Lecture WithAll

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Advanced Quantum Mechanics

Jonas Rademacker

Michaelmas 2005

mailto:[email protected]:[email protected]


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1 Introduction

1.1 Outline

The aim of this course is to take you from special relativity and quantummechanics, via relativistic quantum mechanics, towards quantum electrody-namics (QED) and the calculation of scattering cross sections with Feynmandiagrams in a “pedestrian way”, without using field theory. As we go along,we’ll cover a lot of material useful for other courses.

You probably have done some relativistic quantum mechanics during yourundergraduate course, while you probably didn’t do much QED. The coursestructure will reflect this. While you are not expected to remember anydetails from your undergraduate course beyond basic quantum mechanics andsome relativity (we won’t skip things just because you should have seen thembefore), the pace might be a bit fast if you have never done any relativisticquantum mechanics before and see everything for the first time. If this is thecase, please let me know. On the other hand, this is not meant to be simplya repetition of what you’ve done before, and I hope it will be interesting evenfor those, who remember every detail from the undergraduate course.

The second part will take you from relativistic quantum mechanics, towardscalculating Feynman diagrams for QED processes, in a pedestrian but math-ematically reasonably sound way, avoiding field theory.

1.2 Books

No single book will cover the entire course. Here I’ll list a few books I usedwhen I prepared the course - by no means do I recommend you buy themall, or in fact any. One of the most useful books that covers a lot of material

from this and other courses is:

• Aitchison& Hey: Gauge Theories in Particle Physics; published in theGraduate Student Series In Physics by the Institute of Physics.

Their first volume takes you from relativistic quantum mechanics to QED,and, from that point of view, covers the subject of this course. The approachto the derivation of Feynman diagrams is however quite different (using fieldtheory). The only books I know of that use a similar approach to ours are:

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• Greiner, Reinhardt: “Quantum Electrodynamics”, published by Springer

- good for step-by-step cross section calculations.

• Bjorken& Drell: “Relativistic Quantum Mechanics”. Same approachas in Greiner and Reinhardt (Bjorken& Drell came first), but fewerexample calculations.

• Richard P. Feynman: “Quantum Electrodynamics”; the one I have waspublished in the “Advanced Book Classics” series of Westview Press.The original.

Our approach to the Dirac equation is covered in

• Lewis H. Ryder: “Quantum Field Theory”; Cambridge University Press.

The bit on classical relativistic electrodynamics will mainly be based on

• Jackson “Classical Electrodynamics”; Wiley.

1.3 Lecture Notes

You’ll find lecture notes (both from this and last year’s course - the new noteswill appear as I write them) on the web: http://www-pnp.physics.ox.ac.uk/∼rademack.I’ll write them as we go along and bring a copy to each lecture, so you won’tneed to print them out yourself.

1.4 Problem Sheets

You’ll find exercise questions within the lecture notes. These will constitutethe problem sheets. They’ll be grouped into 2 problem sets, that we willdiscuss in two classes - one this term, one at the beginning of next term.Please hand your problem sheets in at least a week before the classes, soI have some time to read them. However, there is no reason not to do theexercises as they appear in the notes. I hope the excercises will help you withyour understanding of the lecture, and it is at the point when they appear inthe notes that they will be most useful to you. If you get stuck with them,consult a book, ask your friends, ask me. Do collaborate and help each other.

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http://www-pnp.physics.ox.ac.uk/~rademackhttp://www-pnp.physics.ox.ac.uk/~rademackhttp://www-pnp.physics.ox.ac.uk/~rademackhttp://www-pnp.physics.ox.ac.uk/~rademack


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1.5 Feedback

My email address is: [email protected]. Don’t hesitate to cometo my office (604 A) or write an email if you have a question or a comment.The group is quite small for a lecture - this means that it is comparably easyto take your feedback into account. Take advantage of that!

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mailto:[email protected]:[email protected]


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2 Quick Reminder: Lorentz Transformations

2.1 Notation

We will label our co-ordinates with indices from 0 to 4, where 0 is the timecomponent, and 1-3 the space components

(x0, x1, x2, x3) = (t,x,y,z )

The x without an arrow on top represents the entire 4-vector:

x = (x0, x1, x2, x3)

while x represents the space components only:

x = (x1, x2, x3) = (x,y,z )

Whether a symbol without an arrow, like x or p, refers to a 4-vector or ascalar has to be deduced from the context.

2.2 Invariant 4-interval

We want a constant speed of light.

c2 = (∆x)2 + (∆y)2 + (∆z )2

(∆t)2 = const (1)

Leavingc2(∆t)2 −

(∆x)2 + (∆y)2 + (∆z )2

(2)

invariant clearly fulfils this requirement. Poincaré transformations are de-

fined as those that leave

c2(∆t)2 −

(∆x)2 + (∆y)2 + (∆z )2

(3)

invariant. These transformations form a group, the Poincaré Group. Thesubgroup of Poincaré transformations that exclude translations and thereforeeven leave

c2t2 −

x2 + y2 + z 2

(4)

invariant is the Lorentz Group.

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Rotation Lorentz Transformationx2 + y2 + z 2 =const c2t2 − (x2 + y2 + z 2) =const

(x y z )

xyz

= const (ct − x − y − z )

ctxyz

= const

define g =

1

−1−1

−1

rtr = const = rtr (gr)t r = const

rtr = rtr = rtU tUr rtgr = const = rtgr = rtLtgLr

U tU = 1 LtgL = gdet(U tU ) = det(U )2 = 1 det(LgL) = det(g)

det(L)2 = det(g)2 = 1U has 9 entries L has 16 entriesU tU = 1 ⇒ 6 constraints LtgL = g ⇒ 10 constraints3 independent variables 6 independent variables3 Euler angles α, β, γ 3 Euler angles α, β, γ

3 boost variables (vx, vy, vz)

(5)

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2.4 Lorentz boost in x-direction

Let’s take, for simplicity, a boost in x direction and ignore the y and z coordinates, i.e. our coordinates are (ct x). We require

LtgL = g (6) a cb d

1 00 −1

a bc d

=

1 00 −1

(7)

⇒ a2 − c2 = 1, ab − cd = 0, b2 − d2 = −1 (8)

These relations are fulfilled if we set (we are looking for a single free variable):

L =

cosh(φ) − sinh(φ)− sinh(φ) cosh(φ)

(9)

Note again the similarity to space rotations!

Let us now relate the boost angle φ to a more physically meaningful quantity,the relative speed of the two frames of reference. The origin in frame A shouldshould move with speed −v in the frame A if A moves with speed v in frameA.

ctx = cosh(φ) − sinh(φ)− sinh(φ) cosh(φ) ct0 = cosh(φ)ct− sinh(φ)ct (10)

So we know that a particle moving with speed −v along the x axis is repre-sented by (cosh(φ)ct, − sinh(φ)ct) which should allow us to relate φ and v.Using that speed is distance by time, we get from 10:

v

c = − x

ct = − cosh(φ)− sin(φ) = tanh φ (11)

Let’s define β

≡ bc. Then:

β cosh(φ) = sinh(φ) (12)

using sinh2(φ) − cosh2(φ) ≡ 1

cosh2(φ) = 1

1 − β 2 (13)

sin(φ) = β √

1 − β 2 (14)

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Defining

γ ≡ 1√ 1 − β 2 (15)

we arrive at the familiar expression:

L =

γ −βγ −βγ γ

(16)

or, writing out all coordinates:

L = γ −βγ

−βγ γ

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(17)The boost in an arbitrary direction is a bit more cumbersome. The resultfor a pure Lorentz boost with a velocity v and β ≡ v/c is:

L =

γ −β 1γ −β 2γ −β 3γ −β 1γ 1 + (γ −1)β 1β 1β 2 (γ −1)β 1β 2β 2 (γ −1)β 1β 3β 2−β 2γ (γ −1)β 2β 1β 2 1 + (γ −1)β 2β 2β 2 (γ −1)β 2β 3β 2

−β 3γ

(γ −1)β 3β 1β 2

(γ −1)β 3β 2β 2

1 + (γ −1)β 3β 3β 2

(18)

A general Lorentz transformation is this combined with a rotation and/orreflection. Those transformations form a group, the Lorentz group. Notethat boosts by themselves do not form a group, i.e. two sequential boosts indifferent directions cannot usually be described by a single boost matrix, butwill also have the effect of rotating the co-ordinate system. We will discussthis in more detail later in the course. If we also allow changes of origin(t, x) → (t, x) + (∆t, ∆x), rather than only rotations and boosts, we obtainanother group, the Poincaré group. Such a shift does however not effect“real vectors”, only the co-ordinates themselves (which is a different thing

altogether), so it doesn’t change the matrix describing the transformationof Lorentz vectors. We deal with this subtlety in the lecture on co- andcontravariant vectors.

We will usually restrict ourselves to “proper” Lorentz transformations. ProperLorentz transformations are defined as those that can be obtained by a con-tinuous succession of infinitesimally small iterations, like a rotation, andexclude therefore operations like mirror reflection. They form again a group.Proper Lorentz transformations have det(L) = +1. Improper Lorentz trans-formations can have det(L) = −1 or det(L) = +1.

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2.5 Summary

1. Proper Lorentz transformations leave (cdt)2−(dx2+dy2+dz 2) invariant.A subset of these are rotations, that separately leave ct and (x2+y2+z 2)invariant.

2. Lorentz transformations must satisfy LtgL = g with g = diag(1, −1, −1, −1).3. L has 6 independent variables, which can be taken to be the 3 Euler

angles for the rotation, and three components of the relative velocity.

4. A boost along the x−axis (one parameter) is described by:

L =

cosh(φ) − sinh(φ)

− sinh(φ) cosh(φ)1

1

=


11

(19)

5. Note: In future, we’ll set c = 1. See Appendix A for more details.

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3 Vectors & Tensors

In the remainder of this course, and many others, you will come across vectorsand tensors of rank 0, 1, and 2. Tensors have very nice transformationproperties, such that equations written in tensorial form are automaticallyvalid in all co-ordinate systems. This is what people mean when they sayan equation is co-variant, or more often “obviously covariant” or “manifestlycovariant”. The aim of the following is to make these concepts a bit clearer,in particular the distinction between

• co-ordinates• covariant vectors/tensors• contravariant vectors/tensors

We start with a rather general treatment (which would work for generalrelativity), where these distinctions are more obvious, but move on quicklyto the friendlier, special case of special relativity, where the same differencesstill exist, but the motivation for them is somewhat less obvious.

3.1 Conventions

We will label our co-ordinates and vectors with indices from 0 to 4, where 0is the time component, and 1-3 the space components

(x0, x1, x2, x3) = (t,x,y,z )

Instead of writing the full vector, we’ll write one component, which representsall:

xµ

= (x0

, x1

, x2

, x3

)

Einstein summation convention: Repeated indices imply summation over thisindex (“dummy index”):

xµxµ ≡30

xµxµ

We use the convention that Greek indices run from 0 to 3, while Latin indicesgo from 1 to 3, i.e. only over the spatial part. Note that some books use thisconvention, while others use exactly the opposite.

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3.2 Co and Contravariant Vectors and Tensors

Tensors and vectors are defined by their transformation properties underco-ordinate transformations alone. There are co- and contravariant tensors,and mixed tensors. Vectors are a special kind of tensor, a tensor of rank 1.Contravariant tensors of rank 1, or contravariant vectors, are like the normalvectors we are used to.

We define a contravariant vector (Tensor of rank 1) as an object that behavesunder co-ordinate transformation like this:

Aµ = ∂xµ

∂xν Aν (20)

The classic example for this is the total differential:

dx

µ

=

∂xµ

∂xν dx

ν

(21)

We will see later that contravariant vectors are simply what we would nor-mally consider a vector in space-time, like the 4-momentum. This is becausefor linear co-ordinate transformations, contravariant vectors transform ex-actly like the co-ordinates (see the following aside), i.e. points in space.

3.2.1 Aside: Relation to more familiar notation

Let’s take a Lorentz transformation of the 4-momentum for a boost in x-

direction. The co-ordinate transformation is:x0

x1

x2

x3

=


11

x0

x1

x2

x3

(22)This is a linear transformation, which means that the partial derivatives ∂x

µ

∂xν

are simply the entries in that matrix, Lµν = ∂xµ

∂xν (try it out). So that some-

thing behaves like a vector under this particular transformation is equivalent

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to saying it transforms by multiplying it with this the same Lorentz trans-

formation, like for example the 4-momentum:E

p1

p2

p3

=


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E p1

p2

p3

(23)

Note however that this formula:

Aµ = ∂xµ

∂xν Aν (24)

is much more general and is valid for all co-ordinate transformations, whichincludes those where ∂x

µ

∂xν are not constants (like they are here), but depend

on space and time co-ordinates themselves.

3.3 Co-ordinates and contravariant vectors

From the following example it will be clear that co-ordinates (“position vec-tors”) are not vectors. Possible co-ordinate transformations include shifts:

xµ = xµ + ∆µ

Now for this transformation, ∂xµ

∂xν = δ µν and hence, for a vector according to

our definition, we demandAµ = Aµ

Therefore, position co-ordinates are not vectors! They only behave like vec-tors under linear transformations like the Lorentz transformation used above,i.e. those that can be expressed by a matrix of constant numbers.

3.3.1 Geometric interpretation

Let’s say we live on some manifold, which is some space with a co-ordinatesystem (or more than one). Importantly: it is not a vector space. You mightargue that, by (successfully) identifying 3-vectors with arrows in space, wedo live in a vector space. But this is only true if the space we live in is indeedthe flat Euclidean IR3, or if you include the time dimension, IR3+1. Accordingto General Relativity, space might be curved - in which case it is not usuallypossible to define vectors in an analogous way (you could for example imagine

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drawing arrows along a curved surface and then, when you add them up, find

that they point outside the surface). The point about a manifold is, that itlooks locally like a flat space, which is what we observe in every day life(Euclidean geometry works very well locally), but not necessarily globally.

Let’s take a sphere for visualisation purposes. Of course a sphere is 2 dimen-sional, and we live in a 3+1-dimensional space, but 4 dimensional manifoldsare notoriously difficult to visualise. On this sphere you have co-ordinates,and you might have curves:

xµ(u)

where u is some parameter. The derivative of this is a tangent-vector to that

curve: dxµ

du .

All tangent vectors of all possible curves going through a given point form avector space: the tangent vector space at this given point on the manifold.Every point has one, and it looks exactly like our familiar, flat IRn. Sothis tangent vector space is the mathematical formulation of the statementthat the manifold looks “locally like IRn”. And those tangent vectors in thetangent vector space transform like:

dxµ

du =

∂xµ

∂xν dxν

du

- they are our prototypes for contravariant vectors.

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A manifold with curves, and tangent vector space

Of course we live on the manifold, not the tangent space, but if you stay veryclose to the point where the tangent vector space is defined, the manifoldlooks very much like the vector space and in the limit of infinitesimally smalldistances, they become indistinguishable. And if in fact the manifold is notcurved like a sphere, but flat, like Minkovsky space in special relativity, themanifold and the tangent vector space become the same. And then, we canindeed identify differences between co-ordinates as vectors, and if we makethe further condition that we don’t allow a change of origin (no shifts!),we can indeed talk about co-ordinates as vectors. But these requirements

are quite strict (especially the last one), so usually we have to distinguishco-ordinates and vectors.

3.4 Tensors of Rank 2 and higher

A contravariant tensor of rank 2 transforms like:

Aµν = ∂xµ

∂xα∂xν

∂xβ Aαβ (25)

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... of rank 3 transforms like:

Aµνλ = ∂xµ

∂xα∂xν

∂xβ ∂xλ

∂xγ Aαβγ (26)

etc.

An example of a tensor of rank 2 is the combination of two vectors:

Aµν = X µY ν .

3.5 Tensors of Rank 0 (scalars)

A contravariant tensor of rank 0 is simply a number, which is the same in allcoordinate system:

Φ = Φ (27)

3.6 Covariant vectors / tensors

3.6.1 Covariant vectors /tensors of rank 1

Covariant vectors live in a different vector space than contravariant vectors,and their task is to act on them. When a covariant vector acts on a con-travariant vector, they produce an invariant scalar. The important point hereis, that the same covariant vector and contravariant vector always combineto produce the same number, irrespective of the coordinate system used.

A covariant vector transforms like this:

A

µ =

∂xα

∂xµ Aα (28)

Note that covariant vectors have lower indices, while contravariant vectorshave upper indices. Also note that this does not imply that everything withupper or lower indices is a tensor!

Exercise 1. Show that, with this definition, Aν Bν is invariant under co-

ordinate transformations if A and B are tensors. Hint: Use the chain rule of differentiation and the fact that ∂x

µ

∂xν = δ µν . Write down this (short) calculation

twice: Once using Einstein’s summation convention, once using

symbols.

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3.6.2 Covariant tensors of rank 2, ..

Similarly, a covariant tensor of rank 2 transforms like this:

X µν = ∂xα

∂xµ∂xβ

∂xν X αβ (29)

etc.

3.6.3 Mixed tensors of rank 2

Tensors can be of mixed type, for example a mixed tensor of type (1,1)transforms like:

X ν µ = ∂xα

∂xµ∂xν

∂xβ X β α

3.7 Contracting Tensors

The operation of combining a covariant and a contravariant tensor of rank 1(vectors) to a tensor of rank 0, that we met above, is an example of contract-

ing a tensor, that is letting one tensor act on another to produce a tensorof lower rank. The important point here is that such an operation alwaysresults in a new tensor, i.e. an object with the well-defined transformationproperties of a tensor.

Example: Contracting a contravariant tensor of rank 2 with a covarianttensor of rank 1 to a tensor of rank 1:

Aν = BµC µν (30)

Example: Contracting a contravariant tensor of rank 5 with a covariant

tensor of rank 2 to a tensor of rank 3:

K µνσ = Lαβ M αβµνσ (31)

In the above examples we always ended up with contravariant tensors. Youcan do the same thing resulting in co-variant tensors (just swap upper withlower indices in the above example) and mixed tenors. In all cases, it isalways a covariant (lower) index that goes with a contravariant (upper) one.

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This means in practise:

YOU SHOULD ONLY ENCOUNTER REPEATED INDICES WHERE ONE IS AN UPPER, AND ONE IS A LOWER INDEX . If you encountertwo repeated lower indices, or two repeated upper indices, probably some-thing has gone wrong! (Unless no summation is implied, or you are reading abook by an author who does not distinguish between upper and lower indices,which you find frequently especially in older textbooks.)

Exercise 2. Show that K µνσ in equation 31 is a contravariant tensor of rank3 if L and M are tensors.

3.8 1st Summary of Tensor Operations

The trick with tensor operations is that if you put in tensors, you get tensorsout. So when you write an equation in tensorial form, using the operationssummarised below, you can be sure that the left hand and the right handside of your equation transform in the same way, i.e. your equation is validin all co-ordinate systems (covariant). This will become very important lateron, and is the reason we are going through all this pain here.

Adding two tensors gives a new tensor of the same rank:C µν...αβ... = A

µν...αβ... + B

µν...αβ... (32)

For example, if A and B are vectors:

C µ = Aµ + Bµ (33)

You get a tensor of higher rank by multiplying two tenors:

C µν...αβ... = Aµ...

β... Bν...

α... (34)

For example, if A and B are vectors:

C µν = AµBν (35)

Contracting two tensors also gives you a tensor:

Aν = BµC µν (36)

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3.9 Tensors and Derivatives

Now that we can add, subtract and multiply, we’d like to differentiate in atensorial form. We will use the following notation:

∂

∂xα ≡ ∂ α (37)

The partial derivative ∂ α behaves like a covariant vector when applied to ascalar field. But it doesn’t behave like a covariant vector when applied toa vector or tensor, i.e. ∂ β X

α is not a tensor (of type (1,1)), in general (see

exercises).

It is possible to define a quantity Γασβ , which is not a tensor, such that thequantity Dβ defined to act on on a vector field like this

Dβ X α = ∂ β X

α + Γασβ X σ (38)

is a tensor. This is called the covariant derivative. Equation 38 defines the

covariant derivative for vectors, acting on scalars it is defined to be

Dβ φ ≡ ∂ β φ (39)

The concept of a covariant derivative will become important in a differ-ent context, later, which is the main reason for bringing it up here. How-ever, we will soon move towards considering only Lorentz transformations inMinkovsky space, rather than general co-ordinate transformations on arbi-trary manifolds. In this special case, Γασβ can be shown to vanish, and we cantreat ∂ α as if it were a covariant vector. The connection between differenttypes of co-ordinate transformations and covariant derivatives is a conceptthat we will also meet again.

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Exercise 3.

• Show that ∂ β φ is a covariant tensor, if φ is a scalar (strictly: a scalarfield φ(x)).

• Show that ∂ β X α, is not a tensor, where X α = X α(x) is a tensor field.• Define the covariant derivative as in equation 38. How does Γασβ trans-

form in order for Dβ X α to be a tensor?

• There are co-ordinate systems where Γασβ = 0. In such a co-ordinatesystem, let’s call it S, Dβ = ∂ β . Show that Dβ = ∂ β in all systems

related to S by “global” co-ordinate transformation that do not differat different points in space-time (for example Lorentz transformations).

It can be shown that Γασβ = 0 if the metric gµν (you probably already knowwhat a metric is, but we’ll define it later) is constant, i.e. ∂ σgµν = 0. Thisis the case in Minkovsky space-time. This means that in the near future,when we restrict ourselves to Lorentz transformations in Minkovsky space,we can forget about all this and treat ∂ β as a covariant vector, because insuch co-ordinate systems it happens to be identical to Dβ .

3.10 The metric, scalar products, distances

The metric defines how we measure distances (hence its name)

ds2 = gµν dxµdxν (40)

(the differential line element, squared), and define the length of vectors:

X 2 = gµν X µX ν (41)

and the scalar product between two vectors:

X · Y = gµν X µY ν (42)which we require to be independent of the co-ordinate system, which impliesthat the metric is a covariant tensor of rank 2. We’ll only consider cases inwhich the metric has an inverse. This inverse is gµν :

gµν gνλ = δ λµ (43)

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From these definitions it follows that gµν is a covariant tensor of rank 2 and

gνλ is a contravariant tensor of rank 2.

3.11 Raising and lowering indices, relating co- and con-travariant vectors

So far, co- and contravariant vectors lived in completely different spaces.We viewed contravariant vectors as normal vectors, and covariant vectors asobjects that act on contravariant vectors to produce a number (always thesame number independent of the co-ordinate system). The metric relatesthose two spaces. The object

gµν X µ

in the equationgµν X

µY ν = a (44)

is (by the definition of gµν ) an covariant vector that acts on the contravariantvector Y ν and produces a number, that is the same in all co-ordinate systems.We therefore have for each contravariant vector X µ a corresponding covariantvector, gµν X

µ, which we will call X µ:

gµν X ν ≡ X µ (45)

The contravariant metric takes us back:

gνµX µ = gνµgµσX

σ = δ ν σX σ = X ν (46)

We can use the metric to transform contravariant vectors to covariant vectors,and back. There is a true one-to-one relationship between the two vectorspaces. For many practical purposes, we can consider them the same. InEuclidean space they certainly are the same. There the line-element is ds2 =dx2 + dy2 + dz 2, so the metric in is simply the identity matrix.

Also note that

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gµν X µY ν = gµν X µY ν = δ

µν X µY

ν = X · Y (47) justifying the notation X · Y for the scalar product without specifying whichis the co- and which the contravariant vector - it doesn’t matter, it’s alwaysthe same scalar product.

What the metric can do for vectors, it can also do for tensors, i.e. it canraise an lower indices of tensors of any rank, for example:

gµν T ανβ = T α β µ (48)

Note that now we can raise and lower indices, the order of the indices matters:

T α β µ = T αβ µ (49)

3.11.1 The metric and the covariant derivative

Coming back to the covariant derivative of a vector field:

Dβ X α = ∂ β X

α + Γασβ X σ (50)

If a metric is defined on the manifold, it can be shown that the followingdefinition for Γαβγ has the correct transformation properties that you derivedin the previous exercise:

Γαβγ = 1

2gασ (∂ β gσγ + ∂ γ gσβ − ∂ σgβγ ) (51)

This is a good way to check that your result was correct. This result showsthat if you have a metric that is constant in space and time, then Dβ = ∂ β .

3.12 Summary

• We defined co- and contravariant tensors by their transformation prop-erties.

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• The most important ones are tensors of rank 1, called vectors. Con-

travariant vectors transform like

Aµ = ∂xµ

∂xν Aν

and covariant vectors transform like

Aµ = ∂xα

∂xµAα

Tensors of higher ranks are generalisations of these. One way to re-member the transformation rules is to memorise that the prototype for

a contravariant vector is dxα (total differential), and covariant vectoris ∂ αφ (partial derivative acting on a scalar field). Then use the chainrule of differentiation.

• Tensors of rank 0 are scalars. They are the same in all co-ordinatesystems. Contracting a co- with a contravariant vector gives a scalar:

AµBµ = φ = AµBµ

• Vectors are not co-ordinates.

• Using tensors we can write equations that are indeed manifestly covari-ant.

• Among the operations defined for tensors are addition, contraction,external product, and the covariant derivative Dµ.

• In some special, but important cases, the covariant derivative is thesame as the simple partial derivative ∂ α. In the following this willalways be the case.

• The metric allows us to define distance, length and a scalar product.

Note that these quantities are scalars, and therefore invariant underco-ordinate transformations, e.g. ds2 = gµν dx

µdxν is the same in allco-ordinate systems.

• With an invertible metric, we can create a one-to-one relationship be-tween co- and contravariant vectors/tensors. Vectors/tensors in thetwo spaces become different representations of what is essentially thesame object. The metric lets us move freely between the spaces, byraising and lowering indices.

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4 Special Relativity

Source (mainly): Jackson, Classical Electrodynamics Now we move from general co-ordinate transformations to the special case of Lorentz transformation. We will first define them, and then investigate thetransformation properties of some important vectors. This will be a good mo-ment to remind ourselves that we are physicists, not mathematicians. Thereis nothing wrong with defining objects called tensors by their transformationproperties. But we will have to find those objects in nature, where trans-formation properties are ultimately defined by what we measure in differentframes of reference.

4.1 The Metric in Special Relativity

The requirement of a constant speed of light leads to:

dt2 − dx2 − dy2 − dz 2 = Lorentz-invariant (52)Above we defined the invariant differential interval by

ds2

= gµν dxµ

dxν

= Lorentz-invariant (53)

Defining our metric such that

ds2 = dt2 − dx2 − dy2 − dz 2 (54)therefore fulfils our requirements. This means that our metric is

gµν =

1 0 0 00 −1 0 00 0 −1 00 0 0

−1

(55)

It is easy to show that the requirement that

gµν gνλ = δ λµ

yield to:

gµν =

1 0 0 00 −1 0 00 0 −1 00 0 0 −1

(56)

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Because we want

ds2 = dt2 − dx2 − dy2 − dz 2in all frames of reference, Lorentz transformations are those transformationsLαµ =

∂xα

∂xµ that leave this metric invariant:

Lαµgαβ Lβ ν = gµν with gµν = diag(1, −1, −1, −1). (57)

The same in matrix notation:

LtgL = g (58)

Some properties of Lorentz transformations are a direct consequence of thisis that:

det(LgL) = det (g) (59)

det(L)2 det (g) = det (g) (60)

det(L) = ±1 (61)

The matrix form of a Lorentz boost, and more, can be found in the quicksummary on Lorentz transformation (page 4). Let’s now see how physicalquantities behave under Lorentz transformations.

4.2 Invariants and 4 vectors

The “prototype” contravariant vector:

dxµ = (dt,dx,dy,dz ) (62)

The “prototype” covariant vector is the covariant derivative, which coincideswith the normal partial derivative for flat Minkovsky space where Γ αβγ van-ishes:

Dµ = ∂ µ = ∂

∂t, ∂

∂x, ∂

∂y, ∂

∂z (63)

Each of them can have its index lowered or raised, making it a co- or con-travariant vector respectively:

dxµ = (dt, −dx, −dy, −dz ) (64)

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∂ µ =

∂

∂t, − ∂

∂x, − ∂

∂y, − ∂

∂z

(65)

In the following we will look for physical quantities that transform either as4-vectors, as scalars (Lorentz-invariants), or as tensors of rank 2. Of course,each 4-vector X µ we find immediately gives us a Lorentz-invariant scalarX · X .

Transformation of Lorentz 4-vectors

A co- and a contravariant vector with the metric g=diag(1,-1,-1,-1), and howthey transform.

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4-interval Obviously

ds2 = dt2 − dx2 − dy2 − dz 2

and, for finite differences

(∆s)2 = (t2 − t1)2 − | x2 − x1|2

are invariant - the theory is set up that way. If we exclude changes of origin,then

s2 = t2 − |x|2

is also invariant.

Proper time Consider a particle moving along with velocity u, i.e dx =udt. Clearly neither u nor t are Lorentz invariant. But

ds2 = dt2 − |dx|2 = dt2

1 − β 2

is, where β = u/c with u the speed of the particle. Going to the referenceframe where the particle is at rest, i.e. dx = 0 we find another invariant:

dτ = dt 1 − β 2 = dt

γ u (66)

the proper time of the particle, which is the same as the time of the particle atrest. Time intervals observed from a moving co-ordinate system are stretchedby a factor of γ relative to the time in the restframe - the famous timedilatation.

4-dimensional volume element is invariant because of |det(L)| = 1:d4x =

|det(L)

|dx4 = dx4 (67)

Together with the time dilatation this implies that the 3-volume contractsby a factor of 1/γ .

Velocity The derivative with respect to time of the 4-position clearly doesnot transform like a 4 vector (restoring c for a moment):

d

dt

ct

x

=

dt

dtd

dt

ctx

= L d

dt

ctx

(68)

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because dtdt

= 1. This follows directly from the time dilatation earlier, but

let’s nevertheless prove it quickly for boost in the x direction. In the boostedframe, the time component is given by

t = γt − βγ xDifferentiating both sides with respect to t, and remembering that

d

dt =

dt

dtd

dt =

dt

dt

∂

∂t +

dx

dt

∂

∂x +

dy

dt

∂

∂y +

dz

dt

∂

∂z

we finddt

dt =

1

γ 1 + βγ dxdt = 1.However, the derivative with respect to the proper time (which is a Lorentzscalar, so d

dτ = d

dτ ) does transform like a Lorentz vector:

d

dτ

ct

x

=

d

dτ

ct

x

=

d

dτ L

ct

x

d

dτ

ct

x

= L d

dτ

ctx

(69)

with d

dτ = γ d

dt (70)

Hence:

cγ vγ vv

= L

cγ vγ vv

(71)

So the 4 velocity

U = cγ vγ vv (72)is a Lorentz vector. The corresponding invariant is

√ U 2 = c.

Energy and Momentum Multiplying the 4-velocity with the mass of aparticle gives another, more important 4-vector:

p ≡ mU =

γ vmcγ vmv

(73)

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In the small velocity limit (γ

≈1 + 1

2β 2) we get:

p0 = mc

1 − v2/c2= mc +

1

2mv2/c + O

v4

c4

(74)

p = mv

1 − v2/c2≈ mv + O

v2

c2

(75)

(76)

We see that, in the small velocity limit, to first order (restoring c for once),cp0 = (mc

2 + 12

mv2) which is a constant plus the non-relativistic kinetic en-ergy, and p is simply the classical momentum. So it makes sense to identifycp0 = γmc

2 as the relativistic energy and p = γ vmv as the relativistic mo-mentum. However, the most important property of momentum and energyare their conservation laws. They apply, but this has not been shown here.See Jackson, Sect 11.5, or, for a different approach, section 5.6.1 (page 35).

From the Lorentz invariance of the 4-momentum, we get the important rela-tionship between energy, restmass and momentum of a particle:

p20 − p · p = m2 (77)

⇒E =

p2 + m2 (78)

(The equivalent in non-relativistic mechanics is E non−relativistic = p2

2m.)

5 Electrodynamics

5.1 Maxwell’s equations

inhomogeneous:

∇ · E = ρ (79) ∇ × B − ∂

E ∂t

= (80)

homogeneous:

∇ · B = 0 (81) ∇ × E + ∂

B∂t

= 0 (82)

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5.2 Scalar and Vector Potential

The homogeneous Maxwell equations are automatically fulfilled if we definethe scalar potential Φ and the vector potential A such that1

E = − ∇Φ − ∂ A

∂t (83)

B = − ∇ × A (84)

In terms Φ and A, Maxwell’s equations are:

∇

2

Φ + ∂ t ∇ · A = −ρ (85) ∇2 A − ∂ 2t A − ∇

∇ · A + ∂ tΦ

= − (86)

Equations 83 and 84 do not not determine A and Φ uniquely, we have somegauge freedom: simultaneously transforming A and Φ according to:

Φ → Φ − ∂ ∂t χ A → A + ∇ · χ (87)

where χ is sufficiently differentiable scalar function, results in the same phys-ical fields E and B. We come to that in more detail later, for the time beingwe use this gauge freedom to demand the following relationship between Aand Φ:

∂ Φ

∂t + ∇ · A = 0 (88)

which will turn out to be a convenient choice. Maxwell’s equation in terms

of the scalar and vector potentials are, with the Lorentz gauge:

∂ 2 Φ

∂t2 − ∇2Φ = ρ (89)

∂ 2 A

∂t2 − ∇2 A = (90)

1Don’t get confused by the usage of the terms scalar and vector in this section, wherewe are dealing dealing with 3-vectors: Φ is not a scalar under Lorentz transformations,neither is the 3-vector A a Lorentz vector.

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5.3 Maxwell’s equation in covariant form

Historically, Maxwell’s equations were first, then came special relativity, ex-actly because it was found experimentally, that they are invariant underLorentz transformations. The task is now, to make this Lorentz invarianceexplicit, i.e. we want to formulate Maxwell’s equations in terms of tensors(mostly vectors).

To start with, consider the continuity equation, which expresses the conser-vation of charge:

∂ρ

∂t + ∇ · = 0 (91)

where ρ is the charge density and the current density.

The invariance of charge under Lorentz transformations (which we take fromexperiment) implies

ρd3x = const (92)

so ρ transforms like dx0, the 4th component of a 4-vector. This implies that

ργ

= const (93)

If we multiply the 4-velocity by the invariant ρ/γ , we get another 4-vector,which we will interpret as the 4-dimensional current, the 0 th component of which is the charge density:

ρ

γ U =

ρ

γ

γ vγ vv

=

ρρv

=

ρ

(94)

So we define the contravariant 4-vector

jα

= ρ (95)And can now write the continuity equation 91 in (now we have it) obviouslyco-variant form:

∂ α jα = 0 (96)

Now we define the 4-vector potential Aα as

Aα =

Φ

A

(97)

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Maxwell’s equation, in terms of Aα, with the Lorentz gauge, are given by:

∂ µ∂ µAα = j α (98)

If we take it as an experimental fact that the Maxwell equations are Lorentzinvariant, then the above equation 98 implies that Aα is a 4-vector (up to

some gauge transformation - Maxwell’s equations are for E and B), since ∂ µ∂ µis a scalar, and on the right-hand-side is a 4-vector. We now realise that wechose a sensible, because Lorentz invariant, gauge condition 88, which nowreads

∂ αAα = 0 (99)

For completeness, we note that Maxwell’s equations in terms of Aα

withoutany particular choice of gauge are given by:

∂ µ∂ µAα − ∂ α (∂ µAµ) = j α (100)

To recover the E and B from this in a covariant form, we define the followingtensor:

F αβ ≡ ∂ αAβ − ∂ β Aα (101)This tensor is obviously antisymmetric

F αβ = −F βα (102)So it has (42 − 4)/2 = 6 degrees of freedom, which turn out to be the com-ponents of the E field and the B field:

F αβ =

0 −E x −E y −E zE x 0 −Bz ByE y Bz 0 −BxE z

−By Bx 0

(103)

In terms of F αβ the inhomogeneous Maxwell equations 80 can be written as

∂ αF αβ = j α (104)

and the homogeneous Maxwell equations 82

∂ αF βγ + ∂ β F γα + ∂ γ F αβ = 0 (105)

(for a more elegant version of the homogeneous equations: Jackson page 551).

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5.4 The Lorentz Force

The electromagnetic force acting on a moving particle with charge q is:

d p

dt = q

E + v × B

(106)

The first step towards re-writing this in obviously Lorentz covariant, i.e.tensorial form, is to re-write it in terms of the (invariant) proper time, ratherthan the changeable time, i.e. we use

d p

dt =

dτ

dt

d p

dτ =

1

γ

d p

dτ (107)

Rembering that the 4-velocity is given by U α = (γ, γv), we can write 106 as

d p

dτ = q

U 0 E + U × B

(108)

Where the left hand side is now the space part of a 4-vector. The correspond-ing time-part describes the rate of change of energy of the particle movingtrough an electric field (see Jackson, eq 6.110, page 237 and page 551, eq11.144 for details):

d p

0

dτ = q U · E (109)Using the definitions above, we can write this in obviously covariant form:

d pα

dτ = m

dU α

dτ = qF αβ U β (110)

5.5 Lagrangian Formalism

5.5.1 None-relativistic

We define an action S,

S =

t2 t1

dtL (q 1, q 2, ..., q̇ 1, q̇ 2, t) (111)

and demand that the variation of S be stationary

δS = 0 (112)

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Which yields to the Euler Lagrange equations for L:

∂L

∂q i− d

dt

∂L

∂ q̇ i= 0 (113)

The canonical momenta associated with the generalised co-ordinates q i are

πi = ∂L

∂ q̇ i(114)

The Hamiltonian is:H =

i

πi q̇ i − L (q ν , q̇ ν , t) (115)

The Hamiltonian equations

π̇i = −∂H ∂q i

, q̇ i = ∂H

∂πi(116)

are equivalent to 113.

5.5.2 Non-relativistic, charged particle in e/m field

The following Lagrangian leads to the equation of motion for a charged par-ticle in an electromagnetic field 106.

L = T − V (117)T =

p2

2m (118)

V = q Φ − q A · v (119)L =

p2

2m − q Φ + q A · v (120)

The conjugate momentum in Cartesian co-ordinates is, for this Lagrangian:

πi = ∂L∂vi

= mvi + qAi (121)

π = mv + q A (122)

We get the following Hamiltonian:

H =

a

πa q̇ a − L

= 1

2m

π − q A

2+ q Φ (123)

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Note that this can be obtained from the free Hamiltonian

H free = π2

2m (124)

by what is called “minimal substitution”:

E → E − q Φπ

→ π

−q A (125)

Hence

H = 1

2mπ2 (no e/m field) (126)

−→ H − q Φ = 12m

π − q A

2(with e/m field) (127)

Where π is defined by πi = dLdvi

, so it is different for the free Hamiltonian andthe one with an electromagnetic field.

5.6 Relativistic Lagrangian

We want an invariant Lagrangian, so we re-write the action terms of aninvariant integration variable, the proper time, and an invariant L:

S =

τ 2 τ 1

L

xµ,

dxµ

dτ , τ

dτ (128)

Where dxµ

dτ is the 4-velocity. So all we did is substitute t with the Lorentz

invariant variable τ . The Euler Lagrange equations become:

∂L

∂xµ − d

dτ

∂L

∂

dxµ

dτ

= 0 (129)

The relativistic Lagrangian for a free particle is:

Lfree = −m

gαβ dxα

dτ

dxβ

dτ (130)

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Adding in the effect of an external electromagnetic field Aµ:

L = −m

gαβ dxα

dτ

dxβ

dτ − q dx

α

dτ Aα (131)

There are various ways of making this appear like a sensible choice, see forexample Jackson. The ultimate proof of course is that we get the correctequations of motion out.

Exercise 4. Use either the non-relativistic Lagrangian 120 or the relativisticone 131 to get the equations of motion of a particle in an external e/m field,106, or 110

The conjugate momenta for the Lagrangian in equation 131 are

πα = − ∂L∂

dxαdτ

= mU α + qAα (132)where the minus sign has been introduced to make this conform with thenon-relativistic case.

The Hamiltonian is:

H = 1

2 (π

αU α + L) (133)

= 1

2m (πα − qAα) (πα − qAα) − 1

2m (134)

Hamilton’s equations:

dxα

dτ =

∂H

∂πα=

1

m (πα − qAα)

dπα

dτ = − ∂H

∂xα=

q

m (πβ − qAβ ) ∂ αAβ (135)

5.6.1 Aside: Lagrangian & Geodesics

This aside gives some insight in the geometry of the action principle, and its

relation to a more general understanding of the metric as a length measure in

configuration space. It also motivates the form of the free relativistic Lagrangian

130 . But it is a bit too far off our course towards QED to be considered in the

lecture.

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A general Lagrangian (without velocity-dependent potential, i.e. no e/m)

can be written as:

L = T − V (136)T = gαβ ẋ

αẋβ (137)

V = V (x) (138)

L = gαβ ẋαẋβ − V (x) (139)

where x are now generalised co-ordinates. Note that the x are not necessarilypositions, and that gαβ is not the space-time metric, but a metric in n-dimensional configuration space. The equation of motion resulting from this

is:ẍ + Γαβγ ẋ

β ẋγ = F α (140)

where F α are the generalised forces,

F α = − ∂V ∂xα

(141)

If the potential vanish, the action integral to be minimised, is:

S =

t2

t1 gαβ ẋαẋβ dt (142)

It can be shown, that this leads to the same equations of motion as

S =

t2 t1

gαβ ẋαẋβ dt =

t2 t1

ds

dtdt =

t2 t1

ds (143)

so requiring δS = 0 means we we are looking for the shortest (or longest)path between two points in configuration space, the geodesic.

For a single particle with mass m in 3 dimensions (non-relativistic), we find

gαβ = gαβ =

m 0 00 m 00 0 m

= m 1 0 00 1 0

0 0 1

(144)and the free-particle equations of motion are the familiar d

dt(mẋa) = 0. This

result is a particle moving in a straight line, which is geodesic in Euclideanspace.

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In 4-dimensional Minkovsky space, we simply generalise this concept of par-

ticles moving on geodesics, and define the following action:

S = −ατ 2

τ 1

ds (145)

where τ is the proper time α is a constant to be determined such that inthe limiting case v/c 1, we recover the non-relativistic Lagrangian. Forthis comparison, we replace the relativistic integration variable s with theclassical time t:

ds = dt√

1

−v2 (146)

In terms of this parameter, the action can be written as:

S = −αt2

t1

√ 1 − v2dt (147)

So our Lagrangian (not a Lorentz scalar if we integrate over t) is

L = −α√

1 − v2 ≈ −α + 12

αv2 + · · · (148)

To make the energy term 1

2αv2

compatible with the non-relativistic La-grangian, we therefore identify α with the rest mass of the particle. Sothe fully covariant action is now:

S = −mτ 2

τ 1

ds =

gαβ ẋαẋβ dτ (149)

From which we get for the momenta (while L was not a Lorentz scalar, Lis, because it is the Lagrangian defined by the action integral over dτ ):

pα = −∂L

∂ ẋα = mU α (150)

So the momenta of this Lagrangian are in fact the relavistic Energy andmomentum, and, as in non-relativistic mechanics, translational symmetry int and xa result in the conservation of the relativistic energy and momentum.

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6 Wave Equations

6.1 Non-relativistic

6.1.1 No Spin: Schrödinger Equation

Let’s remind ourselves of the Schrödinger equation:

Ĥ Ψ(x, t) = i∂ tΨ(x, t)

ˆ p2

2m + V (x)Ψ(x, t) = i∂ tΨ(x, t)− ∇22m

+ V (x)

Ψ(x, t) = i∂ tΨ(x, t) (151)Making the Ansatz

Ψ(x, t) = ψ(x) · θ(t)we can separate the equation in a time-dependent and time-independent part.The time-dependent equation is

i∂ tθ(t) = Eθ(t) (152)

where E is an energy eigenvalue of the Hamiltonian Ĥ . We get

θ(t) = e−iEt (153)

The time-independent part is the eigenvalue equation for Ĥ :− ∇22m

+ V (x)

ψ(x, t) = Eψ(x, t) (154)

6.1.2 Schrödinger Equation with Electromagnetism

We apply our result from the classical theory to the Hamiltonian in theSchrödinger Equation. The recipe (that works for both, the non-relativisticand relativistic Hamiltonian) was: Take the free equation with V (x) = 0 andthen substitute:

π → π̂ − q AE → E − φ (155)

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In this context it is important to remember that in quantum mechanics, the

momentum is defined as “that what is conjugate to the position”, and theoperator ∇ means the conjugate momentum to the Cartesian co-ordinates,whatever that is, not necessarily the linear momentum. In the classical case,in the presence of a magnetic field, the momentum π as defined by

πi = ∂L

∂ q̇ ichanged like this in the presence of a magnetic field:

π = mv −→ π = mv + q AOur operator relation is between

∇ and π, not between

∇ and mv:

∇ ↔ π = ∂L∂ ̇x

.

So the Schrödinger equation with an electromagnetic field is:

1

2m

−i ∇ − q A

2ψ = (E − eφ)ψ

1

2m

−i ∇ − q A

2+ eφ

ψ = Eψ (156)

6.1.3 Spin in a non-relativistic equation: The Pauli Equation

Experimentally, particles have spin. The spin projection of fermions can havetwo values with respect to any given direction in space (we’ll take z ): +1

2, −1

2

(spin up and spin down). We can take this into account by changing froma single-valued scalar wave function to a two-component spinor, where theupper component represents spin up, the lower spin down:

Ψ =

ψ↑ψ↓

(157)

Now the task is to write down a Schr ödinger-type equation for this two-component spinor. We shall demand that it becomes the free Schrödingerequation in the absence of magnetic fields. We’ll use an ad-hoc approach, bycombining the spinless Schrödinger equation with the 2 × 2 Pauli matrices:

σx ≡ σ1 =

0 11 0

σy ≡ σ2 =

0 −ii 0

σz ≡ σ3 =

1 00 −1

(158)

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The most important properties of the Pauli matrices are their commutation

and anti-commutation relations. The commutator is defined as:

[A, B] = AB − BA (159)

The anticommutator:{A, B} = AB + BA (160)

For the Pauli matrices:1

2σx,

1

2σy

=

1

2σz and cyclic (161)

{σi, σ j} = 2δ ij 11 (162)where δ ij = 1 if i = j and 0 otherwise. We recognise the commutation rela-tions 161 as those of angular momentum, and clearly, in our representation,the operator for the spin in z -direction is 1

2σz. In general

S = 1

2σ =

1

2

σxσyσz

(163)where we introduced the somewhat awkward notation of a vector of matrices,

that we’ll use just as any other vector. For example the dot product withsome arbitrary vector p, which can have numbers or matrices as entries, isdefined as:

p · σ = pxσx + pyσy + pzσz (164)where the result will be a matrix.

Now consider the following equation for a free particle:

1

2m

σ ·−i ∇2 ψ↑

ψ↓

= E

ψ↑ψ↓

(165)

Let us look at the factor

σ · P 2 where P could be any vector. Writing thisout, we get:

σ · P 2

= (P xσx + P yσy + P zσz)2

=

P z P x − iP y

P x + iP y −P z

·

P z P x − iP yP x + iP y −P z

(166)

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If the different components of P commute, it is easy to show that

σ · P 2

=

P z P x − iP y

P x + iP y −P z

·

P z P x − iP yP x + iP y −P z

=

P 2 0

0 P 2

(167)

The components of ∇ do commute, so for the free equation 165, we simplyrecover the Schrödinger equation, or in fact a rather redundant combinationof two identical, independent Schrödinger equations:

1

2m σ · −i ∇2 ψ↑ψ↓ = E ψ↑ψ↓ ⇔ 1

2m

− ∇2 00 − ∇2

ψ↑ψ↓

= E

ψ↑ψ↓

(168)

Which is exactly what is required of course, because so far the effect of spinshould not be noticeable since we haven’t got anything for the spin to interactwith. Let us now switch on the electro magnetic field:

1

2m σ ·−i ∇ − q A

2

ψ↑ψ↓

= (E − qφ)

ψ↑ψ↓

(169)

Now components of P do not commute anymore, andσ · P 2 = P 2 11

Instead, for P = −i ∇ − q A, we getσ ·−i ∇ − q A

2=−i ∇ − q A

211 − qσ ·

∇ × A

(170)

So the Pauli equation is:

12m−i ∇ − q A2 11 − q

2mσ · B + qφ11 ψ↑ψ↓ = E ψ↑ψ↓ (171)

The energy coming from the interaction of the spin with the field is:

∆E spin = − q 2m

σ · B = − 2q 2m

S · B with S = 12

σ (172)

In general, the energy of a particle with magnetic moment µ in a magneticfield B is

∆E = µ · B

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with

µ = gq 2m S

where S is the angular momentum or spin of the particle. For classicalangular momenta (and orbital angular moment in QM), g = 1. Comparingthe above expressions for ∆E and µ, we come to the famous result that forthe spin of a particle

g = 2 (173)

This result is normally associated with the relativistic Dirac equation. Herewe did not exactly derive it, but we obtained it in a “natural” way, by makingthe simplest possible Ansatz and carrying on from there. Note in particular

that, had we chosen any other factor in front of the σ in equation 165, forexample 1

2, we would not have recovered the Schrödinger equation for the

case without a magnetic field.

We shall keep the notion of spinors from this section in mind until we get tothe Dirac equation. But first we’ll go to the relativistic wave equation thatdescribes spinless particles, the Klein-Gordon equation.

6.2 Relativistic Wave Equations

6.2.1 The Klein Gordon Equation

The relativistic expression for the energy in terms of the momentum is:

E =

p2 + m2 (174)

Translating this into quantum mechanical operators leaves us with the diffi-culty of defining the square-root of an operator. We can avoid this difficultyby squaring equation 174:

E 2 = p2 + m2 (175)

Squares of operators are well-defined. By substituting

p → −i ∇E → i ∂

∂t (176)

we get the following wave equation, known as the Klein-Gordon equation fora free particle.

− ∇2 + m2

ψ = − ∂ 2

∂t2Ψ (177)

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which is better expressed in 4-vector notation as∂ µ∂

µ + m2

ψ = 0 (178)

This is now, for once, manifestly covariant.

In the small velocity limit, where E kin = p2

2m, we recover the free-particle,

spinless Schrödinger equation:

− 12m

∇2ψ = i ∂ ∂t

ψ (179)

So we can interpret the Schrödinger Equation as the non-relativistic ap-

proximation to the Klein-Gordon Equation. The probability density for theSchrödinger Equation is

ρ = ψ∗ψ (180)

And the probability current:

= − i2m

ψ∗ ∇ψ − ψ ∇ψ∗

(181)

They obey the continuity equation

∂ρ

∂t

+

∇ · = 0 (182)

Which can be shown by using the Schrödinger Equation and its complexconjugate.

For the Klein-Gordon Equation, we can derive a similar expression, wherenow we don’t want the probability density to transform as a Lorentz scalar,as in equation 180 but, like the charge density in classical electrodynam-ics, as the 0thcomponent of a 4-vector whose spatial components are givenby in equation 181. Now we know that the time component correspond-ing to ∂ a = − ∂

∂xa is ∂

∂t (note that we want to interpret the in 181 as a

contravariant vector, therefore the upper index and the sign change, since∂ α = ∂

∂t, − ∂

∂x, − ∂

∂y, − ∂

∂z), so a ρ that would satisfy our requirement is ob-

tained by replacing − ∇ in 180 with ∂ ∂t

:

ρ = i

2m

ψ∗

∂

∂tψ − ψ ∂

∂tψ∗

(183)

We now have the following 4-vector current:

jµ = i

2m (ψ∗∂ µψ − ψ∂ µψ∗) (184)

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which obeys the usual continuity equation:

∂ µ jµ =

i

2m ((∂ µψ

∗) ∂ µψ + ψ∗∂ µ∂ µψ − (∂ µψ) ∂ µψ∗ − ψ∂ µ∂ µψ∗)

= i

2m (ψ∗∂ µ∂

µψ − ψ∂ µ∂ µψ∗)

= i

2m

ψ∗m2ψ − ψm2ψ∗

= 0 (185)

While this seems to behave like the probability current and density that wewant, there is a problem, and that is the possibility of negative probability

densities. This is a consequence of the fact that the Klein-Gordon Equationis 2nd order in time, which means that we can fix arbitrary initial conditionsfor both ψ and ∂ψ

∂t , so there is nothing preventing us from choosing one

where ρ = i2m

ψ∗ ∂

∂tψ − ψ ∂

∂tψ∗

is negative. It turns out that this problemcan be over-come by re-interpreting jµ as a charge, rather than probabilitycurrent. Charged particles are then represented by complex ψ = ψ∗, whileneutral particle are represented by real ψ = ψ∗. It is obvious that for real ψ ,equation 184 leads to jµ = 0.

The other problem is that the equation we based the Klein-Gordon Equationon

E 2 = p2 + m2

allows positive as well as negative solutions for E :

E = ±

p2 + m2 (186)

This is not so problematic for a free particle, whose energy is constant anyway,but is very problematic for interacting particles that can exchange energywith their environment. There is nothing from stopping the particle fromemitting an infinite amount of energy while cascading down to an infinitelylow energy state. In field theory, these problems do go away. We howeverwill resolve this in a different way, which we’ll postpone until later.

For now, we move towards spin 12

particles.

6.3 The Dirac Equation

We all know how to arrive at the Dirac equation by making the simplestpossible Ansatz for a wave equation that is 1st order in time (thus avoid-ing the problem with the negative probability densities), and that, when

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squared, fulfils the relativistic momentum-energy relation, i.e. the Klein-

Gordon Equation, which must be fulfilled by any relativistic wave equation.

We will go a different route though (as in Lewis H. Ryder, “Quantum FieldTheory”), which is somewhat less straight forward and requires some prepa-ration, but turns out to be very insightful.

Our strategy is to have a look at Lorentz transformations from a grouptheory point of view. We know the representation of the Lorentz group ina 4-dimensional real vector space. We’ll look for one in a two-dimensionalcomplex spinor space. From that we’ll derive the transformation propertiesof spinors. We’ll find that there are two inequivalent representations of theproper Lorentz group in the 2-dimensional spinor space, that are related bythe parity transformation. This will lead us to considering “double”, i.e.4-component spinors. At the end the Dirac equation will pop out.

Proper Lorentz transformations are characterised by 6 variables, which wecan take to be three velocity components for the boost, and 3 Euler anglesfor rotations. We’ll start our work by looking at rotations, only.

6.3.1 Rotations, O(3)

In three real dimensions, rotations are characterised by preserving distances: x

y

z

= R xy

z

with x2 + y2 + z 2 = x2 + y2 + z 2 (187)or

r = Rr with rtr = rtr

(Rr)t Rr = rtr

rtRtRr = rtrRtR = 11 (188)

matrices with this property are called orthogonal, and they form a group(two rotations form a rotation, there is a unit element, there is an inverse foreach rotation) called O(3), the orthogonal group in 3 dimensions. It is easyto show that the above condition leads to

det(R) = ±1 (189)

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If we additionally demand

det(R) = +1 (190)

we exclude mirror reflections and allow only real rotations. This group is thespecial orthogonal group in 3 dimensions, SO(3).

Rotations can be parametrised by three parameters, since we have 3 × 3entries in the matrix, and the orthogonality condition imposes 6 constraints.All possible rotation matrices can be obtained by sequentially performingrotations by some angle about the x, the y, and the z axis. We’ll use thoseangles to parametrise rotations. The 3 rotation matrices are:

Rx(φ) = 1 0 00 cos φ sin φ

0 − sin φ cos φ

Ry(ψ) =

cos ψ 0 − sin ψ0 1 0sin ψ 0 cos ψ

Rz(θ) =

cos θ sin θ 0− sin θ cos θ 00 0 1

(191)

Those rotation matrices do not commute: Rx(φ)Ry(ψ) = Ry(ψ)Rx(φ), soO(3) is a non-Abelian group. It is also a Lie group. This means it isa continuous group, with an infinite number of elements (because the an-gles can have an infinite number of values). It is parametrised by a finitenumber of parameters, though. Here it’s three. The good thing aboutLie groups is, that we get most of the information about a Lie group byconsidering its generators, only. There is one generator for each parame-ter. So instead of an infinite number of elements, we only need to look atthree - much better! The generators are the differentials associated with theparameters, multiplied with −i to get Hermitian matrices out. They are:

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J x = 1

i

dRx(φ)

dφ

φ=0

=

0 0 00 0 −i0 i 0

J y = 1

i

dRy(ψ)

dψ

ψ=0

=

0 0 i0 0 0

−i 0 0

J z =

1

i

dRz(θ)

dθ

θ=0

=

0 −i 0i 0 00 0 0

(192)Infinitesimal rotations are given by:

Rz(δθ) = 11 + iJ zδθ (193)

The generators obey the following commutation relations:

[J x, J y] = iJ z and cyclic (194)

which can be written in a more compact form as

εabcJ aJ b = iJ c (195)

We recognise these as the commutation relations for angular momentum.The angular momentum operators are the generators of rotations. Here itcomes in handy that those generators are Hermitian, so that they can be QMobservables.

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We can recover finite rotations by performing an infinite number of infinitely

small infinitesimal rotations:

Rz(θ) = limN →∞

1 + iJ z

θ

N

N = eiJ zθ (196)

which can be easily verified to be the same as in 191.Exercise 5. Easily verify this.

A general rotation around an axis n through and angle θ is given by

Rz(θ) = ei J ·nθ = ei

J · θ (197)

6.3.2 SU(2)

The special unitary group is the group of complex 2 × 2 matrices with theproperties

U U † = 1 , det(U ) = 1 (198)

Unitary matrices have 4 complex = 8 real parameters, and 2(4) constraintfrom UU † = 1. So there are two complex or 4 real parameters. A generalunitary matrix can be written like this:

U = a b−b∗ a∗ , |a|2 + |b|2 = 1 (199)So we have 2 complex, 4 real numbers left, and one more condition, whichgives us (again) three free real parameters, and we’ll expect three generators.

Our strategy is now to relate SU (2) transformations to O(3) transformations.We’ll proceed by considering the transformation properties of the matrix

h = r · σ =

z x − iyx + iy −z

(200)

(where σ are again the Pauli matrices) under SU (2) transformations. Thiswill relate our 3-dimensional position vector (x,y,z ) to SU (2) transforma-tions. This matrix is Hermitian and traceless and any Hermitian and tracelessmatrix can be parametrised that way. The determinant of the above matrixis

det(h) = −

x2 + y2 + z 2

(201)

Unitary transformations will preserve all three of these properties: Hermitic-ity, tracelessness, and the determinant, so under a unitary transformation:

h = U hU † (202)

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we end up with a matrix

h =

z x − iy

x + iy −z

with x2 + y2 + z 2 = x2 + y2 + z 2 (203)

so the effect of SU (2) on a 2 × 2 Hermitian, traceless matrix is exactly thatof a O(3) on a 3-vector.

Let us now relate matrices like h to spinors, and their transformation prop-erties. We consider spinors, and their Hermitian conjugate:

ξ = ξ 1

ξ 2 and ξ † = (ξ ∗1 , ξ ∗2) (204)They transform like

ξ = Uξ

ξ † = ξ †U † (205)

The inner product:

ξ †ξ = (ξ ∗1 , ξ ∗2)

ξ 1ξ 2

= |ξ 1|2 + |ξ 2|2 (206)

is obviously invariant since UU † = 11. The outer product

ξξ † =

ξ 1ξ 2

(ξ ∗1 , ξ

∗2) =

|ξ 1|2 ξ 1ξ ∗2ξ 2ξ

∗1 |ξ 2|2

(207)

transforms likeξ ξ † = U ξξ †U † (208)

So there we have our spinors now in matrix form. We want however relate thetransformation properties of one spinor (and not a spinor and its Hermitianconjugate) to the transformation properties of 3-vectors, via a hermitian andtraceless matrix. For that purpose, observe that, due to the unitarity of U ,

ξ 1ξ 2

and

ξ ∗2−ξ ∗1

transform in the same way. Using equation 199:

ξ 1 = aξ 1 + bξ 2

ξ 2 = −b∗ξ 1 + a∗ξ 2 (209)

Complex conjugating equations 209, swapping the lines, multiplying one of them with −1, and a bit of re-ordering, yields:

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ξ ∗2 = aξ ∗2 + b(−ξ ∗1)

−ξ ∗1 = −b∗ξ ∗2 + a∗(−ξ ∗1) (210)

which proves the claim. Hence, ξ ∗2−ξ ∗1

transforms like ξ =

ξ 1ξ 2

(ξ 2, −ξ 1) transforms like ξ † = (ξ ∗1 , ξ ∗2) (211)

Therefore, when we now form the outer product of (ξ 2, −ξ 1) with ξ 1ξ 2 :

H = (ξ 2, −ξ 1)

ξ 1ξ 2

=

ξ 1ξ 2 −ξ 21

ξ 22 −ξ 1ξ 2

(212)

We know it transforms like ξ †ξ , i.e.

H = U HU † (213)

H is also traceless. We summarise

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• The transformation properties of a Hermitian, traceless 2

×2 matrix

under under SU (2) can be identified with those of a 3-vector underO(3).

h =

z x − iy

x + iy −z

• The transformation properties of a 2-spinor under SU (2) can also beidentified with those of traceless 2 × 2 matrix

H =

ξ 1ξ 2 −ξ 21

ξ 22 −ξ 1ξ 2

We conclude, identifying h and H

SU (2) transformation on

ξ 1ξ 2

≡ O(3) transformation on

xyz

with

x = 1

2 ξ 22

−ξ 21 , y =

1

2i ξ 21 + ξ

22 , z = ξ 1ξ 2 (214)

We note in passing that by identifying h with H , we put some restrictionon the possible ξ , −ξ 21 = ξ 2∗2 , which we wouldn’t put on spinors that aresupposed to describe our wave functions. This is less of a problem than itmight seem at first sight. We are not really interested in identifying spinors(with 4 real parameters) with 3-vectors (with 3 real parameters). We are

interested in finding a relation (if possible 1-to-1, we will see below that wenearly get that, but not quite) between the group objects that act on thespinors and 3-vector, which are the matrices U ∈ SU (2) and R ∈ O(3). Theabove relation between certain types of spinors, and 3-vectors, shows that wecan indeed identify matrices in O(3) with matrices in SU (2) by their identicaleffect on x, y,z . After we’ve done that, we can let our U ∈ SU (2) matricesact on whatever spinor we like.

We will now use the relationships found above, to identify members of SU (2)with members of O(3). Let’s write down the transformation of a spinor under

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SU (2) again:

ξ 1 = aξ 1 + bξ 2

ξ 2 = −b∗ξ 1 + a∗ξ 2 (215)

Squaring both lines, and taking the product between them:

ξ 21 = a2ξ 21 + b

2ξ 22 + 2abξ 1ξ 2

ξ 22 = b∗2ξ 21 + a

∗2ξ 22 − 2a∗b∗ξ 1ξ 2ξ 1ξ

2 = −ab∗ξ 21 + a∗bξ 22 + |a|2 ξ 1ξ 2 − |a|2 ξ 1ξ 2 (216)

Solving equations 214 for ξ 21 , ξ

22 , ξ 1ξ 2 gives

ξ 21 = −x + iyξ 22 = x + iy

ξ 1ξ 2 = z (217)

Putting this into equation 216, yields:

x = 1

2

a2 + a∗2 −

b2 + b∗2

x− i

2

a2 − a∗2 +

b2 − b∗2

y − (a∗b∗ + ab) z

y = i

2 a2 − a∗2 − b2 − b∗2x + 1

2 a2 + a∗2 + b2 + b∗2 y − i (ab− a∗b∗) zz = (ab∗ + a∗b)x + i (a∗b− ab∗) y +

|a|2 − |b|2

z (218)

which we can write as

x = Re (a2 − b2) x + Im (a2 + b2) y − 2Re (ab) z y = −Im (a2 − b2) x + Re (a2 + b2) y + 2Im (ab) z z = 2Re (a∗b) x − 2Im (a∗b) y +

|a|2 − |b|2

z

(219)

Now our condition on a and b is, that |a|2 + |b|2 = 1. There are of coursemany possible choices, but to make things easier for us, we would like to find

the correspondence between our basic rotation matrices Rx, Ry, Rz, definedin 191. Let’s start with Rz, i.e. we want z

= z . This is achieved by choosingb = 0 and a = eiθ/2, where we put in a factor of 1

2 in the exponent for

convenience, anticipating that a will be squared. We get:

a = eiθ/2

b = 0 ⇒

x = x cos θ + y sin θy = −x sin θ + y cos θz = z

(220)

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Therefore we get the following correspondence between U

∈SU (2) (equation

199) and R ∈ O(3):

U =

eiθ/2 0

0 eiθ/2

↔ Rz(θ) =

cos θ sin θ 0− sin θ cos θ 00 0 1

(221)In terms of generators, we get:

U = eiσzθ/2 ↔ R = eiJ zθ (222)

Similarly for Ry with a = cos ψ2

, b = sin ψ2

:

U =

cos ψ/2 sin ψ/2− sin ψ/2 cos ψ/2

↔ Ry(ψ) =

cos ψ 0 − sin ψ0 1 0sin ψ 0 cos ψ

(223)U = eiσyψ/2 ↔ R = eiJ yψ (224)

and Rx with a = cos φ2

, b = i sin φ2

:

U =

cos φ/2 i sin φ/2i sin φ/2 cos φ/2

↔ Rx(φ) =

1 0 00 cos φ sin φ

0 − sin φ cos φ

(225)

U = eiσxφ/2 ↔ R = eiJ xφ (226)

So, in general, we get:

SU (2) U = eiσ· θ/2 ↔ R = ei J · θ ∈ O(3) (227)

We might expect, since there is a mapping between the two groups, thegenerators of the group obey the same commutation relations. In fact, thefamiliar Pauli matrices σx,y,z obey the following commutation relations:

1

2σx,

1

2σy

= i

1

2σz and cyclic (228)

Which are the same as for J, which we will repeat here:

[J x, J y] = iJ z and cyclic (229)

The factors 12

before the Pauli matrices in 228, and 227, show that the spinorrotates through half the angle the vector rotates through. This means that

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we did not quite get a one-to-one mapping between SU (2) and O(3) (then

they would be identical), but we got pretty close:There is a two-to-one mapping of the elements of S U (2) to those of O(3).

In particular, increasing the rotation angle by α = 2π maps an O(3) matrixonto itself: R(θ) = R(θ + 2π), while it maps U to −U : U (θ) = −U (θ + 2π).We will have to increase the angle by 4π to come back to the original element.Graphically, this can be illustrated by:

The 2 → 1 mapping between SU(2) and O(3)Now we have the correspondence between rotations in three real dimensions,and two complex ones, we extend this to the full Lorentz group.

6.3.3 SL(2,C) and the Lorentz group

We’ve done the rotations, now we do the boosts. There will be three basicboost matrices according to a boost in x, y, and z direction. We start witha boost in z direction:

Lx(χ) =

cosh χ sinh χ 0 0sinh χ cosh χ 0 0

0 0 1 00 0 0 1

=

cos iχ − sin iχ 0 0− sin iχ cos iχ 0 0

0 0 1 00 0 0 1

(230)

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The corresponding generator is:

K x = 1

i

∂Lx∂χ

χ=0

= −i

0 1 0 01 0 0 00 0 0 00 0 0 0

(231)

Similarly for y :

K y = −i

0 0 1 00 0 0 01 0 0 00 0 0 0

(232)

... and for z :

K z = −i

0 0 0 10 0 0 00 0 0 01 0 0 0

(233)We get all 6 generators of the Lorentz group, if we combine the above withthe 3 generators for rotations, this time written out in 4 dimensions:

J x = −i0 0 0 00 0 0 00 0 0 10 0 −1 0

(234)

J y = −i

0 0 0 00 0 0 −10 0 0 00 1 0 0

(235)

J z =

−i

0 0 0 00 0 −1 00 1 0 00 0 0 0

(236)We can calculate the commutation relations between all 6 generators, whichyields:

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[K x, K y] = −iJ z and cyclic[J x, J y] = iJ z and cyclic

[J x, K x] = 0 etc

[J x, K y] = iK z and cyclic (237)

The first commutation relation means thatPure Lorentz boosts do not form a group .

So if you boost a system say first with speed vx, then vy, then −vx and then−vy, for example, you end up with a rotated system.Regarding the two-dimensional complex equivalents of the of K x,y,z, we nowtake an educated guess, inspired by the suggestive form of the Lorentz trans-formation looking nearly like a rotation by a complex angle. We guess:

K 2,C = ±i 12

σ (238)

It is easy to show that both, +i σ2

and

−i σ2

satisfy the commutation relations

237. (Note that this is not trivial. For example in the case of rotations, only12

σ satisfy the angular momentum commutation relations, not −12

σ.)

That there are two sets of generators in spinor space that satisfy the commu-tation relations in 237, suggests that there are two types of spinors, behavingdifferently under Lorentz transformations.

From the K and J , we can form two sets of generators, that each have a

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closed algebra:

A ≡ 12

(J + iK )

B ≡ 12

(J − iK ) (239)

These don’t mix anymore, and their respective commutation relations arethose of the angular momentum:

[Ax, Ay] = iAz and cyclic

[Bx, By] = iBz and cyclic

[Ai, B j] = 0 for all i, j

(240)

So A and B each generate the group SU (2). This means, that the Lorentzgroup is essentially SU (2)⊗SU (2). A spinor state can therefore be labelled bytwo (abstract equivalents of) angular momenta, ( j, j), the first corresponding

to A, the second to B. Two special cases are those where one of them is zero:

( j, 0) → K ( j) = −i J ( j) B = 0(0, j)

→ K ( j) = i J ( j) A = 0

(241)

which corresponds to the two possibilities K = ±i σ2

. Clearly, in our 2 di-mensional spinor space, the angular momentum is 1

2. For ( j, 0) states:

Az = 1

2 (1 + (−i)i) 1

2σz =

12

00 −1

2

(242)

For (0, j) states:

Bz = 12 (1 − (+i)i) 12σz = 1

2 00 −12

(243)We can now define two types of spinors. Those that transform like ( j =12

, j = 0) and those that transform like ( j = 0, j = 12

). We call them ξ andη:

(12

, 0) = ξ : J = σ2

, K = −i σ2

(0, 12

) = η : J = σ2

, K = i σ2

(244)

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ξ -type spinors transforms like this under Lorentz transformations:

ξ = expiσ2 · θ − iσ

2 · χ ξ

= exp

i

σ

2 · θ + σ

2 · χ

ξ

= exp

i

σ

2 ·

θ − i χ

ξ

≡ Mξ (245)

η type spinors like this:

η = expiσ2 · θ + iσ2 · χ η= exp

i

σ

2 · θ − σ

2 · χ

η

= exp

i

σ

2 ·

θ + i χ

η

≡ Nη (246)

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Note that M (θ, χ) and N (θ, χ) are inequivalent representations of the Lorentz

group, i.e. you can’t get from one to the other by a simple basis transforma-tion, or, mathematically, there is no matrix S such that N = S MS −1. Theyare however related by

N = ζ M ∗ζ −1 with ζ = −iσ2 (247)

The matrices N and M are complex 2 × 2 matrices, with unit determinant.These form a group, called S L(2, C ). This group has 4 complex parameters,and the determinant corresponds to one complex and two real conditions:M 11M 22

−M 12M 21 = 1 (in the S U (2) case the determinant contributed only

one condition, because we U U † = 1 implies that |det(U )| = 1, and hencedet(U ) = 1 just fixes a phase). This has therefore 6 real parameters, asexpected. These six parameters can be related to three real angles and threeLorentz boosts, in two inequivalent ways.

Summary

• We looked for a representation of the Lorentz group for spinors.

• Using the correspondence between the groups O(3) and SU (2), we

found the behaviour of spinors under rotations.

• Adding in Lorentz boosts, we found that there are two inequivalentrepresentations of the Lorentz group in spinor space, ( 1

2, 0), (0, 1

2).

• So there are two different types of 2-component spinors out there, trans-forming according to different representations of the Lorentz group in2 complex dimensions, SL(2, C ).

6.3.4 Parity, 4-spinors

Let us now relate those two types of spinors. The operations that relatesthem, is parity. So far we neglected parity, because we treated Lorentz trans-formations as a Lie group, allowing therefore only transformations that canbe built up out of infinitesimal elements, like rotations and boosts. Undersuch transformations, as we have seen, ξ and η type spinors behave inde-pendently. Under parity however, they transform into one another: Underparity, velocities (boosts) change sign: v → −v. Hence the corresponding

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generators K change sign. Angular momenta however rem

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