Lecture 5: QED •  The Bohr Magneton •  Off-Shell Electrons •  Vacuum Polarization •  Divergences, Running Coupling &

Renormalization •  Yukawa Scattering and the Propagator

Section 5.1, Section 5.2, Section 7.1.2

Useful Sections in Martin & Shaw:

Bohr Magneton

Bremsstrahlung

Ze ~ Zα3

Internal electron line since real electron cannot emit a real photon and still conserve energy and momentum:

me = meʹ′ + pʹ′e2/2meʹ′ + Eγ (conservation of energy)

⏐pʹ′e⏐ = ⏐pγ ⏐ = Eγ (conservation of momentum)

So, for a positive energy photon to be produced, the electron has to effectively ''lose mass"

me - meʹ′ = Eγ2/2meʹ′ + E γ ⇒

Pair Production

Ze Ze

''twisted Bremsstrahlung" Matrix element will be the same; Cross-section will only differ by a kinematic factor

Ze

+

⇒ virtual electron goes ''off mass shell"

in electron rest frame

consider

e-

- q

Thus, we never actually ever see a ''bare" charge, only an effective charge shielded by polarized virtual electron/positron pairs. A larger charge (or, equivalently, α) will be seen in interactions involving a high momemtum transfer as they probe closer to the central charge.

In QED, the bare charge of an electron is actually infinite !!!

Note: due to the field-energy near an infinite charge, the bare mass of the electron (E=mc2) is also infinite, but the effective mass is brought back into line by the virtual pairs again !!

Vacuum Polarization

⇒ ''running coupling constant"

e-

Another way...

Consider the integral corresponding to the loop diagram above:

At large q the product of the propagators will go as ~ 1/q4

Integration is over the 4-momenta of all internal lines where d4q = q3 dq dΩ (just like the 2-D area integral is r dr dθ and

the 3-D volume integral is r2 dr dcosθ dφ )

∫ (1/q4) q3 dq ∞

qmin

so the q integral goes like: ⇒ logarithmically

divergent !!

With some fiddling, these divergences can always be shuffled around to be associated with a bare charge or mass. But we only ever measure a ''dressed" (screened) charge or mass, which is finite. Thus, we ''renormalize" by replacing the bare values of α0 and m0 by running parameters.

∞

qmin

= ∫ dq/q ∞

qmin

= ln q ]

Analogy for Renormalization:

Instead of changing the equations of motion, you could instead (in principle) find the ''effective" mass of the cannonball by shaking it back and forth in the water to see how much force it takes to accelerate it. This ''mass" would no longer be a true constant as it would clearly depend on how quickly you shake the ball.

Still unhappy? Well, if it’s any comfort, note that the electrostatic potential of a classical point charge, e2/r, is also infinite as r → 0

(perhaps this all just means that there are really no true point particles... strings?? something else?? )

Does an electron feel it’s own field ???

''Ven you svet, you smell you’re own stink, Ja? Zo..."

M. Veltman

Steve’s Tips for Becoming a Particle Physicist

1) Be Lazy

3) Sweat Freely

2) Start Lying

e-

W -

νe

e -

e+

γ γ

Is A Particle?

No: All Are But Shadows Of The Field

Spin (Dirac Equation)

Klein-Gordon Equation Antimatter

Recap:

Yukawa Potential

New ''charge" + limited range

Strong Nuclear Force

Prediction of pion

''virtual" particles

exchange quanta

Feynman Diagrams (QED)

(Nobel Prize opportunity missed) Asymmetry?!

Relativity

Relativistic Wave Equation

Central Reality of the FIELD

Subjectivity of position, time, colour, shape,...

Consider the scattering of one nucleon by another via the Yukawa potential: V(r) = e-Mcr/ħ - g2

4π r

In the CM, both nucleons have equal energy and equal & opposite momenta, p. Take the incoming state of the 1st nucleon (normalised per unit volume) to be: Ψo = eip•r/ħ

After scattering, the final state will have a new momentum vector, pʹ′, where |p| = |pʹ′ | : Ψf = eipʹ′•r/ħ

If θ is the angle between p and pʹ′, write an expression for the momentum transfer, q , in terms of p .

p

pʹ′ θ q = pʹ′ - p magnitude of q = 2p sin(θ/2)

6 sheet 2

Compute the matrix element for the transition.

<Ψf|V(r)|Ψo> = ∫ e-ipʹ′•r/ħ eip•r/ħ e- Mcr/ħ - g2 4πr ( ) r2 dr dφ d(cosθ)

∫ e-iq•r/ħ e- Mcr/ħ r dr d(cosθ) - g2 2 =

∫ e-iqrcosθ/ħ e- Mcr/ħ r dr d(cosθ) - g2 2 =

∫ e-(iq+Mc)r/ħ dr -iħg2 2q = ( - ) e (iq-Mc)r/ħ

e-(iq+Mc)r/ħ iħ2g2 2q = [ + ] e(iq-Mc)r/ħ

iq + Mc iq-Mc r = ∞

r = 0

e-(iq+Mc)r/ħ iħ2g2 2q = [ + ] e (iq-Mc)r/ħ

iq + Mc iq-Mc

r = ∞

r = 0

This will go to zero as r → ∞, so we’re left with:

<Ψf|V(r)|Ψo> = 1 - iħ2g2 2q [ + ] 1

iq + Mc iq-Mc iq-Mc - iħ2g2

2q [ ] + iq + Mc (iq + Mc)( iq-Mc)

=

ħ2g2 (q2+M2c2) = 1

q2+M2 ~ (natural units)

NOTE: This is all really a semi-relativistic approximation!

Known as the “propagator” associated with the field characterised by an exchange particle of mass M

Really, we want to consider a 4-momentum transfer

Recall that P2 = E2 - p2

1 q2+M2 so, 1

q42 - M2

(choosing appropriate sign convention)

Show that the relation dp/dE = 1/v , where v is the velocity, holds for both relativistic and non-relativistic limits.

Also show that, for the present case: p2d(cosθ) = - ½ dq2

Classically:

E = p2/2m dE = (p/m) dp

dp/dE = (m/mv) = 1/v

Relativistically:

E2 = p2 + m2

dp/dE = E/p = (γm/γmv) = 1/v

q = 2p sin(θ/2)

q2 = 4p2 sin2(θ/2)

cosθ = cos2(θ/2) - sin2(θ/2) = 1 - 2sin2(θ/2)

sin2(θ/2) = (1 – cosθ)/2

q2 = 2p2 (1 - cosθ) dq2 = -2p2 dcosθ

p2d(cosθ) = - ½ dq2

Now, from the definition of cross-section in terms of a rate, the relative velocities of the nucleons in the CM, and using Fermi’s Golden Rule, derive the differential cross-section dσ /dΩ .

Rate = vbeamNbeamNtarget σ Volume

in our case Nbeam = Ntarget = 1 vbeam = 2v (relative velocity in CM)

Normalised Volume (= 1) so, dσ = dRate 1

2v given by FGR

dσ = 1 2v

2π ħ

|<Ψf|V(r)|Ψo>|2 1 (2πħ)3

p2dp dφ dcosθ dEtot

2 = ( )

1 (2πħ)3 dφ

ħ2g2 (q2+M2c2)

1 2v (- ½ dq2) 1

2v 2π ħ

= ( ) dq2dφ

(2πħ)3 ħ2g2 (q2+M2c2)

2 π 4v2ħ

= ( ) dq2dφ

(2πħ)3 ħ2g2 (q2+M2c2)

2 π 4v2ħ dσ

= dq2dφ

8π3ħ3 ħ4g4 (q2+M2c2)2

π 4v2ħ =

dq2dφ (q2+M2c2)2

1 (4π)2

g4 2v2

σ = dq2dφ (q2+M2c2)2

g4 2v2(4π)2 ∫ =

dq2 (q2+M2c2)2

πg4 v2(4π)2 ∫

= 1 (q2+M2c2)

πg4 v2(4π)2 - [ ] q

2 = 0

q2 = ∞

since integral was from cosθ = -1 to 1 (where 1 = zero momentum transfer)

taking v ∼ c

σ = πg4 Mc2(4π)2

ħ Mc = π g2

4πħc ( ) ( ) 2 2

strong coupling constant (~1)

“range” of Yukawa potential (~1fm)

σ = π × 10-30 m2

~ 30 mb

(within a factor of 2-3)

Integrate this expression and take the limit as v → c