Lecture 4: Antiparticles &

Virtual Particles

• Klein-Gordon Equation

• Antiparticles & Their Asymmetry in Nature

• Yukawa Potential & The Pion

• The Bound State of the Deuteron

• Virtual Particles

• Feynman Diagrams

Chapter 1

Useful Sections in Martin & Shaw:

Ψ = Aei(kx-ωt) = Ae (px-Et)ℏi

Free particle⇒

iℏ Ψ = Ε∂t∂Note that: -iℏ Ψ = p

∂x∂&

So define: E ≡ iℏ∂t∂ p ≡ -iℏ ∇

E=p2/2m ⇒ ∇2Ψ ℏ2m∂t

∂Ψ = i

Schrodinger Equation

(non-relativistic)

E2 = p2c2 + m2c4

To make relativistic, try the same trick with

c2 ∂t2∂ 2Ψ ∇2Ψ − Ψ =

m2c2

ℏ2

Klein-Gordon Equation

⇒1 first proposed

by de Broglie

in 1924

Ψ = Ae (px-Et)ℏi

For every plane-wave solution of the form

Ψ∗ = Ae (-px+Et)ℏi

There is another solution of the form

(momentum p and positive E)

(momentum −p and negative E)

Try again, but attempt to force a linear form:

Where αnand β are determined by requiring that solutions

of this equation also satisfy the Klein-Gordon equation

iℏ Ψ = ∂t∂

-iℏ Σ Ψ ∂x

n

∂ + βmc2 Ψ

3

n=1

c αn

Dirac Equation

⇒ α and β need to be 4x4 matrices and

Ψ1Ψ2Ψ3Ψ4

Ψ =still have positive

and negative energy

states but now also

have spin!

How do you prevent transitions into ''negative energy" states?

0

E

m

−m

Dirac

''Hole" Theory

''sea" of negative

energy states

Nowdays we don’t

think of it this way!

Instead we can say that

energy always remains

positive, but solutions

exist with time reversed

(Feynman-Stukleberg)..

Antimatter

Anderson

1933

The Earth →

The Moon →

The Planets →

Ouside the Solar System →

Another Part of the Galaxy →

Other Galaxies→

Larger Scales →

Spontaneous combustion is relatively rare

Neil Armstrong survived

Space probes, solar wind...

Comets...

Cosmic Rays...

Mergers, cosmic rays...

Diffuse γ−ray background

Where’s the Antimatter ???

∇2Ψ = Ψ m2c4

ℏ2

For a static solution,

Klein-Gordon reduces to

in this case the solution is

Ψ ≡ V(r) = −g2 e-r/R

4π r

Yukawa Potential

note that if m=0, we would

have the equivalent of an

electromagnetic potential:

∇2Φ=0whose solution is

V(r) = eΦ(r)= −e2 1

4πεο r

where R ≡≡≡≡ ℏℏℏℏ/mc So this gives us a

new ''charge" g and

an effective range Rhmmm... sounds like the

''neutron-proton" problem

n pnp

whatever keeps them together must

be very strong and short-ranged⇒

EM ⇒ ''carrier" of electromagnetic field = photon (massless boson)

Strong nuclear force ⇒ ''carrier" of field must be some massive boson

R ∼ 10-15m ≡ 1fmℏc = 197 MeV fm

⇓mc2 = 100 MeV ''meson"

Yukawa (1934)

µ-meson (muon) Anderson & Neddermeyer (1936)

mµ = 105.6 MeV ! ...but a fermion, doesn’t interact strongly

(looks like a heavy electron)

''Who ordered that ?!" (I. I. Rabi)

e− = 0.511 MeV ''lepton"

p = 938 MeV ''baryon"

π-meson (pion), mπ=140 MeV Powell et al. (1947)

Cecil Powell Marietta BrauDon Perkins

ED= p2/ 2µ + V(r)

reduced mass

assume mp ≃ m

n ≡ M, so

µ = (MM)/(M+M) = M/2also take p ≃ ℏ/r(de Broglie wavelength)

''Bohr Condition"

ED= − exp(−mπcr/ℏ)ℏ2 g2

Mr2 4πr

let: x ≡ mπcr/ℏ

E = − e-xmπ

2c2 g2mπc

Mx2 4πℏx

n pThe Bound State of the Deuteron

ED= − e-x

mπ2c2 g2mπc

Mx2 4πℏx

for a bound state to exist, ED< 0

( ) e-x g2

4πℏcmπM

1

x

mπM

1

x2> ( )2

ex g2

4πℏc1

x >

mπM( )

this is a

⇒ minimumwhen x=1

g2

4πℏc >140 MeV

938 MeV( )(2.718)

αs≡

g2

4πℏc > 0.4 compare with α ≡ = e2 1

4πℏc 137

= Mc ( ) − ( ) e-xg2

4πℏcmπM

mπM

1

x

1

x2[ ]2 2

What does ''carrier of the field" mean ??

Note: the time it would take for the carrier

of the strong force to propagate over

the distance R is ∆t ∼ R/c

Heisenberg uncertainty ⇒ ∆E ∆t ℏ∼>

so R ~ ℏc/∆E

if we associate ∆E with the rest mass energy of the pion, then

R ~ ℏ/mπc

which is what enters into the Yukawa potential !

This implies we are ''borrowing" energy over a ''Heisenberg time"

⇒ ''virtual particle"

+ −

EM

(infinite range)

p n

Strong Nuclear Force

(finite range)

''Field Lines"

e+ e+

e- e-

e+

e-

e+

e-

p1

p2

p3

p4

p1

p2

p3

p4

qq

Leading order diagrams for Bhabha Scattering

e+ + e− → e+ + e−

x

t

Feynman Diagrams

e+ e+

e- e-

e+

e-

e+

e-

p1

p2

p3

p4

p1

p2

p3

p4

qq

Leading order diagrams for Bhabha Scattering: e+ + e− → e+ + e−

x

t

1) Energy & momentum are conserved at each vertex

2) Charge is conserved

3) Straight lines with arrows pointing towards increasing time represent

fermions. Those pointing backwards in time represent anti-fermions

4) Broken, wavy or curly lines represent bosons

5) External lines (one end free) represent real particles

6) Internal lines generally represent virtual particles

Some Rules for the Construction & Interpretation of Feynman Diagrams

e+ e+

e- e-

e+

e-

e+

e-

p1

p2

p3

p4

p1

p2

p3

p4

qq

Leading order diagrams for Bhabha Scattering: e+ + e− → e+ + e−

x

t

7) Time ordering of internal lines is unobservable and, quantum

mechanically, all possibilities must be summed together. However, by

convention, only one unordered diagram is actually drawn

8) Incoming/outgoing particles typically have their 4-momenta labelled as

pnand internal lines as q

n

9) Associate each vertex with the square root of the appropriate

coupling constant, √αx, so when the amplitude is squared to yield a

cross-section, there will be a factor of αxn , where n is the number of

vertices (also known as the ''order" of the diagram)

Some Rules for the Construction & Interpretation of Feynman Diagrams

e+ e+

e- e-

e+

e-

e+

e-

p1

p2

p3

p4

p1

p2

p3

p4

qq

Leading order diagrams for Bhabha Scattering: e+ + e− → e+ + e−

x

t

10) Associate an appropriate propagator of the general form 1/(q2 + M2)

with each internal line, where M is the mass of mediating boson

11) Source vertices of indistinguishable particles may be re-associated to

form new diagrams (often implied) which are added to the sum

Thus, the leading orderdiagrams for pair annihilation

( e- + e+→ γ + γ ) are: and

Some Rules for the Construction & Interpretation of Feynman Diagrams

The ''play catch" idea seems to work intuitively when

it comes to understanding how like charges repel.

But what about attractive forces between dissimilar charges??

Are you somehow exchanging ''negative momentum" ???!

The ''play catch" idea seems to work intuitively when

it comes to understanding how like charges repel.

But what about attractive forces between dissimilar charges??

Are you somehow exchanging ''negative momentum" ???!

The best I can offer: Note from Feynman diagrams (and later CPT)

that a particle travelling forward in time is

equivalent to an anti-particle, going in the

opposite direction, travelling backwards in time.e+

e-

⇓Feynman-Stuckelberg interpretation is that

the photon scatters the electron back in time!

..

More Bhabha Scattering...

So this basically a perturbative expansion in powers of the coupling constant. You

can see how this will work well for QED since α ~ 1/137, but things are going to get dicey with the strong interaction, where α

s~ 1 !!

Richard Feynman

(Baron) Ernest Stuckelberg..

von Breidenbach zu Breidenstein und Melsbach