1. ATOMIC PHYSICS
I’ve seen that the hydrogen atom (one electron, one proton) has an infinite set of
energy levels. These levels are characterized by quantum numbers n, �, and m�. These
have the values
n = 1, 2, 3, 4 . . .∞ � = 0, 1, 2, . . . n− 1 m� = −�,−(�− 1) . . . �− 1, �
Energy levels are shown in Fig. 1.
Fig. 1.
Write down where you can see it that � = 0, 1, 2, 3, 4 is also called s, p, d, f, g.
1
I’ve seen that there is a fourth number that characterizes the electron: spin s = 1/2.
This is an angular momentum, and the z component is
ms = +1/2 or − 1/2
or
ms = ↑ or ↓I’ve seen that spin half particles are fermions and obey the Pauli exclusion principle:
• No two fermions may occupy the same state.
• No two electrons may have the same set of quantum numbers, n, �, m�, and ms.
• Spin up and spin down are distinguishable.
• Spin up and spin up are indistinguishable.
These facts are all I need to understand the periodic table
Fig. 2. Periodic table. All 118 atoms shown.
2
Fig 2. Periodic table. All 118 atoms shown.
• Hydrogen has 1 electron; it goes into the 1s state. The spin can be ↑ or ↓.• Helium has 2 electrons. Both go into the 1s state. One has spin ↑ andone ↓.Comment 1: Actually, the two particle wave function has space parts for the two
electrons that look like the product of two hydrogen n = 1, � = 0 (1s) wave functions
multiplied by a spin state that is antisymmetric: |↑, ↓〉 − |↓, ↑〉.Comment 2: Helium has Z = 2, so the ground state 1s level is Z2 = 4 times lower than
in Hydrogen and it is occupied by two electrons. So neglecting interactions, the energy
is E = −108 eV. But the electrons repel each other and this adds a Coulomb energy
Vc = +1
4πε0
e2
|r1 − r2|
or +34 eV. So E = −74 eV.
3
• Lithium has Z = 3. Two electrons go into the 1s state, one with spin ↑ and one ↓.The third electron cannot occupy 1s (Pauli principle) so it goes into the n = 2 level,
the 2s orbital. The 2p electrons are higher in energy than 2s ones due to screening
effects that result from electron-electron interactions, so the 2s state is the one used.
Fig. 3. Li orbital filling.
The 2s electron’s wave function looks like the 1s wave function of hydrogen,
ψ∼(1− r
2a)e−r/2aY00,
except that the radius a differs from the Bohr radius on account of Z = 3 but also
on account of the two 1s electrons.
Fig. 4. Atomic sizes.
4
I can see how it is going. Each heavier atom will add one more electron to the wave
function. The electron will go into the lowest possible energy level.
• Beryllium has Z = 4. Two electrons in 1s, two in 2s.
• Boron has Z = 5. Two electrons in 1s, two in 2s, one in 2p.
Fig. 5. Be and B orbital filling.
There are (2� + 1) orbitals (m� values). For � = 1 there ares 3. Each can hold 2
electrons, one with spin ↑ and one ↓. for a total of 6. It turns out that the Coulomb
energy is less if I put 1 each into two orbitals rather than 2 into one orbital. This happens
for carbon and nitrogen, then the next 3 go to filling up all the 2p orbitals by neon.
5
Fig. 6. Nitrogen and oxygen orbital filling.
• Sodium has Z = 11, one more than neon, Z = 10. The 1s, 2s, and 2p orbitals of Ne
are all full. The 11th electron goes into 3s.
• The story continues for Z = 12 (Mg) . . . Z = 18 (Ar). With Ar the 3s and 3p orbitals
are all full.
6
The story partially breaks down with potassium, Z = 19. The n = 3 hydrogen wave
function has 3d levels (3d means � = 2) also. For the same reason that E2p > E2s, I’d
expect E3d > E3p > E3s. And it is. But even better, E3d > E4s
• Potassium, Z = 19 has “inner shells” like Argon with one extra electron in the 4s
orbital. Calcium has two in the 4s orbital.
• Then, electrons start to fill the 3d orbitals. The atoms from Sc to Zn are called
transition metals or d-electron atoms. With � = 2, there are 10 of them. The same
story (more or less) is told about the row 5 atoms, Rb. . .Xe.
• In row 6, after filling 6s and 1 5d level, there are the rare earths, Ce. . .Lu, where the
fourteen 4f orbitals are filled. In row 7, the actinides, Th. . .Lr, filling the fourteen 5f
orbitals. (Lr == Lawrencium)
The notation used in defining these is to list in energy order the levels, and then to
specify the “filling” with a superscript. Adding all the superscripts gives Z.
He is 1s2.Li is 1s22s1.Be is 1s22s2.B is 1s22s22p1.O is 1s22s22p4.
Ne is 1s22s22p6.Na is 1s22s22p63s1. Sometimes called [Ne]3s1.Fe is 1s22s22p63s23p64s23d6.Xe is [Kr] 5s24d105p6
U is 1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p67s26d15f3
7
Finally: Oganesson, Z = 118.
8
2. MOLECULES
2.1 Ionic bonds
The rightmost column is occupied by the “noble gases,” He, Ne, Ar, Kr, Xe, Rn, Og(!)
These are very inert, rarely participating in chemical reactions. The closed shell of, say,
Ne, 1s22s22p6 or Ar, 1s22s22p63s23p6 is very stable. Now consider Na, [Ne]3s1, and Cl,
[Ne]3s23p5. Na has one more electron than Ne and Cl has one less than Ar. The alkali
metal atoms, column 1, and the halogens, column 17, are very reactive.
If I take one electron off Na, forming Na+, and give it to Cl, forming Cl−, I have two
ions, now both closed shell, like the noble gases Ne and Ar. I can join these two ions to
form an NaCl molecule. I benefit from the Coulomb attraction of the oppositely charged
ions.
9
For the molecule to exist, the energy of NaCl must be lower than the energy of the two
atoms when they are far apart as neutral atoms. Indeed, it must be lower than the energy
of the two ions far apart.
The cost to remove the electron from Na is the “ionization energy” which for Na is
5.14 eV. There is a gain to be had by putting the electron onto Cl, the “electron affinity,”
which for Cl is 3.62 eV. If I do this, I am still in the hole by 1.52 eV.
But I more than break even by bringing the ions together. The attractive potential is
V (r) = − 1
4πε0
e2
r.
10
There is a repulsive “hard core” contribution, often written +A/r12. For small r this is
dominant.
11
The result is
For NaCl, the potential minimum is at 0.236 nm. The minimum energy (the
“dissociation energy” or “binding energy”) is 5.78 eV. The net energy gain is 4.26 eV;
the molecule is stable.
Most molecules (even very complex ones)∗ have ionic character to their bonds. In
addition to NaCl, there are the other alkali halides, KCl, LiF, KBr, CsI, NaH, HCl,† etc.There are alkaline earth molecules, BaF2. There is a raft of oxides, MgO, N2O, etc.
∗ And solids too!
† Hydrogen can act as a donor or acceptor.
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2.2 Covalent bond
Covalent bonds involve sharing of the electrons between two (or more) molecules. It
allows bonds like what occurs in diatomic molecules of the same kind, H2 and O2, and in
organic molecules, CH3CH.
Consider H2. Hydrogen gas is made of H2 molecules. (As are the liquid and the solid.)
Start with two hydrogens far apart. Each is in a 1s state ψ1s(r)∼e−r/a0 where r is
measured from the nucleus. When they are set side by side, the fact that electrons are
indistinguishable must be honored. So I can write the combined wave function as either
symmetric or antisymmetry linear combinations.
ψB = ψ1s(r1) + ψ1s(r2)
ψA = ψ1s(r1)− ψ1s(r2)
where now r1 and r2 point from some common origin to the electrons, the protons are
located at R1 and R2, and each electron sees the Coulomb potential of both protons.
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In the bonding (+) state, the wave functions add, and the amplitude and probability
are larger in the space between the protons than in either case alone. In the antibonding
(−) state, the wave function subtract; there is a zero in the middle, and amplitude and
probability are much smaller in the space between the protons than in either case alone.
Because the protons repel each other and attract the electrons, the antibonding state is
higher in energy than the bonding state. The bonding state is lower in energy that that of
the two isolated atoms; the molecule is stable.
These are called molecular orbitals; they are the analog of the atomic orbitals of
multi-electron atoms.
• Note again: the electrons are indistinguishable. I cannot know which atom they came
from.
• When you think about it, it is clear that some of the time both electrons are in the
“same” 1s orbital, especially in the bonding state. This is OK, it can hold 2 electrons.
• If the protons get too close, their +e2/R12 repulsion dominates. (The electron orbits
are always the size of a0 ≈ 0.05 nm.)
• MO theory tells us that the energy is a minimum when R12 = 0.106 nm. The binding
energy is 2.7 eV.
• The spins of the bonding state are anti-parallel, in accord with the Pauli principal.
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• The total spin is S = s1 + s2 = 0. The term for this state is that it is a singlet . A
magnetic field does not separate the level into two or more.
2.3 Other types of bonds
Google says: “U.S. Treasury, municipal, and corporate.”
There are polar, hydrogen, and metallic bonds.
• Polar == van der Waals, at least in liquids and solids. Dipoles attract, even if due to
fluctuations.
• Hydrogen. In water, H2O, the H of one molecule is attracted to the O of a different
one. It’s a long and a weak bond. (Water has a some ionic character, H is positive, O
negative. There is covalent character too!)
• Metallic. In metals like Na, K, Ag, the outermost s electron gets “delocalized” to move
through the metal. the remaining positive ions sit in the “sea of electrons” and are
bound. Mostly solids (and liquids).
2.4 Quantum mechanics of molecules
Molecules have a richer set of physics than atoms. The electrons can be excited
to higher levels. The two levels shown above are not the only ones. These come from
the 1s atomic orbitals but there are 2s, 2p, 3s, 3p, 3d, . . . levels, with “bonding” and
“antibonding” combinations of all. The excited electrons can be placed in any of these,
not necessarily the same. Moreover, bonding between electrons in states with angular
dependence (p, d etc) allow for additional richness. This is what allows organic chemists
to make almost anything, combining carbon (with two 2s electrons and two 2p electrons)
and hydrogen (only one 1s electron) into complex and useful molecules.
The optical spectra of organic and inorganic molecules are to the quantum mechanics
of molecules as are the optical spectra of atoms to the quantum mechanics of the atom.
I want to discuss things that don’t occur in atoms:
• The bond length, the distance between the nuclei, can change. The molecule can
15
vibrate. (It is a harmonic oscillator, in fact.)
16
• Classically, the molecule is two masses on the ends of a stick. The stick is the bond
joining them. The lingo is that it is a “free rotor”
The molecule can rotate in three dimensions (axis along x, y, or z). But rotation
along the molecule’s axis has no inertia, because the all mass is in the protons. Thus, mR2
is small if R ≈ 10−15 meters compared to 10−10 m for the proton-proton distance, the
length of the other two axes. For the diatomic molecule (H2, HCl) I’ll consider rotation
around only two axes, both perpendicular to the bond. More complex molecules (H2, CH4,
C10H4N2) have 3 axes.
2.5 Rotational energy levels
The free rotor has no potential energy∗ The kinetic energy of a rotor with moment of
inertia I rotating at angular frequency ω is
K =1
2Iω2 =
(Iω)2
2I=
L2
2I
with L the angular momentum.
In quantum mechanics L is an operator†; I’ve seen it in the quantum mechanics of the
hydrogen atom, where I = mer2.
So I write
Hψ = Eψ
with H = K.
Then:1
2IL
2ψ = Eψ
But I know the eigenfunctions and eigenvalues of L2, they are the spherical harmonics
Y�m that also give the angular behavior of the electron in hydrogen. the eigenvalues are
L2Y = �(�+ 1)h2Y
with � = 1, 2, 3 . . . So, set ψ = Y .
∗ Otherwise it would be called a “hindered rotor.”
† I put the hat on to emphasize this. I have done it before. I may not do it again. H is an operator too and
I did not bother there.
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Then
E� = h2�(�+ 1)
2I.
The energies (and, hence, the rotational velocities) of the quantum rotor are
quantized. The ground state energy is for � = 0 and is E0 = 0. Then next level is for
� = 1, E1 = h2/I. The energies increase as the rotor spins up. The energy difference
between � and �+ 1 is
ΔE�+1,� =h2
2I[(�+ 1)(�+ 2)− �(�+ 1)] =
h2
2I(�+ 1).
The moment of inertia for a diatomic molecule is I = m1R21 +m2R
22 with the distances
measured from the center of mass. But the center of mass is set by m1R1 = m2R2. Now
let R0 = R1 +R2 and you get
I = μR20,
with1
μ=
1
M1+
1
M2.
The energy scale is rather low. The first rotational level of HCl, with � = 1 is 2.6 meV.
• In reality, the separation (and the moment of inertia) is not constant. It is a spring
and it stretches as the molecule is spun up, increasing the moment of inertia.
• If I equate thermal and rotational energies, 2.6 meV = kT when T = 30 K. At room
temperature, 300 K, many rotational levels are occupied.
• Precision spectroscopy can measure the rotational transitional energies to 1 part in
1000, allowing for detection of a wide range of molecular species.
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2.6 Vibrational energies
The potential energy of the molecule can be represented as a harmonic oscillator.
I know how to solve the harmonic oscillator, especially as this is a one-dimensional
problem. The wave functions are the Hermite polynomials. And the energies are
En = (n+1
2)hω0
where ω0 =√
K/μ, with K the “spring constant” and μ the reduced mass. Typical
vibrational energies of diatomic (and other) molecules are in the 0.1–0.5 eV range. For
HCl, the energy spacing is hω0 = 0.36 eV. (2 eV is red light, so the vibrational energy
corresponds to infrared frequencies.)
2.7 Vibrational-rotational energies
The selection rule for optical transitions is Δ� = ±1. So the absorption spectrum of
molecules involves both rotational and vibrational energies. Because (at 300 K) a number
of rotational levels are populated, there are many absorption peaks.
HCl spectrum. The two peaks are there because there are two Cl isotopes.
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3. SOLIDS
To consider solids, I take a huge step in the complexity of the system. Instead of 1 or
2 or even 92 electrons in an atom or molecule, a cubic centimeter of a solid may contain
1022 electrons. The Schrodinger equation still applies, and I’ll be able to make progress.
But, as opposed to the accuracy of the solution for hydrogen, I’ll rely on making some
simplifications.
As in molecules, solids have bonds that hold their atoms together. They may be ionic,
covalent, metallic, or van der Waals. These bind the atoms to each other. The lowest
energy state is that of a crystal, a periodic arrangement of atoms.
The textbook has some discussion of crystal bonding. Here I’ll discuss crystal
symmetry.
Translational symmetry is the basic symmetry that defines a crystal. Symmetry is
a fundamental concept in physics. The basic idea is that any physical system which is
unchanged—the correct word is “invariant”—under some operation will have observables
that must respect the symmetry.
3.1 Translational symmetry
Each crystal structure can be constructed by starting at some point in space and
moving by the translation vector (actually a set of vectors) that defines the crystal to
generate all the points of the crystal lattice. For example, I can define the NaCl lattice by
starting at the center of an Na atom and then moving by a translation vector to the center
of a neighboring Na atom, then by a translation vector to the center of another Na atom,
and so on. Now, if I move to the nearest Cl atom, the same set of translation vectors will
trace out all the Cl atoms in the crystal.
The translation vector is usually written as
T = na+mb+ pc
where n, m, and p are (positive and negative) integers and the vectors a, b, and c define
the unit cell of the crystal. The set of points defined by T as the integers run over the
range from minus to plus infinity is known as a Bravais lattice. To each point of the
Bravais lattice one attaches a basis of atoms. (The attachments is done in an identical
fashion for each point in order to preserve translational symmetry.) The basis in the
example above is one Na and one Cl atom.
The vectors a, b, and c define a rectangular prism with edges of length a, b, and
c; there are angles α, β, and γ defined respectively in the following way. α is the angle
between b and c. β is the angle between c and a. γ is the angle between a and b.
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Fig. 7. A sketch of a face centered cubic lattice (for a monatomic crystal), showing the
primitive translation vectors (red) and the conventional translation vectors (blue).
3.1.1 Seven crystal systems
Every crystal belongs to one of seven crystal systems. Each system is defined in
terms of the relative lengths of the unit vectors a, b, and c of the lattice unit cell, as well
as the angles between these vectors. The seven crystal systems are cubic, tetragonal,
orthorhombic, monoclinic, triclinic, hexagonal, and trigonal. They are illustrated in Fig. 8.
Fig. 8. The seven crystal systems.
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Cubic. I start with the system with the highest symmetry: cubic. The cubic crystal
has all axes equal (a = b = c) and all angles 90◦. The rectangular prism of the cubic
system is, not surprisingly, a cube.
Tetragonal. If I take the cubic crystal and expand (or compress) it along one axis, I
get a tetragonal crystal. Here, a = b �= c; all angles are 90◦.Orthorhombic. If I take the tetragonal crystal and expand (or compress) it along a
second axis, I get an orthorhombic crystal. Here, a �= b �= c; all angles are 90◦.Monoclinic. If now I take the orthorhombic crystal and tip one axis away from the
normal to the plane defined by the other two in a way that preserves two right angles, I
get a monoclinic crystal. For the monoclinic crystal, a �= b �= c; the angle β �= 90◦ while
the other two are 90◦.Triclinic. To make triclinic, I take monoclinic and tip the other axes so that none are
perpendicular to any other, making α �= β �= γ �= 90◦. It remains true that a �= b �= c also.
I’ve now made the lattice as low in symmetry as I can.
Hexagonal. I return to the tetragonal system and open the angle γ to 120◦. The
formerly square ab plane is now a hexagonal figure. The hexagonal crystal has a = b �= c
and α = β = 90◦; γ = 120◦.Trigonal. Finally, I return to the cube and pull it along one body diagonal, rendering
all angles different from 90◦ but all still equal. The axes also remain equal, a = b = c.
Trigonal sometimes called rhombohedral.
Fig. 8. The seven crystal systems.
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3.2 Metals
The free-electron theory of metals was developed in the late 1920s and early 1930s.
The basic idea is that one or more electrons of each atom in the solid are detached from
their atomic location and free to move throughout the crystal as a free gas. The electrons
are fermions, subject to the Pauli principle. The electrons are said to form a “Fermi gas”
with a Fermi surface that is the constant-energy surface of the most energetic electrons.
The Pauli principal overwhelms other forces. Thus, the free-electron model treats the
potential energy in which the electron moves (the ion potential) as a constant. Including
the potential can be done of course; this is the realm of electronic band structure. The
band structure—in particular the presence of the filled d-electron levels below the Fermi
level—is the reason why gold, silver and copper appear different. The basic low-energy
physics is however very close to what the free-electron model predicts.
The electron-electron interaction is also left out of the free-electron model, so that one
individual electron state is independent of the other electron states.
3.2.1 Schrodinger equation for free electrons
The simple free-electron model is surprisingly successful in explaining many
experimental phenomena, including electrical conductivity, heat conductivity, the
Wiedemann-Franz law (a relation of the ratio of electrical conductivity to thermal
conductivity), the electronic contribution to the heat capacity, Hall effect, and optical
properties. I’ll start with the Hamiltonian, a sum of kinetic and potential energies:
H =p2
2m+ V ,
where p is the momentum, m the mass, and V the potential. In general I would have
V = V(r) where r is the location of the electron and the potential would be periodic in the
lattice structure and account for the attraction of the electron to the positive ions in the
crystal. But in the free electron model I take V = constant and I might as well take it to
be zero.∗
I’ve already defined the momentum operator: p = −ih∇. The free electron
Hamiltonian is
H = − h2∇2
2m(1)
and the time independent Schrodinger equation is
Hψ = Eψ, (2)
∗ I need a wall at the surface to keep the electrons in the crystal, but the choice of V’s constant value has
no effect. I take the wall height to be infinite. (And then I eliminate the walls by using periodic boundary
conditions.)
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where E is the energy eigenvalue and ψ is the wave function of the single electron I am
considering.
3.2.2 Wave function
Consider a plane-wave wave function:
ψ = ψ0eik·r, (3)
with ψ0 a complex amplitude and k a wave vector to be determined by solving the
Schrodinger equation.
Of course ∇2ψ = −k2ψ. I need to normalize the wave function:
1 = 〈ψ|ψ〉 =∫VdV ψ∗(r)ψ(r) =
∫VdV |ψ0|2 = |ψ0|2V
where V is the entire volume of the crystal. Then,
ψ0 =eiφ√V
with φ an arbitrary phase. (The time-dependent Schrodinger equation leads to a phasor
e−iωt where the frequency ω is related to the energy E by Planck’s constant: E = hω.)
Substituting Eq. 3 into Eq. 2 with the Hamiltonian of Eq. 1 yields
+h2k2
2mψ = Eψ
Multiply from the left by ψ∗ and integrate to find
E =h2k2
2m. (4)
Eq. 4 is just the kinetic energy of a free particle. The momentum operator is p = −ih∇and so pψ = hkψ and so I can identify hk as the momentum. The plane-wave wave
functions are eigenfunctions of both the Hamiltonian and of momentum. The energy is
quadratic in k, with zero energy∗ at k = 0. The probability density to find the electron at
any position in the crystal is constant, P (r) = ψ∗ψ = 1/V . For conceptual clarity, let me
assume a cube of edge length L = Ja where a is the lattice constant of the crystal and J is
a (large) integer. There is one free electron per atom. The volume is V = L3 ≈ J3a3. If L
is 1 cm, and a ≈ 4.6 A, then the solid contains about 1022 electrons and J ≈ 2.2× 107.
∗ Recall: the potential energy is at zero.
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3.2.3 Exclusion principle and boundary conditions
The electrons are fermions; only one electron can have a particular set of quantum
numbers. The quantum numbers here are the three components of the wave vector k
and the spin: {kx, ky, kz, σ} where σ =↑ or ↓. The allowed values of the wave vector
components are set by the boundary conditions. If I consider particle-in-a-box conditions,
then the wave function will be zero outside the solid, and the solutions are sines and/or
cosines depending on the choice of coordinate-system origin.
It is much better to use periodic boundary conditions. I’ll write these mathematically
as
ψ(r+ L) = ψ(r), (5)
where L is oriented along either the x, y, or z directions. One may think of this as
illustrated in Fig. 9. I take the block at the center and make many copies of it and lay
them all next to each other, filling space with replicas of the center block. The system
boundary is moved off to infinity, but the fact that the copies are exact means that Eq. 5
is obeyed. For example the electron with the gray center about to leave the block on the
right will reappear from its neighbor on the left.
Fig. 9. A square system, with edge L, containing N electrons is repeated many times and
the copies laid as tiles to cover the plane. The system is periodic in both directions with
period L.
I will apply periodic boundary conditions to the free-electron wave function of Eq. 3
for L = Jax:
ψ(r) = ψ0eik·r = ψ(r+ L) = ψ0e
ik·(r+Jax)
with a the lattice constant. Now, k · x = kx. The periodic boundary conditions for this
specific value of L then yield
eikxJa = 1 (6)
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This equation has a large number of solutions; the phase of the complex exponential kxJa
must be an integer (positive or negative) times 2π radians. Hence
kx =2π
Jajx where jx = 0, ±1, ±2, . . .
Applying periodic boundary conditions in the y and z directions give equivalent
quantization of ky and kz. Because the components of the wave vector are set by the
integer values of {jx, jy, jz} and are no longer continuous variables, the energy, Eq. 4 is
also quantized:
E(k) = h2k2
2m=
h2
2m
(2π
Ja
)2
(j2x + j2y + j2z )
I note that there are many degeneracies in the energy spectrum of the free electrons. The
two values of spin (up and down) are degenerate as well as many permutations of integers
(positive and negative). These permutations are distinguishable because the momentum is
distinguishable:
k = (2π/Ja)(jxx+ jyy + jzz). (7)
The Pauli principle says that no two electrons can have the same quantum numbers. I
can call the set {jx, jy, jz, σ} the quantum numbers. I now pour electrons into the metal,
stopping when I reach electrical neutrality.∗ The first few energies are
E0 = 0 {0, 0, 0, ↑} {0, 0, 0, ↓}E1 = δ {1, 0, 0, ↑} {0, 1, 0, ↑} . . . {0, 0, 1, ↓}E2 = 2δ {1, 1, 0, ↑} {1, 0, 1, ↑} . . . {0, 1, 1, ↓}
where δ = (2πh)2/2mJ2a2 is a typical spacing between levels.
3.2.4 The Fermi energy
The values of k allowed by the boundary conditions form a grid of points in when
plotted in three-dimensional space† with axes kx, ky, kz. Because the energy E is
quadratic in the wave vector k, the surfaces of constant energy are spheres of radius
k. Indeed,
E(k) = h2
2m
(2π
Ja
)2
j2
where j2 = j2x + j2y + j2z is an integer.
∗ One to a few electrons per atom in simple metals; one to a few per formula unit in metallic compounds.
† This space is called “momentum space,” “k-space,” or “reciprocal space.” It has units of inverse length.
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The highest energy of any of these electrons (at T = 0) is called the Fermi energy:
EF =h2
2mk2F =
h2
2m
(2π
Ja
)2
j2F
where kF is the radius of the sphere corresponding to the highest energy electrons and jF
is the radius of an equivalent sphere in coordinates of the integers jx, jy, jz. A diagram of
this Fermi sphere is shown in Fig. 10.
Fig. 10. The Fermi sphere of the free-electron metal. The surface of this sphere is the Fermi
surface.
Now let me calculate the radius kF of the Fermi surface. There are N electrons in the
metal. Because of the spin degeneracy, two of these electrons occupy each point on the
grid of points indexed by jx, jy, and jz. The total number of points inside the sphere is
just the volume of the sphere in “j-space.” Hence,
N
2=
4π
3j3F (8)
or
jF =
(3N
8π
)1/3
.
The Fermi wave vector is
kF =2π
JajF =
1
Ja
(3π2N
)1/3.
But (Ja)3 = V or Ja = V 1/3. Hence∗
kF =
(3π2N
V
)1/3
=(3π2n
)1/3(9)
where n = N/V is the number density of electrons.
∗ I can now forget ji and J . It was helpful for me to introduce them; it was not necessary. In fact, the
periodic boundary conditions may also be left behind at this point.
27
Finally, the Fermi energy is
EF =h2
2m
(3π2n
)2/3(10)
Related to the Fermi energy is the Fermi temperature TF , defined as EF /kB, where kB
is Boltzmann’s constant. I also define the Fermi momentum pF and Fermi velocity vF .
pF = hkF =√2mEF
and
vF =pFm
.
These quantities correspond respectively to the momentum and group velocity of an
electron at the Fermi surface. Fermi surface parameters for many metals are shown in
Table 1, after Ref. 1
Table 1. Free-electron Fermi surface parameters for simple metals
Valence Metal n kF vF EF TF
1022 cm−3 108 cm−1 108 cm/s eV 104 K
1 Li 4.70 1.11 1.29 4.72 5.48
Na 2.65 0.92 1.07 3.23 3.75
K 1.40 0.75 0.86 2.12 2.46
Rb 1.15 0.70 0.81 1.85 2.15
Cs 0.91 0.64 0.75 1.58 1.83
Cu 8.45 1.36 1.57 7.00 8.12
Ag 5.85 1.20 1.39 5.48 6.36
Au 5.90 1.20 1.39 5.51 6.39
2 Be 24.2 1.93 2.23 14.14 16.41
Mg 8.60 1.37 1.58 7.13 8.27
Ca 4.60 1.11 1.28 4.68 5.43
Sr 3.56 1.02 1.18 3.95 4.58
Ba 3.20 0.98 1.13 3.65 4.24
Zn 13.10 1.57 1.82 9.39 10.90
Cd 9.28 1.40 1.62 7.46 8.66
3 Al 18.06 1.75 2.02 11.63 13.49
Ga 15.30 1.65 1.91 10.35 12.01
In 11.49 1.50 1.74 8.60 9.98
4 Pb 13.20 1.57 1.82 9.37 10.87
Sn(w) 14.48 1.62 1.88 10.03 11.64
28
Fermi energies are in the range of 1.5–15 eV; Fermi temperatures are thus 18,000–
180,000 K. These temperatures are much above 300 K, the temperature of the room. I did
not make a bad mistake by considering T = 0. The Fermi velocities are surprisingly large;
they are in the range 0.8–2×108 cm/sec. The average is about c/200. For example, silver
has EF = 5.5 eV, TF = 64, 000 K and vF = 1.4× 108 cm/s.
3.2.5 The density of states
I may need the density of states, so let me work it out now. I can write an equation
for NE , the number of electrons with energy between the zero of energy and energy E , bynoting that Eq. 10 is such an equation, so
E =h2
2m
(3π2NE
V
)2/3
so that
NE =
(2mEh2
)3/2V
3π2
I now raise E by just a little bit dE and find that the number goes up by dN .
NE + dN =
(2m(E + dE)
h2
)3/2V
3π2
Because dE is infinitesimal, I can expand (E + dE)3/2 = E3/2(1 + 3dE/2E + . . .) to find
NE + dN =
(2mEh2
)3/2V
3π2(1 +
3dE2E )
The first term on the right is just NE and the second is dN . The density of states (the
number of states between E and E + dE) is
dN
dE = D(E) =(2m
h2
)3/2V
2π2
√E .
The density of states∗ grows as the energy increases as the square root of the energy.†
∗ Sometimes the symbol N(E) is used for the density of states.
† The√E behavior is correct for three-dimensional metals, where the energy depends on three—x, y, and
z—components of k. In two dimensions D(E) is constant and in one dimension D(E) ∼ 1/√E .
29
3.2.6 Electrical conductivity
In the absence of an external field, the metal does not carry current. The electrons are
moving in all directions and the average or net motion is zero.
Consider now the effect of an electric field on these electrons. To be definite, I will
specify the field as a uniform field oriented in the y direction.
E = yE0. (11)
The field exerts a force on each and every charged electron:
Fext = −eE = −eyE0.
Newton’s second law, F = ma is best written
∑F =
dp
dt= h
dk
dt(12)
So the external force and any internal forces will change the k-state of the electron. The
resultant response is not hindered by the Pauli exclusion principle: as the electron in state
k evolves to k′, the one at k′ has gotten out of the way by shifting to k′′.What other forces act on the electrons? Well, the electrons are not bound; indeed,
the potential energy has been set to zero. But the electrons can suffer collisions with
impurities, other defects, lattice vibrations, and the surface. These collisions have a mean
free time τ and relax the system back towards equilibrium. This is Drude’s model of
conduction. I will consider that the relaxation generates an impulse Fcollτ which changes
the momentum hk back to the equilibrium value∗ hk0. Thus,
Fcoll = −hδk
τ= −h
k− k0
τ. (13)
The collisions enter the picture as abrupt changes in momentum that occur every τ
seconds. Note that hδk = mδv = mvdrift where vdrift is the average velocity acquired by
the collection of electrons, known as the drift velocity.
Let me think for a moment about how the applied electric field affects the motion of
an electron. The electron has wave vector k0 or velocity v0 = hk0/m; this velocity will be
changed (in magnitude, direction, or both) by the field. I will write the k of my electron
in the presence of the field as
k = k0 + δk, (14)
with δk in the direction of the field, y. δk represents the linear response of the electron
to the external field. k0 is the wave vector in the absence of the field; it does not depend
∗ Drude considered the equilibrium value to be zero, but I know that it is close to the Fermi momentum.
30
on the time. δk contains the amplitude and phase of the driven motion of the electron.
Newton’s second law becomes
hdk
dt= −h
k− k0
τ− eE. (15)
In the steady state, nothing is changing, and dkdt = 0. On substituting for k, I get
0 =−hδk
τ− eE0 (16)
A little algebra then yields this relation between the shift of k and the electric field:
δk = −eτ
hE0, (17)
where all vectors are in the y direction. Notice that h appears. The appearance will be
fleeting, however, and the theory is at best semiclassical.
The picture of what happen in the metal due to the external electric field is not
intuitive. Every electron has changed its k-vector under the influence of the electric field.
This displacement occurs alike and by the same amount to the deeply buried electrons
with small k values and low speeds and to the Fermi-surface electrons which are whizzing
along at c/200. The Fermi surface, which is centered at k = 0 in equilibrium is moved as a
whole by δk, as illustrated in Fig. 11.
Fig. 11. The momentum-space location of the Fermi sphere in equilibrium is within the
dashed line that indicates the Fermi surface. When an electric field is applied along y, the
electrons are accelerated downward. (They are negatively charged.) The displaced sphere
is shown in blue. There is a current, because more electrons are going down than are going
up. The acceleration continues until it is balanced by the scattering of electrons; one such
event is shown.
31
Were there no scattering, the displacement of the Fermi sphere would continue to
grow; the electric field exerts a force on the electrons, causing acceleration. At zero
frequency, both δk and the velocity of each electron would change linearly with the time,
increasing for left-moving electrons and decreasing for right-moving electrons. With
scattering, the relaxation eventually balances the acceleration and there is a fixed, steady-
state∗ displacement. Because scattering events are uncorrelated and affect at random
one or the other of the independent and non-interacting electrons, the relaxation process,
shown schematically in Fig. 11, must respect the Pauli principal, taking the electron from
a filled state to an empty state. If the scattering is elastic, then the energy of the initial
and final states should be the same. So the relaxation process takes an electron from the
leading edge of the displaced Fermi surface, where electrons occupy states that are empty
in equilibrium, and deposits it just outside the trailing edge, where states that once were
occupied have been emptied.
The electrons that are relaxed are therefore very close to the Fermi energy and move
at the Fermi velocity.† Thus the relation between the mean free time τ and mean free
path � is
� = vF τ (18)
The electrical current is j = −nevd where n is the electron density and vd is an
average or drift velocity. This velocity is not the Fermi velocity, but the average of all
electrons’ responses to the field. The averaging process goes like:
vd = 〈vk〉 = 〈v0 + δv〉 = 〈δv〉 = δv
where the last equality comes from the fact that every electron feels the same force. In
the above, the velocity of an electron with wave vector k is vk = pk/m = hk/m and
vd = hδk/m, making
j = −neh
mδk. (19)
I then substitute δk from Eq. 17 to obtain
j = −(neh
m
)(−eτ
h
)yE0.
Using Ohm’s law, j = σE, I find finally
σ =ne2τ
m. (20)
∗ But not equilibrium.
† The realization that electron relaxation involves the fast-moving Fermi-surface electrons is where quantum
mechanics (and quantum statistical mechanics) leads to new concepts in the Drude theory.
32
3.2.7 Uncertainty principle
Uncertainty principle arguments are always worth exploring. An electron with
quantum numbers k travels freely between collisions for a time τ . Whether the collision
is elastic or inelastic, the length of time the electron occupies state ψ(k) is τ . The
consequence is that the energy is not known better than ΔE = h/τ . I use Eq. 4, E =
h2k2/2m, to find (by differentiation) that the energy blurring of ΔE means a momentum
spread of Δk. ΔE = h2kΔk/m or (because the relevant energy is the Fermi energy)
Δk =ΔEkF2EF =
hkF2EF τ =
m
hkF τ=
1
vF τ
But! vF τ is the mean free path �! So the position uncertainty is the mean free path
and the momentum uncertainty is hΔk = h/�. This calculation should encourage you to
stop thinking about the electron as a point-like particle skating around like a billiard ball
on table filled with obstacles and think of it as a quantum-mechanical wave packet that is
delocalized over a range given by the mean free path.∗
∗ If the idea of the physical extent of the wave function being the mean free path bothers you, do not forget
that the wave function of the plane-wave state eik·r has finite amplitude everywhere in the solid.
33
3.3 Semiconductors and insulators
This section is about the band structure of semiconductors and insulators. All solids
show the effects of what are called electronic “bands.” The bands appear for two reasons
• The material is a crystal with translation symmetry.
• There is an attractive potential energy for electrons that follows the periodicity of the
lattice.
V (x+ na) = V (x)
with n an integer.
I know or assume some other things:
• I’ll do the discussion for a one-dimensional chain or line of atoms. Its length is
L = Na.∗ The physics is all there and one can extend to two of three dimensions
pretty easily.
• The electrons are non-interacting . I neglect the repulsive potential among the
electrons and consider only the attractive potential of the atoms.
• I’ll use k = p/h as the quantum numbers of the electrons. I’ll add spin 1/2 as well.
• The Pauli principal reigns supreme. Each energy level can hold one spin up electron
and one spin down electron.
• I’ll use periodic boundary conditions. Then, I will have quantization of k as in metals:
k =2π
Nan where n = 0, ±1, ±2, . . .
3.3.1 Band structure
I first should say something about what these “bands” really are. Free electrons do
not have bands. The energy is solely the kinetic energy h2k2/2m; I take the electrons and
pour them into these free-electron states until I have run out of electrons.
In reality of course, no electron is actually free. All interact with the atomic nuclei
(and with other electrons). The innermost ones are deeply bound in atomic-like very
low-energy states. The most energetic electrons may be nearly free, with the details
determined by the chemical and crystallographic structure of the solid. Even states above
the Fermi energy (which are unoccupied—or empty—at low temperatures) will be affected
by the band structure. The electron energies turn out to be organized into a range of
quasi-continuous energies, called “bands,” separated by forbidden energy gaps where no
states exist.
∗ I used J in the metals section (3d) so that J3 = N . For the chain, J = N ; so use N . N is the number of
atoms in the chain of atoms.
34
Fig. 12. Upper panel: Hydrogen ground-state wave function for two atoms: A and B. Lower
panel: the two wave functions can either be added or subtracted. Addition (purple) leads
to the bonding orbital; subtraction (green) to the antibonding orbital.
Fig. 13. Splitting of the bonding and antibonding levels in H2.
3.3.2 Some qualitative arguments
Let me start by considering our friends, the H atom and the H2 molecule. I know
that the energies of electrons bound to this atom are quantized into discrete levels and
that electromagnetic absorption (photons) will promote electrons from one level to
another. Consider the ground state wave function of a simple atom, such as hydrogen.
Two H atoms and their 1s orbitals are shown in the upper panel of Fig. 12. The wave
function falls off quickly away from the central nucleus. When the two hydrogens are
allowed to interact, the two wave functions can add either constructively or destructively.
Constructive addition leads to the bonding orbital; destructive to the antibonding orbital.
The energy levels are shown in Fig. 13. The bonding level holds the two electrons and the
antibonding level—which could hold two more—is empty.
Now consider an long chain of such atoms, containing N atoms, all separated by
lattice constant a. There will be many ways to form linear combinations of the wave
functions and so there will be many levels, ranging from one where all are added, so all in
phase, to one where each is 180◦ out of phase with its two neighbors. Indeed there will be
N such levels.
If there is one electron/atom, half will be full and half empty. I claim that this linear
chain would be a metal.
If there were two electrons per atom, the uniform chain would have every level filled
35
with two electrons, one spin up and one spin down. It seems natural to claim that this
chain would be an insulator.
I have gone about as far as I can without doing some calculations. Let me return to
my free electron model and add a periodic potential.
3.3.3 Nearly free electrons
The nuclei in the solid form a periodic lattice, so that the potential satisfies
V (r) = V (r+T) (21)
where T is the translation vector of the crystal. In the one-dimensional case I will
consider, T = nax where the lattice constant is a and n is an integer. The one-
dimensional chain has length L = Na.
Now if V = 0, E = h2k2/2m (free electrons) with k = 2πn/L (periodic boundary
conditions). What is the effect of the periodic potential? Well, the free electron states are
plane waves,
ψ =eikx√L
and plane waves traveling in a periodic medium (e.g., a crystal or a diffraction grating)
are diffracted. In one dimension, the incident wave and the diffracted wave must both
travel along ±x. Diffraction effects thus produce waves traveling in both directions along
the chain of atoms; they are analogous to a partial reflection. (As in the potential barrier
problem.
For certain values of k, I can get strong interference effects. Consider a partial wave
which travels forward for distance a, is reflected and travels backward for distance a, and
is reflected again. If the total distance traveled is an integer times the wavelength, there
will be constructive interference. So the condition for diffraction is 2a = λ. But of course
k = 2π/λ so the condition is also
k = mπ
a
where m, the diffraction order, is a positive or negative integer. Look at the case where
m = ±1. (Other orders have higher energies.) Because of diffraction and constructive
interference, the wave functions become standing waves:
ψ1 =eiπx/a + e−iπx/a
√2L
=
√2
Lcos(
πx
a)
and
ψ2 =eiπx/a − e−iπx/a
√2L
= i
√2
Lsin(
πx
a)
36
where I have done a side calculation for the normalization. The standing waves lead to a
non-uniform probability density,
|ψ1|2 = 2
Lcos2(
πx
a)
and
|ψ2|2 = 2
Lsin2(
πx
a)
One probability density has crests at x = 0, ±a, ±2a, . . . and the other has crests half a
lattice constant away.
Because the potential V (x) is periodic, it can be written as a Fourier series; because
the period is a, the longest wavelength component is a, making the lowest Fourier
component 2π/a. Let me consider that component alone,
V (x) = −V0 cos(2πx
a)
with a minimum at the lattice sites 0, ±a, etc. and a maximum between the atoms. For
ψ1 ∼ cos(kx),
E (1)1 =
∫ L
0dxψ∗
1V ψ1
= − 2
LV0
∫ L
0dx cos(2π
x
a) cos2(π
x
a)
but cos2(φ) = [1 + cos(2φ)]/2 and so
= − 1
LV0
∫ L
0dx[cos(2π
x
a) + cos2(2π
x
a)
The first term in the integral is zero and the second is L/2, so that
E (1)1 = −V0
2and, for ψ2 ∼ sin(kx)
E (1)2 = − 2
LV0
∫ L
0dx cos(2π
x
a) sin2(π
x
a)
= +V02.
For the states with k = ±π/a, there is an energy difference between the two wave
functions. One (the one that goes as sin(kx)) is raised and the other (that goes as
cos(kx)) is lowered. That this happens is what one might have expected: the cos(kx)
function has the electron most likely to be at the nuclei of the atoms, where the potential
37
is a minimum while the sin(kx) function has the electron between the nuclei, where the
potential is maximum.
The jump in energy of magnitude V0 at k = ±π/a means that there are is a range
of energies for which there are no solutions to the Schrodinger equation. Note that the
number of k states is unchanged, for spin-1/2 electrons there are 2 2π/a2π/L = 2N states
between k = −π/a and k = +π/a.∗ Moreover, there are no disallowed values of k—
only disallowed values of E(k). These disallowed energies are call forbidden energy gaps.
Additional gaps appear at k = ±2π/a, ±3π/a, . . . .
The interference/diffraction effect affects ψ(k = ±π/a), which has λ = 2a, the most
but states with nearby wave vectors are also affected. A calculated dispersion relation is
shown in Fig. 14. Three bands and part of a fourth are shown; there are gaps between
the bands. The underlying parabolic free-electron curve is also shown; it is close to these
curves in the middle of the bands, but passes below or above them near the band edges.
Fig. 14. Energy as a function of the wave vector for a solid with a periodic lattice potential.
There are 3 bands shown completely, plus part of a fourth. There are gaps between the
bands where no allowed energies exist. The free-electron parabola, E = h2k2/2m, is shown
as a dashed line.
I can now understand the reason why some materials are metals and some insulators.
Suppose the linear chain is made up of hydrogen, which has 1 electron per atom. Or
suppose it is potassium or silver, with effectively one valence electron per atom. Then
there are N electrons and 2N valid k values, so the Fermi wave vector is kF = ±π/2a; half
the states in the lowest-energy band are occupied at T = 0 and half unoccupied.
If there are two electrons per atom (helium), this one-dimensional solid would have
the first band full and a gap in energy between the highest occupied state (the most
energetic state in the first band) and the lowest unoccupied state (the least energetic state
in the second band). Then kF = ±π/a and the electrons cannot be accelerated (with
∗ Plus one. The range is more properly stated as −π/a < k ≤ π/a.
38
a corresponding increase in momentum) without acquiring additional energy. The solid
would be an insulator.
If there are 3 valence electrons per atom (aluminum), the one-dimensional Fermi wave
vector is kF = ±3π/2a. The first band is full but the second is half-full. Metal.
If 4 (silicon, germanium), then kF = ±2π/a and the first and second bands are both
full. The material would be an insulator.∗
3.3.4 The Brillouin zone
Bloch’s theorem is the basis for much of the picture solid-state physics has for
electrons, phonons, and other quantum-mechanical objects in a crystal with translational
invariance. I won’t prove it here, but state the result of the theorem:
ψk(r) = eik·ruk(r), (22)
where uk(r) is periodic in T, as in Eq. 21. So uk(r) describes the variation of the wave
function within the unit cell and the plane-wave exponential function gives the phase
between unit cells. If uk = constant, then Eq. 22 is the wave function for free electrons.
Otherwise,
u(x+ a) = u(x).
Bloch’s theorem, the periodic lattice, and the boundary conditions all conspire
to limit what values k can take. I began with it as a real number (a quantity along
a continuous line) taking any value from negative to positive infinity. Then, periodic
boundary conditions restricted it to a countable infinity, represented by integers. Now
I will show that only a finite range of these integer-based k values has distinct physical
significance.
I start by once again translating by a single lattice constant
ψ(x+ a) = uk(x+ a)eik(x+a) = uk(x)eikxeika
= eikaψ(x). (23)
Now, define a wave vector k′ by
k = k′ +2πm
a,
with m an integer. Substitute into Eq. 23 to find
eikaψ(x) = eik′ae2πimψ(x).
But e2πim = 1 so
eikaψ(x) = eik′aψ(x).
The state with wave vector k and wave vector k+2πm/a, with m an integer, are the same.
∗ As indeed are Si and Ge. They are called semiconductors for historical reasons. The band gap is not so
large in these materials as in, for example, Al2O3 or diamond.
39
Fig. 15. Left: Energy as a function of the wave vector in the reduced zone scheme. Right:
Energy as a function of the wave vector in the repeated zone scheme. Both diagrams show
three bands completely, plus part of a fourth.
The wave function is the same as is the energy eigenvalue. k is periodic with period 2π/a.
Values of k in the range
−π
a< k ≤ π
a(24)
are all I need.
I am free to modify Fig. 14 by adding or subtracting integer times 2π/a as many times
as needed to bring the wave vector into the range specified by Eq. 24. This translation
produces the dispersion relations shown in the left-hand panel of Fig. 15. The small price
that I have to pay for this reduction is that there are many values of energy at for any k
in range, so that I have to supply a band index to specify about which I speak.
The range of k specified by Eq. 24 is said to be in the first Brillouin zone. The bands
lie one above another. I’ll have to supply a band index n to specify the wave function and
the energy En. Despite this drastic change, the basic discussion of page 38 as to whether
a given material is metal or insulator is unchanged. I may also apply translations by the
range of the Brillouin zone, 2πm/a, m = 1, 2. . . , to extend each of the bands, as shown
in the right panel of Fig. 15. The curves join smoothly, repeating once per Brillouin zone.
All three representations (extended, reduced, and repeated) are equally valid.
40
3.3.5 Band gaps of semiconductors
Table 2, based on data in Wikipedia, lists a large number of semiconductors, showing
the group, material, chemical formula, band gap value, and type of gap.
Table 2. Semiconductor materials.
Group Formula Gap type
eV
IV C 5.47 indirect
IV Si 1.11 indirect
IV Ge 0.67 indirect
IV 6H-SiC 3 indirect
VI Se 1.74
VI Te 0.33
III-V cubic BN 6.36 indirect
III-V AlN 6.28 direct
III-V GaP 2.26 indirect
III-V GaAs 1.43 direct
III-V InSb 0.17 direct
II-VI CdSe 1.74 direct
II-VI CdTe 1.49
II-VI ZnSe 2.7 direct
IV-VI PbS 0.37
41
Figure 16 (left) shows the energy bands of GaAs. The horizontal axis is marked
with special values of k in two directions. Γ is the zone center, k = 0 while X is the
zone boundary in the kx direction (or, on account of the cubic symmetry, the ky or kz
direction). L is the zone boundary in the body diagonal of the cubic lattice. The three,
concave down, lower bands are the (filled) valence bands and the upper band is the
(empty) conduction band. The forbidden energy gap is the range from the top of the
valence band (at Γ) to the bottom of the conduction band (also at Γ). The left panel
shows a sketch of the band structure of a noble metal (Ag, Cu, or Au) for conduction-
band energies. The horizontal axis is the wave vector. The Fermi energy is shown. (Note
that the Fermi surface in these metals is not a sphere, and the Fermi wave vector in some
directions—the so-called “necks”—is at the zone boundary.) The d band is the narrow
band well below the Fermi level. Optical transitions from the d band to the conduction
band occur for conduction band states above EF .
(a) (b)
Fig. 16. (a) Band structure, E(k), of GaAs for energies around the band gap. (b) Schematic
band structure, E(k), of a noble metal.
42
3.3.6 Effective mass
The free electron kinetic energy depends on the electron mass m,
E =h2k2
2m=
p2
2m
The kinetic energy is parabolic in the momentum, with the curvature of the function
determined by m−1, so∂2E∂k2
=h2
m,
or, observing (Fig. 15) that the second derivative may not be the same for all vales of k,
m∗k =
h2
∂2E(k)∂k2
.
The free-electron parabola is concave upward, and the mass is everywhere equal
to the free-electron mass. Looking at Fig. 14, I see that the curvature at the bottom
of the lowest band is nearly that of the free-electron band, but that near k = ±π/a,
k = ±2π/a, or other zone-boundary values of k the curvature is considerably larger,
making m∗k smaller than the free-electron mass. Moreover, the top half of each band is
concave downward, changing the sign of the second derivative, so that the effective mass of
these electrons is negative!
Fig. 14. Energy as a function of the wave vector for a solid with a periodic lattice potential.
There are 3 bands shown completely, plus part of a fourth. There are gaps between the
bands where no allowed energies exist. The free-electron parabola, E = h2k2/2m, is shown
as a dashed line.
Technologically important semiconductors (Si, Ge, GaAs, InSb, . . . ) all have four
electrons per atom (or three on one and five on the other) so the first two bands are
full and the third empty. The second band is called the valence band and the third the
43
conduction band. Now, in these semiconductors,∗ the gap between the top of the valence
band and the bottom of the conduction band is small enough that at finite temperatures
there are thermally-excited carriers in the conduction band.† Because the total number of
electrons is fixed, empty states are left behind when thermal energies promote carriers to
the conduction band. These vacant states are called holes . The absence of a negatively
charged electron is a positive hole; the electron mass in the top half of the band is negative
so the hole mass is positive. Hole energy is measured downwards from the top of the band.
Table 3 Effective mass in selected semiconductors
Group Material Electron Hole
IV Si (4K) 1.06 0.59
Si (300K) 1.09 1.15
Ge 0.55 0.37
III-V GaAs 0.067 0.45
InSb 0.013 0.6
II-VI ZnO 0.29 1.21
ZnSe 0.17 1.44
Holes contribute to the total electrical and optical properties. I write
j = −neve + pevh
where n is the electron number density, p is the hole number density, and ve (vh) are the
electron (hole) drift velocity.
∗ See also the footnote on p. 39.
† In doped semiconductors, the majority carriers come from impurity levels located in the forbidden band
gap, but here I consider intrinsic (undoped) semiconductors and insulators.
44
3.3.7 Impurities or doping
Fig 2. Periodic table.
Silicon, GaAs, Ge, and other semiconductors are a dominant force in the world
economy for one reason: their electrical properties may be determined by incorporating
controlled amounts of other atoms. These atoms are called “impurities” and the process is
called “doping.” Neither term is particularly well chosen; in particular the dopants are not
undesirable impurities but instead are carefully controlled as to type, concentration, and
location.
Si is typically doped with atoms from periodic table columns on either side of Si
itself. Si (group IVA) has four valence electrons (3s2 3p2) which bond covalently to four
neighboring Si atoms, forming a tetrahedrally coordinated cubic crystal structure. Atoms
from the neighboring columns have one more or one less valence electron. B, to the
left (group IIIA), has three and P, to the right (group VA), has five. When a B atom
substitutes for Si it accepts an electron from the valence band to complete the bonding,
leaving a hole behind. When a P atom is substituted it uses four of its electrons for the
covalent bonds and donates the fifth to the conduction band. Thus doping controls the
free-carrier density, and hence the electrical conductivity. Hole doped Si (e.g., with B, Ga,
In, N, and Al) has positive carriers and is called p-type silicon. Electron doped Si (e.g.,
with P, As, and Sb) is called n-type silicon.
The dopant atoms have a net charge. P and other donors have given up one electron
and so have a charge +e. B, and other acceptors, have accepted one, and have a charge
45
−e. There is a Coulomb interaction between the dopant ions and an electron or hole in
the solid. The result is another hydrogen atom problem.
I’ll start by looking at the quantum problem at T = 0. The Hamiltonian is
H = − h2
2m∗∇2 − e2
4πε0κ|r| , (25)
where m∗ is the effective mass, either hole or electron depending on which case is in play, r
is the separation of the electron or hole from the impurity, and κ is the relative dielectric
constant of the semiconductor. Equation 25 is a scaled version of the Hamiltonian for the
hydrogen atom, which I know how to solve. In fact, shallow impurities are an interesting
analog to the hydrogen atom. The energy levels for the extra electron bound to the donor
atom are
Ee = − R
n2m∗
e
mκ2= −Re
n2. (26)
Here R = me4/32π2ε20h2 = 13.6 eV is the Rydberg energy, n = 1, 2, 3 . . . is the principal
quantum number, and Re is the effective Rydberg. The effective mass m∗e = 0.26m for
Si. If I use κ = 11.7 as the Si dielectric constant, then Re = 25.8 meV. The energy
is measured from the bottom of the conduction band, so the donor level is about 26
meV below the conduction band edge. At the same time, the radius for n = 1 is ae =
a0mκ/m∗e ≈ 22 A, with a0 the hydrogen Bohr radius. The electron wave function extends
over many lattice constants.
For the hole bound to the acceptor atom, Eq. 25 gives
Ev = − R
n2m∗
h
mκ2= −Rh
n2,
Fig. 17. Calculated and measured donor levels in Si.
46
Here m∗h(= 0.39m for Si) is an effective average of light-hole and heavy hole masses (a
detail of the band structure of Si). Using these numbers then Rh = 38.7 meV. The
energy of the hole states are measured downward from the valence-band maximum, so this
negative energy level is about 39 meV above the top of the valence band.
47
3.3.8 The p-n junction
To repeat, semiconductors may be doped either with donors or acceptors. Donor-
doped materials conduct via electrons in the conduction band. The materials are called
n-type because the electrons have negative charge. Acceptor-doped materials conduct
via holes (an empty electron state) in the valence band. The materials are called p-type
because the holes have positive charge. In general, donor levels lie close to the edge of the
conduction band and acceptor levels lie close to the edge of the valence band; the carrier
density is equal to the donor or acceptor density.
The simplest semiconductor device is the p-n junction diode. The diode energy band
diagram is shown in Fig. 18 for the unbiased, forward-biased, and reverse-biased cases.
Fig. 18 Diode band diagram: the unbiased, the forward-biased, and the reverse-biased cases.
48
One side of the crystal is doped p type, the other n type. A thin (∼ few μm wide)
junction separates these two sides. Away from the junction region on the p side are
negatively-charged acceptor ions and an equal number of free holes. On the n side are
positively-charged donor ions and an equal number of mobile electrons. In addition both
sides have a small number of thermally generated “minority” carriers of the other type
(holes in the n region and electrons in the p region).
Both holes and electrons tend to diffuse through the crystal. Electrons from the n
region diffuse across the junction and recombine with holes in the p region (and vice
versa). These recombinations leave the n-region depleted of electron carriers and positively
charged and they leave the p-region depleted of hole carriers and negatively charged.
The recombinations occur over a thin depletion layer around the junction. This charged,
double layer grows until the electric field it produces is strong enough to inhibit any
further diffusion of electrons across the junction.
The inhibition is not complete (at least at finite temperatures) and the most energetic
electrons and holes (those in the tails of the Boltzmann distribution function) can cross
the barrier. At zero applied voltage, the current due to electrons diffusing into the p
region is canceled by the minority electrons diffusing in the opposite direction. Similarly,
there is a cancellation between the currents due to holes and minority holes.
A voltage applied across the junction can either increase or decrease the height of
the potential barrier. If the applied voltage is such that the p region is positive (the p
region is connected to the positive electrode of a battery and the n region to the negative
electrode), the junction is said to be biased in the forward direction. If the voltage is
applied in the opposite direction, the junction is said to be biased in the reverse direction.
When the junction is biased in the forward direction, the barrier height is reduced and
the current increases rapidly. In contrast, a reverse bias increases the barrier height and
produces only a small reverse current up to a saturation limit −Is set by the thermal
generation of minority carriers. To create a minority carrier (hole) in the n-region, an
electron must overcome the band gap energy Eg (in silicon about 1.2 eV) via its thermal
energy and governing Boltzmann distribution. Consequently Is is very temperature
dependent and given by
Is = I0e−Eg/kT
where k is Boltzmann’s constant and T is the temperature.
The current I as a function of applied bias voltage V is predicted by the Shockley
diode equation.
I = Is
(eeV/kT − 1
)
49
where Is is called the reverse saturation current because I = −Is when the diode is reverse
biased (V < 0) sufficiently that eeV/kT << 1.
Real diodes have a series resistance Rs, typically on the order of an ohm or so. Real
diodes also have a large parallel resistance, typically over 10 kΩ, which adds a small
additional ohmic current to the measurements.
Diode I-V curve. The voltage to “turn on” the diode is about 0.6 V. The breakdown voltage
in the reverse direction can be as high as 1000 V or more.
50
Fig. 19. Resistance vs. temperature for Hg. (After Onnes.)
3.4 Superconductivity
All discussions of superconductivity are supposed to start in 1911, in Leiden, when and
where Kamerlingh Onnes and his assistants discovered that the resistance of solid mercury
abruptly falls to zero at a temperature of 4.2 K. Shown in Fig. 19, the resistance is falling
with temperature, similar to other pure metals. (See Fig. 20, for example.) But then, the
resistance takes a sudden drop to zero at 4.2 K. Onnes reported that the resistance was
smaller than 10−5 Ω compared to 0.11 Ω just above Tc; it is now believed to be identically
zero. Evidence for zero resistance includes experiments, done on rings of superconducting
material, in which continuous and steady currents are sustained for years with no applied
voltage.
In the slightly more than a century since its discovery, superconductivity has been
found to be widely distributed, occurring in elements, alloys, intermetallic compounds,
oxides, ionic compounds, organic charge-transfer compounds, polymers, and many other
systems. It seems that the only two prerequisites are (1) that the material be conductive,
so that there are carriers to condense into the charged superfluid, and (2) that it not
be a magnet; strong magnetic fields destroy superconductivity. But I should note that
the conductivity does not need to be that of free-electron metals and that there are
counterexamples to prerequisite 2.
Figure 21 shows the “records” for superconductivity over a 107 year period, 1911–
2018. Several classes of materials are shown. The blue circles are metals, alloys and
intermetallic compounds that are superconducting due to electron phonon interactions.
The record as of 2018 is held by H3S under enormous (155 GPa) pressure; it is thought to
be an electron-phonon superconductor. The high-Tc cuprates are shown as red squares;
organic conductors as magenta pincushions; fullerenes as purple triangles; heavy-Fermion
51
Fig. 21. Record transition temperatures in many classes of superconductors over a little more
than a century. Note that there is a change of scale in the time axis at 1980; the first
third covers 80 years and the second two-thirds covers 40 years. The materials with an
asterisk attached require pressure to become superconducting (or to reach the Tc shown).
The material with a dagger (FeSe) shows the Tc given only as a monolayer film.
metals as gold diamonds; finally, the Fe-based superconductors are in green oh-plus
symbols. The diagram is probably neither complete nor accurate. It does however indicate
the enormous range of material types where superconductivity is observed.
The second key property of superconductors was discovered in 1929, when Meissner
observed that a superconductor expels magnetic flux, making the condition for the field
inside the superconductor be that B = 0, where B is the magnetic field. The flux
expulsion is sketched in Fig. 22. Start with the constitutive equations,
0 = B = H +M = (1 + χ)H
B is zero but H is not zero. Hence,
χ = −1. (27)
In the Meissner state, the superconductor is a perfect diamagnet. The magnetic
moment of the sphere is generated by currents which circulate on the surface of the
52
Fig. 22. Two spheres in a uniform external magnetic field B. The sphere on the left is a
normal metal; the field penetrates uniformly. The sphere on the right is a superconductor
(below Tc). The field is expelled, making B = 0 inside. At the same time, the sphere
acquires a uniform magnetization.
sphere parallel to the equator.∗ The flux exclusion continues in all or in part as the field
increases, until an upper critical field, which I’ll call Hci, is reached and superconductivity
is destroyed.
Superconductors come in two flavors, illustrated in Fig. 23. Type I superconductors
function as described above, with perfect diamagnetic response, following Eq. 27, up to
the critical field Hc when superconductivity ceases to exist. Type II superconductors
are perfect diamagnets up to a (typically relatively low) lower critical field called Hc1.
Above this field, an inhomogeneous state occurs, with a magnetic flux lattice existing in
the superconductor. The density of flux in the material grows with increasing field. At
a (typically relatively high) upper critical field, Hc2, the flux penetration is complete.
Superconductivity (zero resistance) persists up to Hc2. The materials that remain
superconducting at high magnetic fields are all type II and some of these (NbTi, Nb3Sn,
YBa2Cu3O7, etc.) are used in superconducting magnets.
Fig. 23. Magnetization versus applied field for type I and type II superconductors. Parameters
are chosen to represent Pb. Clean Pb is a type I superconductor with Hc = 800 G (0.08
T). Adding impurities to Pb renders it a type II superconductor.
∗ Actually the currents are not on the surface but decay exponentially into the superconductor with a
characteristic penetration depth. I will calculate this depth shortly.
53
3.4.1 Superconducting materials
A list of superconductors is given in Table 4. The table lists the material, transition
temperature, field required to destroy superconductivity, and the type (I or II). Transition
temperatures can be seen to vary from 15 mK (α-W) to to 200–210 K. The current
record∗ is held by H3S at very high pressure.
Table 4. A list of superconducting materials. The materials are organized into
elements, compounds, alloys, organics, cuprates, and pnictides. Hci is the
maximum field at which zero resistance is observed. The classification of I or II
identifies the way magnetic flux is expelled.
Formula Tc (K) Hci (T) Type
Metals
α-Hg 4.15 0.04 I
Al 1.20 0.01 I
In 3.4 0.03 I
Sn 3.72 0.03 I
Ta 4.48 0.09 I
Pb 7.19 0.08 I
Nb 9.26 0.82 II
α-W 0.015 0.00012 I
α-U 0.68 I
Zn 0.855 0.005 I
V 5.03 1 II
Compounds
C6K 1.5 II
C60K3 19.8 0.013 II
C60Rbx 28 II
In2O3 3.3 II
V3Si 17.1 23 II
Nb3Sn 18.3 24.5 II
Nb3Ge 23.2 37 II
MgB2 39 74 II
H3S 203 (at 1500 kbar)
∗ There are Tc’s below 15 mK but none—so far—at 300 K.
54
Table 5, continued.
Formula Tc (K) Hci (T) Type
Binary alloys
NbO 1.38 II
TiN 5.6
ZrN 10
NbTi 10 15 II
NbN 16 15.3 II
Charge-transfer salts
(TMTSF)2PF6 1.1 (at 6.5 kbar) II
(TMTSF)2ReO4 1.2 (at 9.5 kbar) II
(TMTSF)2ClO4 1.4 II
β-(ET)2I3 1.5 II
β′′-(ET)2SF5CH2CF2SO3 5.3 II
κ-(ET)2Cu(NCS)2 10.4 10 II
Cuprates (high-Tc)
La1.85Sr0.15CuO4 38 45 II
La2CuO4.11 43 II
YBa2Cu3O7 (123) 92 140 II
Bi2Sr2CuO6 (Bi-2201) 20 II
Bi2Sr2CaCu2O8 (Bi-2212) 92 107 II
Bi2Sr2Ca2Cu3O10 (Bi-2223) 106 II
Tl2Ba2CaCu2O8 (Tl-2212) 108 II
Tl2Ba2Ca2Cu3O10 (Tl-2223) 125 75 II
HgBa2CuO4 (Hg-1201) 94 II
HgBa2Ca2Cu3O8 (Hg-1223) 134 190 II
HgBa2Ca2Cu3O8 (Hg-1223) 164 (at 300 kbar) II
Nd1.7Ce0.3CuO4 24 II
Iron-based (pnictide)
LaO0.9F0.2FeAs 28.5 II
PrFeAsO0.89F0.11 52 II
GdFeAsO0.85 53.5 II
BaFe1.8Co0.2As2 25.3 II
SmFeAsO0.85 55 II
Ba0.6K0.4Fe2As2 38 II
CaFe0.9Co0.1AsF 22 II
Sr0.5Sm0.5FeAsF 56 II
FeSe 27 II
55
3.4.2 Theoretical background
Here I will list some of the results of BCS theory and the phenomenological Ginzburg-
Landau theory.
BCS theory is the microscopic theory of metallic superconductors. These
superconductors have the following properties
• Below Tc, electrons form Cooper pairs. The Cooper state is a superposition of two
electrons of the normal metal into a zero-momentum, spin-singlet state
ψ(r1, r2) = eik·r1 |↑〉 e−ik·r2 |↓〉
• The superconducting condensate is a coherent superposition of Cooper pairs, involving
all the conduction electrons in the metal.
• Because the pairs are made up of two electrons, the number density of Cooper pairs is
nc = n/2; the charge is Qc = −2e; and the mass is mc = 2m.
• The Cooper pairs exists because of an effective attractive interaction between the two
electrons.
• In metallic superconductors, this attraction arises through the interaction with the
lattice (the phonons).
• The attractive interaction reduces the energy of the electrons. In fact, the Fermi
surface is unstable towards the formation of a Cooper pair even if there is only an
infinitesimal attractive interaction between the electrons.
• To break a Cooper pair at the Fermi surface requires energy
2Δ = 3.5kBTc (28)
to be supplied. This energy is known as the superconducting energy gap.
• In metallic (called “conventional”) superconductors the gap is small compared to the
Fermi energy. The gap scale is a few meV; the Fermi energy several eV.
56
• Thermal energies break Cooper pairs and, because the gap is a collective effect, reduce
the gap. It eventually reaches zero at Tc. The behavior (from both experiment and
theory) is shown in Fig. 24.
• Excitations above the gap are called quasiparticles . They have many of the properties
of normal electrons, being spin-half Fermions, but are affected by the BCS coherence
factors and have an energy-wave vector dispersion relation that differs significantly
from that of ordinary electrons in the solid.
Fig. 24. Superconducting energy gap measured by tunneling as a function of temperature.
57
• Ginzburg-Landau theory is a phenomenological theory that is based on the Landau
theory of second-order phase transitions. The free energy of a superconductor near Tc
is written in terms of a complex order parameter, ψ, which is nonzero below Tc and
zero above.
• The theory (when folded into BCS theory) makes the following identifications,
ψ =√nc(T )e
iφ = Δ(T ), (29)
where ψ represents a coherent superposition of all Cooper pairs, the pair density is nc,
φ is the phase of the order parameter, and Δ is the BCS energy gap.
• London’s two-fluid model, in which there is a normal fluid intermixed with the
superfluid, is consistent with the Ginzburg-Landau theory. As nc is reduced
approaching Tc, (Fig. 24) nn grows so that n = 2nc + nn.
58
3.4.3 London penetration depth
The Meissner effect (B = 0) requires a transition length from the strong field outside
the superconductor to the zero field inside. This region is where the electrical currents
flow that magnetize the superconductor (M = χH = −H). The “London penetration
depth,” λL =√
mc2/4πnse2, is typically in the 10–200 nm range. Some values are shown
in Table 5. Metals with high electron density (Al) have shorter penetration depths than
those with lower electron density.
Table 5. London penetration depth of superconductors.
Material λL nm
Al 16
Sn 34
Pb 37
Nb 39
YBa2Cu3O7 144
The London penetration depth is also the “skin depth” of the superconductor. Light
falls of in the interior with characteristic decay length λL.
3.4.4 The coherence length
In addition to the penetration depth discussed already, a second length scale exists. It
is the coherence length, first introduced by Pippard. The coherence length also appears
in the Ginzburg-Landau theory and BCS theory. The three lengths are similar but not
identical, either in concept or in value. I’ll give a heuristic argument as follows. The
Cooper-pair wave function extends over a finite distance; at longer distances the phase
coherence is lost. The wave function ψ thus has an envelope
ψ∼e−r/ξ0
where ξ0 is the coherence length. The uncertainty principle tells me that a the finite
length for the wave function requires a range of wave vectors. The range is
δksc = kF ± Δp
h= kF ± 1
ξ0.
The range of wave vectors must be related to the superconducting binding energy Δ,
leading to a spread of kinetic energies,
h2(δksc)2
2m=
h2k2F2m
± h2kFmξ0
= EF ± 2Δ,
where I have neglected the term containing 1/ξ20 . Note that this argument basically
invokes the uncertainty principle: limiting the space part to ξ0 leads to a spread of
59
momenta and corresponding spread of energies. Subtracting EF from both sides gives
h2kF /mξ0 = 2Δ or
ξ0 =h2kF2mΔ
=hvF2Δ
, (30)
using hkF = mvF .
Two materials parameters appear in Eq. 30. In the numerator is the Fermi velocity
(or Fermi wave vector), which increases with carrier density; the denominator holds the
gap (or, equivalently, the transition temperature: 2Δ = 3.5kTc in BCS theory). Thus high
electron density, low Tc materials (think Al) will have long coherence lengths; low electron
density, high Tc materials (think YBa2Cu3O7) will have short coherence lengths. This
expectation is in accord with observation. See Table 6.
Table 6. Coherence length of superconducting materials.
Materials Tc (K) ξ0 (nm)
Aluminum 1.19 1200
Indium 3.40 330
Tin 3.72 260
Gallium 5.90 160
Lead 7.22 80
Niobium 9.25 35
PbMo 15 2.5
Nb3Sn 17 4.0
C60K3 19 3.0
C60Rb3 31 2.3
YBa2Cu3O7 93 1.5
60
3.4.5 Quantum mechanics
The London and BCS models consider the superconducting state to be a coherent
superposition of all the electrons in the superfluid, described by a single quantum-
mechanical wave function.
• Analogy: the two electron superposition in the bonding orbital of H2.
• Reminder: the classical equation for the electrical current carried by a set of Cooper
pairs woud be
j = ncQcv =ncQc
mcp, (31)
where v is the average velocity and p the average momentum of the superconducting
particles (Cooper pairs), nc is their number density, Qc is their charge, and mc is their
mass.
• Quantum mechanically:
js =Qc
2mc(ψ∗pψ − ψpψ∗), (32)
with p the momentum operator and ψ the wave function of the superconducting state.
• The momentum operator is
p = −ih∇, (33)
• The wave function itself is related to the density of superfluid
ψ =√nc(T )e
iφ,
so that∫dV ψ∗ψ = nc
• I invoke charge neutrality: nc is constant in space. Hence,
−ih∇ψ = −ih√nce
iφi∇φ,
and Eq. 32 becomes
js =ncQc
mch∇φ. (34)
• The relations describing how I count Cooper pairs and ordinary electrons are
Qc = −2e mc = 2m and nc =ns2,
with ns the superfluid density in a two-fluid model. At zero temperature in clean
metals, ns = n. Thus ncQc/mc = −nse/2m.
• With these replacements I arrive at
js = −nse
2mh∇φ. (35)
• The current is proportional to the gradient of the quantum mechanical phase
61
3.4.6 Josephson tunneling
The Josephson effect is a truly quantum effect. It was predicted theoretically by Brian
Josephson (as a student) 1962 and observed shortly after.
It is a tunneling effect. Remember when I did tunneling through a square barrier, the
particle (electron) was a plane wave with energy E = h2k2/2m incident on the barrier and
it transmitted through (with some probability depending on V0 − E and left with the same
energy.
The Josephson effect is observed in a device with a “weak link” or barrier between two
superconductors. A current or voltage is applied to the device. The observation is that
current can flow with no voltage across the superconducting device. (This is quite different
from a p-n junction.)
The superconductor has a time-dependent wave function Ψ. Before making the device.
Ψ satisfies a Schrodinger equation
ih∂Ψ
∂t= UΨ
where U is an external potential. The kinetic energy is taken as zero because the time-
independent wave function for the Cooper pairs, ψ = eik·r1 |↑〉 e−ik·r2 |↓〉, is a zero-
momentum state.
After making the junction, I have a chance for superfluid (Cooper pairs) to tunnel
through, affecting the ave function on the other side. I’ll model this with a linear term in
the Schrodinger equations for both sides:
ih∂ΨL
∂t= ULΨL +KΨR
ih∂ΨR
∂t= URΨR +KΨL
where “L” and “R” refer to the left and right sides of the junction and K is the coupling
across the barrier. I’ll let the potentials be related to the voltage difference V across the
junction as UL − UR = −2eV . (−2e is the charge of the Cooper pair.)
62
With
ΨL =√nLe
iφL and ΨR =√nRe
iφR
with φi the phase on side i = L or i = R and ni the superfluid densities.
Then, calculating the derivatives, there are two equations
ih∂ΨL
∂t=
ih
2
1√nL
eiφL∂nL∂t
+ ih√nLe
iφL(i)∂φL∂t
= UL√nLe
iφL +K√nRe
iφR
ih∂ΨR
∂t=
ih
2
1√nR
eiφR∂nR∂t
+ ih√nRe
iφR(i)∂φR∂t
= UR√nRe
iφR +K√nLe
iφL
Actually there are four equations because the real and imaginary parts must be
separately equal. I multiply through by√nLe
−iφL in the first and√nRe
−iφR in the
second, identify
e±i(φL−φR) = cos(φL − φR)± i sin(φL − φR) ≡ cos(δφ)± i sin(δφ)
with δφ = φL − φR, I get these 4 equations
• Reals:
−hnL∂φL∂t
= ULnL +K√nLnR cos(δφ)
−hnR∂φR∂t
= URnR +K√nLnR cos(δφ)
• Imaginaries
h∂nL∂t
= 2K√nLnR sin(δφ)
h∂nR∂t
= −2K√nLnR sin(δφ) = −h
∂nL∂t
• The next step is to combine these. Subtract the results from the imaginaries.
h∂(nL − nR)
∂t= 4K
√nLnR sin(δφ)
• I’ll rewrite ∂(nL − nR)/∂t ≈ Δn/tT where tT is the time to tunnel through the
junction.
63
Recall the experiment
• The current through the junction is
j = −(Δn)(2e)v = −4K
h(eD)nc sin(δφ)
where v = D/tT is the velocity (equal to junction thickness D divided by time tT to
tunnel through) and I use√nLnR = nc.
• I can lump everything into a materials-and-junction specific constant j0 and write
j = j0 sin(δφ)
as the equation for the dc Josephson effect. The result is a supercurrent tunneling
through the junction at junction voltage V = 0. J0 is the critical current of the
junction.
Note that δφ/D = ∇φ.
64
Next, I take the difference of the real equations:
• Reals:
−hnL∂φL∂t
= ULnL +K√nLnR cos(δφ)
−hnR∂φR∂t
= URnR +K√nLnR cos(δφ)
• I use again nL ≈ nR ≈ nc and UL − UR = 2eV with V the voltage across the junction.
I find∂δφ
∂t=
2eV
h
• Integrating
δφ(t) =2e
h
∫ t
0dt′ V (t′)
• If V (t) = V0, a constant,
δφ(t)2eV0h
t
I go back to j = j0 sin(δφ) to find
j = j0 sin(2eV0h
t)
• This is an alternating current in response to a dc voltage. It is called the ac Josephson
effect. The frequency is
ω =2e
hV0 or f =
2e
hV0
with value 4.835979× 1014 Hz/V.
• Frequencies can be measured with great precision. The fundamental constants e and h
are nowadays defined to be exact numbers.
• So the Josephson frequency is now the definition of the standard Volt.
V =hω
2e
Much more accurate than a standard current through a standard resistor.
• Practically speaking, the frequency may be in the range of 50 GHz; the voltage is then
in the range of 0.1 mV.
65
4. ELEMENTARY PARTICLES
In this last section, the plan is to drill down to size scales smaller, indeed much
smaller, than the size of an atom. I’ll discuss briefly the nuclei of atoms, made up of
neutrons and protons. And then quarks, components of neutrons and protons, and the
forces among them.
In the end, the inventory of the most “elementary” things in the universe makes up
a relatively short list. But I should note that as one constructs nuclei, atoms, molecules,
solids, and the cosmos as a whole, there are new emergent phenomena—some understood
and some not understood—that make up the overall picture of modern physics.
As a final comment, this section will be less mathematical than others. The reason is
that although there is an excellent theory for the details of the hydrogen atom, based on
the Coulomb attraction of electron to proton, there is no corresponding equations
4.1 Atomic Nuclei
Except for hydrogen, the nuclei of all atoms contain both protons and neutrons. The
proton has charge +e, spin 1/2, and a mass about 1800 times heavier than the electron:
mpc2 = 938 MeV. The neutron has charge 0, spin 1/2, and a mass slightly higher than
than the proton: mnc2 = 940 MeV. (The electron has mc2 = 0.511 MeV.)
All elements have a number of isotopes. Each element has a specific number of protons
(and an equal number of electrons in the atomic state). But the number of neutrons
varies. So the approximate mass—in units of the mass of Carbon-twelve (6 protons + 6
neutrons) divided by 12—is attached to the chemical symbol thusly: 4He, 12C, 235U.
Some isotopes are stable; others are not and emit “rays,” decaying to other nuclei or
otherwise changing their energy. At the heavy end of the periodic table all isotopes are
radioactive. The heaviest stable nucleus is 209Bi.
Light atoms (He, C, O) have stable isotope with equal numbers of protons and
neutrons; there are 2 of each in 4He, 6 of each in 12C. But somewhere around Z = 12
the number of neutrons begins to grow faster than the number of protons. 209Bi has 83
protons and 126 neutrons. 235U hs 92 and 143.
The size of a nucleus is really small: diameters of 10−15 m are typical, compared to
10−10 m for atoms.
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4.2 The strong force
With the protons and neutrons packed together in a tiny volume, a strong force is
required to bind them together. And bind them it does.
The force is known as the strong force or strong nuclear force. I don’t have a formula
for the force or for the potential. (The force is the gradient of the potential, remember?)
It is short ranged (3 fm = 3×10−15 m). It is deep enough that the Coulomb repulsion
of one proton to another is overcome. It is also deep enough that the kinetic energy
generated by the uncertainty principal (p2/2m) is overcome.
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The function for the proton-proton potential contains a repulsive +e2/4πε0r Coulomb
term. It is absent for the neutron-neutron or neutron-proton potential.
There is a very strong repulsive term right near r = 0 that keeps the particles from
occupying the same location.
4.2.1 Nuclear reactions
Radioactive decay is one form of a nuclear reaction. The nucleus is unstable and
emits one or more “rays”: an alpha (a 4He nucleus, reducing Z by 2), beta (an electron or
positron, changing Z by ±1) or a gamma (a high energy photon, changing the energy but
not Z.
But nuclear reactions can also occur in nature (the Sun) or in labs or facilities (fission
or fusion reactors).
• Fusion: combination of two light nuclei to produce one heavier nucleus. It occurs in a
star: hydrogen nucleii fuse to form helium, lithium, and a variety of heavier nuclei. In
supernovae and in the collision of neutron stars at the end of a binary inspiral, almost
all nuclei are produced.
• Fission: A heavy nucleus is split into two lighter ones, releasing energy (at least until
the maximum of the binding energy curve). Thermal neutrons do the splitting; if the
process is efficient a chain reaction may occur. This is the basis of nuclear reactors for
generating energy.
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4.3 Concepts of particle physics
Protons and neutrons are not “elementary”; they are composed of three particles
within a sort of “bag” that makes up the proton/neutron. These interior particles are
called quarks .
In contrast, the electron does seem to be elementary. As far as I know, it is a point
like particle, charge −e and spin 1/2. Scattering of electrons on electrons (an example
of Rutherford scattering) down to very close approaches observe only the Coulomb force
between them. (Rutherford observed Coulomb scattering for alpha particles on nuclei
down to a few femtometers but not closer.)
The absence of a theory for the strong force impacts the study of quarks and related
topics in a big way. Instead of calculating energy levels and wave functions, one is
reduced to identifying symmetries and quantum numbers. A rather beautiful picture
however has been built up in great detail; it is called the “standard model” or “quantum
chromodynamics”∗
Here I list some of the concepts:
• Every particle has an antiparticle. The antiparticle to the electron is the positron.
The positron has charge +e, spin half, the same mass as the electron, and interacts via
the Coulomb force. A positron would bind to an anti-proton (which does exist; it is a
particle with charge −e, spin half, and the same mass as the proton). The spectrum of
light emission from anti-hydrogen would be the same as from hydrogen.
• A particle an its antiparticle can annihilate with a release of energy. An electron and
positron annihilate with the emission of two gamma rays (photons). Together they
carry energy 2hω equal to the sum of the rest mass energies 2mc2 of electron and
positron.†
• Forces between particles are exerted or mediated by other particles. For instance the
Coulomb force between two electrons is described as due to an exchange of photons.
The photon is the mediating particle for electromagnetic forces.
• This should not be completely new: I know that the photoelectric effect and Compton
scattering involved the interpretation of electromagnetic waves as particles. The wave
had associated electric and magnetic fields, so the photon must also be associated with
such fields. The fields exert forces on charged particles.
• The force carriers are spin 1 or spin 2 objects, and thus bosons, not subject to the
exclusion principle. Some have zero mass; some finite mass. The things that make up
matter are spin half and thus Fermions, obeying the exclusion principle.
∗ These are not exactly the same thing, but they are often used interchangeably.
† They have equal frequencies ω and leave in opposite directions. Do you understand why?
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4.4 Building blocks of the standard model
4.4.1 Leptons
• The electron and the neutrino (called the “electron neutrino) are leptons . The name
means “small” or “thin”; it was chosen to distinguish electrons from nucleons.
• The electron and electron neutrino make up the first of three generations of leptons.
They are followed by the muon and tau, with associated neutrinos.
4.4.2 Quarks
• Quarks are rather heavy (compared to leptons), spin 1/2, charged particles. They are
members of the family baryons. (Leptons are not baryons.) The name comes from
James Joyce’s book Finnegans Wake:
Three quarks for Muster Mark!
Sure he hasn’t got much of a bark
And sure any he has it’s all beside the mark.
• There are three generations of quarks, 6 quarks in total.
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• Protons and neutrons are composed of three quarks: two up +(2/3)e and one down
−(1/3)e for the proton and one up and two down for the neutron.
Left: proton uud, charge +e. Right: neutron udd, charge 0
• Other “elementary” particles that are called “baryons” or “mesons” are made up of
quarks. Baryons contain 3 quarks and have half-integer spin: 1/2, 3/2, 5/2. . . Mesons
contain 2 quarks and have integer spin: 0, 1, 2 . . . Baryons + mesons = hadrons.
• Free quarks are not observed; it is thought that the forces between them actually
increase as they are separated.
• Quarks are said to be colored, or to have “color charge.” There are 3 values, red,
green, and blue. A proton and neutron contain one of each color. (I think they
actually contain linear combinations of the colors, antisymmetric under exchange, so
that the sum of all colors, viewed as vectors pointing at 120◦ around a circle, adds to
zero.)
• There is an anti-quark to each quark, and anti-color as well.
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4.4.3 Force carriers
• There are four (and only four) types of forces in nature:
1. Strong
2. Electromagnetic
3. Weak
4. Gravity
4.4.4 Relative strengths
• The list above is organized from “strongest” to “weakest.” How do I define the what is
strong and what is weak?
• The idea is that one makes the factor between the potential V (r) and the space
dependence r dimensionless. For the Coulomb potential, I’ll write
V (r) =α
r
where α is just a number. In SI the potential between two identical charges is
(1/4πε0)(e2/r) and is in Joules when r is in meters. Divide this by hc and I can
write
V (r) =e2
4πε0hc
1
r
The quantity e2/4πε0hc is just a number (work it out!) and is called the fine structure
constant, α. α = 1/137. With r in meters, the unit for potential energy (and energy in
general) in this “natural system of units” is in m−1.
• The Schrodinger equation can be divided by hc; the wave functions are not changed
because this factor affects equally the kinetic, potential, and total energies; all would
appear in m−1.
• The strong force has αS ≈ 1. Electromagnetism has α = 1/137. The weak force has
αW ≈ 10−5. And gravity has αG ≈ 10−38.
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4.5 Force carriers
4.5.1 Strong force
The strong force binds the protons and neutron in the nucleus. I know it is “stronger”
than electromagnetism, because it overcomes the repulsive force among the protons at
10−15 distances.
The mediator is called the gluon. The gluon works on and modifies the color of
quarks and antiquarks. Gluons only act on the “color charge.” Consequently, electrons for
example, which have no color charge, do not experience the strong force.
Gluons have zero mass, spin 1, and zero electrical charge.
The energy is high, so that the characteristic time over which he gluons are exchanged
between two quarks is τ ≈ 10−23 sec.
This time is comparable to the size over which the particles stay in contact even if
they are moving very fast: cτ ≈ 3× 10−15 m, the size of a nucleus.
4.5.2 Electromagnetic force
This is the force between electron and proton in hydrogen. It is exerted between any
particles with electric charge.
It does not matter whether they are leptons or baryons. Or fermions or bosons. But
they must have electrical charge.
The mediator is the photon. This identification makes this force consistent with
relativity: photons move at the speed of light, as does the “news” of charged particle
movement between the particles.
Photons have zero mass, spin 1, and zero electrical charge.
4.5.3 Weak force
This interaction can affect all quarks and leptons. It does not matter if they are
charged or neutral.
It is responsible for some radioactive decays in which there is a change from one
element to a neighbor element in the periodic table.
An example is beta decay: emission of electron or positron (and neutrino).
A down quark within a neutron is changed into an up quark, thus converting the
neutron to a proton. The charge of the nucleus (the atomic number) is increased from
A to A + 1, An electron and an electron (anti)neutrino is emitted. Other decays involve
positron and electron neutrino emission.
The weak interaction is very short ranged, 10−18 m but relatively slow: 10−16–10−10 s.
There are two mediating particles (force carriers): the neutral Z boson and the
charged W∓ bosons. (The W− boson is the particle while the W+ boson is the
antiparticle. The Z boson is its own antiparticle.)
The Z and W bosons have mc2 of 92 and 80 GeV respectively, making them almost
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100 times more massive than the proton. The Z has charge 0; the W’s have charge ∓e.
The weak interaction is responsible for neutron decay. Neutrons in a nucleus, bound
by the strong force, are stable. But an isolated neutron all by itself in space decays with
a lifetime of about 15 minutes. The neutron decays into a proton, an electron, and an
electron anti-neutrino
n → p+ + e− + νe,
or, including what is happening within the neutron,
udd → uud+W− → uud+ e− + νe.
The neutron is more massive than the sum of masses of proton + electron (+ anti-
neutrino). The remaining mc2 energy gives the particles a significant kinetic energy. The
momentum and energy conservation equations are complicated (because they involve
motion in three-dimensional space) but the outcome is that the electron leaves with
relativistic energies and the proton is nonrelativistic.
4.5.4 Gravitational force
Gravity is the weakest of the four forces of nature. It is a factor of 1038 smaller than
the strong force. That’s 100,000,000,000,000,000,000,000,000,000,000,000,000 times weaker.
Gravity regulates the motion of the planets, the orbit of the sun within the galaxy,
and the motions of galaxies in the universe. So how can such a weak force be dominant
at large distances? Why is it not overwhelmed by the strong, electromagnetic, or weak
force? The answers are that the strong and weak forces only operate at short distances,
and are completely negligible for two nuclei at meters or light years of distance. The
electromagnetic force is as long-ranged as gravity, but macroscopic objects are almost
completely neutral. Newton’s gravitational force between earth and moon is much larger
than the Coulomb force because the charges on earth and moon is not very large but the
masses of earth and moon are large.
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The (hypothetical) mediating particle of gravity is the graviton, a massless, spin two
boson. It interacts with the mass of a material object.
The evidence for zero mass is compelling. In 2017, the LIGO detectors observed a
burst of gravitational waves that arrived within 2 seconds of a gamma-ray burst detected
by the Fermi and INTEGRAL satellites. The source was 130 million light years away, so
the gravitons and photons had speeds that were the same within Δv/v = 5× 10−16.
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4.5.5 A chart of the standard model
The standard model addresses the weak, electromagnetic, and strong forces. (A goal
of what is generally called a “Theory of Everything” is to incorporate gravity into the
standard model. It is not there yet.) The figure below shows all the particles in the
standard model.
The chart shows the quarks on the top left, leptons on the bottom left, and the force
carrier particles in a column on the right.
The mass, charge and spin of the particles is shown in the diagram.
There is one additional particle shown in the chart: the Higgs boson. The Higgs
is a consequence of a theory about the origin of mass in the universe. The Higgs was
discovered a the CERN collider in Switzerland in 2012. (The Florida high-energy physics
group played an important role in the discovery.)
Peter Higgs and Franois Englert were awarded the 2013 Nobel Prize for proposing this
mass-generating mechanism. The mechanism involves a field, the Higgs field, permeating
all of space. The Higgs particle is a quantum fluctuation of this Higgs field. Quantum
mechanics mixes the particles and the force-mediating bosons; the presence of the Higgs
field spills over into other quantum fields; its this coupling that gives their associated
particles mass.
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4.5.6 Beyond the standard model
There are several observations in nature for which the standard model does not
account. (Not including gravity.)
• The universe is made of matter and not antimatter. It should have equal amounts of
both but it does not.
• The motion of stars in galaxies and of galaxies within clusters of galaxies show that
there is much more matter present than is accounted for by the stars and planets.
It is not known what this “dark matter” is. It is not ordinary matter.
One candidate is the “axion.” Florida is part of a project to search for the axion
by a method proposed by Pierre Sikivie in 1983.
The universe’s inventory contains 4.9% ordinary matter, 26.8% dark matter, and
68.3% something even more mysterious, called “dark energy.”
• Some symmetries of nature (next section) are violated (slightly) by the weak
interaction but not by the strong interaction. The reason is also unknown. (But
the axion, if found would provide a reason.)
• A very appealing theory (which would allow gravity to be unified with the other
forces) is called “supersymmetry.” One prediction is that for every fermion in the
standard model, there is a supersymmetric “partner” that is a boson. And vice versa.
The quark is paired with the “squark,” the electron with the “selectron,” and so
forth. The photon is paired with the “photino,” the W with the “wino,” and so
forth.
It was expected that “the lightest supersymmetric” partner would be observed
at CERN. If it were found, it probably would solve the dark matter problem too.
But it has not been found; and doubts are spreading about the correctness of the
theory.
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Review:
Leptons: 3 generations, starting with electron and neutrino.
Quarks: 3 generations, starting with up and down. Proton is uud. Neutron is udd.
Forces: strong, electromagnetic, weak, gravity, in that order of strength
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4.6 More on forces
4.6.1 Range
As already mentioned, the four forces have different ranges. Each involves the
exchange of force mediators between the two particles. If the force mediator has mass,
and interesting questions arise in this case. Even if it does not have mass, it has energy
(hω). It also has momentum. Momentum is needed because the exchange of particles is
the way forces are exerted.
But still, let me discuss the case of force mediators with mass, such as the W and the
Z. One could ask this question: the neutron decays by beta decay as shown here:
udd → uud+W− → uud+ e− + νe.
The energy balance between the initial state and final state is fine: the rest mass
energies of proton, electron, and antineutrino are less than the rest mass energy of the
neutron. But in the middle there is the W−. It has a rest mass energy of 80.5 GeV, 86
times larger than the neutrons rest mass energy of 0.94 GeV and 165,000 times larger than
the difference between neutron and proton rest mass energies.
How can this be? It appears that energy is violated on a huge scale. The uncertainty
principal comes to the rescue. On very short time scales, energy uncertainty is large:
ΔEΔt > h
I set ΔE = mc2 and the range R = cΔt. Using c gives an upper limit on the range.
Setting then inequality to an appriximate equality gives a lower limit. Then
R ≈ ch
mc2=
h
mc
If mc2 = 80.5 GeV, R = 1.23× 10−18 m.
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Recall the phenomenon of Compton scattering, where a photon scatters off a charged
particles. The electron recoiled at angle θ and the photon decreased its energy (increased
its wavelength) by
Δλ =h
mc(1− cos θ) ≡ λc(1− cos θ)
so R = λc/2π.
4.6.2 Range of electromagnetism and gravity
If the force mediators have zero mass, the force has infinite range. The rest masses of
the photon and the graviton are zero and indeed these are long-range forces.
4.6.3 Range of the strong force
The force between quarks, mediated by massless gluons, is an interaction between
color and particles that possess color. Quarks possess one of three colors, green, red, or
blue and the strong force among quarks is an attractive force between these mediated by
gluons. This strong force has no theoretical limit to its range.
The force is so strong however the color-charged gluons and quarks are bound tightly
together into color neutral baryons and mesons. Baryons consist of three quarks of the
three colors, which cancel to color-neutrality. Mesons consist of a quark and antiquark
(with corresponding color and anticolor).
Because color does not appear outside of any baryon or meson, the strong force only
directly has effects inside a hadron. The strong force that bind protons and neutrons is a
residual effect. Color-neutral baryons and mesons can interact with the strong force due to
their color-charged constituents. The force carriers are the mesons: pions, kaons, rhos, Ds,
etas, and many others. The neutral pion has a rest mass energy of 135 MeV; the range is
R = 0.73× 10−15 m.
4.6.4 Quark confinement
The force between quarks has analogies to a spring: if a quark is pulled away from
its equilibrium position, the color-force field “stretches.” In so doing, more and more
energy (1/2kx2) is added. At some point, it is energetically cheaper for the color-force
field to “snap,” creating a quark-antiquark pair. In so doing, energy is conserved because
the energy of the color-force field is converted into the mass of the new quarks, and the
color-force field can relax back to an unstretched state.
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4.7 Conservation laws and symmetries
Physics relies on conservation laws in all of its aspects. In closed systems energy is
conserved; momentum is conserved; angular momentum is conserved; charge is conserved.
Physics also relies on symmetries. There is translational symmetry in crystals.
Molecules have certain symmetries; these affect their rotational and vibrational motions.
Fermions are antisymmetric under exchange; bosons are symmetric. Relativity imposes
symmetries between clocks and meter sticks in frames moving relative to each other.
The hydrogen atom is symmetric under rotation; this affects the way I wrote the wave
functions of hydrogen.
Experiments on elementary particles have revealed these symmetries and confirmed
these conservation laws. Here I list a few.
4.7.1 Lepton number
Lepton number is conserved. The lepton number for lepton particles is +1. The lepton
number for lepton antiparticles is −1. All other particles have lepton number 0.
For example, one way by which the antimuon, μ+ (lepton number −1) can decay is
μ+ → e+ + νe + νμ
The decay is to a positron (−1), an electron neutrino (+1), and a muon
antineutrino (−1). So
−1 = −1 + 1− 1
Now consider
μ+ → e+ + γ
A gamma ray, has lepton number 0, so this decay would seem to conserve lepton number.
−1 = −1
But this decay channel has never been observed, although the detection (an electron
and a photon) would seem to be simple. Note that the muon is in the second generation of
leptons and the positron is in the first. Lepton number is conserved for each generation.
Recall the decay of the neutron:
n → p+ e− + γ + νe
I get a proton, an electron, and an electron antineutrino. It has to be an electron
antineutrino in order to conserve lepton number.
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4.7.2 Baryon number
Baryon number is conserved. The baryon number for baryon particles is +1. The
baryon number for baryon antiparticles is −1. All other particles have baryon number 0.
In neutron decay, both neutron and proton have baryon number +1.
Conservation of baryon number along with conservation of energy explains why the
proton is a stable particle. It has the lowest mass among the baryons and so cannot
decay into another baryon (energy) or into something else, meson or positron for example
(baryon number).
As I said, the preponderance of matter over antimatter is a current mystery in physics.
The universe right after the the big bang had an enormously high temperature and is
said to have been dominated by photons. These photons can convert to a particle plus its
antiparticle. An example is pair production, observable today as
γ → e+ + e−
where a photon converts to a positron and electron. We can also have
γ → p+ + p−
where a photon converts to an antiproton and a proton. These reactions are just matter-
antimatter annihilation running backwards.
In each case I must have a large enough photon energy: hω > 2mc2. But this is no
problem in the very early universe.
The first reaction conserves lepton number and the second baryon number. And an
equal amount of matter and antimatter is produced. But these equal amounts are not
what is observed.
4.7.3 Quark quantum numbers
The quark numbers re
Quarks have baryon number 1/3. They have something called “isospin”. It is not
actually a spin, but obeys the rules of spin. The up, charm, and top quarks have isospin
+1/2. The down, strange, and bottom quarks have isospin −1/2.
For the other quantum numbers, charm, strangeness, topness (also called truth) and
bottomness (aka beauty), the corresponding quark has quantum number +1 and the
antiquark has quantum number −1. These quantum numbers are conserved by both the
electromagnetic and strong interactions (but not the weak interaction).
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4.7.4 Parity
Parity (P) involves a transformation that changes the algebraic sign of the coordinate
system.
The parity transformation converts a right-handed coordinate system into a left
handed one. Two such transformations put it back the way it was.
A vector, like momentum p is converted to −p by parity. (It is the axis that
is flipped.) Initially with all three components positive, the vector after the parity
transformation has all three components negative. The vector is said to have odd parity.
Angular momentum is unchanged under a parity transformation. A particle orbiting in
the x-y plane has a velocity that during the orbit points along both plus and minus x and
plus and minus y. This is still true after the parity transformation. Or, one can say that
L = r× p. If I do a parity transformation on r× p, I get (−r)× (−p) = L, unchanged.
Many discussions use the reversal by a reflection in a mirror as an example of a
parity transformation. The mirror reflection certainly turns a right-handed coordinate
system into a left handed system. (A mirror in the x-y plane converts z → −z.) But one
also needs a rotation to complete the parity transformation. (Mirror reflection converts
L → −L.
In general, if a system is identical to the original system after a parity transformation,
the system is said to have even parity. If the final formulation is the negative of the
original, its parity is odd. For either parity the physical observables, which depend on the
square of the wave function, are unchanged. A complex system has an overall parity that
is the product of the parities of its components.
The strong and electromagnetic interactions are unchanged by parity. The weak
interaction has been shown to violate parity. (Left handed phenomena differ from right
handed ones.)
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4.7.5 Charge conjugation
Charge conjugation (C) replaces all particles with their antiparticles. But the
quantum mechanics of an electron interacting with a proton in the form of a hydrogen
atom is the same as a positron interacting with an antiproton. So the wave function is
unchanged by charge conjugation.
In classical electromagnetism, charge conjugation is straightforward. Replace positive
charges by negative charges and vice versa. Because electric and magnetic fields have
charges as their sources, these fields are reversed. Mass, energy, momentum, and spin are
unaffected.
In quantum mechanical systems, charge conjugation also requires reversing all the
internal quantum numbers like those for lepton number and baryon number.
The strong and electromagnetic interactions obey charge conjugation symmetry; the
weak interaction does not. As an example, neutrinos have intrinsic parities: neutrinos
have negative parity and antineutrinos positive. Now charge conjugation would leave the
spatial coordinates untouched, then if I applied charge conjugation to a neutrino, I would
produce a negative parity antineutrino. But there is no experimental evidence for such
a particle; all antineutrinos appear to have positive parity. This is parity violation by
neutrinos, i.e., the weak force.
The combination of the parity operation P and the charge conjugation operation C
on a neutrino does produce a right-handed antineutrino, in accordance with observation.
Although beta decay does not obey parity or charge conjugation symmetry separately, it is
invariant under the combination CP.
4.7.6 Time reversal
In simple classical terms, time reversal just means replacing t by −t, inverting the
direction of the flow of time. Reversing time also reverses the time derivatives of spatial
quantities, so it reverses momentum and angular momentum. Newton’s second law is
quadratic in time (F = md2x/dt2) and is invariant under time reversal. Its invariance
under time reversal holds for both gravitational or electromagnetic forces.
Very sensitive experimental tests have been done to put upper bounds on any violation
of time-reversal symmetry. One experiment is the search for an electric dipole moment
for the neutron. Even though the neutron is neutral, it is made up of three quarks: one
up (charge +2/3) and two down (charge -1/3) and therefore could conceivably have an
electric dipole moment. Experimental evidence is consistent with zero dipole moment, so
time reversal symmetry seems to hold in this case.
The small violation of CP symmetry suggests some departure from T symmetry in
some weak interaction process since CPT invariance seems to be on very firm ground.
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4.7.7 A zoo of quarks
This chart shows the properties of some baryon (three quark) particles.
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Lets think about the Delta. There are 4, consisting of only up and down quarks. The
Δ+ is uud, just like the proton,but the spin is 3/2, so all spins are parallel. The Δ0 us
udd, like the neutron.
The two with no spin 1/2 counterparts are the Δ++, charge = 3 · (2/3) and the Δ−,charge = 3 · (−1/3).
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This chart shows the properties of some meson (two quark) particles.
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The pions are the lightest mesons. There are 3, consisting of only up and down
quarks. The π+ is ud. (The antiparticle has the opposite charge from the particle, so 2/3
+ 1/3 = 1.) The π−, ud, is the antiparticle to the π+. The π0 us a linear combination of
uu and dd.
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