Particle Physics of the early Universe

Particle Physics of the early Universe

Alexey BoyarskySpring semester 2015

Why do you need that?

Things I consider known

Beginning of the XXth century: attempts to explain atomic and nuclearphysics, taking into account Quantum Mechanics and Relativity.

The state of the system is described by the complex wavefunction ψ(x)

|ψ(x)|2 is interpreted as probability density and∫d3x|ψ(x)|2 = const

as the full probability

~J = i~2m(ψ∗∇ψ−ψ∇ψ∗) = |ψ|2

m ∇ argψ is interpreted as a probabilitydensity current so that

∂|ψ|2

∂t+ div ~J = 0 (1)

The quantum system is described by Schrodinger equation:

i~∂ψ(x, t)

∂t= Hψ(x, t) (2)

Alexey Boyarsky PPEU 1

Things I consider known

where the operator H is called Hamiltonian

The operator Hamiltonian is built based on correspondenceprinciple: if a classical system is described by the HamiltonianH(x, p) then the quantum system is described by H(x,−i~∂x). Forexample for a particle of mass m in external field we get

Classical H(x, p) =p2

2m+V (x)⇒ Quantum H = − ~2

2m

∂2

∂x2+V (x) (3)

Uncertainty principle ∆x∆p ∼ ~

Solution of the Schodinger equation for the hydrogen atom waspredicting the energy spectrum

En = −mee4

2~2n2, n = 1, 2, . . . (4)


QM perturbation theory

(Stationary) perturbation theory:

– Full Hamiltonian H = H0 + V– Unperturbed spectrum H0ψ

(0)n = E

(0)n ψ

(0)n

– Corrections to energy due to perturbation: E(1)n = E

(0)n +〈n|V |n〉

– Corrections to the wave function:

ψ(1)n =

∑k 6=n

VknEk − En

ψ(0)k +

∫dν

VνnEν − En

ψ(0)ν (5)

if there are states of continuous spectrum

The probability of transition between the states n and k are givenby

Pkn =∣∣∣⟨ψ(1)

n

∣∣∣ ψ(0)k

⟩∣∣∣2 =

∣∣∣∣ VknEk − En

∣∣∣∣2 (6)


The time-dependent perturbation theory

Let perturbation be time-dependent V = V (t). In this case theprobability to transition between the states k and n is given by

Pkn(t) =

∣∣∣∣∫ t

−∞dt′⟨ψ

(0)k

∣∣∣V (t′)∣∣∣ψ(0)n

⟩ei(E

(0)k−E(0)

n

)t′∣∣∣∣2 (7)

Important case: the energy spectrum is continuous. For example:H0 = ~p 2

2m or H0 =√~p2 +m2 – free particle

Degenerate spectrum: for each energy there is a continuum ofallowed momenta, each defines an eigenstate for the same E

dWi→f(Ω) ≡ |Pif |2dνf = |Pif |2p2dpdΩ

(2π)3(8)


The time-dependent perturbation theory

Consider interaction V (t) that is non-zero only during the finite timeperiod. Then at t → +∞ and t → −∞ the system is free and itsenergy is defined by the same H0. Expect conservation of energy

Consider a periodic V (t) = V0eiωt−λ|t|. The number of transitions

per unit time

dPkn(t)

dt=d

dt

∣∣∣∣∫ t

−∞dt′ 〈ψf |V (t

′) |ψi〉 ei(Ef−Ei)t

′∣∣∣∣2 →λ→0

2π|Vif |2δ(Ef−Ei−ω)dνf

(9)(Landau & Lifshitz, vol. 3, § 43)

Formula for transitions between the states of continuous spectrumis given by a formal limit ω → 0:

dwif =2π

~|Vif |2δ(Ei − Ef)dνf (10)


Cross-section

Formula (10), dwif = 2π~ |Vif |

2δ(Ei − Ef)dνf , has a differentinterpretation if in-state is normalized in such a way that it gives anumber of particles per unit time per unit area (flux). For example,for non-relativistic particle

ψi(x) =

√m

piei~pi·~x =⇒ Flux =

|ψ|2

m∇ argψ = 1 (11)

Final states, ~pf , are normalized as∫dx 〈~p1|~p2〉 = (2π)3δ(3)(~p1 − ~p2)

with this normalization Eq. (10) has the dimension of length2 andis called differential cross-section

dσ

dΩ= 2π

∫4πp2

fdpf

(2π)3δ(Ei − Ef(pf)

)|Vif |2 (12)

where dΩ is a solid angle of a scattered particle


Second order perturbation theory

If the transition i → f is forbidden, the analog of Eq. (6) reads (c.f.Landau & Lifshitz, vol. 3, § 43):

Pif =

∣∣∣∣∑′

n

〈i|V |n〉〈n|V |f〉(Ei − En)(En − Ef)

∣∣∣∣2 (13)

For the continuous spectrum

dwif =2π

~

∣∣∣∣∫ dν〈i|V |ν〉〈ν|V |f〉

Ei − Eν

∣∣∣∣2 δ(Ei − Ef)dνf (14)

where∫dν is the integral over any basis of intermediate states.


Pauli Hamiltonian

In order to explain the emission spectra of elements Pauli introduceda spin (“a two-valued quantum degree of freedom” ). Its interpretationwas not known. Pauli postulated Pauli (Pauli-Schrodinger) equation

HPauli =1

2m

[(−i~~∇− e ~A

)2

− e~~σ · ~B]

+ eA01 (15)

where Pauli matrices ~σ = (σx, σy, σz) are defined as

σx =

(0 1

1 0

), σy =

(0 −ii 0

), σz =

(1 0

0 −1

)(16)

and 1 is 2× 2 identity matrix. Pauli Hamiltonian acts on the 2-componentwave-function

i~∂

∂t

(ψ↑ψ↓

)= HPauli

(ψ↑ψ↓

)(17)

Eq. (15) can be rewritten as

HPauli =1

2m

[~σ ·(−i~~∇− e ~A

)]2+ eA01 (18)


Quantum mechanics and relativity

Attempts to blend quantum mechanics and special relativity led to theproblem of infinities.1

Electromagnetic energy of uniformly charged ball of radius reshould be smaller than the total mass of electron (mec

2)?

mec2 >

e2

re=⇒ re >

e2

mc2= 3× 10−13 cm (19)

– larger than the size of atomic nucleus (Hydrogen rH ∼ 1.75 ×10−13 m)

Pauli’s idea of spin meant that electron possesses magneticmoment: µe = e~

2mec. The energy of magnetic sphere with radius

re would be µ2er3e

which would exceed mec2 for r ∼ 1

me

(e~c4

)2/3(even

larger distances than (19)!

1See an interesting exposition of the historical perspectives at http://people.bu.edu/gorelik/cGh_Bronstein_UFN-200510_Engl.htm, Sec. 4


http://people.bu.edu/gorelik/cGh_Bronstein_UFN-200510_Engl.htm

http://people.bu.edu/gorelik/cGh_Bronstein_UFN-200510_Engl.htm

Relativistic Schodinger equation 2

Recall that the Schrodinger equation can be built using thecorrespondence principle:

E =p2

2m⇒

E → i~∂

∂t

p→ −i~∇

⇒ i~∂ψ

∂t=

(i~∇)2

2mψ (20)

Free Schrodinger equation is not Lorentz invariant. Idea: Replacer.h.s. by relativistic dispersion relation E =

√p2c2 +m2c4

Relativistic Schrodinger equation ? i~∂ψ

∂t=

√−c2~2~∇2 +m2c4 ψ (21)

1The presentation of this topic follows Bjorken & Drell, Chap. 1, Sec. 1.1–1.3


Relativistic Schodinger equation 3

How to make sense of square root? This is a non-local operator:√−c2~2~∇2 +m2c4ψ(x) =

∫d3x′

∫d3p eip(x−x

′)√p2 +m2ψ(x′)

(22)non-compatible with the finite speed of light propagation (ψ(x) isdefined via values of ψ(x′) in arbitrary far points in space)


Klein-Gordon equation

Try to take square of Eq. (21)?

− ~2∂2ψ

∂t2=(−~2c2∇2 +m2c4

)ψ ⇔

( +

(mc~

)2)ψ = 0 (23)

This is Klein-Gordon equation.

By taking square of (21) we have included negative energy states(with E = −

√p2c2 +m2c4)

Probabilistic interpretation is gone? Indeed, let us checkprobability density |ψ|2:

∂

∂t

∫d3x |ψ(x)|2 6= 0 (24)

show that (24) is conserved as a consequence of the 1st order intime dynamics


Klein-Gordon equation

The following current is conserved (show!) as a consequence ofKlein-Gordon equation:

Jµ = i

(ψ∗

∂ψ

∂xµ− ψ∂ψ

∗

∂xµ

)(25)

Can we use J0 as the probability density? No!

J0 = i

(ψ∗∂ψ

∂t− ψ∂ψ

∗

∂t

)= −|ψ|2∂(argψ)

∂t< 0 (26)

for any negative energy solution ψ = e+i|E|t we could have chosena different sign in Eq. (74), but for any choice some of the configurations wouldhave negative probability interpretation.


Dirac equation 4

Dirac (1928) had constructed linear in time evolution. Indeed, ifi~∂tψ = Hψ then the probability density is conserved for anyHermitian Hamiltonian

Dirac proposed the form

i~∂ψ

∂t= Hψ (27)

where the function H is a linear function of p and m and is aHermitian matrix :

HDirac = α · p+ βm (28)

where matrices α = (αx, αy, αz) and β are Hermitian matrices thatshould obey the following identities (for the plane wave function

ψ(x) = u(p)e−ipx (29)3The presentation of this topic similar to that of Bjorken & Drell, Chap. 1, Sec. 1.1–1.3


Dirac equation 6

with p2 = m2 to be a solution (for some choice of a vector u(p))

This means that for i, j = 1, 2, 3

αiαj + αjαi = 2δij (30)

αiβ + βαi = 0 (31)

and additionally β2 = 1

⇒Both αi nor β are not ordinary numbers (they do not commute).

Eq. (30) looks very much like the relation for the Pauli matrices. Butguess that αi = σi is wrong there are only 3 Pauli matrices and weneed 4


Dirac equation 7

Another guess:5

αi =

(0 σiσi 0

); β =

(1 00 −1

)(32)

Show that properties (30)–(31) are satisfied by (32). Find different set ofmatrices αi, β

i~∂ψ

∂t=[−i~c

(αx

∂

∂x+ αy

∂

∂y+ αz

∂

∂z

)+ βmc2

]︸︷︷︸

Dirac Hamiltonian

ψ (33)

5Show that αi, β are even-dimensional matrices


Dirac equation

The Dirac equation(i∂

∂xµ(γµ)αβ−m δαβ

)ψβ = 0 (34)

is the Euler-Lagrange equation of the following action:

SDirac[ψ, ψ] =

∫d4x ψα

(i∂

∂xµ(γµ)αβ−m δαβ

)ψβ (35)

where ψ ≡ ψ†γ0 – independent spinor

involves 4 Dirac matrices γµ (index µ = 0, 1, 2, 3), each matrix havingsize 4× 4 (indices α, β run over their dimensions)

γ0 =

(1 00 −1

); γi =

(0 σi−σi 0

)(36)


Dirac equation

The “plane wave solution” of the Dirac equation

ψ(x) = u(p)e−ipx (37)

where 4-component complex spinor

u(p) =

u1

u2

u3

u4

obeys the equation

(pµγµ −m)u(p) = 0 (38)

show that det(pµγµ−m) = p2−m2 and therefore non-trivial solution

u(p) exists only if particle has E2 = p2 +m2


Spinors

Lorentz transformation Λµ′ν preserves xµxµ in Minkowski space

xµ′= Λµ

′ν x

ν (39)

Matrix Λ is a 4× 4 matrix

Let us define a 2× 2 matrix:

x = x01 + ~x~σ =

(x0 + x3 x1 − ix2

x1 + ix2 x0 − x3

)(40)

Notice that det(x) = (x0)2 − (x1)2 − (x2)2 − (x3)2 = xµxµ

Transformation (39) will induce a transformation x→ x′ via

xΛ−→ x′ = UxU+ (41)

where U(Λ) is a complex matrix 2× 2


Spinors

Lorentz transformation preserves the length of the vector xµ

⇒det(x) = det(x′)

det(UxU†) = det(x)|det(U)|2 (42)

U acts as a linear operator on the space of the complex vectors ψof length 2:

ψΛ−→ ψ′ = Uψ (43)

Thus for every matrix Λ that rotates a vector xµ we haveconstructed a matrix U that rotates a 2-component complexvector ψ. Every such a mapping (from Λ to U ) is called arepresentation. The representation (43) is known as spinorrepresentation. Vector ψ is called spinor


Lorentz transformation

Consider an infinitesimal (very small) Lorentz transformation:

Λ = 1− β~n1~α+ ϕ~n2~Σ (44)

where β = arctanh(|~v|) is a boost parameter, ϕ is a rotation angle,~n1 and ~n2 are unit vectors in the boost direction and rotation axisrespectively. ~α and ~Σ are vectors of matrices:

~α =

( 0 11 0

0

0 0

),

0 1 00 0

1 00 0

0

,

0 0 10 0

0 01 0

0

~Σ =

( 0 0

0 0 −11 0

),

0 0 00 −1

0 00 1

0

,

0 0 0−1 0

0 10 0

0


Lorentz transformation

The infinitesimal transformation can be written in 3D manner:

x0′ = x0 − β~n1~x (45)

~x′ = ~x− β~n1x0 + ϕ[~n2 × ~x] (46)

Finite Λ(β, ~n1, ϕ, ~n2) can be found using exponential formula:

Λ(β, ~n1, ϕ, ~n2) = e−β~n1~α+ϕ~n2~Σ (47)

In particular the boost along x axis and rotation around z axis:

Λ(β, ~nx, 0,~0) =

cosh β − sinh β 0 0− sinh β cosh β 0 0

0 0 1 00 0 0 1

; Λ(0,~0, ϕ, ~nz) =

1 0 0 00 cosϕ − sinϕ 00 sinϕ cosϕ 00 0 0 1


Matrix U

Let’s search for U corresponding to the infinitesimal Lorentztransformation (44):

U = 1 + ~c · ~σ (48)

U+ = 1 + ~c∗~σ (49)

We choose this form because Pauli matrices (with unity matrix) forma basis in the space of 2× 2 complex matrices.

The transformation is given by

x′ = UxU+ = x+(~c ·~σ)(x0 +~x ·~σ)+(x0 +~x ·~σ)(~c∗ ·~σ)+O(c2) (50)

For product of 2 σ matrices there exist a nice formula:

σiσj = δij + iεijkσk (51)


Matrix U

Using it we can rewrite (50) as:

x′ = x+ (~c+ ~c ∗)~σx0 + (~c+ ~c ∗)~x+ i[(~c− ~c ∗)× ~x]~σ (52)

Comparing (50) with (45) and (46) it’s easy to see, that:

~c+ ~c ∗ = −β~n1 (53)

~c− ~c ∗ = −iϕ~n2 (54)

so:U = 1− β

2~n1~σ − i

ϕ

2~n2~σ (55)

Finally, we write the finite transformations for 2 important cases:boosts and rotations

– For boosts:

Uboost = e−β2~n~σ = cosh

(β

2

)1− sinh

(β

2

)~n~σ (56)


Matrix U

– For rotations:

Urot = e−iϕ2~n~σ = cos

(φ

2

)1− i sin

(φ

2

)~n~σ (57)

It’s easy to see that rotation on 2π angle is not a unitytransformation; Urot(ϕ = 2π) = −1. This indicates, that spinorrepresentation is different from vector one, for which rotation on2π angle is always a unity transformation.


Left and right spinors

We can now build a Lorentz-invariant equation for ψR(x, t) ≡ ψ(x, t).

i(∂t + σi∂i)ψR(t, xi) = 0 (58)

If ψR transforms under Lorentz transformations as we defined above,this equation is Lorentz invariant. Another equation we could write ,for another function ψL(x, t) is

i(∂t − σi∂i)ψL(t, xi) = 0 (59)

If now we define that parity ( mirror image) acts on ψR as

PψR = ψL (60)

we can write the system that is invariant under whole Lorentz group,including parity:


Left and right spinors

i(∂t + σi∂i)ψR(t, xi) = mψL (61)

i(∂t − σi∂i)ψL(t, xi) = mψR (62)

Introducing wavefunction ψ =

(ψLψR

)we obtain Dirac equation:

(iγµ∂µ −m)ψ = 0 (63)

in another basis of γ-matrices, called Weyl basis:

γ0 =

(0 11 0

)γi =

(0 σi

−σi 0

)(64)

show that these matrices obey relation γµ, γν = gνµ, as it should befor any γ-matrices.


Interaction with electromagnetic field 8

In the same paper [Proc. R. Soc. Lond. (1928) 610, 24] Dirac introducedcoupling of spinors to electromagnetic field

Recall in non-relativistic quantum mechanics coupling to theelectromagnetic field was via “minimal coupling” (i.e. the momentumpµ → Pµ = (pµ − eAµ) , Pµ is sometimes called “generalized momentum”)

p2

2m→ 1

2m

(p− e ~A

)2

+ eA0(x) (65)

In Dirac equation, we make similar substitution

(pµγµ − eAµγµ −m)ψ = 0 (66)

Dirac Hamiltonian in the electromagnetic field

H = H0 +A0 + γ0~γ · ~A (67)7Bjorken-Drell, Chap. 1, Sec. 1.4


http://www.lorentz.leidenuniv.nl/~boyarsky/media/Proc.R.Soc.Lond.-1928-Dirac-610-24.pdf

Probability and current density

Dirac has constructed an equation of the form (i∂t −H)ψ = 0

where H is Hermitian

Therefore the quantity ∫d3xψ+ψ (68)

is conserved

The quantity

ψ†ψ = |ψ1|2 + |ψ2|2 + |ψ3|2 + |ψ4|2 ≥ 0 (69)

can be interpreted as the probability density. Contrary to theKlein-Gordon case, it is non-negative by construction.

The density ψ†ψ is a part of the current vector

jµ = ψ†γ0γµψ (70)


Probability and current density

As a consequence of the Dirac equation, this current is conserved9

∂µjµ = 0 (71)

therefore the spatial part of jµ has the meaning of the currentdensity.

9Check this


Negative energy states

As for any relativistic equation, Dirac equation has two branches ofdispersion relation

E =√p2 +m2 and E = −

√p2 +m2 (72)

Consider a state with negative energy (pµ = (−|E|,p)):

ψ(x) = u(p)e−ip·x = u(p)e+i|E|t+ip·x

(pµγµ −m)u(p) = 0

(73)

Naively, if the system contains negative energy states – it isunstable, since it will try to choose the state with the lowest possibleenergy, but there is no natural lower bound on the value of thenegative energy.

What should be there interpretation?



Observe that complex conjugated spinor ψ∗ = u∗e+ip·x haspositive energy:

ψ∗ = u∗e−i|E|t−ip·x (74)

Idea: build a positive energy state, ψc, corresponding to thenegative energy solution (73) and satisfying the Dirac equationwhenever ψ does.

Guess 1: Does the spinor (74) obey the Dirac equation?

(/∂ −m)ψ∗ = e+ip·x(−pµγµ −m)u∗?= 0 (75)

Notice that as a consequence of Eq. (73) we have(pµ(γµ)∗ −m

)u∗(p) = 0 (76)

This could work (as a consequence of (76)) if gamma-matrices



were imaginary:(γµ)∗

?= −γµ (77)

This is not the case. In the representation that we use– γ0, γ1, γ3 are real– γ2 is imaginary (show it)

Guess 2: In addition to complex conjugation rotate the spinor u(p):

ψc = Cψ∗ = (Cu)e+ip·x (78)

(/∂ −m)ψc = e+ip·x(−pµγµ −m)Cu∗?= 0 (79)

where matrix C is chosen in such a way that

− γµC = C(γµ)∗ (80)

(and also C2 = 1). Exercise: demonstrate that this matrix is givenby

C = iγ2 (81)(trivial consequence of γµ, γν = ηµν and γ2 being imaginary).



Now consider Dirac equation in the external electromagnetic field:(/∂−e/A−m

)ψ = 0 (82)

The spinor ψc = Cψ∗ obeys the equation:(/∂+e/A−m

)ψc = 0 (83)

Spinor ψc describes a state with the opposite charge


Dirac sea

The interpretation of Dirac: Since fermions obey the Pauli principle,the occupation number of each energy level cannot exceed 110

Assume that we start fromthe many-fermion system,where all the energy stateswith E < 0 are fullyoccupied (the Dirac sea).

Momentum p

Ener

gy

Ep

Unoccupied levels

Occupied levels

This fully-occupied state is interpreted as the vacuum

Such vacuum is stable due to Pauli exclusion princple10Actually it cannot exceed 2, when we take into account two possible spin states


Dirac sea

When we add an electron to the vacuum, it increases overall energyby E > 0. The dynamics of this additional particle is described bythe positive-energy solutions of the Dirac equation.

We may also remove one particle from the vacuum. The resultingstate is called “hole”, and behaves like the absence of electron withE = −

√p2 +m2. The removal of fermion with negative energy

increases the energy of the system. The vacuum has the lowestenergy


Holes

Therefore, the hole (i.e. the spinor ψc = Cψ∗ where ψ = ue−ip·x withthe negative energy) has11

positive charge ehole = +|e|

positive energy Ehole = +√p2 +m2

opposite momentum phole = −p

11The analogy is an air bubble in water, compared to the drop of water in air: effectively, the bubblebehaves like a drop with negative density.


Prediction of antiparticles

The hole state is called the antiparticle

For electron, there should exist a positively-charged antiparticle –positron. It was first predicted by Dirac.

In 1932, positron was discovered by Anderson in cosmic raysPhys. Rev. 43, 491494 (1933) “The positive electron”http://link.aps.org/doi/10.1103/PhysRev.43.491


http://link.aps.org/doi/10.1103/PhysRev.43.491

Prediction of antiparticles

From Phys. Rev. 43, 491494 (1933) “The positive electron”http://link.aps.org/doi/10.1103/PhysRev.43.491

Abstract. Out of a group of 1300 photographs ofcosmic-ray tracks in a vertical Wilson chamber 15tracks were of positive particles which could nothave a mass as great as that of the proton. Froman examination of the energy-loss and ionizationproduced it is concluded that the charge is lessthan twice, and is probably exactly equal to,that of the proton. If these particles carry unitpositive charge the curvatures and ionizationsproduced require the mass to be less than twentytimes the electron mass. These particles will becalled positrons. Because they occur in groupsassociated with other tracks it is concluded thatthey must be secondary particles ejected fromatomic nuclei.


http://link.aps.org/doi/10.1103/PhysRev.43.491

Quantum mechanics of weak interactions


Fermi theory of β-decay 12

Continuumspectrum ofelectrons(1927)

Prediction ofneutrino(1930, 1934)

Fermi theory(1934)

Universality ofFermiinteractions(1949)

Decay of nucleus N(A,Z) →N(A,Z + 1) and

Q = MA,Zc2 −MA,Z+1c

2 > 0

Neutron decay n→ p+ e− + νe

Two papers by E. Fermi:

An attempt of a theory of beta radiation. 1. (InGerman) Z.Phys. 88 (1934) 161-177

DOI: 10.1007/BF01351864

Trends to a Theory of beta Radiation. (InItalian) Nuovo Cim. 11 (1934) 1-19

DOI: 10.1007/BF02959820

11History of β-decay (see [hep-ph/0001283], Sec. 1,1); Cheng & Li, Chap. 11, Sec. 11.1)


http://dx.doi.org/10.1007/BF01351864

http://dx.doi.org/10.1007/BF02959820

Fermi theory of β-decay The emitted electrons are

(quasi) relativistic (kineticenergies from keV tofew MeV). Neutrinos arerelativistic

Typical nuclear size rnucl ∼10−13 cm.

Typical wavelengths:

~pe∼ ~pνe∼ ~

100 keV∼ 2×10−10 cm rnucl

(where ~ = 6.6× 10−19 keV·sec)

Fermi: assume that there are some new interactions VW . Theseinteractions are small perturbations around free particle statesp, n, e, νe

Compute perturbation between an initial state |i〉 = |0〉 and the final


Fermi theory of β-decay

state |f〉 = |e−, νe〉 (the matrix element of the operator that convertselectron into a proton is already included into the definition of thematrix element of VW ).

As the wave lengths of two particles are much larger than thenuclear size, one can take

Vif = 〈i|VW |f〉 ≈ const = GF

GF is a new constant with a dimension EL3 and the probability ofdecay is given by

dWif(pe, pνe) = 2π|Vif |2δ(Q− Ee − Eνe)d3pe(2π)3

d3pνe(2π)3

(84)

Integrate over the momentum of νe (it is not detected) and build a



spectrum of electrons, using d3pe = 4πp2edpe = 4π

√E2e −m2

eEedEe

dW (Ee) =

∫d3pνe(2π)3

dWif(pe, pνe) =1

2π3G2F

√E2e −m2

eEe(Q−Ee)2dEe

(85)

Fermi 4-fermion theory:

LFermi = −GF√2

[p(x)γµn(x)][e(x)γµν(x)]

GF – Fermi coupling constant. Value determined experimentally tobe GF ≈ 10−5 GeV−2

Other weak processes: µ− → e− + νe + νµ or π+ → µ+ + νµ, led togeneralization of Fermi Lagrangian:

LFermi = −GF√2

[J†lepton(x) + J†hadron(x)

]·[Jlepton(x) + Jhadron(x)

]Alexey Boyarsky PPEU 44


where current Jµ has leptonic (e±, µ±, ν, ν) and hadronic (p, n, π)parts


Particle Physics of the early Universe

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