The WorldParticle content

qe,

qe,

Interactions

Schrodinger Wave Equation

He started with the energy-momentum relation for a particle

he made the quantum mechanical replacement:

How about a relativistic particle？

Expecting them to act on plane waves

ipxrpiiEt eee

The Quantum mechanical replacement can be made in a covariant form. Just remember the plane wave can be written in a covariant form:

As a wave equation, it does not work.It doesn’t have a conserved probability density.It has negative energy solutions.

ipxrpiiEt eee

The proper way to interpret KG equation is it is actually a field equation just like Maxwell’s Equations.

Consider we try to solve this eq as a field equation with a source.

)(2 xjm

We can solve it by Green Function.

')',()',( 2 xxxxmxxG

G is the solution for a point-like source at x’.

By superposition, we can get a solution for source j.

)'()',(')()( 40 xjxxGxdxx

Green Function for KG Equation:

')',()',( 42 xxxxGmxxG

By translation invariance, G is only a function of coordinate difference:

)'()',( xxGxxG

The Equation becomes algebraic after a Fourier transformation.

)(

~

2)'( )'(

4

4

pGepd

xxG xxip

1)(~22 pGmp

)'(4

44

2)'( xxipe

pdxx

22

1)(

~

mppG

This is the propagator!

'x x

Green function is the effect at x of a source at x’.

That is exactly what is represented in this diagram.

KG Propagation

The tricky part is actually the boundary condition.

B

A

B

A

For those amplitude where time 1 is ahead of time 2, propagation is from 1 to 2.

C1

2

B

A

B

A

For those amplitude where time 2 is ahead of time 1, propagation is from 2 to 1.

C

1

2

is actually the sum of the above two diagrams!

To accomplish this, 22

1)(

~

mppG

imp

pG

22

1)(

~

Blaming the negative energy problem on the second time derivative of KG Eq., Dirac set out to find a first order differential equation.

This Eq. still needs to give the proper energy momentum relation. So Dirac propose to factor the relation!

For example, in the rest frame:

mct

i

Made the replacement

First order diff. Eq.

and we need:

Now put in 3-momenta:

Suppose the momentum relation can be factored into linear combinations of p’s:

Expand the right hand side:

We get

Dirac propose it could be true for matrices.

It’s easier to see by writing out explicitly:

Oops! What!

00110 0110 or

No numbers can accomplish this!

2 by 2 Pauli Matrices come very close

1i jiji 0,

ijji ,

10

01

0

0

01

10321

i

i

Dirac find it’s possible for 4 by 4 matrices

We need:

that is

He found a set of solutions:

Dirac Matrices

Dirac find it’s possible for 4 by 4 matrices

Pick the first order factor:

Make the replacement and put in the wave function:

If γ’s are 4 by 4 matrices, Ψ must be a 4 component column:

03210

mczyxt

i

It consists of 4 Equations.

The above could be done for 2 by 2 matrices if there is no mass.

Massless fermion contains only half the degrees of freedom.

and we need:

Now put in 3-momenta:

Suppose the momentum relation can be factored into linear combinations of p’s:

Expand the right hand side:

We get

3211 k 321 ---1 k

交叉項抵銷

0222 cmp

0p E

There are two solutions for each 3 momentum p (one for +E and one for –E )

xpitipipx eaeax

0

)(

Plane wave solutions for KG Eq.

ipxipxipx eameapeap 2220

2220 mpp

p

xpiiEtxpiiEt

p

ipxipx eaeaeaeax

)(

Expansion of a solution by plane wave solutions for KG Eq.

p

xpiiEtxpiiEt

p

ipxipx ebeaebeax

)(

p

xpiiEtxpiiEt

p

ipxipx eaeaeaeax

)(

If Φ is a real function, the coefficients are related:

Multiply on the left with

mcp

0222 ucmp 0222 cmp 0p E

There are two sets of solutions for each 3 momentum p (one for +E and one for –E )

ueaueax xpitipipx 0

)(

2

1

2

1

B

B

A

A

B

A

u

u

u

u

u

uu

Plane wave solutions for Dirac Eq.

How about u?

0p 0p0p

0p

0p

0p

0p

You may think these are two conditions, but no.

We need:

Multiply the first by

So one of the above is not independent if 0222 cmp

0p 0p

20p

0p

0p

0p

0p 20p

22 pp

1

We need:

or 0p0p

How many solutions for every p?Go to the rest frame!

)0,(or)0,(

mmp

0p

0p

0p

0p

)0,(

mp

020

00

B

A

u

u

m

0Bu uA is arbitrary

1

0,

0

1Au

Two solution (spin up spin down)

0p

0p

0p

0p

)0,(

mp

000

02

B

A

u

um

0Au uB is arbitrary

1

0,

0

1Bu

Two solution (spin down and spin up antiparticle)

It’s not hard to find four independent solutions.

There are four solutions for each 3 momentum p (two for particle and two for antiparticle)

or

-

We got two positive and two negative energy solutions!Negative energy is still here!In fact, they are antiparticles.

0p0p

0p

0p

0p

0p

0p

0p

Electron solutions:

ueauea rpiiEtipx

Positron solutions:

0),(4,3 pEumcp

0),(2,1 pEvmcp

veaveauea rpiiEtipxrpiiEt

)2,1()4,3(

Expansion of a solution by plane wave solutions for KG Eq.

p

xpiiEtxpiiEt

p

ipxipx ebeaebeax

)(

p

ipxipx evbeuax

)(

Expansion of a solution by plane wave solutions for Dirac Eq.

Bilinear Covariants

Ψ transforms under Lorentz Transformation:

Interaction vertices must be Lorentz invariant.

The weak vertices of leptons coupling with W

e

We

ee ?

eWeg

Bilinear Covariants

Ψ transforms under Lorentz Transformation:

Interaction vertices must be Lorentz invariant.

How do we build invariants from two Ψ’s ?

A first guess:

Maybe you need to change some of the signs:

It turns out to be right!

We can define a new adjoint spinor:

is invariant!

In fact all bilinears can be classified according to their behavior under Lorentz Transformation:

A

p

ipxipx evbeuax

)(

)( 1pe

0)( 11 upeau p

p

ipxipx euaevbx

)(

0u

)(0 313133peuuuau ppppp

Feynman Rules for external lines

Find the Green Function of Dirac Eq.

')',()',( 4 xxIxxmcGxxGi

')',()',( 42 xxxxGmxxG

Now the Green Function G is a 4 ˣ 4 matrix

')',()',( 4 xxIxxmcGxxGi

How about internal lines?

')',()',( 4 xxIixxmcGxxGi

)(

~

2)'( )'(

4

4

pGepd

xxG xxip

ipGmp )(~

)'(4

44

2)'( xxipe

pdxx

mp

ipG

)(~

Using the Fourier Transformation

22 mp

mpi

Fermion Propagator

p

Photons:

It’s easier using potentials:

forms a four vector.

4-vector again

Charge conservation

Now the deep part:

E and B are observable, but A’s are not!

A can be changed by a gauge transformation without changing E and B the observable:

So we can use this freedom to choose a gauge, a condition for A:

J

cA

4

For free photons:

0 A

Almost like 4 KG Eq.

ipxeaxA )(

Energy-Momentum Relation

Polarization needs to satisfy Lorentz Condition:

We can further choose

Lorentz Condition does not kill all the freedom:

then Coulomb Guage

The photon is transversely polarized.

For p in the z direction:

For every p that satisfy

there are two solutions!

Massless spin 1 particle has two degrees of freedom.

A

),( q

0)( i

Feynman Rules for external photon lines

2,1,

)()()(ip

ipxipp

ipxipp eaeaxA

Gauge Invariance

Classically, E and B are observable, but A’s are not!A can be changed by a gauge transformation without changing E and B the observable:

But in Qunatum Mechanics, it is A that appear in wave equation:

),(' xtAA Transformation parameter λ is a

function of spacetime.

AAAAAA ''

In a EM field, charged particle couple directly with A.

Classically it’s force that affects particles. EM force is written in E, B.But in Hamiltonian formalism, H is written in terms of A.

txetxAc

ep

mH ,,

2

12

Quantum Mechanics or wave equation is written by quantizing the Hamiltonian formalism:

eAie

mti

2

2

1

Is there still gauge invariance?

B does not exist outside.

eAie

mti

2

2

1

Gauge invariance in Quantum Mechanics:

In QM, there is an additional Phase factor invariance:

),(),( txetx i

It is quite a surprise this phase invariance is linked to EM gauge invariance when the phase is time dependent.

),(),( ),( txetx txie

),( txAA

This space-time dependent phase transformation is not an invariance of QM unless it’s coupled with EM gauge transformation!

),(),( ),( txetx txie

),( txAA

),(),(),(),(),( txietxietxietxietxie eeeee

Derivatives of wave function doesn’t transform like wave function itself.

),(),(),( txietxietxie eieeieieAeieA

ieAeieA txie ),(

2

2

1

mti

Wave Equation is not invariant!

But if we put in A and link the two transformations:

This “derivative” transforms like wave function.

AieeAie txie

),(

ie

teie

ttxie ),(

eAie

mti

2

2

1

2

2

1Aie

mie

ti

2

2

1Aie

meie

tei ieie

In space and time components:

The wave equation:

can be written as

It is invariant!

),(),( ),( txetx txie

),( txAA

ieAeieAD txie ),(

Your theory would be easily invariant.

This combination will be called “Gauge Transformation”It’s a localized phase transformation.

Write your theory with this “Covariant Derivative”.

There is a duality between E and B.

Without charge, Maxwell is invariant under:

Maybe there exist magnetic charges: monopole

Magnetic Monopole

0 A

The curl of B is non-zero. The vector potential does not exist.

03 AxdadB

If A exists,

there can be no monopole.

But quantum mechanics can not do without A.

Maybe magnetic monopole is incompatible with QM.

But Dirac did find a Monopole solution:

sin

cos1

r

gA

rr ArA

rrrA

r

A

r

AA

rrA

1

sin

11

sinsin

1

Dirac Monopole

sin

cos1

r

gA

It is singular at θ = π. Dirac String

It can be thought of as an infinitely thin solenoid that confines magnetic field lines into the monopole.

sin

cos1

r

gA

Dirac String doesn’t seem to observe the

symmetry

But a monopole is rotationally symmetric.

In fact we can also choose the string to go upwards (or any direction):

sin

cos1'

r

gA

They are related by a gauge

transformation!

It has to!

Charge Quantization

sin

cos1

r

gA

Since the position of the string is arbitrary, it’s unphysical.

4eg

Since the string is unphysical. 14 iege

g

ne

2

BadeAade

Using any charge particle, we can perform a Aharonov like interference around the string. The effects of the string to the phase is just like a thin solenoid:

Finally…. Feynman Rules for QED

e

e

e

e

4-columns4-rows

4 ˣ 4 matrices

4 ˣ 4 matrices

1 ˣ 1 in Dirac index

Dirac index flow,from left to right!

kp

'kp

using

The first term vanish!

Photon polarization has no time component.

0 ump

0),( spump

The third term vanish!

Numerator simplification

In the Lab frame of e

Denominator simplification

in low energy

Assuming low energy limit:

0k

2nd term:

k1

00 00

Amplitude squared

Finally Amplitude:

00

02 kkk

旋轉帶電粒子所產生之磁偶極

磁偶極矩與角動量成正比

Lm

epr

m

eervr

vr

eriiA

222222

Lm

e 2

Sm

egS

m

es

2

帶電粒子自旋形成的磁偶極

Anomalous magnetic moment

12theory 108.89.11596521871

2

1 g

12experi 103.44.11596521881

2

1 g