particle physics of the early universe€¦ · lecture to the electrodynamics – a system of...

Particle Physics of the early Universe

Alexey BoyarskySpring semester 2015

Why do you need that?

Plan of today’s lecture

Let us apply the formalism that we developed in the previousLecture to the electrodynamics – a system of interacting fermionsplus dynamical electromagnetic field

We will consider three different cases:

– Electron is scattering in the external electromagnetic field– Electron is scattering on the (dynamical) proton– Electron is scattering on its own anti-particle (positron)

Alexey Boyarsky PPEU 1

Electron scattering in Coulomb field

In non-relativistic quantum mechanics if Hamiltonian has the formH = H0+ V then the probability of transition between an initial stateψi(x) and the final state ψf(x) of unperturbed Hamiltonian H0 isgiven by (Landau & Lifshitz, vol. 3, § 43):

dwif =2π

~|Vif |2δ(Ei − Ef)dνf (1)

where |Vif | is the matrix element between initial and final states; and dnf is thenumber of final states with the energy Ef (degeneracy of the energy level).

The interaction that perturbs the Hamiltonian is given by

Vint = eγ0γµAµ(x) (2)

recall that electric current jµ = eψ(x)γµψ(x)

0Following Bjoren & Drell, Sec. 7.1

If we consider static point source with the Coulomb field

A0(x) =Ze

4π|x|, ~A = 0 (3)

and wave-functions1

ψi(x) = C1us(pi)eipi·x , ψf(x) = C2ur(pf)e

ipf ·x (4)

Find C1 that ψi in (4) has the correct normalization per unit flux in the relativisticcase. Normalization of u is given by (8)

Using (2–4) we write the matrix element

Vif = C1C2ur(pf)γ0us(pf)

ix·(pi−pf )A0(x) (5)

1Here us, ur are 4-component spinors – solution of the Dirac equations (γ · p − m)us = 0,ur(γ · p+m) = 0, s = ±, r = ± – polarizations of spin.

Degeneracy of a final state with Ef is given by

dνf = 2× d3pf(2π)3

∫p0>0

dp0 δ(p2 −m2)

︸︷︷︸density of states

(2π)3Ef(6)

Eq. (6) and normalization of the final state ψf should agree witheach other in such a way that∫

dνf ψf(x)ψf(x′) = δ(3)(x− x′) (7)

Notice that this condition fixes the normalization of the spinor uf inEq. (4).

us(pi) and ur(pf) carry the information about spin-polarizations ofinitial and final states. We can sum over these states (i.e. the

experiment does not measure the polarizations). This can be doneusing the identity

us(p)us(p) =

(take into account (/p−m)(/p+m) = p2 −m2)

As a result we get

dwif = 2π|Vif |2δ(Ei − Ef)d3pf

(2π)3Ef

= Z2(4πα)

2|C1|2|C2|2|ur(pf)γ0us(pi)|2

|pi − pf |4d3pf

(2π)3Efδ(Ei − Ef)

as a result the sum over initial and final states in (9) becomes

∑r,s

|ur(pi)γ0us(pf)|2 =

∑r,s

urγ0ususγ

0ur = Tr

0/pi +m

0/pf +m

the spin sum rule leads toaveraging over initial polarizations

|urγ0us|2 ≡ F (pi, pf ,m) = (EiEf + pipf +m

2 (10)

summing over the final polarizations

Using () and representing d3pf = dΩp2fdpf we find the differential

cross-sectiondσ

Z2α2mEi

4|pi|4 sin4(θ2)F (pi, pf ,m) (11)

– which coincides with the familiar Rutherford scattering (up tothe function F (. . . ) that depends on the spins/polarizations of theparticles

Electron scattering on proton

Consider next the situation when the electromagnetic field iscreated by other particle (“proton”)

While the formulas (4)–(6) remain true, the expression for Aµchanges

Notice that in general Aµ(x) created by the moving proton will betime-dependent so instead of Eq. (1) the formula

dwif =2π

~|Vif |2δ(Ei − Ef − q0)dnf (12)

where q0 is the frequency of the perturbation Aµ(t,x) ∝ e−iq0t

If proton is described by a spinor

Ψ = U(P )e−iP ·x, U(P )–4 component spinor (13)

then its electric current is

Jµ(y) = Ψf(y)γ

µΨi(y) (14)

(the form of Ψi and Ψf is the same as Eq. (4) with m → Mp anddifferent momenta)

The electromagnetic field obeys the Klein-Gordon equation

Aµ = Jµ or Aµ(x) =1

Jµ(y) (15)

Very naively, the operator −1 (inverse to the Klein-Gordonoperator) can be easily constructed if one considers (15) in Fourierspace:2

p2Aµ(q) = Jµ(q) or Aµ(q) =

q2Jµ(q) (16)

2We denote by Aµ(p) and Jµ(p) Fourier transform

Jµ(q) ≡∫d

4x e−iq·x

Ψf(x)γµΨi(x) (17)

√√√√ M2p

E(p)i E

Ur(Pi)γµUs(Pf)

4x e−i(q−Pf−Pf )·x (18)

√√√√ M2p

E(p)i E

Ur(Pi)γµUs(Pf)δ(4)

(q − Pf − Pf) (19)

Notice that in Eq. (5) we only need Aµ(pi − pf). The resultingexpression is then equivalent to (5) if one substitutes

|q|2→ γ

√√√√ M2p

E(p)i E

Ur(Pi)γµUs(Pf)δ(4)

(q − Pf − Pf) (20)

where q = pf − pi – transferred 4-momentum and q is its spatial component.4-spinors Us and Ur are the in- and out- 4-spinors of a proton.

From Eq. (5) and (20) we find

∫d3x

∫d3q (uγµu)e−i(q−pi+pf)·xAµ(q) (21)

where Aµ(q) is given by r.h.s. of (20):

The resulting Vif is proportional to

Vif ∝ δ3(pi − pf + P i − P f) (22)

That is the result looks like a scattering of electron in externalfield (9) where the external field Aµ is the field created by the proton(14).

As a result the probability dwif has the form very similar to Eq. (9)

for Z = 1 and

dwif =(. . . )

∣∣ur(pf)γµus(pi)∣∣2∣∣Ur′(Pf)γµUs′(Pi)∣∣2|pi − pf |4

× δ(Ei + E(p)i − Ef − E

(p)f )δ

3(pi − pf + P i − P f)

Interaction of light with the Dirac sea

Interaction of charged particles goes via exchange of virtual photonquanta

Electron-positron scattering3

+Electron-positron scattering

Two differences betweenpositron and proton. One istrivial – mass (not important forour computations so far).

blue diagram is similar to whatwe have seen before (electron-proton scattering with proton →positron).

Another difference: electron plus positron can convert into aphoton

red diagram : electron+positron disappear in the intermediate stateand only a new particle (photon) stays

⇒need a quantum theory of photon2See Bjoren & Drell, Sec. 7.9

Electromagnetic field as collection ofoscillators 4

Consider the solution of wave equation for vector potential ~A ( impose

A0 = 0 and div ~A = 0)

c2∂2A

∂t2−∆A = 0 (24)

Solution

A(x, t) =

∫d3k

(2π)3

[ak(t)eik·x + a∗k(t)e−ik·x

where the complex functions ak(t) have the following time dependence

ak(t) = ake−iωkt, ωk = |k| (26)

3See Landau & Lifshitz, Vol. 4, §2

Generalized coordinate and momentum

ak(t) and a∗k(t) obey the following equations:

ak(t) = −iωkak(t) , a∗k(t) = iωka∗k(t) (27)

Notice that we re-wrote partial differential equation (25) secondorder in time into a set of ordinary differential equations for infiniteset of functions ak(t) and a∗k(t). Any solution of the freeMaxwell’s equation is parametrized by the (infinite) set of complexnumber ak,a

To make the meaning of Eqs. (30) clear, let us introduce theirimaginary and real parts.

Qk ≡ ak + a∗k

P k ≡ −iωk

(ak − a∗k

) dynamics=⇒

Qk = P k

P k = −ω2kQk

Hamiltonian of electromagnetic field

Hamiltonian (total energy) of electromagnetic field is given by

∫d3x

[E2 +B2

∫d3k

(2π)3

k +B2k

Using mode expansion (25) and definition (28) we can write withfrequencies ωk

H[Qk,P k

∫d3k

(2π)3

2 + ω2kQk

Therefore dynamical equations (28) are nothing by the Hamiltonianequations

Qk =∂H∂Pk

, Pk = − ∂H∂Qk

with Hamiltonian (30)

Hamiltonian of electromagnetic field

Eqs. (30)–(31) describe Hamiltonian dynamics of a sum ofindependent oscillators with frequencies ωk

Classical electromagnetic field can be consideredas an infinite sum of oscillators with frequencies ωk

Recall: for quantum mechanical oscillator, described by theHamiltonian

Hosc = − ~2

dx2+mω2

2x2 (32)

one can introduce creation and annihilation operators:

a† =1√

2~mω(mωx+ ~∂x) ; a =

1√2~mω

(mωx− ~∂x) (33)

Commutation [a, a†] = 1

Hamiltonian can be rewritten as Hosc = ~ω(a†a+ 12)

Properties of creation/annihilationoperators

Commutation [a, a†] = 1

If one defines a vacuum |0〉, such that a |0〉 = 0 (Fock vacuum) thena state |n〉 ≡ (a†)n |0〉 is the eigenstate of the Hamiltonian (32) withEn = ~ω(n+ 1

2), n = 0, 1, . . .

Given |n〉, n > 0, a† |n〉 =√n+ 1 |n+ 1〉 and a |n〉 =

√n |n− 1〉

Time evolution of the operators a, a†:

i~∂a

∂t= [Hosc, a] (34)

and Hermitian conjugated for a†

Birth of quantum field theory

Dirac (1927) proposes to treat radiation as a collection of quantumoscillators

Paul A.M. Dirac Quantum theory of emission and absorption of radiation

Proc.Roy.Soc.Lond. A114 (1927) 243

Take the classical solution (25)

A(x, t) =

∫d3k

(2π)3

[ak(t)eik·x + a∗k(t)e−ik·x

Introduce creation/annihilation operators ak, a†k

[ak, a†p] = ~δk,p [ak, ap] = 0 (35)

Replace Eq. (25) with a quantum operator

A(x, t) =

∫d3k

(2π)3

[ak(t)eik·x + a†k(t)e−ik·x

Operator a†k creates photon with momentum k and frequencyωk

Operator ak destroys photon with momentum k andfrequency ωk (if exists in the initial state)

State without photons↔ Fock vacuum:

ak |0〉 = 0 ∀k (37)

State with N photons with momenta k1,k2, . . . ,kN :

|k1,k2, . . . ,kN〉 = a†k1a†k2

. . . a†kN |0〉 (38)

Let us find the average of the operator (36) between a vacuum andthe 1-photon state |k〉 ≡ a†k |0〉 (recall that we are working in the gauge A0 = 0)

〈0| A(x)(a†k |0〉

)=ε(k)√2ωk

e−iωkt+ik·x (39)

where the polarization 3-vectors ε(k) are such that ε · k = 0 andε2 = 1'

Quantum electrodynamics (QED) – first quantum field theory hasbeen created. Free fields with interaction treated perturbatively infine-structure constant: α = e2

Second order perturbation theory

The Dirac Hamiltonian H0 is perturbed by V given byV = γ0γµAµ(x)

Let us start with the blue diagram

We could have proceeded as in the case of electron-protoninteraction – find the electromagnetic field, created by the positronand compute the scattering of electron in this field. Instead, we dothis computation in a different way, and then demonstrate that theresult is the same

Computation of the matrix element

u(p1) v(q1)

u(p′1) v(q′1)

Intermediate 3-particle state

– the initial state |i〉 = |e−(p1), e+(q1)〉 ⊗ |0〉.

The energy of initial state: Ei = E(q1) + E(p1)

– The intermediate state |n〉 = |e−(p′1), e+(q1)〉⊗|k〉.

The energy of intermediate state: En =E(q1) + E(p′1)± ωk

– The final state |f〉 = |e−(p′1), e+(q′1)〉 ⊗ |0〉.

The matrix element for the process can be described as follows:

∫d3k

(2π)3〈i| V |n〉 1

Ei − En〈n| V |f〉 (40)

The matrix element Vin ≡(〈0| ⊗ 〈e+e−|

)V(|e+e−〉 ⊗ |k〉

)is given

by the following expression

Computation of the blue diagram

Vin =u(p′1)γ

µu(p1)

εµ√2ωk

i(p1−p′1−k)·x

ei(E(p1)−ωk−E(p′1)

)t+ ωk → −ωk

=u(p′1)γ

µu(p1)

εµ√2ωk

(p1 − p′1 − k)e

i(E(p1)−ωk−E(p′1)

)t+ ωk → −ωk

(41)(the positron state is the same in |i〉 and |n〉 and therefore

⟨e+∣∣ ∣∣e+

⟩= 1)

Notice that the expression (41) contains explicit time dependence as E(p1) 6=ωk + E(p′1). This is the indication of the fact that 3-momentum and energycannot be conserved at the same time.

Similarly Vnf = δ(3)(q1−q′1 +k)v(q1)γµv(q′1)εµ√2ωk

ei(E(q1)+ωk−E(q′1))t

As a result Mif is given by (after the integral over dn ≡ d3k(2π)3):

∫d3k

(2π)3

VinVnfEi − En

Computation of the blue diagram

Mif = ei(Ei−Ef)tδ(3)(p′1 + q′1 − p1 − q1)(v(q1)γµv(q′1))

(u(p′1)γµu(p1)

)(E(p1)− E(p′1)

)2 − ω2k

∣∣∣∣k=p1−p′1

Show that the denominator of (42) is equal to 1k2 where kµ is 4-vector equal to

the difference (p1 − p′1)µ.

notice, that time dependence will cancel out from (42) when we takeinto account δ(Ei − Ef)

Blue diagram can be thought of as 2nd order perturbation theorywith the 3-particle intermediate state |n〉, containing photon.

Notice that the matrix element (42) coincides with the matrixelement (22) (taking into account (21))

Computation of the red diagram

There is another intermediate state where there is only photon —red diagram

In this case we have different intermediate state: |n〉 = |0〉 ⊗ |k〉.The energy of intermediate state: En = ±ωk

Matrix element

Vin ≡⟨e+e−

∣∣ V |k〉 = v(q1)γµu(p1)εµ√2ωk

∫d3x ei(p1+q1−k)·xei(Ei−ωk)t

=v(q1)γµu(p1)εµ√2ωk

δ(3)(p1 + q1 − k)ei(Ei−ωk)t + ωk ↔ −ωk(43)

where the initial energy Ei = E(p1) + E(q1) .

Similarly, for the Vnf element we get

Vnf = u(p′1)γµv(q′1)εµ√2ωk

δ(3)(p′1 + q′1 − k)ei(ωk−Ef)t (44)

Computation of the red diagram

As a result, we get:

VinVnf

Ei − En

= ei(Ei−Ef )t

(p′1 + q

′1 − p1 − q1)

(v(q1)γµu(p1))(u(p′1)γ

µv(q′1))

E2i − ω2

∣∣∣∣k=p1+q1

. . . here Ei = E(p1) + E(q1) – initial energy of electron+positron,k = p1 + q1 – their total momentum of the pair

. . . again one can show that E2i −ω2

k = k2 where k is a 4-momentumof the intermediate photon, k = p1 + q1

Total matrix element

Amplitudes of two processes (blue and red on the Figure) shouldbe added together before | . . . |2 is taken. That is the probability ofthe process is proportional to |M1+M2|2 rather than |M1|2+|M2|2(interference terms are present)

|M|2 = (. . . )

∣∣∣∣∣∣∣∣∣∣i

(u(p′1)γµu(p1)

)(v(q1)γµv(q′1)

)(p1 − p′1)2︸︷︷︸blue diagram

− i(u(p′1)γµv(q′1)

)(v(q1)γµu(p1)

)(p1 + q1)2︸︷︷︸red diagram

∣∣∣∣∣∣∣∣∣∣

where . . . is a prefactor, depending on energies/masses of particles.

Perturbative series in V

Mif = M(1)if +M

(2)if + . . .

We saw that M (1)if is equal to zero (Eq. (1)). If non-zero, M (1)

if ∝ e

(charge in V )

Total matrix element

We found M (2)if (2nd order perturbation theory in V )

This expression M (2)if is proportional to e2

This parameter e is known experimentally to be small. Theexpansion parameter of electron-photon interaction is known as thefine structure constant α ≡ e2

~c ≈ 1137

Indeed, consider next order in expansion. It includes moreintermediate states – higher order in e

Next order

There are e+e− scattering processes that go through severalintermediate states:

M(3)if =

∫dn1

∫dn2

Vin1Vn1n2Vn2f

(Ei − En1)(En1 − En2)

M(4)if =

∫dn1

∫dn2

∫dn3

Vin1Vn1n2Vn2n3Vn3f

(Ei − En1)(En1 − En2)(En2 − En3)

Virtual particles

In the above computations we have explicitly separate space andtime.

The intermediate states were “physical” (energy and momentumwere related via E2 = p2 +m2), but only 3-momentum conservationwas imposed in every vertex. The total (initial - final) energy wasconserved, but for intermediate states it was not

It is possible to construct explicitly Lorentz-invariant technique forcomputation of such matrix elements

This is called Feynman technique. Its rules are presented in

Feynman rules

There are three types of objects in constructing Feynman graphs:

– external lines (real particles)– internal lines (virtual particles)– vertices (interaction points)

To each external fermion line one associates a spinor u, v , etc.according to the following rule:

To each external photon line one associates a polarization vector

Feynman rules

Each virtual line adds a propagator:

– Virtual fermion:

SF (p) =i(/p+m)

p2 −m2 + iε

– Virtual photon:

Dµν(p) =−iηµνp2 + iε

Each vertex (two fermion lines plus one photon line) receives afactor −ieγµ

Feynman rules

Energy-momentum conservation is imposed at every vertex

Electron-positron scattering

Let us repeat the computation of electron-positron scattering

u(p1) v(q1)

u(p′1) v(q′1)

Dµν(p1 − p′1)

(u(p′1)(−ieγµ)u(p1)

)(v(q1)(−ieγµ)v(q′1)

)(p1 − p′1)2

Similarly

(u(p′1)(−ieγµ)v(q′1)

)(v(q1)(−ieγµ)u(p1)

)(p1 + q1)2

Compton scattering

Write general form of matrix elements for Compton scattering

Derive differential cross-section in non-relativistic case (photonenergy ω me) and in ultra-relativistic case (ω me)

Peskin & Schroeder, Sec. 5.5

Pair creation

Compute cross-section of γ + γ → e+ + e− (pair production)

Derive differential cross-section in ultra-relativistic case (ω me)

Peskin & Schroeder, Sec. 5.5

Onther consequences of Dirac theory ofpositrons

Photons are bosons (particles of spin = 1). Electrons/positrons arefermions particles of spin = 1/2. Therefore, angular momentumconservation means that photon couples to electron + positron

Photons could produce electron-positron pairs. However, the

process γ → e+e− is not possible if all particle are “real” (i.e.

photon obeys E = cp, electron/positron E =√p2c2 +m2

ec4 – “on-shell

conditions”)

Instead, a pair of photons can produce electron-positron pair via

γ + γ → e+e− :γ

where (k1 + k2)2 ≥ 4m2e

Similarly, electron-positron pair can annihilate into a pair of

Onther consequences of Dirac theory ofpositrons

photons

Kinematically, the red electron is virtual (i.e. for it E 6=√p2c2 +m2c4

– check this)

k′1If energies of incoming photons aresmaller than twice the electron mass (i.e.(k1 + k2)2 < 4m2

e) photons produceonly virtual electron-positron pair whichcan then “annihilate” into another pair ofphotons – light-on-light scattering

Charge screening:

Systems with many particles

Presence of the negative-energy levels means that you can createparticle-antiparticle pairs out of “nowhere”

Particles in the pair can be real, but they can be also virtual (i.e.E2 − p2 6= m2)

According to the Heisenberg uncertainty relation ∆E ∆t & 1, ifone measures the state of system two times, separated by a shortperiod ∆t 1/m, one will find a state with 1, 2, 3, ... additional pairs.

It means that we no longer work with definite number of particles:number of particles may change! (Contrary to non-relativisticquantum mechanics)

We need an approach that naturally takes into account states withdifferent number of particles (we will return to this point in thisLecture)

Quantum electrodynamics (QED) – first quantum field theory.Free fields with interaction treated perturbatively in fine-structureconstant: α = e2

Divergencies? Many answers beyond tree-level (1st order)perturbation theory were infinite, because one had to sum upcontributions from infinite number of virtual particles with growingenergies Ep = ±

√m2 + p2

particle physics of the early universe€¦ · lecture to the electrodynamics – a system of...

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