Download F34TPP Theoretical Particle Physics notes by Paul Saffin Contents

F34TPP Theoretical Particle Physics
notes by Paul Saffin
Contents
1 Lecture one
1.1 Natural units. . . . . . . . . .
1.2 converting back to SI units . .
1.3 relativistic notation . . . . . .
1.3.1 raising/lowering indices
1.4 examples . . . . . . . . . . . .
1.4.1 energy-mass relation .
1.4.2 quantum operators . .
1.4.3 Klein-Gordon equation
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2 Lecture two - Dirac equation
2.1 Dirac equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 solving the Dirac equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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3 Lecture three - spin, chirality
3.1 rotation generators . . . . .
3.2 rotating a fermion . . . . . .
3.3 chirality and helicity . . . .
3.4 Weyl fermions . . . . . . . .
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5 Lecture five - Dirac equation and magnetic fields
5.1 magnetic moments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2 magnetic moment of electron - Zeeman interaction . . . . . . . . . . . . . . . . .
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6 Lecture six - Feynman rules
6.1 propogator theory . . . . . . . . . . . .
6.1.1 Schrödinger propogator . . . . .
6.1.2 Klein-Gordon propogator . . . .
6.2 Feynman rules - including interactions
6.3 diagramatic expansion . . . . . . . . .
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and helicity
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4 Lecture four - gauge symmetry and electromagnetism
4.1 electromagnetism . . . . . . . . . . . . . . . . . . . . . .
4.2 gauge invariance of the Dirac equation . . . . . . . . . .
4.3 covariant derivatives . . . . . . . . . . . . . . . . . . . .
4.4 the standard model and gauge symmetries . . . . . . . .
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7 Lecture seven - action functionals and Feynman rules
7.1 action and equations of motion . . . . . . . . . . . . . . . . . .
7.1.1 Klein-Gordon action . . . . . . . . . . . . . . . . . . . .
7.1.2 Dirac action . . . . . . . . . . . . . . . . . . . . . . . . .
7.2 the structure of an action . . . . . . . . . . . . . . . . . . . . .
7.2.1 example, real KG action I . . . . . . . . . . . . . . . . .
7.2.2 example, real KG action II . . . . . . . . . . . . . . . . .
7.2.3 example, complex KG action . . . . . . . . . . . . . . . .
7.3 example, Dirac particle coupled to photons . . . . . . . . . . . .
7.4 complex KG equation coupled to photons . . . . . . . . . . . . .
7.5 Feynman rules for the electroweak sector of the standard model
7.6 Fermi’s golden rule . . . . . . . . . . . . . . . . . . . . . . . . .
7.6.1 ee scattering . . . . . . . . . . . . . . . . . . . . . . . . .
7.6.2 µ decay . . . . . . . . . . . . . . . . . . . . . . . . . . .
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8 Lecture eight - group theory
8.1 definition of a group . . . . . . . . . .
8.2 direct product of groups . . . . . . . .
8.3 Abelian groups . . . . . . . . . . . . .
8.4 Lie groups (pronounced ”Lee groups”)
8.4.1 general linear group GL(n,R) .
8.5 special linear group SL(n,R) . . . . . .
8.6 orthogonal group O(n,R) . . . . . . . .
8.7 special orthogonal group SO(n,R) . . .
8.8 O(p,q,R) . . . . . . . . . . . . . . . . .
8.9 symplectic group Sp(2n,R) . . . . . . .
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10 Lecture ten - some representation theory
10.1 SU(2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.2 normalizing the states . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.3 general SU(2) represenation . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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9 Lecture nine - non-Abelian gauge theory
9.1 unitary groups . . . . . . . . . . . . . . .
9.2 unitary group U(n) . . . . . . . . . . . .
9.3 special unitary groups . . . . . . . . . .
9.4 unitary groups and the standard model .
9.5 generators and Lie algebras . . . . . . .
9.5.1 o(n), the Lie algebra of O(n) . . .
9.5.2 su(n), the Lie algebra of SU(n) .
9.6 non-Abelian gauge theory . . . . . . . .
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11 Lecture eleven - isospin
11.1 neutrons, protons and isospin . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11.2 pions and isospin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11.3 using isospin in scattering calculations . . . . . . . . . . . . . . . . . . . . . . .
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12 Lecture twelve - the quark model
12.1 more multiplets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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13 Lecture thirteen - charge conjugation
13.1 Charge conjugation . . . . . . . . . .
13.1.1 photons . . . . . . . . . . . .
13.1.2 fermion-antifermion pairs . . .
13.1.3 C violation . . . . . . . . . .
13.2 parity . . . . . . . . . . . . . . . . .
13.2.1 electromagnetism . . . . . . .
13.2.2 fermions . . . . . . . . . . . .
13.3 linearity of parity . . . . . . . . . . .
13.4 electron dipole moment (edm) . . . .
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14 Lecture fourteen - parity violation and time reversal
14.1 parity violation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14.2 time reversal symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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15 Lecture fifteen - CP symmetry
15.1 CP symmetry . . . . . . . . .
15.2 neutral kaons . . . . . . . . .
15.3 kaons and C, P and CP . . .
15.4 Klong and Kshort decays . . . .
15.5 matter vs anti-matter . . . . .
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16 Lecture sixteen - quark mixing
16.1 three generations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
16.2 quark mixing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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17 Lecture seventeen - neutrino oscillations
17.1 Solar neutrinos . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
17.2 neutrino oscillations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
17.3 atmospheric neutrinos . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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18 Lecture eighteen - Higgs mechanism and
18.1 symmetry restoration/breaking . . . . .
18.2 superconductivity . . . . . . . . . . . . .
18.3 massive fields . . . . . . . . . . . . . . .
18.4 breaking the symmetry . . . . . . . . . .
18.5 non-Abelian Higgs mechanism . . . . . .
18.6 Yukawa coupling and fermion masses . .
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19 Lecture nineteen - the real standard model
19.1 electroweak sector, part I . . . . . . . . . . .
19.2 chiral theories . . . . . . . . . . . . . . . . .
19.3 electroweak sector, part II . . . . . . . . . .
19.4 electroweak sector, part III . . . . . . . . . .
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List of Figures
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2
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The predicted flux of solar neutrinos, Bahcall et. al. ApJ, 621, L85 (2005). . . .
The typical shower of particles produced by a cosmic ray. . . . . . . . . . . . . .
This shows a superconductor hovering above a magnet. . . . . . . . . . . . . . .
A typical wine bottle, where the dregs of the wine tend to form at the bottom,
around the circular minimum. . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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List of Tables
1
Table showing quark charges . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
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F34TPP Theoretical Particle Physics
1
Lecture one
Aim: To introduce the notion of natural units, dimensional analysis, and revise some relativistic
notation.
Learning outcomes: At the end of this lecture you should
• be able to convert quantities between different systems of units
• know how to use Einstein summation convention
1.1
Natural units.
The first thing to note is that the numerical value of a dimensionful quantity does not have any
intrinsic value. For example, if you were told that the speed of light was 7.9×1014 , that would be
meaningless; the reason for this is that the numerical value depends on the system of measuring
rods and clocks you use. If you use metres and seconds you would get c = 3 × 108 ms−1 , if you
use miles and hours you get c = 6.7 × 108 miles hour−1 , and if you use cubits and fortnights
you get c = 7.9 × 1014 cubits f ortnight−1 . Similarly, if we use kilograms, pounds or grains
to measure measure mass we find that ~ = 10−34 m2 kg s−1 , 3 × 10−37 miles2 lb hour−1 , 9 ×
10−24 cubits2 grain f ortnight−1 . So, by picking our units appropriately, we may essentially get
any value we wish for dimensionful quantities. As we are interested here in relativistic quantum
field theory, it makes sense to choose units where
c = 1,
~ = 1,
(1.1)
(1.2)
these are called natural units. For example we may use the year as our measure of time, and
the light-year as our measure of distance, then light travels one light-year per year, i.e. c = 1
in these units. In effect, we should think of c as the quantity that allows us to relate distances
to times, and ~ is the quantity that relates masses to times and distances.
Using these natural units we find that the basic relations
E 2 = m2 c2 + p2 c2 ,
E = ~ω
2π
p = ~
λ
(1.3)
(1.4)
E 2 = m2 + p2 ,
E = ω,
2π
p =
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λ
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(1.7)
(1.5)
become
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(1.8)
Now, we still need to pick our basic unit, and in particle physics the convention is to use
electron-Volts, eV, to measure energy. Then, from (1.6), we have that mass is measured in eV ,
as is momentum. Now (1.8) shows that distances are measured in (eV )−1 , and (1.7) gives that
time is measured in (eV )−1 .
1.2
converting back to SI units
At some point an experimenter will want to know how big their machine should be, and rulers
don’t come with eV notches on them, so we need to able to convert from our eV numbers back
to SI units. The best way to see how this is done is to use an example, so consider the following.
The scattering cross section, σ, of a target particle, and the scattering rate, Γ, are related by
Γ = nσv
(1.9)
where n is the number density of particles that are being sent toward the target, and v their
speed, so Γ is the number of scattering events we expect per unit time, the dimensions are as
follows
[σ] = [L2 ],
[Γ] = L−3 L2 LT −1 = T −1
(1.10)
(1.11)
So, suppose we are told that the cross section associated with Compton-scattering (ei γ → e− γ),
in natural units, is
σ =
8π 2 α2
.
3m2e
(1.12)
This clearly cannot be in SI units, because if it were, the left-hand-side would be m2 , while the
right-hand-side is in kg −2 . Note that the fine structure constant α ∼ 1/137 is dimensionless.
So, to make sense of this expression we need to sprinkle some factors of ~ and c around, so write
8π 2 α2 a b
σ =
~ c.
3m2e
(1.13)
where a and b are unknown constants. Note that this gives the same expression as (1.12) in
natural units where ~ = 1, c = 1. Now we do some dimensional analysis
[L2 ] =
[M L2 T −1 ]a [LT −1 ]b
[M 2 ]
(1.14)
For this to be consistent, we see that a = 2, b = −2, so
σ =
8π 2 α2 ~2
.
3m2e c2
6
(1.15)
1.3
relativistic notation
First of all, recall that the object variously called the line element/proper distance/proper
time/invariant interval... is given by
ds2 = −dt2 + dx2 + dy 2 + dz 2 .
(1.16)
Rather than writing this out each time we introduce a matrix quantity called the metric, with
components ηµν , where the first index µ labels the rows, and the second index ν the columns of
the matrix. In (t, x, y, z) := (x0 , x1 , x2 , x3 ) co-ordinates the matrix is diagonal, with entries
ηµν = diag(−1, 1, 1, 1).
(1.17)
Given this, the line element is
ds2 =
X
ηµν dxµ dxν .
(1.18)
µν
where we have introduced the spacetime four-vector with components dxµ = (dt, dx, dy, dz).
The summation convention simply says that if an index appears twice in a single expression,
then that index must be summed over. So, for example, the line element becomes
(1.19)
ds2 = ηµν dxµ dxν .
P
There is no need to explicitly write the µν symbol, because both µ and ν appear twice, so we
know they must be summed over.
1.3.1
raising/lowering indices
Along with the metric ηµν we have the inverse metric η µν , which is just the inverse matrix of
the metric. So, in (t, x, y, z) co-ordinates we have
η µν = diag(−1, 1, 1, 1).
(1.20)
This allows us to ”raise” and ”lower” indices of spacetime vectors, for example
V µ = η µν Vν
Vµ = ηµν V ν .
(1.21)
(1.22)
This can be useful as it often further simplifies certain expressions, meaning that we don’t even
need to write ηµν , on top of dropping the summation symbols.
1.4
1.4.1
examples
energy-mass relation
recall that the energy-momentum four-vector is
P µ = (E, p)
7
(1.23)
i.e. the zero-component of four-momentum is just the eneryg, P 0 = E. Using Pµ = ηµν P ν we
also have
Pµ = (−E, p)
(1.24)
Now we take the energy-momentum relation
E 2 = p2 + m2
(1.25)
and rewrite this as
−E 2 + p2 + m2 = 0
−(P 0 )2 + (P 1 )2 + (P 2 )2 + (P 3 )2 + m2
η00 P 0 P 0 + η11 P 1 P 1 + η22 P 2 P 2 + η33 P 3 P 3 + m2
ηµν P µ P ν + m2
P µ Pµ + m 2
=
=
=
=
0
0
0
0
(1.26)
(1.27)
(1.28)
(1.29)
(1.30)
As this final expression contains only spacetime indices it is manifestly a relativistic equation.
1.4.2
quantum operators
From quantum mechanics we know that
Ê = i∂t
P̂ x = −i∂x
(1.31)
(1.32)
where we denote
∂x =
∂
.
∂x
(1.33)
Using the fact that P0 = −P 0 and P1 = P 1 we see
P̂0 = −i∂0 , P̂1 = −i∂1 P̂2 = −i∂2 P̂3 = −i∂3
(1.34)
or, using our relativistic notation
1.4.3
P̂µ = −i∂µ
(1.35)
P µ Pµ + m 2 = 0
(1.36)
P̂µ = −i∂µ
(1.37)
Klein-Gordon equation
We have already seen that
and
8
so we may combine them to give
∂ µ ∂µ φ − m2 φ = 0
(1.38)
This is the Klein-Gordon equation. If we unravel the index structure we find
η µν ∂µ ∂ν φ − m2 φ = 0
−∂0 ∂0 φ + ∂1 ∂1 φ + ∂2 ∂2 φ + ∂3 ∂3 φ − m2 φ = 0
−φ̈ + ∇2 φ − m2 φ = 0
(1.39)
(1.40)
(1.41)
where a dot denotes time derivative
φ̇ =
2
∂φ
∂t
(1.42)
Lecture two - Dirac equation
Aim: To introduce the Dirac equation.
Learning outcomes: At the end of this lecture you should
• know how to derive the Dirac equation
• know what the Dirac algebra is
• be able to solve the Dirac equation
2.1
Dirac equation
The Schrödinger equation
~2 2
∂ φ − V (x)ψ = 0
i∂t ψ +
2m x
(2.43)
treats time and space on an unequal footing, and so cannot be relativistic. The Klein-Gordon
equation on the other hand is manifestly relativistic, however, it is second-order in time derivatives, unlike the first-order time derivatives of the Schr”odinger equation. Dirac wanted to
construct a relativistic wave equation that was first-order in time derivatives, as well as being
relativistic, so he wrote down the following
γ 0 ∂0 ψ + γ 1 ∂1 ψ + γ 2 ∂2 ψ + γ 3 ∂3 ψ + mψ = 0
(2.44)
which is the most general linear, homogeneous, first-order equation, where the γ are as yet
unknown. This equation is written in relativistic form as
γ µ ∂µ ψ + mψ = 0
(2.45)
In order to find out what the γ are we use ∂µ = iP̂µ and find
(iγ µ P̂µ + m)ψ = 0
9
(2.46)
Given that this vanishes, we can also act on it with (iγ ν P̂ν − m), which when acting on zero
gives zero, so
⇒
(iγ ν P̂ν − m)(iγ µ P̂µ + m)ψ = 0
(2.47)
⇒ (−γ ν γ µ P̂ν P̂µ − m2 )ψ = 0
(2.48)
1 µ ν
[γ γ + γ ν γ µ ]P̂µ P̂ν ψ + m2 ψ = 0
2
(2.49)
Now, we denote the anti-commutator with curly braces
{γ µ , γ ν } := γ µ γ ν + γ ν γ µ
(2.50)
1 µ ν
2
{γ , γ }P̂µ P̂ν + m ψ = 0
2
(2.51)
so we find
which we should compare to the relativistic energy-momentum equation
η µν Pµ Pν + m2 = 0
(2.52)
{γ µ , γ ν } = 2η µν .
(2.53)
leading to the conclusion that
So, how do we use this to find out what the γ µ are? Well, let’s start with
{γ 0 , γ 0 } = 2η 00 ⇒ (γ 0 )2 = −1 ⇒ γ 0 = ±i
(2.54)
{γ 1 , γ 1 } = 2η 11 ⇒ (γ 1 )2 = 1 ⇒ γ 1 = ±1
(2.55)
{γ 0 , γ 1 } = 2η 01 ⇒ γ 0 γ 1 = 0 ⇒ ±i = 0
(2.56)
Now lets try
And now we go for
and we run into a contradiction. The reason this has gone wrong is that we assumed the γ µ are
numbers, they are not, they are matrices, and so we should really write
{γ µ , γ ν } = 2η µν I.
(2.57)
This is known as the Dirac algebra. In fact, the smallest set of matrices that can satisfy the
Dirac algebra are 4 × 4 matrices, meaning that ψ is actually a four-component object - four
complex components - as opposed to the Schrödinger wave function which is a single complex
function.
In fact, there are infinitely many choices of matrices that satisfy (2.57). This is not a
problem, and in fact is a good thing; we can think of each of these different choices as being
required for the infinite variety of observers, all travelling at different velocities. For example,
10
suppose that a set γ µ solves (2.57), then so will Γµ = −γ µ , and so will Γµ = ±γ µT , and so will
Γµ = ±γ µ∗ . In fact, if we take a general invertible matrix R, we have that Γµ = Rγ µ R−1 will
also satisfy the Dirac algebra, e.g.
{Γµ , Γν } =
=
=
=
=
=
Γµ Γν + Γν Γµ
Rγ µ R−1 Rγ ν R−1 + Rγ ν R−1 Rγ µ R−1
Rγ µ γ ν R−1 + Rγ ν γ µ R−1
R(γ µ γ ν + γ ν γ µ )R−1
R(2η µν )R−1
2η µν
(2.58)
(2.59)
(2.60)
(2.61)
(2.62)
(2.63)
Although each of these different representations will lead to the same physical results, there are
certain choices that can make calculations easier, two common choices are the:
Dirac representation
iI 0
0 iσ
0 I
0
5
γ =
, γ=
, γ =
(2.64)
0 −iI
−iσ 0
I 0
where the σ are the Pauli matrices. And the
Weyl representation
0 iI
0 iσ
−I 0
0
5
γ =
, γ=
, γ =
iI 0
−iσ 0
0 I
(2.65)
Now, you will have noticed that I sneaked in a new matrix, γ 5 . It turns out that γ 5 plays a
crucial role in much of the physics of the standard model, and it is defined as follows,
γ 5 = iγ 0 γ 1 γ 2 γ 3 .
(2.66)
And the remarkable property that it satisfies, making it such a useful object, is
γ 5 γ µ = −γ µ γ 5 .
(2.67)
We shall have more to say about this later.
2.2
solving the Dirac equation
The Dirac equation is linear, so if we are to look for wave soluitons it is natural to look at the
Fourier modes; that means we look for solutions of the form
UA
ipµ xµ
ψ=e
(2.68)
UB
where UA,B are two-component column vectors.
Given this ansatz we note that
i∂t ψ = −P0 ψ = Eψ
11
(2.69)
and if we use the Dirac representation we find that
EUA − σ.P UB + mUA = 0
−EUB + σ.P UA + mUB = 0
(2.70)
(2.71)
In particular, if we focus on the case of a wave in the z-direction, P = (0, 0, P z ), we have
(E + m)UA = P z σ 3 UB
(E − m)UB = P z σ 3 UA
(2.72)
(2.73)
So that we find
(E 2 − m2 )UB = (E + m)(E − m)UB = P z σ 3 (E + m)UA = P z σ 3 P z σ 3 UB = (P z )2 (σ 3 )2 UB = (P z )2 UB
(E 2 − m2 − P 2 )UB = 0
(2.74)
which is just the relativistic energy-momentum equation, as expected. As UB is a two-component
object, we have two linearly independent solutions:
solution one has
 z

P
/(E
+
m)


Pz
0
1
1

⇒ ψ↑ = eiP.x 
(2.75)
UB =
⇒ UA =


1
0
E+m 0
0
solution two has

UB

0
 −P z /(E + m) 
Pz
0
0

=
⇒ UA =
⇒ ψ↓ = eiP.x 


1
0
E + m −1
1
(2.76)
In other words, for a particle travelling in the z− direction there are two possible solutions,
these are just the spin-up and spin-down posibilities, as we shall see.
3
Lecture three - spin, chirality and helicity
Aim: To see where angular momentum fits into the Dirac equation
Learning outcomes: At the end of this lecture you should
• know what the angular momentum operators for a Dirac fermion are
• know what chirality and helicity are
12
3.1
rotation generators
By now you will have come across the phrase ”a fermion is spin-half” a number of times over
the course of your degree, here we shall see why. To do this we need to understand how to
rotate objects, in particular fermions. The key is the angular momentum algebra, as it is the
angular momentum operators that are resonsible for rotating objects,
h
i
(3.77)
L̂x , L̂y , = iL̂z
This is typically derived using
Lz = xPy − yPx → L̂z = −ix∂y + iy∂x
(3.78)
relevant for orbital angular momentum, but in fact the L̂ appearing in (3.77) can take a number
of different forms. To see this, let’s consider how we rotate a three-vector V ,
V0 =R V
(3.79)
where R is a matrix satisfying RRT = I. For example, a
is accomplished by

cos θ − sin θ
Rz =  sin θ cos θ
0
0
rotation about the z-axis with angle θ

0
0 ,
1
whereas a rotation about the x and y axes come from




1
0
0
cos θ 0 − sin θ
,
1
0
Rx =  0 cos θ − sin θ  , Ry =  0
0 sin θ cos θ
sin θ 0 cos θ
(3.80)
(3.81)
Now we know how to rotate vectors (we just use the R matrices), what about rotating fermions?
We cannot use R to rotate a four-component fermion field ψ, as R is a 3x3 matrix. In order to
progress we need something more general, and the route to generality comes by again looking
at the rotation of vectors, but from a different perspective. We write, for example,
Rz = e−iθT
3
(3.82)
where

T3

0 −i 0
=  i 0 0 ,
0 0 0
(3.83)
and the exponential of a matrix is defined through the taylor expansion of the exponential
function,
1
1
eM := I + M + M 2 + M 3 + ...
2
3!
13
(3.84)
One makes similar definitions for Rx

0
1

0
T =
0
1
= e−iθT and Ry = e−iθT


0
0 0
0 −i  , T 2 =  0
−i
i 0
2
where

0 i
0 0 ,
0 0
(3.85)
This may seem like an obscure thing to do, but the point is that general group elements, g,
α α
can be written as g = e−iθ T , where the set of matrices T α are known as the generators of the
group. The magic thing is that, by explicit computation,
1 2
(3.86)
T ,T
= iT 3 ,
which is just the angular momentum algebra, (3.77). The point is that it is the angular momentum algebra generators that are resposnsible for rotations, be they orbital angular momentum,
in which case one should use the generators given by (3.78), or vector rotation, in which case
one should use (3.83,3.85). Our task now is to find a set of 4x4 matrices that satisfy (3.77) so
that we can rotate fermions.
3.2
rotating a fermion
What we need to do is find some set of matrices, S that obey the angular momentum algebra
1 2
S ,S
= iS 3 ,
(3.87)
these should be 4x4 matrices, in order to be able to act on the four-component fermion field
ψ. As we already have a set of 4x4 matrices - the Dirac matrices - this is the natural place to
start looking. In fact, we shall simply state that these generators are given by defining a set of
matrices Σjk , where
Σjk := −
S i :=
i j k
γ ,γ ,
4
1 ijk jk
Σ
2
(3.88)
(3.89)
for example
S3 =
1 312 12 1 321 21
i
i
Σ + Σ = Σ12 = − (γ 1 γ 2 − γ 2 γ 1 ) = − γ 1 γ 2
2
2
4
2
(3.90)
along with
i
S 2 = − γ3γ1
2
i
S 1 = − γ2γ3
2
(3.91)
(3.92)
One may check, by explicit computation using te Dirac algebra, that (3.87) is satisfied. So, we
now know how to rotate a fermion. For example, in the representations given previously we
find that the generator of a rotation about the z-axis is
1 σ3 0
3
S =
,
(3.93)
0 σ3
2
14
so, for example, our plane-wave solutions (2.75), (2.76) have
1
S 3 ψ↑ = + ψ↑
2
1
S 3 ψ↓ = − ψ↓
2
(3.94)
(3.95)
(3.96)
which, as expected, shows that ψ↑ has spin + 12 , while ψ↓ has spin − 12 .
By analogy with vectors, a rotation of a fermion about the z-axis through angle θ is accomplished by
3
ψ 0 = e−iθS ψ,
(3.97)
which means we may check explicitly the common phrase that ”a fermion picks up a minus sign
when rotated through 2π”, by noting that
−iθ/2 σ3
e
0
−iθS 3
e
(3.98)
=
,
3
0
e−iθ/2 σ
I cos(θ/2) − iσ 3 sin(θ/2)
0
=
(3.99)
0
I cos(θ/2) − iσ 3 sin(θ/2)
so, for θ = 2π,
ψ 0 = −ψ
3.3
(3.100)
chirality and helicity
due to the presence of a γ 5 matrix we have a neat way of splitting up a Dirac fermion using the
projectors
PR =
1
1
(I + γ 5 ), PL = (I − γ 5 )
2
2
(3.101)
and we define the left and right chiral components by
ψR = PR ψ, ψL = PL ψ
(3.102)
This can always be done, because
1
1
1
1
ψ + γ5ψ + ψ − γ5ψ
2
2
2
2
= ψR + ψL
ψ =
(3.103)
(3.104)
we then note that the chiral components are eigenfunctions of the γ 5 matrix
γ 5 ψR = +ψR
γ 5 ψL = −ψL
15
(3.105)
(3.106)
At first sight, there seems to be little reason to label these chiral components as left and right,
why not white and black, chalk and cheese...? To understand this we need to introduce the
notion of helicity. We now know that a fermion has two vector associated with it, its momentum,
and its spin. Helicity tells us whether they are pointing in the same direction or not. If the
momentum and spin are aligned (positive helicity) then the fermion is spinning in the direction
given by the usual right-hand-rule, so we call it right-handed. We shall now see that this ties
in exactly with the chirality of massless fields.
The Dirac equation for a wave mode of a massless fermion may be manipulated as follows
γ µ ∂µ ψ = 0
⇒ γ µ Pµ ψ = 0
⇒ γ 0 P0 ψ + γ.P ψ = 0
×γ 0 ⇒ −P0 ψ − γ.P γ 0 ψ = 0
×γ 5 ⇒ −P0 γ 5 ψ − γ.P γ 0 γ 5 ψ = 0
⇒ −P0 γ 5 ψ + γ.P iγ 1 γ 2 γ 3 ψ = 0
⇒ −P0 γ 5 ψ + iP x γ 2 γ 3 ψ + iP y γ 3 γ 1 ψ + iP z γ 1 γ 2 ψ
⇒ −P0 γ 5 ψ − 2P x S x ψ − 2P y S y ψ − 2P z S z ψ
⇒ Eγ 5 ψ − 2P .Sψ
P .S
ψ
E = |P | ⇒
|P |
= 0
= 0
= 0
1 5
=
γ ψ
2
(3.107)
so we define the helicity operator
h :=
P .S
|P |
(3.108)
which measure whether the momentum is aligned with the spin. So, we see that right and left
massless chiral fields have
1
hψR = + ψR
(3.109)
2
1
hψL = − ψL
(3.110)
2
which is why they are called right and left, they follow the right hand rule, or left hand rule.
In the above derivation we made critical use of the masslessness of the fermion, this is because
helicity is not Lorentz invariant for massive fields. For example, suppose we see a fermion with
helicity + 12 , we can simply convert it to − 12 by travelling faster than it in its direction of motion,
then P will change sign, but S will remain the same. We cannot overtake massless fields, and
so the helicity of massless particles is a Lorentz invariant. However, the notion of chirality, as
defined by γ 5 , is Lorentz invariant. Chirality and helicity only coincide for massless particles,
nevertheless, we still call chiral fermions left and right, irrespective of whether they are massless.
3.4
Weyl fermions
We have just seen that mass plays an important role for chiral fermions, this can be seen in
another context - Weyl fermions. Taking the Dirac equation and operating on it with PR , and
16
PL we find
PR [γ µ ∂µ ψ + mψ] = 0
⇒ γ µ ∂µ ψL + mψR = 0
(3.111)
(3.112)
PL [γ µ ∂µ ψ + mψ] = 0
⇒ γ µ ∂µ ψR + mψL = 0
(3.113)
(3.114)
and
If m = 0 then these are two independent equations, and we may consistently set either ψR = 0
and ψL 6= 0, or ψL = 0 and ψR 6= 0. However, if the fermion has non-zero mass we need both
left and right components1 .
4
Lecture four - gauge symmetry and electromagnetism
Aim: To see what electromagnetism has to do with symmetry
Learning outcomes: At the end of this lecture you should
• be able to turn a global symmetry into a local one
• know what role gauge fields play in local symmetries.
4.1
electromagnetism
In the standard model, the list of particles can be split up into matter particles and force
carriers. Matter particles are the bits that stuff is made out of, such as electrons, protons and
quarks, whereas force carriers are what bind the matter particles together, such as photons and
gluons. Here we shall see that the origin of the photon is a symmetry principle.
First of all, what do we mean by a symmetry? The type of symmetry that we will be looking
at is the analogue of a symmetry in Newton’s second law,
F = −∇V.
(4.115)
We can see quite clearly that this system of equations is unchanged if we take
V → V + V0 , V0 ∈ R
(4.116)
i.e. V0 is a real constant. This is a symmetry; we can shift the potential by any number, and
none of the physics changes.
A similar thing happens in electrical circuits, the current between two points is determined
by the difference in potential between those points,
I=
∆V
R
1
(4.117)
There is another type of mass term that we will not be considering, a Majarona mass, for which these
conclusions are a bit premature.
17
so we can again shift the potential by a constant amount, everywhere on the circuit, and the
physics remains the same.
In electromagnetism we have that
B = ∇ × A, E = −∇φ − Ȧ.
(4.118)
This system is again governed by derivatives, so we are free to shift both φ and A by constants
without changing the physics; in fact, we can do much more than that. Rather than just shifting
by a constant, we can shift by an amount determined by an arbitrary function, α(t, x)
φ → φ − α̇, A → A + ∇α
(4.119)
and the physics remains the same, i.e. under these shifts
E → E, B → B.
(4.120)
These shifts are known as gauge transformations. The fact that φ and A are undefined up to
this arbitrary function highlights that they are not physical quantities, only the electric and
magnetic field can be measured.
In line with our relativistic notation we shall now introduce the four-vector potential, by
combining φ and A into
Aµ = (φ, A), ⇒ Aµ = (−φ, A)
(4.121)
Then, in this unified language, we see that our gauge transformations become
Aµ → Aµ + ∂µ α.
(4.122)
This will make it easier when we come to the relativistic Dirac equation - which is now.
4.2
gauge invariance of the Dirac equation
The Dirac equation
γ µ ∂µ ψ + mψ = 0
(4.123)
has what is called a U(1) symmetry - more of which later. Simply put, this means the equation
for ψ and ψ 0 are exactly the same if
ψ 0 = e−iΛ ψ, Λ ∈ R
(4.124)
so, Λ is a real constant. This just means that we are allowed to perform a phase rotation by
angle Λ, so long as we do it everywhere in spacetime. This is what is termed a global symmetry
- we do it everywhere. Now we ask the question, can we make this symmetry local, i.e. let
Λ → Λ(x)? The naive answer is no. For
ψ 0 = e−iΛ(x) ψ
18
(4.125)
we find that the Dirac equation for ψ implies
γ µ ∂µ ψ 0 + mψ 0 = −iγ µ ∂µ Λ ψ 0 ,
(4.126)
which is not the Dirac equation. Rather than just giving up, we notice that if, instead of starting
with the Dirac equation, we start with
γ µ (∂µ − iqAµ )ψ + mψ = 0
(4.127)
and
ψ 0 = e−iΛ(x) ψ,
1
A0µ = Aµ − ∂µ Λ
q
(4.128)
(4.129)
then we end up with
γ µ (∂µ − iqA0µ )ψ 0 + mψ 0 = 0.
(4.130)
i.e. the (ψ, Aµ ) variables satisfy the same equation as the (ψ 0 , A0µ ) variables, moreover, the
transformation of the four-vector Aµ is exactly what is required by electromagnetism if we
identify α = −Λ/q.
Let us recap. We now have something that looks a bit like the Dirac equation, but it
generalizes the global U(1) symmetry of the basic Dirac equation to a local U(1), i.e. we can
now perform a different U(1) transformation at each point in spacetime and the physics is
unchanged. The penalty for achieving this is that we had to introduce a new quantity Aµ ,
but that’s just electromagnetism, so it’s more of a prize than a penalty! U(1) gauge symmetry
implies the existence of a photon, Aµ .
4.3
covariant derivatives
The best way to think about gauge invariance is through the notion of a covariant derivative.
The reason that the original Dirac equation does not have a local U(1) symmetry is that when
we take ψ → e−iΛ(x) ψ, ψ and ∂µ ψ transform differently. What we really want is some derivative,
Dµ , that transforms as
Dµ ψ → e−iΛ Dµ ψ,
(4.131)
because then the derivative piece of the enhanced Dirac equation (γ µ Dµ ψ) will transform in the
same way as the mass term. This is just what we have constructed,
Dµ = ∂µ − iqAµ
(4.132)
and we have
ψ 0 = e−iΛ(x) ψ,
A0µ = Aµ + ∂µ α
(4.133)
(4.134)
(Dµ ψ)0 = e−iΛ(x) (Dµ ψ).
(4.135)
19
With this covariant derivative we are now in a position to construct new objects, the most
important of which is the field strength, Fµν . This is define by
[Dµ , Dν ] ψ = −iqFµν ψ.
(4.136)
This is a nice quantity because it is gauge invariant, and so could be of physical relevance. To see
that it is gauge invariant we note that as Dµ ψ transforms covariantly, do does Dν Dµ ψ, that is
one of the strengths of using covariant derivatives. This implies that [Dµ , Dν ] ψ also transforms
covariantly (i.e. [Dµ , Dν ] ψ → e−iΛ(x) [Dµ , Dν ] ψ). But, the right-hand-side of (4.136) also picks
0
up a e−iΛ(x) , meaning that Fµν must be unchanged, Fµν
= Fµν .
To be more explicit note that
ψ 0 = U ψ,
(Dµ ψ)0 = U (Dµ ψ),
(Dµ Dν ψ)0 = U (Dµ Dν ψ)
(4.137)
where U = e−iΛ(x) . So, the primed field strength is just
([Dµ , Dν ] ψ)0
U [Dµ , Dν ] ψ
U (−iqFµν ψ)
U Fµν ψ
=
=
=
=
0
−iqFµν
ψ0
0
−iqFµν
Uψ
0
−iqFµν
Uψ
0
Fµν U ψ
(4.138)
(4.139)
(4.140)
(4.141)
Now, this is required to hold for all ψ so
0
U Fµν = Fµν
U
(4.142)
and now note that U U † = 1 = U † so multiply on the right by U † .
0
U Fµν U † = Fµν
(4.143)
then note that in this case the gauge theory is Abelian, i.e. U(1), so the U , U † are just numbers
and can move through the Fµν to cancel each other
0
Fµν = Fµν
(4.144)
This proves the gauge invariance of the field strength. Now we need to see what the field
strength actually is, to do this, just calculate it
−iqFµν ψ = Dµ Dν ψ − Dν Dµ ψ
= (∂µ − iqAµ )(∂ν ψ − iqAν ψ) − (µ ↔ ν)
= −iq(∂µ Aν − ∂ν Aµ )ψ
(4.145)
(4.146)
(4.147)
Fµν = ∂µ Aν − ∂ν Aµ
(4.148)
so we find
and now we are nearly there, just a few examples should convince you that Fµν is nothing more
than a re-packaging of the electric and magnetic fields.
e.g. F01 = ∂0 A1 − ∂1 A0 = Ȧx + ∂x φ = −E x .
e.g. F12 = ∂1 A2 − ∂2 A1 = ∂x Ay − ∂y Ax = B z .
It is a useful excercise to check the other components.
20
4.4
the standard model and gauge symmetries
In later lectures we shall come across more general symmetry groups than U(1); ones of particular importance are SU(2) and SU(3). The reason for their significance is that the standard
model is a gauge theory of U(1)×SU(2)×SU(3), with the U(1)×SU(2) supplying the photon,
the Z0 , and the W± gauge bosons, and the SU(3) giving us the gluons.
5
Lecture five - Dirac equation and magnetic fields
Aim: To see how electromagnetism fits in with the Dirac equation
Learning outcomes: At the end of this lecture you should
• know what a magnetic moment of a particle is
• know how to derive the magnetic moment of a non-relativistic Dirac particle
5.1
magnetic moments
Recall from electromagnetism that the magnetic moment of a circular, planar loop of area A,
with current I, is given by
µ = IA.
(5.149)
Now, for a total charge of q running around the loop, in period T we have
v
(5.150)
I = q/T = q
2πr
so we find
q
q
mvr =
L
µ =
(5.151)
2m
2m
where L is the angular momentum of the current, and m its mass. Now, suppose we place such
a magnetic moment in a magnetic field, one finds that they prefer to align with the background
field. This is because aligned magnets have a lower energy. In fact, classical electromagnetism
says that the energy of a magnetic moment, µ, within a magnetic field B is given by the
interaction Hamiltonian
q
Hint = −µ.B = −
B.L
(5.152)
2m
This is the starting point for writing down the interaction energy of a particle of spin S within
a magnetic field. As we expect there to be differences with the basic loop of wire we adopt a
fudge-factor, g, to account for differences, giving the interaction Hamiltonian as
q
H = −g
B.S
(5.153)
2m
where g for the electron is found, by experiment, to be
gelectron ∼ 2.002
It is our task to derive the magnetic moment of the Dirac particle.
21
(5.154)
5.2
magnetic moment of electron - Zeeman interaction
Now that we have learned all about gauge invariance, we shall see one of its consequences,
namely a prediction for the magnetic moment of Dirac particles. We start with the gauged
Dirac equation
γ µ (∂µ − iqAµ )ψ + mψ = 0
and we write the Dirac field as
ψ =
ψ1
ψ2
(5.155)
Taking the Dirac representation of gamma matrices (2.64), the definition of the four-vector
potential (4.121), and the relation between momentum and derivatives (1.35) we find
i∂t ψ1 = (−m + qφ)ψ1 + (p − qA).σψ2
i∂t ψ2 = (m + qφ)ψ2 + (p − qA).σψ1 .
(5.156)
(5.157)
For a free wave, we have that the wave function varies as e−iEt , and now we take the nonrelativistic approximation that
E ∼ m,
m >> qφ
(5.158)
1
(p − qA).σψ2
2m
(5.159)
and use this in (5.156) to find
ψ1 '
showing that ψ1 is small compared to ψ2 , and also we find that (5.157) now becomes
i∂t ψ2 = (m + qφ)ψ2 +
1
[(p − qA).σ][(p − qA).σ]ψ2
2m
(5.160)
and we then use that
[(p − qA).σ]2 = (p − qA)2 − qB.σ
(5.161)
which gives the final equation
(p − qA)2
q
ψ2 + (m + qφ)ψ2 − B.Sψ2 = i∂t ψ2
2m
m
(5.162)
which is just the Schrödinger equation for a particle in a magnetic field, whose magnetic interaction is
q
Hzeeman = − B.S
(5.163)
m
giving a g-factor for the Dirac particle of
gDirac = 2
22
(5.164)
As this is so close to the observed value for the electron, we identify the electron as a Dirac
particle. The corrections to the value of 2 is explained using quantum field theory - very precisely
g − 2 = (1159.65218111 ± 0.00000074) × 10−6
(5.165)
2 experiment
g − 2 = (1159.65213 ± 0.000003) × 10−6
(5.166)
2 theory
(5.167)
6
Lecture six - Feynman rules
Aim: To uncover the origin of a particuarly useful perturbative technique
Learning outcomes: At the end of this lecture you should
• know what a propogator is
• be familiar with the route from Huygen’s principle to the propogator equation
• be able to solve the propogator equation when there is a small perturbation present
• be able to derive the propogtor for the Klein-Gordon equation
6.1
propogator theory
Based on an idea of Dirac, Feynman decided to formulate his own version of quantum mechanics,
and he based it on Huygen’s principle for calculating the propogation of wavefronts. This should
not come as a great surprise, given that the Schrödinger equation is a wave equation. The
question we want to ask is, what is the probability amplitude at some location xf , at time tf ,
given the amplitde at some earlier time t?
To answer this, we introduce the notion of the propogator, G(tf , xf ; t, x), which tells us how
an amplitude at (t, x) evolves into an amplitude at the late time (tf , xf ).
X
(6.168)
ψ(tf , xf ) =
iG(tf , xf ; t, x) ∆xψ(t, x),
tf > t
x
What this equation does, is split space into blocks of size ∆x, then the block at location x
contributes ∆xψ(t, x) to the amplitude at the initial time t. This disturbance then propogates
to (tf , xf ) with the help of the propogator. We then have to add up all such contributions. The
continuum version changes the summation to an integral, and we may incorporate the tf > t
condition by including a theta-function
Z
Θ(tf − t)ψ(tf , xf ) = dx iG(tf , xf ; t, x)ψ(t, x),
(6.169)
To see how this works in practise, let’s take write the Schrödinger equation in the following
form, introducing the operator Ô
Ô(tf , xf )ψ(tf , xf ) = [i∂tf − H(tf , xf )]ψ(tf , xf ) = 0
23
(6.170)
Now operate on both sides of (6.169) with Ô, using
∂tf Θ(tf − t) = δ(tf − t)
(6.171)
to find
Z
iδ(tf − t)ψ(tf , xf ) =
i
h
dx Ô(tf , xf )iG(tf , xf ; t, x) ψ(t, x)
(6.172)
which implies that
Ô(tf , xf )G(tf , xf ; t, x) = δ(tf − t)δ(xf − x).
(6.173)
So, this just tells us that the propogator, G(tf , xf ; t, x), is the Green-function associated to the
Schrödinger equation. What this means is that the propogator is, in some sense, the inverse of
the Schrödinger equation, i.e. ÔÔ−1 = I.
6.1.1
Schrödinger propogator
The propogator for the free (no interaction terms in the Hamiltonian) Schrödinger equation
comes from solving
[i∂tf − H0 (tf , xf )]G0 (tf , xf ; t, x) = δ(tf − t)δ(xf − x)
(6.174)
As this is a linear equation we consider a solution using the Fourier transform by writing
Z
dE dp
G0 (tf , xf ; t, x) =
G̃0 (E, p)e−iE(tf −t)+ip(xf −x)
(6.175)
2π 2π
and then we use
Z
dxe−ix(y−a) = 2πδ(y − a)
(6.176)
to find
G̃0 (E, p) =
1
E − p2 /2m
(6.177)
In fact, it is more common to call this, the Fourier transformed version, the propogator.
6.1.2
Klein-Gordon propogator
The Klein-Gordon equation is
(−∂µ ∂ µ + m2 )φ = 0
(6.178)
and to find the KG propogator we simply replace φ with G0 (tf , xf ; t, x), and the zero on the
right-hand-side by δ(tf − t)δ(xf − x).
(−∂µ ∂ µ + m2 )G0 (tf , xf ; t, x) = δ(tf − t)δ(xf − x)
24
(6.179)
Again, the equation is linear so we look for the Fourier solution
Z
d4 p
G0 (tf , xf ; t, x) =
G̃0 (p)eip.(xf −x)
(2π)4
(6.180)
and now we find the propogator for the KG equation to be
G0 (p) =
1
.
p2 + m2
(6.181)
where p2 = −E 2 + p2 is the spacetime-square of the momentum four-vector. Now we start to
spot a pattern, the (momentum-space) propogator is just 1/(equation of motion), both for the
Schrödinger equation, and the KG equation - this is rather generic, especially given that we
expect the propogator to be the inverse of the equation of motion, ÔÔ−1 = I.
6.2
Feynman rules - including interactions
So far we have not discussed the propogators for interacting theories, only the free versions,
now we shall add a small perturbation to the Schrödinger Hamiltonian to see how we may
accomodate interactions. The starting point is
Ô(tf , xf ) = i∂t − Ĥ = i∂t − Ĥ0 − VI = Ô0 (tf , xf ) − VI
Ô(tf , xf )G(tf , xf ; t, x) = δ(tf − t)δ(xf − x)
(6.182)
(6.183)
These may be rewritten as
Ô0 (tf , xf )G(tf , xf ; t, x) = δ(tf − t)δ(xf − x) + VI (tf , xf )G(tf , xf ; t, x)
(6.184)
the solution of which is an integral equation
Z
G(tf , xf ; t, x) = G0 (tf , xf ; t, x) + dt1 dx1 G0 (tf , xf ; t1 , x1 )VI (t1 , x1 )G(t1 , x1 ; t, x)
(6.185)
This may not be apparent at first sight, and is easier to confirm by working backwards. If
we operate on (6.185) with O0 (tf , xf ), the left-hand-side simply recovers the left-hand-side of
(6.184), while the first term on the right-hand-side gives the first term on the right-hand-side of
(6.184). To understand the rest, note that O0 (tf , xf ) only sees tf and xf , it’s not interested in
t, t1 , x, x1 , so when O0 (tf , xf ) hits the integral on the right-hand-side of (6.185) it just turns
the G0 (tf , xf ; t1 , x1 ) into δ(tf − t1 )δ(xf − x1 ), allowing the integral to be performed.
We now use an unusual trick, that usually results in 0=0 when tried for most equations, we
substitute the equation (6.185) into itself, i.e. the replace the G(t1 , x1 ; t, x) inside the integral
on the right-hand-side of (6.185) by the solution for G(t1 , x1 ; t, x). Dropping the details we have
Z
G = G0 + G0 VI G
Z
Z
G = G0 + G0 VI G0 + G0 VI G
Z
Z Z
G = G0 + G0 VI G0 +
G0 VI G0 VI G
25
Now we repeat what we have just done, by replacing the the G on the right-hand-side of this
expression with the initial solution
Z
Z Z
Z
G0 VI G0 VI G0 + G0 VI G
G = G0 + G0 VI G0 +
Z
Z Z
Z Z Z
= G0 + G0 VI G0 +
G0 VI G0 VI G0 +
G0 VI G0 VI G0 VI G
at which point we start to see what’s going on. In principle we should do this an infinite amount
of times, but as we assume VI is small, the later terms will be negligible, meaning that we have
an expression for the full propogater G in terms of the much simpler free-propogater G0 .
6.3
diagramatic expansion
As is common with such perturbative expansions, it is useful to use diagrams as mneumonics
for the integrals, and objects appearing in the integrals. The diagramatic expansion for the
above solution to the integral propogator equation is
f ull propogator, G =
+
+
+ ...
(6.186)
where the line represents the free propogator G0 and the vertices (dots) represent the interaction
term. We then think of a given diagram in physical terms as the particle moving for a bit, then
interacting with the external potential, then moving a bit, and so on. Quantum mechanics is
a single-particle theory, so the diagramatic technique is not really needed. But in the standard
model we shall see that it allows for a much simpler visualization of what is going on.
7
Lecture seven - action functionals and Feynman rules
Aim: To introduce action functionals for the basic theories, and see how to read off the Feynman
rules.
Learning outcomes: At the end of this lecture you should
• know how to derive equations of motion from an action
• know how to write down the Feynman rules, once an action has been given.
7.1
action and equations of motion
A neat way of characterizing the dynamics of a system is through the action principle, which
states that a system evolves in such a way as to extremize the action of a system. Such
statements for static systems should be obvious by now, in that a static system has extremized
its potential energy - a stationary ball on a slope must be either at the top of a hill, at the
bottom of a valley, or at a saddle point. The action is the next step up, to include dynamical
systems. This is not course on action principles, so we shall just proceed by example.
26
7.1.1
Klein-Gordon action
The action for the real Klein-Gordon scalar field is given by
Z
1
1
S[φ] =
d4 x [− ∂µ φ∂ µ φ − m2 φ2 ]
2
2
(7.187)
and it is this that we need to extremize. Now, when we find the extrema of a function f (x) we
solve the following equation
lim f (x + δx) − f (x) = 0
δx→0
(7.188)
which is just the first-principles way of saying df /dx = 0. Here what we do is vary the field φ,
rather than x, and so we write
lim S[φ + δφ] − S[φ] = 0
δφ→0
(7.189)
In the case of the real KG field we find
Z
Z
1
1 2
1
1
4
µ
2
lim
d x [− ∂µ (φ + δφ)∂ (φ + δφ) − m (φ + δφ) ] − d4 x [− ∂µ φ∂ µ φ − m2 φ2 ] = 0
δφ→0
2
2
2
2
We now expand this out, dropping the δφ2 as they are subdominant in the limit. We find
Z
d4 x [−∂µ φ∂ µ δφ − m2 φδφ] = 0
(7.190)
Now we integrate by parts, dropping the boundary term, i.e. use
∂µ φ∂ µ δφ = ∂ µ (δφ∂µ φ) − δφ∂µ ∂ µ φ
(7.191)
R
and ignore the d4 x ∂ µ (δφ∂µ φ) term. We shall always assume this vanishes; it amounts to
saying there is no current at infinity. Then
Z
d4 x [∂µ ∂ µ φ − m2 φ]δφ = 0
(7.192)
Now, we want this integral to vanish for all possible deformations δφ, and the only way to
achieve this is for the integrand itself to vanish,
∂µ ∂ µ φ − m2 φ = 0.
(7.193)
This is just the Klein-Gordon equation (1.38). The point is that the action contains the same
dynamical information as the field equations, but it turns out that the action is easier to handle,
especially when there are lots of field. For example, the standard model contains 37 fields, so if
we wanted to write it down in terms of equations of motion, that would be 37 rather complicated
equations. However, in terms of the action, there is just one.
27
7.1.2
Dirac action
The Dirac action is given by
Z
S = −
d4 x[ψ̄γ µ ∂µ ψ + mψ̄ψ]
(7.194)
where ψ̄ = −iψ † γ0 . This new object is because ψ is a fermion, and so it needs this peculiar
version of Hermitian conjugation. Now, despite all the complications one would expect from
the Dirac equation, the variation of the action is remarkably simple. We treat ψ and ψ̄ as
independent objects, in which case the action must be extremized under variation of ψ and ψ̄
separately. If we vary ψ̄ then we find
Z
δSψ̄ = S[ψ, ψ̄ + δ ψ̄] − S[ψ, ψ̄] = − d4 x δ ψ̄[γ µ ∂µ ψ + mψ] = 0
(7.195)
and this must hold for all variations δ ψ̄, in which case we must have
γ µ ∂µ ψ + mψ = 0,
(7.196)
which is just the Dirac equation (2.45). One can check that varying ψ gives the same results,
after a bit of work.
7.2
the structure of an action
The action, S, is usually written in terms of a Lagrangian density, L, where they are related by
Z
d4 x L
(7.197)
S =
and L takes the form
L = T −V
(7.198)
where T is the ”kinetic” term, and V is the potential, which contins interaction term. There
are a few useful things about the action that help to restrict the possible theories we may write
down, they are
• L is real
• L is a Lorentz scalar - there are no free spacetime indices, they all come in pairs
• L is a scalar with respect to internal symmetries
From the examples given it is now clear that pieces in the action that are quadratic in fields
will produce terms in the field equations that are linear. So, the interaction terms that will
be the perturbation away from the linear (free) theory must be cubic (or higher) in the action,
allowing us to write
L = L0 − VI
28
(7.199)
where VI is the interaction potential, and is cubic (or higher) in the fields. In terms of Feynman
diagrams, it is the L0 that leads to the free propogator, and the VI that produces the vertices.
The vertices are constructed rather simply by looking at each term in VI : if VI is cubic in the
field, then the vertex has three free-propogators meeting at a point;
(7.200)
if VI is quartic in the field, then the vertex has four free-propogators meeting at a point;
(7.201)
if VI is quadratic in one field, but also quadratic in another, then the vertex has two freepropogators of each species.
(7.202)
In fact, each term in VI will come with a parameter in front of it, for example λ in VI = λφ4 ,
in which case we would associate λ with the vertex, just as we would associate the propogator
with a basic line.
7.2.1
example, real KG action I
1
1
1
L = − ∂µ φ∂ µ φ − m2 φ2 − λφ4
2
2
4
(7.203)
has
VI =
1 4
λφ
4
(7.204)
so the equation of motion for this theory is
∂µ ∂ µ φ − m2 φ = λφ3
and the Feynman rules are
(7.205)
1
p2 + m2
(7.206)
λ
(7.207)
Note that we shall not be worrying about factors of 2, 14 ,... for the vertices, we only want the
parametric scale.
29
7.2.2
example, real KG action II
1
1
1
L = − ∂µ φ∂ µ φ − m2 φ2 − ζφ3
2
2
3
(7.208)
has
1 3
ζφ
3
VI =
(7.209)
so the equation of motion for this theory is
∂µ ∂ µ φ − m2 φ = ζφ2
and the Feynman rules are
7.2.3
p2
1
+ m2
ζ
(7.210)
(7.211)
(7.212)
example, complex KG action
1
L = −∂µ φ∂ µ φ? − m2 φφ? − λ(φφ? )2
2
has
VI =
and the Feynman rules are
1
λ(φφ? )2
2
(7.213)
(7.214)
1
p2 + m2
(7.215)
λ
(7.216)
here we see something new, arrows on the lines. This is because there is a conserved quantity
associated to the field, due to the global U(1) symmetry. As we have seen before, such a U(1)
symmetry is leads to an electric charge for the particle. When we have such a symmetry then,
as charge cannot just dissapear, there must be the same amount of charge entering a vertex as
leaving it, the arrows help us to keep track of where the charge is.
30
7.3
example, Dirac particle coupled to photons
We have seen the Lagrangian for a basic Dirac particle, to couple in electromagnetism we simply
replace ∂µ → Dµ (4.132),
so
L = −ψ̄γ µ Dµ ψ − mψ̄ψ = −ψ̄γ µ ∂µ ψ − mψ̄ψ + iq ψ̄γ µ ψAµ
(7.217)
VI = −iq ψ̄γ µ ψAµ
(7.218)
which is something different to those we have seen before, as there are now two fields involved.
This just means that the vertex has two fermion lines, and one photon
−iγ µ pµ + m
p2 + m2
(7.219)
q
(7.220)
Again, we put an arrow on the fermion line as there is a U(1) symmetry, and so charge is
conserved.
7.4
complex KG equation coupled to photons
To couple the complex KG equation to electromagnetism we simply replace ∂µ → Dµ , giving
the Lagrangian
L = −(Dµ φ)(Dµ φ)? − m2 φφ?
= −∂µ φ∂ µ φ? − m2 φφ? + iqAµ (φ? ∂ µ φ − φ∂ µ φ? ) + q 2 Aµ Aµ φφ?
(7.221)
and then the Feynman rules are
7.5
p2
1
+ m2
(7.222)
q
(7.223)
q2
(7.224)
Feynman rules for the electroweak sector of the standard model
Now that we have a grasp of where the Feynman rules come from we shall write down the rules
of one of the sectors of the standard model, the electrowek sector, i.e. the part that tells us
31
about the electromagnetic force and the weak nuclear force. The matter particles involved in
this are the leptons
νe
νµ
ντ
,
,
,
(7.225)
e
µ
τ
and the gauge bosons (force carriers)
Z 0 , W ± , γ.
(7.226)
It is also believed to contain the Higgs scalar, but this has not been observed, yet. The basic
set of diagrams are as follows
e
γ
e
(7.227)
γ
e
(7.228)
Z0
gw
(7.229)
W−
gw
(7.230)
Z0
gw
(7.231)
e
W−
W+
νe
νe
νe
e
W+
W−
32
νe
Z0
gw
(7.232)
Z0
gw
(7.233)
νe
e
e
Although we have only explicitly given the rules for the
νe
e
doublet, they also hold for the
muon and the tau lepton doublets.
7.6
Fermi’s golden rule
Now that we have our Feynman rules we need to do something with them. Each diagram tells
us about the amplitude for a process to happen, for example the amplitude for an electron to
anihilate an anti-electron producing a photon is given by diagram 7.227, and has amplitude e.
However, we are interested in physical quantities, such as rates of events, and so we actually
need the probability, rather than the amplitude, and this is given by the modulus-square of the
amplitude. There is one extra factor to take into account, the density of final states ρ, which
gives us the number of states that are available to be occupied. The final result is Fermi’s golden
rule
Γ(i → j) =
2π
|Tij |2 ρ
~
(7.234)
where Tij is the probability amplitude.
7.6.1
ee scattering
e
e
∼ e2
γ
(7.235)
e
e
This diagram shows the scattering of two electron by a photon. The diagram contains two
vertices, each of strength e, and so the amplitude of the diagram is given parametrically by e2 .
2
1
so the rate of this process is
Recall that the fine structure constant, α = 4πe0 ~c ' 137
Γ ∼ α2
33
(7.236)
As the diagram contiains two vertices we say it is a second-order process. Another diagram that
contributes to electron-electron scattering is
e
e
γ
∼ e4
γ
(7.237)
e
e
but this is fourth-order (four vertices) and subleading so can be ignored at the first approximation.
7.6.2
µ decay
The muon is the heavy version of the electron and, as it is heavy, is liable to decay. The leading
order diagram responsible for the decay is
νe
µ
W
e
∼ gw gw
1
2
MW
(7.238)
νµ
and here we also include the internal propogator of the W boson, which we have approximated
as M12 , which is fine for low-momentum scattering. In fact, we can use this to give us an
W
expression for the Fermi coupling constant.
Before the discovery of the W ± , and Z 0 it was known that a muon could decay into an
electron and two neutrinos, so Fermi proposed an interaction term of the form
VI (F ermi) ∼ GF (ēνe )(µ̄νµ )
which has the diagram
(7.239)
νe
µ
e
∼ GF
(7.240)
νµ
It was this GF that was measured first (with a value of 10−5 (GeV )−2 ), and we can see from our
2
electroweak diagram that GF ∼ Mgw2 .
W
8
Lecture eight - group theory
Aim: To introduce the concept of a group, and some important examples of groups.
Learning outcomes: At the end of this lecture you should
34
• know the definition of a group
• know how to check whether a set, in combination with a composition law, forms a group
• know what defines Abelian groups, Lie groups and orthogonal groups.
8.1
definition of a group
A group is simply a set of objects, supplied with a rule that allows you to take two of these
objects and make another one. This law of composition is denoted with a dot,
g1 · g2 = g3 ,
(8.241)
where gi are members of the set. For the set and composition law to be a group, G, it is required
to satisfy the following
• it must be closed: If g1 , g2 ∈ G, then g1 · g2 ∈ G
• it must have an identity, denoted e: e · g = g · e = g, ∀g ∈ G.
• each g must have an inverse, denoted g −1 : g −1 · g = g · g −1 = e.
• the composition must be associative: g1 · (g2 · g3 ) = (g1 · g2 ) · g3 .
Some examples should help to clarify this.
• Consider the set {1, i}, where the composition law is multiplication. This does not form
a group because i · i = −1, and -1 is not part of the set
• Consider the set {1, i, −1, −i}, where the composition law is multiplication. This does
form a group.
• Consider the set {1, i, −1, −i}, where the composition law is addition. This does not form
a group as, for example, 1 · 1 = 1 + 1 = 2 is not part of the set
• Consider the set of integers, Z, where the composition law is addition. This does form a
group
• Consider the set of integers, Z, where the composition law is multiplication. This does
not form a group as, for example, 2 does not have an inverse within the set, the inverse
of 2 is 1/2, which is not an integer.
The above examples should really be called representations of groups, which just means that
they are explicit examples of the more abstract group itself. For example, we could say that
the group we are interested in (called Z2 ) consists of elements {e, a}, such that a · a = e. In this
definition we make no comment about what a, e actually are, nor do we say what the composition
law is. An explicit representation of this group could be {e → 1, a → −1, · → multiplication}
but there are many others. We shall see more consequences of different representions later.
35
8.2
direct product of groups
In the standard model of particle physics there are a number of distinct groups, acting as
independent internal symmetries, so we need a way of combining groups together, the direct
product does just that. Suppose we have two groups G1 and G2 , we may form a group G3 =
G1 ×G2 with elements g3 = (g1 , g2 ) and composition law g3 ·g30 = (g1 ·g10 , g2 ·g20 ), and it is simple to
show that G3 is indeed a group. For example, consider the group D2 = Z2 ×Z2 , this is composed
of elements {(e, e), (a, e), (e, a), (a, a)} where, for example (e, a) · (a, a) = (e · a, a · a) = (a, e).
8.3
Abelian groups
Abelian groups are the simplest, they are the ones for which
g1 · g2 = g2 · g1 .
(8.242)
For example, the group formed by multiplication within the rational numbers is Abelian; addition is also Abelian. However, there are important examples that are not Abelian, with
the non-Abelian nature coming typically because matrix multiplication is not commutative,
AB 6= BA.
8.4
Lie groups (pronounced ”Lee groups”)
The previous examples were all discrete groups, meaning that there was a countable number of
elements in the set; the next step is to think about non-countable sets. For example the set of
rational numbers, excluding zero, forms a group under multiplication (zero is excluded because
it has no inverse under multiplication). More common examples are those associated to some
sort of rotation. Consider the unitary group U(1), or rather its fundamental representation,
defined by the elements
gα = eiα
(8.243)
with the composition law being multiplication (it is easily checked that this forms a group).
This just corresponds to rotations in the complex plane by angle α. and as α is a continuous
parameter this is a Lie group.
8.4.1
general linear group GL(n,R)
The fundamental representation of GL(n,R) is defined to consist of elements A where
• A is a real n × n matrix
• det(A) 6= 0
The dimension of this group (the number of indepenedent numbers within an element) is n2 , as
there are n rows and n columns.
36
8.5
special linear group SL(n,R)
The fundamental representation of SL(n,R) is defined to consist of elements A where
• A ∈GL(n,R)
• det(A) = 1
The dimension of this group is n2 −1, as there are n rows and n columns, and one real constraint
coming from det(A) = 1.
8.6
orthogonal group O(n,R)
The fundamental representation of O(n,R) is defined to consist of elements A where
• A ∈GL(n,R)
• AAT = I = AT A
The dimension of this group is 12 n(n − 1), which we now explain. Think of the matrix B, where
B = AT A. By constraining B = I we are requiring the n terms along its diagonal to be 1, i.e. n
constaraints. The same constraint also sets all n2 − n off-diagonal terms to zero, however, as B
is symmetric this only imposes 12 (n2 − n) independent constraints. The total number of degrees
of freedom is then n2 − [n + 21 (n2 − n)] = 12 n(n − 1).
Note that
AAT
⇒ det(AAT )
⇒ det(A) det(AT )
⇒ det(A) det(A)
⇒ det(A)
8.7
=
=
=
=
=
I
1
1
1
±1
(8.244)
(8.245)
(8.246)
(8.247)
(8.248)
special orthogonal group SO(n,R)
The fundamental representation of SO(n,R) is defined to consist of elements A where
• A ∈O(n,R)
• det(A) = 1
The dimension of this group is the same as that of O(N,R), 21 n(n−1), because imposing the unit
determinant only selects a sector of O(N,R) rather than reducing its dimension, as we already
have det(A) = ±1 for orthogonal matrices.
37
8.8
O(p,q,R)
The notion of orthogonal matrices is easily extended to O(p,q,R) by thinking of the defining
relation of O(n,R) as
AIAT = I
(8.249)
Then we define elements of O(p,q,R) by requiring
AηAT = η
(8.250)
where η − diag(−1, −1, −1, ..., 1, 1, 1...), with p minus ones, and q plus ones. This may seem like
an obscure thing to do, but in fact this is just the Lorentz group of special relativity if we take
p = 1, q = 3.
8.9
symplectic group Sp(2n,R)
Define the symplectic matrix
Ω :=
0 I
−I 0
(8.251)
where I is the n × n identity matrix, then the symplectic group is defined to be those elements
that satisfy
• A ∈GL(n,R)
• AT ΩA = Ω
9
Lecture nine - non-Abelian gauge theory
Aim: To introduce the concept of a Lie algebras, and non-Abelian gauge theory
Learning outcomes: At the end of this lecture you should
• know the definition of unitary groups
• know what group generators are, and how to calculate their basic properties
• know how to gauge a non-Abelian symmetry
9.1
unitary groups
Although the groups presented in the previous lecture all have their role to play in some area of
physics, we are particularly interested in a different set of groups, the unitary groups, composed
of complex matrices satisfying certain properties as follows
38
9.2
unitary group U(n)
We have already seen a unitary group, U(1), here we generalize that to matrices by saying
U ∈U(n) if
• U †U = I = U U †
where
U † = (U T )∗ .
(9.252)
The dimension of U(n) is n2 , which we see as follows. First note that there are n2 complex
entries in U , corresponding to 2n2 real degrees of freedom. Now the diagonal part of U † U = I
gives us n real constraints, while the off-diagonal part gives 2 21 (n2 − 1) real constraints (the 12
is because U † U is symmetric and the 2 because the entries are complex). This gives a total of
2n2 − [n + (n2 − n)] = n2 real degrees of freedom for each U .
9.3
special unitary groups
U ∈SU(n) if
• U ∈U(n)
• det(U ) = 1
The dimension of U(n) is n2 − 1, because the determinant imposes one real constraint, which
we see by considering
U †U
⇒ det(U † U )
⇒ det(U † ) det(U )
⇒ |det(U )|2
⇒ det(U )
=
=
=
=
=
I
1
1
1
eiα
(9.253)
(9.254)
(9.255)
(9.256)
(9.257)
so, if we impose det(U ) = 1, all we are doing is requiring the single real constraint α = 0.
9.4
unitary groups and the standard model
In the standard model one finds that the paricles are described by their transformation properties
under three different unitary groups, each with its own physical interpretation
• U(1) describes electromegnetism. It has dimension one, and gives us the single photon
• SU(2) describes the weak nuclear force. It has dimension three, and gives us the W ± and
Z 0 bosons.
• SU(3) describes the strong nuclear force. It has dimension eight, and gives us the eight
gluons.
39
9.5
generators and Lie algebras
We have already introduced the notion of generators when we were thinking about how to rotate
a fermion field; the concept of generators is also useful for internal symmetries. Simply stated,
the generators X a and the group elements g of a Lie group G are related by
g = e−iθ
aXa
(9.258)
.
Once we have fixed our choice of generators, the parameters θa are what give us our different
group elements. The reason that the generators are studied, rather than the group elements, is
that for small θa this just corresponds to looking at group elements near the identity,
g = e−iθ
aXa
' I − iθa X a + ...
(9.259)
and that, it turns out, makes our life simpler. The generators X a form what is called a Lie
algebra.
9.5.1
o(n), the Lie algebra of O(n)
The defining property of O(n) is enough to tell us that the generators of O(n) are skewsymmetric, we see this as follows
OT O
⇒ (I − iθa T a + ...)T (I − iθa T a + ...)
⇒ θa [T a + (T a )T ]
⇒ Ta
=
=
=
=
I
I
0
−(T a )T
(9.260)
(9.261)
(9.262)
(9.263)
Now, recall from the question sheet that O(n) is just the set of matrices associated with rotations
(actually also reflections), but we already have a set of generators for three-dimensional rotations
given by (3.83), (3.85), which we note are indeed skew-symmetric. Although we chose these
generators to do our calculations, we could have in fact chosen any set of linearly independent
skew-symmetric matrices - it would just hve made the calculations harder and less intuitive.
9.5.2
su(n), the Lie algebra of SU(n)
Again, we may use the defineing property of SU(n) to tell us something about its generators.
U †U
⇒ (I − iθa X a + ...)† (I − iθa X a + ...)
⇒ θa [X a − (X a )† ]
⇒ Xa
=
=
=
=
I
I
0
(X a )†
(9.264)
(9.265)
(9.266)
(9.267)
So, the generators are Hermitian. However, we also need to impose unit determinant
det(I − iθa X a ) = 1
⇒ 1 − iθa T r(X a ) = 1
⇒ T r(X a ) = 0
40
(9.268)
(9.269)
(9.270)
so, as well as being Hermitian, the generators of SU(n) are traceless. Now, again, we already
know of a set of matrices that are Hermitian and traceless - the Pauli matrices. And they do
indeed generate SU(2). However, any set of traceless and Hermitian matrices would do, it’s just
that Pauli matrices make life easier.
9.6
non-Abelian gauge theory
This is what the standard model is all about, and fortunately it’s not much different, conceptually, from electromagnetism - the maths is a little bit trickier though. To see how it works,
consider the theory described by the Lagrangian
L = −∂µ φ† ∂ µ φ − m2 φ† φ
where φ is not just a complex scalar, but is a complex vector of scalars


φ1
 φ2 


φ =  .. 
 . 
φn
(9.271)
(9.272)
Now, given the definition of U(n), it is clear that this Lagrangian is invariant under φ → U φ, for
U ∈SU(n), and where the matrix U does not depend on spacetime, i.e. it is a constant matrix.
We can see this because if φ0 = U φ then
φ0† φ0 = (U φ)† (U φ) = φ† U † U φ = φ† φ,
∂µ φ0† ∂ µ φ0 = ∂µ (U φ)† ∂ µ (U φ) = ∂µ φ† U † U ∂ µ φ = ∂µ φ† ∂ µ φ
(9.273)
(9.274)
Now we make the same grand statements as we did for U(1) electromagnetism, and say that
we want this SU(n) symmetry to be local, and not just global. To do that we again introduce
a covariant derivative
Dµ φ = ∂µ φ − iqAµ φ
(9.275)
(Dµ φ)0 = U (x)Dµ φ
(9.276)
and require
when we promote the constant U matrix to a spacetime dependent matrix U (x). The only
difference is that now the vector potential Aµ is actually a matrix, so we have to be a bit careful
about how we manipulate expressions involving it, for example Aµ Aν 6= Aν Aµ and U Aµ 6= Aµ U .
With the introduction of the gauge-covariant derivative Dµ we may easily construct a gauge
invariant Lagrangian
L = −(Dµ φ)† (Dµ φ) − m2 φ† φ
41
(9.277)
The next step is to determine how Aµ changes under the gauge transformation, which we do
with the following calculation
(Dµ φ)0
⇒ ∂µ φ0 − iqA0µ φ0
⇒ ∂µ (U (x)φ) − iqA0µ U (x)φ
⇒ ∂µ U (x)φ + U (x)∂µ φ − iqA0µ U (x)φ
=
=
=
=
U (x)Dµ φ
U (x)(∂µ φ − iqAµ φ)
U (x)∂µ φ − iqU (x)Aµ φ
U (x)∂µ φ − iqU (x)Aµ φ
i
⇒ A0µ U (x) = U (x)Aµ − ∂µ U (x)
q
i
⇒ A0µ = U (x)Aµ U † (x) − [∂µ U (x)]U † (x)
q
(9.278)
(9.279)
(9.280)
(9.281)
(9.282)
(9.283)
which joins φ0 = U φ to give us the full set of non-Abelian gauge trnsformations.
We should compare this to the U(1) case (4.128, 4.129) to see that it does at least reduce
to electromagnetism in the Abelian case.
10
Lecture ten - some representation theory
Aim: To reintroduce the idea of SU(2) representations
Learning outcomes: At the end of this lecture you should
• be aware that there are many different guises for the same symmetry group
• know how to use ladder operators to construct the states transforming in a given representation of SU(2)
10.1
SU(2)
Because SU(2) is such an important group we shall spend a lecture going into some more of its
properties. By now we have seen the algebra
[j 1 , j 2 ] = ij 3 ,
+cyc. perm
(10.284)
many times - this is the definition of the algebra of the generators of SU(2). At this point
however, we have made no statement about what the j a are, nor what there composition law is.
We have seen examples where the j a are matrices, and the composition is matrix multiplication,
but they need not be matrices. Another important representation of this algebra comes from
the differential operators
L̂z = −ix∂y + iy∂x ,
+cyc. perm
(10.285)
which also satisfy this algebra. This should not come as a surprise, because these are just the
orbital angular momentum operators. Even if the j a are matrices, there are an infinite number
of different matrix representations. the three-vector rotation generators (3.83,3.85) satisfy
[T 1 , T 2 ] = iT 3 ,
+cyc. perm
42
(10.286)
and the Pauli matrices satisfy
[σ 1 /2, σ 2 /2] = iσ 3 /2,
+cyc. perm
(10.287)
And, as these two matrix representations consist of 2x2 and 3x3 matrices, they are not trivially
related, and form distinct representations. In fact, we can continue, and find a 4x4 representation
√


3/2
0
0
√0
 3/2

0
1
√0 
R1 = 
(10.288)
 0
1
3/2 
√0
3/2
0
0
0
√


0
i
3/2
0
0
√
 −i 3/2

0
i
2

√0 
R = 
(10.289)
0
−i
i 3/2 
√0
0
0
−i 3/2
0
R3 = diag(−3/2, −1/2, 1/2, 3/2)
(10.290)
For later convenience we note that
1
2
2
2
3
2
(σ /2) + (σ /2) + (σ /2)
(T 1 /2)2 + (T 2 /2)2 + (T 3 /2)2
(R1 /2)2 + (R2 /2)2 + (R3 /2)2
1 1
=
+1 I
2 2
= 1 (1 + 1) I
3 3
+1 I
=
2 2
(10.291)
(10.292)
(10.293)
This is all very interesting, but it would not be very convenient if we had to go through all of this
for each representation, we need a way of unifying the calcuation. The way to acheive this is to
find a set of commuting operators, and use their eigenvalues to classify the representations. We
have just seen that in the three cases we have looked at, (j 1 )2 + (j 2 )2 + (j 3 )2 is proportional to
the identity, and so commutes with all the j a . In fact, one can show using the defining algebra
that
[(j 1 )2 + (j 2 )2 + (j 3 )2 , j a ] = 0.
(10.294)
This gives us our first commuting operator. The next one is a matter of choice, and it’s
conventional to choose j 3 as the second commuting operator, giving us our commuting set of
{j 2 = (j 1 )2 + (j 2 )2 + (j 3 )2 , j 3 }
(10.295)
We are not saying that these commute with everything, only that everything in the set {j 2 , j 3 }
commutes. For example, we could not add j 1 to the set, as it does not commute with j 3 .
As is well known, for example from quantum mechanics, we are free to assign simultaneous
eigenvalues to operators that commute, and that is what we shall now do.
We define j 3 to have eigenvalue m, and j 2 to have eigenvalue β, and the eigenstate will be
denoted |β, mi.
j 3 |β, mi = m|β, mi
j 2 |β, mi = β|β, mi
43
(10.296)
(10.297)
Now, rather than working with j 1 and j 2 directly, we introduce the raising and lowering operators
j ± defined by
and note that the algebra gives us
j ± := j 1 ± ij 2
(10.298)
[j 3 , j ± ] = ±j ±
(10.299)
2
The reason for calling these raising/lowering operators is that
j 3 [j ± |β, mi] =
=
=
=
[j ± j 3 + (j 3 j ± − j ± j 3 )]|β, mi
[j ± j 3 ± j ± ]|β, mi
j ± [j 3 ± 1]|β, mi
(m ± 1)[j ± |β, mi]
(10.300)
(10.301)
(10.302)
(10.303)
Let’s explain what has just happened. We start with a state |β, mi that has a j 3 eigenvalue
of m, then we showed that the state j ± |β, mi has a j 3 eigenvalue of m ± 1, i.e. j + raises the
eigenvalue by one, and j − lowers the eigenvalue by one. What this means is that we now have
a way of constructing all the eigenstates, once one has been given
Now, suppose we are given a value of β, how should we go about constructing all the
eigenstates? First of all we notice that if we choose the normalization hβ, m|β, mi = 1 then
β = hβ, m|j 2 |β, mi
= hβ, m|(j 1 )2 + (j 2 )2 + (j 3 )2 |β, mi
> m2
(10.304)
(10.305)
(10.306)
which tells us that there is a maximum, and minimum, value for m. So, at some value of m,
usually denoted j, we will find
j + |β, ji = 0,
(10.307)
because there are no more states with a higher eigenvalue than j by assumption.
Now, a little algebra shows that j 2 = j − j + + j 3 (j 3 + 1) so we find
j 2 |β, ji = [j − j + + j 3 (j 3 + 1)]|β, ji
= j(j + 1)|β, ji
(10.308)
(10.309)
but j 2 |β, ji = β|β, ji so
β = j(j + 1).
(10.310)
Just as j is the largest possible eigenvalue for j 3 (by assumption), −j is the smallest eigenvalue.
We see this by noting that j 2 = j + j − + j 3 (j 3 − 1), so
j 2 |β, −ji = [j + j − + j 3 (j 3 − 1)]|β, ji
= [j + j − + j(j + 1)]|β, ji
2
(10.311)
(10.312)
this technique has already been seen in the quantum mechanics of a harmonic oscillator, eg lec 20 of quantum
World
44
but j 2 |β, −ji = j(j + 1)|β, ji so
j − |β, −ji = 0.
(10.313)
To sum up, a given representation is specified by its j 2 eigenvalue, and a particular state
within that representation is specified by its j 3 eigenvalue. As a matter of notation, it is more
standard to denote states by |j, mi.
Now, what are the possible values for m? We know that there are an integer number of
states, with j 3 eigenvalues ranging from −j, −j + 1, ...j − 1, j, i.e. a total 2j + 1 states. As
there is an integer number of states then 2j + 1 is an integer, then j must be either an integer,
or half-integer. Here are some examples
• j = 0 rep: The only possible value for m is zero. An example of this is the s-state of the
hydrogen atom, corresponding to l = 0 angular momentum.
• j = 12 rep: The possible states for this are m = ± 21 , an example being the two spin-states
of an electron. We also note that this rep’ has j 2 = j(j + 1) = 12 ( 12 + 1), which matches
the Pauli matrices (10.291)
• j = 1 rep: The possible states for this are m = −1, 0, 1, with an example being the l = 1
p-state of the hydrogen atom. Again, note that (10.292) has j 2 = j(j + 1) = 1(1 + 1), and
so is a j = 1 rep’.
• j=
3
2
rep: the possible states are m = − 32 , − 12 , 21 , 23 , an example being the Ra rep (10.293)
• j = 2 rep: this has m = −2, −1, 0, 1, 2, and an example is the l = 2 d-state of the hydrogen
atom.
• in fact, for orbital angular momentum we just make the identification j → l, and note
that l must be integer
10.2
normalizing the states
Although we have seen that j ± takes us from |j, mi to |j, m ± 1i, we have to be careful of
normalization. From our eigenvalue calculation, all we can actually say is that
j ± |j, mi ∝ |j, m ± 1i
(10.314)
It is convenient to choose a normalization such that
hj, m|j, mi = 1
(10.315)
for all m and j. So, we do this by
|j, m + 1i = N j + |j, mi
(j ± )† = f ∓ ⇒ hj, m + 1|j, m + 1i = |N |2 hj, m|j − j + |j, mi
⇒ 1 = |N |2 hj, m|j 2 − j 3 (j 3 + 1)|j, mi
(10.316)
(10.317)
(10.318)
⇒ 1 = |N |2 [j(j + 1) − m(m + 1)]hj, m|j, mi (10.319)
⇒ 1 = |N |2 [j(j + 1) − m(m + 1)]
(10.320)
45
we are at liberty to choose the phase of N , and we pick it to be zero, so we end up with
p
j + |j, mi = j(j + 1) − m(m + 1)|j, m + 1i
(10.321)
if we perform a similar calculation starting with |j, m − 1i = Ñ j − |j, mi we find
p
j − |j, mi = j(j + 1) − m(m − 1)|j, m − 1i
10.3
(10.322)
general SU(2) represenation
it may interest you, but most probably not, to have a general expression for SU(2) generators
of all dimensions


0 dl
0

 dl 0 dl−1




0
d
0
1
l−1

D1 =
(10.323)
,

.
.

2
.



0
d−l+1 
d−l+1
0


0 −dl
0
 dl 0 −dl−1




0
i
 0 dl−1

D2 =
(10.324)

,
.
.

2
.



0
−d−l+1 
d−l+1
0


l


l−1




l
−
2


D3 = 
(10.325)
,
..


.




−l + 1
−l
where
dk =
11
p
l(l + 1) − k(k − 1)
Lecture eleven - isospin
Aim: To introduce the concept of isospin
Learning outcomes: At the end of this lecture you should
• know how isospin relates to neutrons and protons
• be able to use isospin considerations in basic scattering calculations
46
(10.326)
11.1
neutrons, protons and isospin
At a very basic level, we note that the mass of a proton, p(938MeV), and neutron, n(940MeV),
are very close. We also note that the excitation energy levels of mirror nuclei (where p and n
13
7
7
11
are swapped) are very similar, e.g. 13
6 C and 7 N are mirror, as are 3 Li and 4 Be, and also 5 B
11
and 6 C. Heisenberg suggested that this points to the idea that neutrons and protons are, from
the perspective of nuclear dynamics, the same particle, or rather different states of the same
particle. This is not such a wild idea, we do not think of the spin up and spin down states of
an electron as separate particles. In fact, that analogy is rather apt, as the spin 1/2 property
of an electron is going to become an isospin 1/2 property of the neutron/proton.
So, we introduce the nucleon
a
|N i =
(11.327)
b
with |a|2 + |b|2 = 1 to get the normalization, and identify
1
0
|pi =
,
|ni =
0
1
(11.328)
So that
|N i = a|pi + b|ni
(11.329)
|N⊥ i = −b? |pi + a? |ni
(11.330)
along with the orthogonal state
which we can combine into
|N i
a
b
|pi
|pi
=
=U
|N⊥ i
−b? a?
|ni
|ni
(11.331)
where we see that U is in fact an SU(2) matrix. And that is it. The point about doing all this
is to show that, once again, SU(2) is at the heart of things. The proton and neutron form a
doublet, i.e. a j = 12 rep, of SU(2); and we call that SU(2) isospin, with the proton being the
m = + 21 state of isospin and the neutron being the m = − 12 isospin state.
Because we have given this SU(2) a special name, it is also convential to use the notation
j → I,
m → I 3.
(11.332)
1 1
|ni = | , − i.
2 2
(11.333)
along with our standard notation for states
1 1
|pi = | , i,
2 2
47
11.2
pions and isospin
So now we know what the I = 12 states of isospin are, what happens next? Well, from our
analysis of SU(2) in general, we know that there are an infinitude of SU(2) representations, how
do they make an appearnce here? For example, what do the I = 1 states correspond to? Well,
the first thing we know is that there should be three particles in this multiplet, corresponding
to I 3 = −1, 0, 1, and that they should all have roughly the same mass - as they are all related
by an SU(2) transformation. Fortunately such a set of particles exists, the pions, π + (140MeV),
π 0 (135MeV), π − (140MeV), and we identify these with the I 3 = 1, 0, −1 states of the I = 1
isospin representation.
The isospin system does more for us than just classify particles, it tells us about the isospin
composition of the elements within a multiplet. This concept will be of great value when we
come to quarks, so we shall take a close look at how, from the isospin perspective, the pion
triplet is constructed from the nucleon doublet.
Given the two states |pi and |ni there are four independent combinations of their product that we may construct: |pi|pi, |pi|ni, |ni|pi, |ni|ni, and these may be arranged into the
symmetric states
|pi|pi,
1
√ (|pi|ni + |ni|pi),
2
|ni|ni,
I=1
(11.334)
to form the I = 1 triplet of pions, and
1
√ (|pi|ni − |ni|pi),
2
I=0
(11.335)
to form the I = 0 singlet. From the point of view of group theory, what we have shown is that
2⊗2 = 1⊕3
(11.336)
i.e. the direct product of two doublets gives us an SU(2) triplet and an SU(2) singet.
It may not be obvious why we should look at the symmetric and anti-symmetric decompositions, but it is easily checked that they are the correct ones by using the generators on the
product space
3
Iprod
= I3 ⊗ I + I ⊗ I3
±
Iprod
= I± ⊗ I + I ⊗ I±
(11.337)
(11.338)
where the product of generators acts in the natural way, for example
(I 3 ⊗ I + I ⊗ I 3 )|pi|pi = (I 3 ⊗ I)|pi|pi + (I ⊗ I 3 )|pi|pi
= (I 3 |pi) ⊗ (I|pi) + (I|pi) ⊗ (I 3 |pi)
1
1
= ( |pi) ⊗ (|pi) + (|pi) ⊗ ( |pi)
2
2
1
=
(|pi|pi + |pi|pi)
2
= 1 |pi|pi
48
(11.339)
(11.340)
(11.341)
(11.342)
(11.343)
i.e. the state |pi|pi has eigenvalue I 3 = 1, which is what we claimed.
All of this may be stated as follows for the pion triplet
1 1 1 1
|π + i = |1, 1i = | , i| , i = |pi|pi
(11.344)
2 2 2 2
1 1 1 1
1
1 1 1 1
1
0
|π i = |1, 0i = √ | , i| , − i + | , − i| , i = √ (|pi|ni + |ni|pi)(11.345)
2 2 2 2
2 2 2 2 2
2
1 1 1 1
|π − i = |1, −1i = | , − i| , − i = |ni|ni
(11.346)
2 2 2 2
(11.347)
so, we have the nice results that the pions are constructed out of nucleons, but we get an extra
state, the singlet. This is associated with the deuteron, |di, coming from the anti-symmetric
product of nucleon states.
1
1 1 1 1
1 1 1 1
1
|di = |0, 0i = √ | , i| , − i − | , − i| , i = √ (|pi|ni − |ni|pi) (11.348)
2 2 2 2
2 2 2 2 2
2
(11.349)
Or, equivalently
|pi|pi = |π + i
1
|pi|ni = √ |π 0 i + |di
2
1
|ni|pi = √ |π 0 i − |di
2
−
|ni|ni = |π i
11.3
(11.350)
(11.351)
(11.352)
(11.353)
using isospin in scattering calculations
This is all very nice, but it is actually also quite useful, as it tells us something about the
scattering of particles. If isopspin really is a useful quantum number, then it should be conserved,
so we can use this to rule out, or constrain certain scattering rates. For example, let’s consider
the two scattering processes
pp → dπ + ,
pn → dπ 0
(11.354)
The amplitude of a scattering event is given by
amplitude(initial → f inal) = hf inal|Ĥint |initiali
(11.355)
so we need the bras and kets associated with pp, pn, dπ + , dπ 0 .
1 1 1 1
pp → |pi|pi = | , i| , i
2 2 2 2
1 1 1 1
pn → |pi|ni = | , i| , − i
2 2 2 2
+
+
dπ → |di|π i = |0, 0i|1, 1i
dπ 0 → |di|π 0 i = |0, 0i|1, 0i
49
= |1, 1i,
1
= √ (|1, 0i + |0, 0i) ,
2
= |1, 1i
= |1, 0i
(11.356)
(11.357)
(11.358)
(11.359)
Now, given that isospin is conserved, it commutes with the Hamiltonian, meaning that the
initial and final states must have the same isospin eigenvalues, or else the amplitude will vanish.
It is then immediately clear, for example, that we cannot have pp → dπ 0 , because the pp state
and the dπ 0 state have different isospin eigenvalues. We can, however, have pn → dπ 0 , because
the pn state contains a bit of |1, 0i. In fact, we can make a prediction for the ratio of cross
sections as follows
amp(pn → dπ 0 )
=
amp(pp → dπ + )
√1
2
(h1, 0| + h0, 0|) Ĥint |1, 0i
(11.360)
h1, 1|Ĥint |1, 1i
1
= √
2
h1, 0|Ĥint |1, 0i
h1, 1|Ĥint |1, 1i
+
h0, 0|Ĥint |1, 0i
!
h1, 1|Ĥint |1, 1i
1 h1, 0|Ĥint |1, 0i
= √
2 h1, 1|Ĥint |1, 1i
1
= √
2
(11.361)
(11.362)
(11.363)
the last step may not be obvious, but recall that (10.321)
1
|1, 1i = √ I + |1, 0i
2
2
− +
I I
= I − I 3 (I 3 + 1).
(11.364)
(11.365)
Now that we have the ratio of the amplitudes, we find the ratio of cross sections by squaring,
giving
σ(pn → dπ 0 )
1
=
,
+
σ(pp → dπ )
2
(11.366)
which matches observations.
12
Lecture twelve - the quark model
Aim: To introduce where quarks came from
Learning outcomes: At the end of this lecture you should
• know how to construct the baryon octet and decuplet
• know the particles in the baryon octet and decuplet, and their quark content
• be able to state the Gell-Mann Nishijima formula
12.1
more multiplets
The success of isospin continued with the identification of the I =
of particles
∆++ (1230M eV ),
∆+ (1231M eV ),
∆0 (1232M eV ),
50
3
2
quadruplet as the ∆-series
∆− (1232M eV ).
(12.367)
and it was noted that the known multiplets (nucleons, pions, deltas) safisfied
1
Q = I 3 + B,
2
(12.368)
where Q is the electric charge, and B the baryon number. e.g. π − has I 3 = −1, B = 0, Q = −1.
This was all very nice, until more particles came along. The troublesome particle was one called
te lambda, Λ, which appeared without any charged partners, so was an isospin singlet with
I = 0. That is fine, but it decayed via
Λ → π−p
(12.369)
and the state on the right hand side has non-vanishing isospin (lec 11 questions). Actually
it was not not a total disaster, because the decay time was ∼ 10−10 s, compared with the
expected timescale for strong interactions of 10−23 s. So, although it was produced by the
strong interaction, it did not decay by it. Nevertheless, as it could be produced by the strong
interaction in π − p collisions it would have to be produced in pairs, so that the collision could
conserve isospin. This meant that along with the Λ we also had another particle, the K 0 . In
fact there were a host of pair-produced particles
 − +
 Σ K
Σ0 K 0
π − p+ →
(12.370)

0
ΛK
Gell-Mann and Nishijima independently suggested that the new data could be explained by
introducing a new quantum number, preserved by the strong interaction, strangeness, and
assigned S(K + ) = 1, S(Σ− ) = −1.
It was discovered that the lowest mass spin- 12 baryons fitted neatly into the following octet
S=0
p(938M eV )
S = −1 Σ+ (1189M eV )
S = −2
↑
I3 = 1
n(940M eV )
Σ− (1197M eV )
Σ0 (1192M eV )
Λ(1116M eV )
Ξ0 (1315M eV )
↑
I 3 = 1/2
↑
I3 = 0
Ξ− (940M eV )
↑
I 3 = −1/2
↑
I 3 = −3/2
This new quantum number extended (12.368) to the Gell-Mann Nishijima formula
1
Q = I 3 + (B + S).
2
(12.371)
Along with the spin- 12 it was also discovered that the lowest-mass spin- 32 baryons fit nicely into
51
the following decuplet
S=0
S = −1
∆++
∆+
Σ∗+
S = −3
Σ?0
Ξ?0
S = −2
∆−
∆0
Σ?−
Ξ?−
Ω−
↑
↑
↑
↑
↑
↑
↑
I 3 = 3/2 I 3 = 1 I 3 = 1/2 I 3 = 0 I 3 = −1/2 I 3 = −1 I 3 = −3/2
In fact, the Ω− was actually a prediction, but they were right.
Gell-Mann has subsequently commented that the next step was obvious, but he believed it
to be more a mathematical trick, than a physical picture. The observation is that triangles are
the base shape of both the hexagonal pattern for spin- 21 baryons, and the triangular pattern of
the spin- 12 baryons; so he introduced three quarks3 . He gave them the labels u, d, c, standing
for up, down and strange. Just as |ni and |pi were identified as the different states of the ”same
particle”, with an SU(2) symmetry relating them, so |ui, |di, |si were considered the ”same”,
but with an SU(3) symmetry relating them, called flavour; Baryons were then formed of three
such quarks.
We are then led to ask what we can form out of three quarks, and there are 27 possibilities:
uuu,uud,udu,... However, rather than just list them in the basic form we should think bck to
how our SU(2) isospin multiplets worked, for example Qu 5 of lecture 11. The multiplets form
out of particular combinations of states that have symmetry under the interchange of pairs
• totally anti-symmetric, ψf lavour (1):
uds + dsu + sud − usd − sdu − dus
(12.372)
• anti-symmetric on the first pair, ψf lavour (8)12 :
(ud − du)q, (us − su)q, (ds − sd)q
(12.373)
where q is one of u, d, s.
• anti-symmetric on the second pair, ψf lavour (8)23 :
q(ud − du), q(us − su), q(ds − sd)
(12.374)
where q is one of u, d, s.
• totally symmetric, ψf lavour (10):
uuu, ddd, sss, uud + udu + duu, uus + usu + suu, ddu + dud + udd,
(12.375)
dds + dsd + sdd, ssu + sus + uss, ssd + sds + dss
3
Zweig had a similar idea, calling them aces instead.
52
On the face of it, this gives us too many states, 29 instead of 27; however, we have over-counted.
A linear combination of the states in the list that are anti-symmetric in the first pair is also
toally anti-symmetric, but that one has already been counted, the same goes for the list of states
that are anti-symmetric in the second pair. So, we find the final result
3 ⊗ 3 ⊗ 3 = 1 ⊕ 8 ⊕ 8 ⊕ 10
(12.376)
and that is more like what we need, explaining the origin of the decuplet and the octet of
baryons. However, we seem to have an embarrassment of riches, as there appears to be an
extra octet and singlet that are not observed. To explain these we need to remember that
quarks are spin 21 particles, and baryons are fermions so their overall wavefunction needs to be
anti-symmetric. The total wavefunction consists of
Ψtotal = ψspace ψspin ψcolour ψf lavour
(12.377)
and we know that physical states are colourless so the colour wavefunction is the totally antisymmetric colour singlet. For lowest enery states we have no angular momentum (l = 0), which
is symmetric, so we are left with requiring ψspin ψf lavour to be totally symmetric. We calculated
in question 5 of lecture 11 that the product of three spin- 12 states gave 2 ⊗ 2 ⊗ 2 = 4 ⊕ 2 ⊕ 2
totally anti − symmetric
1
anti − symmetric on f irst pair, ψspin ( )12
2
1
anti − symmetric on second pair, ψspin ( )23
2
3
totally symmetric, ψspin ( )
2
:
none
: (↑↓ − ↓↑) ↑, (↑↓ − ↓↑) ↓
: ↑ (↑↓ − ↓↑), ↓ (↑↓ − ↓↑)
: ↑↑↑, ↓↑↑ + ↑↓↑ + ↑↑↓, ↓↓↑ + ↓↑↓ + ↑↓↓, ↓↓↓
which means that we may form a totally symmetric ψspin ψf lavour out of
3
ψf lavour (10) × ψspin ( )
2
(12.378)
as each of the components are totally symmetric, or we can form the totally symmetric state
1
1
1
ψf lavour (8)12 ψspin ( )12 + ψf lavour (8)23 ψspin ( )23 + ψf lavour (8)31 ψspin ( )31
2
2
2
(12.379)
and that is all. So, the ”extra” singlet and octet are in fact not allowed because the quarks are
spin- 12 . Moreover, this explains why the baryon decuplet is spin- 32 (it is made out of the spin- 32
part of 2 ⊗ 2 ⊗ 2, i.e. the 4 in 2 ⊗ 2 ⊗ 2 = 4 ⊕ 2 ⊕ 2) and the baryon octet is spin- 12 (it is made
out of the spin- 21 part of 2 ⊗ 2 ⊗ 2, i.e. the 2s in 2 ⊗ 2 ⊗ 2 = 4 ⊕ 2 ⊕ 2).
Since the early days the quark model has been extended, as more particles were discovered
due to higher energy colliders coming online, we are now at the stage where the quark model is
given as in table 1 and the Gell-Mann Nishijima formula has extended to
1
Q = I 3 + (B + S + C + B 0 + T ).
2
53
(12.380)
13
Lecture thirteen - charge conjugation and parity
Aim: To introduce discrete symmetries
Learning outcomes: At the end of this lecture you should
• know how to find the parity of photons and mesons
• know the parity of fermions, anti-fermions and fermion-antifermion pairs
• be able to argue that electric dipole moments of particles would violate parity
13.1
Charge conjugation
We have seen that the standard model relies on continuous Lie symmetries to give us vector
bosons, such as photons and gluons, but then we also need to think about discrete symmetries.
For example, does the Universe care about a distinction between left and right - can we flip
the Universe in a mirror, and leave it looking the same? Can we swap all particles in the
Universe with their anti-particles without anyone noticing? The second of these is known as
charge conjugation, C, and it acts on particle states as
C|pi = |p̄i
(13.381)
where |p̄i is the anti-particle of |pi. We can find the eigenvalues of C by noting that C2 = 1
CC|pi = C|p̄i = |pi
(13.382)
so C has eigenvalues ±1. We also see that only particle that are there own anti-particles can
be C eigenstates, i.e. the eigenvalue equation
(13.383)
C|pi = λ|pi
implies that |p̄i = λ|pi.
quark
Q
B
I
I3
S C
d
-1/3 1/3 1/2 -1/2 0 0
u
2/3 1/3 1/2 1/2 0 0
s
-1/3 1/3 0
0
-1 0
c
2/3 1/3 0
0
0 1
b
-1/3 1/3 0
0
0 0
t
2/3 1/3 0
0
0 0
B’
0
0
0
0
-1
0
T
0
0
0
0
0
1
mass (MeV)
5
2
100
1,200
4,200
174,000
Table 1: Table showing quark charges
54
13.1.1
photons
Recal that the photon is just the quantum version of electromagnetism, and that one, actually
two, of Maxwell’s equations are
∂µ F µν = −µ0 j ν
(13.384)
Now, it is clear that under charge conjugation the sign of the current will change
C : j µ → −j µ
(13.385)
so to maintain Maxwell’s equations we better have
C : Aµ → −Aµ
(13.386)
Showing that the photon has negative intrinsic parity, and that electromagnetism preserves C.
13.1.2
fermion-antifermion pairs
• Consider the decay of a lowest energy state scalar meson, i.e. l = s = 0, into two photons
As electromagnetism preserves C, then the initial and final states must have
the same C value, and we know that the photon has C = −1, so a final state of two
photons has C = (−1)(−1) = 1, meaning that the l = s = 0 meson has C = 1.
• Now suppose we consider the spin-flip of a meson, i.e. an s = 1 meson decays to an s = 0
meson plus a photon. As the photon has C = −1, and the s = 0 meson has C = 1, then
the s = 1 meson must have C = (1)(−1) = −1 = (−1)s
• Now suppose we have a meson in a zero spin, but non-zero angular momentum state, l,
decaying to an l − 1 angular state via the emmision of a photon, the same argument as
above shows that it has C = (−1)l .
We combine all the above results into the single relation that
C|f f¯, l, si = (−1)l+s |f f¯, l, si
(13.387)
This has important consequences for particle decays, and explains why certain decays do not
occurr, for example the scalar meson π 0 in the l = 0 state has C = 1 so
π 0 → γγ
π 0 → γγγ
is allowed
is not allowed
55
(13.388)
(13.389)
13.1.3
C violation
Unfortunately, or fortunately depending on your point of view, the charge conjugation symmetry
is not a symmetry of the full standard model. C is only a symmetry of the strong sector and
the electromagnetic sector, with the weak nuclear force breaking charge conjugation symmetry.
For example, the following decay is observed
π + → µ+
(L) νµ(L)
(13.390)
with the muons always coming out left-handed. If we now perform charge conjugation on both
sides we get
π − → µ−
(L) ν̄µ(L)
(13.391)
which looks fine, until you realize that experimentally all the muons from π − decay actually
come out right-handed, so this weak sector decay (13.390) violates C.
13.2
parity
The question of whether the Universe looks the same in a mirror is answered by considering
parity symmetry, where we take the co-ordinates x → −x. When we do this we need to know
how physical quantities change, for example the momentum p → −p, and so is an example of a
polar vector - one that changes sign along with the co-ordinates. Angular momentum, however,
changes according to L → L, because L = r × p; this is an example of a pseudo vector. This
behaviour is not limited to vectors, scalars also split up into scalars and pseudo scalars. Electric
charge is an example of a scalar as ρ → ρ, whereas the volume of a parallelipiped defined by
vectors a, b and c, a · (b × c), is a pseudo scalar, changing sign under parity.
13.2.1
electromagnetism
Electromagnetism is invariant under parity, and this can be seen by assigning the photon negative intrinsic parity as follows. From Maxwell’s equation
∇ · E = ρ/0
(13.392)
we see that invariance of this equation under P : ρ → ρ, P : ∇ → −∇ requires
P : E → −E.
(13.393)
However,
E = −∇φ −
∂
A
∂t
(13.394)
showing that
P : φ → φ,
A → −A.
56
(13.395)
Now recall that the four-vector potential Aµ = (φ, A) and we see that
P : Aµ → −Aµ
(13.396)
so we associate a photon state with negative intrinsic parity
P |γi = −|γi
(13.397)
Now that we know how the vector-potential behaves we can see that the magnetic field is a
pseudo vector, as
B = ∇ × A.
13.2.2
(13.398)
fermions
Fermions are a little trickier to figure out, but the result of quantum field theory is that fermions
and anti-fermions satisfy
P |f i = |f i,
P |f¯i = −|f¯i
(13.399)
so fermion and anti-fermion have opposite parity. In particular, the s-state (vanishing angular
momentum) of the scalar meson π 0 has negative intrinsic parity, so is actually a pseudo scalar
meson. But now we seem to have a puzzle. If a meson has negative intrinsic parity, then why
is the following allowed
π 0 → γγ
(13.400)
as the final state, which consists of two photons, appears to have positive parity.
The point is that the parity must take into account the wave function, as well as the intrinsic
parity, so if the wavefunction of the two-photon state has negative parity, this leaves a negative
parity state. For example, write a wavefunction in terms of spherical harmonics
ψ(x) = φ(r)Ylm (θ, φ)
(13.401)
where parity corresponds to
x → −x
≡
θ → π − θ, φ → φ + π.
(13.402)
then
P : Ylm (θ, φ) → (−1)l Ylm (θ, φ).
(13.403)
For example, Y11 ∼ sin θeiφ → sin θ(−eiφ ). So, if the two photons are in an l = 1 state, then
the overall parity is negative. Another way to think about it is to ask how the pions couple to
the electromagnetic field. We know from our Feynman diagram work that if we want a single
pion to decay to two photons then the Lagrangian needs to have a term containing a single pion
field, and two photons, and that such a term must be a scalar. There are two choices
L1int ∼ π 0 E · B
L2int ∼ π 0 (E · E + BB)
57
(13.404)
(13.405)
In the first choice, as E is a vector and B is a pseudo vector, we must have the pion as a pseudo
scalar for L1int to be a scalar. The second choice leads to the pion being a scalar, which is not
what is observed. In fact, the E · B coupling actually tells us that the photons come out of the
experiment with orthogonal polarizations, which is how it was figured out experimentally.
All of this argument about including the spatial wavefunction into the parity assignment
goes through for the fermion-antifermion pair, so in fact the parity for a meson state is
P |f f¯i = (−1)l+1 |f f¯i
13.3
(13.406)
linearity of parity
We now do something that may seem strange, which is to calculate the parity of i, the square
root of -1. Although it will give us what we think of as an obvious result, it is worth doing as
we will see that things change when we look at time reversal. So, let’s start with the canonical
commutator
[x̂, p̂] = i~
(13.407)
Now we perform a parity transformation, remembering that operators transform as P x̂P −1 if
states transform as P |i.
P x̂P −1 = −x̂,
P p̂P −1 = −p̂
(13.408)
to find
P [x̂, p̂] P −1 = i~
[x̂, p̂] = P i~P −1
(13.409)
(13.410)
so for the commutation law to be preserved we must have
P i = iP
⇒ P iP −1 = i
(13.411)
which is just the statement that the parity transformation acts linearly. As we usually think
of i as a number this result is what we would normally write down without thinking, so that’s
good, but we need to take care later on when we consider time reversal.
13.4
electron dipole moment (edm)
We now think of ways that parity could be violated, or rather how we could observe its violation.
Suppose that a fundamental particle, which has a non-zero spin, had an electric dipole moment,
this would violate parity symmetry. There are now two possibilities, the electric dipole moment
could either align or anti-align with the spin; we shall go through the argument for the aligned
case, but the same holds for anti-aligned. So, we place such a particle in a state where the spin
(and electric dipole moment) both point up, now perform a parity transformation. The parity
transformed state would have the spin pointing up (spin is a pseudo vector so doesn’t change)
and electric dipole moment pointing down. However, such a particle is not in our theory, we said
58
the particle had aligned spin and electric diole moment. So, the presence of an e.d.m. implies
the violation of parity.
For example, the edm of a neutrons and electrons are bounded by
dneutron < 2.9 × 10−26 e cm
dneutron < 1.6 × 10−27 e cm
(13.412)
(13.413)
which, to put into perspective, compares to the size of a neutron of around 10−13 cm, i.e.
dneutron < 2.9×10−13 e lneutron . So, imagine the neutron is the size of the Earth, then 10−13 e learth
is 0.01e mm, corresponding to displacing an electron by 0.01mm in the centre of the Earth!
14
Lecture fourteen - parity violation and time reversal
Aim: To introduce discrete symmetries, and their breaking
Learning outcomes: At the end of this lecture you should
• know how parity was found not to be a symmetry
• know what time reversal symmetry is
14.1
parity violation
In experiments that observed the decays of cosmic rays it was noted that two particles, which
were otherwise identical, had two different decay channels, these were called the θ and τ
θ+ → π + π 0
+ 0 0
π π π
+
τ
→
π−π+π−
(14.414)
(14.415)
If these were indeed distinct particles then this would be fine, but if not, there’s a problem with
parity conservation. As pions have negative intrinsic parity then the parity of the θ+ final-state
is positive, implying that θ+ has positive parity, but the parity of the τ + final-state is negative,
implying the τ + parity is negative. For these to be the same particle (now caled K+ ), then the
decays would have to violate parity - nobody wanted that.
In 1956 Lee and Yang noted that there was an absence of data showing that parity ws
conserved in the weak interaction, in contradistinction to data for the strong nuclear force
and the electromagnetic force. In the same year Wu performed an experiment looking at the
following decay of cobalt to nickel
60
Co →
60
N i e− ν̄e
(14.416)
Wu and collaborators aligned the spins of the cobalt nuclei by utilising their magnetic moment,
which point in the same direction as their spin. The experiment revealed that most of the
electrons were emitted in a direction opposite to the spin vector of the nucleus, a result which
means that parity is not a symmetry in these decays. To understand that, think about the
correlation between the direction of nuclear spin, and the momentum of the electrons, Corr =
59
hs · pi. If parity is a symmetry then Corr should have the same value before and after a parity
transformation, however, P : Corr → −Corr as s is a pseudo vector, and momentum a polar
vector. Another way to see this is to imagine aligning the nuclear spins pointing up, then the
experiment says that the electrons are mostly emmitted downwards. If we perform a parity
transformation the spin still points up, but now the electrons have reversed their momentum,
and are also moving upwards. So, the actual experiment and its parity-reversed version are
different, so explicitly violating parity4 .
Another place that parity is seen to be broken by the weak sector is in the decay of pions
to muons.
π + → µ+
L νµ,L
(14.417)
where it is observed that the muon always comes out left-handed. If parity were a good symmetry
of this decay then we would expect π + → µ+
R νµ,R , as this is the parity reversed version of the
observed decay. Similarly, we observe
π − → µ−
R ν̄µ,R
(14.418)
but not its parity-reversed partner.
The underlying reason for this is that all neutrinos are left-handed, and all anti-neutrinos
are righthanded - that’s about as parity violating as you can get!
14.2
time reversal symmetry
Although it may appear that at first-sight that nature does not have a time-reversal symmetry,
shattered glass never recombines to become whole again, this is really just because of entropy,
rather then an asymmetry of the fundamental interactions. If we think about Newton’s laws
there is no problem in taking t → −t. Now we want to ask whether quantum mechanics has
anything to say about it.
Suppose that we have some map that takes un-primed states |ψi to primed states |ψ 0 i, for
all states. Then Wigner showed that if the inner product is preserved
|hφ|ψi|2 = |hφ0 |ψ 0 i|2
(14.419)
we must have either
|ψ 0 i = U |ψi,
or
|ψ 0 i = A|ψi
(14.420)
where U is a unitary and linear operator, and A is anti-linear and anti-unitary. The terms
unitary and so on are defined by
U [α|ψi + β|φi]
A[α|ψi + β|φi]
hU ψ|U φi
hAψ|Aφi
=
=
=
=
αU |ψi + βU |φi,
α? A|ψi + β ? A|φi,
hψ|φi,
hφ|ψi = hψ|φi? ,
4
linear
anti − linear
unitary
anti − unitary
(14.421)
(14.422)
(14.423)
(14.424)
note that a lot of texts get this the wrong way around, using arguments like the spin changes direction in a
mirror - be careful.
60
So, the unusual thing about an anti-linear operator is that when you move it past a complex
number, you have to complex-conjugate it, i.e. they don’t commute with numbers!
To see why this makes an appearance in quantum mechanics consider again the canonical
commutator
[x̂, p̂] = i~
(14.425)
and now we perform a time-reversal transformation with operator T̂ ,
T̂ x̂T̂ −1 = x̂,
T̂ p̂T̂ −1 = −p̂
(14.426)
so, whereas parity changes the sign of both x and p, time reversal changes only the momentum,
meaning that
T̂ [x̂, p̂] T̂ −1 = T̂ i~T̂ −1
[x̂, p̂] = −T̂ i~T̂ −1
(14.427)
(14.428)
So, if we are to maintain a time reversal symmetry in quantum mechanics, i.e. keep the commutation relations, then we must have
T̂ iT̂ −1 = −i,
T̂ i = −iT̂
(14.429)
so the time reversal operator is anti-linear.
This result also manifests itself in the behaviour of wavefunctions under time reversal, consider for example te plane wave
T : ψ = e−i(Et−px) → ei(Et−px) = ψ ?
(14.430)
because t → −t and p → −p. So, we see that effectively i → −i.
Under time reversal the spin changes sign, but the electric dipole moment does not so,
just as a non-zero edm signals a breakdown of parity symmetry, it also implies a violation if
time-reversal invariance. T may also be checked using a principle called detailed balance, for
example, the scatterings
np → Dγ,
and Dγ → np
should occur with the same rate if T is a symmetry.
15
Lecture fifteen - CP symmetry
Aim: To introduce a product discrete symmetries, and its breaking
Learning outcomes: At the end of this lecture you should
• recognise how the electroweak sector (almost) preserves CP
• know how C and P act on the neutral kaon system
• be able to calculate the CP violation in te neutral kaon system.
61
(14.431)
15.1
CP symmetry
We have seen that the weak sector violates both C and P individually, but there is still the
possibility that together they are preserved. For example, the observed decay π + → µ+
L νµ,L
becomes the following under C and P and the product CP
P
+
+
π + → µ+
L νµ,L −−−→ π → µR νµ,R




yC
yC
(15.432)
P
−
−
π − → µ−
L ν̄µ,L −−−→ π → µR ν̄µ,R
And in fact the CP (or PC) transformed decay is observed, so maybe the weak sector preserves
CP even though each discrete symmetry is broken.
In fact, it was argued by Landau that CP, rather than P, is the correct version of the discrete
symmetry associated to mirror reflections (see the questions associated with this lecture), and
if we take that view, then maybe the Universe is mirror-symmetric afterall.
15.2
neutral kaons
A good laboratory for studying CP symmetry is the neutral kaon system, it is interesting because
the the K 0 is able transform to its anti-particle, K̄ 0 . This is achieved through the following
processes,
d
s
d
s
u
W
W
,
u
W
u
u
(15.433)
W
s
s
d
d
where we start with a K 0 (a ds̄ bound state), and evolve to a K̄ 0 . What this effectively means
is that from the perspective of the weak interaction the K 0 and K̄ 0 are sort of the sme particle.
However, as far as the strong force is concerned, they are definately different, for example
K 0p → K +n
(15.434)
is allowed through the strong force as it conserves strangeness, however, the strong force does
not allow K̄ 0 p → K + n, as this would require a change in strangeness.
15.3
kaons and C, P and CP
as kaons are mesons (quark-antiquark pairs) then they have negative intrinsic parity, so we have
P |K 0 i
P |K̄ 0 i
C|K 0 i
C|K̄ 0 i
=
=
=
=
62
−|K 0 i
−|K̄ 0 i
−|K̄ 0 i
−|K 0 i
(15.435)
(15.436)
(15.437)
(15.438)
the sign change in the charge-conjugation operation is just convention. In particular, we note
that kaons are not CP eigenstates
CP |K 0 i = |K̄ 0 i
CP |K̄ 0 i = |K 0 i
(15.439)
(15.440)
We now consider the view that the weak interaction preserves CP and see if that is consistent
with experiment. So, the picture is that kaons are produced by the strong force (so are eigenstates of strangeness), and then they decay via the weak force, which by assumption means
they will be CP eigenstates. Although the kaons themselves are not CP eigenstates, it is easy
to construct them, we call the CP eigenstates |K1 i and |K2 i,
1
|K1 i = √ (|K 0 i + |K̄ 0 i)
2
1
|K2 i = √ (|K 0 i − |K̄ 0 i)
2
(15.441)
(15.442)
so
CP |K1 i = |K1 i
CP |K2 i = −|K2 i
(15.443)
(15.444)
These are the quantum states that are relevant for the weak decay of kaons, assuming the weak
force preserves CP. Kaons decay mostly into pions, and CP conservation implies
K1 → ππ
K2 → πππ
(15.445)
(15.446)
as pions have negative intrinsic parity. This is useful because the decay into three particles is
slower (0.5 × 10−7 s) than into two pions (0.9 × 10−10 s). So, Cronin, Fitch and collaborators
decided to set up a beam of kaons and watch what they decayed to. Given that the K1 decays
so much faster than K2 , if we have a long enogh beam, then all of the k1 will have decayed by
the end of the beam-pipe. So, we should expect only πππ decays at the end of the beam - if CP
is conserved in the weak sector. However, experiment showed that about 1 in 500 of the kaon
decays at the end of the pipe was into two pions, direct evidence of CP violation in the weak
sector.
15.4
Klong and Kshort decays
So, now that we know CP is violated, it does not make sense to analyse the system in terms of
weak eigenstates, but rather we should use mass (or lifetime) eigenstates. Given the smallness
of CP violation however, we know that the mass eigenstates are close to the CP eigenstates, so
we write
1
|KS i = p
(|K1 i − |K̄2 i)
1 + ||2
1
(|K2 i + |K̄1 i)
|KL i = p
1 + ||2
63
(15.447)
(15.448)
Now we look for decays of KL into final states that are CP-conjugates of each other, for example
the decays
KL → π − e+ νe ,
KL → π + e− ν̄e ,
rate Γ1
rate Γ2
(15.449)
(15.450)
are allowed and, as we shall now see, allow us to distinguish between matter, and antimatter.
• Γ1 : K 0 can decay to π − e+ νe , but not to π + e− ν̄e , via
W
s
d
0
+ −
• Γ2 : K̄ can decay to π e ν̄e , but not to π − e+ νe , via
W
s
e
νe
(15.451)
u
d
e
νe
(15.452)
u
d
d
What all this means is that the rate Γ1 is controlled by the amount of K 0 in KL , and the rate
Γ2 is controlled by the amount of K̄ 0 in KL , but we know that
1
1
|KL i = √ p
(1 + )|K 0 i − (1 − )|K̄ 0 i
2 1 + ||2
(15.453)
so if we define
∆ :=
Γ1 − Γ2
Γ1 + Γ2
(15.454)
to distinguish between the rates we find
∆ =
|1 + |2 − |1 − |2
∼ 2 Re()
|1 + |2 + |1 − |2
(15.455)
experiments give ∼ 10−3 , so
Γ1 > Γ2
64
(15.456)
15.5
matter vs anti-matter
We now have a way of distinguishing matter from anti-matter. This is really quite remarkable, it
would seem entirely natural to assume that our Universe could have been made from anti-matter
rather than matter, without making any difference, but this is not the case - an anti-matter
Universe is different to a matter Universe. The way we can define anti-matter uniquely is to
say that it is the stuff associated with the more common lighter charged product of KL decay,
i.e. e+ .
In fact, that there is an anti-symmetry between matter and anti-matter is a good thing for
us, otherwise we would expect our Universe to be half matter and half anti-matter, in fact we
observe a baryon to photon ratio of
bb − nb̄
∼ 6 × 10−10
nγ
(15.457)
compared to around 10−40 that we would expect simply from statistical fluctuations.
16
Lecture sixteen - quark mixing
Aim: To extend the idea of state mixing to quarks
Learning outcomes: At the end of this lecture you should
• know about the three generations of the standard model
• know how the Cabbibo mechanism allows for generation-changing processes
• be able to estimate the ratio of rates for certain basic processes.
16.1
three generations
The lepton sector of the standard model neatly splits into three generations,
ντ
νµ
νe
.
,
,
τ−
e−
µ−
(16.458)
and in terms of Feynman diagrams the basic interactions occur with strength gw
νµ
νe
u
gw
W
gw
W
gw
.
W (16.459)
e
µ
d0
In particular, there is no weak decay that allows an electron to turn into a W and a νµ , weak
interactions do not mix the lepton generations. The next observation is that the quark sector
also has three generations
u
c
t
,
,
.
(16.460)
d
s
b
65
which makes us think there could be a correspondence, i.e. does the e− → W − νe , with strength
gw , diagram imply a d → W − u diagram of strength gw ? Moreover, does the non-existence of
weak-sector generation process imply the impossibility of d → W − c? It turns out that this
intuition is close, but not quite right. For example, along with
π − → W − → µ− ν̄µ
(16.461)
which just involves u, d quarks, we also find
K − → W − → µ− ν̄µ
(16.462)
which shows that the W must couple to both u and s, as K − is a s̄u meson.
16.2
quark mixing
We have already seen the importance of mixing quantum states in particle physics, this was how
we explained the neutral kaon decays; here we just apply that notion to the quark sector. The
point is that the weak and the strong forces are distinct, the interaction eigenstates of one need
not correspond to the interaction eigenstates of the other. The correct quark weak-interaction
eigenstates are actually a linear combination of the strong-interaction eigenstates, i.e. the weak
nuclear force sees d0 , s0 , t0 rather than d, s and t. To keep things simpler, let’s just think about
the first two generations, in which case one has
0 d
d
cos θc sin θc
(16.463)
=
0
s
− sin θc cos θc
s
where θc is the Cabbibo angle, i.e. they are related by a unitary matrix. The correct Feynman
diagrams are then
c
u
gw
gw
W
W
(16.464)
d0
s0
We may now, for example, test this idea through measurements of rates such as
Γ(K − → W − → µ− ν̄µ )
Γ(π − → W − → µ− ν̄µ )
these rates involves the diagrams
µ
u
µ
u
W
(16.465)
W
(16.466)
νµ
νµ
d0 sin θc + s0 cos θc
d0 cos θc − s0 sin θc
−
The point being that the K is a particle in an eigen state of the strong interaction, K − =
us̄ = u(d¯0 sin θc + s̄0 cos θc ), and the pion is also a strong-force eigenstate, π − = dū = (d0 cos θc −
s0 sin θc )ū. Now, in the first diagram there is an us0 W vertex, but there is no such coupling, so
66
the first diagram contributes gw sin θc to the amplitude, as there is a ud0 W vertex in the standard
model and it has strength gw . In the second diagram the ūW s0 vertex gives no contribution,
meaning that the whole diagram contributes gw cos θc to the amplitude. So,
Γ(K − → W − → µ− ν̄µ )
(gw2 sin θc )2
=
Γ(π − → W − → µ− ν̄µ )
(gw cos θc )2
= tan2 θc
(16.467)
(16.468)
and experiments show that
θc ∼ 13◦
(16.469)
Another way of stating this is to say the weak coupling strengths for quarks are
gud = gcs = gw cos θc
gus = −gcd = gw sin θc
(16.470)
(16.471)
Now that we understand the coupling of quarks toνthe W boson, we use the weak interactions
e
e
gz
Z
gz
Z
(16.472)
gz
Z
(16.473)
νe
u
e
to predict the strong interactions
d0
gz
Z
u
d0
i.e. an interaction Lagrangian
of the form
Lint = ... + gz Z(uu + d0 d0 + cc + s0 s0 )
= ... + gz Z(uu + dd + cc + ss)
(16.474)
(16.475)
and therefore Feynman diagrams of the form
d
gz
Z
d
There are no neutral weak interactions
that change, for example, strangeness.
17
Lecture seventeen - neutrino oscillations
Aim: To extend the idea of state mixing to neutrinos
Learning outcomes: At the end of this lecture you should
• know about the solar neutrino problem
• know about atmospheric neutrinos
• be able to calculate neutrino oscillations due to their mass difference
67
(16.476)
17.1
Solar neutrinos
Continuing the theme of state mixing we now examine neutrinos. Ever since their introduction
by Pauli, in order to conserve energy, neutrinos hve been rather mysterious, and there are still
some rather basic properties of neutrinos that are not understood. Our story here, however,
goes back to 1968 with the Homestake experiment of Ray Davis. He and his team put in a
herculean effort to measure the number of neutrinos coming from the nuclear fusion events in
the Sun, in an effort to test the prediction of Bahcall and collaborators; a more recent version
of the predicted flux is given in Fig. 1. There are a number of fusion processes in the Sun that
Figure 1: The predicted flux of solar neutrinos, Bahcall et. al. ApJ, 621, L85 (2005).
lead to neutrinos, the largest one being the pp fusion
pp → de+ νe
ppe− → dνe
(17.477)
(17.478)
another key one, from an experimental point of view, is
8
B →8 Be∗ e+ νe
(17.479)
What we see from Fig. 1 is some rather extreme numbers, with around 60 billion neutrinos
going through your small fingernail each second! And yet we do not notice them. The reason
we do not feel some huge pressure due to this flux is that they don’t interact very strongly with
68
matter. To a very good approximation they al go straight through the Earth without noticing
it is there - so a rather delicate experiment is need to measure them.
Although most of the solar neutrinos do go through the planet, some of them hit stuff, and
Davis’ team relied on the reaction
νe
37
17 Cl
→ e−
37
18 Ar
(17.480)
to detect them. They used 615 tons of chlorine (essentially just cleaning fuid), waited for a few
weeks, then measured how much of the chlorine had turned into argon. They ended up with
33 argon atoms. This is an astonishing acheivement, and when they reported that this was a
factor of three smaller than predicted, nobody was particularly surprised - it would be quite
easy to miss a few atoms in 615 tons of cleaning fluid.
There were, however, some people who took the experiment seriously, one of whom was
Pontecorvo who, in 1968, suggested that the neutrinos were all there, but the electron neutrinos
had rotated to muon neutrinos. This is just the same basic physics as that behind the kaon
decays, and the Cabbibo model - the physical neutrino states are really a superposition of
electron, muon and tau neutrinos. And, if they have a different mass, then the state oscillates
between them.
Since the homestake experiment the tanks have got bigger, and the liquids have changed.
There are two key experiments that really made people believe that Davis was right, the Solar
Neutrino Observatory at Sudbury, using heavy water in its tank which is sensitive to all three
neutrinos via
νe d → ppe−
νe,µ,τ d → npνe,µ,τ
νe,µ,τ e− → νe,µ,τ e−
(17.481)
(17.482)
(17.483)
and the superK observatory in Japan, with 50,000 tons of water which is also sensitive to all
three neutrino flavours through elastic scattering
νe,µ,τ e− → νe,µ,τ e−
17.2
(17.484)
neutrino oscillations
For simplicity we shall shall only consider two neutrino flavours, electron and muon, and instead
of writing the neutrinos as flavour eigenstates, we consider the Hamiltonian eigenstates ν1 and
ν2 , as these are the objects whose evolution is simplest.
ν1
cos θ − sin θ
νe
=
(17.485)
ν2
sin θ cos θ
νµ
then we know that
ν1 (t) = ν1 (t = 0)e−iE1 t ,
ν2 (t) = ν2 (t = 0)e−iE2 t
(17.486)
We know that the solar neutrinos start life as electron neutrinos, as those are what the nuclear
fusion events produce, so
νe (t = 0) = 1,
νµ (t = 0) = 0
⇒ ν1 (t = 0) = cos θ,
ν2 (t = 0) = sin θ
69
(17.487)
(17.488)
so we find
νµ (t) = sin θ cos θ −e−iE1 t + e−iE2 t
(E2 − E1 )t
2
2
2
⇒ |νµ (t)| = sin (2θ) sin
2
(17.489)
(17.490)
so we see explicitly that the amount of muon neutrino oscillates with time. Now we may make
some approximations, namely that because neutrinos are very light compared to their energy
we have
q
q
E2 − E1 =
p2 + m22 − p2 + m21
(17.491)
1
1
' |p|(1 + m22 /|p|2 ) − |p|(1 + m21 /|p|2 )
2
2
2
2
m2 − m1
'
2|p|
(17.492)
(17.493)
As the neutrinos are highly relavistic we have |p| ∼ E, and t ∼ x, so
P (νµ ) ' sin2 (2θ) sin2
L =
π x 2L
,
2πE
∆m2
(17.494)
(17.495)
where we should be careful to note that ∆m2 = m22 − m21 , not (m2 − m1 )2 .
Thus we find that for oscillations to occur we need to properties
• there must be mixing between the neutrino flavours, as happens in the quark sector, i.e.
θ = 0.
• the neutrinos must have different masses, ∆m2 6= 0
For three neutrinos the basic physics is the same, but the maths gets more complicated, in
particular, the 2x2 matrix relating mass eigenstates to flavour eigenstates becomes a 3x3 unitary
matrix. This matrix is called the MNS (Maki, Nakagawa, Sakata) matrix, and is parametrized
by three angles5




cos θ13 0 sin θ13
cos θ12 sin θ12 0
1
0
0
0
1
0   − sin θ12 cos θ12 0 
UM N S =  0 cos θ23 sin θ23  
0 − sin θ23 cos θ23
− sin θ13 0 sin θ13
0
0
1
The results of all the experiments to date reveal that
θ12 = 34.5◦ ± 1.4◦ ,
θ23 = 43.1◦ ± 4◦ ,
θ13 < 10◦
∆m223 = (2.4 ± 0.5) × 10−3 eV 2
∆m212 = (8 ± 0.5) × 10−5 eV 2 ,
5
actually there’s also a phase, but I shall ignore it.
70
(17.496)
(17.497)
17.3
atmospheric neutrinos
As well as solar neutrinos there is another abundant source of freely-available neutrinos, sourced
by cosmic rays. These arise from cosmic rays striking a nucleus of some atom in the atmosphere,
producing a shower of (mostly) pions. These pions then decay to muons and muon neutrinos,
and these muons then decay to electrons as well as muon and electron neutrino as shown in Fig.
2. The first, immediately obvious, thing to note from Fig. 2 is that there are twice as many
cosmic ray
pions
stuff
µ−
νµ
νe νµ
e−
Figure 2: The typical shower of particles produced by a cosmic ray.
muon neutrinos as electron neutrinos, however, the observation of solar neutrinos by superK
shows that roughly equal number of νe and νµ are actually observed, with roughly the correct
amout of νe . Our knowledge of mixing then tells us that the muon neutrinos must be oscillating
mostly into tau neutrinos.
18
Lecture eighteen - Higgs mechanism and symmetry
breaking
Aim: To discovery a way of giving particles a mass
Learning outcomes: At the end of this lecture you should
• be able to show how symmetry-breaking gives a mass to gauge bosons, and to fermions
18.1
symmetry restoration/breaking
A nice intuitive way to think about symmetry breaking is to think back to ferromagnets, such
as a bar magnet, which is a lump of material composed of many atoms, each of which is a tiny
magnetic dipole. At high temperatures all these little dipoles point in random directions as the
interaction energy between dipoles is overwhelmed by termal energy. This means no direction
is picked out, and so we say that the system has a rotational symmetry - when averaged over a
macroscopic volume. One way to phrase this is to ask what is the average angle of the dipoles,
φ̄ = hφi, and then for this symmetric state we have hφi = 0.
Now suppose we reduce the temperature, such that the thermal energy is below the energy
scale of the dipole-dipole interaction. In this situation it is energetically favourable for the little
71
dipoles to point in the same direction and then, as a direction has been selected, we say the
rotational symmetry is broken. In terms of the mean angle of the dipoles we thus have hφi =
6 0
in the symmetry-broken phase.
This may not seem relevant to mass generation, but stick with me, I want to highlight the
fact that symmetry breaking, which will eventually give us the standard model masses, is in
fact a rather common effect.
18.2
superconductivity
Another place where symmetry breaking plays an important role is in superconductors. Here,
however, it is not the rotation angle of a magnetic moment that’s important, but a rather more
abstract quantity - the boson made by pairing two electrons together. In a metal one has lots
of electrons floating around and, as they are fermions, they occupy different quantum states.
If we think of these electrons as being in a lattice, then there can actually be an attractive
force between electrons, due to lattice vibrations6 . This allows pairs of electrons to become
associated, forming a boson, which can then Bose-condense. We can loosely think of the order
parameter in this case as φ̄ = heei, with a non-zero value in the superconducting phase, and a
zero value in the normal phase.
Here is where the hint comes that symmetry breaking, φ̄ 6= 0, has something to do with
mass. It was known for some time that superconductors repel magnetic fields, they cannot
penetrate a superconductor, this is why superconductors can hover above magnet, as in Fig. 3.
The physical reason for this is that photons aquire an effective mass inside the superconductor,
Figure 3: This shows a superconductor hovering above a magnet.
and so it costs them energy to be inside. The idea behind the Higgs mechanism is to give gauge
bosons, and other particles, a mass by exploiting a non-zero order parameter, just as photons
get an effective mass inside a superconductor when φ̄ 6= 0.
6
Superconductivity is not part of this course, so I will not go into detail about the pairing mechanism.
72
18.3
massive fields
Before we give a mass to particles we should figure out what we mean by mass. Going back to
the Klein-Gordon Lagrangian we see that the mass of the KG field comes from the quadratic,
non-derivative term of the Lagrangian,
1
1
LKG = − ∂µ φ∂ µ φ − m2 φ2
2
2
(18.498)
For the complex KG field the mass also came form the quadratic, non-derivative piece
LKG,cpx = −∂µ φ∂ µ φ? − m2 φφ?
(18.499)
And, finally, we have the Dirac field, whose mass similarly comes from the quadratic, nonderivative piece of the Dirac Lagrangian
Lψ = −ψ̄γ µ ∂µ ψ − mψ̄ψ
(18.500)
So, the theme is that mass is just the quadratic, non-derivative piece of the Lagrangian. At
which point we just write down
1
LA = ... − m2 Aµ Aµ
2
(18.501)
in order to give the photon a mass. Easy.
The problem with doing that is that this term breaks the gauge invariance that we worked
so hard to acheive, i.e. if we take Aµ → Aµ − 1q ∂µ Λ, then this mass term is not invariant.
However, we have sort of already solved this problem, in the form of covariant derivatives.
Recall from lecture 4 that to construct a gauge invariant action, all we need to do is replace
derivatives with their covariant counterparts, so the gauge invariant complex KG Lagrangian
becomes
LKG,cpx = −Dµ φDµ φ? − m2 φφ? − Vint
= ∂µ φ∂ µ φ̄ − iqAµ (φ̄∂µ φ − φ∂µ φ̄) − (q 2 φ̄φ)Aµ Aµ − m2 φφ̄ − Vint
(18.502)
(18.503)
and the important term is this one
LKG,cpx = ... − (q 2 φφ? )Aµ Aµ ...
(18.504)
So, if we can arrange for φ̄φ 6= 0, this leads to an effective mass for the photon of
1 2
m = q 2 φφ? .
2 γ
18.4
(18.505)
breaking the symmetry
Before we break the symmetry in the complex KG system, we should make sure we understand
what that symmetry is. Recall that the construction of electromagnetism involved a U(1)
symmetry,
φ → φ0 = e−iΛ φ
73
(18.506)
so, for example, if we had φ = 0 then we would get φ0 = φ, as both vanish. This is what we
mean by the symmetry not being broken. On the other hand, if we had φ = η, for some non-zero
constant η, then clearly φ0 6= φ, which is what we mean by the symmetry being broken. So we
see that, as with magnetism and superconductivity, the non-vanishing of the order parameter
is just a statement about the symmetry being broken.
The way we get the symmetry to break is by writing down a suitable interaction potential,
the typical choice of the full potential, i.e. quadratic piece plus interaction piece, is
λ
V (φ) =
(18.507)
(|φ|2 − η 2 )2
4
which has a shape somewhat reminiscent of a wine bottle bottom as in Fig. 4. As we can see,
Figure 4: A typical wine bottle, where the dregs of the wine tend to form at the bottom, around
the circular minimum.
this potential is positive, and is minimized at
|φ| = η.
(18.508)
So, the potential is minimized at a particular value for the modulus of φ, but the angle is not
given. This is just like the ferromagnet, the little dipoles in the broken phase could have pointed
anywhere, but they must make a choice, and when they do the symmetry is broken.
What we have found, then, is a way of creating φ? φ 6= 0, which is precisely what we needed
in (18.505) to give the photon an effective mass - and we have done it in a way the preserves
gauge invariance.
74
18.5
non-Abelian Higgs mechanism
Once the Abelian - U(1) - version is understood it is easy to generalize it to the non-Abelian
case, just use the covariant derivatives (9.275), where φ is now a column vector (9.272). Recall
that the Lagrangian has the form
L = −(Dµ φ)† Dµ φ + ...
= −φ† A†µ Aµ φ + ...
(18.509)
(18.510)
and again, for a non-zero value of φ this term just looks like a mass term for the gauge boson
(i.e. quadratic in Aµ ).
In the weak sector of the standard model one has to work a little bit harder because the
gauge group is a product of groups, SU(2)×U(1), but the principle is the same. In fact it
is the product structure that allows the W and Z to have a different mass, Mw ' 80GeV ,
Mz ' 91GeV , and the photon is not quite the U(1) that appears in SU(2)×U(1), but it lives
partly in the U(1) and partly in the SU(2). The SU(3) gluons remain massless.
18.6
Yukawa coupling and fermion masses
The Higgs mechanism is not just limited to giving mass to gauge bosons, it also provides a
natural way of generating fermion mass. Consider, for simplicity, a model that has a real KG
field and a Dirac field, and no gauge bosons. We could then couple the KG scalar to the fermion
by an interaction term of the form
Vint = gφψ̄ψ.
(18.511)
This type of interaction (linear in scalar and quadratic in fermion) is called a Yukawa interaction.
Again, we see immediately that if we had φ = η, then
mψ = gη.
(18.512)
so, once more, if φ 6== 0 we generate a mass term for some species - in this case a fermion.
This highlights the importance of the Higgs field. In the standard model the Higgs field is
responsible not just for the masses of the gauge bosons (the W and Z), but also for the fermions;
it has a lot of work to do for something we don’t actually know exists!
19
Lecture nineteen - the real standard model
Aim: To have a glimpse of what the standard model is actually like
Learning outcomes: At the end of this lecture you should
• be scared
75
19.1
electroweak sector, part I
Over the course of this module we have introduced a number of different ideas, all of them with
the purpose of understanding the standard model. As you may expect, although the standard
model is simple compared to what it could have been, it is still rather complicated. In this
lecture we shall bring together some of the concepts from earlier lectures and see how they fit
into the standard model, or at least the electroweak sector.
The first thing we note is that all the neutrinos that have been observed have been leftchiral, and anti-neutrinos are all right-chiral. We also know that there is a Feynman diagram
connecting e νe W − , but this can’t be the full story if the neutrinos are only left-chiral,
the
ν
e
actual Feynman diagram connects e(L) νe(L) W − . This means that the doublet
should
e
νe(L)
and Re = e(R) . The fact that the right-hand component
really be split up into Le =
e(L)
of the electron couples only to the photon and Z0 whereas the left-hand component couples to
photon, W and Z, means that Le and Re have different covariant derivatives. Moreover, it also
complicates the possible mass terms as we are not allowed to have L̄e Le or R̄e Re because they
both vanish - mass terms must contain one of each chirality. However, we are also not allowed
to have R̄e Le as Re contains two fermi fields, but Le contains only one so the coupling does not
make sense. Given that the electron does have a mass there must be a resolution to this.
19.2
chiral theories
The term chiral theory refers to theories where the left and right component of the same fermi
field have different gauge symmetries. To see how this works let’s consider a fermion coupled
to a complex KG scalar. The starting point is the ungauged Lagrangian
g
L = −ψ̄γ µ ∂µ ψ − ∂µ φ? ∂ µ φ − φψ̄ψ
(19.513)
2
which we write as
L = −ψ̄L γ µ ∂µ ψL − ψ̄R γ µ ∂µ ψR − ∂µ φ? ∂ µ φ − gφψ̄L ψR
(19.514)
Now that we have it in this form we start thinking about the symmetries. If it weren’t for the
Yukawa term there would be three undependent U(1) symmetries, one each for ψL , φR and φ.
However, for the Yukawa term to be invariant we find that one of these three is not independent,
leading to the full set of symmetry transformations
ψL → eiqL α ψL
ψR → eiqR α ψR
φ → eiqφ α φ
(19.515)
(19.516)
(19.517)
under which the Lagrangian remains invariant if qφ = qL − qR . So, we are free to give the left
and right components a different charge when we gauge this symmetry, so long as the scalar
has the correct charge. This means that if we promote α to α(x) then
L = −ψ̄L γ µ (∂µ − iqL Aµ )ψL − ψ̄R γ µ (∂µ − iqR Aµ )ψR
−(∂µ φ − iqφ Aµ φ)? (∂ µ φ − iqφ Aµ φ) − gφψ̄L ψR
76
(19.518)
is invariant.
This model then shows us how to make it work in the standard model, even though the
left and right fermions transform differently, we soak up that difference in the mass term (the
Yukawa piece) with a scalar field that transforms appropriately.
19.3
electroweak sector, part II
We know follow the same proceedure as above, but for the standard model, and start by writing
down the ungauged version
L = −L̄e γ µ ∂µ Le − R̄e γ µ ∂µ Re − ∂µ Φ† ∂ µ Φ − g L̄e ΦRe
(19.519)
(19.520)
where, at this point, we don’t really know what Φ needs to be. There are, again, two symmetries
in this lagrangian, but now they are an SU(2) and a U(1) symmetry. The SU(2) comes from
Le being a doublet, composed of both electron and neutrino fields; under this symmetry Re
remains inert. The correct charge assignments for the standard model are
SU (2) :
U (1) :
Φ → UΦ
Le → U L e
Re → Re
Φ → e−iα/2 Φ
Le → eiα/2 Le
Re → eiα Re
(19.521)
where U ∈ SU (2), so that when we gauge the symmetry we need the following covariant
derivatives
i
(19.522)
Dµ Φ = ∂µ Φ + g 0 Bµ Φ + igAµ Φ
2
i
Dµ Le = ∂µ Le − g 0 Bµ Le + igAµ Le
(19.523)
2
Dµ Re = ∂µ Re − ig 0 Bµ Re
(19.524)
φ1
- this is the Higgs boson
and we find that Φ is a two component complex scalar, Φ =
φ2
that LHC is hoping to find. Note also that Aµ is the non-Abelian gauge boson associated to
i
SU(2) - and so is a matrix given by Aµ = σ2 Aiµ - while Bµ is the Abelian gauge boson.
19.4
electroweak sector, part III
We are still not quite there as we haven’t come across the photon, the W± or the Z0 yet.; to
do that we need to break the SU(2)×U(1) symmetry. This is achieved, as usual, by giving
2
2
an expectation
value to the scalar field, conventionally taken as |Φ| = ν /2. So, if we take
0√
Φ=
the symmetry will be broken, and we find
ν/ 2
ν2 (Dµ Φ) Dµ Φ = ... +
(gA3µ − g 0 Bµ )(gA3µ − g 0 B µ ) + g 2 (A1µ A1µ + A2µ A2µ )
8
?
77
and by introducing
g 0 = q tan θw
e = g sin θw = g 0 cos θw
õ = cos θw Bµ + sin θw A3µ
(19.525)
(19.526)
(19.527)
Zµ = − sin θw Bµ + cos θw A3µ
1
Wµ± = √ (A1µ ± iA2µ )
2
(19.528)
(19.529)
we see that this becomes
1
1
(Dµ Φ)? Dµ Φ = ... − m2z Z µ Zµ − m2w Wµ+ W −µ − mγ õ õ
2
2
(19.530)
where
1 p 2
ν g + g 02
2
1
gν
=
2
= 0
mA =
(19.531)
mW
(19.532)
mγ
(19.533)
i.e. we have just found a theory that gives us a massless photon, Ã, two W± vector bosons of
the same mass, and a single heavier Z vector boson. Moreover, we may calculate the fermion
covariant derivatives to find
Dµ Re = ∂µ Re − ig 0 cos θw õ Re + ig 0 sin θw Zµ Re
(19.534)
1 +
−
+
−
2
i
σ Wµ + Wµ
iσ Wµ − Wµ
√
√
Dµ Le = ∂µ Le − g 0 cos θw õ − sin θw Zµ Le + ig
−
Le
2
2
2
2
2
σ3 +ig
(19.535)
sin θw õ + cos θw Zµ Le
2
i g
1
sin2 θw I + cos2 θw σ 3 Zµ Le
= ∂µ Le − ie (I − σ 3 )õ Le +
2
2 cos θw
ig
ig
+ √ (σ 1 − iσ 2 )Wµ+ Le + √ (σ 1 + iσ 2 )Wµ− Le
(19.536)
2 2
2 2
and so we find that, as Re = eR ,
R̄e γ µ Dµ Re = ēR γ µ ∂µ eR − ieēR γ µ eR õ + ig 0 sin θw ēR γ µ eR Z µ
(19.537)
(19.538)
showing that the right-handed fermion, Re couples to the photon with strength e, and to
the Z-boson
with
strength g 0 sin θw , but not to the W, as required. We also have that, as
νe(L)
Le =
, the left-chiral part gives
e(L)
g
ν̄L γ µ νL Zµ
(19.539)
2 cos θw
g(2 sin2 θw − 1)
g
g
+i
ēL γ µ eL Zµ + i √ ēL γ µ νL Wµ+ + i √ ν̄L γ µ eL Wµ−
2 cos θw
2
2
L̄e γ µ Dµ Le = ν̄L γ µ ∂µ νL + ēL γ µ ∂µ eL − ieēL γ µ eL õ + i
78
and this one shows that the left-chiral part of the electron couples to the photon with the same
strength as the right-chiral part, i.e. e. As well as this we see that the neutrino does not couple
to the photon, as should be expected - it has no electric charge. The Z-boson couples to both
the electron and the neutrino, and we have a coupling between electron-W-neutrino.
What you should take away from this is that although it’s a bit scary at first sight, on
reflection it just contains physics that we have seen before in the module: chirality; Abelian and
non-Abelian gauge symmetry; Higgs mechanism. Given that the above Lagrangian is supposed
to describe the Universe, I think we have got away with a remarkably simple theory, it could
have been a lot worse.
79