topics in standard model - universiteit leidenboyarsky/media/lecture2.pdf · +ea0(x) (9) ¥ in...

Topics in Standard Model

Alexey BoyarskyAutumn 2013

Natural units

A slight detour. . .

¥ In particle physics it is convenient to work in the “natural units”:~ = c = kB = 1

¥ This means that

• [Distance] = [Time](× c)• [Energy] = [Mass](×c2) = Momentum(×c) =[Temperature](×kB)• [Distance] = [Energy]−1

¥ Exercises: try to convert:

– Velocity v = 0.1 in cm/sec– Temperature = 1 eV; 1 keV; 1 MeV; 1 GeV in Kelvins– Distance 1 GeV−1 in cm– 1 second in GeV−1

– 1 gram in GeV

– 1 g/cm3 in GeV4 and in GeV/cm3

Alexey Boyarsky STRUCTURE OF THE STANDARD MODEL 1

Spinors

¥ The Dirac equation(i∂

∂xµ(γµ

)αβ−m δαβ

)ψβ = 0 (1)

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

γ0 =(1 00 −1

); γi =

(0 σi−σi 0

)(2)

¥ Therefore, the wave-function ψβ has actually 4 components

ψ =

0BB@

ψ1

ψ2

ψ3

ψ4

1CCA (3)

1We use the same representation, as in Lecture 1


Spinors

and is called spinor (or 4-component spinor)

¥ Recall: if we square this equation we obtain Klein-Gordon equationfor each component:

(¤−m2)ψβ = 0 ∀ β (4)


Probability and current density

¥ As planned, we have constructed an equation of the form (i∂t −H)ψ = 0

¥ Therefore the quantity ∫d3xψ+ψ (5)

is conserved

¥ The quantity

ψ†ψ = |ψ1|2 + |ψ2|2 + |ψ3|2 + |ψ4|2 ≥ 0 (6)

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 (7)


Probability and current density

¥ As a consequence of the Dirac equation, this current is conserved2

∂µjµ = 0 (8)

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

2Check this


Interaction with electromagnetic field 3

¥ 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) (9)

¥ In Dirac equation, we make similar substitution

(pµγµ − eAµγµ −m)ψ = 0 (10)

¥ Problem: check that the Dirac equation (10) is invariant under the gaugetransformation Aµ → Aµ − ∂µf, ψ → ψeief .

2Bjorken-Drell, Chap. 1, Sec. 1.4


Interaction with electromagnetic field. Problems

¥ Write the Dirac Hamiltonian in the constant magnetic field pointing in the z

direction.¥ Find the energy levels (“Landau levels”) of this Hamiltonian.¥ Compare them with non-relativistic Landau levels

(see e.g. Landau & Lifshitz, vol. 3, § 112)


Lorentz transformation of the Dirac equation

¥ Free Dirac equation is linear and describes translational invarianttheory ⇒Let us look for the plane-wave solutions of the Diracequation4

ψ(x) = u(p)e−ipx, px ≡ pµxµ ≡ Et− px (11)

¥ Such a Fourier transform of the Dirac equation (i∂µγµ − m)ψ = 0becomes

(pµγµ −m)ψ = 0 (12)

¥ If we change a reference frame 4-momentum changes

pµ → pµ = Λ νµ pν (13)

where Λ νµ is a 4× 4 matrix of Lorentz transformation.

4From now on, to make notations simpler, we will use bold letters on 3-dimensional quantitiesinstead of drawing vectors. So ~A→ A, coordinates ~x→ x and momentum ~p→ p. Pauli matricesσ = {σx, σy, σz}.



¥ show that the matrix

Λ =

0BB@

coshω − sinhω 0 0sinhω coshω 0 0

0 0 1 00 0 0 1

1CCA (14)

Defines a boost along the x axis with velocity v such that tanhω = vc

¥ We could introduce a new set of γ-matrices, defined via

γµ = (Λ−1)µνγν (15)

so thatγµpµ = γµpµ (16)

¥ Check (16) based on the definition (15).

Show that the new set of γ also satisfies {γµ, γν} = 2ηµν

¥ As a consequence of (16) the same spinor u(p) satisfied thetransformed equation

(pµγµ −m)u(p) = 0, (17)



where u(p) = u(p(p)) and p is expressed via p by inverse of Eq. (13)

¥ However, if in the original frame of reference the form of the Diracequation was (

E −m −p · σp · σ −E −m

)u = 0 (18)

then after the Lorentz transformation the Eq. (18) would have adifferent form.

¥ Instead, we would prefer to work with the same equation, i.e. tohave (

E −m −p · σp · σ −E −m

)u(p) = 0 (19)

instead of Eq. (18).

¥ In order to get the same form of equation (with exactly the same setof γ-matrices) the spinor u(p) should be modified (U is a 4× 4 matrix)

u = Uu (20)



¥ The matrix U should possess the following property:

(γµpµ −m)Uu = U(γµpµ −m)u =⇒ U−1γµU = γµ (21)

¥ It can be shown5 that the property (21) is satisfied by

U = exp(i

4φµνσ

µν

)(22)

where φµν is an anti-symmetric 4×4 matrix parametrizing 6 Lorentztransformations (3 boosts + 3 spatial rotations)

¥ Such multiplication mixes all 4 components of spinor

¥ 4-component spinor u changes in a different way than 4-vectors5See Bjoren & Drell, Sec. 2.2; Peskin & Schroeder, Sec. 3.2. See the original Dirac’s paper (§3)

for a different proof of this fact.



¥ Show that there are 16 matrices 4 × 4 linearly independent matrices that onecan build from γ-matrices by multiplication


Non-relativistic limit of the Dirac equation6

¥ Divide the spinor u(p) in two two-component spinors

u =(χ1

χ2

)(23)

¥ The Dirac equation (18) then becomes:

{(E −m)χ1 − (p · σ)χ2 = 0

(p · σ)χ1 − (E +m)χ2 = 0(24)

¥ Express χ2 via χ1 from the second line of Eq. (24)

χ2 =p · σE +m

χ1 (25)

5Bjorken-Drell, Chap. 1, Sec. 1.4; See also the original paper by Dirac, Proc. R. Soc. Lond. (1928)610, 24



¥ If we plug (25) into the first line of Eq. (24), we will see that each ofthe spinors obeys Klein-Gordon equation:

(E2 − p2 −m2)χ1,2(p) = 0,

so both positive and negative energy solutions exist:

E = ±√

p2 +m2 (26)

¥ Consider positive energies, E > 0. In the non-relativistic limit, E −m¿ m and |~p| ¿ m we can write

χ2 ≈ p · σ2m

χ1 (27)

(we neglected v2/c2 terms in Eq. (27)).

¥ As a result the equation for χ1 then becomes just the Schrodinger



equation for a two-component spinor that can be re-written as

i∂χ1

∂t= mχ1 +

(p · σ)(p · σ)2m

χ1 =(m+

p2

2m

)χ1 (28)

¥ Introducing a trivial time dependence χ1 → e−imtχ′1, we find for χ′1the ordinary Schrodinger

¥ Derive Pauli Hamiltonian

i∂χ1

∂t=

»1

2m(p− eA)

2 − e

2mσ ·B + eA0

–χ1 (29)

as a non-relativistic limit of the Dirac equation (10) with electromagnetic field. AsAµ depends on x, one cannot do the Fourier transformation and a differentialoperator will be acting on χ1 to obtain χ2


Spin and degrees of freedom

¥ By applying the rotations in 3-dimensional space (x, y, z) to χ1, weextract the operator of total angular momentum9

J = x× p +12σ (30)

¥ The first term on the right-hand side is the operator of orbitalangular momentum L, the second term is therefore the spinoperator S.

¥ It means that for the state

χ1 =(

10

)→ S χ1 =

(+

12

)χ1 (31)

Therefore, this state has definite projection of spin on axis z, sz =+1/2.

9See e.g Peskin & Schroeder, Sec. 3.5


Spin and degrees of freedom

In analogy, for the state

χ1 =(

01

), S χ1 =

(−1

2

)χ1. (32)

In total, positive-energy solutions have two degrees of freedom, thatcorrespond to spin-1/2 particles with definite spin projections.


Dirac sea11

¥ Now consider states with negative energy

E = −√

p2 +m2 (33)

¥ The system is unstable, since it will try to choose the state with thelowest possible energy, but there is no natural lower bound on thevalue of the negative energy.

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

¥ Assume that we start from the many-fermion system, where all theenergy states with E < 0 are fully occupied (the Dirac sea).

9See Bjoren & Drell, Sec. 5.1–5.310Actually it cannot exceed 2, when we take into account two possible spin states


Dirac sea

¥ This fully-occupied state was interpreted as the vacuum

¥ Such vacuum is stable

¥ 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.


Holes

Therefore, the hole has12

¥ positive charge ehole = +|e|

¥ positive energy Ehole = +√

p2 +m2

¥ opposite momentum phole = −p

12The analogy is an air bubble in water, compared to the drop of water in air: effectively, the bubblebehaves like a drop with negative density.


Holes

The removal of fermion with negative energy increases the energy ofthe system. The vacuum has the lowest energy


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


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.


Other consequences of the Dirac sea¥ Presence of the negative-energy levels means that you can create

particle-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)


Interaction of light with the Dirac sea

Since vacuum is not “empty”, electromagnetic waves act on it non-trivially:

¥ the virtual particle-antiparticle pairs are excited

¥ the pairs are polarized by the electric field of the wave

¥ polarization changes the propapagation of the wave (vacuumpolarization)

Two different electromagnetic waves can act on each other, throughthe interaction of the polarized virtual pairs. Light can scatter off lighteven in the vacuum! See V. Dunne

1202.1557

The vacuum behaves like a medium.


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