PHYSICS EXAMINATION PROBLEMS SOLUTIONS AND HINTS …newton.ex.ac.uk/teaching/resources/past-exams/SolnsHints_2013-14... · PHYSICS EXAMINATION PROBLEMS SOLUTIONS AND HINTS FOR STUDENT

1

PHYSICS EXAMINATION PROBLEMS SOLUTIONS AND HINTS FOR STUDENT SELF-STUDY

Module Code PHY1024

Name of module Properties of Matter

Date of examination May / June 2014

1. (i) The pressure increases as V decreases until

the dew-point is reached and condensate forms. As V is reduced further more molecules enter the liquid state until the sample is all liquid. At this point the pressure starts to rise again at a rate governed by the compressibility.

ddv

fv(v)⎛

⎝⎜

⎞

⎠⎟vp= 0 ⇒ Mv2 = 2kBT ⇒ vp = 2kBT M

U = 32nRT = 3

2× 0.75mol×8.314JK−1mol−2 × 350K=3.3kJ

M = 2kBT

vp2 = 2 ×1.38×10

−23 JK−1 × 350K

390ms−1( )2= 64 ×10−27 kg

2 (a) Y = F⊥ A0Δl l0

=60MPa0.04%

=150GPa

(b) Δlmax = 0.09%×250mm = 225 µm

(c) πd2 4( )× 60MPa = 7.3kN ⇒ d = 4 × 7.3kN π × 60MPa( ) =12.4mm

(d) Eel Δl( ) = F x( )0Δl∫ dx = YA

l0

x2

2

#

$%%

&

'((0

Δl

=AYΔl2

2l0

(e) Pmax =max(stress× area)× rateof extension

= 90MPa ×π 12.4mm( )2

4× 250mm × 0.09%

3× 60s=14mW

P

V

T < TC

vapour liquid

2-phase region

2


Module Code PHY1024

Name of module Properties of Matter

Date of examination May / June 2014

3. (i) ddrUBM r( ) = 0 ⇒ e2

4πε0r02 =12 B

r013 ⇒ B = e2r0

11

48πε0

F r( ) = ddrUBM r( ) = −e2

4πε0r2 +

e2r011

4πε0r13 =

−e2

4πε0r2 1− r0

11

r11⎛

⎝⎜⎞

⎠⎟

4. (i)

(ii) I = m1m2m1 +m2

r02 = 2.326 ×10

−26 kg × 2.657×10−26 kg2.326 ×10−26 kg + 2.657×10−26 kg

0.115nm( )2

=1.64 ×10−46 kgm2

Erot l =1( )− Erot l = 0( ) = !2

2I1× 2 − 0 ×1( ) =

1.054 ×10−34 Js( )21.64 ×10−46 kgm2

= 6.77×10−23 J

ab

120°

(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)! De Moivre: r cosθ + isinθ( )⎡⎣ ⎤⎦n= rn cos nθ( ) + isin nθ( )( )

! ! sin 3θ( ) = im cosθ + isinθ( )3( ) . Expand the right hand side and select the imaginary

part.

! (ii)! −i3 = exp − iπ6

⎛⎝⎜

⎞⎠⎟ , exp

iπ2

⎛⎝⎜

⎞⎠⎟ , exp − 5iπ

6⎛⎝⎜

⎞⎠⎟

! (iii)!cosα + isinα = exp iα( )

−cosα + isinα = −exp −iα( )! ! For the last part, write the left hand side in terms of complex exponentials using the

above results, then take a common factor of exp iα 2( ) out the numerator, and a com-

mon factor of exp −iα 2( ) out of the denominator leaving behind something that is

proportional to cot α 2( ) . Then tidy up what is left, writing the i which appears in the denominator as a complex exponential.

2.! (i)! For the chain rule, see your lecture notes.

! ! ∂T∂ϕ ρ

= 3ρ 3 cosϕ sinϕ sinϕ − cosϕ( )

! (ii)! Use circular polars, for the ρ integral use the substitution u = ρ 2 a2 (or recognise the

form of the integral – it is one that we have discussed several times in lectures). Mass

is πm0a2 1− 1

e⎛⎝⎜

⎞⎠⎟ .

3.! !

! !

! ! Surface integral has to be calculated in six parts, one for each face. From AOCB there

is no contribution; from FEDG the contribution is 1/3; from ABGF it is 3/2; from

OCDE it is 1/2; form AOEF no contribution; and from BCDG, -1/2. Total 11/6. For

the volume integral ∇⋅A = 2 + x2 − 2xz , and it is then a simple 3D integral which

confirms the value 11/6.

! ! 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.! ! a0 =8π 2

3, an =

4n2, bn = − 4π

n

SOLUTION TO DEGREE EXAMINATION QUESTIONfor Coordinator

Name of setter Usher Paper Question

Name of module Mathematics for Physicists

Year of examina-tion

2014 PHY1026 3

Initials of checker

5.! (i)! x = exp − t2

⎛⎝⎜

⎞⎠⎟ Acos 7 t

2⎛⎝⎜

⎞⎠⎟+ Bsin 7 t

2⎛⎝⎜

⎞⎠⎟

⎛

⎝⎜⎞

⎠⎟

! (ii)! Use the substitution provided to produce a differential equation for v in terms of x (eliminating y completely). This equation is

! ! ! dvdx

− 2vx= −2 .

! ! This can be solved using either the homogeneous or the integrating factor method –

they are equally straightforward. After converting back, y = 12x + cx2

.


Module Code PHY2023

Name of module Thermal Physics

Date of examination May 2014

1. (i) .0universereversible =S∆ (ii) Melting: 15.273== constT K; 2446ice ==⇒==

TmL

SmLSTQ ff ∆∆ JK-1.

239215.27315.363ln42002lnd

1

2water

2

1

=××=== ∫ TTmc

TTmcS

T

T

∆ JK-1.

(iii) Monoatomic ideal gas: RTU ν23

= and RTPV ν= : ( ) ( )112212 23

23 VPVPTTRU −=−=⇒ ν∆ .

(iv) Adiabatic: .ddddd;dd0 22 TCVVAPV

VATCU

VATCUUVPQ VVV −=+⇒+=⇒−=+==d

UseBV

RTVAP

−=+ 2 ⇒ ( )[ ] ( ) .lnlndd

21 const

VAPBVRTBV

TT

RC

BVV

VV CR

CRV =+−=−⇒−=

−+

Thus, ( ) 22 constBVVAP =−+ γ , where

V

V

CCR +

=γ .

2. (i) Bookwork (see notes). (ii) Refrigerator: .CH

CCR QQ

QWQ

−==η Ideal (Carnot): ⇒

−=

CH

CR TT

TCarnotη

.1+

=R

RHC TTηη

Initially, K270=oldCT and K300=old

HT .930/270 ==⇒ Rη

K27919

9)10300(1

=+

×+=+

=R

RnewH

newC TT

ηη

or 8.25°C.

(iii) ( ) ( ) .2

00 212111

TTTTTCTTCQQ ffVfV+

=⇒=−+−⇒=+ ∆∆

( ).0

41lnlndd

21

221

21

2

21

≥−

+==+= ∫∫ TTTTC

TTT

CTTC

TTCS V

fV

T

TV

T

TV

ff

∆

(iv) For maximum possible work extraction, BAfBA

fBAtotal TTT

TTT

CSSS =⇒==+⇒= 0ln02

∆∆∆ .

3. (i) TSUF −= . WFTSVPF d−=⇒−−= dddd for constT = .

(ii) B

E kωθ !

= . Using ∑∞

=

−=0 B

expi

i

TkZ

e and the hint yields:

)2exp()2exp(1

EE TTZ

θθ −−= .

The factor ‘3’ is due to the three spatial directions of oscillations.V

V TUC∂∂

= ; differentiating leads to the

required result. For ( ) ( ) 0Texp31Texp, E

2E

BEE →−≈⇒>><< θθ

θθT

NkCT V .

For ( ) BE

EE 31Texp, NkCT

T V ≈⇒+≈>>θ

θθ – the classical limit (Dulong-Petit law).

4. (i) Ω= lnBkS , see course notes. For the most probable macrostate: ( ) ( )!2!2!NN

Nmp =Ω .

For ,1>>N using Stirling’s formula yields: .2lnln Nmp →Ω Thus, ( ) ( ) .2ln21

212 BNkNSNSmp ==

(ii) The total number of Fermions nppLkkLNp

!!

pp

pp

=⇒== ∫ F2F2

2

0

2 F

d22

4 , where 2LNn = .

npEE !αpαα ==⇒= FFp .

SOLUTIONS AND HINTS 1 PHY2024

PHY2024

UNIVERSITY OF EXETER

PHYSICS

MAY / JUNE 2014

CONDENSED MATTER I

SOLUTIONS AND HINTS


1. Consider a linear monatomic chain that consists of atoms of mass m, with nearest

neighbour separation a and force constant C.

Considering the chain as a crystal structure, sketch its direct space and reciprocal

space lattices and label clearly the lattice constants in each lattice. [4]

HINT: A reciprocal lattice of a "linear monotmic chain" real space lattice is

again a linear chain with lattice constant 2π/a, where a is the lattice constant of

the real space lattice

Assuming harmonic approximation, use the plane wave ansatz to show that the

dispersion relation, kZ , of the longitudinal vibrational waves, with frequency ω

and wave number k, propagating along the chain is

¸¹·

¨©§ 2

sin2 kamCkZ . [5]

HINT: See the lecture notes

Define the first Brillouin zone and use the dispersion relation above to show that

the vibrational waves with wave numbers at the Brillouin zone edge cannot

propagate along the chain. [2, 4]

HINT: "The first Brillouin zone is the Wigner-Zeitz unit cell in the reciprocal

space" is the definition, although you could also define the BZ using a sketch

(as in the lecture notes) with all notations corresponding to the case of a

monatomic chain.

For the second part, one needed to calculate the group velocity and to show that

it is equal to zero at the BZ boundary (both as in the lecture notes)


Use this dispersion relation to derive the density of states, ZD , as a function of

frequency, ω, of the longitudinal vibrational waves in the chain, consisting of N unit

cells. [5]


Based on the known results for dispersion of longitudinal vibrational waves in

monatomic and diatomic chains, deduce the longitudinal sound velocity in a

material with lattice constant a and three different atoms of mass m1, m2 and m3 in

the basis, all coupled with the same force constant C. [5]

HINT: To begin with, one had to appreciate that the sound velocity does not

depend on the microscopic details but is expressed in terms of the

macroscopically defined mass density and appropriate elastic moduli (e.g.

Young's modulus in case of sound propagation along a thin long rod of the

material).

Then, the simplest way to solve this question was to replace each of the three

masses with their average value while using the same force constant and

distance between atoms and then to use the equation given above for this

monatomic chain.

Alternatively, one could assume that the entire mass of the three atoms was

concentrated in one and the same point. Note however that a three times

smaller force constant had to be used in this case, so that the same force

induced the same total displacement.

Finally, one could deduce the necessary expression by induction, using the

expressions derived for monatomic and diatomic chains in the lecture notes


2. (i) Define the Fermi surface and explain why it has spherical shape in the free electron

model. [2, 2]


Sketch the effect of uniform electric field upon the Fermi sphere in the free electron

model and explain how the details of the sketch are related to the electron scattering

time. [2, 2]


Consider a cube-shaped sample with side L. The sample is made of a material with

one valence electron per simple cubic unit cell. The lattice constant is a (a << L).

The temperature of the sample is F01.0 TT , where TF is the Fermi temperature of

the sample.

(a) Calculate the total number of electrons that are assumed to contribute to

electrical transport in the sample within the Drude theory. [1]

HINT: (L/a)3

(b) Calculate the total number of electronic states in the conduction band of the

sample. [2]

HINT: 2(L/a)3

(c) Calculate the Fermi wave number. [2]


(d) Estimate the total number of electrons in the sample that are assumed to

contribute to thermal transport within the Sommerfeld theory. [5]

HINT: Firstly, one had to note that only electrons near the Fermi surface


contribute. Secondly, recalling the lecture notes, one had to times the total

number of valence electrons in the sample by factor of T/TF

(e) Express the effective electron mass in terms of fundamental constants, the

lattice constant and the Fermi temperature of the sample. [3]

HINT: One need to combine the answer from (c) above with the free electron

dispersion, and then to appropriately rearrange the result

(ii) Use a sketch to show that every phonon-phonon umklapp scattering event results in

the reversal of the energy transport direction. [4]



3. (i) Consider a square Brillouin zone to describe a two-dimensional electron gas. Show

that the kinetic energy of a free electron with its k-vector at the corner of the

Brillouin zone is a factor of two greater than that of an electron with its k-vector in

the middle of the Brillouin zone edge. [4]

HINT: For free electrons, the kinetic energy is equal to the total energy and

scales quadratically with the linear momentum, i.e. k-vector. The length of the

k-vector "at the corner of the Brillouin zone" is related to that of the k-vector

"in the middle of the Brillouin zone edge" by the Pythagoras theorem

(ii) The one-dimensional electronic dispersion relation, kE , near the Brillouin zone

edge could be approximated as

¸¸¹

·¨¨©

§¸¹·

¨©§rr¸

¹·

¨©§

222222

2221

222)( g

mUmkUg

mkE !!!

,

where E is the energy of an electron with wave number k relative to the Brillouin

zone edge, g is the reciprocal lattice constant, m is the electron mass, ħ is the Planck

constant and U is a parameter proportional to the depth of modulation of the

periodic potential. The Fermi level lies in the band gap.

(a) What is the width of the band gap in the electronic spectrum? [2]

HINT: 2U

(b) What are the effective masses of electrons and holes in the system? [4]

HINT: One had either to differentiate twice the expression wrt k or simply to

compare it with a standard free electron parabolic dispersion

(c) What is the real space periodicity of the electron potential giving rise to this

dispersion? [2]

HINT: 2π/g


(d) What is the mobile charge carrier density in the sample at zero temperature? [2]

HINT: Zero, since the Fermi level is in the band gap

(iii) Sketch the temperature dependence of the electronic contribution to the thermal

conductivity of a metal. [4]

On your sketch, indicate the Debye temperature and the functional form of the

temperature dependence of the electronic thermal conductivity in the low and high

temperature regimes. [3]

Explain the shape of the sketched curve in terms of the temperature dependence of

the electron scattering time and the number of thermally excited electrons. [4]



4. The average mark (out of 25) was 10.5 and the highest mark was 24.

(i) What happens to the conductivity of an intrinsic group IV semiconductor when it is

doped with equal numbers of donor and acceptor impurities? [2]

HINT: Nothing as the effects cancel, as correctly stated by many

State the functional relation between equilibrium densities of electrons and holes

dictated by the law of mass-action? [2]


Explain why relatively small concentrations of impurities have a drastic effect on

the conductivity of semiconductors. [4]


(ii) Consider a pn-junction between n- and p-type semiconductors with doping densities

of nD and nA, respectively.

Calculate the thicknesses of the positively and negatively charged regions at the

junction at zero bias voltage if the total thickness of the depletion layer is d. [5]

HINT: The answer is given by the solution of the system of two equations:

1. The total thickness is d. 2. The total charge is zero.

Sketch the dependence of the total charge density, electric field and voltage across

the junction at zero bias. [6]



Sketch the electron energy bands across the junction

(a) in forward bias, [2]

(b) in reverse bias. [2]

Sketch the total electro-chemical potential in both case (a) and (b). [1, 1]



Module Code PHY2030Name of module Observing the UniverseDate of examination May/June 2014

1. proper motion 0.05 arcsec/yr , parallax 0.05 arcsec distance 20 pc , transverse velocity 4.74 km/s

measured (relative) parallax = true parallax of target - true parallax of reference measured parallax = 1/20 - 1/500 = 0.048 measured distance = 20.83 pc (overestimate by 0.83 pc)

GAIA will use absolute parallaxes (bookwork)

CMD: de-redden the MS, then find distance modulus, convert to distance distance = 1 kpc

approx 3 mag difference between tip of upper MS and Sun luminosity is approx 16 solar luminosities L ~ M^3.5 , so T_MS ~ L/M ~ M^-2.5 MS lifetime approx 0.14 solar MS lifetimes (approx 1.4 Gyr)

2. eccentricity > 0 (not sinusoidal) symmetry: peri- and ap-astron orbital velocities oriented along line of sight at extrema, subject to sin(i) projection Vp sin (i) / Va sin(i) = Vp / Va = (1+e)/(1-e) = 180/25 re-arrange to find e = 0.75

sketch: bookwork photometric monitoring for transits to test whether inclination is approx 90 degrees.

semiamplitude K1 approx 102.5 km/s (measure from plot)

mass function equation, with appropriate approximation to find secondary mass approx 2.1 MJ

KIII to find a approx 0.95 a.u.

max separation approx a(1+e) = 1.66 a.u. at 46 pc distance, this is 0.036 arcseconds Rayleigh criterion, min diameter 3.85 m longer wavelength PRO star/companion flux contrast ratio smaller CON larger telescope required

3. specific intensity, Planck function, flux density: bookwork Bnu(T) = 3.14 x 10^-16 W m^-2 Hz^-1 sr^-1 RJ approx: not valid (require hnu/kT << 1) mass calculation (bookwork) approx 0.5 solar masses core mean density approx 3.8 x 10^-16 kg m^-3 free-fall time approx 0.1 Myr

4. adaptive optics / interferometry: bookwork J mag approx 26.8

with background approx 21.3 (approx 156 times brighter)

higher resolution --> reduce aperture --> reduce sky background count

1


Module Code PHY2031

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

Date of examination May 2014

2.

4. Number of longitudinal HeNe modes lasing = 3

1 PHY3052


Module Code PHY3052

Name of module NUCLEAR & HIGH ENERGY PHYSICS

Date of examination June 2014

1. (i) (a) 6- o /0 + e- + ve (d) /0 + p o + e+ + ve

(b) W o S- + S0 + vW (e) ve + p o n + e+

(c) S + e- o S+ ve (f) W o P + vP + vW

Weak interaction

Lorentz factor = J = 14.5 / 1.784 = 8.13

From this, speed = 0.992 c

Length travelled = 0.992 × c × J × W = 6.29 × 10-2 cm

(ii) (a) 1/2, - (b) 3/2, + (c) 0, + (d) eQo

-2517

8177 e )( O N

(e) eQo e)( Cl Ar 2337

173718

(f) eQo -4821

4820 e)(4 *Sc Ca

Bookwork

2. (i) 3H o 3He + e- + ve Bookwork

(ii) Bookwork is a magic nucleus. This means product state has high BE, so that reaction will give out greater than expected energy. Released as KE of alpha particle due to light mass.

Q – mvc2

W

vW vP

P

W

2 PHY3052

vD = 0.0518 c = 1.55 x 107 m/s

Aparent = 210, so R= 9.09 fm

Zdaughter = 82

24 8Zc mc RZG v c

SD D = 145.2 – 81.10 = 64.10

O = 1.23x 10-7 s-1

t1/2 = ln(2)/O = 5.6 x 106 s = 64.8 days

(iii) Neutral pion has quark – antiquark structure:

First need to calculate energy of pion particle in lab frame, assume stationary

proton. From energy and momentum conservation:

ES = (Mp2 + MS

2 – Me2) c2 / (2 Mp)

From this, Lorentz factor = JS = 3.548

Speed = vS = 0.9595 c

Energy of photons in pion frame = MS c2 / 2

Then need to calculate Lorentz/Doppler shift for moving pion.

Maximum/minimum energy will be for photon travelling towards/away from

observer: Emax = 469.3 MeV, Emin = 9.7 MeV

3 (i) Bookwork

P = Fcas / A

0 12

uu ddS

3 PHY3052

0.0013 Pa for 1Pm.

Bookwork

(ii) (a) weak

(b) strong

(c) weak

(d) EM and weak

(e) weak

(f) weak

(g) EM and weak

(h) weak

Bookwork

4. (i)

Bookwork

(ii) Bookwork

2H + 3H o 4He + n

Q = BE (4-He) – BE(2-H) – BE (3-H) = 17.6 MeV

Bookwork

Using Q value from above, total nuclear density = 6.8 x 1018 m-3.

Density of deuterium = 3.4 x 1018 m-3. 7 orders smaller than density of air.

1s

1p1/2

2s 1d5/2 1d3/2

1p3/2

1


Module Code PHY3065Name of module Quantum optics and photonicsDate of examination May/June 2014

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

2.

3.

4.

5.

1. (i)

a, a†

= 1, [a, a] = 0,a†, a†

= 0.

Also acceptable:a, a†

= 1

Action on number states:

a|ni =

pn|n 1i, a†|ni =

pn + 1|n + 1i.

Matrix elements of a† and a:

hm|a|ni =

pnhm|n 1i =

pn m,n1, hm|a†|ni =

pn + 1hm|n + 1i =

pn + 1 m,n+1,

(where m,n1 is the Kronecker delta). Hence a† and a have the following expressions in terms of their matrix elements in thenumber-state basis:

a =

1X

n,m=0

hm|a|ni|mihn| =

1X

n=0

pn|n 1ihn|, a†

=

1X

n,m=0

hm|a†|ni|mihn| =

1X

n=0

pn + 1|n + 1ihn|.

(ii)Hamiltonian of single-mode quantum light:

ˆH = ~!

a†a +

1

2

.

Schrodinger equation:

i~ d

dt| (t)i =

ˆH| (t)i

with solution:

| (t)i = eiHt/~| (0)i.

| (0)i =

1p2

(|4i + |1i).

Applying time evolution equation above, and the single-mode Hamiltonian:

| (t)i = ei!(a†a+1/2)t 1p2

(|4i + |1i) = ei!t/2 1p2

(e4i!t|4i + ei!t|1i).

Thus

| (/!)i = ei/2 1p2

(e4i|4i + ei|1i) = ip2

(|4i |1i).

2

Inner product of | (0)i and | (/!)i is

h (/!)| (0)i =

ip2

(h4| h1|)

1p2

(|4i + |1i)

=

i

2

(1 1) = 0,

so | (0)i and | (/!)i are orthogonal.(iii)

Calculate h↵|i and find condition for it to vanish.

h↵|i = e|↵|2/21X

n=0

(↵)

n

pn!

hn| (c1|1i + c2|2i) = e|↵|2/21X

n=0

(↵)

n

pn!

(c1n1 + c2n2)

= e|↵|2/2

c1↵p

1!

+ c2(↵

)

2

p2!

= e|↵|2/2↵

c1 + c2

↵p

2

We thus have h↵|i = 0 if

c1 = c2↵p

2

.

(iv)A density operator given by

=

X

i

pi| iih i|

describes a system that has probability pi of being in the state | ii.For any vector | i,

h | Ô†Ô| i =

Ô| i

† Ô| i

0

because this is the norm squared of the vector Ô| i, which is a positive number. Thus Ô†Ô is positive.

2. (i)Energy of state |gi is ~!/2. Energy of state |ei is ~!/2.In the coupling term, a|eihg| takes the atom from the ground to the excited state and annihilates a photon; thus absorption of a

photon by the atom. a†|gihe| takes the atom from the excited to the ground state and creates a photon; thus emission of a photonby the atom.

ˆH| +i =

ˆH1p2

(|1, gi + |0, ei) =

1p2

~! 3

2

|1, gi 1

2

~!|1, gi +

1

2

~|0, ei + ~! 1

2

|0, ei +

1

2

~!|0, ei +

1

2

~|1, gi

=

1p2

~! +

2

|1, gi + ~

! +

2

|0, ei

= ~

! +

2

1p2

(|1, gi + |0, ei)

= ~! +

2

| +i

Thus | +i is an eigenstate of ˆH with eigenvalue ~! +

2

.

ˆH| i =

ˆH1p2

(|1, gi + |0, ei) =

1p2

~! 3

2

|1, gi +

1

2

~!|1, gi 1

2

~|0, ei + ~! 1

2

|0, ei +

1

2

~!|0, ei +

1

2

~|1, gi

=

1p2

~! +

2

|1, gi + ~

!

2

|0, ei

= ~

!

2

1p2

(|1, gi + |0, ei)

= ~!

2

| i

Thus | i is an eigenstate of ˆH with eigenvalue ~!

2

.

3

Time evolution is given by the operator eiHt/~, which gives the time evolution

ei(!+/2)t| +i and ei(!/2)t| i.

Initial state at t = 0 is

1p2

(| +i | i) = |1, gi

so the light mode has one photon and the atom is in the ground state.State at time t is

1p2

(ei(!+/2)t| +i ei(!/2)t| i) = ei!t 1p2

(eit/2| +i eit/2| i)

At t = / this is

ei!/ 1p2

(ei/2| +i ei/2| i) = ei!/ ip2

(| +i | i) = iei!/|0, ei

so the mode is in the vacuum state and the atom is in its excited state.At t = 2/ state is

e2i!/ 1p2

(ei| +i ei| i) = e2i!/ 1p2

(| +i + | i) = e2i!/|1, gi

so the mode has one photon and the atom is in the ground state.Probability amplitude of finding atom in state |gi is coefficient of |1, gi in total state at time t, which is

ei!t 1

2

(eit/2+ eit/2

) = ei!tcos(t/2),

so probability is cos

2(t/2).

(ii)Photons of frequency ! have energy ~! so the probability P (n) of having n photons is

P (n) = A exp

n~!

kBT

The normalization constant is such thatP

n P (n) = 1, so

A1=

1X

n=0

exp

n~!

kBT

=

1X

n=0

exp( ~!

kBT)

n=

1

1 exp

~!

kBT

Thus

P (n) =

1 exp

~!

kBT

exp

n~!

kBT

.

3. (i)In linear media 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 media 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

(2)ijkEj(r, t)Ek(r, t)

Also acceptable:

PNL(z, t) = "0(2)

[E(z, t)]2

4

In parametric down-conversion a pump photon of frequency !p is converted to two photons of frequencies ! and !p!. Thegenerated photon with the higher frequency is usually called the signal, the one with the lower frequency is called the idler.

The sum of the signal and idler frequencies is equal to the pump frequency. Pump wavelength p = 800 nm, signal wavelength = 1400 nm. Idler frequency is

c

p c

so idler wavelength is

1

p 1

1

=

1

800

1

1400

1

nm =

5600

3

nm = 1867 nm.

(ii)

[aL, aL] =

"p5

2

a0 +

1

2

e5i a†0,

p5

2

a0 +

1

2

e5i a†0

#=

p5

2

1

2

e5i[a0, a

†0] +

1

2

e5i

p5

2

[a†0, a0] =

p5

2

1

2

e5i 1

2

e5i

p5

2

= 0,

[a†L, a†

L] =

"p5

2

a†0 +

1

2

e5i a0,

p5

2

a†0 +

1

2

e5i a0

#=

p5

2

1

2

e5i[a†

0, a0] +

1

2

e5i

p5

2

[a0, a†0] = 0,

[aL, a†L] =

"p5

2

a0 +

1

2

e5i a†0,

p5

2

a†0 +

1

2

e5i a0

#=

5

4

[a0, a†0] +

1

4

[a†0, a0] =

5

4

1

4

= 1.

State is input vacuum state, given by a0|0i. Average obtained is the expectation value of the output number operator a†LaL in

the state |0i:

h0|a†LaL|0i =

1

4

h0|a0a†0|0i =

1

4

h1|1i =

1

4

.

Average of 1/4 photons is measured.For uncertainty, need expectation value of square of number operator a†

LaL:

h0|a†LaLa†

LaL|0i..

Only terms with same number of a†0 and a0, and a†

0 at the end, and a0 at the beginning, contribute:

h0|a†LaLa†

LaL|0i = h0|12

e5i a0

5

4

a0a†0 +

1

4

a†0a0

1

2

e5i a†0|0i =

1

16

h1|5a0a†0 + a†

0a0|1i =

1

16

h1|5 + 6a†0a0|1i =

11

16

.

Square of uncertainty in photon number is

11

16

1

4

2

=

5

8

so uncertainty isp

5/8.Parametric down-conversion creates photons in pairs, so number measurements will always find an even number of photons

in the output beam (assuming perfect detection efficiency).In a squeezed state the uncertainty in one of the quadratures is less than the vacuum value (1/2). The uncertainty in the

conjugate quadrature is increased compared to the vacuum value so that the quadrature uncertainty relation is satisfied.4. (i)

a†3a3 =

1p2

a†1

ip2

a†2

1p2

a1 +

ip2

a2

=

1

2

a†1a1 + a†

2a2

a†4a4 =

1p2

a†2

ip2

a†1

1p2

a2 +

ip2

a1

=

1

2

a†2a2 + a†

1a1

5

a4

a3

a2

a1

Thus

a†3a3 + a†

4a4 = a†1a1 + a†

2a2

so number of photons in output beams equals number of photons in input beams.We have

a1 =

1p2

a3 ip2

a4, a2 =

1p2

a4 ip2

a3.

Input state is

|1i1 |1i2 = a†1a

†2|0i =

1

2

a†3 + ia†

4

a†4 + ia†

3

|0i =

i

2

a†3a

†3 + a†

4a†4

|0i =

i

2

(|2i3 + |2i4)

So output state is i((|2i3 + |2i4)/2.Two photons will be found in arm 3, with probability 1/2, OR two photons will be found in arm 4, with probability 1/2. One

photon in each of arms 3 and 4 has probability 0.Photons with a narrow frequency spread are wave packets of finite length. For output arms of equal length, the photon wave

packets in each input arm must exactly overlap at the beam splitter in order for the probability of one photon in each output armto vanish. If the arrival time of the photons at the beam splitter is varied then there are zero coincidences in the output detectorswhen the arrival times are the same (zero probability of one photon in each output arm), whereas when the arrival times aredifferent there are coincidences (non-zero probability of one photon in each output arm).

The decrease in the coincidence rate when the photon wave packets overlap at the beam splitter is called the Hong-Ou-Mandeldip.

For classical particles with probability 1/2 of reflection and probability 1/2 of transmission, the probability of detecting bothphotons in output arm 4 is (1/2)(1/2) = 1/4, which is also the probability of detecting both photons in output arm 3 . Thus theprobability of detecting both photons in same output arm is 1/2. The probability of detecting one photon in each output arm is(1/2)(1/2) + (1/2)(1/2) = 1/2.

(ii)Hadamard gate on photon 1:

1p2

(|Hi1 1hH| + |HihV | + |V i1 1hH| |V i1 1hV |) |Hi1 |Hi2 =

1p2

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

Now a NOT gate:

1p2

(|Hi1 |Hi2 + |V i1 |Hi2) !1p2

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

so state is (|Hi1 |Hi2 + |V i1 |V i2)/p

2.The polarization measurement on photon 1 yields either H with probability 1/2, after which photon 2 is in the state |Hi2, OR

it yields V with probability 1/2, after which photon 2 is in the state |V i2. Photon 2 thus has probability 1/2 of being in state|Hi2 and probability 1/2 of being in state |V i2 .

Density operator for state of photon 2 is

=

1

2

|Hi2 2hH| + 1

2

|V i2 2hV |.

PHY3068: Principles of Theoretical Physics

Hints and tips for the 2013/14 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 Lagrange function is the difference of potential and kinetic energies. Inthe moving reference frame, it is sufficient to change the velocity as explainedin the question. (The potential energy depends only upon the absolute valueof r, which is the same in both reference frames.) Expanding the kinetic energyresults in two terms, mv · Ω × r and m

2(Ω × r)2. The variation can be easily

computed; the respective contributions of the two terms in the action to theforce are the Coriolis force, 2mΩ×v, and the centrifugal one: −mΩ× (Ω×r).

One can notice that the cross-term in the action, mv · Ω × r has the formof vector potential term, ev · A(r). The magnetic field can be found fromthe standard expression, B = curlA, which gives B = 2mΩ/e. This analogycan be employed to write Schrodinger equation, by introducing the covariantderivative, D = ∇ + ie

hA, with eA(r) = mΩ × r. The gradient term in the

Schrodinger equation then takes the form D2ψ.

2. Spontaneous symmetry breaking occurs when the Lagrangian possesses certainsymmetry, but its individual minima break the symmetry. Goldstone theoremstates that the spectrum of such a system must include a massless (or gapless)mode.

Obviously, this Lagrangian is invariant with respect to rotations in the internalspace of fields φ1,2, as the quantity φ2

1 + φ22 is invariant. Steady-state config-

urations are provided by minima of the Lagrangian. Broken-symmetry statesoccur if a > 0, so that the potential energy is minimal when φ2

1 + φ22 = a/b.

When the Lagrangian is expanded to the second order in χ, it takes the form ofLagrangian describing two decoupled scalar fields. The field χ1 represents themassive radial mode, with the mass m1 = 2a, while the field χ2 represents themassless Goldstone mode.

3. The rate of quantum tunneling is proportional to the exponential of imaginary-time action. The turning points can be found from the condition U(x∗) =E = 0; the central domain, U(x) > 0, is classically forbidden. The imaginary-time action can be obtained by changing the sign of U(x), which results inthe action of a quantum harmonic oscillator with frequency Ω. The trajectorythat connects the two classical turning points is a simple harmonic motion

x(τ) = x∗ sinΩτ , with the amplitude x∗ =!

2ϵmΩ2 . Thus, the travel time is

1

given by half-period of the oscillations, τ∗ = π/Ω. The two terms in the action,x2 and x2, cancel each other after averaging over a half-period, so that the actionis ϵτ∗. This gives the tunneling action πϵ/Ω, and tunneling rate ∼ exp

"

−2πϵhΩ

#

.The analysis is applicable if ϵ≫ hΩ.

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 forthat trajectory. In the presence of an external electromagnetic field, this actionis modified due to the term ev ·A in the Lagrangian. The integral of this termover the full trajectory is proportional to the line integral

$

A · dr. For a loop,this integral can be related to the flux of the magnetic field through an arbitrarysurface bounded by the loop. The orientation of the normal to the loop is takenso that the loop is traversed clockwise, as seen along the normal. The flux ofa monopole through the full sphere is 4πg, each hemisphere contributing 2πg.The loop shown in the figure can be viewed as a boundary of either northernor southern hemisphere. In the first case, the flux is 2πg, in the second casethe flux is −2πg, due to different orientation of the loop with respect to thesurface. In quantum theory, this would result in ambiguity in phase factor dueto the monopole: the phase factor is equal to exp

"

±2πigeh

#

, depending on thechoice of the hemisphere. However, when the phase is a multiple of π, the twoexpressions for the phase factor give identical values. This gives eg = 1

2hn,

where n is an integer.

2

1 PHYM007

PHYM007

UNIVERSITY OF EXETER

PHYSICS

MAY / JUNE 2014

ULTRAFAST PHYSICS

SOLUTIONS AND HINTS

2 PHYM007

1. (i) Sketch a diagram of an all-optical two-colour pump-probe experiment for

measurements of the intensity and polarisation response of samples deposited on

opaque substrates. [5]

On the sketch (diagram) indicate clearly:

(a) the ultrafast laser; [1]

(b) the optical paths of the pump and probe beams; [2, 2]

(c) the focusing and polarisation optics; [2, 2]

(d) the optical bridge detector; [1]

(e) the sample. [1]


Describe the factors limiting the temporal resolution of the experiment. [5]

HINT: (a) durations of the pump and probe pulses; (b) the minimal step,

accuracy and repeatability of the motion of the mechanical delay line.

Explain why time-resolved pump-probe experiments are limited to measurements of

repeatable phenomena? [4]

HINT: Each point in a time-resolved signal is an average of millions of

measurements of the sample's response for the specific pump-probe delay. In

addition, the signals measured for different pump-probe delay values must be

consistent among themselves in order to constitute a meaningful time-resolved

signal.

(ii) State the Nyquist-Shannon sampling theorem. [3]


3 PHYM007

Consider a time resolved experiment designed to study coherent phonons with

frequencies in the range 100 GHz – 1 THz and linewidths in the range 1 – 5 GHz.

Estimate the maximum time step and the minimum length of the time resolved

signals that need to be acquired in order to resolve peaks associated with the

phonons in the Fourier spectra of the time resolved signals. [3, 3]

HINT: Applying the Nyquist-Shannon theorem, obtain 0.5 ps and 1 ns for the

maximum time step and minimum signal length, respectively.

4 PHYM007

2. (i) Consider a solid state sample pumped by a high frequency electromagnetic wave

with the electric field rkEE tt Zcos, 0r .

Using a Taylor series expansion with respect to small atomic displacements, define

the Raman contribution to the polarisation of the sample induced by the incident

electromagnetic wave and label the Raman tensor and terms corresponding to the

Stokes and anti-Stokes scattering processes. [4, 1, 1, 1]

Name the other two main classes of optical processes that contribute to the optically

induced electrical polarisation of the sample. [1, 1]


(ii) Describe the mechanism of displacive optical excitation of coherent phonons in

solids and the conditions in which the mechanism could be expected to be active. [4]


The contribution to the reflectivity signal induced by the optical phonon, δR(td),

measured in a pump-probe experiment has been successfully fitted by the following

functional form

> @dddd

d sinerf1exp21)( tbtctttR »

¼

º«¬

ª¸¹·

¨©§ ¸

¹·

¨©§v Z

WVWG ,

where td is the pump-probe time delay, ω and τ are the phonon's frequency and life

time, erf (x) is the error function, and σ and c are parameters.

(a) Identify the likely mechanism of excitation of the phonon and justify your

answer in terms of the temporal character of the signal. [1, 2]

(b) Name parameter b and describe the physical implications of its non-zero value

for interpretation of the signal. [1, 2]

HINT: (a) Impulsive: note the phase at td = 0; (b) See the lecture notes

5 PHYM007

Is it possible to observe excitation of acoustic phonons in an all-optical pump-probe

experiment? Justify your answer. [1, 3]

HINT: Yes. Indeed, although acoustic phonons do not couple to the optical

field directly, they could still be excited if the optical pump action is non-

uniform (e.g. due to finite skin depth or tight focusing) or if the sample is itself

non-uniform (e.g. a multlayer)

(iii) Describe the concept of meta-materials. [3]

Explain the significance of resonant phenomena for creation of negative refractive

index electromagnetic meta-materials. [3]

Sketch the ray paths from vacuum through a perfect lens made of a meta-material

with refractive index of -1. [3]


(iv) Find the frequency separation between longitudinal cavity modes in a laser with

cavity length of 1 m. [2]

Numerical answer: 0.15 GHz

6 PHYM007

3. (i) What polarisation of the pump beam results in a specular optical Kerr effect (SOKE)

of maximal strength? [2]

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

and the temporal dependence of the pump-induced specular inverse Faraday effect

(SIFE) signal, for the case when the characteristic life time of the pump induced

angular polarisation is comparable to the pump pulse duration. [1, 2]


(ii) Consider a sample excited by an optical pump pulse with duration much shorter than

the electronic thermalisation time. The electronic density of states in the sample can

be assumed to be constant over the energy range affected by the excitation. The

central angular frequency of the pump pulse is ω0.

Sketch on the same graph the electron distribution functions, f (E), where E is the

electron energy,

(a) at equilibrium before the excitation; [2]

(b) just after the excitation when the electrons are still "hot"; [3]

(c) shortly after the excitation when the electron sub-system has thermalised. [2]

Neglecting the temperature dependence of the Fermi energy, clearly label on the

sketch

(d) the Fermi energy level, EF; [1]

(e) the energies EF±ħω0, where ħ is the Planck constant; [1]

(f) the energies EF±kBT0 and EF±kBT, where kB is the Boltzmann constant, T0 is the

equilibrium temperature and T is the temperature of thermalised electrons. [2]


(iii) Write the equations of the two-temperature model in one dimension, naming and

7 PHYM007

explaining physical meaning of all variables and material parameters in the

equations. [10]

State the main assumption made when the two-temperature model is applied to the

interpretation of ultrafast pump-probe measurements of metals. [2]


(iv) In order to generate ultrashort electrical pulses, some pump-probe experiments use

GaAs-based Auston photoconductive switches gated by optical pulses that are

shorter than the electron-hole recombination time. Consider an Auston switch made

from intrinsic GaAs, which has a band gap of 1.5 eV and room temperature carrier

density of 2.5∙1014 m-3.

Estimate the maximum value of the resistivity ratio ρ(T = 300 K) / ρ photo-induced

induced in the Auston switch by 50% absorption of an ultrashort optical pulse tuned

to the GaAs interband absorption edge. Assume an excitation density of 1 pJ / µm2

and an optical skin depth of 100 nm. [6]

Numerical answer: 1011

8 PHYM007

4. (i) In the interpretation of pump-probe measurements of ultrafast demagnetisation, it is

sometimes assumed that the ferromagnetic ordering is partly preserved even when

the transient electron temperature exceeds the Curie temperature of the material that

is being studied. Explain how this assumption could be justified. [4]

HINT: Note that the standard value of the Curie temperature applies to

equilibrium. However, upon ultrafast excitation, when the electron subsystem

is heated above Tc, the full demagnetisation might not occur if there has been

not enough time for the angular momentum associated with the magnetic

ordering to be transferred to the environment, e.g. the lattice sub-system.

Estimate the value of the exchange integral of a ferromagnetic material with Curie

temperature of 1000 K. [4]

Numerical answer: ≈ 1.4·10-21 J

Using a sketch, explain how and under what conditions ultrafast demagnetisation of

a thin ferromagnetic film could initiate precession of its magnetisation. [6]


(ii) Write the Landau-Lifshitz equation of motion of the magnetisation in the absence of

dissipation. [3]

Write the Landau and Gilbert magnetic damping terms and demonstrate their

equivalence in the limit of small dissipation. [3, 3, 4]


(iii) A meta-material exploits resonant coupling of the high frequency magnetic field of

9 PHYM007

an electromagnetic wave to standing spin waves in a magnetic film that has its

magnetisation pinned at one or both surfaces.

Assuming perfect pinning at both film surfaces, calculate the ratio of the overlap

integrals between the high frequency magnetic field and the first two standing spin

wave modes that can couple to the electromagnetic field. [7]

Numerical answer: 3

1


Module Code PHYM008

Name of module Physical Methods in Biology and Medicine Date of examination May/June 2014

1. (a) see notes on muscle structure, (b) e.g., confocal for nucleus, optical super-res for

mitochondria, using respective resolution formulae from notes, (c) e.g. x-ray crystallography, single particle cryo electron microscopy, (d) L10ms = 375 nm, L1s = 3.75 um

2. (a) see, for example, typical dual colour filter sets from Semrock; (b) thin film interference;

manufacturing processes include evaporative deposition, ion-assisted deposition, ion beam sputtering; show schematics, advantages (c) OD ~ 6 for emitter in excitation bands; (d) Iscatter/Iem ~ 1785

3. (a) NA 1.3, mag 63, variable immersion, plan/flat field correction, coverslip 0.15-0.19 mm,

DIC; (b) rlateral = 235 nm with water immersion; (c) Mtotal should be 68.1 for sampling, requires coupling magnification of 1.08; total field 117.5x117.5 um^2; (d) change mag by 500/400, otherwise may result in aliasing; (e) solid angle fraction based on NA: 0.24, Ndet = 1800 photons, SNR = 42.4

4. live cell conditions, contrast method (e.g. DIC, phase contrast or fluorescence), movement with

spatial sensor (camera), shape may require high-resolution (optical super-res, EM); temporal sampling based on characteristic time from mito size and velocity, determines required frame rate, spatial sampling ½ of smallest detail and requires appropriate magnification on camera; mechanisms: ATP, acto-myosin, microtubule & motors, test by specific inhibition of possible mechanisms

Documents

PHYSICS EXAMINATION PROBLEMS SOLUTIONS AND HINTS …newton.ex.ac.uk/teaching/resources/past-exams/SolnsHints_2013-14... · PHYSICS EXAMINATION PROBLEMS SOLUTIONS AND HINTS FOR STUDENT