Lecture From Dirac equation to Feynman diagrammssvogel/lecture_ss_2012/notes... · Lecture From Dirac equation to Feynman diagramms ... is then obtained by summing the potentials

1

Lecture Lecture

From Dirac equation to From Dirac equation to

Feynman diagrammsFeynman diagramms

SS2012SS2012: : ‚‚Introduction to Nuclear and Particle Physics, Part 2Introduction to Nuclear and Particle Physics, Part 2‘‘

2

Part I: Part I:

Dirac equationDirac equation


The Dirac equation - the wave-equation for free relativistic fermions follows the

requirements :

1) that the wave-equation – as in case of the Schrödinger equation – should be of

1st order in ∂/∂t ≡∂/∂x0

2) to allow for a continuity equation with a positive density ψ*ψ:

3) relativistic covariance (with respect to Lorentz transformations) then requires

that the wave-equation also has to be of 1st order in the spacial derivatives

∂/∂xk (k = 1, 2, 3) , i.e.:

3


(1)

This equation can be rewritten in covariant notation:

4


(2)

(3)

The covariant form of the

Dirac equation:

4-momentum

4-coordinate

covariant derivative

electromagnetic 4-potential

4-current

(4)

then involves

with the four-vector coefficients

Further four-vectors are given by:

5


(7)

where we have employed the pseudometric (Lorentz invariant) tensor:

Scalar products are Lorentz invariant, e.g. the invariant mass

with

Thus we have:

(5)

(6)

6


(11)

Including the interaction with vector fields Aµµµµ implies:

Then the Dirac equation reads:

The (anti-commutator) algebra of the γ-matrices has to follow:

with the properties:

By counting the number of boundary conditions the γ-matrices have to be 4x4

matrices and consequently the wavefunctions ΨΨΨΨ(x)must have 4 components

(8)

(9)

(10)

7


Then we get:

which reads explicitly:

(12)

(13)

(14)

(15)

The solution of the Dirac equation are plane waves with positive and negative energies

separate the four components wave vector ψψψψ into two vectors with 2 components ϕ, χϕ, χϕ, χϕ, χfor spin ‚up‘ and ‚down‘ (relative to the z-direction = direction of motion):

using

8

Dirac equation: fermionsDirac equation: fermions

The solution of the coupled equation (15) reads:

where σk (k=1,2,3) are the Pauli matrices.

(16)

I. Consider the positive energy

Since the components are two-vectors, we may expand them as

spin ‚up‘ spin ‚down‘

N is the normalization factor

(17)

9

Dirac equation: fermionsDirac equation: fermions

Then

In matrix notation:

The solutions of the Dirac equation then read explicitly

for fermions with spin ‚up‘ and spin ‚down‘:

(18)

(19)

(20)

8

Dirac equation: antiDirac equation: anti--fermionsfermions

II. Consider the negative energy states = anti-fermions

Using

we obtain for the anti-fermion components with spin ‚up‘ and ‚down‘:

Accordingly free (anti-)fermions are fully defined by the spinors specified above!

(21)

Commonly one uses 2 ways of normalization:

1) as in Bjorken, Drell *:

thus,

in the rest-frame (E=m):

8

Dirac equation: normalizationDirac equation: normalization

2) the normalization used here (e.g. as Aitchison, Hey):

(22)

(23)

Constrain for the normalization:

(24)

8

Dirac spinorsDirac spinors

Free (anti-)fermions are fully defined by the spinors specified above (with normalization (24)):

1) Spinors with positive energy (fermions):

2) Spinors with negative energy (anti-fermions):

Wave vector ψ ψ ψ ψ :::: fermions

anti-fermions

spin ‚up‘ : spin ‚down‘

(27)

(26)

(25)

Dirac equation: positive and negative energy statesDirac equation: positive and negative energy states

Interpretation of the solutions with positive and negative energies:

1) Dirac (1930): particle-hole picture

E > 0: particles

E < 0: hole states =anti-paticles

Dirac sea

particles

anti-particles=holes2) Feynman picture:

E<0, e<0 anti-particles: travelling back in time

Emission of an antiparticle with 4-momentum pµµµµ is equivalent

to the absorption of a particle with 4-momentum - pµµµµ

Absorption of an antiparticle with 4-momentum pµµµµ is equivalent

to the emission of a particle with 4-momentum - pµµµµ

14

ππππ+ scattering

1)1)1)1) ππππ+ - scattering

on a time dependent electromagnetic

potential V(t)~e-iωωωωt Interaction by

electromagnetic potential

Matrix element:

at time t : ππππ+ -meson absorbs the photon of energy and increases its energyωωωωh

absorbtion at time t of the photon

of energy ωωωωh

15

π π π π - scattering

2)2)2)2) π π π π - - scattering

Matrix element:

π- - scattering π π π π + - scattering

with positive energy with negative energy

Energy of π π π π – is equal to the energy of ππππ+ -meson

16

ππππ+ π π π π - -pair production

3)3)3)3) ππππ+ π π π π - - pair production/creation

Matrix element:

The sum of π π π π – and ππππ+ meson energies is equal to the energy of the absorbed photon

17

ππππ+ π π π π - -pair annihilation

4)4)4)4) ππππ+ π π π π - - pair annihilation

Matrix element:

The energy of π π π π – and ππππ+ mesons is equal to the energy of the produced photon

In order to see how to solve the inhomogenuous Dirac equation (30) for electrons in

an electromagnetic field let‘s first consider the example from electrostatics -

solution of Poisson equation:

18

Dirac equation: Green functionsDirac equation: Green functions

The Dirac equation for electrons in an electromagnetic field can be obtained from the free

Dirac equation (2) by the substitution (minimal coupling)

Notation:

(28)

(29)

(30)

(31)

Here ρρρρ(x) is the free charge density.

• For a pointlike charge, i.e.

the static Coulomb potential - solution of (31) - is known:

(32)

(33)

19


For a continuous charge distribution ρρρρ(x), the solution of (31) is then obtained by summing

the potentials for all particles:

(34)

(35)

The Poisson equation may be solved also using a Green's function

which is obtained by solving the point source equation:

Then (36)

Using (36), the Poisson eq. (31) can be re-written as

(37)

20


(38)

The solution of eq. (36) is the spatial Green‘s function:

To solve the Dirac eq. (30) one defines the Green function K(x,x‘) (where x,x‘ are 4-vectors)

by the requirement(39)

(40)

Thus, the solution of the inhomogenuous Dirac eq. (30) reads

Green function = integration kernel

Indeed:

21

Electron propagatorElectron propagator

The general solution of the inhomogenuous Dirac eq. (30) reads

homogenuous solution of

free Dirac equation

inhomogenuous solution of Dirac

equation with electromagnetic potential

Since the coupling constant is weak,

one can use the perturbation theory:

(41)

(42)

22


The explicit form of the Green‘s function can be written as a Fourier transform

(42)

(43)

Substitute (42) in the eq. for the Green‘s function (39)

Multiply (43) by and one gets

(44)forElectron propagator:

Note: propagator (44) is defined only for virtual electrons, since for real electrons

23


Thus, the Green‘s function is

for positive energy states:

(45)

In (45) integral over p0 has 2 poles:

The integral in (45) can be evaluated by the method of residues by closing the

contour in the lower(upper) half of the p0 -plane

24


Method of residues:

the integral is equal to 2ππππ i times the residue of the integrand at the poles:

1) For

(46)

(47)

(48)

2) For

25


The integral (45) can be evaluated also by integration along Re(p0) line, however, by shifting

the poles by an infinitesimal positive value ε (ε ε (ε ε (ε ε (ε 0)0)0)0):

(49)

Thus, the electron propagator reads:

(50)

26


The name ‚propagator‘ is also used for the Green‘s function since K(x,x‘) describes

the propagation of the particles from x to x‘ :

•The wavefunction at the final space time point x‘ w.f. of a free particle with

positive energy = free plane waves:

•The wavefunction at space-time point x:

(51)

(52)

Indeed, for t > t‘ using eq.(47)

(53)

27


A wave function of positive energy will spread only forward in time and not backward in

time, i.e. for t < t’ one gets:

Thus, for the wave function with positive energy (k0 > 0):

=0 from Dirac eq.

In a similar way one can show that for negative energy (k0 < 0) by virtue of K(x-x’) the

wave function only propagates backward in time.

(54)

(55)

28

Part II:Part II:

Feynman diagrammsFeynman diagramms


29

Photon propagatorPhoton propagator

(1)

Electron-proton scattering by an exchange of virtual photons

(‚Dirac-photons‘)

(2)

The photon vector field Aµµµµ follows the wave equation:

where Jµµµµ is the proton 4-current and

(3)

Solve the inhomogenuous wave equation (2) using the Green function:

The inhomogenuous solution of equation (2) can be written as

(4)

(5)

in Lorentz gauge.

Indeed, using (4) and (3) one obtaines eq. (2) again:

e-

e-

p

p

virtual photon

30

Photon propagatorPhoton propagator

(6)

The Green‘s function can be written as a Fourier transform

(7)

From (7) we obtain for the photon propagator:

(8)

Consequences: A wave function of positive energy will spread only forward in

time t > t’ and of negative energy - backward in time t < t’

31

Charge 4Charge 4--currentcurrent

(9)

The proton charge 4-current Jµµµµ can be written as:

where ψψψψi and ψψψψf are the spinors of the proton in the initial and final states:

(10)

(11)

From (10) and (9) we get

substitute (8) into (4)

(12)

(13)

The photon vector field Aµµµµ::::

0uu γγγγ++++≡≡≡≡

32

Feynmann diagrammsFeynmann diagramms

(14)

For the free electron:

(15)

(16)

electron virtual photon proton

The matrix element (16) can be presented as a Feynman diagramm:

S-matrix element

33


For the electron:

For the photon:

For the incoming and outgoing photon we have:

34


Electron propagator:

S-matrix element:

Electron current :

Proton current:

35


emission absorption of photon

absorption emission

of virtual photon

1) Scattering of charged particles

2) Compton scattering emission of real photon

virtual electron

absorption of real photon

emission of real photon

virtual electron

absorption of real photon

36


3) Bremsstrahlung emission of real photon

virtual electron

Ze

absorption of virtual photon

virtual photon (time-like)

4) Pair creation and annihilation

37

Scattering of electrons on an external potentialScattering of electrons on an external potential

Consider the scattering of electrons on an external Coulomb potential

In first order perturbation theory : particles = plane waves

photon vector field Aµ µ µ µ ::::

S-matrix element:

use that

(16a)

Ei, Ef are total initial and final energy of the system energy conservation to ∞∞∞∞→→→→t

38

Scattering of electrons on an external potentialScattering of electrons on an external potential

S-matrix element is

(17)

(18)

(19)

• Potential acts during the time period T (i.e. T is the interaction time) : -T/2 < t < T/2

Off-shell function f(ωωωω):

maximum at ωωωω=0 (Ef=Ei), the amplitude ~T2;

width ~1/T

matrix element

During the time -T/2 < t < T/2 the system can be in a

state within the energy interval Ef , Ef+δδδδE f

ρ(ρ(ρ(ρ(Ef )δδδδE f – number of energy levels in this interval

ρ(ρ(ρ(ρ(Ef ) – level density = number of states per energy

interval

39

Feynmann diagramms:Feynmann diagramms: scattering of electrons scattering of electrons

(20)

(21)

Uncertainty relation:

If t >>T, f(ω)ω)ω)ω) δδδδ-function

(22)

Thus, transition probability W is obtained by an integration of dW over dE f :

Differential transition probability dW is

40

Feynmann diagramms:Feynmann diagramms: scattering of electronsscattering of electrons

For t >>T, f(ω)ω)ω)ω) δδδδ-function

(22)

(23)

Gold rule from Fermi:

w is the transition probability per unit time: w = W/T

Total cross section:

= transition probability per unit time over jein - initial current:

(24)

41

Feynmann diagramms:Feynmann diagramms: scattering of electronsscattering of electrons

The scattering probability (17) is then

(25)

(26)

(27)

Furthermore, perfom the integration over the phase space:

i.e. over the number of levels in the energy interval Ef , Ef+δδδδE f and the direction of

the particle in the solid angle element dΩΩΩΩf

Use that

factor 1/2Ef due to the normalization of 2Ef particles per volume V

)(2T ωωωωπδπδπδπδ⇒⇒⇒⇒∞∞∞∞→→→→

S-matrix element

42

Feynmann diagramms: Feynmann diagramms: scattering of electronsscattering of electrons

Then, the differential transition probability is

matrix element

(28)

(29)Initial current:

Differential cross section:

Using the explicit form for the matrix element (17), we obtain:

(30)

(31)

where since

43

Feynmann diagramms: nonFeynmann diagramms: non--relativistic caserelativistic case

Consider the non-relativistic limit:

(32)

(33)

From (31) Rutherford formula

44

Feynmann diagramms: relativistic caseFeynmann diagramms: relativistic case

Consider the scattering of relativistic electrons

For fermions:

1) avaraging over the spin of initial fermions

2) summation over the spin of the final fermions

∑∑∑∑iS2

1

∑∑∑∑fS

∑∑∑∑fi S,S2

1

Consider

Spin avaraging:

(34)

(35)

(36)

For different components α,β:α,β:α,β:α,β:(37)

8

Dirac spinorsDirac spinors

Free (anti-)fermions are fully defined by the spinors specified above (with normalization (L7.24)):

1) Spinors with positive energy (fermions):

2) Spinors with negative energy (anti-fermions):

Wave vector ψ ψ ψ ψ :::: fermions

anti-fermions

spin ‚up‘ : spin ‚down‘

(L7.27)

(L7.26)

(L7.25)

46

Spinor tensorsSpinor tensors

(37)

Summing over spin gives:

0uu γγγγ++++≡≡≡≡Notation:

47


Summing over spins of the initial and final fermions – we may write in matrix form:

Notation: Spur=Sp=Tr

(38)

(39)

(40)

(41)

with

Result:

48


(42)where I is the 4x4 unitary matrix

2) Spur (even number of γγγγµµµµ) =0

Notation: a≡≡≡≡µµµµµµµµµµµµ γγγγγγγγγγγγγγγγγγγγ ========

++++ 00Use that

Using (42), averaging over the spin of the initial fermions and summation

over the spin of final fermions leads to:

(42)

(43)

mass of electron

49


Evaluate the scalar product in the cms of the process i+f:

(44)

(45)

Note: non-relativisticely:

(46)

E

p≡≡≡≡ββββ

Thus, substitute (45) in (31) :

eq. (46) describes the scattering of a relativistic spin ½ particle on a spin 0 target

(e.g. nucleus) with large mass M !

50


For the high energy electrons, i.e. E>>m, ββββ1

we obtain the Mott cross section including the backscattering of the target of finite mass M:

eq. (47) describes the scattering of a relativistic spin ½ electron on a spin 0 target

with large mass M (e.g. nuclei, pion)

(47)

51

Feynmann diagramms: electronFeynmann diagramms: electron--muon scatteringmuon scattering

Electron-muon scattering by an exchange of virtual photons

S-matrix element

(1)

(2)

(3)

(4)

Matrix element

S-matrix element squared is proportional to the 4-volume V0T,

V0 is spacial volume of interaction, T is interaction time :

The differential cross section is proportional to S-matrix element per 4-volume V0T:

)p(u)p(uq

1)p(u)p(u

V

1ie)p(u)p(u

iq

g)p(u)p(u

V

1ieM 242132

2

242132

2)1(

fi

µµµµµµµµνννν

µνµνµνµν

µµµµ γγγγγγγγγγγγεεεε

γγγγ ====++++

====

electron current muon current

photon propagator

52

Differential cross sectionDifferential cross section

The definition of the differential cross section dσσσσ for 2 particle scattering:

(5)

a) matrix element x δδδδ4444-function

b) initial current

c) number of target particles per volume (consider a stationary target)

d) number of different final states of the 2-body system c,d

υ υ υ υ is relative velocity of

initial particles a and b : 21 υυυυυυυυυυυυrrr

−−−−====

53

Differential cross section Differential cross section

The product of the factors in b) and c) can be written in a Lorentz invariant way

and is denoted as F = flux :

F = (6)

Thus, the definition of the differential cross section dσσσσ for 2 particle scattering:

Flux x matrix element x Lorentz invariant phase space

Lorentz invariant phase space:

(7)

(98)

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Lecture From Dirac equation to Feynman diagrammssvogel/lecture_ss_2012/notes... · Lecture From Dirac equation to Feynman diagramms ... is then obtained by summing the potentials