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Relativistic

Contents

Review of and Relativity

Klein Gordon Equation

Feynman Stueckelburg Interpretation

An tiparticles

Fermion

variant notation Co

Massless fermions

Learning Outcomes

Be able to derive KG equation and explain physical meaning of ve E solutions

i

Be able to derive Dirac eqn and algebra for

Be able to solve the free Dirac equation and interpret the solutions in terms of

and spin

Be able to use the covariant notation

Reference Bo oks

Halzen and Martin

many other QM b o oks cover this material

In this section of the course we will study relativistic quantum mechanics In a

for example the interactions of highly ener particle accelerator we are interested in

getic so the need to combine relativity and quantum mechanics is pressing

Some remarkable results will come out of this synthesis In particular we will theo

retically predict the existence of antiparticles and also fermion spin Sit back and

enjo y

NonRelativistic Equation Review

Lets review how wave equations describ e nonrelativistic quantum particles Exp er

imentally we learnt that a particle with denite p and energy E could b e

asso ciated with a plane wave

p E

i(k xw t)

e k w

h h

To extract E and p from the wave we use operators

d

E ih p ih r

dt

2

Classically we know that E p m V so we arrive at the Schro edinger equation

2

d h

2

ih r V

dt m

Equally we can write the Schro edinger equation quite generally with h as

ih H

t

where H is the Hamiltonian ie the energy op erator

Density

dx as the probability that the particle will b e found Rememb er that we interpret

b etween x and x x Probability is a conserved quantity there is always probability

one that the particle will b e somewhere

This means that if the probability of the particle b eing in some area decreases

w then the probability that it lies outside must increase In other words there is a o

of probability satisfying the usual conservation equation cf electric

charge

I Z ρ Z

q = d V

dV d J A

t

S

R R

Using the rA dV we have the continuity equa AdS

tion

rJ

t

wherein J is the probability densitycurrent Now we can show using the Schro edinger

equation that satises such a relation We add two copies of the Schro edinger

equation as follows

i SE SE i

This gives

 2

i h

2

h h r i V

t t 2m

2

ih

2

r i V

2m

and hence

ih

r r r

t m

which indeed has the form of a conservation equation with

Relativity Review

In relativity an event is describ ed by the four co ordinates of a fourvector

x x ct

Under Lorentz transformations LT it transforms a familiar example of a LT is a

b o ost along the z axis for which

B C

B C

B C

A

2 2 1

LTs can b e thought of as generalized with as usual v c and

rotations

x is then a contravariant vector since it transforms as

x x x

f g denote Lorentz indices and the summation The Greek lab els

convention is used

2 2 2

The length of the vector c t jxj is invariant to LTs In general we dene

the Minkowski scalar pro duct of two vectors x and y as

x y x y g x y

where the metric

if

g g g diag g

if

has b een intro duced The last step in is nothing but the denition of a covariant

vector sometimes referred to as covector

g x x

To formulate a coherent relativistic theory of dynamics we dene kinematic vari

ables that are also vectors ie transform as describ ed ab ove For example we dene

a velo city

dx

u

d

where is the proper time measured by a clo ck moving with the particle Everyone

will agree as to what the clo ck says at a particular event so this measure of time is

Lorentz invariant and u transforms as x Note

dx dt

c v u

d dt

and has invariant length

2 2 2 2

u u c jv j c

Similarly momentum provides a relativistic denition of energy and momentum

p mu E c p

The invariant length gives us the crucial relation

2 2 2 2 2

p p E c jpj m c

Note that is a covariant vector

x

x

i i 0

so r and r

I will use in this course This rstly means redening the unit of

distance so that c Secondly I shall redene the unit of energy so that E h

ie seth So energy inverse length and inverse time all have the same

Generally think of energy E as the basic unit eg mass m has units of

1

GeV and distance x has unit GeV

The Klein Gordon Equation

For a free the total energy E is no longer given by the equation we

used to derive the Schro edinger equation Instead it is given by the Einstein equation

2 2 2

E p m

In p osition space we write the energymomentum op erator as

ir p i E p i

t

so that the KG equation b ecomes

2

2 m x

where we have intro duced the b ox notation

2 2 2

2 t r

and x is the vector t x

The KleinGordon equation has plane wave solutions

i(E tpx)

x N e

p

2 2

where N is a normalization constant and E p m

There are two problems with this equation though Firstly there are b oth p ositive

and solutions The negative energy solutions p ose a severe problem

if you try to interpret as a as indeed we are trying to do The

sp ectrum is no longer b ounded from b elow and you can extract arbitrarily large

amounts of energy from the system by driving it into ever more negative energy

states Any external p erturbation capable of pushing a particle across the energy

gap of m b etween the p ositive and negative energy continuum of states can uncover

y Furthermore we cannot just throw away these solutions as unphysical this dicult

since they app ear as Fourier mo des in any realistic solution of

A second problem with the wave function interpretation arises when trying to

2

nd a probability density Since is Lorentz invariant jj do es not transform like a

density so we will not have a Lorentz covariant r J

t

To search for a candidate we derive such a continuity equation Start with the

KleinGordon equation multiplied by and subtract the complex conjugate of the

KG equation multiplied by Dening and J by

i

t t

J i r r

you obtain a covariant conservation equation

J

where J is the vector J It is thus natural to interpret as a probability density

2

and J as a However for a plane wave solution jN j E

so is not p ositive denite since weve already found E can b e negative

Feynman Stueckelb erg Interpretation

We have found that the KleinGordon equation a candidate for describing the quan

tum mechanics of spinless particles admits unacceptable negative energy states when

is interpreted as the single particle wave function There is another way forward

this is the way followed in the textb o ok of Halzen Martin due to Stueckelb erg

and Feynman Causality forces us to ensure that p ositive energy states propagate for

wards in time But if we force the negative energy states only to propagate backwards

in time then we nd a theory that is consistent with the requirements of causality

and that has none of the aforementioned problems In fact the negative energy states

cause us problems only so long as we think of them as real physical states propagat

ing forwards in time Therefore we should interpret the emission absorption of a

negative energy particle with momentum p as the absorption emission of a p ositive

energy with momentum p

+

In order to get more familiar with this picture consider a pro cess with a and a

+

in the initial state and nal state In gure a the starts from the p oint

+

A and at a later time t emits a photon at the p oint x If the energy of the is

1 1

still p ositive it travels on forwards in time and eventually will absorb the initial state

photon at t at the p oint x The nal state is then again a photon and a p ositive

2 2

+

energy

There is another pro cess however with the same initial and nal state shown in

+

gure b Again the starts from the p oint A and at a later time t emits a

1

photon at the p oint x But this time the energy of the photon emitted is bigger

1

+ +

than the energy of the initial Thus the energy of the b ecomes negative and it

time t it absorbs the initial is forced to travel backwards in time Then at an earlier

2

state photon at the p oint x thereby rendering its energy p ositive again From there

2

it travels forward in time and the nal state is the same as in gure a namely a

+

photon and a p ositive energy

B B

time (t 1, x 1) (t 2, x 2) (t 1, x 1) (t 2, x 2)

A A

(a) (b)

Figure Interpretation of negative energy states

In to days language the pro cess in gure b would b e describ ed as follows in

+

the initial state we have an and a photon At time t and at the p oint x the

2 2

+ +

photon creates a pair Both propagate forwards in time The ends up in

the nal state whereas the is annihilated at a later time t at the p oint x by

1 1

+

the initial state thereby pro ducing the nal state photon To someone observing

in real time the negative energy state moving backwards in time lo oks to all intents

and purp oses like a negatively charged with p ositive energy moving forwards in

time We have discovered antimatter

Dirac Equation

Historically the Klein Gordon equation was b elieved to b e sick although now we

understand it is telling us ab out antiparticles To try to solve the problem of negative

energy solutions Dirac wanted an equation rst order in time derivatives His starting

p oint was to assume a Hamiltonian of the form

P H P P m

1 2 2 3 D 1 3

where P are the three comp onents of the momentum op erator p and and are

i i

some unknown constant quantities which as will b e seen b elow cannot simply b e

commuting numb ers

We should write the momentum op erators explicitly in terms of their dieren tial

op erators using equation Then the Dirac equation b ecomes using the Dirac

Hamiltonian in equation

i r m i

t

which is the p osition space Dirac equation

If is to describ e a it must though satisfy the KleinGordon equation

so that it has the correct energymomentum relation This requirement imp oses

relationships among and To see these apply the op erator on each side

1 2 3

of equation twice ie iterate the equation

2

i j i j i i i 2 2

r r i mr m

2

t

from to The KleinGordon equation by with an implicit sum over i and j

comparison is

2

i i 2

r r m

2

t

i

If we do not assume that the and commute then the KG will b e satised if

i j j i ij

i i

2

for i j The and cannot b e ordinary numb ers but it is p ossible to give

i

them a realization as matrices In this case must b e a multicomp onent on

which these matrices act

In two dimensions a natural set of matrices for the would b e the

i

3 2 1

i

However there is no other indep endent with the right prop erties for

so the smallest numb er of dimensions for which the Dirac matrices can b e realized is

four One choice is the Dirac representation

Note that each entry ab ove denotes a twobytwo blo ck and that the denotes the

There is a theorem due to Pauli that states that all sets of matrices ob eying the

y y

and clearly relations in are equivalent Since the hermitian conjugates

ob ey the relations you can by a change of basis if necessary assume that and

are hermitian All the common choices of basis have this prop erty Furthermore we

would like and to b e hermitian so that the Dirac Hamiltonian is hermitian

i

Equation Continuity

If we dene

0 y y

J J

then it is a simple exercise using the Dirac equation to show that this satises the

E DE and rearrange it continuity equation J Simply add D

Note that is now p ositive denite this seemed like a ma jor achievement to

Dirac

Solutions to the Dirac Equation

The wave function in the Dirac equation is a four comp onent vector To shed light

on what this means lets lo ok at free particle solutions

We lo ok for plane wave solutions of the form

p

i(E tpx)

e

p

where p and p are twocomp onent that dep end on momentum p but

are indep endent of x Using the Dirac representation of the matrices and inserting

the trial solution into the Dirac equation gives the pair of simultaneous equations

m p

E

p m

There are two simple cases for which equation can readily b e solved namely

p m which might represent an in its rest frame

m p which describ es a massless particle or a particle in the ultra

m relativistic limit E

Particle at Rest

For case an electron in its rest frame the equations decouple and b ecome

simply

E m E m

So in this case we see that corresp onds to solutions with E m while cor

resp onds to solutions with E m Dirac had therefore failed to remove these

solutions In light of our earlier discussions we no longer need to recoil in horror at

the app earance of these negative energy states

Diracs Interpretation of Negative Energy

As an aside it is interesting to see how Dirac cop ed with the negative energy states

Dirac interpreted the negative energy solutions by p ostulating the existence of a

sea of negative energy states The or has all the negative

energy states full An additional electron must now o ccupy a p ositive energy state

since the Pauli exclusion principle forbids it from falling into one of the lled negative

energy states On promoting one of these negative energy states to a p ositive energy

one by supplying energy an electronhole pair is created ie a p ositive energy

electron and a hole in the negative energy sea The hole is seen in nature as a

p ositive energy p ositron This was a radical new idea and brought pair creation and

into physics m pair creation energy=0 by 2m energy −m

−ve levels filled

Figure Diracs lled negative energy states

The problem with Diracs hole theory is that it do es not work for b osons Such

particles have no exclusion principle to stop them falling into the negative energy

states releasing their energy

General Solutions

The negative energy solutions p ersist for an electron with p for which the solutions

to equation are

p p

E m E m

2 2

Now we can substitute one of these equations into the other and use p p

Explicitly

2

p p ip

3 1 2

2

p

p ip p

1 2 3

2 2 2

p p p

1 2 3

2

p I

2 2 2

p p p

1 2 3

We nd that

2 2

p p

2 2 2 2

E m E m

p

2 2

p m j medskip from which we deduce that E j

p

2 2

We write the p ositive energy solutions with E j p m j as

i(E tpx)

x e

p

E +m

p

2 2

m j are p while the general negative energy solutions with E j

p

i(E tpx)

E m

e x

for arbitrary constant and Clearly when p these solutions reduce to the

p ositive and negative energy solutions discussed previously

We want to rewrite the solutions eqs and intro ducing the spinors

s p and v sp The lab el f g is a spinor index that often will b e u

suppressed Take the p ositive energy solution equation and dene

p

r

ipx ipx

e us pe E m

p

r

E +m

For the negative energy solution of equation change the sign of the energy

E E and the threemomentum p p to obtain

p

p

r

ipx ipx

E +m

E m e v s pe

r

In these two solutions E is now and for the rest of the course always p ositive and

2 2 12

given by E p m The argument s takes the values with

2 1

The uspinor solutions will corresp ond to particles and the v spinor solutions to

antiparticles The role of the two s will b ecome clear in the following section where

it will b e shown that the two choices of s are spin lab els Note that each spinor

0

solution dep ends on the threemomentum p so it is implicit that p E

Orthogonalit y and Completeness

Our solutions to the Dirac equation take the form

ipx ipx

N u e N v e r s

s r

The N is a normalization factor We have already included a normalization factor

p

E m in our spinors With this factor

y y r s

u r pus p v r pv s p E

This corresp onds to the standard relativistic normalization of E particles p er unit

y

volume this makes u u and hence transforms like the time comp onent of a

vector under Lorentz transformations as it must to b e the zeroth comp onent of J

Note that the spinors are ortho gonal

We must further normalize the spatial wave functions In fact a plane wave is not

normalizable in an innite space so we will work in a large b ox of volume V

Z

3 2 y

d x E N V

b ab

a

b run over the p ossible values of r s and the value of p Note again the where a

orthogonality of the states To normalize to E particles p er unit volume we must

p

R

y 3

V Sometimes its helpful to normalize so that d x so that set N

b ab

a

p

there is one particle p er unit volume in which case N EV this is not a

Lorentz invariant normalization so must b e done in a particular frame

Rememb er that the solutions to the wave equation form a complete set of states

meaning that we can expand like a Fourier expansion an arbitrary function x in

terms of them

X

a x x

n n

n

The a are the equivalent of Fourier co ecients and if is a wave function in some

n

2

quantum mixed state then ja j is the probability of b eing in the state

n n

Spin

Now it is time to justify the statements we have b een making that the Dirac equation

describ es spin particles We will see that the two comp onents of each of the

p ositive and negative energy solutions describ e spin up and spin down states

Conserved Quantities Rememb er that a conserved quantity in quantum mechan

ics is describ ed by a time indep endent op erator that commutes with the Hamiltonian

To prove this we evaluate

R

dhf i

d

F dx

dt dt

R R



F dx F dx

t t

Note that F is time indep endent here Now we use the Schro edinger equation

H ih

t

to nd

Z

dh f i i

HF FH dx

dt h

So if the F H vanishes the exp ectation value is conserved

Now the Dirac Hamiltonian in momentum space is given in equation as

H p m

D

and the orbital op erator is

R p L

L and H may not commute b ecause they contain x and p which do not commute

D

x p i Evaluating the commutator of L with H

i j ij D

L H R p p

D

R p p

i p

we see that the orbital angular momentum is not conserved otherwise the commu

tator would b e zero

Now we would like to nd a total angular momentum J that is conserved by

adding an additional op erator S to L

J L S J H

D

To this end consider the three matrices

i

1 2 3

where the rst equivalence is merely a denition of and the last equality can b e

veried by an explicit calculation The have the correct commutation relations

to represent angular momentum since the Pauli matrices do and their

with and are

i

i j ij k k

Here is a totally antisymmetric which is zero if any of the three indices

ij k

are the same is and we get a minus sign if we interchange any two indices

123

so From the relations in we nd that

213

H i p

D

Comparing equation with the commutator of L with H in equation

D

you see that

H L

D

and we can identify

S

as the additional quantity that when added to L in equation yields a conserved

total angular momentum J We interpret S as an angular momentum intrinsic to

the particle Now

2

S

2

and recalling that the eigenvalue of J for spin j is j j we conclude that S

represents spin and the solutions of the Dirac equation have spin as promised

We worked in the Dirac representation of the matrices for convenience but the result

is necessarily indep endent of the representation

Now consider the uspinor solutions us p of equation Cho ose p p

z

and write

p

E m

p

C B C B

E m

C B C B

p

C u u p B C u u p B

A A

E m

p

E m

With these denitions we get

u S u u S u

z z

So these two spinors represent spin up and spin down along the z axis resp ectively

For the v spinors with the same choice for p write

p

E m

p

C B B C

E m

C B B C

p

C v v p B v v p B C

A A

E m

p

E m

where now

v S v v S v

z z

This apparently p erverse choice of up and down for the v s is actually quite sensible

when one realizes that a negative energy electron carrying spin backwards in

time lo oks just like a p ositive energy p ositron carrying spin forwards in time

tz Covariant Notation Loren

There is a more compact way of writing the Dirac equation which requires that we

get to grips with some more notation Dene the matrices

0

In the Dirac representation

0

In terms of these the relations b etween the and in equation can b e written

compactly as

f g g

Combinations like a o ccur frequently and are conventionally written as

a a a

pronounced a slash

Observe that using the matrices the Dirac equation b ecomes

i m

or in momentum space

p m

The spinors u and v satisfy

p mus p

p mv s p

since for v s p E E and p p

We want the Dirac equation to preserve its form under Lorentz transforma

tions eq Weve just naively written the matrices in the Dirac equation as

however this do es not make them a vector They are just a set of numb ers in four

matrices and theres no reason they should change when we do a b o ost However

the notation is delib erately suggestive for when combined with Dirac elds you can

construct quantities that transform like vectors and other Lorentz we wont

show this here Since do es transform for the equation to b e Lorentz covariant

we are led to prop ose that transforms to o As an example of such a transformation

lets lo ok at transformations

Parity

Consider parity space inversion transformations P t x t x

We wouldnt exp ect physics to change b ecause of such a redenition of our axes

lab elling For the Dirac equation to remain the same though we must also transform

as

To see this works note that under Parity

r r

t t

r r

t t

So

r

t

r

t

This means we can write the parity transformed Dirac equation

i m

as

i m

which has the same solutions as the Dirac equation b efore the transformation as we

require

The upshot is that we have discovered that particles and antiparticles have op

p osite intrinsic parity

Massless Ultrarelativi stic Fermions

2 2

At very high we may neglect the of particles E jpj Let us lo ok

therefore at solutions of the Dirac equation with m on the basis that this will

b e an extremely go o d approximation for many situations

From equation we have in this case

E p E p

These equations can easily b e decoupled by taking linear combinations and dening

and N the two comp onent spinors N

R L

N N

R L

which leads to

EN p N EN p N

R R L L

The system is in fact describ ed by two entirely separated two comp onent spinors If

we take them to b e moving in the z direction and noting that diag we

3

see that there is one p ositive and one negative energy solution in each

Further since E jp j for massless particles these equations may b e written

p p

N N N N

L L R R

jpj jpj

p

1

Now is known as the helicity op erator ie it is the spin op erator pro jected

2 jpj

in the direction of of the momentum of the particle We see that the N

L

corresp onds to solutions with negative helicity while N corresp onds to solutions

R

with p ositive helicity In other words N describ es a lefthanded particle while N

L R

describ es a righthanded particle and each typ e is describ ed by a twocomp onent

spinor

Note that under parity transformations like R p p p therefore

p p ie the spinors transform into each other

N N

R L

So a theory in which N has dierent interactions to N such as the standard mo del

L R

in which the weak force only acts on left handed particles manifestly violates parity

Although massless particles can b e describ ed very simply using two comp onent

spinors as ab ove they may also b e incorp orated into the fourcomp onent formalism

as follows From equation we have in momentum space

jp j p

For such a solution

S p p

5 5

jpj jpj

1 1

5

using the spin op erator S with dened in equation But S p jp j

2 2

is the pro jection of spin onto the direction of motion ie the helicity and is equal to

5

Thus pro jects out the particle with helicity right handed and

5

pro jects out the particle with helicity left handed

5 5

R L

dene the fourcomp onent spinors and

R L