ysics ph4a02 lagrangian field theory & symmetry ysics

PH4A02 Lagrangian Field Theory& Symmetry

Or Physics is Symmetry

by Dennis Dunn

Version date: Wednesday, 3 October, 2007 at 08:49

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Copyright c©2003 Dennis Dunn.

Preface

PH4A02 LAGRANGIAN FIELD THEORY & SYMMETRYMODULE DESCRIPTION

Aims

I aim to show that:

• the fundamental equations of Physics can be derived from variation principles;• each (continuous) symmetry property of these equations give rise to a conservation law;• symmetry properties of space-time give an indication of the sort of fields we should be study-

ing;• symmetry predicts the equations of these fields;• the equations of electromagnetism can be predicted from local gauge symmetry.

Summary

Classical Field Theory

A reformulation of Classical Mechanics via a Principle of Least Action

A derivation of the form of the classical Lagrangian from symmetry principles.

Action and Lagrangian density for fields

Euler-Lagrange equations.

Examples of Lagrangian densities and the resulting field equations.

Symmetry transformations

Transformations of space-time variables and fields.

Condition for a transformation to be a symmetry.

Examples

Noether’s theorem and current densities

It is shown that each continuous symmetry gives rise to a conservation law. current density.

Symmetry Properties of Space-Time translations

Rotations

Lorentz Transformations

Relations between symmetry generators.

Representations of the Lorentz-Rotation Symmetry

Scalars

3

4

Left- & Right-Handed Spinors

Vectors

Lagrangians for Left- & Right-Handed Spinors

Global & Local Gauge Symmetry

Electromagnetism

Learning Outcomes

After the unit each student should be able to:

• Discuss the Principle of Least Action and use it to solve simple problems in classical mechan-ics;

• use symmetry principles to determine the form of the Lagrangian for (non-relativistic) parti-cles;

• describe the Action Principle for fields ( in terms of a Lagrangian density);• use the Euler-Lagrange equations to determine the dynamical properties of fields;• specify transformation properties of space-time variables and fields;• specify the condition for a transformation to be a symmetry;• use Noether’s theorem to relate symmetries to conserved currents and conservation laws;• discuss the relationship between local gauge symmetries and interactions• derive the interactions arising from specific local symmetries

Module Prerequisites

• Basic skills in calculus• Concepts of Rotational and Lorentz transformations• Basic ideas of classical mechanics• Basic ideas of fields such as in Waves and Optics; in Electromagnetism and in Quantum Me-

chanics Waves.

Teaching, Learning and Assessment Strategy

There will be TWO assessed problem sheets issued during the term. These account for 30% ofthe assessment of the unit. Such marked assignments provide a means of assessing each student’sprogress.

The overall understanding developed during the course is assessed through a single open-book Uni-versity examination of 2 hours duration. The final grade for this unit comprises the open-bookexamination (70%) and the assessed problems (30%).

Thursday, 29 June 2006 at 15:40

Contents

Preface 3

Contents 5

1 PLAN OF ACTION 7

2 CLASSICAL MECHANICS & THE PRINCIPLE OF LEAST ACTION 9

2.1 Classical Mechanics & Interactions 13

3 ACTION PRINCIPLE AND LAGRANGIAN FOR FIELDS 14

3.1 Field Theory 14

3.2 Action Principle for Fields 14

3.3 Euler-Lagrange Equations 15

3.4 Example: Vibrations of stretched string 16

3.5 Problems 1 16

4 TRANSFORMATION OF VARIABLES AND NOETHER’S THEOREM 19

4.1 Noether’s Theorem 19

4.2 Transformations 19

4.3 Example 20

4.4 Symmetry Transformations 21

4.5 Infinitesimal Symmetry Transformations 21

4.6 Example 1 23

4.7 Example 2 23

4.8 Current Densities 24

4.9 Symmetry and Conservation Laws 25

4.10 Time translation invariance & energy conservation 26

4.11 Example 3 26

4.12 Summary of General Method 27

4.13 Problems 3 27

5 SPACE-TIME SYMMETRIES & CONSERVATION LAWS 29

5.1 Space-Time Translation Symmetry 29

5.2 Space-Time Rotational Symmetry 30

6

6 SPACE-TIME TRANSFORMATIONS 33

6.1 Space-Time Translations 33

6.2 Eigenvalues and eigenfields 34

6.3 Rotations 35

6.4 Lorentz Transformations 40

6.5 Basic Fields 45

6.6 Space-time rotations 45

7 FIELDS AND LAGRANGIANS 47

7.1 Complex scalar field 48

7.2 Pauli Spinor Fields 48

7.3 Dirac spinor 51

7.4 Problems 53

8 LOCAL GAUGE INVARIANCE & INTERACTIONS 54

8.1 Complex Scalar Field & Local U(1) Gauge Invariance 55

8.2 Complex Spinor Field & Local U(1) Gauge Invariance 58

8.3 Local SU(N) Gauge Invariance 58

9 EXAMPLES OF SYMMETRIES 61

9.1 Type I spinor ΨR. 61

10 NON-UNIQUENESS OF L AND AR 64

10.1 Example 64

11 HYDROGEN ATOM 65

Index 66

Chapter 1

PLAN OF ACTION

The module will be organized as follows:

• A reformulation of the laws of classical mechanics as a Principle of Least Action.This formulation seeks to answer the question ”What is special about the path taken by aparticle?”.I will look at some examples of the use of the new formulation of Classical Mechanics (whichhas some additional bonuses) and I will derive an expression for the form of the Lagrangianfor a set of particles.

• The extension of the Least Action Principle to fields.”Most” of Physics is a field theory so this extension is important.The dynamical equations of the fields (that is, how they vary with space and time) is determinedby the Euler-Lagrange equations.At this point, the fields are arbitrary functions of space and time. The answer to the question”What are the relevant fields?” will be supplied by the theory later.

• Symmetry Transformations & Noether’s TheoremI will look at transformations of the co-ordinates and fields (for example rotations of co-ordinate frames or Lorentz transformations). Transformations which leave the Lagrangianunchanged are called symmetry transformations.Emmy Noether was the first to show that each such symmetry transformation corresponds to aconservation law.The relation between the symmetry and the conservation will explicitly expressed.Examples of space-time symmetries and the corresponding conservation laws will be studied.

• Space-Time PropertiesI will take a more detailed look at the symmetry properties of space-time. In particular :– group properties of symmetry transformations;– group generators;– eigenvalues & eigenvectors of group generators;– space- & time-translations and Fourier Analysis;– group generators of the rotation group O(3) or SU(2);

Eigenvalues and eigenvectors.– Lorentz-Rotation Group SU(2)×SU(2).

Complete solution for this group. The solution provides us with the fields that we need inPhysics.

• Fields and LangrangiansI take fields which arise from the Lorentz-Rotation group and determine (again using symmetryprinciples) the Lagrangian for each of the fields.

• Gauge SymmetriesGauge transformation symmetries have had an enormous impact on field theory and particlephysics. (If you continue on to study PH4B04 Particle Physics and the Standard Model youwill see how field theory evolves to become a theory of particles).The conservation law which follows from the simplest gauge transformation symmetry ischarge conservation.

7

8

Moving from global to local gauge symmetry introduces a new vector fields which interactwith the original fields; and predicts the Lagrangian and equation for these vector fields. Thesimplest of these gauge fields is the Electromagnetic Field.Maxwell’s equations are predicted exactly and in their entirety from a simple symmetryprinciple.More complicated local gauge symmetries give rise to the W- & Z-boson fields which areinvolved in Weak Interactions and to the 8 gluons fields which are involved in quark (strong)interactions.

Chapter 2

CLASSICAL MECHANICS & THE PRINCIPLEOF LEAST ACTION

Before I delve into field theory, I take a look at an area of physics with which you should be morefamiliar: classical mechanics. I shall look at an alternative formulation of the laws of classicalmechanics which I will then extend to include fields.

I first consider a single particle and I denote its co-ordinate vector by q. (It will become clear laterwhy I don’t use x!)

The normal expression of the physical principle describing this situation is via Newton’s laws. Inthis case the equation which specifies the motion of the particle is

md2qdt2

= F(q, t) (2.1)

The theory has to specify the particular form of the force F(q, t).

Specific solutions of this equation require either

• the co-ordinate q and its derivativedqdt

to be specified at some (usually initial) time t0; or• the co-ordinate q to be specified at two (usually initial and final) times t0 and t1

The solution is therefore a particular path in the q−t space starting at (q0, t0) and ending at (q1, t1).

There are an infinity of paths starting at (q0, t0) and ending at (q1, t1). What is special about thepath which the particle actually takes?

t

q

(q0, t0)

(q1, t1)

This can be answered in the following way:

Out of all the paths which start at (q0, t0) and end at (q1, t1), the physical path, is the one for whicha quantity called the Action is a minimum.

9

10

This Action is defined as an integral from t0 to t1 of the Lagrangian which is some function of q,dqdt

and possibly time t.

A =

∫ t1

t0

L (q, q, t) dt (2.2)

q is shorthand fordqdt

which is, of course, the velocity.

The Lagrangian in this case is

L =12mq2 − V (q, t) (2.3)

The function V (q, t) is called the potential and has to be specified as part of the theory.

This version of the theory of classical mechanics is known as the Principle of Least Action. Thereis a very good discussion of this in Feynman’s ”Lectures in Physics Volume 1, Chapter 19.

2.0.1 Using the Action Principle

The Action principle says what is special about the physical path but how do we actually make usethis information?

Suppose q(t) is the required physical path: this minimizes the Action. Now make a small changeq(t) → q(t) + δq(t) where δq(t) is arbitrary except that it vanishes at the times t0 and t1 and is avery small change to the path. The fact that q(t) minimizes the Action means that if I work out theAction using q(t) + δq(t) the changes are quadratic or higher in δq(t): That is, there are no linearterms.

If I insert q(t) + δq(t) into the expression for action, I get

A =∫ t1

t0

(12m(q + δq

)2

− V (q + δq, t))dt (2.4)

δq is again shorthand ford δqdt

.

Ignoring anything higher than first order in δq and its derivative gives:

A =∫ t1

t0

[(12mq2

− V (q, t))

+mq • δq− δq • dV (q, t)dq

]dt (2.5)

I can integrate the term mq • δq by parts to finally yield:

A =∫ t1

t0

[(12mq2 − V (q, t)

)− δq •

(mq +

dV (q, t)dq

)]dt (2.6)

You should check this step yourself carefully!

The requirement that the Action be a minimum means that the terms linear in δq must vanish. Sinceδq is (almost) arbitrary the requirement must be that the coefficient of δq in the integrand must bezero. That is,

md2qdt2

= −dV (q, t)dq

(2.7)

11

If I identify −dV (q, t)dq

as the force F (q, t) then this is just Newton’s equation but derived in a

very different way.

2.0.2 Lagrangian via symmetry

The form of the Lagrangian (2.3) was specifically chosen to reproduce Newton’s equations. How-ever, suppose we had never heard of Newton’s equations: How could we choose the Lagrangian?

Consider a free particle in space. Suppose I choose the Lagrangian to be a function of q anddqdt

. Iwant the Lagrangian to be a scalar quantity: it should be the same if I rotate the reference frame orif I shift the origin of the reference frame. (Note: I have not gone relativistic yet!).

The scalar quantities I can form from q and q are

q • q q • q and q • q

However the first and second of these change if I change the origin – and there is no combinationof the two which does not change. So the conclusion is that the Lagrangian for a free particle mustonly be a function of

q • q

The simplest such function is

L =12mq • q (2.8)

The coefficient has been chosen to be12m so that the Lagrangian is what is traditionally known as

the kinetic energy but this quantity has been derived entirely from symmetry.

I cannot determine the expression (2.3) which involves the potential V from symmetry because insuch a case different positions in space are not equivalent.

In general if the Lagrangian is a function of n co-ordinates q1 . . . qn, not necessarily Cartesian co-ordinates, and the corresponding velocities q1 . . . qn the equations of motion are

∂L

∂qj

=d

dt

∂L

∂qj

j = 1, . . . , n (2.9)

2.0.3 Classical mechanics example

In the above the Action Principle and the Lagrangian yield exactly the same equations as would adirect application of Newton’s Laws. However there are situations where the Lagrangian method ismuch more efficient than directly using Newton’s equations.

This is where the particle is constrained. Consider a particle subjected to a uniform gravitationalforce and constrained to move on the inside of a spherical bowl of radius a.

12

mg

R

In the Newtonian method we would have to introduce a force of constraint R which is the force thatthe surface exerts on the particle and is perpendicular to the surface. We would then have to solvefor this force and then substitute the solution back into the equations.

In the Lagrangian method there is no need to introduce this force of constraint.

I introduce spherical polar co-ordinates for the particle:

qx = a sin(θ)cos(φ) qy = a sin(θ)sin(φ) qz = a cos(θ) (2.10)

Because a is not a variable, but the fixed radius of the spherical bowl, this builds in the constraint.

The corresponding expressions for the components of the velocity are

qx = a(−sin(θ)sin(φ)φ+ cos(θ)cos(φ)θ

)qy = a

(sin(θ)cos(φ)φ+ cos(θ)sin(φ)θ

)qz = a sin(θ)θ

(2.11)

and hence the kinetic energy part of the Lagrangian is

12mq • q =

12ma2

(θ2 + sin2(θ)φ2

)The potential energy part is

mgqz = mga cos(θ)

and so the complete Lagrangian is

L =12ma2

(θ2 + sin2(θ)φ2

)−mga cos(θ)

The equations which follow from directly applying equation (2.9) are

2.1 Classical Mechanics & Interactions 13

θ : ma2θ = ma2sin(θ)cos(θ)φ2 +mga sin(θ)

φ :d

dtma2sin2(θ)φ = 0

The second equation, of course, implies that sin2(θ)φ is constant. We still have to solve a rathercomplicated equation to determine the solution but there has been no need to introduce, and theneliminate, the force which the surface exerts on the particle.

2.1 Classical Mechanics & Interactions

Suppose I have two particles with co-ordinates q1 and q2. A simple generalization of the La-grangian for a single free particle, equation (2.8), is

L =12

(m1q1 • q1 +m2q2 • q2)

This describes two non-interacting particles. How can I introduce interactions? Suppose that thisis done via some interaction potential U which is a function of q1 and q2. I suppose, as in thederivation of equation (2.8), that the quantity must be a scalar. The scalar quantities that I can formfrom q1 and q2 are:

q1 • q1 q1 • q2 q2 • q2

These all change if I shift the origin but there is a combination (and only one combination) whichremains invariant:

(q1 − q2) • (q1 − q2)

A suitable potential is therefore any function of this combination. So the resulting Lagrangian whichdescribes two interacting particles is

L =12

(m1q1 • q1 +m2q2 • q2)− U ((q1 − q2) • (q1 − q2)) (2.12)

This does not completely specify the form of the interaction between two particles but it does ruleout any interaction which is not simply a function of the square of the separation of the particles.

You need to be able to distinguish the case where particles interact (via a potential U) from the casewhere particles are subject to an external force. All the symmetry properties of space apply to theformer case but not to the latter.

The above Lagrangian describing two interacting particles can be extended to an arbitrary numberof particles

L =12

N∑r=1

mrqr • qr −N∑r=1

r−1∑s=1

U ((qr − qs) • (qr − qs)) (2.13)

The form of the rs summation is to ensure that each pair of particles is only counted once.

Chapter 3

ACTION PRINCIPLE AND LAGRANGIAN FORFIELDS

3.1 Field Theory

At a fundamental level Physics is a theory of fields. A field is some physical quantity which existsthroughout space and time (or at least throughout some region of space and time).

Ultimately the fields we need to consider will be those that describe fundamental particles and thosethat mediate the interactions between these particles. However the nature of these fields will emergefrom symmetry properties of space-time.

In particular I want to show:

• the fundamental fields can be determined from the symmetry properties of space-time;• the dynamic properties of these fields (how they change with space and time) can be determined

by the same symmetry properties; and• to every (continuous) symmetry there corresponds a conservation law.

3.2 Action Principle for Fields

The function q(t) which is a function of just the time variable t, in classical mechanics, will bereplaced by some quantities which are functions of space x and time t. (You should see now why Idid not use x as the name of the particle co-ordinate!)

I consider some set of fields fj(x), j = 1, .., n where x is shorthand for the space-time variablesx0 = xt = ct,x1 = x, x2 = y, x3 = z. In some illustrative examples the space may be reduced tobe just x and t or even just t.

The Action Principle enables us to derive the dynamic equations for a set of fields from a simplevariation principle.

The Action Principle is: The fields fj(x) are such that if they are specified on the boundary of anarbitrary closed space-time region R then in the interior of R they are determined by the conditionthat the action AR is an extremum.

In a multi-dimensional space an extremum can, and usually is, more complicated than just a maxi-mum or minimum.

The action AR over a space-time region R is defined by

AR =∫R

Ldx (3.1)

14

3.3 Euler-Lagrange Equations 15

where dx is the space-time volume element and L is called the Lagrangian. This Lagrangian is afunction of

xµ; fj(x);dfjdxµ

(x) (3.2)

3.3 Euler-Lagrange Equations

The Action Principle gives rise to the field equations

∂L

∂fj(x)=∑

µ

d

dxµ

∂L

∂

(dfj(x)

dxµ

) (3.3)

These are known as the Euler-Lagrange equations. These equations determine the dynamic proper-ties of the fields. That is how the fields change with changes in space and/or time.

Notes:

• in most cases the fields are defined over all of space and time and are simply specified to bezero on the ’boundary’, that is at infinity.

• the derivatives of the Lagrangian∂L

∂fj(x)and

∂L

∂

(dfj(x)dxµ

) are evaluated as if fj anddfjdxµ

are

completely independent quantities.

Proof: In order to prove this I form δAR which is the change in the action produced by smallchanges δfj(x) in the fields:

δAR =∫R

dx

[∂L

∂fj(x)

]δfj(x) +

∫R

dx

∂L

∂

(dfj(x)dxµ

) δ(dfj(x)dxµ

)(3.4)

The second integral can be converted by using the four-dimensional version of Gauss’s theorem:This may be expressed as

∫R

dx(∇ • G

)=∮S

dS • G. (3.5)

where the vectors are space-time vectors (or four-vectors). Using this result with G replaced by

δfj(x)∂L

∂dfjdxµ

and discarding the surface terms (because, on the surface, fj(x) is specified and

hence δfj(x) is zero) gives the required result.

In equation (3.3) I use the complete derivative d/dxµ : In detail, when applied to a function of x, fj

3.4 Example: Vibrations of stretched string 16

anddfjdxµ

, this has the form

d

dxµG(f, fj , x) =

∂ G

∂xµ+∑k

∂ G

∂ fkdµfk +

∑k,ν

∂ G

∂

(dfkdxν

) dµνfk (3.6)

where I have used dµνfj to denoted 2fj(x)d xµd xν

If I make use of this then equation (3.3) can be written in detail as:

∂L

∂fj(x)=∑µ

∂2L

∂xµ∂

(dfj(x)dxµ

) +∑k

∂2L

∂fk(x)∂(dfj(x)dxµ

) (dµfk) +∑k,ν

∂2L

∂

(dfk(x)dxν

)∂

(dfj(x)dxµ

)dµνfk

(3.7)

Don’t worry about the details of this proof: This won’t form part of the assessment. However youdo need to make sure that you can use equation (3.3).

3.4 Example: Vibrations of stretched string

Consider a field S(x, t) which is a function of a 1D space variable x and time t. The Lagrangianand action are

L =12

[ρ(x)

(dS(x, t)dt

)2

− T (x)(dS(x, t)dx

)2]

(3.8)

AR =∫∫

12

[ρ(x)

(dS(x, t)dt

)2

− T (x)(dS(x, t)dx

)2]dxdt (3.9)

This model represents the motion of a stretched string. S(x, t) is the displacement of a string atposition x and time t; ρ(x) is the mass density of the string and T (x) is the tension.

You might like to consider how a string could have a non-uniform tension.

According to (3.3) the dynamical equation is

ρ(x)d2S(x, t)dt2

=d

dx

[T (x)

dS(x, t)dx

](3.10)

3.5 Problems 1

(i) Consider a classical mechanical system with two particles moving in one dimension. TheLagrangian is

L =12m1

(dq1dt

)2

+12m2

(dq2dt

)2

+Gm1m2

|q1 − q2|(3.11)

Determine the two dynamical equations and comment on the system they describe.

(ii) A particle with co-ordinate q moves under the influence of a scalar potential φ (x, t) and avector potential A (x, t). The Lagrangian is

3.5 Problems 1 17

L =12mq2 − φ (q, t) + q •A (q, t) (3.12)

Determine the equation of motion of the particle. Your solution should only involve the fields

in the combinationsd

dq∧ A (q) and

∂A (q)∂ t

+dφ (q)dq

. Note that you will need to use the

relation

dA (q, t)d t

=∂A (q, t)

∂ t+ qx

∂A (q, t)∂ qx

+ qy∂A (q, t)∂ qy

+ qz∂A (q, t)∂ qz

(iii) A particle is moving under the influence of a uniform gravitational force but is constrained tobe on the surface of a cone. This cone has its axis of symmetry vertical, with the vertex at thebottom, and the cone angle is α.Use variables z along the vertical axis and φ the angle about the vertical axis. Since the particleis constrained to move on the cone the radial co-ordinate (distance from the vertical axis) isz tan (α).Show that the square of the velocity is

z2

cos2 (α)+ z2 tan2 (α) φ2

Write down the Lagrangian and hence determine the equations of motion for the particle.

(iv) Consider a complex field f (x, t) defined over the full four dimensional space-time. The La-grangian is

L =i

2

(f∗d f

d t− f

d f∗

d t

)

−12

[(d f∗

d x

)(d f

d x

)+(d f∗

d y

)(d f

d y

)+(∂f∗

∂z

)(∂f

∂z

)]−V (x, y, z) f∗f

(3.13)

Determine the dynamic equation for this complex field. There are two ways to deal withcomplex fields: One is to treat the real and imaginary parts as two independent fields; theother, and much more useful, approach is to treat field f (x, t) and its complex conjugate fieldf∗ (x, t) as the two independent fields. Comment on this dynamical system.

(v) Electromagnetism: Consider a space-time vector field A with components At, Ax, Ay andAz . I will write the space components collectively as A. These fields are functions of the fullspace-time variables xt = ct, x, y and z. The Lagrangian is

L =12

[(dAdxt

+ OAt

)2

− (O ∧A)2]

(3.14)

Determine the equations for these fields and show that these equations can be put into the form

O •E = 0

O ∧B =dEdxt

(3.15)

where E and B are

E = −(dAdxt

+ OAt

)B = O ∧A

(3.16)

3.5 Problems 1 18

These are the equations of electromagnetism. I have used different units to what is conven-tional. You should show how these electric and magnetic fields are related to the conventionaldefinitions.

Chapter 4

TRANSFORMATION OF VARIABLES ANDNOETHER’S THEOREM

The Lagrangian formalism, via the Euler-Lagrange equations, is useful in determining the dynam-ical equations of the fields but there are often other ways of getting at these equations. Howeverthe most powerful feature of the Lagrangian formalism is its capacity to bring out the relationshipbetween symmetries and conservation rules.

A symmetry is a transformation of the co-ordinates and fields which leaves the equations of motionthe same.

4.1 Noether’s Theorem

This states:

For every (continuous) symmetry there is a corresponding conservation law.

4.2 Transformations

Consider first a transformation that is not necessarily a symmetry: I make a transformation in whichthe co-ordinates xµ are replaced by x′µ and the fields replaced by f ′j(x

′):

xµ → x′µ

fj(x) → f ′j(x′)

d fj(x)d xµ

→d f ′j(x

′)d x′µ

(4.1)

The new action is defined by

A′R′ =

∫R′

L(x′, f ′(x′), dµf ′(x′))dx′ (4.2)

In this expressionL has the same functional form as previously andR′ is the mapping ofR producedby the change of co-ordinates.

In general I can express the new Action in terms of the original co-ordinates:

A′R′ =

∫R

L(x′(x), f ′(x′(x)), dµf ′(x′(x)))[dx′

dx

]dx (4.3)

19

4.3 Example 20

That is, the new co-ordinates are written as functions of the original co-ordinates.[dx′

dx

]is short-

hand for the Jacobian which transforms the volume element from one set of co-ordinates to theother.

In the case of just two variables x and t this Jacobian is the determinant

[dx′

dx

]=

∣∣∣∣∣∣∣dx′

dx

dx′

dtdt′

dx

dt′

dt

∣∣∣∣∣∣∣ (4.4)

I can use the above expression for A′R′ to introduce a new Lagrangian defined by

L′(x, fj ,dfjdxµ

) = L(x′, f ′j , d′µf

′j)[dx′

dx

](4.5)

In detail this new Lagrangian is

L′(x, fj(x),dfj

dxµ

(x)) = L(x′(x), f ′

j(x′(x)), d′

µf′j(x

′(x))) [dx′

dx

](4.6)

4.3 Example

Consider the stretched string example defined above.

New co-ordinates could be defined by

x′ = x+ δx; t′ = tS′(x′) = S(x) (4.7)

That is the only change in this case is a translation of the space variable x.

The new Action is then

A′R′ =

∫∫R′

12

[ρ(x′)

(dS′(x′, t′)

dt′

)2

− T (x′)(dS′(x′, t′)

dx′

)2]dx′dt′ (4.8)

and if I express this new action in terms of the original co-ordinates

A′R′ =

∫∫R′

12

[ρ(x+ δx)

(dS(x, t)d t

)2

− T (x+ δx)(dS(x, t)d x

)2]dxdt (4.9)

In this example this Jacobian is simply 1 and the new Lagrangian is

L′ = 12

[ρ(x+ δx)

(dS(x, t)d t

)2

− T (x+ δx)(dS(x, t)d x

)2]

(4.10)

4.4 Symmetry Transformations 21

4.4 Symmetry Transformations

A transformation is a symmetry if exactly the same field equations result from this new L′ as fromL. Hence a transformation is a symmetry if

L′(x, fj ,dfjdxµ

) = L(x′, f ′j , f′j,µ)

dx′

dx= L(x, fj ,

dfjdxµ

) (4.11)

That is the Lagrangian remains unchanged.

In fact there is a more complicated possibility:

L′(x, fj ,dfjdxµ

) = L(x′, f ′j , f′j,µ)

dx′

dx= kL(x, fj ,

dfjdxµ

) +∑µ

dFµdxµ

(4.12)

for some constant k and some functions (of x and fj)Fµ

However the simplest cases, and the only ones we shall need, are those for which k is 1 and Fµ iszero.

4.5 Infinitesimal Symmetry Transformations

It is simpler if I consider infinitesimal transformations. For example:

xµ → x′µ = xµ + ∆xµ(x)fj → f ′j = fj + ∆fj(x)

dfjdxµ

→df ′jdxµ

=dfjdxµ

+ ∆(dfjdxµ

) (4.13)

where ∆xµ(x) and ∆fj(x) are infinitesimal functions. Note that, in general, ∆xµ(x) varies with x.That is the space-time translation is different at each space-time point.

The space-time volume ratio that appears above is simply[dx′

dx

]= 1 +

∑µ

d(∆xµ(x))dxµ

(4.14)

The actual changes in the fields have two parts: One due to the change in the field itself, which Ihave denoted by ∆fj(x) , and one due to the change in the co-ordinate. I denote this total changein the field by δfj .

The relationship between δfj and ∆fj can be written as

∆fj (x) = δfj(x)−∑µ

∆xµ(x)dfj(x)dxµ

(4.15)

Similarly the changes in the derivatives can be written as

∆(dfj(x)dxµ

)=df ′j(x)dxµ

− dfj(x)dxµ

=d (∆fj(x))

dxµ

(4.16)

and

4.5 Infinitesimal Symmetry Transformations 22

δ

(dfj(x)dxµ

)=df ′j(x

′)dx′µ

− dfj(x)dxµ

=d (δfj(x))dxµ

−∑ν

dfj(x)dxν

d∆xνdxµ

=d∆fjd xµ

+∑ν

∆xνd2fj

d xµ d xν

(4.17)

Using these relationships the change in the new Lagrangian, to first-order in ∆x and ∆f is

δL = L′(x, fj(x),

dfjdxµ

(x))− L

(x, fj ,

dfjdxµ

)

= L

(x, fj ,

dfjdxµ

)∑µ

d∆xµ(x)dxµ

+∑µ

∆xµ(x)∂ L

∂ xµ

+∑j

∂ L

∂fj

(∆fj(x) +

∑µ

∆xµ(x)dfjdxµ

(x)

)+∑j,µ

∂ L

∂

(dfjdxµ

) (∆(dfjdxµ

)(x) +

∑ν

∆xµ(x)dµνfj(x))

(4.18)

The expression for the change in the Lagrangian can be simplified by using the total change in thefield δfj (x) = f ′j(x

′) − fj(x) and the corresponding total change in the field derivatives. Thechange in the Lagrangian is

δL = L

(x, fj ,

dfjdxµ

)∑µ

d∆xµ(x)dxµ

+∑µ

∆xµ(x)∂ L

∂ xµ

+∑j

∂ L

∂ fjδfj(x) +

∑j,µ

∂ L

∂

(dfjdxµ

)δ( dfjdxµ

(x))

(4.19)

The transformation is therefore a symmetry if and only if

L∑µ

d∆xµ(x)dxµ

+∑µ

∆xµ(x)∂ L

∂ xµ

+∑j

∂ L

∂fjδfj(x) +

∑j,µ

∂ L

∂dfjdxµ

δdfjdxµ

(x)

= ∆kL(x, fj ,dfjdxµ

) +∑µ

d∆Fµ(x, fj)dxµ

(4.20)

This is the most general case but I shall have need only of the simpler case in which ∆k = 0 and∆Fµ = 0. The criterion for a symmetry in this restricted sense is

δ L = L∑µ

d∆xµ(x)

dxµ

+∑µ

∆xµ(x)∂ L

∂ xµ

+∑j

∂ L

∂ fj

δfj(x) +∑j,µ

∂ L

∂dfj

dxµ

δdfj

dxµ

(x)

= 0

(4.21)

4.6 Example 1 23

This can also be written as

δ L = L∑µ

d∆xµ(x)

dxµ

+∑µ

∆xµ(x)∂ L

∂ xµ

+∑j

∂ L

∂ fj

δfj(x) +∑j,µ

∂ L

∂

(dfj

dxµ

) (d δfj

dxµ

(x)−∑ν

dfj

dxν

d∆xν

dxµ

)= 0

(4.22)

4.6 Example 1

Consider the stretched string example defined above.

New co-ordinates could be defined by

x′ = x+ δx; t′ = tS′(x′, t′) = S(x, t) (4.23)

That is the only change in this case is a translation of the space variable x.

The total change in the field which includes the change in the field itself plus the change due to thechange in the co-ordinate is

δS(x, t) = 0 (4.24)

This can be used to determine the change in the Lagrangian:

δL = δx

[12

(dρ

dx

)(dS

dt

)2

− 12

(dT

dx

)(dS

dx

)2]

(4.25)

This is not zero and cannot be written in the form ∆k L(x, fj ,

dfjdxµ

)+∑µ

d∆Fµ(x, fj)d xµ

. Hence

this transformation is NOT a symmetry.

(This statement is true in general but there are particular choices for the functions ρ(x) and T (x)for which δL is in the required form.)

4.7 Example 2

Now consider the transformation defined by

x′ = x; t′ = t+ δtS′(x′, t′) = S(x, t) (4.26)

That is the only change in this case is a translation of the time variable t.

The total change in the field is again zero and so the change in the Lagrangian is

δL = 0 (4.27)

Hence this transformation IS a symmetry

4.8 Current Densities 24

4.8 Current Densities

I first show how a conserved quantity gives rise to a density and a current density with a specialrelation between these.

Suppose Q is some quantity which is conserved. That is, Q can neither be created nor destroyed.

Let q(x, t) be the density of Q: That is, the amount of Q per unit volume.

Consider a volume V . The amount of Q inside V is

QV =∫V

q(x, t)dV (4.28)

and the rate of change of QV with time is

dQVdt

=∫V

dq(x, t)dt

dV (4.29)

Since Q can neither be created nor destroyed, the only way QV can change is by means of a flowacross the surface S of V . If I define J(x, t) to be the current density for Q, that is the flow of Qper unit surface area per unit time, then the conservation of Q requires that

dQVdt

=∫V

dq(x, t)dt

dV = −∮S

J(x, t) • dS (4.30)

The surface integral can be converted using Gauss’s theorem into

∫V

∇ • J(x, t) dV (4.31)

Since this is true for any volume V, I get the relation

dq(x, t)dt

+ ∇ • J(x, t) = 0 (4.32)

Hence conservation of a physical quantity Q gives rise to a density q(x, t) and a currentdensity J(x, t) which must be related by (4.32).

Conversely given a density q(x, t) and a current density J(x, t) which are related by (4.32) then

Q =∫q(x, t)dV (4.33)

is a conserved physical quantity.

Equation (4.32) can be expressed in a way which exhibits relativistic symmetry better if I defineJt = c q. In this case the conservation equation becomes∑

µ

dJµ

dxµ

= 0 (4.34)

4.9 Symmetry and Conservation Laws 25

4.9 Symmetry and Conservation Laws

I now want to use expression (4.21) that defines the condition for an infinitesimal symmetry in orderto generate a corresponding density and current density.

First, if I write δf j in terms of ∆f j and collect together the terms in (4.32) which do not involve∆fj , I obtain:

∑µ

d

dxµ[L ∆xµ(x)] +

∑j

∂ L

∂ fj(∆fj(x)) +

∑j,µ

∂ L

∂ (dfjdxµ

)

(∆dfjdxµ

(x))

= 0

(4.35)

Next I use the Euler-Lagrange equations (3.3) to eliminate ∂ L/∂ fj : and this gives

∑µ

d

dxµ

L∆xµ(x) +∑j

∂ L

∂

(dfjdxµ

) ∆fj(x)

= 0

(4.36)

The problem in making use of this equation is that ∆xµ, ∆fj etc are not independent.

I need to introduce a notation for the transformation which makes the relation between ∆xµ, ∆fjclear.

Suppose that a transformation is defined by one (constant) infinitesimal parameter δ ω and that theinfinitesimal quantities in (4.36) are given by

∆xµ = χµδω ; ∆fj = ϕj δω (4.37)

Then from (4.32) or (4.34) I can identify a density and current:

Jµ(x) = −

L χµ +∑

j

∂ L

∂

(dfj

dxµ

) ϕj

(4.38)

These current densities satisfy the equation∑µ

dJµdxµ

= 0 (4.39)

and the conserved quantity is:

Q =

∫Jt(x, t) dV = −

∫ Lχt +∑

j

∂ L

∂

(d fj

d xt

) ϕj

dV (4.40)

where the integral is over 3D space.

4.10 Time translation invariance & energy conservation 26

Note that I could change the sign of both the density and the current and still have the basic differ-ential equation satisfied. It is the convention that wherever possible the time-like component of thespace-time current density should be chosen to be positive.

4.10 Time translation invariance & energy conservation

Consider the (infinitesimal) time translation transformation:

∆x(x) = 0; ∆xt(x) = δxt

δfj(x) = 0; ∆fj = −δxtdfjdxt

(4.41)

Suppose that the Lagrangian is such that this is a symmetry transformation. Then I can use (4.37)and (4.38) to determine the corresponding conservation rule.

From (4.37) I have

χt = 1; χx = 0; χy = 0; χz = 0;

ϕj = − dfjdxt

(4.42)

Inserting these values into (4.38) gives the corresponding conserved current densities as

Jt(x) = −

L −

∂ L

∂

[d fj(x)d xt

] d fj(x)d xt

(4.43)

Jr(x) =

∂ L

∂

[d fj(x)d xr

] d fj(x)d xt

r = x, y, z (4.44)

Equation (4.43) defines the energy density for the system specified by the Lagrangian and equation(4.44) defines the corresponding energy current density.

The energy, itself, which is a conserved quantity is the integral of (4.43) over all (3D) space.

4.11 Example 3

Consider the transformation in example 2. In this case we have

∆x(x, t) = 0; ∆t(x, t) = δt

∆S(x, t) = −δtdS(x, t)dt

(4.45)

Applying the general result (equations (4.37) and (4.38)) to the case of the vibrating string andcalculating the derivatives gives the two components of the physical current as

Jt(x, t) = 12

[ρ(x)

(dS(x, t)d t

)2

+ T (x)(dS(x, t)d x

)2]

(4.46)

4.12 Summary of General Method 27

Jx(x, t) = −T (x)(dS(x, t)d x

)(dS(x, t)d t

)(4.47)

Note that the signs of these two quantities could be changed without affecting the conservation rule.The sign is normally chosen so as to make the first density positive. The first of these densities isthe energy density and the second is the energy current density which describes the flow of energydensity.

The energy density and energy current density, in one spatial dimension, are related by

d Jtd t

+d Jxd x

= 0 (4.48)

The total energy can be obtained by integrating this equation over x and assuming that no energyflows into or out of the system:

E =∫

12

[ρ(x)

(dS(x, t)d t

)2

+ T (x)(dS(x, t)d x

)2]dx (4.49)

This quantity is conserved.

To summarize what has been done:

(i) A time-shift is a symmetry for the ‘1D stretched string’;

(ii) This symmetry gives rise to an energy density and an energy current density;

(iii) The total energy, which is the integral of the energy density is a conserved quantity.

4.12 Summary of General Method

The general method is as follows:

(i) Determine ∆xµ(x) and ∆fj(x) (or better still δ fj(x)) for the transformation in terms of someparameter δ ω

(ii) Insert these values into (4.21) to check whether the the change in L is zero.

(iii) If the change in the Lagrangian is zero then the transformation is a symmetry.

(iv) If the change in the Lagrangian is not zero but can be expressed as ∆k L(x, fj ,dfjdxµ

) +∑µ

d∆Fµ(x, fj)dxµ

, for some constant ∆k and some function ∆Fµ(x, fj) , then the transformation

is also a symmetry. (This is rather rare but I have included it for completeness.)

(v) If the change in the Lagrangian is zero, use equations (4.38) to determine the density and currentdensity. The conserved quantity, associated with the symmetry transformation is given by (4.40).

This general procedure relating current densities and conserved quantities with symmetry transfor-mations is known as Noether’s Theorem

4.13 Problems 3

(i) Derive the relations (4.15), (4.16) and (4.17) for the changes in the field derivatives.

4.13 Problems 3 28

(ii) Consider the Lagrangian (3.13) which is a function of a complex field f(x, t) in four-dimensionalspace-time. Show that a time-translation is a symmetry and determine the corresponding con-servation law.

Chapter 5

SPACE-TIME SYMMETRIES & CONSERVATIONLAWS

In this chapter I look at the conservation laws which are related to space-time transformations.

5.1 Space-Time Translation Symmetry

If we assume that space-time is homogeneous the we should expect the equations of physics tobe invariant to a space-time translation.

In the case of space-time translation the total change in each field is zero: δfj(x, t) = 0. Thiscomprises a component from the change in the field itself and a component resulting from thechange in the co-ordinate:

δfj(x, t) = 0 = ∆fj(x, t) +∑µ

(d fjd xµ

∆xµ

)(5.1)

This gives the change in the field itself as

∆fj(x, t) = −∑µ

(d fjd xµ

∆xµ

)(5.2)

I now consider the various space-time translations separately:

5.1.1 Time Translation Symmetry

I have already done this but I will include it again so that all the results are in one place.

The changes to the co-ordinate and field are

∆x(x, t) = 0; ∆xt(x, t) = δxt

δfj(x, t) = 0; ∆fj(x, t) = −δtdfj(x, t)d xt

(5.3)

A system for which this transformation is a symmetry has conserved (energy) current densities:

29


Jt(x) =∑j

∂L

∂

[d fj(x, t)d xt

] d fj(x, t)d xt

− L (5.4)

J(x) =∑j

∂L

∂

[d fj(x, t)dx

] d fj(x, t)d xt

(5.5)

The first of these is the energy density and the second is the energy current density.

5.1.2 Space Translation Symmetry

I shall take the case in which the translation is along the x-axis: The y and z-translations givecorresponding results. In this case the changes to the co-ordinate and field are

∆x(x, t) = δx; ∆y(x, t) = 0; ∆z(x, t) = 0; ∆xt(x, t) = 0

δfj(x, t) = 0; ∆fj(x, t) = −δxdfj(x, t)dx

(5.6)

Appealing to the general result gives the current densities

J (x)t(x) =

∑j

∂L

∂

[d fj(x, t)d xt

] d fj(x, t)d x

(5.7)

J (x)x(x) =

∑j

∂L

∂

[d fj(x, t)d x

] d fj(x, t)d x

− L (5.8)

J (x)y(x) =

∑j

∂L

∂

[d fj(x, t)d y

] d fj(x, t)d x

(5.9)

J (x)z(x) =

∑j

∂L

∂

[d fj(x, t)d z

] d fj(x, t)d x

(5.10)

The first of these set of currents is the x-momentum density and the next three represent the x-momentum current density.

Notice that the x-momentum current density has components in the y- and z-directions. Clearly thisis quite complicated!

5.2 Space-Time Rotational Symmetry

If we believe that space-time is isotropic then the equations of physics should be invariantunder space-time rotations.


There are two aspects of space-time ’rotational’ symmetry: The first is a rotation of 3D space; andthe second is a Lorentz transformation which is a sort of rotation involving the time dimension anda space dimension.

5.2.1 Rotational Symmetry

If an infinitesimal rotation is represented by a vector δΘ which specifies the magnitude and direction(axis) of rotation then

∆xt(x, t) = 0; ∆x(x, t) = −δΘ ∧ x (5.11)

In a rotation the total change in each field is not (necessarily) zero. The way each field changesdepends on the type of field. We will, later, explore various types of fields. The simplest is a scalarfield. If the fields fj are scalars then the total change in each field is zero:

δfj(x, t) = 0; ∆fj(x, t) =d fjdx

• (δΘ ∧ x) (5.12)

For simplicity, suppose that the rotation vector δΘ is in the z-direction with magnitude δΘ.

In this case the changes in the co-ordinates and fields (assumed to be scalar) are:

∆xt(x, t) = 0; ∆z = 0 ∆x = y δΘ ∆y = −x δΘ

δfj(x, t) = 0; ∆fj(x, t) = δΘ(xd fjd y

− yd fjd x

) (5.13)

If I insert these expressions into the general forms for the current densities, I get

J(z)t =

∑j

(yd fjd x

− xd fjd y

)∂ L

∂

(d fjd xt

)

J(z)x =

∑j

(yd fjd x

− xd fjd y

)∂ L

∂

(d fjd x

) − y L

J(z)y =

∑j

(yd fjd x

− xd fjd y

)∂ L

∂

(d fjd y

) + xL

J(z)z =

∑j

(yd fjd x

− xd fjd y

)∂ L

∂

(d fjd z

)

(5.14)

J(z)t is the density of the z-component of angular momentum density. J(z)

x , J(z)y and J(z)

z are the threecomponents of a (space) vector which gives the flow of this quantity. As in the case of momentumdensity, the current density for the z−component of angular momentum can have components inthe x- and y-directions.


The conserved quantity is

∫dV

∑j

(yd fjd x

− xd fjd y

)∂ L

∂

(d fjd xt

) (5.15)

Chapter 6

SPACE-TIME TRANSFORMATIONS

I now intend to look at space-time transformations in more detail.

This will reveal various types of fields and also the Lagrangians for the fields.

Each of the types of transformation constitutes a group of transformations (technically a Lie Group)and each of these groups is characterized by a set of group generators

6.1 Space-Time Translations

I introduce a space-time (or four) vector by

x = xtet + xex + yey + zez (6.1)

where et , ex , ey , ezare unit vectors along the four axes and xt is c t. The scalar products of theseunit vectors are defined by

eµ • eν = 0 ; µ 6= νex • ex = 1 ey • ey = 1 ez • ez = 1 et • et = −1 (6.2)

No, the last scalar product is not a mistake!

The form of these scalar products is dictated by the Lorentz transformation which requires that(x2 + y2 + z2 − x2

t ) is invariant and that (x2 + y2 + z2 + x2t ) is not.

These definitions mean that the scalar product of two space-time vectors is

A • B = AxBx +AyBy +AzBz −AtBt (6.3)

and this is an invariant quantity. This means that this quantity does not change if we apply a rotationor a Lorentz transformation.

A space-time translation is

x → x′ = x + ∆x(x) (6.4)

where

∆x (x) = δ x constant (6.5)

This transformation can be written as a space-time vector operation on x:

∆x (x) = i

(δx • P) x (6.6)

33

6.2 Eigenvalues and eigenfields 34

The components of the vector operator P are called the generators of the group of space-timetranslations. These components are

Px = −i dd x

; Py = −i dd y

; Pz = −i dd z

; Pt = id

d xt(6.7)

This apparently strange procedure (introducing complex numbers !) is done so that the resultinggenerators are hermitian (or self-adjoint). This means that

∫dxF ∗(x)PµG(x) =

∫dx(PµF (x)

)∗G(x) (6.8)

for any F (x) and G(x) for which the integrals are defined.

The effect of a finite space-time translation on a function can be written in terms of these operators:

F (x′) = F (x + a) = exp(ia • P)F (x) (6.9)

These 4 generators commute with each other

PµPν − Pν Pµ = 0 ; µ, ν = t, x, y, z (6.10)

Note the 4D vector P can be written in terms of the 4D gradient operator

P = −i∇ = −i(−et

d

d xt+ ex

d

d x+ ey

d

d y+ ez

d

d z

)(6.11)

6.2 Eigenvalues and eigenfields

Consider a generator G of some (group of) transformations. Can G be assigned a value?

In general, the answer is no. G is an operator which operates on some field to give, in general, acompletely different field. This is what operators do!

However in the special case in which G operates on some particular field ϕ(x) such that the newfield is simply a multiple of ϕ(x), the multiplier could be said to be the value of G.

This special case is defined by

Gϕg (x) = gϕg (x) (6.12)

g is called an eigenvalue of G and ϕg (x) is an eigenfunction.

For this particular field G does indeed have a value and it is g the eigenvalue.

6.2.1 Self-Adjoint (or Hermitian) Operators

If G was chosen to be self-adjoint (hermitian) then the eigenvalues and eigenfields have specialproperties.

• every eigenvalue g is real ;• the eigenfields form a complete set. This means that the eigenfields can be used as the ‘building

blocks’ of all other functions.

6.3 Rotations 35

Completeness means that for a general ϕ(x) I have:

ϕ(x) =∑g

cgϕg(x) (6.13)

where cg is some set of (possibly complex) numbers. If the eigenvalues are continuous then thesummation has to be replaced by an integral:

ϕ(x) =∫dg cg ϕg(x) (6.14)

Consider the generator Px as an example. The eigenvalue equation is

Pxϕk (x) = −idϕk (x)d x

= kxϕk (x) (6.15)

and its solution is

ϕk(x) = Nk exp (ikx) (6.16)

where k is any real value and Nk is some normalization constant. Since Px was chosen to beself-adjoint the eigenfields are a complete set. This is Fourier analysis!

If two or more generators commute with each other then a common set of eigenfields exists. So forexample Pt, Px, Py, Pz have eigenfields

ϕK(x) = NK exp(iK • x

)(6.17)

where the space-time vector K has components kt, kx, ky, kz . This gives rise to 4D Fourier Analy-sis.

6.3 Rotations

Suppose we are using a reference frame with axes x, y and z and xt and we make a rotation ofϑx about the x-axis to get a new reference frame with axes axes x′, y′ and z′. This rotation canbe represented by a 4 × 4 matrix operator Rx operating on the column vector containing the fourcomponents of the vector:

xt

′

x′

y′

z′

=

1 0 0 00 1 0 00 0 cosϑx sinϑx0 0 − sinϑx cosϑx

xtxyz

(6.18)

This can be done in the same way for rotations about other axes. You should note that the rotationmatrices do not commute. That is, the result of two rotations depends on the order in which they arecarried out.

If I consider infinitesimal rotations then the co-ordinate transformation is

x′ = x + iδϑxJx x (6.19)

where Jx is the matrix operator

6.3 Rotations 36

Jx =

0 0 0 00 0 0 00 0 0 −i0 0 i 0

(6.20)

Similar matrix operators can be introduced for rotations about the y-and z axes:

Jy =

0 0 0 00 0 0 i0 0 0 00 −i 0 0

Jz =

0 0 0 00 0 −i 00 i 0 00 0 0 0

(6.21)

A general infinitesimal rotation can then be written as

x′ = x + i(δΘ • J

)x (6.22)

where δΘ is a vector describing the direction and magnitude of the infinitesimal rotation.

The generators Jx, Jy, Jz satisfy the following commutation rules

JxJy − JyJx = iJzJyJz − JzJy = iJxJzJx − JxJz = iJy

(6.23)

A finite rotation can be expressed as

x′ = exp(iΘ • J

)x (6.24)

where Θ is a vector describing the direction and magnitude of the finite rotation.

These generators can also be written as differential operators:

Jx = −i(yd

d z− z

d

d y

)Jy = −i

(zd

d x− x

d

d z

)Jz = −i

(xd

d y− y

d

d x

) (6.25)

Using this form or the matrix form I can show that Js and the Ps commute as follows

JxPx − PxJx = 0 = JyPy − PyJy = JzPz − PzJzJxPy − PyJx = iPzJyPz − PzJy = iPxJzPx − PxJz = iPyJxPt − PtJx = 0 = JyPt − PtJy = JzPt − PtJz

(6.26)

6.3.1 Generalizations of the rotation generators

It is assumed that the fundamental quantities are the generator commutation relations (6.23) and(6.26) and not the particular expressions for the operators.

6.3 Rotations 37

Since these three operators Jx, Jy and Jz do not commute, I cannot find a common set of eigenfieldsas in the case of the operators Px, Py and Pz .

I want to show that there are many representations of the rotation operators.

To do this I start from the commutation relations (6.23). From these it follows that the operator

J2 = J2x + J2

y + J2z (6.27)

commutes with each of the three generators Jx, Jy and Jz .

Since J2 and Jz commute I can find a common set of eigenfields for this pair:

J2ϕJm = J2ϕJmJzϕJm = mϕJm

(6.28)

where J and m are real (but unknown at the moment) constants.

It is useful to form the two operators

J± = Jx ± i Jy (6.29)

which are called raising (+) and lowering (−) operators.

The commutation relations involving these operators are:

J+ J− − J− J+ = 2 JzJ+ Jz − Jz J+ = − J+

J− Jz − Jz J− = J−

(6.30)

I use these raising and lowering operators to define two new fields

ϕ±Jm = J±ϕJm (6.31)

Next I operate with J2 and Jz on these new fields. This shows that(J±ϕJm

)satisfies

J2(J±ϕJm

)= J2

(J±ϕJm

)Jz

(J±ϕJm

)= (m± 1)

(J±ϕJm

) (6.32)

That is, it is an eigenfield of J2 with the same eigenvalue J2 and an eigenfield of Jz with nowdifferent eigenvalues (m±1).

It is because of this property that J± are called raising and lowering operators: they produce neweigenfields of Jz with the eigenvalues raised or lowered by 1.

J+ is the adjoint (hermitian conjugate) of J− and vice versa. This means that the operators J+J−and J−J+ are operators whose eigenvalues are real and non-negative. These products can be ex-pressed as

J+J− = J2 − J2z + Jz

J−J+ = J2 − J2z − Jz

(6.33)

Applying these two operators to ϕJm gives

6.3 Rotations 38

J+J−ϕJm =(J2 −m(m− 1)

)ϕJm

J−J+ϕJm =(J2 −m(m+ 1)

)ϕJm

(6.34)

The eigenvalues on the right-hand sides must be greater or equal to zero. These properties can beused to put restrictions on m:

−

(√J2 +

14− 1

2

)≤ m ≤

(√J2 +

14− 1

2

)(6.35)

These restrictions seem to contradict (6.32) which suggests that we can keep using the raising (orlowering) operator to keep increasing (or decreasing) the eigenvalue m by 1.

The only resolution of this is that the must be upper and lower eigenvectors, with eigenvalues mU

and mL which satisfy

(J+ϕJmU

)= 0

(J−ϕJmL

)= 0 (6.36)

These relations lead to

mU =

(−1

2+√J2 +

14

)

mL = −

(−1

2+√J2 +

14

)= −mU

(6.37)

These must differ by a whole number of integers. So

(−1

2+√J2 +

14

)must be a non-negative

half integer j.

This means the eigenvalues of J2 and Jz are

J2 = j(j + 1); j = 0,12, 1,

32, . . .

m; m = −j, 1− j, 2− j, . . . , j − 1, j(6.38)

For a given j the eigenfields ϕJmcan be written as a column vector with (2j + 1) elements.

The rotations I started with correspond to j = 1 and hence a 3-dimensional matrix representationbut the matrix representation was different to that derived here.

In the matrix representation derived here the matrices representing J2 and Jz are, of course, diago-nal:

J2 = j(j + 1)I; Jz =

j 0 0 · · · 0 0 00 j − 1 0 · · · 0 0 00 0 j − 2 · · · 0 0 0...

......

. . ....

......

0 0 0 · · · 2− j 0 00 0 0 · · · 0 1− j 00 0 0 · · · 0 0 −j

(6.39)

The properties of J± = Jx ± iJy can be used to determine the matrices of Jx and Jy . First I needto use the relations (6.34) in order to normalize the eigenvectors

6.3 Rotations 39

ϕjm+1 =1√

j (j + 1)−m (m+ 1)J+ϕjm

ϕjm−1 =1√

j (j + 1)−m (m− 1)J−ϕjm

(6.40)

These relations can be used to get the matrix representations of first J+ and J−. and hence of Jxand Jy .

The only non-zero elements of the two matrices J+ and J− are :[J−

]m−1,m =

√j (j + 1)−m (m− 1)

[J+

]m+1,m

=√j (j + 1)−m (m+ 1)

(6.41)

Hence, using (6.29), the only non-zero elements of the two matrices Jx and Jy are[Jx

]m+1,m =

12

√j (j + 1)−m (m+ 1)

[Jy

]m+1,m

=−i2

√j (j + 1)−m (m+ 1)

[Jx

]m−1,m

=12

√j (j + 1)−m (m− 1)

[Jy

]m−1,m =

i

2

√j (j + 1)−m (m− 1)

(6.42)

The commutation relations (6.23) or (6.30) we have just analyzed form what is known as the SU(2)Lie algebra.

6.3.2 Spinor representation

The lowest dimensional (non-trivial) representation of the SU(2) Lie algebra occurs for j = 1/2. Inthis case the space is two-dimensional and the three matrix operators are

Jz =[

1/2 00 −1/2

]Jx =

[0 1/2

1/2 0

]Jy =

[0 −i/2i/2 0

](6.43)

These are very simple operators and in this representation a finite rotation can be expressed as (see(6.24)):

exp(iΘ • J) = cos(ϑ/2)

+ 2i sin(ϑ/2)(

n • J)

(6.44)

where n is a unit vector in the direction of the rotation Θ and ϑ is the magnitude of Θ.

This transformation has the very interesting property that a rotation by 2π results in a multipli-cation by -1.

The elements of the 2-dimensional space in which these operators act are called spinors. These areimportant in particle physics: electrons, neutrinos, quarks etc are all represented by spinors.

The Lie algebra we have just studied, with generators Jx, Jy and Jz , is known as SU(2).


6.4 Lorentz Transformations

Lorentz transformations are ’rotations’ in a space-time plane.

A Lorentz transformation involving motion along the x-axis is

x′ =x− Vx

cxt√

1− V 2x

c2

y′ = y z′ = z

x′t =xt −

Vxcx√

1− V 2x

c2

(6.45)

This can be written in matrix notation as

x′tx′

y′

z′

=

1√1− V 2

x/c2− Vx/c√

1− V 2x/c2

0 0

− Vx/c√1− V 2

x/c2

1√1− V 2

x/c20 0

0 0 1 00 0 0 1

xtxyz

(6.46)

This can be made to look very similar to the matrices involved in rotations if I introduce a quantityϕx defined by

cosh (ϕx) =1√

1− V 2x/c2

; sinh (ϕx) =Vx/c√

1− V 2x/c2

; tanh(ϕx) = Vx/c (6.47)

In terms of this parameter the transformation is

ct′

x′

y′

z′

=

cosh (ϕx) −sinh (ϕx) 0 0−sinh (ϕx) cosh (ϕx) 0 00 0 1 00 0 0 1

ctxyz

(6.48)

If I consider an infinitesimal change, this becomes

ct′

x′

y′

z′

=

1 −δϕx 0 0−δϕx 1 0 00 0 1 00 0 0 1

ctxyz

(6.49)

In space-time vector form this is

x′ = x + iδϕ xKxx (6.50)

where Kx is the matrix operator

Kx =

0 i 0 0i 0 0 00 0 0 00 0 0 0

(6.51)


Note that Kxturns out to be anti-hermitian. Anti-hermitian operators have purely imaginaryeigenvalues.

The reason for this is the strange nature of the scalar product of space-time vectors. In fact in termsof this scalar product the operator is hermitian:

For complex space-time vectors a and b, and using the scalar product, the operator Kxis self-adjoint(or hermitian) if

a∗ •(Kxb

)=(Kxa

)∗• b (6.52)

That is, the operator can be applied to either component of a (4D) scalar product and the result isthe same. The form (6.51) ensures that this is satisfied.

However I shall continue to use the conventional matrix nomenclature and describe Kx as anti-hermitian.

I can also introduce operators corresponding to Lorentz transformations along the y- and z-axes:

Ky =

0 0 i 00 0 0 0i 0 0 00 0 0 0

Kz =

0 0 0 i0 0 0 00 0 0 0i 0 0 0

(6.53)

A general finite Lorentz transformation can be written as

x′ = exp(iΦ • K

)x (6.54)

Note that Φ , K are 3D vectors whereas x is a 4D vector. Φ is a vector with components ϕx, ϕy, ϕz;see (6.47).

As with the rotation generators in section (4.3) I can express these Lorentz generators as differentialoperators:

Kx = i

(ctd

d x+ x

d

d xt

)Ky = i

(ctd

d y+ y

d

d xt

)Kz = i

(ctd

d z+ z

d

d xt

) (6.55)

The commutation properties of these Ks amongst themselves is

KxKy − KyKx = −iJzKyKz − KzKy = −iJxKzKx − KxKz = −iJy

(6.56)

This is a strange result. It says that the generators of Lorentz transformations, by themselves do notform a group: Rotations have to be included.

The complete set of commutation of this set of six generators is


JxJy − JyJx = iJzJyJz − JzJy = iJxJzJx − JxJz = iJy

KxKy − KyKx = −iJzKyKz − KzKy = −iJxKzKx − KxKz = −iJy

JxKx − KxJx = 0JyKy − KyJy = 0JzKz − KzJz = 0

JxKy − KyJx = iKz

JyKz − KzJy = iKx

JzKx − KxJz = iKy

KxJy − JyKx = iKz

KyJz − JzKy = iKx

KzJx − JxKz = iKy

(6.57)

The properties of the combined set of generators can be simplified if I introduce the combinations:

A =12

(J + iK

)B =

12

(J− iK

) (6.58)

These new operators have the following commutation properties:

AiBj − BjAi = 0 ; i, j = x, y, z

AxAy − AyAx = iAzBxBy − ByBx = iBz

(6.59)

The last two equations also apply with cyclic permutations made to the indices.

These equations imply that I have two sets of SU(2) generators (compare (6.23)) which commutewith each other. I can therefore use the results of section (5.3.1) to immediately write down generalmatrix representations of the two sets of operators.

The Lie Algebra in this case is SU(2)×SU(2) and is called the Lorentz group algebra.

I can choose eigenfieldsϕama;bmb

(6.60)

where

a = 0,12, 1,

32, · · ·

ma = −a, 1− a, · · · , a− 1, a

b = 0,12, 1,

32, · · ·

mb = −b, 1− b, · · · , b− 1, b

(6.61)

In this representation the matrices corresponding to A and B are given by the equations (6.39) and(6.41).

The z-components are diagonal


[Az

]mamb,mamb

= ma[Bz

]mamb,mamb

= mb

(6.62)

The only non-zero elements of the A−, A+ and B−, B+ are

[A−

]ma−1mb,mamb

=√a (a+ 1)−ma (ma − 1)

[A+

]ma+1mb,mamb

=√a (a+ 1)−ma (ma + 1)

[B−

]mamb−1,mamb

=√b (b+ 1)−mb (mb − 1)

[B+

]mamb+1,mamb

=√b (b+ 1)−mb (mb + 1)

(6.63)

These relations can be used to get the matrices for J and K.

The z-components are diagonal and the diagonal elements are

[Jz

]mamb,mamb

= (ma +mb)[Kz

]mamb,mamb

= −i (ma −mb)(6.64)

The only non-zero elements of the x- and y-components of J are

[Jx

]ma+1mb,mamb

=12

√a (a+ 1)−ma (ma + 1)

[Jy

]ma+1mb,mamb

= − i2

√a (a+ 1)−ma (ma + 1)

[Jx

]mamb+1,mamb

=12

√b (b+ 1)−mb (mb + 1)

[Jy

]mamb+1,mamb

= − i2

√b (b+ 1)−mb (mb + 1)

(6.65)

and

[Jx

]ma−1mb,mamb

=12

√a (a+ 1)−ma (ma − 1)

[Jy

]ma−1mb,mamb

=i

2

√a (a+ 1)−ma (ma − 1)

[Jx

]mamb−1,mamb

=12

√b (b+ 1)−mb (mb − 1)

[Jy

]mamb−1,mamb

=i

2

√b (b+ 1)−mb (mb − 1)

(6.66)

The only non-zero elements of the x- and y-components of K are


[Kx

]ma+1mb,mamb

=−i2

√a (a+ 1)−ma (ma + 1)

[Ky

]ma+1mb,mamb

= −12

√a (a+ 1)−ma (ma + 1)

[Kx

]mamb+1,mamb

=i

2

√b (b+ 1)−mb (mb + 1)

[Ky

]mamb+1,mamb

=12

√b (b+ 1)−mb (mb + 1)

(6.67)

and [Kx

]ma−1mb,mamb

=−i2

√a (a+ 1)−ma (ma − 1)

[Ky

]ma−1mb,mamb

=12

√a (a+ 1)−ma (ma − 1)

[Kx

]mamb−1,mamb

=i

2

√b (b+ 1)−mb (mb − 1)

[Ky

]mamb−1,mamb

=−12

√b (b+ 1)−mb (mb − 1)

(6.68)

Note that a and b are not affected in these relations and so, effectively, specify the representation.

The above matrix notation needs some explanation! Suppose we have a matrix M. The elements ofthe matrix are denoted by

Mi,j

where index i denotes the row of the matrix and index j denotes the column. In the above matricesthe row index is a composites constructed from the pair mamb. Consider as an example the casewhere a = 1 and b = 1. In this case ma = −1, 0, 1 and mb = −1, 0, 1. There are therefore9(= 3 × 3) row and column indices. These can be labelled in a variety of ways. Here is a simpleexample:

row index ma mb

1 1 12 0 13 -1 14 1 05 0 06 -1 07 1 -18 0 -19 -1 -1

Any other method would be just as valid: we just need a way to map from the pair (mamb) to therow index. The column index is treated in exactly the same way.

The simplest representations can be listed as

6.5 Basic Fields 45

Lorentz Group Algebra RepresentationsType a bScalar S 0 0spinor I ΨR 1/2 0spinor II ΨL 0 1/2Vector A 1/2 1/2

If an infinitesimal rotation is represented by δΘ and an infinitesimal Lorentz transformation by δΦthen the total change in a field is given by

δf(x, t) = iδΘ • Jf + iδΦ • Kf (6.69)

where J and K are the appropriate operators for the type of field f .

For a scalar field the operators J and K are both zero.

For a type I spinor field the operators are

J =12σ K =

−i2

σ (6.70)

and for a type II spinor they are

J =12σ K =

12σ (6.71)

In these equations σ is the vector 2 × 2 Pauli spin matrix. That is σ is a vector in which eachcomponent is a 2× 2 matrix.

σx =(

0 11 0

)σy =

(0 −ii 0

)σz =

(1 00 −1

) (6.72)

6.5 Basic Fields

The fields in our field theories will be chosen to correspond to (simple) representations of theLorentz-Rotation group of transformations. So that the fields themselves are determined by thesymmetry group.

6.6 Space-time rotations

J and K can be looked on as rotations in four-dimensional space-time. In order to see this moreclearly it is better to change the notation:

Jx → Tyz = −TzyJy → Tzx = −TxzJz → Txy = −Tyx

(6.73)

So that we no longer regard Jx as a rotation about the x-axis but as a rotation Tyz in the yz-plane:Tzy is just the rotation in the opposite sense.

Similarly we can re-label the K operators as

6.6 Space-time rotations 46

Kx → Ttx = −TxtKy → Tty = −TytKz → Ttz = −Tzt

(6.74)

So, for example, Ttx is a transformation (rotation ?) of the tx-plane. This ‘rotation’ preserves(x2− c2t2) whereas the normal rotation Txy preserves (x2 +y2).Written in this form the quantitiesTµν are the components of an anti-symmetric space-time dyadic:

T =∑µν

eµTµν eν (6.75)

where the diagonal elements are zero.

A dyadic is a particular case of a a tensor.

Chapter 7

FIELDS AND LAGRANGIANS

I am going to insist that the fields themselves and the Lagrangians specifying the dynamics of thefields are determined in some way by the symmetry properties of space-time.

I choose the basic fields to be simple representations of the Lorentz-Rotation transformations andtry to choose simple Lagrangians which are invariant.

The simplest invariants are:

(i) A product of scalar quantities formed from the fields. In the case of a (complex) scalar field asimple example is;

S∗S (7.1)

or a product of scalar quantities formed from the fields.

(ii) A scalar product of two space-time vector combinations of the fields. If S is a scalar field thenPS is a space-time vector field and so we could form a scalar quantity as(PS(x))∗

•(PS(x)

)= −∇S∗(x) • ∇S(x) (7.2)

(iii) A scalar product of two dyadic combinations of the fields. If ˜D and ˜E are two such dyadics,then I define the dyadic scalar product as follows:

˜D =∑µν

eµDµν eν˜E =∑µν

eµEµν eν˜D • ˜E =∑µν

gµDµνEνµgν

(7.3)

where the quantity gν is defined as gt = −1; gx = 1; gy = 1; gz = 1.

If the field is complex I will only take combinations which are gauge invariant. This means thatchanging the field f → eiχf should leave the Lagrangian unchanged. ( There will be more aboutgauge transformations later.)

If I make an infinitesimal rotation and an infinitesimal Lorentz transformation characterized by (3D)vectors δϑ and δϕ then a scalar S and a space-time vector V transform like:

S → SV t → V t + (δϕxVx + δϕyVy + δϕzVz)V x → V x + (δϕxVt + δϑzVy − δϑyVz)V y → V y + (δϕyVt + δϑxVz − δϑzVx)V z → V z + (δϕzVt + δϑyVx − δϑxVy)

(7.4)

47

7.1 Complex scalar field 48

I am therefore looking for combinations of the field variables which behave in one of these ways. Ishall return to the dyadic later.

7.1 Complex scalar field

This is the (0, 0) category in the Lorentz-Rotation representations. If I call the field S then a simpleinvariant is

S∗S (7.5)

I can form a space-time vector using the vector operator(⇒ PS)and so I can form another invari-

ant by taking the scalar product of the vectors:

(PS(x))∗

•(PS(x)

)= −∇S∗(x) • ∇S(x). (7.6)

A suitable Lagrangian density is

L =−12

(∇S∗ • ∇S +m2S∗S

)=−12

(∇S∗ •∇S − dS∗

d xt

dS

dxt+m2 S∗S

) (7.7)

The Euler-Lagrange equation is:

∇ • ∇S(x)−m2 S(x) =(

∇2 − d2

d xt2

)S(x)−m2S(x) = 0 (7.8)

This is known as the Klein-Gordon equation. In the conventional notation m →(mc

~

)2

and m isa mass.

7.2 Pauli Spinor Fields

There are two types of Pauli spinor fields corresponding to (a=1/2, b=0) and (a=0, b=1/2). I shalldenote these by ΨR and ΨL (for right and left). The significance of the L and R symbols is thatone of the fields has a right-handed (or positive) helicity and one has a left-handed helicity. A right-handed helicity means that the field has a spin component in the direction of the field current. I shallnot, in this course, prove these statements.

Type I spinor ΨR

The dimensionality of the field is given in general by (2a+1)*(2b+1). In this case the fields aretwo-dimensional and the rotation and Lorentz transformation matrices can be written in terms ofthe Pauli spin matrices:

σz =[

1 00 −1

]σx =

[0 11 0

]σy =

[0 −ii 0

](7.9)

In terms of these spin matrices:


J = 1/2σ K = −i/2σ (7.10)

The components of the spinor field are properly labelled by the value of m (see section 4.3.1) andin this case these are +1/2 and −1/2. However since there are only two values I shall just label these1 & 2. That is m= +1/2 → 1 and m=−1/2 → 2.

In an infinitesimal rotation, defined by vector δϑ , and an infinitesimal Lorentz transformation ,defined by vector δϕ, the components of the field transform as

ψR1 → ψR1 +(i

2(ϑx − iϕx)ψR2 +

12

(ϑy − iϕy)ψR2 +i

2(ϑz − iϕz)ψR1

)ψR2 → ψR2 +

(i

2(ϑx − iϕx)ψR1 −

12

(ϑy − iϕy)ψR1 −i

2(ϑz − iϕz)ψR2

) (7.11)

These are necessarily complex fields so I shall only consider gauge invariant combinations of theform:

ψR∗1 ψR1 ψR∗2 ψR2 ψR∗1 ψR2 ψR∗2 ψR1 (7.12)

I need to check how each of these combinations changes under the Lorentz-Rotation transforma-tions:

ψR∗1 ψR1 → ψR∗1 ψR1 +

(δϕz)ψR∗1 ψR1 +(i

2(δϑx − iδϕx) +

12

(δϑy − iδϕy))ψR∗1 ψR2 +(

−i2

(δϑx + iδϕx) +12

(δϑy + iδϕy))ψR∗2 ψR1

(7.13)

ψR∗2 ψR2 → ψR∗2 ψR2 +

(−δϕz)ψR∗2 ψR2 +(i

2(δϑx − iδϕx)−

12


−i2

(δϑx + iδϕx)−12


(7.14)

ψR∗1 ψR2 → ψR∗1 ψR2 +

(−iδϑz)ψR∗1 ψR2 +(i

2(δϑx − iδϕx)−

12


−i2

(δϑx + iδϕx) +12


(7.15)

ψR∗2 ψR1 → ψR∗2 ψR1 +

(iδϑz)ψR∗2 ψR1 +(i

2(δϑx − iδϕx) +

12


−i2

(δϑx + iδϕx)−12


(7.16)

It is somewhat surprising that it is not possible to choose a combination of these that is invariant.However it is possible to choose a combination which transforms like (and therefore is !) a space-time vector. The time-like component of this vector is


ΨR†ΨR =(ψR∗1 ψR∗2

)( ψR1ψR2

)=(ψR∗1 ψR1 + ψR∗2 ψR2

)(7.17)

The space-like part can be written as

ΨR†σ ΨR = ex

(ψR∗1 ψR2 + ψR∗2 ψR1

)− i ey

(ψR∗1 ψR2 − ψR∗2 ψR1

)+ ez

(ψR∗1 ψR1 − ψR∗2 ψR2

)(7.18)

A simple invariant Lagrangian is

L =12

(ΨR†PtΨR −ΨR†P • σ ΨR +

(PtΨR

)†ΨR −

(PΨR

)†• σ ΨR

)(7.19)

which can be written as

L =i

2

(ΨR† dΨR

d xt−(dΨR

d xt

)†ΨR + ΨR†σ •∇ΨR −

(∇ΨR

)† • σ ΨR

)(7.20)

and the equation of motion is

idΨR

d xt= −iσ •∇ΨR (7.21)

This is believed to be the wave equation of an anti-neutrino.

Type II spinor ΨL

In this case the relevant matrices are:

J = 1/2σ K = i/2σ (7.22)

In an infinitesimal rotation, defined by vector δϑ , and an infinitesimal Lorentz transformation ,defined by vector δϕ, the components of the field transform as

ψL1 → ψL1 +(i

2(ϑx + iϕx)ψL2 +

12

(ϑy + iϕy)ψL2 +i

2(ϑz + iϕz)ψL1

)ψL2 → ψL2 +

(i

2(ϑx + iϕx)ψL1 −

12

(ϑy + iϕy)ψL1 −i

2(ϑz + iϕz)ψL2

) (7.23)

These are necessarily complex fields so I shall only consider gauge invariant combinations of theform:

ψL∗1 ψL1 ψL∗2 ψL2 ψL∗1 ψL2 ψL∗2 ψL1 (7.24)

I need to check how each of these combinations changes under the Lorentz-Rotation transforma-tions:

ψL∗1 ψL1 → ψL∗1 ψL1 +

(−δϕz)ψL∗1 ψL1 +(i

2(δϑx + iδϕx) +

12

(δϑy + iδϕy))ψL∗1 ψL2 +(

−i2

(δϑx − iδϕx) +12

(δϑy − iδϕy))ψL∗2 ψL1

(7.25)

7.3 Dirac spinor 51

ψL∗2 ψL2 → ψL∗2 ψL2 +

(δϕz)ψL∗2 ψL2 +(i

2(δϑx + iδϕx)−

12


−i2

(δϑx − iδϕx)−12


(7.26)

ψL∗1 ψL2 → ψL∗1 ψL2 +

(−iδϑz)ψL∗1 ψL2 +(i

2(δϑx + iδϕx)−

12


−i2

(δϑx − iδϕx) +12


(7.27)

ψL∗2 ψL1 → ψL∗2 ψL1 +

(iδϑz)ψL∗2 ψL1 +(i

2(δϑx + iδϕx) +

12


−i2

(δϑx − iδϕx)−12


(7.28)

Again it is not possible to choose a combination of these that is invariant. However it is still possibleto choose a combination which transforms like (and therefore is !) a space-time vector. The time-like component is

ΨL†ΨL =(ψL∗1 ψL∗2

)( ψL1ψL2

)=(ψL∗1 ψL1 + ψL∗2 ψL2

)(7.29)

The space-like part can be written as

−ΨL†σ ΨL = −ex

(ψL∗1 ψL2 + ψL∗2 ψL1

)+ i ey

(ψL∗1 ψL2 − ψL∗2 ψL1

)− ez

(ψL∗1 ψL1 − ψL∗2 ψL2

)(7.30)

A simple invariant Lagrangian is

L =12

(ΨL†PtΨL + ΨL†P • σ ΨL +

(PtΨL

)†ΨL +

(PΨL

)†• σ ΨL

)(7.31)

and the equation of motion is

idΨL

d xt= iσ •∇ΨL (7.32)

This is believed to be the equation of a neutrino.

7.3 Dirac spinor

In this case I allow symmetric combinations of both ΨRand ΨL. . Formally this is written as the

choice (1/2,0)⊕(0,1/2).

In addition to combinations of (7.13–7.16) and (7.25–7.28), I need the following:

7.3 Dirac spinor 52

ψR∗1 ψL1 → ψR∗1 ψL1 +(i

2(ϑx + iϕx)ψR∗1 ψL2 +

12

(ϑy + iϕy)ψR∗1 ψL2 +i

2(ϑz + iϕz)ψR∗1 ψL1

)+(− i


12

(ϑy + iϕy)ψR∗2 ψL1 −i


)(7.33)

ψR∗2 ψL2 → ψR∗2 ψL2 +(i

2(ϑx + iϕx)ψR∗2 ψL1 −

12



)+(− i


12



)(7.34)

ψR∗1 ψL2 → ψR∗1 ψL2 +(i


12



)+(− i


12



)(7.35)

ψR∗2 ψL1 → ψR∗2 ψL1 +(i


12



)+(− i


12



)(7.36)

With this additional set of terms it is possible to construct a scalar invariant that is also symmetricin R & L:

ΨR†ΨL + ΨL†ΨR (7.37)

Therefore an invariant Lagrangian can be constructed in the form

L =12

(ΨR†PtΨR −ΨR†P • σ ΨR +

(PtΨR

)†ΨR −

(PΨR

)†• σ ΨR

)12

(ΨL†PtΨL + ΨL†P • σ ΨL −

(PtΨL

)†ΨL +

(PΨL

)†• σ ΨL

)m(ΨR†ΨL + ΨL† ΨR

) (7.38)

The first two terms are just those used in (7.19) and (7.31); the additional term is the new invariantmade possible by considering both left- and right-spinors.

The new equations of motion are

idΨR

d xt= −iσ •∇ΨR −mΨL (7.39)

idΨL

d xt= iσ •∇ΨL −mΨR (7.40)

This pair of equations constitute the Dirac equation and describes the electron-positron field. In theconventional notation m→

(mc/~).

7.4 Problems 53

7.4 Problems

Chapter 8

LOCAL GAUGE INVARIANCE &INTERACTIONS

Gauge invariance is a symmetry property which just involves the fields: there is no transformationof the co-ordinates. The fields must be complex fields and the Lagrangian must contain fields incomplex conjugate pairs: fj and fj∗. For the purposes of the field theory these fields can be consid-ered to be independent. We could also consider the real and imaginary parts to be the independentvariables; however this (equivalent) approach is always more complicated.

The simplest gauge transformation, called a U(1) gauge symmetry, assumes that only the magnitudeof a field is physically relevant. This means that a transformation of the form

fj(x) → eiθfj(x)fj

∗(x) → e−iθfj∗(x) (8.1)

should leave the Lagrangian unchanged.

In order for this to be a symmetry we require:

δL = iδθ

[∑j

∂L

dfjfj(x) +

∑j,µ

∂L

dfj,µfj,µ(x)

]

−iδθ

[∑j

∂L

dfj∗ fj

∗(x) +∑j,µ

∂L

df∗j,µfj,µ

∗(x)

]= 0

(8.2)

If this is satisfied then there are conserved currents of the form:

Jµ(x) = i

∑j

(∂L

∂ fj,µfj

)−∑j

(∂L

∂ fj,µ∗ fj

∗) (8.3)

and the conserved quantity is the volume integral of

Jt(x) = i

∑j

(∂L

∂ fj,tfj

)−∑j

(∂L

∂ fj,t∗ fj

∗) (8.4)

This integral is called the charge.

54


8.1 Complex Scalar Field & Local U(1) Gauge Invariance

In the case of a complex scalar field S the Lagrangian is a function of S, S∗, ∇S and ∇S∗. Thesequantities change as

S → eiθSS∗ → e−iθS∗

∇S → eiθ∇S

∇S∗ → e−iθ∇S∗

(8.5)

and, if this is to be a symmetry, they must appear in the Lagrangian in such a way that the termsinvolving θ cancel.

An example of a Lagrangian involving a complex scalar field for which this is a symmetry is:

LS = −12

[∇S∗ • ∇S + µS∗S

](8.6)

The gauge transformation (8.1) and (8.5) is called a global gauge transformation. The global tagsignifies that θ is required to be a constant throughout the universe!

A local gauge transformation is one in which this (unrealistic?) restriction is relaxed. That is θ isallowed to vary through space. In this case the change in the fields (8.5) has to be modified to takeaccount of the variation of θ :

S → eiθ(x)SS∗ → e−iθ(x)S∗

∇S → eiθ(x)(∇ + i∇θ

)S

∇S∗ → e−iθ(x)(∇− i∇θ

)S∗

(8.7)

Clearly we have now done more than simply multiply by a factor. In general the terms involving∇θ will not cancel and so the local transformation will not be a symmetry.

If I apply this transformation S(x) → eiθ(x)S(x) to the above Lagrangian I get

LS → −12

[(∇− i∇θ

)S∗ •

(∇ + i∇θ

)S + µS∗S

]LS → LS +

i

2∇θ(x) •

(S∗∇S − S∇S∗

)− 1

2

(∇θ(x) • ∇θ(x)

)S∗S

(8.8)

This is clearly not a symmetry.

If I try to force this local gauge invariance to be a symmetry then I need to modify the Lagrangianin some way.

This can be done by introducing a real space-time vector field A (x), called a gauge field, whichcouples to the field S in such a way that everywhere in the Lagrangian we make the changes:

∇S →(∇ + iA

)S

∇S∗ →(∇− iA

)S∗

(8.9)

and which under the local gauge transformation changes as

A (x) → A (x)− ∇θ (x) (8.10)

In this way the change induced in A (x) by the local gauge transformation cancels the terms involv-ing ∇θ.


With this modification the Lagrangian is assumed to be a function of S, S∗,(∇ + iA

)S and(

∇− iA)S∗ and under the local gauge transformation these four quantities change as follows:

S → eiθ(x)SS∗ → e−iθ(x)S∗(

∇ + iA)S → eiθ(x)

(∇ + iA

)S(

∇− iA)S∗ → e−iθ(x)

(∇− iA

)S∗

(8.11)

The changes now simply involve the exponential factors as in the global transformation and sincewe have have assumed the global transformation to be a symmetry these factors must cancel.

In the above example the Lagrangian has now become

−12

[(∇− iA

)S∗ •

(∇ + iA

)S + µS∗S

](8.12)

This can be expressed as LS +LAS where LS is the Lagrangian for the field S and LAS representsthe interaction of S with the new field A. To this Lagrangian we must add the Lagrangian LA forthe field A itself. This must be such that it does not change under the local gauge transformation.That is it must not change under the transformation A (x) → A (x)− ∇θ (x).

This can be done if LA contains only derivatives of A (x) in the anti-symmetric combinations

Aµν =∂Aµ∂xν

− ∂Aν∂xµ

(8.13)

The effect of (8.10) on such a term is

∂Aµ∂xν

− ∂Aν∂xµ

→ ∂Aµ∂xν

− ∂Aν∂xµ

−(

∂

∂xν

∂ θ

∂xµ− ∂

∂xµ

∂ θ

∂xν

)=∂Aµ∂xν

− ∂Aν∂xµ

(8.14)

That is it has no effect.

In order to form a Lorentz invariant Lagrangian I first make a dyadic out of Aµν (see page 28):

Ã =∑µν

eµAµν eν (8.15)

Then I take the dyadic scalar product of Ã with itself (see (7.3)) – this is guaranteed to be Lorentzinvariant.

A simple Lorentz invariant Lagrangian can then be constructed as

LA = −14Ã • Ã = −1

4∑µν

gµAµνAνµgν

=12

[(∇At +

∂Ac∂t

)2

− (∇ ∧A) • (∇ ∧A)

] (8.16)

At and A are the time and space components of A (x).

The complete Lagrangian for the above example, which is Lorentz invariant and gauge invariant, is


L = −12

[− 1c2

(∂S∗

∂ t

∂S

∂ t

)+ ∇S∗ •∇S + µS∗S

]−1

2

[i

cAt

(S∗∂S

∂t− S

∂S∗

∂t

)− iA • (S∗∇S − S∇S∗) + S∗S

(A •A−A2

t

)]+

12

[(∇At +

∂Ac∂t

)2

− (∇ ∧A) • (∇ ∧A)

] (8.17)

There are two sets of Euler-Lagrange equations, one for the original complex scalar field S and one(pair) for A (x):

[∂

c∂ t− iAt

]2S − [∇ + iA] • [∇ + iA]S = µS (8.18)

∇2At + 1c

∂

∂ t(∇ •A) =

−i2c

(S∗∂S

∂ t− S

∂S∗

∂ t

)−AtS

∗S (8.19)

∇ ∧ (∇ ∧A) + 1c2∂2A∂ t2

=−i2

(S∗∇S − S∇S∗) + AS∗S (8.20)

These last two equations are essentially Maxwell’s equations of electromagnetism.

This can be seen more clearly if I define new variables

E = −(

∇At +∂Ac∂ t

); B = ∇ ∧A (8.21)

In terms of these new variables equations (8.19) and (8.20) become

∇ •E =i

2c

(S∗∂S

∂ t− S

∂S∗

∂ t

)+AtS

∗S (8.22)

∇ ∧B = 1c

∂E∂ t

− i

2(S∗ (∇ + iA)S − S (∇− iA)S∗) (8.23)

which are clearly two of Maxwell’s equations and define the charge density and current. The othertwo of Maxwell’s equations are satisfied automatically.

The charge density is specified to be

ρ(x) =i

2c

(S∗∂S

∂ t− S

∂S∗

∂ t

)+AtS

∗S (8.24)

and the corresponding current density

J(x) = − i2

(S∗ (∇ + iA)S − S (∇− iA)S∗) (8.25)

So this procedure has shown that insisting on local gauge invariance forces an interaction withanother (electromagnetic) field and specifies the equation for this (electromagnetic) field.

Hence local gauge symmetry gives rise to interactions and to electromagnetism.

8.2 Complex Spinor Field & Local U(1) Gauge Invariance 58

8.2 Complex Spinor Field & Local U(1) Gauge Invariance

I shall now apply the same procedure to the case of the complex spinor field ΨR. The Lagrangianfor this field is (from (7.19):

Lψ =i

2

(ΨR∗ ∂

c∂tΨR + ΨR∗σ • ∂

∂xΨR −

(∂

c∂tΨR

)∗ΨR −

(∂

∂xΨR

)∗• σΨR

)(8.26)

If I apply a transformation ΨR(x) → eiθ(x)ΨR(x) the Langrangian changes as

Lψ → Lψ + ΨR†ΨR ∂θ

c∂t+ ΨR†σΨR • ∂θ

∂x(8.27)

and so the transformation is not a symmetry. However the symmetry can be restored if I introducea space-time vector field A (x) that under the local gauge transformation changes as

A (x) → A (x)− ∇θ (x) (8.28)

and that is included in the Lagrangian as

L = Lψ −[ΨR†σΨR •A−ΨR†ΨRAt

]+ LA (8.29)

Again the modification to Lψ can be achieved by changing ∇Ψ to(∇ + iA

)Ψ.

Again LA is chosen so that it remains invariant under the gauge transformation of A (x) and againthe simplest form is as above and again represents the electromagnetic field.

This does however present a problem: Since the field ΨR(x) interacts with the electromagneticfield it must be ‘charged’ and so cannot represent a neutrino! This is only resolved within theWeinberg-Salam model of electro-weak interaction.

The same discussion can be applied to the field ΨL(x).

8.3 Local SU(N) Gauge Invariance

I now consider a more complicated form of gauge invariance. In this case the complex field is acomposite object with N components. This type of gauge invariance was first discussed by Yangand Mills in the case N = 2 and the two components of the field were supposed to represent thefields of a neutron and a proton. However I shall consider the more general case.

The suggestion for the, at first, global gauge invariance comes from the assumption that only themagnitude of the composite field

N∑n=1

S∗nSn (8.30)

is of physical significance.

In this case the Lagrangian is a function of S, S∗, ∇S and ∇S∗ where S is a column vector with Ncomponents and S∗ is a row vector with N components. In this type of global gauge transformationthese quantities change as

S → eiθUSS∗ → e−iθS∗U∗

∇S → eiθ∇US∇S∗ → e−iθ∇S∗U∗

(8.31)


U is a unitary N × N matrix and U∗ is its hermitian conjugate. These matrices have determinantequal to 1 and satisfy the equations

UU∗ = I = U∗U (8.32)

where I is the N ×N unit matrix.

If this is to be a symmetry, these fields must appear in the Lagrangian in such a way that the termsinvolving θ and the matrices U and U∗ cancel.

Strictly the SU(N) gauge transformation is just the part involving the N ×N unitary matrices (thiswhat the notation means) but I have included the U(1) symmetry, discussed above.

If, as before, we promote the transformation to be local θ and the matrices are allowed to be func-tions of x. In this case the change in the variables becomes

S → eiθ(x)U (x) SS∗ → e−iθ(x)S∗U∗ (x)

∇S → eiθU (x)(∇ + i∇θ + U∗∇U

)S

∇S∗ → e−iθ[U (x)

(∇ + i∇θ + U∗∇U

)S]∗ (8.33)

Again it is the terms involving the derivatives that no longer cancel and again I attempt to remedythis by introducing new fields.

In this case the required fields are A, which is introduced as before to eliminate ∇θ, and W which isintroduced to eliminate the matrix U∗∇U. This field W is rather complicated: It is both a space-timevector and an N ×N matrix. In detail there are 4 matrix fields Wt, Wx, Wy and Wz .

These new fields are introduced into the Lagrangian by making the replacements

∇S →(∇ + iA+ iW

)S

∇S∗ →[(

∇ + iA+ iW)

S]∗ (8.34)

Again I choose the change in the new fields under the local gauge transformation to cancel exactlythe changes in S. The required transformation properties are

A (x) → A (x)− ∇θ (x)W → UWU∗ + i

(∇U

)U∗ (8.35)

With the introduction of these new fields the quantities in the Lagrangian change under the localgauge transformation as

S → eiθ(x)U (x) SS∗ → e−iθ(x)S∗U∗ (x)(

∇ + iA+ iW)

S → eiθU(∇ + iA+ iW

)S[(

∇ + iA+ iW)

S]∗→ e−iθ

[(∇ + iA+ iW

)S]∗

U∗

(8.36)

The gradients of θ and U have all been eliminated and the other factors all cancel because of theassumed global symmetry.

Again the resulting Lagrangian can be written as

L = LS + LSA + LSW (8.37)

where LSA describes the interaction with the A field and LSW describes the interaction with the Wfields. Again I need to add the Lagrangians of fields A and W.


LA is exactly as before and is the Lagrangian for the electromagnetic field.

LW has to be chosen in such a way that it is invariant under the change

W → UWU∗ + i(∇U

)U∗

In order to see how to do this I first look at the change, under the local gauge transformation, of thequantities (∇µWν −∇νWµ). This is

(∇µWν −∇νWµ) → U (∇µWν −∇νWµ) U∗

+UWν∇µU∗ − UWµ∇νU∗

+(∇µU) WνU∗ − (∇νU) WµU∗

−i∇µU∇νU∗ + i∇νU∇µU∗

(8.38)

This does not look very hopeful! However if I look at the change in the quantity i (WµWν −WνWµ)I get

i (WµWν −WνWµ) → iU (WµWν −WνWµ) U∗

−UWν∇µU∗ + UWµ∇νU∗

− (∇µU) WνU∗ + (∇νU) WµU∗

+i∇µU∇νU∗ − i∇νU∇µU∗

(8.39)

it is clear that the last three lines of (8.39) are exactly equal and opposite to the last three line of(8.38). Hence the transformation properties of the combination

Wµν = (∇µWν −∇νWµ) + i (WµWν −WνWµ) (8.40)

is much simpler:

(∇µWν −∇νWµ)+i (WµWν −WνWµ) → U [(∇µWν −∇νWµ) + i (WµWν −WνWµ)] U∗

(8.41)Apart from multiplication by the U matrices, it is unaltered.

Again I can form a Lorentz invariant by first forming a dyadic and then taking the dyadic scalarproduct as before. The problem of the the pre- and post-multiplication by the U matrices can besolved by taking the trace of the resulting matrices. The trace of a matrix is the sum of the diagonalelement. Trace has the following property:

Trace [XYZ] = Trace [YZX] = Trace [ZXY] (8.42)

for any (square) matrices X, Y and Z. Hence, applying this to the current situation,

Trace [UYU∗] = Trace [YU∗U] = Trace [Y] (8.43)

The resulting Lagrangians for the Gauge fields are

LA = −14˜A • ˜A = −1

4∑µν

gµAµνAνµgν

LW = − 18Trace

[˜W • ˜W] = − 18

∑µνTrace [gµWµνWνµgν ]

(8.44)

In SU(N) the matrix field W can be represented by(N2 − 1

)independent (space-time) vector fields.

There are two particularly important cases.

SU(2): Here the matrix field is represented by three vector fields. These fields lead to the threefields responsible for the weak interaction (W+,W−, Z).

SU(3): Here the matrix field is represented by eight vector fields. These fields lead to the eightfields responsible for the strong interaction – the gluons.

Chapter 9

EXAMPLES OF SYMMETRIES

In this section I am going to take the general results of section 3 and apply them to the space-timetransformations and gauge transformations discussed in section 4.

9.1 Type I spinor ΨR.

I shall take the type I spinor field (without electromagnetic interactions) as an example and derive,in detail, the conservation laws associated with gauge invariance and with space-time translationsymmetry.

The Lagrangian in this case is

L =i

2

(ΨR† ∂ΨR

c∂ t− ∂ΨR†

c∂ tΨR + ΨR†σ •∇ΨR −

(∇ΨR

)† • σ ΨR

)(9.1)

9.1.1 Gauge invariance

The Lagrangian is clearly invariant under the (global) gauge tranformation

ΨR → eiθΨR ΨR† → e−iθΨR† (9.2)

If I consider only an infinitesimal transformation then the relevant changes are

∆x = 0 ∆t = 0 ∆ΨR = iδθΨR ∆ΨR† = −iδθΨR† (9.3)

and so the equations for the densities are for the time- and space components:

Jt =∂L

∂

(∂ΨR

c ∂ t

) iΨR +∂L

∂

(∂ΨR†

c ∂ t

) (−iΨR†)= ΨR†ΨR

(9.4)

J =∂L

∂ (∇ΨR)iΨR +

∂L

∂ (∇ΨR†)(−iΨR†)

= ΨR†σ ΨR(9.5)

The conserved quantity is the space-integral of (9.4):

∫ΨR†ΨR dV (9.6)

61


9.1.2 Space-time translation symmetry

In an infinitesimal space-time translation. The co-ordinate transformation is

The transformation of the fields is determined by

ΨR′(x + ∆x) = ΨR(x) (9.7)

which gives

δΨR(x) = 0∆ΨR(x) = −δx • ∇ΨR(x)

(9.8)

The first step is to determine if this is a symmetry. In this case all the differentials in the expressionfor δ L are zero. So L is unchanged and the transformation is a symmetry.

The next step is to determine the corresponding conservation laws. It simpler to do this consideringone component of the translation at a time.

• time translation

∆x = 0 ∆ct = cδ t ∆ΨR = −cδ t∂ΨR

c∂ t∆ΨR† = −cδ t∂ΨR†

c∂ t(9.9)

The density is:

Jt = L+∂L

∂

(∂ΨR

c ∂ t

) (−∂ΨR

c∂t

)+

∂L

∂

(∂ΨR†

c ∂ t

) (−∂ΨR†

c∂t

)

= − i2(ΨR†σ •∇ΨR −∇ΨR† • σΨR

) (9.10)

This quantity (times c) is the energy density and the conserved energy is

E = − i2

∫ (ΨR†σ •∇ΨR −∇ΨR† • σΨR

)dV (9.11)

The corresponding energy current density (that is the flow of energy) is:

J =∂L

∂ (∇ΨR)

(−∂Ψ

R

c∂t

)+

∂L

∂ (∇ΨR†)

(−∂Ψ

R†

c∂t

)=i

2

(ΨR†σ

∂ΨR

c∂t−ΨRσ

∂ΨR†

c∂t

) (9.12)

Now consider one of the space-translation components

• space translation in x-direction

∆x = δx ∆y = 0 ∆z = 0 ∆ct = 0 ∆ΨR = −δx∂ΨR

∂x∆ΨR† = −δx∂Ψ

R†

∂x(9.13)

The corresponding density is given by:


Jxt =∂L

∂

(∂ΨR

c ∂ t

) (−∂ΨR

∂x

)+

∂L

∂

(∂ΨR†

c ∂ t

) (−∂ΨR†

∂x

)

=i

2

(ΨR† ∂ΨR

∂x−ΨR ∂ΨR†

∂x

) (9.14)

The corresponding current density components are:

Jxy =∂L

∂

(∂ΨR

∂y

) (−∂ΨR

∂x

)+

∂L

∂

(∂ΨR†

∂y

) (−∂ΨR†

∂x

)

=i

2

(ΨR†σy

∂ΨR

∂x−ΨRσy

∂ΨR†

∂x

) (9.15)

Jxz =∂L

∂

(∂ΨR

∂z

) (−∂ΨR

∂x

)+

∂L

∂

(∂ΨR†

∂z

) (−∂ΨR†

∂x

)

=i

2

(ΨR†σz

∂ΨR

∂x−ΨRσz

∂ΨR†

∂x

) (9.16)

These are all components of what is known as the stress tensor. They correspond to momentumdensity (equation(9.14)) and to the flow of momentum (equations (9.15)-9.16))

Chapter 10

NON-UNIQUENESS OF L AND AR

Another Lagrangian L′, that is a different function of xµ; fj(x); dµfj(x), will lead to exactly thesame field equations if, and only if,

L′ = kL+∑µ

dFµdxµ

(10.1)

k is a constant and Fµ is any function of x and fj(x) but not of dµfj(x). (This can be prove but theproof is not particularly illuminating). Clearly L and L′ are equivalent.

Note that in (10.1) a complete expression for the derivative is

dF (f(x), x)dxµ

=∂ F (f(x), x)

∂ xµ+∂ F (f(x), x)

∂ f

d f(x)d xµ

(10.2)

10.1 Example

Consider the same field S(x, t) as in the example above but with the Lagrangian

L = 12

[ρ(x)

(dS(x, t)d t

)2

− T (x)(dS(x, t)d x

)2

+ a(x)S(x, t)dS(x, t)d t

](10.3)

The Euler-Lagrange equations for this case yield:

a(x)dS(x, t)d t

= ρ(x)d 2S(x, t)d t2

+ a(x)dS(x, t)d t

− d

d x

[T (x)

dS(x, t)d x

](10.4)

The term on the left cancels with a corresponding term on the right and the resulting equations areexactly the same as given by the original Lagrangian.

We should have expected this because the difference in the two Lagrangian densities can be ex-

pressed as a complete derivatived

dt

(14a(x)S

2(x, t)).

64

Chapter 11

HYDROGEN ATOM

Assuming the proton can be described by a simple Dirac spinor I can write the Lagrangian for theelectron-proton-electromagnetic system as

L = −12

(ΨL†

(Pt +

e

~At

)ΨL +

((Pt +

e

~At

)ΨL)†

ΨL

)−1

2

(ΨL†

(P +

e

~A)• σ PsiL +

((P +

e

~A)ΨL)†• σ ΨL

)−1

2

(ΨR†

(Pt +

e

~At

)ΨR +

((Pt +

e

~At

)ΨR)†

ΨR

)−1

2

(ΨR†

(P +

e

~A)• σ ΨR +

((P +

e

~A)ΨR)†• σ ΨR

)−mec

~

(ΨR†ΨL + ΨL†ΨR

)−1

2

(ΦL†

(Pt −

e

~At

)ΦL +

((Pt −

e

~At

)ΦL)†

ΦL

)−1

2

(ΦL†

(P− e

~A)• σ ΦL +

((P− e

~A)ΦL)†• σ ΦL

)−1

2

(ΦR†

(Pt −

e

~At

)ΦR +

((Pt −

e

~At

)ΦR)†

ΦR

)−1

2

(ΦR†

(P− e

~A)• σ ΦR +

((P− e

~A)ΦR)†• σ ΦR

)−mpc

~

(ΦR†ΦL + ΦL†ΦR

)

+12

((∇At +

∂A∂t

)2

− (∇ ∧A) • (∇ ∧A)

)

(11.1)

In this Lagrangian Ψ describes the electron field and Φ describes the proton field.

The resulting Euler-Lagrange equations are:

idΨR

d xt+e

~AtΨR = −iσ •∇ΨR +

e

~σ •AΨR − mec

~ΨL (11.2)

idΨL

d xt+e

~AtΨL = +iσ •∇ΨL − e

~σ •AΨL − mec

~ΨR (11.3)

idΦR

d xt− e

~AtΦR = −iσ •∇ΦR − e

~σ •AΦR − mpc

~ΦL (11.4)

idΦL

d xt− e

~AtΦL = +iσ •∇ΦL +

e

~σ •AΦL − mpc

~ΦR (11.5)

65

ysics ph4a02 lagrangian field theory & symmetry ysics

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