Physics for Information Science. Module 1, electrical conductivity to superconductivity.

Page 1 of 25

Module 1 at a glance

This module answers two questions about electrons in solids.

1. Why does a copper wire conduct, glass does not, and silicon conducts only a little? We build the answer in three steps. First, a classical picture, in which the electrons in a metal behave like a gas. Then, a quantum correction, in which electrons follow the Pauli exclusion principle and Fermi–Dirac statistics. Finally, band theory, in which the regular arrangement of atoms in a crystal splits the electron energy levels into bands, with gaps between the bands.
2. Why do some materials lose all their resistance when cooled? This is superconductivity. We study its critical temperature, its critical magnetic field, the Meissner effect, the two types of superconductor, the BCS theory that explains it, and its applications.

KTU syllabus. Electrical conductivity: classical free electron theory, electrical conductivity in metals, Fermi–Dirac distribution, variation of the Fermi function with temperature, Fermi energy, energy bands, classification of materials into conductor, semiconductor and insulator. Superconductivity: superconductivity, transition temperature, critical field, Meissner effect, Type I and Type II superconductors, BCS theory, applications of superconductors.

What you will be able to do

• Define and use electric field, current density, drift velocity, relaxation time, mean free path, mobility, conductivity and resistivity, with their units.
• Derive $\sigma=\dfrac{ne^2\tau}{m}$ from the classical free electron theory in two ways, and write the factors that affect conductivity.
• Write the assumptions, successes and failures of the classical theory, and explain how the quantum free electron theory improves it.
• Write the Fermi–Dirac function, find its value at $0\ \text{K}$ and at higher temperatures, draw its graph, and define the Fermi energy.
• Explain how energy bands form, and use band diagrams to classify conductors, semiconductors and insulators.
• Describe the properties of superconductors: zero resistance, persistent currents, critical temperature, critical field, flux quantisation and the isotope effect.
• Derive $\chi=-1$ from the Meissner effect, and distinguish Type I from Type II superconductors.
• Explain Cooper pairs and the BCS theory, and list the applications of superconductors.
• Solve the standard numerical problems on conductivity, the Fermi function, the Fermi energy and the critical field.

How to use these notes

• Each topic is a short page: idea → explanation or derivation → worked example → common mistakes.
• Green boxes show key ideas and definitions. Amber boxes marked Watch out show mistakes that make you lose marks. Boxes marked Note on the source show where the lecture notes were corrected.
• Try each numerical problem yourself before reading the solution. Every answer has been checked again.
• The last page has practice problems with answers, a list of exam-type questions and a formula sheet.

Where each topic sits

Before you start. You should know Ohm's law, Newton's second law and the equation of motion $v=u+at$, magnetic field and magnetic flux, and the exponential function $e^x$ with its inverse $\ln x$. You should also be comfortable with powers of ten, because almost every number in this module is either very large (electron density near $10^{28}\ \text{m}^{-3}$) or very small (times near $10^{-14}\ \text{s}$).

Page 2 of 25

Terms you need before you start

Every derivation in this module uses the same small set of quantities. First learn their meanings and units.

1. Electric field $E$

The electric field is a vector field that tells the force a charge would feel at each point in space. It is defined as the force per unit charge:

where $F$ is the force experienced by a charge $q$. Its unit is newton per coulomb, which is the same as volt per metre (V/m).

When a conductor is connected to a battery, the electric field pushes the free electrons and makes them flow. This gives an electric current. For a uniform wire of length $l$ with a voltage $V$ across it, the field inside is $E=V/l$.

2. Current density $J$

The current density is the electric current flowing per unit area of the cross-section of the conductor:

where $I$ is the current and $A$ is the cross-sectional area. Its unit is ampere per square metre ($\text{A/m}^2$).

Current density is more useful than current in theory, because it does not depend on the size of the wire. A thick wire and a thin wire of the same metal in the same field have the same $J$, even though the thick wire carries more current.

3. Drift velocity $v_d$

Even without any applied field, the free electrons in a metal move at high speed in random directions because of their thermal energy. When a field is applied, the electrons get a slow average velocity in the direction opposite to the field, in addition to this random motion. This average velocity is the drift velocity $v_d$.

Drift velocity is proportional to the electric field, and it is what produces the current. It is very small: typically $10^{-4}$ to $10^{-2}\ \text{m/s}$, compared with random speeds of about $10^{6}\ \text{m/s}$.

4. Relaxation time $\tau$

The relaxation time $\tau$ is the average time between two successive collisions of an electron in the conductor. After each collision the electron loses the extra velocity it got from the field, and it starts accelerating again from the beginning.

Relaxation time decides how freely the electrons can move. A longer relaxation time means the electrons are accelerated for a longer time between collisions, so they drift faster and the conductivity is higher. Typical values in metals are about $10^{-14}\ \text{s}$.

5. Mean free path $\lambda$

The mean free path is the average distance an electron travels between two successive collisions. It is related to the relaxation time through the average speed $v$ of the electrons:

The mean free path tells us how far an electron goes before it is scattered, so it also affects the conductivity of the material.

6. Mobility $\mu$

The mobility of a charge carrier tells how easily it responds to an electric field. It is defined as the drift velocity per unit electric field:

Its unit is $\text{m}^2/(\text{V s})$. In the classical theory (derived on a later page) $\mu=\dfrac{e\tau}{m}$, so mobility depends on the relaxation time. Higher mobility means the electrons move more freely when a field is applied.

7. Conductivity $\sigma$

Electrical conductivity $\sigma$ tells how easily a material conducts electric current. It depends on the number of free electrons, their charge and their mobility:

The higher the conductivity, the more easily the material allows current to flow. Its unit is siemens per metre (S/m), also written mho/m or $\Omega^{-1}\text{m}^{-1}$.

8. Resistivity $\rho$

Resistivity is the reciprocal of conductivity. It tells how strongly a material opposes the flow of current:

Its unit is ohm metre ($\Omega\,\text{m}$). Resistivity depends on the material and on its temperature. Materials with low resistivity (metals) are good conductors, and materials with high resistivity (insulators) resist the flow of current.

Summary table

Watch out. Do not mix up current $I$ (a total, in amperes) with current density $J$ (per unit area, in A/m²). Also, do not confuse resistivity $\rho$ (a property of the material) with resistance $R$ (a property of a particular piece of it, which depends on its length and area).

Page 3 of 25

Ohm's law, resistivity and conductivity

Ohm's law

Electrical conductivity is a measure of how easily an electric current can pass through a material. When a current $I$ flows through a conductor, the voltage drop across it is given by Ohm's law:

where $I$ is the current in amperes, $R$ is the resistance in ohms and $V$ is the voltage drop in volts.

Resistance and resistivity

Consider an ideal conductor with uniform cross-section and uniform composition, so that the electric field and current density are constant and parallel everywhere inside it.

Experiments show that the resistance is directly proportional to the length $l$ and inversely proportional to the area $A$:

The constant of proportionality $\rho$ is the electrical resistivity of the material, with SI unit $\Omega\,\text{m}$. The conductivity is its reciprocal:

Point form of Ohm's law

If the shape is more complicated, or the resistivity changes from point to point, the current density and the field become functions of position. Then we use a more general, local statement. With $E(x)$ the field at a point $x$ and $J(x)$ the current density there, the conductivity at that point is

This is the point form of Ohm's law. It is true at each point of a material, whatever its shape.

The point form from $V=IR$

For a uniform conductor the point form follows directly from Ohm's law. Start from

Put $R$ in the first equation:

The field in a uniform conductor is $E=V/l$, so $V/l$ can be replaced by $E$:

Dividing by $A$ gives the current density:

So for a uniform conductor of area $A$ and length $l$, the general expression $J=\sigma E$ becomes the familiar $R=\rho l/A$, and the two descriptions agree.

Classification of materials by conductivity

The conductivities of real materials differ by more than twenty powers of ten. This is one of the widest ranges of any physical property. The theories on the next pages try to explain why.

Example

Example 1. A copper wire is $2\ \text{m}$ long and has cross-section $1\ \text{mm}^2$. Copper has resistivity $1.7\times10^{-8}\ \Omega\,\text{m}$. Find its resistance, and the field and current density when it carries $2\ \text{A}$.

Check with the point form: $\sigma=1/\rho=5.9\times10^{7}\ \text{S/m}$, and $\sigma E=5.9\times10^{7}\times0.034=2\times10^{6}\ \text{A/m}^2$. $\checkmark$

Watch out. Convert areas to square metres before putting the values: $1\ \text{mm}^2=10^{-6}\ \text{m}^2$, not $10^{-3}$.

Page 4 of 25

Classical free electron theory (Drude–Lorentz)

The idea

The classical free electron theory was one of the earliest models to explain the electrical and thermal properties of metals. It was developed by Paul Drude in 1900 and later improved by Hendrik Lorentz, so it is also called the Drude–Lorentz theory.

Its main idea is that the valence electrons in a metal behave like the molecules of an ideal gas trapped in a box. The box is made by the fixed positive ions of the metal. The theory gave the base for understanding conduction in metals, but it also had important limitations that only quantum mechanics could remove.

Assumptions

1. A metal is made of fixed ions and free electrons. The metal has positive ion cores, fixed in a regular lattice, and valence electrons that are free to move through the whole metal. The free electrons are not bound to any particular atom.
2. Electrons are independent particles, like a gas. The free electrons are treated as independent particles moving inside the metal, just as gas molecules move inside a container. The positive ions are treated as fixed.
3. Electrons follow Newtonian mechanics. Their motion is described by classical ideas: velocity, momentum, force and Newton's laws.
4. Random thermal motion. The free electrons move randomly in all directions with speeds of about $10^{6}\ \text{m/s}$, and their speeds follow the Maxwell–Boltzmann distribution, as in the kinetic theory of gases.
5. Collisions. Electrons collide with the ions (and with each other). Between collisions they travel in straight lines. The average time between collisions is the relaxation time $\tau$, and the average distance is the mean free path $\lambda$. These collisions give the material its resistance.
6. Response to a field. The velocity of an electron is decided by two things: the random thermal motion due to temperature, and the extra velocity due to the applied electric field.
7. Carriers of charge and heat. The free electrons carry both charge and heat, so they are responsible for electrical and thermal conduction in metals.

Motion in the absence of a field: thermal motion

Without an applied field, each free electron moves at high speed in a straight line until it collides with an ion. Then it changes direction and moves again. Its path is a random zigzag.

Because the directions are random, for every electron moving in a particular direction there is, on average, another electron moving in the opposite direction. So the thermal motion gives no net flow of charge, and no current flows through the metal.

Motion when a field is applied: drift

Now connect the ends of a piece of metal to the terminals of a battery. An electric field is set up inside the metal. Each free electron feels a force in the direction opposite to the field (because it is negatively charged).

The electrons still move randomly and still collide, but along with the random motion, a slow net movement in one direction starts. This movement of electrons in one direction due to an electric field is called drift. The current is carried by the drift.

Key idea. The electrons in a metal move very fast, but in random directions, so on average they do not go anywhere. A field adds a small, steady drift to this random motion. The current comes only from the drift.

Orders of magnitude

Watch out. The drift speed is very small, so electricity does not "arrive" because individual electrons rush along the wire. The electric field is set up along the whole wire almost at once, and every free electron starts drifting at practically the same time.

Page 5 of 25

Drift velocity and the expression for conductivity

Drift velocity

When a field $E$ is applied, each free electron (charge $-e$, mass $m$) feels a force

The minus sign shows that the force on an electron is opposite to the field. By Newton's second law, $F=ma$, so the acceleration is

Electrons do not keep accelerating forever, because they collide frequently. Assume that after each collision an electron starts again from rest on average ($u=0$). By the equation of motion $v=u+at$, its velocity at a time $t$ after the collision is

The average time between collisions is the relaxation time $\tau$. Taking $t=\tau$ gives the average velocity that the electron gets from the field. This is the drift velocity:

In magnitude, $v_d=\dfrac{eE\tau}{m}$. The minus sign only shows that electrons drift against the field.

Current density from the drift (method 1)

Let $n$ be the number of free electrons per unit volume, called the free electron density or electron concentration. Consider a piece of metal of length $L$ and cross-section $A$.

• The total number of free electrons in it is $N=nAL$.
• Their total charge is $Q=Ne=nALe$.
• These electrons drift through the length $L$ in a time $t=L/v_d$, so the current is

The term $L/t$ is the drift velocity $v_d$, so

and the current density is

(With signs: $J=-nev_d$, and since $v_d$ is itself negative, $J$ comes out in the direction of the field, as conventional current should.)

Expression for conductivity

Put $v_d=\dfrac{eE\tau}{m}$ (in magnitude) in $J=nev_d$:

Compare this with the point form of Ohm's law $J=\sigma E$. The two are the same if

This is the classical expression for the electrical conductivity of a metal. The resistivity is its reciprocal, $\rho=\dfrac{m}{ne^2\tau}$.

Second derivation via Ohm's law (method 2)

We get the same result from the macroscopic route on the previous page. From $V=IR$, $R=\rho L/A$ and $E=V/L$ we found

From the microscopic route we found $J=\dfrac{ne^2\tau}{m}E$. Both describe the same current density in the same field, so the coefficients of $E$ must be equal. This again gives $\sigma=\dfrac{ne^2\tau}{m}$.

Mobility

The equation $v_d=\dfrac{e\tau}{m}E$ shows that the drift velocity is proportional to the field, $v_d\propto E$. Write it as

where the constant $\mu$ is the electron mobility: the drift velocity per unit field,

Putting this in the conductivity formula,

Note on the source. One lecture set wrote $\mu=-e\tau/m$, carrying the negative sign of the electron into the mobility. Mobility is normally written as a positive quantity, $\mu=e\tau/m$, and the direction of drift is stated separately.

Example

Example 1. A copper wire of cross-section $1\ \text{mm}^2$ carries $2\ \text{A}$. Copper has $n=8.5\times10^{28}\ \text{m}^{-3}$. Find the drift velocity.

That is about $0.15\ \text{mm/s}$. An electron would take nearly two hours to drift one metre.

Watch out. In $\sigma=ne^2\tau/m$ the charge is squared. One factor of $e$ comes from the force $eE$ that causes the drift, and the other comes from the charge $e$ carried by each drifting electron.

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Factors affecting conductivity, and what the classical theory achieved

Factors affecting electrical conductivity in metals

From $\sigma=\dfrac{ne^2\tau}{m}$, the conductivity depends on the number of free electrons $n$ and on the relaxation time $\tau$ (which is related to the mean free path).

Number of free electrons $n$. Metals have a high density of free electrons, and conductivity increases with the number of electrons available to carry current. Silver and copper conduct very well partly because of their high free electron density.

Mean free path. The longer an electron travels between collisions, the more it is accelerated and the higher the conductivity. At high temperatures electrons are scattered more often, by collisions with ions, with each other, and with the thermal vibrations of the lattice. So the mean free path becomes shorter and the conductivity falls.

Relaxation time $\tau$. The longer the average time between collisions, the greater the mobility and the higher the conductivity.

Temperature.

• At low temperatures the thermal vibrations of the lattice (called phonons) are weak, so electrons are scattered less. Conductivity increases as temperature decreases.
• At high temperatures the lattice vibrates more strongly, electron–phonon scattering becomes more frequent, and conductivity decreases.

So in a metal the resistivity increases with temperature. Metals have a positive temperature coefficient of resistance.

Impurities and defects. Foreign atoms in the lattice and crystal defects (vacancies, dislocations, grain boundaries) scatter the free electrons. This reduces the relaxation time and therefore the conductivity. This is why alloys usually conduct less than the pure metals they are made from.

Importance of the classical theory

Even with its limitations, the classical free electron theory played an important role in the development of solid-state physics.

• It gave the base for understanding the electrical and thermal properties of metals.
• The model is simple, yet it explains basic phenomena such as electrical conductivity and Ohm's law.
• It gives the correct form of the Wiedemann–Franz law: the ratio of thermal conductivity $\kappa$ to electrical conductivity $\sigma$ is proportional to the temperature, $\kappa/\sigma=LT$, where $L$ is the Lorenz number.
• Its failures showed the need for quantum mechanics to explain the behaviour of electrons in solids. This led to more advanced theories such as the band theory of solids.

Transition to quantum models

The Drude–Lorentz model was a step towards the modern quantum theory of metals. In 1928 Arnold Sommerfeld improved the classical theory by using quantum statistics. In his quantum free electron theory, electrons are treated as fermions that follow the Pauli exclusion principle and move according to quantum mechanics. Their distribution over energies is given by the Fermi–Dirac distribution instead of the classical Maxwell–Boltzmann distribution.

Before studying that theory, the next page shows exactly where the classical theory goes wrong.

Watch out. Positive temperature coefficient (resistance increasing with temperature) is a property of metals. Semiconductors and insulators behave in the opposite way, as the band theory pages explain.

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Failures of the classical free electron theory

The classical theory explains Ohm's law, but it fails badly on several other points. Each failure is a clue that electrons do not behave like a classical gas.

1. Electronic specific heat

If the free electrons were a classical gas, each would have an average kinetic energy $\tfrac32kT$ and so contribute $\tfrac32k$ to the heat capacity. Then the electrons would add a contribution to the specific heat of a metal comparable to that of the lattice vibrations.

In experiments, the electronic contribution to the specific heat is much smaller, only about $1\%$ of the total at room temperature. The classical theory overestimates it.

2. Temperature dependence of conductivity

The theory correctly predicts that conductivity falls as temperature rises, because collisions become more frequent. But the predicted dependence is wrong in value. Classical theory gives

whereas experiment shows $\sigma\propto\dfrac1T$ (the resistivity of a pure metal rises roughly linearly with temperature).

3. Mean free path

To match the measured conductivities, the classical theory needs a mean free path of only about one interatomic spacing. In experiments the mean free path is much longer, often hundreds of atomic spacings, especially at low temperatures.

4. Dependence on valency

Since $\sigma\propto n$, metals with more free electrons per atom should conduct better. Then monovalent metals such as copper, silver and gold should conduct worse than divalent (zinc) and trivalent (aluminium) metals. In fact, the monovalent metals copper, silver and gold are among the best conductors.

5. Wiedemann–Franz law

The classical theory gets the form $\kappa/\sigma=LT$ right, but the value of the Lorenz number wrong. It predicts

while the measured value for most metals is about $2.4\times10^{-8}\ \text{W}\,\Omega\,\text{K}^{-2}$. The quantum theory gives $L=\dfrac{\pi^2}{3}\left(\dfrac ke\right)^2=2.44\times10^{-8}\ \text{W}\,\Omega\,\text{K}^{-2}$, which agrees with experiment.

6. Classification of materials

The theory cannot explain why some materials are conductors, some are semiconductors and some are insulators. If we treat every material as a box of free electrons, there is no reason for the huge range of conductivities.

7. Magnetic properties

It cannot explain magnetic properties such as diamagnetism and ferromagnetism, and it predicts wrong values for the magnetic susceptibility of metals. In particular, it cannot explain Pauli paramagnetism, the weak response of free electrons to a magnetic field.

8. Sign of the charge carriers

It cannot correctly predict the sign of the charge carriers in some materials. Some metals (zinc and cadmium, for example) have a positive Hall coefficient, as if the current is carried by positive charges. A free-electron gas cannot explain this.

9. Quantum phenomena

It cannot explain effects that are quantum in nature: quantised energy levels in solids, the existence of energy bands, the photoelectric effect and superconductivity.

Summary

Note on the source. One lecture set said that classical theory predicts conductivity "decreasing linearly with temperature". The other set said the agreement with experiment is only qualitative. The standard statement is the one above: classical theory gives $\sigma\propto T^{-1/2}$, while experiment gives $\sigma\propto T^{-1}$.

Page 8 of 25

Quantum free electron theory (Sommerfeld)

The idea

Sommerfeld kept the simple picture of free electrons moving inside a metal, but replaced Newtonian mechanics and classical statistics with quantum mechanics and Fermi–Dirac statistics. This one change fixes most of the failures on the previous page.

Basic assumptions

1. Electrons are waves. The classical model treats electrons as particles that follow Newton's laws. The quantum theory treats them as waves that follow the Schrödinger equation.
2. Constant potential inside, infinite at the surface. The potential energy of an electron is taken as constant (zero) everywhere inside the metal and infinite at the boundaries. So the electrons are free particles in a box. This deliberately ignores the detailed interaction between the electrons and the ion lattice.
3. Discrete energy levels. Because the electrons are confined in a box, their energies are quantised. Only certain discrete energy levels are allowed.
4. Pauli exclusion principle. Each energy level can hold at most two electrons, with opposite spins.
5. Fermi–Dirac distribution. The very large number of electrons in the metal are distributed over the energy levels according to the Fermi–Dirac distribution function (next page).

How it explains electrical conductivity

At ordinary temperatures, all the energy levels up to a certain energy, the Fermi level, are almost completely full, with two electrons in each. An electron can gain energy from an applied field only if it can move into an empty level just above its own.

• Electrons deep below the Fermi level cannot do this, because every level just above them is already occupied.
• Only the electrons with energies close to the Fermi level can gain energy from the field and move into the empty levels just above.

So fewer electrons take part in conduction than the classical theory assumes. Those that do take part move at the high Fermi speed (about $10^{6}\ \text{m/s}$), with a relaxation time and mean free path different from the classical values. With these quantum values, the theory gives much better values of electrical conductivity.

Other successes

• Temperature dependence. The predicted change of conductivity with temperature is much closer to the observed change.
• Specific heat. Only electrons within about $kT$ of the Fermi level can absorb thermal energy, because only they have empty levels nearby. This is a very small fraction of all the electrons, roughly $kT/E_F\approx0.004$ for copper at room temperature, so the electronic specific heat is very small, in agreement with experiment.
• Mean free path. It predicts a long mean free path, as observed.
• Wiedemann–Franz law. It gives the correct Lorenz number, $L=\dfrac{\pi^2}{3}\left(\dfrac ke\right)^2$.
• Paramagnetism. It explains the weak Pauli paramagnetism of free electrons.

Drawbacks

• It fails for semiconductors and insulators. Like the classical model, it cannot explain why materials fall into three classes.
• It cannot explain positive charge carriers (positive Hall coefficients) in some materials.
• It ignores the periodic potential of the positive ions in the lattice.

The last drawback is the most important one. When we put back the periodic potential, we get the band theory of solids, covered after the Fermi–Dirac pages.

Watch out. "Free electron" in both theories means not bound to a particular atom. It does not mean free of the metal. In the quantum theory the electrons are free inside the box but cannot leave it.

Page 9 of 25

Fermi–Dirac statistics

Statistical methods

A solid contains an enormous number of electrons, about $10^{28}$ to $10^{29}$ per cubic metre in a metal. We cannot follow them one by one. Statistical methods describe how a large number of particles are distributed among the available energy states. There are three main kinds.

Particles can be divided as follows:

• Distinguishable (classical) particles follow Maxwell–Boltzmann statistics.
• Indistinguishable (quantum) particles are either fermions, which follow the Pauli exclusion principle and Fermi–Dirac statistics, or bosons, which do not follow it and follow Bose–Einstein statistics.

Note on the source. One lecture set divided indistinguishable particles into "fermions" and "mesons". The correct division is fermions and bosons (mesons are just one family of bosons).

In solid-state physics, electrons are what matter, and electrons are fermions. So we use the Fermi–Dirac distribution.

The Fermi–Dirac distribution function

The probability $f(E)$ that a quantum state of energy $E$ is occupied by an electron, at thermal equilibrium at temperature $T$, is

where:

• $f(E)$ is the probability that the energy state $E$ is occupied;
• $E$ is the energy of the state we are considering;
• $E_F$ is a constant called the Fermi energy or Fermi level;
• $k$ is Boltzmann's constant, $1.38\times10^{-23}\ \text{J/K}=8.617\times10^{-5}\ \text{eV/K}$;
• $T$ is the absolute temperature in kelvin.

$E_F$ may or may not coincide with an actual energy level, but it gives a reference with which other energies can be compared.

Reading the formula

Everything depends on the exponent $\dfrac{E-E_F}{kT}$, which compares how far a state is from the Fermi level with the thermal energy $kT$.

• If $E$ is much below $E_F$, the exponent is large and negative, $e^{(\cdot)}\approx0$, and $f\approx1$. The state is almost certainly occupied.
• If $E$ is much above $E_F$, the exponent is large and positive, $e^{(\cdot)}$ is very large, and $f\approx0$. The state is almost certainly empty.
• At $E=E_F$ the exponent is $0$ and $f=\tfrac12$ (for $T>0$).

A probability always lies between $0$ (certainly empty) and $1$ (certainly occupied), and $f(E)$ does too.

A useful symmetry

For energies at equal distance above and below $E_F$, the occupation probabilities add up to one:

So the probability that a state $\Delta E$ above $E_F$ is occupied is equal to the probability that a state $\Delta E$ below $E_F$ is empty.

Example 1. At room temperature ($kT=0.0259\ \text{eV}$), find $f(E)$ for a state $kT$ above and a state $kT$ below the Fermi level.

They add up to $1$, as the symmetry says. $\checkmark$

Watch out. Write the exponent as $(E-E_F)/kT$, with $E$ first. If you change it to $(E_F-E)$, occupied states become empty ones.

Page 10 of 25

Variation of the Fermi function with temperature

The Fermi function changes with temperature. We study three cases: absolute zero, a moderate temperature, and a high temperature.

Case 1: at $T=0\ \text{K}$

There are three possibilities: the energy $E$ is below, above or equal to the Fermi energy.

(a) $E<E_F$. Then $E-E_F$ is negative, and dividing by $kT=k\times0$ makes the exponent $-\infty$:

$f(E)=1$ means every energy level below $E_F$ is occupied.

(b) $E>E_F$. Then $E-E_F$ is positive and the exponent is $+\infty$:

$f(E)=0$ means every energy level above $E_F$ is vacant at $0\ \text{K}$.

(c) $E=E_F$. Then $E-E_F=0$ and the exponent is $\dfrac00$, which is indeterminate at exactly $0\ \text{K}$. The occupancy of the Fermi level itself is not fixed at absolute zero. It can be anywhere from $0$ to $1$. (For any temperature above zero, however small, it is exactly $\tfrac12$, as Case 2 shows.)

In short, at $0\ \text{K}$:

The electrons fill the levels from the bottom upwards, two in each level, up to $E_F$, and nothing is above it.

Fermi energy. The Fermi energy is the highest filled energy level at $0\ \text{K}$.

Case 2: at a finite temperature $T>0\ \text{K}$

When the solid is heated, some electrons are excited to higher levels. In a metal, $E_F$ is several electronvolts, while at room temperature $kT\approx0.026\ \text{eV}$. So $E_F\gg kT$.

• Electrons deep below $E_F$ cannot be excited, because the thermal energy $kT$ is far too small to lift them to an empty level.
• Only electrons in a narrow band of width about $kT$ just below $E_F$ can jump to levels just above $E_F$.

As a result, the probability of finding electrons just below $E_F$ decreases below $1$, and the probability of finding electrons just above $E_F$ increases above $0$. The states above $E_F$ are partly occupied and the states just below are partly empty.

For an electron exactly at the Fermi level, $E=E_F$, so the exponent is $0$:

Key idea. At any temperature above $0\ \text{K}$, the probability of occupation of the Fermi level is $\tfrac12$, or 50 %. So the Fermi level can also be defined as the energy at which the occupation probability is one half.

Far from the Fermi level, not much changes:

• For $E\gg E_F$, $f(E)\approx0$: high levels are still almost always empty.
• For $E\ll E_F$, $f(E)\approx1$: low levels are still almost always full.

Case 3: at high temperature

As the temperature rises further:

• the electrons have more thermal energy and spread over a wider range of levels;
• the change from $f=1$ to $f=0$ becomes more and more gradual;
• a larger fraction of electrons occupies states above $E_F$, and the probability of finding electrons just below $E_F$ falls further.

Far above the Fermi level, where $E-E_F\gg kT$, the $1$ in the denominator can be neglected:

This is the classical Maxwell–Boltzmann form. So at very high temperatures (or high energies), the Fermi–Dirac distribution starts to look like the classical one.

Graphical representation

• At $T=0\ \text{K}$: a step. $f=1$ up to $E_F$, and then a sudden drop to $0$.
• At $T>0\ \text{K}$: the step becomes a smooth curve that falls from $1$ to $0$ over a width of a few $kT$ around $E_F$.
• At higher temperatures: the curve spreads out more and its tail looks like an exponential decay.

Every curve passes through $f=0.5$ at $E=E_F$.

Watch out. At exactly $0\ \text{K}$, $f(E_F)$ is indeterminate. Only for $T>0$ is $f(E_F)=\tfrac12$. Examiners like this difference.

Page 11 of 25

The Fermi energy and Fermi level

Two definitions

1. Fermi energy $E_F$: the energy of the highest occupied level at $0\ \text{K}$.
2. Fermi level: the more general term used at any temperature. It is the energy level whose probability of occupation is $\tfrac12$ (50 %).

At absolute zero the two are the same. Many books use the two names in the same meaning.

Value of the Fermi energy

For a three-dimensional gas of free electrons, such as the conduction electrons of a metal, the Fermi energy is

where $\hbar=h/2\pi$ is the reduced Planck constant, $m$ is the electron mass and $n$ is the free electron density. It depends only on $n$: the more electrons are packed in, the higher the levels they must fill.

From the Fermi energy we can also define the Fermi speed, the speed of an electron at the Fermi level:

Example 1. Copper has $n=8.5\times10^{28}\ \text{m}^{-3}$. Find its Fermi energy and Fermi speed.

With a relaxation time of $2.5\times10^{-14}\ \text{s}$, the mean free path at the Fermi speed is $\lambda=v_F\tau=3.9\times10^{-8}\ \text{m}$, about $40\ \text{nm}$ or more than 100 atomic spacings. This long mean free path is exactly what the classical theory could not explain.

Position of the Fermi level in different materials

Conductors (metals). The Fermi level lies inside the conduction band, or in the region where the conduction and valence bands overlap. Even at absolute zero there are free electrons in the conduction band, so metals conduct easily.

Semiconductors. The Fermi level lies within the band gap, between the valence and conduction bands. Its exact position depends on the type:

• in an intrinsic (pure) semiconductor it is near the middle of the gap;
• in an n-type semiconductor (doped with extra electrons) it moves closer to the conduction band;
• in a p-type semiconductor (doped with holes, or electron acceptors) it moves closer to the valence band.

Insulators. The Fermi level is also in the band gap, but the gap is much wider, so the Fermi level is far from both bands. It is extremely difficult for electrons to cross to the conduction band, so insulators have very few free electrons and do not conduct.

Watch out. "Fermi energy" strictly means the highest filled level at $0\ \text{K}$. "Fermi level" is the $f=\tfrac12$ level at any temperature. Write which definition you are using. Also, $E_F\gg kT$ in metals. Because of this, only a thin slice of electrons near $E_F$ takes part in conduction and heat capacity.

Page 12 of 25

Band theory: how energy bands form

Why a new theory

The free electron theories ignore the ion lattice. Band theory keeps the quantum picture but includes the effect of the crystal.

Basic assumptions

1. Electrons have a wave nature, and the laws of quantum mechanics apply to them.
2. The potential energy of an electron inside the crystal is not constant. It is lowest at the positive ion sites and highest midway between two ions, and this pattern repeats with the periodicity of the crystal.

Formation of energy bands

In a single, isolated atom, electrons occupy discrete energy levels, such as 1s, 2s, 2p, 3s, 3p.

When a very large number $N$ of atoms come close together to form a solid, the electron wave functions of neighbouring atoms overlap. Because of this interaction, and because the Pauli principle does not allow two electrons in a system to share the same state, each discrete atomic level splits into $N$ closely spaced levels. Together, these closely spaced levels form a continuous energy band.

Narrow and wide bands

• The electrons in the outer shells screen the inner (core) electrons from the neighbouring atoms. So the inner levels split only slightly and form narrow bands.
• The valence electrons in the outermost shell interact most strongly, so their levels split the most and form wider bands.

How close are the levels inside a band?

The width of an energy band is typically a few electronvolts. A crystal has an enormous number of atoms ($N\sim10^{23}$ in a small piece), so the band has an enormous number of levels. So the energy gap $\Delta E$ between neighbouring levels is extremely small, of the order of

At room temperature, the thermal energy of an electron is about $kT\approx0.026\ \text{eV}$, which is much larger than this spacing. So inside an allowed band, an electron can easily move to a higher vacant level, either by thermal energy or under a small applied field. In effect, the levels in a band form a continuum.

Note on the source. The lecture notes gave the level spacing as about $10^{-27}\ \text{eV}$. If we divide a band of a few eV among about $10^{23}$ levels, we get about $10^{-23}\ \text{eV}$. In both cases the conclusion is the same: the spacing is negligible compared with $kT$.

Allowed and forbidden energies

Between the allowed bands there are ranges of energy in which no electron states exist. These are the forbidden bands or band gaps. An electron in a crystal can have an energy inside an allowed band, but never inside a gap. The next page looks at the two bands that decide the electrical behaviour.

Watch out. A band is made of levels, not of electrons. A band can be full, partly full or empty, depending on how many electrons are there to fill it.

Page 13 of 25

Valence band, conduction band and band gap

The energy band diagram

An energy band diagram shows the energy levels of the top energy band and the next lower band in a solid, with the gap between them. The vertical axis is electron energy. Bands are drawn as horizontal strips, with the inner bands at the bottom.

The two important bands

Valence band. The range of energies of the valence electrons that take part in bonding in the crystal. These electrons are bound to their atoms and are not free to move. The valence band is usually full or nearly full, so it cannot conduct well. Its top edge is $E_v$, the highest energy an electron can have while still bound.

Conduction band. The range of higher energies of electrons that have broken free from their bonds and can move through the whole solid. Only free electrons take part in conduction, hence the name. Its bottom edge is $E_c$, the smallest energy an electron must have to become free. An electron that reaches $E_c$ is no longer bound to an atom and can move under the effect of an electric field.

Forbidden band (band gap). The two bands are separated by a range of energies where no states exist. Its width is the band gap:

For an electron to take part in conduction, it must gain at least the energy $E_g$ to jump from $E_v$, the top of the valence band, to $E_c$, the bottom of the conduction band.

The band gap decides the behaviour

• If the bands overlap (no gap), electrons move freely between them. The material conducts easily.
• If the gap is moderate, electrons can be lifted across it by heat or light. This is a semiconductor.
• If the gap is large, electrons cannot cross it under normal conditions. This is an insulator.

Holes

When an electron leaves the valence band, it leaves behind an empty state, called a hole. A hole behaves like a positive charge carrier. Electrons in the valence band can move into the hole, so the hole effectively moves in the opposite direction. In a semiconductor, both the electrons in the conduction band and the holes in the valence band carry current.

Watch out. $E_c$ is the bottom of the conduction band and $E_v$ is the top of the valence band. The gap is measured between exactly these edges, not between the middles of the bands.

Page 14 of 25

Conductors, semiconductors and insulators

Conductors

Conductors are materials that allow current to flow easily.

• The valence and conduction bands overlap, or the gap is negligible (less than about $0.1\ \text{eV}$).
• There is no significant energy barrier to stop electrons from entering the conduction band. Many electrons are already in it, and they move freely under an applied field.
• Even at room temperature there are plenty of free electrons, so metals are excellent conductors, and no extra energy is needed to excite electrons.
• The Fermi level lies inside the band.
• Examples: copper, silver, gold, aluminium, iron.

Semiconductors

Semiconductors have a small but important band gap, roughly $1\ \text{eV}$.

• At absolute zero the conduction band is empty and the valence band is full, with all electrons bound. So a semiconductor behaves as an insulator at $0\ \text{K}$.
• At room temperature, or when exposed to heat or light, some electrons get enough energy to cross the gap into the conduction band. Each one leaves a hole in the valence band.
• Both the free electrons and the holes carry current. Conduction is weaker than in a conductor because far fewer electrons are in the conduction band.
• Conductivity increases with temperature, so semiconductors have a negative temperature coefficient of resistance.
• Doping, which means adding small controlled amounts of impurity, can increase the conductivity enormously. Doping can increase the number of free electrons (n-type) or the number of holes (p-type). Because we can control this, semiconductors are very useful in electronic devices.
• Examples: silicon ($E_g\approx1.1\ \text{eV}$), germanium ($E_g\approx0.66\ \text{eV}$), gallium arsenide ($E_g\approx1.43\ \text{eV}$).

Insulators

Insulators have a large band gap, roughly $4\ \text{eV}$ or more.

• Electrons in the valence band cannot get enough energy under normal conditions to jump across the gap, so there are very few or no electrons in the conduction band.
• Even an applied field cannot give enough energy to lift electrons across the gap. Conductivity is extremely poor because there are no free carriers.
• Only at very high temperatures do a few valence electrons reach the conduction band. Then the resistivity falls, so insulators also have a negative temperature coefficient of resistance.
• Because they have no free carriers, they are ideal for stopping the flow of current, for example to cover and enclose electrical wiring.
• Examples: diamond ($E_g\approx5.5\ \text{eV}$), glass, rubber, wood, plastic.

Summary

Note on the source. The two lecture sets gave slightly different gap ranges: "1 to 2 eV" or "about 1 eV" for semiconductors, and "greater than 5 eV" for insulators. These are rough values, not sharp limits. The dividing line between a wide-gap semiconductor and an insulator is a matter of convention (around 3 to 4 eV). The values above are the usual textbook figures.

Page 15 of 25

Numerical problems: conductivity

Constants: $e=1.6\times10^{-19}\ \text{C}$, $m=9.11\times10^{-31}\ \text{kg}$, $N_A=6.022\times10^{23}\ \text{mol}^{-1}$.

Problem 1. A material has conductivity $1.2\times10^{5}\ \text{S/m}$, carrier concentration $2\times10^{24}\ \text{m}^{-3}$ and carrier effective mass $0.2\times9.11\times10^{-31}\ \text{kg}$. Find the relaxation time.

Step 1: effective mass. $m=0.2\times9.11\times10^{-31}=1.822\times10^{-31}\ \text{kg}$.

Step 2: rearrange $\sigma=\dfrac{ne^2\tau}{m}$ to get $\tau=\dfrac{\sigma m}{ne^2}$.

Step 3: put the values. $e^2=(1.6\times10^{-19})^2=2.56\times10^{-38}\ \text{C}^2$, so the denominator is $ne^2=(2\times10^{24})(2.56\times10^{-38})=5.12\times10^{-14}$, and the numerator is $\sigma m=(1.2\times10^{5})(1.822\times10^{-31})=2.186\times10^{-26}$.

Problem 2. A metal has electron concentration $8.5\times10^{28}\ \text{m}^{-3}$ and relaxation time $2.5\times10^{-14}\ \text{s}$. Find its conductivity.

Numerator: $ne^2\tau=(8.5\times10^{28})(2.56\times10^{-38})(2.5\times10^{-14})$. Multiply the numbers, $8.5\times2.56\times2.5=54.4$, and add the powers of ten, $28-38-14=-24$, to get $54.4\times10^{-24}$.

so $\sigma\approx\mathbf{5.97\times10^{7}\ \text{S/m}}$, close to the measured value for copper.

Problem 3. Find the drift velocity of the free electrons in a copper wire of cross-section $10\ \text{mm}^2$ carrying $100\ \text{A}$. Copper has density $8969\ \text{kg/m}^3$ and atomic weight $63.54$, and each atom gives one free electron.

Step 1: electron density. The number of atoms per unit volume is (density × Avogadro's number) / molar mass, with the molar mass in kg/mol:

Step 2: drift velocity from $I=nAev_d$:

Even at $100\ \text{A}$ the electrons drift less than a millimetre per second.

Problem 4. Find the mobility of electrons in copper, with $9\times10^{28}$ valence electrons per $\text{m}^3$ and conductivity $6\times10^{7}\ \text{S/m}$.

Problem 5. Find the relaxation time of conduction electrons in a metal of resistivity $1.54\times10^{-8}\ \Omega\,\text{m}$ with $5.8\times10^{28}$ conduction electrons per $\text{m}^3$.

From $\rho=\dfrac{m}{ne^2\tau}$:

Note on the source. The problem sheet printed the resistivity as $1.54\times10^{-3}\ \Omega\,\text{m}$, which would give $4\times10^{-19}\ \text{s}$. The answer given ($\approx4\times10^{-14}\ \text{s}$) needs $1.54\times10^{-8}\ \Omega\,\text{m}$, which is a typical value for a metal, so we have used that.

Problem 6. Silver has density $10.5\times10^{3}\ \text{kg/m}^3$ and atomic weight $107.9$, with one conduction electron per atom. Its conductivity at $20^\circ\text{C}$ is $6.8\times10^{7}\ \text{S/m}$. Find (a) the density of free electrons and (b) the electron mobility.

(a) $\ n=\dfrac{10.5\times10^{3}\times6.022\times10^{23}}{107.9\times10^{-3}}=\mathbf{5.86\times10^{28}\ \text{m}^{-3}}$

(b) $\ \mu=\dfrac{\sigma}{ne}=\dfrac{6.8\times10^{7}}{(5.86\times10^{28})(1.6\times10^{-19})}=\mathbf{7.25\times10^{-3}\ \text{m}^2/\text{V s}}$

Problem 7. For the metal of Problem 2 carrying a current density of $2\times10^{6}\ \text{A/m}^2$, find the field inside it and the drift velocity.

Watch out. Molar masses must be in kg/mol ($63.54\ \text{g/mol}=63.54\times10^{-3}\ \text{kg/mol}$), and areas in m². Most wrong answers on this page come from these two conversions.

Page 16 of 25

Numerical problems: the Fermi function and Fermi energy

Use $k=8.617\times10^{-5}\ \text{eV/K}$ when energies are in eV.

Problem 1. Find the probability that an electron occupies a state $0.2\ \text{eV}$ above the Fermi level at $300\ \text{K}$.

Step 1. $kT=8.617\times10^{-5}\times300=0.02585\ \text{eV}$.

Step 2. $\dfrac{E-E_F}{kT}=\dfrac{0.2}{0.02585}\approx7.74$.

Step 3.

A state just $0.2\ \text{eV}$ above $E_F$ is almost empty at room temperature.

Problem 2. At what temperature is the probability of occupying a state $0.1\ \text{eV}$ above the Fermi level equal to $0.1$?

Step 1: rearrange. With $x=\dfrac{E-E_F}{kT}=\dfrac{0.1}{kT}$,

Step 2: take logs. $x=\ln9=2.197$.

Step 3: solve for $T$.

Problem 3. At $300\ \text{K}$, find the probability that a state $0.1\ \text{eV}$ below the Fermi level is occupied, and that one $0.1\ \text{eV}$ above it is occupied.

$\dfrac{0.1}{kT}=\dfrac{0.1}{0.02585}=3.87$.

They add up to $1$. The chance of finding a hole $0.1\ \text{eV}$ below $E_F$ is equal to the chance of finding an electron $0.1\ \text{eV}$ above it.

Problem 4. Find the Fermi energy and Fermi speed of copper, $n=8.5\times10^{28}\ \text{m}^{-3}$.

$E_F$ is about 270 times $kT$ at room temperature, which justifies treating only electrons near $E_F$ as active.

Problem 5. At what energy above $E_F$ is the occupation probability $0.01$ at $300\ \text{K}$?

So the Fermi function falls from $0.5$ to $0.01$ within about $0.12\ \text{eV}$ of $E_F$ at room temperature. The "rounded step" is only a few $kT$ wide.

Watch out. Use $k$ in eV/K when energies are in eV, and in J/K when they are in joules. Also remember that $f(E)$ is a probability. Any answer above $1$ or below $0$ is wrong.

Page 17 of 25

Superconductivity: discovery and transition temperature

What superconductivity is

Superconductivity is a phenomenon in which some materials, when cooled below a certain temperature, show two special properties:

1. zero electrical resistance, and
2. expulsion of magnetic fields from their inside.

The temperature below which this happens is called the critical temperature or transition temperature, $T_c$.

Discovery

• Superconductivity was discovered in 1911 by the Dutch physicist Heike Kamerlingh Onnes, who was studying the electrical resistance of mercury at very low temperatures.
• He found that at about 4.2 K the resistance of mercury suddenly dropped to zero. This was the start of superconductivity.
• A superconducting material conducts electric current without any loss of energy, which is very different from any ordinary conductor.

In an ordinary metal the resistance falls as it is cooled but becomes steady at a small finite value (because of impurities and defects). In a superconductor it falls suddenly to exactly zero at $T_c$.

The transition (critical) temperature

Transition temperature $T_c$. The temperature below which a material changes from the normal conducting state to the superconducting state.

• Above $T_c$: the material behaves as a normal conductor, with finite resistance and no expulsion of the magnetic field.
• Below $T_c$: the material is superconducting, with zero resistance and the Meissner effect.
• Each superconductor has its own $T_c$, from near $0\ \text{K}$ to above $100\ \text{K}$ for the high-temperature superconductors.
• High-temperature superconductors, such as some ceramic materials, have $T_c$ above $77\ \text{K}$. They are easier to use in practice because they can be cooled with liquid nitrogen (which boils at $77\ \text{K}$) instead of expensive liquid helium (which boils at $4.2\ \text{K}$).
• The transition is reversible. If the material is warmed above $T_c$, it goes back to the normal state, its finite resistance comes back, and any magnetic field that was pushed out enters it again.

Critical temperatures of some superconductors

History of the transition temperature

For 75 years after the discovery, the highest known $T_c$ increased slowly through niobium compounds to about $23\ \text{K}$. In 1986 the copper-oxide (cuprate) ceramics were discovered, and within a few years $T_c$ went above the boiling point of liquid nitrogen: about $92\ \text{K}$ for yttrium compounds, over $100\ \text{K}$ for bismuth and thallium compounds, and about $135\ \text{K}$ for mercury compounds.

Watch out. Above $T_c$ a superconductor is just an ordinary conductor, often not a very good one. Lead, mercury and tin conduct worse than copper at room temperature.

Page 18 of 25

Properties of superconductors

Some facts about superconducting materials

• Apart from pure elements, many compounds and alloys are superconductors.
• Good conductors of current such as copper, silver and gold are not superconductors.
• Some elements become superconducting only under high pressure, and others only in special forms such as thin films.
• Compounds generally have higher transition temperatures than elements.
• A superconducting binary compound may be made from two superconducting elements, from one superconducting and one non-superconducting element, or even from two elements neither of which is a superconductor.

1. Zero resistance

Below $T_c$ the electrical resistance is exactly zero, so there are no $I^2R$ losses. Most applications are based on this property.

2. Persistent currents

A persistent current is a steady current that flows without diminishing in strength.

• Once such a current is set up in a closed superconducting loop, no external power supply is needed to keep it going, because there are no $I^2R$ losses.
• As long as the loop is kept below $T_c$, the current keeps flowing without decay. Persistent currents have been seen to flow for years with no measurable decrease.

3. Flux quantisation

Consider a current flowing round a superconducting ring.

• The magnetic flux density inside the superconducting material itself is zero, but magnetic flux can exist in the hole in the middle of the ring.
• This flux cannot take any value. It can only be a whole-number multiple of a basic unit, the flux quantum or fluxon:

• Numerically,

This is flux quantisation. The charge $2e$ in the denominator is the charge of a Cooper pair of electrons (see BCS theory), so flux quantisation is direct evidence that the current in a superconductor is carried by pairs.

Example 1. A superconducting ring of area $1\ \text{cm}^2$ has a field of $10^{-4}\ \text{T}$ through it. How many flux quanta does it hold?