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