Relativistic Quantum Mechanics

Four Vectors

The covariant position-time four vector is defined as

x0 =ct, x1 =x, x2 =y, x3 =z

And is written compactly as

x

Lorentz Transformations

2

2

2

1

1

)('

'

'

)('

cv

and

xc

vtt

zz

yy

vtxx

)''(

'

'

)''(

2x

c

vtt

zz

yy

vtxx

Transforms from x,y,z into x’,y’,z’ i.e from S into a moving reference frame S’ which has an uniform velocity of v in the x-hat direction

And back again

Now combining four vectors with Lorentz

c

v

and

xx

xx

xxx

xxx

33

22

011

100

'

'

)('

)(' But we may more compactly write this as

1000

0100

00

00

'3

0

where

xx

Rows are index,And is column

Einstein Summation Convention

Repeated indices are to be summed as one (i.e. leave off the sigmas)

OR When a sub/superscript

(letter) appears twice on one side of the equation, summation with respect to that index is implied

xxx

3

0

'

Lorentz invariance

23222120

23222120

'''' xxxxI

xxxxI

The Metric Tensor, g

xxgI

so

g

Let

1000

0100

0010

0001

Covariant

x=gx

Where the subcripted x is called the covariant four vector

And the superscripted x is called the contravariant four vector

I=x x

Note: a ·b=ab

a · b is different from a 3 dimensional dot product

aaaa 202 )(

If a2>0, then a is timelikeIf a2<0, then a is spacelikeIf a2=0, then a is lightlike

From vectors to tensors

A second rank tensor, s carries two indices, has 42=16 components, and transforms like two factors of :

ss '

ss '

A third rank tensor, s carries 3 indices, has 43=64 components, and transforms like three factors of :

What I’ve used these things for years?

In this hierarchy, a vector is a tensor of rank 1, and scalar (invariant) is a tensor of rank 0

A couple of quick definitions

dt

xdv

and

dtd

Proper time is the time w.r.t. to the moving reference frame.

The differential to the left indicates the proper time interval in the moving reference frame.

It should be getting larger and larger as v approaches c.

This is our old friend velocity.

Proper Velocity

cdt

ctd

d

dx

d

dx

or

vd

xd

1

)(00

22

222

22222

)1(

)(

),,,(

cc

vc

vvvc

vvvc

so

zyx

zyx

Very invariant!

In relativity, momentum is the product of mass and proper velocity

22422

222

2

,,,

cpcmE

so

mcmc

Epp

pppc

Ep

mp

zyx

p

Klein-Gordon Equation

acpcmaE

pandt

iELet

cpcmE

)( 22422

222

22422

accmat

)( 222422

22

Klein-Gordon Equation

The trouble is that this equation does not describe spin; although we have used kets for convenience, these kets cannot describe angular momentum transformations.

Actually, this equation works well for pions…

Deriving Dirac

0))(()( 002220 mcpmcpcmp

Dirac’s basic strategy was to factor pp –m2c2 into a perfect square;

now if p is zero, then

Now either p0-mc or p0+mc guarantees that pp –m2c2 =0

))(

))((22

22

cmpmcpp

mcpmcpcmppk

We don’t want any terms linear in momentum: need perfect squares

322332

311331

211221

300330

200220

100110

2323222221212020

23222120

)(

)()(

)()()(

)()()()()()()()(

)()()()(

0

pp

pppp

pppppp

pppp

pppp

or

pppp

So k

Need to get rid of the cross-terms… Dirac said let the gammas be matrices, not numbers g2,

Recall for Spin,

3

2

1

10

01

0

0

01

10

z

y

x

i

i

Pauli Spin Matrices Dirac noted that

0

0

10

010

i

ii

Note: 1 with an arrow is the identity matrix and therefore, these are 4 x 4 matrices

Now for the last bit

))((22 mcpmcpcmpp

3

2

1

0

0

)(0)(

where

mci

ipthen

conventionmcpLet

The wavefunction is NOT a four vector but is called a Dirac spinor or bi-spinor

Dirac Equation

Ugly little sucker, isn’t it?

A Pretty Form:

I

I

where

amcct

ai

0

0

0

0

)( 2

α

Pα

A “simple” example problem

Consider a particle at rest, p=0

So the Dirac equation becomes

00 mctc

i

B

A

B

A

mci

t

t

2

10

01

Where A carries the upper components and B carries the lower components

4

3

2

1

B

A

SO

BB

AA

mci

t

andmci

t

2

2

)0()(

)0()(2

2

B

tmci

B

A

tmci

A

et

andet

so

tEi

e Recall

is the time dependent solution of SE

So one of these states represents the rest mass of “normal” matter.

The other state represents a “negative” energy state. Interpreted as the rest mass of the positron

Wavefunctions

1

0

0

0

)(

0

1

0

0

)(

0

0

1

0

)(

0

0

0

1

)(

22

22

43

21

tmcit

mci

tmcit

mci

etandet

etandet

What if there is momentum?

Then substitute h-bar * k Then take the inner product with alphaSee your text on pages 803 and 804 for

details

Okay, what about potentials?

0

)()(

2

2

2

mcVEc

cmcVE

so

Construct

mccVE

Let

Pσ

Pσ

Pα

PσPσ

Pσ

Pσ

cmcVE

cmcVE

so

cmcVE

cmcVE

22

2

2

1)(

0)(

0)(

Let E-V+mc2 =2mc2

Vm

PE

Vm

E

s

s

2

2

)(

2

2Pσ

So if the energy is low (less than mc2), then the SE is used otherwise, we must use Dirac Equation