n

jjjzjy

n

jjx pdMdpdpdpMd

2

32

2

2

But the integral is not Lorentz invariant!

A particle one decay: n 321

33

3

3

222 22

1

22

1

pd

E

pd

cmp

We make it more complicated by allowing an indefinite p0 and then fixing it:

0222

4

4

22

pcmppd

But in this form, we can be sure it is Lorentz invariant!We can perform the p0 integration to recover the 3 space form.

22220222 cmppcmp

)(2

122 axaxa

ax

2220

222

022220

2

1cmpp

cmppcmpp

This is Lorentz invariant.

n

jjpdMd

2

32

n

jjj

n

j j mcpE 2 2222 2

1

2

1

n

j

j

j

pd

EMd

23

32

22

1

npppp 321442

No matter what, the overall 4-momenta are conserved!

n

j

j

jn

pd

EppppMd

23

3

321442

22

12

n

j

j

jn

pd

EppppMd

23

3

321442

22

12

But M2 is still not Lorentz invariant.Γ transforms like 1/t1. t1 transforms like E1.

2

2

1cv

t

2

2

2

1cv

mcE

1

1

E

n

j

j

jn

pd

EppppM

Ed

23

3

321442

1 22

12

1

Now we can be sure M2 is Lorentz invariant. It’s called Feynman Amplitude.

Now we apply this to pion two photon decay.

20

Choose the rest frame of pion:

21 cmE 01 p

3

33

332

3

2321

442

1 22

1

22

12

1

pd

E

pd

EpppM

Ed

Two body scattering: n 321

n

j

j

jn

pd

EppppMd

33

3

321442

22

12

as you would expected If we consider only Lorentz transformation along the 1-2 colliding axis, the cross section is invariant! But we do want to pull out a Lorentz invariant factor that reflects the inverse flux that must appear in cross section:

2221

2214 cmmpp

In the rest frame of particle 2

11212

222

1212

2221

221 444 vEEpcmcmEcmcmmcmE

112 v

n

j

j

jn

pd

EppppMd

33

3

321442

22

12

2221

2214

1

cmmpp

n

j

j

jn

pd

EppppM

cmmppd

33

3

321442

2221

221

22

12

4

1

Two body scattering

Feynman Rules

To evaluate the Lorentz Invariant Feynman Amplitude MComponents of Feynman Diagrams

External Lines

Internal Lines

Vertex

ip

iqpropogator

Particle Content (their masses and spins)

Interactions

For a toy ABC model

Three scalar particle with masses mA, mB ,mC

Factors

External Lines

Internal Lines

Vertex

ip

iq

-ig

1

222 cmq

i

jj

321442 kkk

A B

C

1k

2k

3k

Draw all diagrams with the appropriate external lines

Integrate over all internal momenta iqd 442

1

Take out an overall momentum conservation. npppp 321

442

That’s it!

It’s so simple.

Consider first the diagrams with the fewest number of vertices.

CBA

Each extra vertex carries an extra factor of –ig, which is small.

Scattering

To the leading order, there could be more than one diagrams!In the diagrams, the B lines in some vertices are incoming particle.

The striking lesson: Lines in a vertices can be either outgoing or incoming, depending on their p0.p0 for an observed particle is always positive.So this vertex diagram is actually 8 diagrams put together! That’s the simplicity of Feynman diagram.

It’s similar to electron-electron scattering!

24 pp

The momentum conservations can be carried out immediately in the beginning!

The result can be written down right away!

This page also teaches us an important lesson:

After momentum conservation is enforced, the momentum of internal particle c is

24 pp

It does not satisfy momentum mass relation of a particle! 22224 cmpp c

The internal particle is not a real particle, it’s virtual.

媒介粒子

交互作用是透過交換媒介粒子來進行

媒介粒子媒介一個交互作用

γ

Feynman’s Diagram

n

p

p

n

交互作用不只改變粒子的動量，也可能改變粒子的身分！

So heavy particles, though decay fast and don’t existent in nature, do matter. (they could be virtual)

媒介粒子

媒介粒子只有在交互作用進行中存在，如果只觀察物體前後變化，就不會看到它。它幾乎不是一個實在的粒子，而是一個虛粒子。

虛粒子即使不穩定地存在，也會透過它所媒介的交互作用，影響實在的世界

24 pp

It does not satisfy momentum mass relation of a particle! 22224 cmpp c

The internal particle is not a real particle, it’s virtual.

24 pp

In a sense, the relation is only for particles when we “see” them!

Internal lines are by definition “unseen” or “unobserved”.

It’s more like:

It’s more a propagation of fields than particles!

24 pp

Field fluctuation propagation can only proceed forward that is along “time“ from past to future.

This diagram is actually a Fourier Transformation into momentum space of the spacetime diagram.

The vertices can happen at any spacetime location xμ and the “location” it happens need to be integrated over. After all you do not measure where interactions happen and just like double slit interference you need to sum over all possibilities.

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

24 pp is actually the sum of the above two diagrams!Feynman diagram in momentum space is simpler than in space-time!

24 pp

In this diagram, particles seem to have certain momenta. According to uncertainty principle, their position uncertainty is infinity!This diagram is only an approximation of more precise treatment using wave packet!

You need to “superpose” similar diagrams with slightly uncertain momentum t produce incoming and outgoing wave packets.

Lesson 3:

Δk Δx

In most situations, the Feynman amplitudes are not very sensitive to small uncertainties in momenta.We can approxiamte the amplitude for wave packets with a momentum distribution by that for plane waves with the central momentum value:k0

Lifetime of A

CBA

321442 ppp igiM

gM

20 for m

If all the momenta in the diagrams can be determined through momentum conservation, the diagram has no loop and is hence called tree diagram!

24 pp

If there is a loop in the diagram, some internal momentum is not fixed and has to be integrated over!

The WorldParticle content

qe,

qe,

Interactions

Schrodinger Wave Equation

He started with the energy-momentum relation for a particle

made Quantum mechanical replacment:

How about a relativistic particle？

The Quantum mechanical replacement can be made 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.

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 xxxxmxxG

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)(

~

We need:

Dirac propose it could be true for matrices.