Lecture 18: Mar 23th 2012

Reading Griffiths chapter 7HW: 7.15, 7.16, 7.18, 7.24, 7.25

1) Charge conjugation:

Given spin up electron 1 = u1(p)e-ipx and positron 1 = 1(p)eipx We would like to find an operator that takes us from the electron to the anti-particle positron.

Define C as i2 times the complex conjugate of transpose of ,

i2 = (

1−1

−11

)Which will convert particles to antiparticles, since it flips u(p) and takes p -p in the exponential

C = i2T

1elec = u1(p)e-ipx, C = 1

pos = 1(p)eipx

Note the transpose doesn’t do much to a vector except to shift it from column to row. Next we define the ajoint operator.

y = +0

The ajoint and other transposed vectors will be useful in defining bra-ket pairs when calculating probabilities.

for the Dirac spinors u = u+0, and u = +0

where

(kpk - m)u = 0, (kpk + m) = 0, u (kpk - m) = 0, u (kpk + m) = 0

u iuj = 2mij, u ij = -2mij

uu = kpk + m, u = kpk – m

These relationships will be useful for calculating matrix elements.

2) Consider Parity

Parity is something that two component objects can have intrinsically and can be conserved or violated in interactions. Therefore, it makes sense that we should check the parity of combinations of two particles with or without terms to cause interactions.

y scalar, parity 1y 5 psuedoscalar, parity -1y vector, parity -1y 5 psuedovector or axialvector, partity 1y σμν antisymmetric tensor, σμν = (i/2)(γμγν - γμγν)

5 = i0123 = (0 11 0 )

The matrices between the four vectors can be considered as a basis of the 16 possibleways to multiply the two four vectors together including scalar, vector and tensor products. Alternatively you can think of it as a basis of all the 4x4 matrices that could be made from the gamma matrices. Note that all such combinations can be expressed as one or two matrices.

When we start forming propagators for forces they will include several of these gamma matrix terms.

We can test these by forming a parity operator

Lum=(

1−1

−1−1

), Reverses 3 space or momentum dimensions

3) The Neutrino

A simpler solution to the Dirac equation is to consider mass 0 spin ½ fermions. The known nearly mass less spin ½ particle is the neutrino.

Consider the original energy momentum space equation.

pp = 0, and wave equation -h¶ ¶ = 0

To factorize this we need to solve a much simpler problem. With general constants:

(kpk)(lpl) = klpkpl = 0, just using gamma to represent an arbitrary constant, not the necessarily the gamma matrix

Then we only need k = l, kl = lk

These conditions can be satisfied by I for the 0th component and the 2D Pauli matrices i, or -i, for the 1, 2 and 3 components.

(p) = (E + p) = 0, (p) = (E - p) = 0

E = -p, E = p

Where for a massless fermion. E=|p| and

Then considering the helicity operator is

Sz⋅p=h2s⋅p

and Sz⋅p c=h

2s⋅p c=- h

2c

spin -½ and Sz⋅p f=h

2s⋅p f =- h

2f

spin ½

There are only two solutions each with definite helicity.Usually these are put together in a four component form like the spin ½ mass fermions

u(p) = (cf )

To go further we have define the gamma matrices for u, (p)u = 0A valid representation is(different than for the massive fermions):

g0=[0 II 0 ] , gi=[ 0 si

−si 0 ] , 5= (−I

I )Note then if we have a V-A force, such as the weak force: ½(1-5)

½(1-5)u(p) = γ μ( I 0

0 0 )( χϕ )=γ μ ( χ0 )The weak force only interacts with the left handed (negative helicity neutrino). Applying the charge conjugation procedure moves into the top position of the two vector. Thus the right handed antineutrino also interacts with the V-A weak force. Also the two parts of the force have different parties intrinsically violating parity conservation!

4) The photon

Starting from Maxwell’s equations. Find the fields in terms of the scalar and vector potentials

Ñ⋅B=0 , which means B=Ñ´ A

Ñ´ E+ 1c¶B¶t

=0,

Ñ´ E+ 1c¶ (Ñ´ A )

¶ t=0

,Ñ´(E+ 1

c¶ A¶ t )=0

which means E=-ÑV− 1

c¶ A¶t

Including charges and currents

Ñ⋅E=4 pρ ,Ñ´B−1

c¶ E¶ t

=4 pc

J

Thus we can express B and E in terms of the four vector potential A (V,A) and all of Maxwels equations as.

∂μFμν=∂μ∂

μAν−∂ν∂μAμ= 4π

cJ ν

where F if a 4x4 matrix with E and B terms.Am is the scalar potential, V, and the vector potential, A, expressed as a four vector.Jν is the scalar charge, , the vector current, j, expressed as a four vector.¶mFmν

represents Maxwell’s equations expressed as single derivatives of E and B or double derivatives of the potentials.

J the the charge density and current and obeys the charge and current continuity equation:

Jm=(J , ρ)Ñ⋅J=−¶ρ

¶t , ¶m Jm=0 ,

The single derivatives of E and B are given by a four current of charge density and the standard current and are double derivatives in terms of the scalar and vector potential

However since the physics of Maxwell’s equations is given by Fμν=∂μAν−∂ν Aμ

then

we can make the transformation Aμ→ Aμ+∂μ λ ( t , x )without changing F

μν

Consider a case where we use gauge freedom to set ¶m Am=0 and there is no four current i.e. no charges.

Then ∂μF

μν=∂μ∂μAν−∂ν∂μA

μ=∂μ∂μ Aν=boxAμ=4 π

cJ ν=0

, box = ¶m¶

m= 1c2

¶2

¶ t2-Ñ2

This is the Maxwell’s equation for a photon and we also recognize it as the Klein-Gordon equation for a massless particle

box A =0,( 1c2

¶2

¶t2 -Ñ2)Am=0,

However, this time we must consider the more general case of four vector solutions. This will allow for particles with spin. Borrowing from previous solutions

Aν( x )=ae−( i /h) xm pm eν( p )

Plugging into the gauge condition givespm em=0

Which can be solved with: e1=(0,1,0,0 ), and e1=(0,0,1,0 ) for a particle traveling in the z direction. This is appropriate for the photon which can have spin oriented along it’s direction of motion of opposite. Spin projection 1 or -1

Now consider the case with a charges so that J is not zero : ¶m¶

m Aν= 4 pc

J ν

A is the potential caused by charges and currents. If we have an electron in this potential will feel a force proportional to E and B times the charge(and velocity) of the electronor in terms of A, e times derivatives of A. There will also be a A2 term for the energy of the potential. Another way to think of this is that a force will cause a change in

momentum F = dp/dt. Integrating to find the total change Δp=∫F=e∫ ∂ A∝eA

Therefore p→ p+eA . Quantizing i ∂μ →i∂ μ +eA μ

This is how you insert the effect of the potential in the wave equation

For simplicity let’s consider the effect on a massive scalar particles satisfying the Klein-Gordan equation.

(¶m¶m+m2 ) f=0→((¶m−ieAm )(¶m−ieAm)+m2 ) f=0

(¶m¶m+m2− ie(¶m Am+Am¶m)−e2 A2) f =0

Note that this gives the expected form of e times derivatives of A. Now if we treat this in perturbation theory then we can say we have a base wave equation and electromagnetic perturbation potential, V, that can change initial states to final states.

V =- ie(¶m Am+Am¶m)−e2 A2

5) T: The transition rate from an initial to a final state

T =- i∫ f f¿ V f id

4 x

Where the A2 part of V will not contribute since it’s a constant and can not convert initial to final states.

T =- e∫ f f¿ ((¶m Am+Am¶

m )) f id 4 x

integrating by parts ∫ f f¿ ¶m Amf id

4 x =-∫ (¶m f f¿ ) Amf i and

T =- e∫ (f f¿ (¶m f i )−(¶m f f¿ ) f i ) Am d4 x

then for Klein-Gordon equation solutions: f=Ne−ip⋅x

T=−ieN i N f∫ ( pi+ p f )μ ei( pf−pi)⋅x Aμd4 x

A the potential and represents the photon and eN iN f ( p i+ pf )μei( pf− p i)⋅x

one particle

Consider Maxwell’s equation for A. ¶m¶

m Aν=4 pc

J ν

Considering that the potential is caused by the second particle a possible solution for J is Jν=eN i 2N f 2 ( pi 2+ p f 2 )ν e

i( pi 2−p f 2 )⋅x

Plugging in the Klein-Gordon equation solution for A Aν=Ne−iq⋅x eν (q ), q=pf2-pi2

¶m¶m Aν=4 p

cJ ν→−q2 Aν= 4 p

cJ ν→Aν=−1

q24 pc

J ν

T =- i∫ Jm1 1q2

Jm2 d4 x

The integrate out the exponentials over space which forces q = pf2-pi2 = pi1-pf1

T = -iNie ( pi1+ p f 1)μ ( gμν

q2 )ie ( pi 2+ pf 2 )ν (2)44 (pi1 + pi2 - pf1 - pf2)

This is the form of the matrix element in EM theory.

Very similar to our toy theory except that there are the momentum terms. Recall that in the toy theory we found that for units to work out the coupling constants had to have units of momentum. Now that unit of momentum is explicitly in the form and we have an identical propagator to the one we used for the toy theory for a massless propagator.