Comprehensive Solid Physics Exam Solutions and Analysis

This document presents solutions to a solid physics exam, covering various topics including Ginzburg-Landau theory of superconductivity, low-dimensional physics with single-electron transistors and the Coulomb Blockade effect, ferromagnetism including domain theory and Heisenberg exchange interaction, and superconductivity focusing on BCS theory and SQUID devices. The solutions provide detailed explanations and calculations, offering insights into complex concepts within solid-state physics. The exam solutions also delve into practical applications and theoretical underpinnings, making it a valuable resource for students studying advanced physics. Desklib provides access to this and many other solved assignments.

SOLID PHYSICS EXAM
SOLUTIONS TO THE EXAM
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SOLUTIONS TO THE SOLID PHYSICS EXAM
Ginzburg-Lindau Theory of Superconductivity
a) Explain the physical significance of the terms in the equation
The degrees of freedom in the equation vary in space. One part of the equation has values
that explain the superconducting strength. The arbitrary part determines the
superconductivity`s strength and gauge and lastly, the gradient term in the equation
determines the arbitrary motions since, they have restoring forces.
b) Show that Landau equation can be written as ζ 2 d 2/dy 2 f (x)+ f ( y)−f ( y) 3=0
At zero field H = 0, J = 0 and Ψ* ∇Ψ- Ψ∇Ψ* = 0
Since, the superconducting phase is constant of position,
∇Ψ = 0 and αΨ + β|Ψ |2Ψ - h .h
2m∗¿ ∇ ¿ 2Ψ = 0
F = Ψ/Ψ∞ and Ψ∞ = -∞/β
1 dimension system
-h2/2m* d2/dx2f + α f + β|Ψ∞|2f3 = 0
-h2/2m* d2/dx2f = ζ2 and it can be defined as a length scale.
Since, α = 1- t and ζ2 = 1/1 – t
Hence, ζ2 d2/dx2 + f + f – f3 = 0
c) Verify that f(x) = tanh (x/ √2 ζ), sketch the Ψ and explain the significance of the
parameter.
-ζ2 d2/dx2 f + f (1 – f2) = 0
Since f = 1 – g and g is not equal to 0,
-ζ2 d2/dx2 g + (1 + g) – (1 – g) 3 = 0
-ζ2g’’ – 2g = 0

SOLUTIONS TO THE SOLID PHYSICS EXAM
g(x) is thus, not equal to e ± √2 x
ζ
Hence, f = eu – e-u/ eu + e-u = tanh u
U = x/√ 2ζ hence, tanh u = tanh | x/√ 2ζ |, when u is a small value.
The significance of the parameter is to be able to obtain the coherent length.
d) Determine whether type I or type II is a superconductor. Use simple sketches of surface
energy to show this.
J = e*/m* ( h∇ φ−e∗∆ ¿∨Ψ ∨¿2
If ∇ φ = 0 for x >> ζ , J = - e*2/m*|Ψ|2
∇ × J = - e*2/m*|Ψ|2
∇ × A = - e*2/m*|Ψ|2B = -1/μγ2 B
From Ampere`s Law, ∇ × B=μJ

SOLUTIONS TO THE SOLID PHYSICS EXAM
Thus, μ ∇ × J =∇ ×∇ × B=∇ ( ∇ . B )−∇ . ∇ B=−∇ . ∇ B
Since,∇ . B=0, ∇2B = 1/μγ2 B, Bz = Boe-x/γ = - A0/γ e-x/γ
The length scales sketches are:
Low Dimensional Physics
a) Use a sketch to describe the structure of single electron transistors, detailing the
length scales of integral quantum dot elements.

SOLUTIONS TO THE SOLID PHYSICS EXAM
A single electron transistor is a three terminal, nano-electronic tunnel junction device
which utilizes a capacitively-coupled input voltage to modulate a drain-source current.
The SET circuit consists of islands which are promptly connected with tunnel junctions
and capacitors with ideal voltage sources which control the circuit. The regions between
the nodes have tunnel junctions defined by tunnel capacitance and tunnel resistance.
There will be independent tunneling of an electron through the tunnel junction from the
source to the drain over the dot when an empty state is present at the energy level of the
island that lies between the Fermi levels of the electrode. Since the bias voltage is zero,
the Fermi levels of the source and drain are in equilibrium.
b) Explain how the Coulomb Blockade effect is experienced.
Electrons in a SET system require a minimum amount of energy to tunnelthrough the
barrier. When applied external barriers, biases are unable to provide this energy and
hence, an electron cannot tunnel through and the device will go into an off state. This
condition is called the Coulomb Effect. The minimum energy required by the electrons to
tunnel through can also be acquired from existing thermal energy sources. The SET is
operable at room temperature and this theory is expalined by Kulik and Schreiffer who

SOLUTIONS TO THE SOLID PHYSICS EXAM
assumed that the electron energy spectrum is continous within the SET island. Also, the
time taken by tunneling electrons over barriers is negligible. The quantum dot is
connected to the source and drain electrodes through tuneel barriers. The gate electrode
steers the potential in the quantum dot which is also capacitatively coupled at the
quantum dot. The gate voltage requires the current through the dot. The number of
electrons is fixed when the current is zero.Hence, to limit the electrons in the quantum
dot, the tunnel junction resistances must be greater than the quantum resistance. Change
in free energy decides whether the rate at which tunneling of electrons from island to
island would take place. Hence, the variation in conductance is due to the circuit`s free
energy changes as a result of the tunnel event and this change determines the rate at
which tunnel event would take place.
c) Explain the criterion that would be used on how cold and small the device may be in
order to display Coulomb Blockade and single electron tunneling at room
temperature.
The tunneling rate expression can be used to determine the size and temperature of
the device. The idea would be to reduce the dimensions of the SET in order to
decrease the capacitances and increase the amount of energy necessary to add an
electron to the island of a single electron transistor. Scaling down to molecular
dimensions will make the device function at room temperature.
Ferromagnetism
a) Account for the possibility that the existence of domains in a ferromagnet does
not necessarily exhibit magnetization and explain how domains contribute to
hysteresis.

SOLUTIONS TO THE SOLID PHYSICS EXAM
Unmagnetized ferromagnetic materials are as a result of their domains having
random magnetization directions. Aligning the domains in one direction will
cause magnetization. Magnetization curves lack retraceability due to the existence
of magnetic domains. Once the domains are oriented, it takes them energy in the
opposite direction to turn them back.
b) Obtain an expression for the exchange energy in terms of Bloch wall thickness
and interatomic spacing by considering the 1800 domain wall.
Exchange energy of a pair of spins, making an angle ∅ is:
Eex = -2JS2cos∅ ij
cos∅ = 1 - ∅ 2/2 + ∅ 4/24…
For a simple cubic structure with interatomic spacing exchange energy is
Eex = N (1/a2)2 × JS2( π / N ¿2
= JS2 π 2/a2N α 1/N

SOLUTIONS TO THE SOLID PHYSICS EXAM
The deviation of spin directions results in an increase of anisotropy energy given
by
EA = K × Thickness of transition region (Na)
Hence the total energy becomes Eex + EA
c) Obtain the expression for the exchange energy per unit area of the wall.
Assume that z = - ∞, the spin angle is 900 and at z = + ∞, the spin angle is still
900.
Hence, ∆ ∅ =( ∂ ∅
∂ z )2 and the exchange energy per spin pair is Eex = JS2a2( ∂ ∅
∂ z )2
The exchange energy per unit area of the wall is therefore,
1/a2 JS2a2 ∫( ∂∅
∂ z ) 2 1/a dz
= JS2a2/a ∫( ∂∅
∂ z )2 dz
d) Describe the origin and failures of Heisenberg Exchange Interaction and give a
short account of the exchange interaction.
Effective magnetic fielsd of 100T responsible for ferromagnetism are not due to
the atomic magnetic dipoles as the dipole sum for a cubic lattice is zero. The
origin of the internal field, H, is the exchange interaction which reflects the
electrostatic Coulomb repulsion of electrons on neighbouring atoms and the Pauli
principle which foebids the electrons from entering the same quantum state.
Moreover, Weis in 1907 supposed that there was an internal ‘ molecular field’ in
a ferromagnet proportional to its magnetization. The origin of those huge
molecular fields remained a mystery until Heissenberg introduced the idea of the
exchange interaction.

SOLUTIONS TO THE SOLID PHYSICS EXAM
e) Using sketches describe how hysteresis loss, coercivity and remanence
magnitudes determine the suitability of a ferromagnet for its application.
Hysteresis loss
Hysteresis loss is due to the reversal of magnetization of transformer cores
whenever subjected to alternating nature of magnetization force. The loss is
emitted as heat in the transformer core and as a result of this, the the transformer
has coolers to prevent overheating. Coercivity and Remanence are relevant in
magnetoviscosity especially in permanet magnets.
f) Calculate the total flux.
Flux = Br/2 × ( D
√ ( R × R ) +[ ( D+Z ) × ( D+ Z ) ]
−Z
√¿ ¿ ¿
Where, Br is the remanence field
Z is the distance from a pole face on the symetrical axis

SOLUTIONS TO THE SOLID PHYSICS EXAM
D is the thickness
R is the radius
Since circumference = 40cm, the diameter = 40/π, The thickness is πDh=4
Hence, h = 40/πD= 4/40
D = 0.1 cm
Z = 0.5 – 0.2 = 0.3
Remanence field 150 – 100 = 50
Hence, flux = 50/2 (0.1/40.69 – 0.3/40.62)
= - 0.1232 thus, the field lines are directed into a closed surface.
Superconductivity
a) Give a short overview accounting for the key features of the BCS theory, its
successes and failures.
The BCS theory accounts for the pairing of electrons close to the Fermi level
into Cooper pairs through interaction with the crystal lattice. The pairing
results from a slight attraction between the electrons related to the lattice
vibrations in a phonon attarction. The elsctron pairs have a slightly lower
energy and leave n energy gap above them which inhibits collission
interactions that lead to resistivity. This theory had a realization of the
existence of band gaps separating change carriers from the state of normal
conduction. Also, the critical temperature for superconductivity was a
measure of of the band gap and it depended on isotopic mass. Single elsctrons
could be eliminated as charge carriers in superconductivity and lastly the
boson behaviour in the theory was consistent, having coupled electron pairs

SOLUTIONS TO THE SOLID PHYSICS EXAM
with opposite spin. BSC theory however failed to explain the high
temperatures of superconductivity because of the electron-photon coupling,
that could not be made strong since, high temperatures lacked a Fermi surface
above Tc. In addition, the theory was unable to expalin the Meissner Effect
and the rotating superconductors.
b) Obtain an expression for the superconducting current flowing in a Jospehson
junction for zero voltage.
Phase difference φ=θl−θr
The macroscopic variable is regulated by two equations.
I0 = IC sinφ
φ∗¿ 2 e hence, θφ
θt =e∗V /h where V is the voltage drop between the two
electrodes and IC is the critical current.
The supercurrent at φ is equal to that at the point φ+ 2 π
Hence, Is = IC sinφ + ∑
m=2
∞
I C sin(mφ ¿
Since the electrode phase differences changes with time, the energy
U = ∫
ti
ti
I s(t) V(t)dt
Substituting I and V
U = Ec (cos φ ( ti )−cosφ (ti)¿
Where Ec = h/e* Ic is the junction characteristic energy.
c) Use schematic diagrams to describe the operation of superconducting
quantum interference device and their use in foundation for commercial
quantum computers.

SOLUTIONS TO THE SOLID PHYSICS EXAM

SOLUTIONS TO THE SOLID PHYSICS EXAM
Superconducting quantum interference devices measurevthe magnetic flux and output signals
in devices. A SQUID consists of tiny loops of superconductors that employ the Josephson
junctions in order to obtain a superposition. They are made from lead alloy consisting of
tunnel barriers that are sandwiched between base and top electrodes. Oscillating current is
then applied to an external circuit whose voltage changes as an effect of the interaction
between the circuit and the rng. The direct current SQUIDs are much more sensitive. They
are used in detecting weak magnetic signals and hence, used in the manufacture of MRI
machines that are used in brain imaging.
Magnetism: Magnetic Moments and Angular Momentum
a) What is the form of quantum mechanical relationship between magnetic moments and
angular momentum?
Particles do not necessarily move in a circular orbit, however, they have an intrinsic
angular momentum that has nothing to do with orbital motion. Since the spin angular
momentum is h/2, the intrinsic moment is MB = hγ e/ 2. The magnetic dipoles from the

SOLUTIONS TO THE SOLID PHYSICS EXAM
orbital and spin angular moments are responsible for paramagnetism and ferromagnetism.
In some cases, materials also undergo diamagnetism.
b) Provide the paramagnetism equation.
J only values jh, (j – 1)h…. –jh
Energy on field μ . β = M0B, only values g(quh/2m) (j, j – 1…-j)
Spacing g(quh/2m) B = hWr
Wr is the frequency absorbed in transition
M0B is any value from | μ∨Bcosθ
Procession around B at K frequency Wp = MB/J = g (qu/2m)B
c) Explain the cooling in paramagnetic materials.
A paramagnetic salt is cooled with liquid helium and most of the spins are lined up until
the magnetization is nearly saturated. The atomic magnets when shaken randomize their
spins without energy change, hence, there is no temperature change. Despite the atomic
magnets being flipped over by thermal motion, there is still some magnetic field present
working against the field. The opposite energy takes the energy from thermal motions
lowering temperatures. If the strong magnetic field is not removed rapidly, the
temperature of the salt will decrease. When the field is strong, thermal fluctuations knock
atoms to upper states and the atoms absorb more energy that could make the temperature
go to a few1 thousandths of a degree.
London Theory of Superconductivity
a) Develop the second London equation
Consider Hamiltonian presence in magnetic field = p + e/c A
Where A is the magnetic vector potential.

SOLUTIONS TO THE SOLID PHYSICS EXAM
B = ∇ × A
In 2nd quantization, introducing field operator,
H = ∑
δ
∫ d 3 rΨδ ( r ) 1/2m ( p+ e
c A ) 2Ψδ ( r)
p.A α ∇ . A=0
Hence, H = H0 + H1
Where H1= e/mc ∑
δ
∫ d 3 rΨδ ( r ) ( A . P ) Ψ ( r )
Total current operator J = - e/v ∑
δ
∫ d 3 rΨδ ( r ) 1/2m ( p+ e
c A )2Ψδ ( r)
At zero temperature J = Jd + Jp.
Jp is the paramagnetic current however when equated to 0 it gives the London
equation which becomes
-e/mv ∑
δ
∫ d 3 rΨ + pΨ ( r ) =0
b) Derive an expression relating entropy differerence between the normal and
superconducting states.
To calculate change in entropy, thermodynamic relation for superconductor along
with BSC theory is
BC(T) = BC(O) [ 1 – (T/TC)2] …. (i)
Critical field at T = 0K
B(0) = 0.206 T
Entropy drift per unit volume between the normal and superconducting states is
∇ S
v =−∂
∂ T ( B 2
2 μ )… ii
Magnetic density to return the superconductor to the normal state is B2/ 2 μ

SOLUTIONS TO THE SOLID PHYSICS EXAM
These types of superconductors are soft superconductors due to fast loss of their
superconducting states making them good materials to be used for magnetization
of other materials.
c) Explain
Meissner effect
It is when a material makes a transition from normal to superconducting state
excluding magnetic fields from the interior
Isotope effect in superconductors explains the origin of the effective attractive interaction
between the charge carriers which leads to superconducting state. It is as a result of magnetic
field impurities, proximity effect and non-adiabatic charge transistors.
Josephson Tunnelling is when two superconducting regions are isolates from their electron pair
phases in both of their regions. As the regions are brought close together, the electron pairs
tunnels across the gap and the two electron pair waves become coupled.
Low Dimensional Device
a) Describe the energy band structures and how a collimated beam of electrons is created.

SOLUTIONS TO THE SOLID PHYSICS EXAM
2DES are grown from ALGaAs – GaAs heterostructure. Electrones originating from the
gate layer more towards the interface between ALGaAs and GaAs due to the loer
condition band energy. Positively charged atoms are left in the donor layer where they set
up on electric field in the space between. This bends the conduction and valence bonds
such that a triangle potential is formed.
b) Describe principles between detection of ballistic electron motion.
Electrons in 2DES only see a small disturbing coulomb potential as a result of ionized
silicon atoms that are far away from the conducting layer. They then travel som distance
without being scattered. They are highly sensitive to an external disturbing potential such
as lines along for their path.
c) At what magnitudes to the peaks occur and what might limit the peaks?
Magnitude peaks occur at areas where the magnetic attraction is at the highest and when
the magnetic effect is at the lowest the troughs forms. These peaks are limited by the
nature of magnetism in the material, and the frequency applied.

SOLUTIONS TO THE SOLID PHYSICS EXAM
Spin Procession in a Magnetic Field
a) Derive the magnetic moment equation..
A small current loop has a magnetic current of μ=I A
Where I is the current and A is the area of the loop.
The electron`s orbital angular momentum is related to the magnetic moment,
μ= −e
2 m L Where e is the magnitude of electron charge and m is the electron mass.
The relationship suggests that there is a similar relationship between the magnetic
moment and spin angular momentum
μ=−g
2 m S
μ= Hs=−μ . B And H = ΩS hence, = -g/2m
b) Describe the major components of ESR and how they differ from NMR.
The ESR results when molecules of a solid show paramagnetism and the transitions
are induced between spin states by applying a magnetic field of microwave frequency
resulting in an absorption spectra. It studies radicals formed in solid materials giving
information and locations of radiation damage.
It differs from NMR in that while ESR deals with magnetically induced splitting of
electronic spin states, NMR deals with the splitting of nuclear spin states.
ESR uses microwave frequencies while the NMR uses lower radio frequencies.
Curie Paramagnetism
a) Explain the quantum mechanical origin of Curie paramagnetization.

SOLUTIONS TO THE SOLID PHYSICS EXAM
Transition metals lose the ns electrons before losing the d electrons to form
cations, their magnetic susceptibility is observed through Gouy balance methods
and are found to be relative to the mass susceptibility of the known magnetic
moment. The apparent change in weight in the balance is due to paramagnetism
and the relationship between the molar susceptibility and the spin only moments
results in Curie`s Law.
b) Obtain an expression for magnetization of transition metals.
U = g √ J ( J + 1 ) BM
The g = 1 + J(J+1)+S(S+1)-L(L+1)/ 2J(J+1)
Ligands quench orbital moments and hence, the spin is
U = g √S (S+1)
The mass susceptibility equation is v = (∆ Ws/∆ Wr ¿ [Wr/Ws ] Xg
Combining the two equations, magnetization is = NB2U2/ 3k (T+Q)
c) Calculate the total angular momentum
Angular momentum = mv
160.19 × 0.5
= 80.095

SOLUTIONS TO THE SOLID PHYSICS EXAM
H = ΩS
Ω= geB/2m

SOLUTIONS TO THE SOLID PHYSICS EXAM
References
Eschrig, H., (2008). Theory of Superconductivity. Website Online: https://www.ifw-
dresden.de/userfiles/groups/itf_folder/Helmut_Eschrig/sc1.pdf retrieved [17th April, 2018]
Lancaster University, (2009). Advanced Solid State (Physics) and Elements of Nanophysics.
Website Online: http://www.lancaster.ac.uk/users/esqn/phys421/programme.pdf retrieved [17th
April, 2018]