎜

⎞⎠⎟

−3γ

$$

so$

dRdt

= 8πG3

ρ0RR0

⎛⎝⎜

⎞⎠⎟

− 32γ +1

$

from$which$you$can$show$that$ R ∝ t23γ $.$Now$you$should$write$ρ(t)∝ t

23γ⎛

⎝⎜⎞

⎠⎟

−3γ

= t −2 $.$

$

1 PHYM007

PHYM007


PHYSICS

MAY 2015

ULTRAFAST PHYSICS

HINTS and TIPS

2 PHYM007

1. Sketch a diagram of a time-resolved scanning Kerr microscopy (TRSKM) setup

with a simple optical bridge detector for imaging the transient magnetisation

configurations of thin film samples deposited on opaque substrates. Your sketch

must include all relevant optical beam paths and electrical cable connections. [4]

On your sketch indicate clearly:

(a) the ultrafast laser; [1]

(b) the focusing and polarisation optics; [1, 1]

(c) the delay line; [1]

(d) the non-polarising beam splitter; [1]

(e) the optical bridge detector; [1]

(f) the sample; [1]

(g) the directions of the bias and ac magnetic fields; [2]

(h) the pulse or microwave generator; [1]

(i) the clock; [1]

(j) the oscilloscope. [1]

List factors limiting the spatial and temporal resolutions of the measurements. [5]

Give one example each of dynamical magnetic phenomena that could and could not

be studied using this experimental setup. Justify your answers. [4]

Sketch an optical bridge detector that is capable of vectorial measurements of the

magnetisation in the sample under study, naming its essential components. [2]

With the aid of a sketch, explain how the electrical outputs within the detector above

need to be combined in order to retrieve signals proportional to the three orthogonal

components of the magnetisation. [7]

SEE LECTURE NOTES

Turn Over

3 PHYM007

2. (i) Using a clearly labelled sketch, explain why four-level rather than two-level gain

media are used within lasers. [7]

Using a sketch, explain the relationship between the laser gain bandwidth, the cavity

mode structure and the spectrum of the laser output. [6]

State the mathematical relationship between the bandwidth of the pulsed laser

output and the minimum duration of the emitted optical pulses. [2]

(ii) Describe the relevance of Raman scattering to the optical excitation of coherent

phonons in solids and the conditions under which this excitation mechanism could

be expected to be active. [4]

Name the two main mechanisms of excitation of coherent optical phonons and

describe how the specific mechanism could be identified from the time dependence

of a measured reflectivity signal. [2, 4]

Sketch a reflectivity signal that contains both coherent and incoherent phonon

contributions. Label clearly the coherent phonon contribution and that due to

thermalized incoherent phonons and electrons. [2, 2]

Ferromagnetic and ferroelectric materials share many similarities in terms of their

respective magnetic and electric polarisation. However, in contrast to ferromagnets,

rapid heating of some ferroelectric materials above their Curie temperature is

accompanied by efficient excitation of coherent phonons. Explain why. [5]

SEE LECTURE NOTES AND RELEVANT SELF-STUDY MATERIAL

Turn Over

4 PHYM007

3. (i) Explain the physical origin of the specular inverse Faraday effect (SIFE) and

specular optical Kerr effect (SOKE). [2, 2]

Hence, describe in words how the strength of each effect depends on the polarisation

of the pump and probe beams. [2, 2]

Sketch on the same graph the intensity envelope of an ultrashort optical pump pulse

and the temporal dependence of the pump-induced SOKE signal for the cases of

short and long characteristic lifetimes (compared to the pump pulse duration) of the

pump induced linear polarisation of the medium. [3, 3]

Under which conditions can the effective magnetic field resulting from the SIFE

lead to impulsive excitation of magnetisation precession? Justify your answer. [3]

(ii) Describe in words the effect of a femtosecond optical pump pulse on the electron

distribution in a metallic sample. [3]

The first of the three-temperature model equations in one dimension is

� � � � � �xtxT

xTTTTG

tT

C ,ee

esephee

e 4�¸¹

·¨©

§ww

ww

�� ww

NE ,

where Te, Ce and κe are the electron temperature, heat capacity and thermal

conductivity, respectively, Tph and Ts are the lattice and spin temperatures,

respectively, G and β are the electron-phonon and electron-spin coupling constants,

respectively, t is time and x is the coordinate.

Explain the physical meaning of the Θe(t, x) term and describe its typical time and

spatial dependences. [4]

Write the other two equations of the three-temperature model, naming all variables

and material parameters in the equations. [6]

Question continues overleaf

Turn Over

5 PHYM007

Sketch a typical time-resolved reflectivity signal acquired from gold in the strong

excitation regime, identifying the characteristic feature of the signal that

distinguishes it from a signal acquired in weak excitation regime. [4]

SEE LECTURE NOTES

Turn Over

6 PHYM007

4. (i) State the main assumption underpinning the Landau-Lifshitz (LL) equation that is

violated in the process of ultrafast demagnetisation. [2]

Write the Landau-Lifshitz-Bar'yakhtar (LLBar) equation of motion of the

magnetisation, identifying and explaining the physical meaning of the additional (as

compared to the LL equation) contributions to the effective magnetic field and

relaxation terms. [10]

Consider a ferromagnetic material with a simple cubic lattice the magnetic

properties of which are described by the Heisenberg model for spins S coupled by

the exchange integral J. Show that the dispersion relation, ω(k), of exchange

magnons propagating along the [100] direction is

� �� kaJSk cos14)( � Z! ,

where a is the lattice constant and k is the wave number in the [100] direction. [10]

From the dispersion relation above, derive an approximate dispersion relation for

long-wavelength exchange spin waves. [5]

In what respect, could the waves also be referred to as short-wavelength waves? [2]

Comparing your result with the corresponding expression derived from the LL

equation ( 2)( kMk DJZ , where γ is the gyromagnetic ratio, M is the saturation

magnetisation and α is the continuous-medium exchange parameter of the magnetic

material), identify a mathematical expression for α in terms of J, a, the g-factor and

the Bohr magneton. [3]

(ii) Sketch the typical ray path through the boundary between vacuum and a negative

refractive index meta-material. [2]

SEE LECTURE NOTES

1


Module Code PHYM008

Name of module Physical Methods in Biology and Medicine Date of examination May/June 2015

1. (a) show output as a convolution of input and impulse response (integrals), define linearity and

shift-inv with eqqations as given in resources on ELE; (b) in Fourier space convolution is a multiplication; (c) the Fourier transform of a pulse is the sinc function; sketch can be found in notes; (d) construct output by graphically overlaying/adding the impulse response at the location of the delta-functions; (e) without need for actual integrations construct the response from estimating the overlap area by geometrical arguments as we have demonstrated in the lectures; basically results in linear rise – plateau – linear fall. (f) o(t) = f(t) * h(t) + n with n a noise term

2. (a) FRET near field dipole-dipole interaction + criteria for strong FRET (see notes); example

applications inc conformational protein studies & FRET based genetic sensors; (b) show 6th power dependence and explain meaning of Foerster radius; (c) for overlap integral see notes and resources on ELE; (d) i. 61.7 deg; ii. 97 nm; (e) e.g. TIRF illuminated FRET experiment; adv: high z-resolution & contrast; disadv: signals may be small; limited to interface.

3. (a) explain Stokes shift with help of Jablonski diagram; (b) possible choices ex/dic/em: dye1

H(or G)/E/D; dye2 A(or C or D)/G/I; absolutely no crosstalk between exciter and emitter; (c) use control experiments where only dye1 or dye2 are present; measure how much signal of dye1 spills into channel2 and vice versa; (d) rewrite equation as matrix equation; d1 and d2 can be solved if matrix invertible (“unmixing”); (e) situations where matrix may be near singular, low S/N ratio.

4. (a) use convolution and Fourier convolution theorem to show how image originates from

convolution with PSF; this implies (low-pass)filtering; criteria how to obtain formulae from notes, e.g. Rayleigh crit etc; (b) e.g. NA =1.4, n_oil=1.51; lat 132 nm; ax 435 nm; resolve e.g. cells, nucleus, mitochondrial outlines/shape (c) localization of single emitters/molecules; switch molecules on one-by-one; assemble composite image with higher res; (d) localization accuracy: e.g. sigma_STORM = sigma_widefield / sqrt(N_photons); spatial sampling = marker density

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