tyty hhtt

7/28/2019 tyty hhtt

1/100

3.2. Parity, charge conjugation and time reversal 101

Invariance under O means OTsO = Ts, and thereforef|Ts|i = f|OTsO|i = f0|Ts|i0. (3.26)

The above equality is not satisfied if the corresponding symmetryis broken. Thus the consequence of symmetry breaking is thatthe amplitude between two states is not equal to that between thestates transformed by the symmetry operation.

But iff|Ts|i and fo|Ts|i0 differ only by a phase, it does notshow up in the transition probability, and violation of the symme-try is in practice unobservable. In order to retain observability ofsymmetry breaking, we therefore require that unless the matrixelement between the transformed states is different in its absolutevalue, a part of the Ts element should change under O. Interfer-ence between different terms in the amplitudes then shows up in

the observable quantities, such as the transition probability.As an example, consider the decay of a negatively charged pion

into a muon and an antineutrino. The observed fact that only aleft-handed is produced in this process can be translated intothe statement

M( L ) = M( R) = 0.The origin of this can be traced to the appearance of the piondecay matrix element, namely

M = f[(1 5)]p , (3.27)which in turn follows from the (VA) character of the interactionlagrangian. Here, one of the two terms proportional to and 5must violate parity, in spite our freedom of rephasing the fields.

As a corollary, one may say that if the states |i and |f havedifferent eigenvalues (1, 1) for O = C, P and CP, then O invari-ance implies

f|Ts|i = f|OTsO|if0|Ts|i0 = f|Ts|i (3.28)

i.e., f|Ts|i = 0,

7/28/2019 tyty hhtt

2/100

102 3. CP-violation: A Pedagogical Introduction

which implies that the transition i f is not possible.A different condition, however, follows for time reversal. Using

the antiunitary character ofT, it is straightforward to see that the

time reversal invariance means

f|Ts|i = iT|Ts|fT (3.29)In other words, the transition matrix elements between the origi-nal and T-transformed states are equal after the initial and finalstates are exchanged.

3.3 A complex phase in the Lagrangian causes CP-

violation

The central idea in studying CP-violation is that it is caused by

the existence of a complex phase in the Lagrangian. We first wishto convince the reader of the truth in this statement. One wayof looking at it is via the CPT theorem. Without entering intoa rigorous discussion of this theorem, it is sufficient here to notethat a Lorentz invariant local field theory is also invariant underthe combined operations of C, P and T. Since we have alreadyseen that time reversal involves a simultaneous conjugation of allcomplex phases, a Lagrangian containing phases is clearly nottime-reversal invariant. The CPT-theorem then implies that thereshould be a concomitant violation of CP as well, so that overallCPT invariance can be ensured.

The important point to note in the above argument is thatany complex phase occurring as a c-number in the Lagrangiangets reversed under T. But there is no conjugation of the fieldswhich are treated as operators here.

Now, exactly the reverse argument can be applied when oneis thinking in terms of the successive operations of C and P. Thefields and not the complex term multiplying them are conjugatedunder C (and CP). When there are complex coefficients multi-plying the fields in the Lagrangian, there is a palpable change inthe phase relationships of the fields and the coefficients in differentterms, which in turn implies CP-violation.

7/28/2019 tyty hhtt

3/100

3.4. CP-violation in the Standard Model 103

To understand this more clearly, consider the Lagrangian

L = h12 + h21 + g

V12 + V

2

1

, (3.30)

where is a complex scalar, V a vector field, a spin-1/2 fieldand h corresponds to a complex coupling. Under the successiveoperations of parity and charge conjugation, the first two termstransform as follows:

h12P h12 C h21,

h21P h21 C h12.

Hence,

Lh CP h21 + h12 (3.31)

The Lagrangian can be seen to transform in the same way uponT-transformation. If h were not complex, hermiticity would haverestored CP and T invariance. It is furthermore not possibleto restore symmetry by rephasing the fields, because then thatphase will show up in the interaction with V, again destroyinginvariance of the Lagrangian.

Thus hermiticity, coupled with the antiunitary character ofT and the requirement of complex conjugation of fields under C,lead us to conclude that a complex phase in the coefficient of someterm in the Lagrangian is a necessary and sufficient condition forCP- and T-violation.

3.4 CP-violation in the Glashow-Salam-Weinberg

model

The standard electroweak model named after Glashow, Salam andWeinberg (GSW), has no C- or P-violation in the gauge and Higgssectors. It is in the interaction of fermions that the non-invarianceunder parity and charge-conjugation becomes manifest, the firstof these underlining the left-handed nature of weak interaction.We, therefore, begin by considering the coupling of fermions withboth of the above sectors. As far as the basic interactions are

7/28/2019 tyty hhtt

4/100


concerned, the forms are identical for both the quark and leptonsectors; however, quarks exhibit the additional phenomenon ofmixing for which a conclusive evidence is yet to emerge in the

domain of leptons (although recent data on neutrinos stronglysuggest similar phenomena there as well). In view of the above,we start by considering just the first family of quarks:

QL =

uLdL

, uR, dR.

After spontaneous symmetry breaking takes place, the gaugeand Yukawa (i.e., Higgs) interactions of these quarks, with theSU(2) U(1) neutral gauge bosons written in the basis (W3, B),are given by

L(gauge-Yukawa) = QLi + g2

i

2 Wi +

g16 BQL

+ uR

i +2g1

3B

uR

+dR

i g13

B

dR (3.32)

+hu( uLuR + uRuL)(v + )

+hd(dLdR + dRdL)(v + ).

Here is the physical Higgs field and its vacuum expectation value(vev) is denoted by v.

The last two terms of the above equation can be written as

hu(v + )uu + hd(v + )dd (3.33)

The fact that gauge bosons belong to real representations of thegauge group restricts the gauge couplings to be real. Also, anyphase in hu or hd can be absorbed by redefining uR/dR. Suchredefinition should not show up in any physical observable, sincethe right-handed quarks participate only in neutral current (weakor electromagnetic) interactions, and any additional phase in themis liable to get canceled. Thus there is no net CP-violating phasein the set of interactions involving the quarks.

7/28/2019 tyty hhtt

5/100


After introducing the physical gauge bosons Z and A, theabove Lagrangian is rewritten as

L LW + LZ + L+ LY=

g22

u1 5

2dW + h.c.

+g

cos W

u(a + b5)uZ + d(a

+ b5)dZ

+e

quuu + qddd

A

+hu(v + )uu + hd(v + )dd, (3.34)

where, qu =23 , qd = 13 and a,b,a, b are functions of Q and T3.

Clearly, both L and LY are invariant under C and P.Now, take a look at LW. With

LW u(1 5)dW + d(1 5)uW, (3.35)

under P it becomes

LPW u(1 + 5)dW + d(1 + 5)uW. (3.36)

A further application of C yields

LCW d(1 + 5)uW u(1 + 5)dW. (3.37)

On the other hand, successive operations of C and P lead to

LCPW u(1 5)dW d(1 5)uW. (3.38)Thus the one-generation Standard Model violates C and P butconserves CP. A similar conclusion follows on examining the Z-interaction terms as well.

The central point here is that V- and A-type bilinears haveopposite transformation properties both under C and P, and her-

miticity takes care of the change caused by charge conjugation.That is how CP-conservation is ensured under the combined op-erations, which can also be seen clearly from the fact that theinteraction terms contain no complex phase after all.

7/28/2019 tyty hhtt

6/100


We have already learnt that CP-invariance is broken if andonly if there are complex factors multiplying the charged currentterms in LW. Such phases will be reversed under hermitian conju-gation but not under C, breaking down the invariance under CP,as discussed in the previous paragraph.

The question is, how can such complex phases arise? In therest of this section, we shall find out the answer to this ques-tion, which basically lies in the fact that complexity is introducedthrough Yukawa couplings when we have not one but several fam-ilies of quarks, as is indeed the situation in real life.

Let us examine the quark mass terms together with theYukawa couplings:

L(m+Y) hu(v + )uu + hd(v + )dd= hu(uu + dd) + muuu + mddd.

The full Lagrangian, including the mass and kinetic terms, gaugeand Yukawa couplings, will in general involve a sum over all quarkfamilies, of which three have been known to exist so far. Taking

this sum into account, let U0 =

u0c0t0

, D0 = d0s0

b0

. U0, D0define the basis in which LW is flavour diagonal. This basis isgiven various names such as the flavour basis, the weak interactionbasis or the current eigenstate basis.

In this basis,

LW++Z U0 1 52 D0W + h.c.+

U0(a + b5)U0Z + U0U

0A

(3.39)

+

U0 D0, a , b a, b

.

Now consider L(m+Y) in this basis:L(m+Y) = huijU0i U0j (v + ) + hdijD0i D0j (v + ), (3.40)

where huij , hdij are not necessarily diagonal in this basis. Since the

mass matrices Mu = huv and Md = hdv are proportional to the

7/28/2019 tyty hhtt

7/100


Yukawa coupling matrices, the mass terms in the flavour basis arealso non-diagonal:

Lm =

U

0

i M

u

ijU

0

j +

D

0

i M

d

ijD

0

j . (3.41)

In general, Muij is not proportional to Mdij and each is non-zero

for i = j. In this sense, the fields U0, D0 are not physical, sincethey do not propagate as states with definite masses. In orderto obtain the corresponding physical quark fields, we need todiagonalise the mass matrices and go to the mass eigenstatebasis.

VuL MuVuR = Mu, V

dLM

dVdR = Md,

where the Vs are unitary matrices.Now,

Lm = U0VuL MuVuRU0 + D0VdLMdVdRD

0

= UMuU + DMdD (3.42)

and one has diagonal mass terms. Thus the mass eigenstates or physical fields are defined as

U = VuL U0, D = VdLD

0.

The charged current lagrangian can be written in terms of thesefields as

LW = g2 U01 52 D0W + h.c.

=g

2U VuL

1 52

VdLDW + h.c.

=g

2(VuL V

DL)ijUi

1 52

DjW + h.c. (3.43)

where the 3 3 unitary matrix Kij = (VuL VDL )ij is known asthe Cabibbo-Kobayashi-Maskawa (CKM) matrix. It diagonalisesMu in a basis in which Md is already diagonal. It is of coursenot necessary for Mu and Md to be real; thus K can naturally

7/28/2019 tyty hhtt

8/100


contain complex phases. We can also see from above that it ispossible to have off-diagonal charged current interactions betweenmass eigenstates, or, in other words, weak interaction allows a

transition between quark families if we write them in terms of themass eigenstates.

It also follows from above that no mixing matrix like K occursin Z and couplings, since the interactions are proportional to

DD and UU in the flavour space and ViL(R)ViL(R) = 1, (i = u, d).

Therefore, neutral current interaction is diagonal in the masseigenstate basis as well. Such absence of tree-level flavour chang-ing neutral currents (FCNC) is a notable feature of the StandardModel, the underlying principle, as explained above, being knownas the Glashow-Illiopoulos-Maiani (GIM) mechanism.

Using our earlier conclusion, it is now obvious that any com-

plex phase(s) in the CKM matrix is a source of CP-violation inthe GSW model. Before we convince ourselves that this is actuallyexpected, let us take note of the following corollaries of the aboveresults, which can easily be verified:

K is the identity matrix if all Us or all Ds are degenerate.This implies that there is no lepton mixing if neutrinos aremassless.

An SU(2) singlet quark added to either the up or the thedown sector will cause FCNC in the down sector, since itwill have a different coupling with the Z compared to a

doublet, and thus a gauge coupling matrix (diagonal, butnot scalar) will be introduced in the flavour space, whichwill not commute with the diagonalising matrices VuL or V

dL .

Now, let us once more shift our attention to the matrix K, andobserve that K is an n n unitary matrix if there are n familiesof quarks. Such a matrix has n2 independent parameters, withn(n 1)/2 angles and n(n + 1)/2 phases. Since three familiesof quarks have already been discovered, one should expect threeangles and six phases in K. However, K always occurs in theinteraction Lagrangian where it is placed between the up U- and

7/28/2019 tyty hhtt

9/100


down d-type quark fields whose phases can be redefined to absorbsome of the CKM phases. Since there are six quarks altogether,they carry five physically relevant phases (an overall phase being

immaterial). On rephasing, the CKM matrix is thus ultimatelyleft with (6 5) = 1 complex phase, which is the sole CP-violating phase of the electroweak sector. It is straightforward tocheck at this point that a two-family scenario allows one to absorball the phases of the mixing matrix via field redefinition, and thatis why a single Cabibbo angle suffices for the description of sucha scenario, admitting of no CP-violation.

Complex Yukawa interactions, together with the existence of(at least) three families of fermions, thus emerge as the sourcesof CP-violation in the GSW theory. K can in general be param-eterized in different ways which are physically equivalent to eachother. The phase can occur is different elements of the matrix; thefact that these choices are physically equivalent can be understoodby realising that all observable manifestations of CP-violation re-quire interference of CP-conserving and CP-violating amplitudes,and that such amplitudes for any process will include more thanone terms involving various CKM elements, whereby the phase,even though located differently in different conventions, will showup identically through interference terms.

A standard parameterisation consists in K expressed in termsof three angles, 12, 23 and 13, together with the phase , in thefollowing way:

K = c12c13 s12c13 s13eis12c23 c12s23s13 c12c23 s12s23s13ei s23c13

s12s23 c12c23s13ei s23c12 s12c23s13ei c23c13

with

Kus sin c 0.22 [Measured from K-decay]Kud 0.97 [Measured from -decay],

where sin c is the Cabibbo angle.An alternative parameterisation which owes itself to Wolfen-

stein shows clearly the relative magnitudes of the different ele-

7/28/2019 tyty hhtt

10/100


ments:

K = 1 22 A3( i)

1

2

2A2

A3(1 i) A2 1 +O(4), (3.44)

where sin c. This shows that the mixing between the firsttwo families is approximately given by the Cabibbo angle, whereasthat between the second and third, and the first and third familiesis of the order of the square and cube of the angle respectively.

3.5 The unitarity triangles

The unitarity ofK imposes orthonormality conditions on the var-ious pairs of rows and columns. These lead to a set of equations

constraining the elements of K, which correspond to triangularrelationships in the complex plane. The resulting triangles (alto-gether six of them) are called unitarity triangles.

Consider, for example, the relation

KudKub + KcdK

cb + KtdK

tb = 0. (3.45)

Thus we have a triangle defined by the three terms above in thecomplex plane. Such conditions, essentially arising from complexnumbers or planar vectors adding to zero, enable us to draw someimportant conclusions concerning CP-violation, as described be-low.

The area of the triangle is zero if all CKM elements are real(no CP-violation).

The area of the triangle also acts as a measure CP-violation. Sides of a unitary triangle are rephasing invariant. The angles of a unitarity triangle, named , and in the

example given above, are expressible in terms of the rela-tive phases of the products of CKM elements. These canbe measured from the rates of various weak processes. Such

7/28/2019 tyty hhtt

11/100

3.6. The neutral kaon system 111

udVV*VV*VV*VV*VV*VV*VV*VV*VV*VV*VV*

ubVV*td tb

cbcdVV*

Figure 3.1: The unitarity triangle.

measurements not only give us crucial information on themagnitude of the CP-violating phase, but also allow us todetermine whether the three-generation CKM phase is ulti-

mately responsible for CP-violation in nature. If there aremore families or additional CP-violating phases, the sides ofthe unitarity triangle will not close in the manner suggestedabove.

3.6 The neutral kaon system: where CP-violation

was first observed

The neutral kaon states with definite strangeness are

|K0

=

|sd

,

|K0

=

|sd

, (3.46)

with s = 1 and 1 respectively. Their CP-transformation proper-ties are phase convention dependent but mutually related.

K0 and K0 are not CP-eigenstates, which is obvious from thefact that quarks become antiquarks on charge conjugation. Let ususe the convention

CP|K0 = |K0, CP|K0 = |K0.Now, it is well-known that strangeness is not conserved in weakinteraction. Supposing that CP is conserved (in weak interaction),

7/28/2019 tyty hhtt

12/100


we can try to construct CP-eigenstates in the following way, aslinear combinations of K0 and K0:

|K1

=

1

2(|K0

+

|K0

), CP

|K1

=

|K1

,

|K2 = 12

(|K0 |K0), CP|K2 = +|K1. (3.47)

If these were the physical neutral K-states that decay weakly, thedecay products would have to be CP-eigenstates. Experimentally,the physical states having definite decay modes were identified asKL and KS (where L and S stand for long and short), with meanlifetimes

(KL) 5.0 108 sec,(KS) 0.9 1010 sec.

Their principal decay channels were initially identified as

KL 000, 0+ (CP = 1),KS 00, + (CP = +1).

Thus everything was fitting in, with |KL = |K1, |KS = |K2and CP known as a symmetry of nature, until in 1964, Christen-son, Cronin, Fitch and Turley observed the decay KL . Infact, if one defines the following quantities involving the ampli-tudes of the errant modes:

+ =+|Twk |KL

+

|Twk

|KS

00 =

00|Twk|KL00|Twk|KS ,

then it is experimentally found that,

|+| (2.285 0.019) 103,|00| (2.275 0.019) 103.

It may be remembered that ifCP|i = |i, CP|f = +|f, where|i = |KL, and |f = |, then CP-conservation would imply

f|Twk|i = f|(CP)1Twk(CP)|i = f|Twk|i,

7/28/2019 tyty hhtt

13/100

3.7. CP-violation in neutral kaons: observables 113

giving f|Twk|i = 0.Thus physicists were led to the unavoidable conclusion that

weak decays of neutral kaons violate CP. There could be two pos-

sible ways in which this could happen: KL is a mixture of CP = +1 and CP = 1 states (K1, K2)

(indirect CP-violation).

CP-violation takes place in the decay process itself (directCP-violation).

We shall see in the following section that a careful measurementof + and 00 can in principle reveal both these effects.

3.7 CP-violation in neutral kaons: observables

Let us start by considering the time evolution of a K0

K0

-system:

id

dt

K0(t)K0(t)

=

H11 H12H21 H22

K0(t)K0(t)

, (3.48)

where H is the effective hamiltonian of the two-level system. Theoff-diagonal elements of H originate from weak interactions, sincestrangeness is not conserved there.

Thus, for example,

H11 = K0|H|K0, H12 = K0|H|K0. (3.49)CPT-invariance of the hamiltonian further implies that

K0

|H

|K0

=

K0

|(CP T)1H(CP T)

|K0

=

K0

|H

|K0

.

The hamiltonian H, representing an unstable system, in generalcontains a hermitian and an anti-hermitian part:

H = M i2

,

where M and are the mass and decay matrices respectively.

Thus if =

K0

K0

,

(t) = eiHt(0) = e2 eiMt(0)

|(t)|2 = e|(0)|2. (3.50)

7/28/2019 tyty hhtt

14/100


The hermiticity ofM and , together with the condition H11 =H22, means

M11 = M22, 11 = 22,M21 = M

12, 21 =

12.

A non-hermitian H in this case admits non-orthogonal eigenstates.Such eigenstates can be found on diagonalising H, the physicalstates thus obtained being

|KL(S) = p|K0 q|K0|p|2 + |q|2 , (3.51)

where,

q

p = M12

i12

/2

M12 i12/2 = H21H12 . (3.52)The corresponding eigenvalues are

ML(S) = M Re Q, L(S) = 2 Im Q,

with

Q =

M12 i12

2

M12

i122

=

H21H12

M = 2 Re Q, = 4Im Q.

The following observations are in order:

KS = K1, KL = K2 if pq = 1.

The quantities p and q, complex in general, depend on thephase convention. For example, a phase transformation|K0 ei|K0, |K0 ei|K0 will change p and q.

As measures of CP-violation, we need to find quantities thatare independent of phase conventions and are directly re-lated to observable quantities.

7/28/2019 tyty hhtt

15/100


One such quantity satisfying the second criterion above is =|q/p|. CP-conservation means that there is no phase in M12 or12, which in turn gives = 1.

It follows from above that

KL|KS =K0|p K0|q p|K0 + q|K0

p2 + q2

=|p|2 |q|2|p|2 + |q|2 =

1 21 + 2

, (3.53)

= 1 +2|12|2

4|M12|2 + |12|2 Im

M1212

.

One can draw the following conclusions now:

= |q/p| = 1 means that there is no CP-violation throughmixing.

(1 ) is a measure of CP-violation (though we still need torelate it to + and 00).

= 1 implies that KL and KS are non-orthogonal and hencethere is indirect CP-violation.

It is also straightforward to see that the time evolution of thesystem in terms of the mass eigenstates is given by

KL(S)(t) = eL(S)teiML(S)tKL(S)(0). (3.54)

Let us now go back to

+ =+|Twk|KL+|Twk |KS

00 =00|Twk |KL00|Twk |KS

and try to express them in terms of the elements of (M i2 ). Thiswill enable us to ultimately link the measures of CP-violation tothe basic parameters (masses, angles and phase) of our underlyingtheory.

7/28/2019 tyty hhtt

16/100


For such a purpose, we need to write the amplitudes for differ-ent processes in terms of K0 and K0. We should also rememberthat the formation of the final state pions involves not only weak

interactions but also strong interaction effects. The latter is inplay when, say, two pions that are produced within a very shortdistance separate out. Such interactions are called final state in-teraction and they in general introduce additional phases (arisingas phase shifts in the elastic scattering amplitudes). These arethe so-called strong phases and are not reversed, unlike the CKMphase, on charge conjugation, a fact whose origin can be tracedto the charge conjugation invariance of strong interactions. Thecontributions to the amplitudes (and the strong phase shifts) fromsuch interactions depends on the total isospin (I) of the final state( in our case). Therefore, we need to decompose the amplitudesinto those for transitions to different states in the isospin basis.

A two-pion final state can have I = 0, 2, since one must havea symmetric state. With this in mind, we define

()I=0|Twk |K0 = a0ei0 ,()I=2|Twk |K0 = a2ei2 ,

where 0, 2 are the strong phases for transition into the twoallowed isospin states. Any weak phases are assumed to be con-tained within a0 and a2, i.e.,

a0 = |a0|ei0 a2 = |a2|ei2. (3.55)

Using CP|K0

= |K0

and (CP T)1

H(CP T) = H.

()I=0|Twk|K0 = a0ei0 ,()I=2|Twk|K0 = a2ei2 .

CP-conservation implies 1, 2 = 0 or . A relative complex (non-trivial phase between a2 and a0 can thus be linked CP-violationin the decay (direct CP-violation).

Now use Eq. 3.51, to obtain

()I=0|Twk|KL(S) = ei0

|p|2 + |q|2 (pa0 qa

0) = a0L(S),

7/28/2019 tyty hhtt

17/100


()I=2|Twk |KL(S) = ei2|p|2 + |q|2 (pa2 qa2) = a2L(S).

One can at the same time use the standard Clebsch-Gordon coef-ficients:

|00 = 12

|I=0

2

3|I=2

|+ = 12

|+ + |+ = 23

|I=0 + 12

|I=2,

thus obtaining

+ =+|Twk|KL+|Twk|KS =

2a0L + a2L2a0S+ a2S

,

00 = 0

0

|Twk|KL00|Twk|KS = a0L 2a2L

a0S 2a2S.

Expressing a0L, a2L, a0S and a2S in terms of a2, a0, 2, 0, pand q, we finally get:

+ =(1 y)[1 + w cos(2 0)] + i(1 + y)w sin(2 0)(1 + y)[1 + w cos(2 0)] + i(1 y)w sin(2 0) ,

00 =(1 y)[1 2w cos(2 0)] 2i(1 + y)w sin(2 0)(1 + y)[1 2w cos(2 0)] 2i(1 y)w sin(2 0) ,

where y = qpe2i0 , w = 1

2|a2a0 |ei(20), (|y| = ). It should be

noted here that CP can be violated indirectly (through mixing)even if 2 = 0,

Next we make use of some empirical inputs:

Since (1 y) is a measure of CP-violation, and y = 1 corre-sponds to no (indirect) CP-violation, |1 y| 1.

K0 has I = 12 , so that a2 corresponds to I = 3/2 and a0corresponds to I = 1/2.

In nonleptonic kaon decays in general, I = 32 is suppressed com-pared to I = 12 . This is the so-called I =

12 rule.

7/28/2019 tyty hhtt

18/100


To be more specific, it is observed that the decay rate forK+ +0 is much smaller than that for K0S +. Theformer decay involve Iinitial = 0, Ifinal = 0 or 2. Since I3 = 0 for

the final state, the only allowed possibility is Ifinal = 2, whencewe know that the process corresponds to I = 3/2. In case ofthe neutral kaon decay, on the other hand, Iinitial = 0 whereasIfinal can be both 0 or 2. The immediate conclusion, therefore,is that the I = 12 transition is responsible for the enhancementof the rate for (K0S +), thereby vindicating the empiricalrule, for which a likely explanation is larger QCD corrections forthe corresponding final states.

General guidelines based on the I = 12 rule tell us that |w| 125 . Thus, in + and 00, the terms proportional to (1 y)w canbe dropped to a good approximation, and one can write

+ 1 y1 + y

+ iw sin(2 0)1 + w cos(2 0) = +

1 + w

2

,

00 1 y1 + y

2iw sin(2 0)1 2w cos(2 0) =

2

1 + w2

,

where,

=1 y1 + y

=a0La0S

,

=i2

ei(20)a2a0

sin(2 0).

With w 1,+ = + , 00 = 2

gives:

=1

3(2+ + 00), =

1

3(+ 00). (3.56)

and are phase convention independent measures of CP-violation. gives the extent of CP-violation through mixing, con-tributing to both + and 00, while is a measure of directCP-violation, showing up in the splitting between + and 00.

7/28/2019 tyty hhtt

19/100

3.8. CP-violating parameters from the Standard Model 119

Let us elaborate on . The expression

1 y

1 + y

=p qe2i0

p + qe2i0

,

where p and q are expressed in terms ofM12 and 12. If one usesfurther empirical inputs, namely, s 2m, Im 12 Im M12 and tan 20 Im M12/Im M12, one obtains, after somealgebra,

12

Im M12M

ei/4. (3.57)

Using the additional information 2 0 4 (obtained from pionscattering data)

i2a2a0 sin(2 0)ei/4. (3.58)

In the next section, we shall indicate how and can be estimatedin the Weinberg-Salam model.

3.8 CP-violating parameters from the Standard Model

3.8.1 The calculation of

Using elementary perturbation theory, we can write

M i2 12 = 12mk K0|Hw|K0+n

1

2mk

k0|Hw|n

n|Hw|K0

mk En + i ,

plus higher order terms. In the above, |n represents all possibleintermediate states with energy eigenvalues En and

1mkEn+i =

P

1mkEn

i(En mk). Therefore,

M i2

12

=1

2mk

K0|Hw|K0

7/28/2019 tyty hhtt

20/100


+1

2mkP

n

k0|Hw|n

n|Hw|K0

mk En

in (En mk)k

0

|Hw

|nn|Hw|K

0 .Equating the Hermitian and anti-Hermitian parts, we get

M12 =1

2mk

K0|Hw|K0

+n

K0|Hw|n

n|Hw|K0

P

1

mk En

,

12 = 2n

(En mk)

k0|Hw|n

n|Hw|K0

,

where 12 is the absorptive part of the K0K0-transition matrix el-

ement and the second term in M12 is the dispersive part (the longdistance contribution, coming from intermediate hadronic states).The dispersive contribution to can be shown to be small.

Thus, Im M12

K0|Hw|K0

, where Hw is calculable fromSM by obtaining the quark-level effective Hamiltonian. Re M12,on the other hand, has large long-distance contributions. Conse-quently, 2Re M12 is replaced by the measured value of the KL-KSmass difference in the expression for . For

K0|Hw|K0

, we ob-

tain the lowest order quark-level effective interaction for sd sd,and take the matrix element by sandwiching the effective interac-tion between the K0 and K0 states.

It has been shown in the literature that the dominant contri-butions come form the box diagrams shown above. Thus, to ahigh degree of accuracy,

=e1/42m

1

2mkIm

K0|Heff(box)|K0

. (3.59)

The procedure to obtain the relevant matrix element is as follows:

Calculate the amplitude from the Standard Model box dia-grams.

Replace the spinors (u, v) by the fields (s, d).

7/28/2019 tyty hhtt

21/100


(d)(c)

(b)(a)

g

s

s

dw

u,c,t d

u,c,t

w

u,c,t

d

s

u,c,tg

s

d

w

w

s

su,c,t

ww

d

u,c,t dd

sw

s w

d

u,c,tu,c,t

Figure 3.2: Feynman diagrams contributing to K0K0-mixing inthe Standard Model.

Take the dimension-six effective interaction term thus ob-tained, and extract its hadronic matrix element, whose imag-inary part (induced by the phase in the CKM matrix) isrelevant for our purpose.

On explicit evaluation (to be outlined presently), one has

Heff(box) = Is(1 5)ds(1 5)d, (3.60)

with , are colour indices and I is the loop integral. Therefore,

K0|Heff|K0

= I

K0|s(1 5)ds(1 5)d|K0

= IK0|O12O34|K0= I

K0|O12|i

i|O34|K0

. (3.61)

The evaluation of the hadronic matrix element is, of course,model-dependent. Here we demonstrate the results in the so-calledvacuum insertion approximation, which basically consists in thepostulate

|ii| |00|, (3.62)which essentially means that the dominant contribution comesfrom vacuum as the intermediate state.

7/28/2019 tyty hhtt

22/100


Under this approximation,

M12 =I

2mK

K0|s(1 5)d|00|s(1 5)d|K0.

In addition, one needs to take into account the possibility of insert-ing the vacuum in all possible ways between the above current-current product, with every allowed colour combination. Thus,writing

s(1 5)d|00|s(1 5)d = O12O34and remembering its Fierz transformation, namely

s(1 5)d|00|s(1 5)d = O14O32,

we get:

M12 =I

2mK(K0|O12|00|O34|K0

+K0|O34|00|O12|K0+K0|O34|00|O12|K0+K0|O14|00|O32|K0

= K0|O34|00|O12 |K0.

Since the only vector available to characterise the neutral kaonis its four-momentum p (and since no axial vector can be con-structed out of it, in the absence of any polarisation tensor for thescalar particle), we can write

K0|s(1 5)d|0 = fkp, (3.63)

where fk is the kaon decay constant. Thus

K0|O12|00|O34|K0 = f2km2K, (3.64)

and similarly for the term O34O12. For each of the Fierz trans-formed terms, however, there will be a colour suppression factor of

7/28/2019 tyty hhtt

23/100


u,c,t

i

sj

s d

d

k

i,j=

w w

Figure 3.3: The notations relevant for the computation of a typicalbox diagram.

1/3 since colour is a conserved quantum number and the vacuumhas to be colourless. Therefore, one finally obtains:

M12 =I

2mK

2.4

3f2km

2K =

4

3f2KmKI (3.65)

in the vacuum insertion approximation.The model dependence of this result can be encapsulated in

the slightly generalised formula

M12 =4

3f2kBkmkI, (3.66)

where Bk incorporates the uncertainties in the calculation of the

hadronic matrix element. In most estimates, the value of Bk ispredicted to lie between 0.3 and 1.2.

3.8.2 Evaluation ofI

We demonstrate a few salient features of the calculation by tak-ing a sample box diagram, as shown in the figure. All externalmomenta and masses are neglected in this calculation.

Simple power counting shows that the box diagram is finite.We shall evaluate the quark-level transition amplitude in the uni-tarity gauge. In that gauge, with the CKM matrix denoted by V,

7/28/2019 tyty hhtt

24/100


the box diagram amplitude is given by

A =

i,j d4k

(2)4 ig

24

(VisVidVjsVjd) s

(1 5)2

i

k

mi

(1 5)2

d s(1 5)

2

i

k mj (1 5)

2d

ig + kk/m2w

k2 m2wig+ kk/m2w

k2 m2w

= (constant) i,j

ij

d4k

s k(1 5)

2d

s k(1 5)2

d

g + kk/m2w g+ kk/m2wk2 m2i

k2 m2j

(k2 m2w)2 ,

where i = VisVid. The condition V

V = 1 impliesi i = 0.

Therefore, any piece in A that does not depend on both i andj must vanish, just as a manifestation of the GIM mechanismmentioned earlier.

Next, replace 1/(k2 m2i ) by1

k2 m2i 1

k2 m2j m

2j m2i

(k2 m2i )(k2 m2j).

It is obvious that the extra piece has no i-dependence and doesnot contribute since i i = 0. A similar consideration can beapplied to 1/(k2 m2j), leading to an equivalent description of theloop integral, with 1

(k2m2j)(k2m2

j)

replaced by

1

2

1

k2 m2i 1

k2 m2j

1

k2 m2j 1

k2 m2i

,

so that

A = (constant) i,j

ij

d4k

s k 1 5

2d

7/28/2019 tyty hhtt

25/100


s k 1 52

d

m2j

m2i

2

g + kk/m2w g

+ kk/m2wk2 m2i 2 k2 m2j2 (k2 m2w)2 ,

which is manifestly divergence-free.The following points round up the calculation:

I = 0, if mu = mc = mt, i.e., a complete degeneracy of thequarks running in the loops causes the amplitude to vanish.This is again a consequence of the GIM mechanism.

Larger mass splitting among the quarks means larger contri-bution to I, which also means that heavier quarks contribute

more. However, we also have the ij as overall multiplica-tive factors. Thus, diagrams with heavier quarks can stillgive less contributions due to more CKM suppression. Inpractice, Re M12 indeed turns out to be charm-dominated,but Im M12 has the largest contribution from the top quarkloop.

The u-contribution is in any case negligible. Thus one finallyobtains

I

2c1F(xc, xc) + 2t2F(xt, xt) + 2ct3F(xc, xt)

,

where xi = (m2i /m2W) and F(xi, xj) is the function arisingout of the integral in the loop with the ith and jth quarks inthe internal lines. Furthermore, the above contributions aresubject to QCD corrections, thus acquiring the multiplica-tive factors i.

Since Im I and u + c + t = 0, one can writeIm c = Im t since u = VusVud can always be madereal in the CKM matrix. Thus the value of is found to beproportional to Im t. As has been mentioned above, thiscan be contrasted with the real part of the amplitude where

7/28/2019 tyty hhtt

26/100


the charm contribution dominated over the top-induced onesince c t.

The above discussion also demonstrates that can be non-zero only because there are three families of quarks.

The present experimental value of is

= (2.28 0.013) 103. (3.67)

turns out to be a function of s12, s23, s13, fK, BK, mt and .It is the still persisting uncertainty in the values of the differentparameters, particularly the bag factor BK, which prevents usfrom obtaining an estimate of the CP-violating phase in theCKM matrix. It is also to be noted that suffers from a quadraticambiguity, i.e., two values for it emerge by solving the equationin terms of the remaining parameters.

3.8.3 / in the Standard Model

The other relevant quantity, characterising direct CP-violation,is parameterised by /, where has been defined earlier. A non-zero value of this parameter means that 00 and + are not equal.Its current experimental value is / = (28.0 4.1) 104. Fromequation (134), / = 0 means:

(a) a2 = 0 , or in other words, the I = 1/2 rule operates insome way.

(b) sin(2 0) = 0, that is to say, a2 and a0 must not berelatively real.

Due to lack of space, we refrain from entering into a detaileddiscussion on the actual calculation of

/ in the Glashow-Salam-Weinberg model. However, the different diagrams which con-tribute to the process KL are shown in Fig. 3.4.

It should be noted that both tree and loop diagrams are op-erative; however, the loop diagrams, popularly know as penguindiagrams, are required to ensure the relative complexity of a2

7/28/2019 tyty hhtt

27/100

3.9. CP-violation in B-meson decays 127

(d)(c)

(b)(a)

u,d

_

d

w

u

_

_

_

g

us

u,di i

g

s d

u,d

d

u,d

d

_

w

_

_

_

s

_

_

d

u,d_

u

d,u

_uu

_

u

s

d

d

Figure 3.4: Contributions to / in the Standard Model.

and a0. This is because diagrams (a) and (b) (between which(b) is colour-suppressed) are proportional to u as defined earlier,and are therefore are completely real. A similar statement can bemade about diagram (c). Therefore, the contributions from (d)plays a crucial role in the Standard Model contribution. In addi-tion, one has to take into account the electroweak penguins wherethe gluons in (d) are replaced by the Z or the photon. Uncertain-ties related to hadronic matrix elements are of course unavoidablehere also. And finally, the QCD corrections to

/ are extremelyimportant in relating its theoretical predictions to precision mea-

surements, and in any search for the signature of physics beyondthe Standard Model in direct CP-violation. A technical discussionof such corrections is beyond the scope of these lectures.

3.9 CP-violation in B-meson decays

Before we end this discussion, a few remarks on the possibilityof CP-violation in the B-meson system are warranted, since it ison this system that the current experimental efforts are focused.While we shall shall give a brief outline which motivated suchefforts, we consciously avoid entering into a detailed discussion,

7/28/2019 tyty hhtt

28/100


for which a separate course of lectures is certainly required.

The lowest-lying spin-0 mesons containing a b-quark are B0d =bd, B0s = bs, B

+ = bu. Physicists are interested in studying the

B-system to find out more about CP-violation for the followingreasons:

CP-violation requires a specific role of the three-family char-acter of the quark mixing matrix. The K-system, studiedearlier, contains the first two families only. To a first ap-proximation, the FCNC processes involving them may notknow about the third family which is connected with thesd-system only via small CKM elements. The B-meson sys-tem (B0d , for example) can give rise to processes involvingall three quark families, where the role of the full 3

3 CKM

matrix is much more central. Thus one may expect to seelarger CP-violating effects in the B-system. Furthermore, ifit happens to be the case that the CKM phase is actuallylarge but the small effects seen in the K-system is due to thesecondary role of the third family, then, too, one can expectto see enhanced effects by studying B-mesons.

A study of the B-system yields additional information onCKM elements. Such information may complete the con-struction of the unitarity triangle. One may therefore say

with a higher degree of confidence whether the CKM phasealone is responsible for all observed CP-violation, or whethernew physics is responsible for it.

It is with such goals that the the so-called B-factories havestarted operation, the two most recent enterprises being the onesin Stanford, USA, and Tsukuba, Japan. Together with the CornellElectron Synchrotron Ring (CESR), USA and B-related experi-ments at hadron machines (BTeV, LHC-B), these are expected toanswer the questions asked above to a fair degree of precision.

7/28/2019 tyty hhtt

29/100


3.9.1 Charged B-decays

Consider the decayB+

f (3.68)

and the conjugate process

B f . (3.69)

If a CP-violating phase shows up in the decay matrix element,the amplitudes of the above processes are not the same. However,this does not necessarily mean an asymmetry in the correspondingrates. That is to say, if M+ = A + Bei, and M = A + Bei,we shall still have |M+|2 = |M|2.

As we have already discussed, for an asymmetry to show up,it is necessary to have (a) interference of more than one diagrams,

and (b) a phase which does not get reversed on CP-conjugation.This is possible only through the strong interaction phase shifts,or final-state interactions.

To understand the point, let us assume

A(B+ f) = a + a, (3.70)

witha = M eiweis , a = Mei

wei

s ,

where W (W) and s (

s) are respectively the weak phases andthe strong interaction phase shifts for the two amplitudes.

ThenA(B+ f)2 = |a|2 + |a|2 + 2Re (aa)= M2 + M2 + 2M M cos

(w w) + (s s)

.

Similarly, with

A(B f) = M eiweis + Meiweis ,

one obtains:

A(B f)

2

= M2 + M2 + 2M M cos

(w w) (s s)

.

7/28/2019 tyty hhtt

30/100


Thus, (B+ f) (B f) M M sin(w w)sin(s s)and a rate asymmetry shows up.

The problem here is in the experimental extraction of all the

quantities of relevance, before any definitive statements can bemade on the CKM phase. From the above expressions, one basi-cally needs to find the values of M, M, (w w) and (s s).This is not an easy proposition, given the different types of uncer-tainties underlying any process.

3.9.2 Neutral B-decay: cleaner grounds

The neutral B-system offers a relatively easier path to the studyof CP-violation. Consider the B0d B0d system, as these are thelowest-lying neutral B-mesons. Given the composition of B0d ,B0dB

0d-mixing also can take place through box diagrams, much in

the same way as the K0 K0 system. The only difference here isthat the top quark contributions (which are by far the largest onesto the loop integrals) are favoured by a diagonal CKM element inone vertex each.

Analogous to the K0K0-system, the time evolution of the B0dB0d system is governed by,

id

dt

B0(t)B0(t)

=

M i

2

B0(t)B0(t)

. (3.71)

Diagonalising the Hamiltonian, the physical states are found tobe

|BL = p B0 + qB0 ,|BH = p

B0 qB0 . (3.72)BL and BH have similar decay widths (H L) so that can be neglected to a good approximation. Here we note thedifference with the neutral kaon system where KL and KS havewidely disparate lifetimes. This difference makes the separationof the mass eigenstates via their decay channels difficult, and weneed to look for other kinds of observables bearing the signatureof CP-violation.

7/28/2019 tyty hhtt

31/100


After solving the coupled differential equations one finds thethe flavour eigenstates B0 and B0 to evolve in time as follows:

|B0

(t) = g+(t)|B0

(0) +q

p g(t)|

B

0

(0),|B0(t) = p

qg(t)|B0(0) + g+(t)|B0(0), (3.73)

where,

g+(t) = et/2eiMt/2 cos

Mt

2

g(t) = i et/2eiMt/2 sin

M t

2

M =

MH + ML2

and M = MH ML.

As the flavour composition changes in the above manner, let usconsider the production of a B0 B0 pair in electron-positroncollision at the (4s) resonance (where = bb). First, we need toselect a final state f into which both B0 and B0 can decay, f beinga CP-eigenstate. Define A =

f|TW|B0(t)

, A =

f|TW|B0(t)

,

where TW is the weak transition matrix.A measure of CP violation, which is time-dependent, is the

asymmetry

afCP =(B0(t) f) (B0(t) f)(B0(t) f) + (B0(t) f) .

afCP can be the result of

Direct CP-violation (inequality of A and A); CP-violation through mixing ( qp = 1).

The relevant parameter characterising the above two effects is =qA/pA, in terms of which one can write

f|TW|B0(t) = A (g+(t) + g(t)) f|TW|B0(t)=

pA

q(g(t) + g+(t)) . (3.74)

7/28/2019 tyty hhtt

32/100


Now,

(B0(t) f) |f|TW|B0(t)|2

= |A|2et (1 + ||2)2

+ (1 ||2

)2

cos(M t)

Im sin (M t)

, (3.75)

and similarly,

(B0(t) f) |A|2 et(1 + ||2)

2+

(1 ||2)2

cos(M t)

+ Im sin(M t)

, (3.76)

so that

afcp =(1 ||2)cos(M t) 2 Im sin (Mt)

1 + ||2 . (3.77)

Now, | qp | = |M12M12 | = 1 if 12 1, in which case |A| = |A| = 1 and|| = 1. This happens when all amplitudes contributing to thedecay carry the same phase factor, and one has

afCP = Im sin(M t). (3.78)

In general, once || and Im can be calculated for a given f,it is possible to predict the CP-asymmetry in terms of the CKM

parameters. The applicability of this method can be checked inthe decay B0(B0) ks, rather distinct final state observable atB-factories.

The question, however, is, how does one ascertain whether, inan experimental situation, the final state f has come from B0(t) f or B0(t) f?

For that purpose, consider the semileptonic decays of B0(t)and B0(t). While the allowed decay is b cl+l for the former,the latter decays in the channel b cll. Since they are bothspinless, one cannot, out of a B0B0 pair produced at a B-factory,have both B0 or both B0 in course of oscillation at a given time,

7/28/2019 tyty hhtt

33/100


otherwise their combined wavefunction will be antisymmetric andBose statistics will not hold for the system of identical B-mesons.

Therefore, the identity of the meson decaying into the final

state f (say, J/Ks) can be established from whether one has anl+ or l on the other side. Thus

afCP =(no. of events with l+) (no. of events with l)(no. of events with l+) + (no. of events with l)

.

One has to still remember, however, that the decays B l andB l+ at the two ends may occur at different times, say, t andt+. Now, ift = 0 corresponds to the moment when the Bd-pairis formed, The probabilities of one meson decaying to l+ at t+is etl and of the other meson surviving till then are each givenby et. If thus an l+ is detected at t+, the other meson at

that point of time is B0. Taking t as the origin of time, theprobability of this meson decaying to f at t+ is given by

Ptf = e(t+t)

(1 + ||2)

2+ Im sinM(t+ t)

.

Thus, the probability of a B0B0 pair formed at t = 0, and onegoing to l+ at t and the other going to f as B0 at t+ is

Ptt+ = etete(t+t)

(1 + ||2)2

+ Im sinM(t+ t) d

2(B0 f)dt+dt

. (3.79)

In order to see CP-violation, we have to obtain the asymmetrybetween d2(B0 f)/dtfdtl+ and d2(B0 f)/dt+dt. Thisasymmetry turns out to be proportional to Im sinM(t+ t).

It is obvious from above that the above kind of asymmetryvanishes when integrated over time. The need of the day is there-fore to have a situation where the decays at different times can berecorded and thus the double differentials mentioned above can be

7/28/2019 tyty hhtt

34/100


measured to a reliable degree, given the finite resolution of the ex-periments. This becomes extremely difficult at the centre-of-massframe of the colliding e+e system, where the must be produced

at rest. In this frame, the average length each Bd travels beforedecaying is on the order of 30 m, which is too small a distanceto achieve the requisite resolution in time.

The solution lies in obtaining the boosted in the laboratoryframe. One, therefore, aims at an asymmetric beam B-factorywhere the electron and the positron are colliding with unequalmomenta. In this manner, it is possible to have the B-mesonstravel over larger distances, as measured in the laboratory frame,before they decay, whereby it becomes feasible to obtain the thetime-dependence of the final state asymmetry.

7/28/2019 tyty hhtt

35/100

Bibliography

[1] CP-violation, C. Jarlskog (ed), World Scientific (1989).

[2] A. Buras, Lectures given at Lake Louise Winter Insti-tute: Electroweak Physics, Lake Louise, Alberta, Canada(1999). In Lake Louise 1999, Electroweak physics 183 (hep-ph/9905437).

[3] B. Kayser, Talk given at ICTP Summer School in High-energy Physics and Cosmology, Trieste, Italy (1995). Pub-lished in Trieste HEP Cosmology 1995: 432478 (hep-ph/9702264).

[4] B. Winstein and L. Wolfenstein, The search for direct CPviolation, Rev. Mod. Phys. 65, 11131148 (1993).

[5] J. Rosner, Invited talk at 2nd Tropical Workshop on Par-ticle Physics and Cosmology: Neutrino and Flavor Physics,San Juan, Puerto Rico (2000). Published in San Juan 2000,Particle physics and cosmology 283304 (hep-ph/0005258).

[6] CP violation, I. Bigi and A. Sanda, Cambridge Monograph onParticle Physics, Nuclear Physics and Cosmology, 9: 1382,2000.

[7] R. Fleischer, Theoretical review of CP violation, hep-ph/0310313.

Apart from these review articles, some important papers are:

[8] J.H. Christenson, J.W. Cronin, V.L. Fitch, R. Turlay, Phys.Rev. Lett. 13, 138 (1964).

[9] S.L. Glashow, J. Iliopoulos and L. Maiani, Phys. Rev. D2,1285 (1970).

[10] M. Kobayashi and T. Maskawa, Prog. Thoer. Phys. 49, 652(1973).

7/28/2019 tyty hhtt

36/100


[11] I. Bigi and A. Sanda, Nucl. Phys. B193, 85 (1981).

[12] P. Ginsparg and M. Wise, Phys. Lett. B127, 265 (1983).

[13] P. Ginsparg, S. Glashow and M. Wise, Phys. Rev. Lett. 50,1415 (1983).

[14] J.F. Donoghue and B. Holstein, Phys. Rev. D29, 2088 (1984).

[15] J.F. Donoghue, B. Holstein and G. Valencia, Int. J. Mod.Phys. A2, 319 (1987).

7/28/2019 tyty hhtt

37/100

Chapter 4

Lectures on Perturbative Quantum

Chromodynamics

Compiled by Debashis Ghoshal & V. Ravindran

The following notes, meant as a pedagogical introduction tothe subject of perturbative quantum chromodynamics (pQCD),are based on the lectures by Ashoke Sen in the XVI SERCMain School on Theoretical High Energy Physics at the HRI,Allahabad. The notes were taken by Anindya Dutta, DebashisGhoshal, Dileep Jatkar, Swapan Majhi, Partha Konar, SubhenduRakshit and V. Ravindran; and compiled in their final form byDebashis Ghoshal and V. Ravindran.

4.1 Lecture I

Quantum Chromodynamics (QCD) is the theory of strong inter-actions. It is to some extent similar to Quantum Electrodynam-ics (QED), the theory of electrons and photons. We will con-centrate on the perturbative aspects of QCD. Perturbative QCD(pQCD for short) involves a Taylor series expansion in terms of thestrong coupling constant, a parameter which enters in the QCDLagrangian and characterise strong interaction. The perturbativeapproach works only when this coupling constant is small, whichis true for QED. The relevant coupling constant in QED is EM

7/28/2019 tyty hhtt

38/100

138 4. Perturbative Quantum Chromodynamics

which is of the order of 1/137. This is not true for QCD where thecoupling constant is not too small. In addition, QCD being thetheory for quarks and gluons which cannot seen in nature, it is

not only that the application of perturbation theory is non-trivial,but also it is not obvious why perturbation theory is valid.

The strongly interacting particles, collectively called hadronsare made up of quarks, anti-quarks, and gluons. There are sixquark flavours (u,c,t,d,s,b). The first three have electromagneticcharge +2/3 and the last three have 1/3. We assign a spin-1/2Dirac field, is for each of these particles. Here s is the flavourindex which runs from 1 to 6, is the spinor index and i is thecolour index, which runs from 1 to 3. In terms of these quark fields,the proton wave function can be written as: p ijkuiujdk. SinceQCD interaction in the proton changes the colour indices of the

quarks, the actual proton wave-function is a linear combination ofall such terms.

Before going into the details of the QCD interaction (whichis a non-abelian gauge interaction), let us recapitulate QED. Thetwo basic fields which enter here are , a Dirac field describing anelectron and A the vector potential which, on quantisation, givesthe photon. The Lagrangian density for QED can be written as

L = 14

FF+ [i( ieA) m] , (4.1)

where, /x and F = A A. Also F =

F where, = diag(1, 1, 1, 1) in our conventions.

The action is given by S= d4xL, which is Lorentz invariant.The action is also invariant under a global phase transformationon :

(x) eie(x),(x) eie(x),

(x) = (x)0 eie(x),A A. (4.2)

7/28/2019 tyty hhtt

39/100

4.1. Lecture I 139

In fact, action has a larger symmetry called gauge symmetryunderwhich the fields transform as follows,

(x)

eie(x)(x),

(x) eie(x)(x),A A + (x). (4.3)

It can easily be checked that under gauge transformation the ac-tion is invariant. So gauge transformation is a special kind ofphase transformation in which the phase parameter is a functionof space-time. If is independent of space-time, the free Dirac La-grangian itself is invariant under the phase transformation. Butwhen we try to extend this global symmetry to a local one it be-comes necessary to introduce a vector field A which modifies thederivative term. This vector field has a certain transformation

property defined above.

In a similar vein, we start with the free Dirac Lagrangian forthe quarks:

L =i,s

is(i ms)is (4.4)

In the following, we will drop the flavour index s for simplicity andwork with one quark only. Consider, the (colour) transformation

k Ukll, k Ukll.If this is to be a symmetry of the action, we must have

UmiUni = mn,

i.e., U U = 133, in other words U is a unitary matrix. Sincewe want to preserve the norm of the quark fields, we will focuson a subset of U, for which det U = 1. Therefore, U is an SU(3)transformation matrix. It should be emphasised that the choiceof this particular symmetry is solely guided by experiments.

If we choose U to be 1, the transformation of is trivial. Solet us choose U to be close to the unity so that,

k k + k,

7/28/2019 tyty hhtt

40/100


where k is infinitesimal. This is obtained by choosing U suchthat,

U = 1 i ,where is some infinitesimal real parameter and is a 3 3matrix. Now, the condition U U = 133 implies that

= + O(2).Moreover, det U = 1 implies that, to order , Tr = 0. In otherwords, is a traceless hermitian matrix.

Let T1, T2, , T8 be eight linearly independent 3 3 tracelesshermitian matrices. Clearly they form a basis for the SU(3) alge-bra. A particulary convenient choice are the so called Gell-Mannmatrices. Let us choose the normalisation of these matrices suchthat,

Tr (TaTb) = 12

ab.

The commutators of the matrices Ta areTa, Tb

= ifabc Tc. (4.5)

From the definition, fabc is anti-symmetric in its first two in-dices. However, due to our choice of normalisation f is totallyanti-symmetric under the exchange of any two of its indices.

Using the basis, =

aTa and U can be written as,

U = 1

i aT

a.

To the lowest order in

k = ia(Ta)kll, k = ia(Ta)lkl. (4.6)It is easy to check that the free Dirac action is invariant underthe global, i.e., when as are independent of space time, SU(3)transformation.

Now we want to make this global symmetry of the action intoa local symmetry:

k = ia(x)(Ta)kl l. (4.7)

7/28/2019 tyty hhtt

41/100

4.1. Lecture I 141

Under the above transformation Lfree Lfree + Lfree, where

Lfree = (Ta)kl kl a(x). (4.8)

So we have to add something to the Lagrangian to cancel thisextra term. Adding a new interaction term

Lint = g kl Aa (Ta)kland demanding that Aa transforms as

Aa = 1

g

a(x) + fabcb(x)Ac,

it is easy to show the full action is gauge invariant.Now we want the action of the gauge fields in the non-abelian

case. If we define non-abelien F

as in QED, it would not begauge invariant as Aa contains a term proportional to f

abc. Letus define the field strength F in case of non-abelian gauge theoryas

Fa = Aa Aa + gfabcAbAc.

Under the gauge transformation

Fa = gfabcbFc,

where, a = a(x)/g. The field strength so defined is gauge covari-

ant. In order to check that it is so, one has to use the following

identity (Bianchi identity)fabcfckl + faclfckb fackfbcl = 0.

Let us use a convenient notation to summarise these. Define

A = AaTa, F = FaT

a and = aTa . Also, =

123

and = ( 1, 2, 3 ) In term of these matrices, the QCD Lagrangianis written as

LQCD = i

igA

m 12

T r

FF

, (4.9)

7/28/2019 tyty hhtt

42/100


where, = diag (, , ). In the new notation,

F = A A ig A, A

,

= ig , A ,F = ig

, F

, (4.10)

= ig , = ig .

4.2 Lecture II

In this lecture, we shall recapitulate the quantization of field the-ories.

Consider a classical field theory with a set of fields. These canbe a set of scalars or (the components of) vectors or spinor fields.We shall denote these collectively by {1, 2...N}. The action

S=

d4x L(1, , N; 1, , N) (4.11)

is a function of the fields and their first derivatives. The canonicalquantization proceeds as follows. Treat each field i as an operatori. The quantities we are interested in are the Greens functions(also called correlation functions) defined by

0 Ti1(xi1) in(xin) 0 = i1(xi1) in(xin) . (4.12)Recall that one needs to compute the Greens function to obtainthe S-matrix elements.

The path integral prescription to compute the Greens func-tions is given by

i1(x1) in(xn) =

Ni=1

Di(xi) eiSi1(x1) in(xn)Ni=1

Di(xi) eiS,

7/28/2019 tyty hhtt

43/100

4.2. Lecture II 143

where, Di(xi) means the sum over all possible field configura-tions.

The simplest way to interpret the path integral is to discretize

spacetime. We write (x0

, x1

, x2

, x3

) (an0

, an1

, an2

, an2

), wherea is a small number and n0, are arbitrary integers. So {i(x)} iscompletely specified by {i(an0, an1, an2, an3)} for every integersn0, n1, n2, n3. We also have to give a meaning to the derivativesafter discertization:

0i =1

a

i(a(n

0 + 1), an1, an2, an3) i(an0, an1, an2, an3)

,

and similarly for the others. The integral is replaced by a sum:d4x a4

n0,n1,n2,n3, (4.13)

while the delta-function can be written as

(4)(x x) = 1a4

n0n0 n1n1 n2n2 n3n3. (4.14)

The integral over all field configuration now becomes an infinitenumber of ordinary integrals:

Ni=1

[Di(xi)]

i,n0,n1,n2,n3

di(an0, an1, an2, an3) (4.15)

This interpretation of path integral defines the Lattice Field The-ory (see the lectures by Rajiv V. Gavai in this volume).Perturbation theory becomes simpler if one works in momen-

tum space. To this end, we define the Fourier transformation

i(x) =1

(2)4

d4k eikxi(k). (4.16)

We will now write the path integral in momentum space using the(Fourier) conjugate variables. For this we discretize momentumspace:

(k0, k1, k2, k3) (bn0, bn1, bn2, bn2), (4.17)

7/28/2019 tyty hhtt

44/100


where b is a small number and ns are integers. Then

i1(x1) in(xn) = d4k1

(2)4eik1x1

d4kn(2)4

eiknxn

i1(k1) in(kn) , (4.18)where,

i1(k1) in(kn)

=

Ni=1

Di(ki) eiS i1(k1) in(kn)Ni=1

Di(ki) eiS.

Let us introduce source functions and define the generatingfunctional of the path integral,

Z(J1, , Jn) =N

i=1

Di(ki) exp

iS+

d4xJi(x)i(x)

=

Ni=1

Di(ki) exp

iS+

d4k

(2)4Ji(k)i(k)

,

where Ji(k)s are the Fourier transforms of the sources Ji(x)s. Thefunctional derivatives with respect to these sources are definedsuch that

Ji(k)Jj(k

) = 4(k k) ij .

We shall need the analogue of this in terms of the discretizedvariables. Note that

4(k k) ij = 1b4

ij n0n0n1n1n2n2n3n3 1

b4ij nn , (4.19)

and Ji(k)

Jj(k) = ij nn with k (bn0, bn1, bn2, bn2). From the

above equations, it follows that Ji(k)

1b4 Ji(k) . Hence

(2)4

Ji1(k1) (2)

4

Jin(kn)Z(J) (4.20)

7/28/2019 tyty hhtt

45/100

7/28/2019 tyty hhtt

46/100


is defined in terms of its Taylor series. To a given order in pertur-bation theory (say n), one truncates at the nth term, and SI is apolynomial in the fields. Hence,

Z(J) =m=0

1

m!

iSI

i(k) (2)

4

Ji1(k1)m

Z0(J). (4.25)

The computation ofZ0(J) requires the knowledge ofS0. The mostgeneral form ofS0 is

S0 = 12

d4k

(2)4i(k)Mij j(k), (4.26)

where Mij is a matrix. E.g., in case of the 4 theory

Mij = (k2

m2

) ij . (4.27)

It is easy to check, using symmetry arguements, that

Mij(k) = Mji(k). (4.28)

Hence,

Z0(J) =N

i=1

Di(ki) exp

i

2

d4k

(2)4

i(k) Mij j(k)

2iJi(

k)i(k). (4.29)

We can simplify the above. First, let us rewrite the term in theparenthesis

i(k) iJl(k)(M1)li

Mij

j(k) i(M1)jm Jm(k)

+ Jl(k)(M1)lmJm(k) (4.30)

so as to complete the square. Next, we use the shifted variables

i(k) = i(k) i(M1)imJm(k)

7/28/2019 tyty hhtt

47/100

4.3. Lecture III 147

and note that Di(k) = Di(k) to arrive at

Z0(J) =

N

i=1Di(k) exp i

2 d4k

(2)4 i(k) Mij j(k)+Ji(k)M1ij Jj(k)

= N exp

i

2

d4k

(2)4Ji(k) M1ij Jj(k)

. (4.31)

Notice that the normalization constant N, arising out of the integral independent of the Js, drops out of the Greens functions.In the following, we shall ignore this normalization.

The most general form ofSI is

SI = m3

d4k1(2)4

d4km(2)4

(2)4 4(

ki)

i1(k1) im(km) f(m)i1im(k1, k2, , km). (4.32)

Finally, we obtain

i1(k1) in(kn)

=

1

Z(0)(2)4

Ji1(k1) (2)

4

Jin(kn)

s=0

1

s!i

m3 d4q1

(2)4

d4qm(2)4

(2)4(i

qi) f(m)(q1, q2, , qm)

(2)4

Ji1(q1) (2)

4

Jim(qm)

sZ0(J)

J=0

.

4.3 Lecture III

In this lecture we continue our review of quantization of field the-ories and derive the Feynman rules.

7/28/2019 tyty hhtt

48/100


An example of a typical term that we need to calculate is thefollowing.

(2)4

Jj1(k1) (2)4

Jjm(km)

exp

i

2

d4k

(2)4Ji(k)M1ij (k)Jj(k)

J=0

.

This vanishes if m is an odd integer (due to the symmetry of thetheory). We need, therefore, to compute it only for even integersm = 2p. Let ij = M

1ij . Then the above is:

ij1j2(k1) (2)4(4)(k1 + k2)

ij3j4(k3) (2)4(4)(k3 + k4) ij2p1j2p(k2p1) (2)4(4)(k2p1 + k2p)

+ all inequivalent pairings of j1 j2p.The above can be understood in the simpler context of ordinarydifferentiation of matrices:

xi1

xi2pexp

1

2xiKijxj

x=0

= Ki1i2Ki3i4 Ki2p1i2p+inequivalent pairings.

A very effective way to keep track of these terms is through a set a

set of diagrams (Feynman diagrams) and the associated Feynmanrules.

First, each propagator is denoted by a line, sometimes withthe momenta and gauge indices displayed explicitly:

k1

i1

k2

i2= ii1i2 (2)4(k1 + k2).

7/28/2019 tyty hhtt

49/100


Secondly, the interaction vertices:

km, im k3, i3

k1, i1 k2, i2

= if(m)i1i2im(k1, k2, , km).

Finally, for every (k) =

(2)4 Ji(k)

in the correlation function,

we draw an external line:

k= 1.

We impose the rule that all lines ending on vertices must be pairedby propagators. We also need to examine the constraints on mo-menta from momentum conservation.

Let us illustrate these in the example of4-theory. The actionis given by

S =1

2

d4k

(2)4i(k) (k2 m2) j(k)

4! d4k1

()4 d4k4

(2)4(2)4(i ki) (k1) (k4)(4.33)

In this case, Mij(k) = k2 m2, since there is only one field. The

Feynman rules are the following.

k1

i1

k2

i2= (2)4 (k1 + k2).

7/28/2019 tyty hhtt

50/100


The interaction vertex is

k1 k3

k2

k4

= i4!

(2)4(k1 + k2 + k3 + k4).

Let us consider the computation of the four-point function(k1)(k2)(k3)(k4)

.

We begin by drawing four external legs with appropriate momenta.In the lowest order (terms independent of ), we simply join thesepairwise by propagators. There are three possibilities:

k3 k4

k1 k2+

k3 k4

k1 k2

+

k3

k1 k2

k4

The contribution, to O(0), is, therefore

(2)4(k1 + k2)i

k21 m2(2)4(k3 + k4)

i

k21 m2

plus two other terms from the pairings (13)(24) and (14)(23). Inthe next order, it is possible to use one vertex, which is to beconnected to the external legs by four propagators. This gives thefollowing diagram.

7/28/2019 tyty hhtt

51/100


k3

k1 k2

k4

There are actually 4! possible pairings, all of which give the sameresult. Therefore, to O(), we get

4! 4j=1

i

k2j m2

(i)44!

(2)4(j

kj).

Some of the momentum integrals in the above are in fact triv-ial. There are also cancellations between factors of (2)4 appear-ing in the integrals and -functions. It is possible to simplify thecalculations by incorporating these from the beginning by the newFeynman rules.

Feynman rules:

1. The propagator:

k= ii1i2(

k)

.

2. The Vertices:

km, im k3, i3

k1, i1 k2, i2

= if(m)i1i2im(k1, k2, , km).

7/28/2019 tyty hhtt

52/100


3. The external line:

k= 1.

4. At the end we examine the constraints on momenta as usual.For every constraint of the form

j kj = 0, we put a factor

of (2)4(j kj).

5. Every independent momenta , which is not determined, isto be integrated over. Hence,

d4(2)4 for every independent

momenta.

The second example we shall consider is that of the six-pointfunction:

(k1)(k2) (k6)

at O(2). We start by drawing the six external legs and twovertices. Next we join the six external legs with two vertices in allpossible ways. This produces diagrams of the following type:

There are various ways to get at this diagram and we need to putappropriate combinatory factors. One can easily convince oneselfthat the correct combinatorial factor is 8 3 2 4 3 2 =2 (4!)2. Now recall the (/4!)2 term is accompanied by the 1/2!from the expansion of the integral. Thus, we get a miraculouscancellation of the factors for this case. This, of course, will not

7/28/2019 tyty hhtt

53/100


happen all the times. Using usual Feynman rules, the amplitudefor this diagram is:

(i)2 (2)4 6j=1

kj 6j=1

i

k2j m2 i(k1 + k2 + k3)2 m2 .As a final example, let us consider a correlator which has an

undetermined momentum. This flows in a loop in the correspond-ing Feynman diagrams. We consider an O(2) contribution to the

four-point function

(k1) (k4)

. A Feynman diagram in this

order is the following.

1

2

3

4

k1 + k2

We have the constraint k1 + k2 + k3 + k4 = 0 from momentumconsevation, which implies a factor of (2)4 (k1 + k2 + k3 + k4).However, this still leaves the momentum flowing in the loop un-determined. Consequently, we should integrate over this momen-tum. The combinatorial factor for this diagram can be seen to be8 3 4 3 2 = (4!)2. In addition, we have 1/2(4!)2 from theperturbative expansion of the exponential. Combining all these,

together with the external leg factors, we get the amplitude forthis diagram:

(2i)22

d4

(2)4i

2 m2i

(k1 + k2 )2 m2

4j=1

i

k2j m2

In the above, we have suppressed the momentum conservationcondition (

k). (Notice that, unlike the previous example, there

is no accidental cancellation of combinatorial factors here.)

Let us now calculate Z[0] using Feynman rules. In the pertur-bative expansion the following Feynman diagrams contribute.

7/28/2019 tyty hhtt

54/100


1 + + + +

These diagrams are known as the vacuum bubbles.Dividing a correlation function by Z[0] (the generating func-

tion with all sources J set to zero), is equivalent to ignoring all vac-uum bubbles from the Feynman graphs of (k1)(k2) (kn).For example, at O(2), one comes across the graph:

This, and similar graphs, are cancelled when we divide by Z[0].The method of canonical quantization is equivalent to the

quantization via path integrals. However, the latter is bettersuited for non-abelian gauge theories. Indeed, the problem ofcanonical methods is already apparent in QED, where one has toimpose constraints. This is because all the fields are not indepen-dent. The constraints in QED are linear in and hence can be dealtwith simply in either method. Those that appear is QCD, on theother hand, are nonlinear and are very difficult to impose in thecanonical method.

Exercise: Consider the part of the QCD action de-pendent on the gauge fields alone:

S =14

d4x FaF

a .

Show that the quadratic terms in A is of the followingform.

1

2

d4k

Aa(k) ab (k2+ kk)

Ma,b

Ab(k).

7/28/2019 tyty hhtt

55/100

4.4. Lecture IV 155

Show that Ma,b has zero eigenvalues. (Therefore,the operator M is not invertible.)

Exercise: Let us separate the QCD action in free

and interacting parts: S = S0 + SI). Write SI in theFourier basis to show that it takes the following form.

SI =

d4k1

(2)4

d4k3(2)4 (2)4(k1 + ... + k3)

Vabc(k1, k2, k3)Aa(k1)Ab (k2)Ac (k3)

+

d4k1(2)4

d4k4(2)4

(2)4(k1 + ... + k4)

Vabcd(k1, k2, k3, k4)Aa(k1)Ab (k2)Ac (k3)Ad(k4)

where, V

abc

(k1, k2, k3) is symmetric in (, ), (a, b) and(k1, k2). Find the expressions for Vabc(k1, k2, k3) and

Vabcd(k1, k2, k3, k4).

4.4 Lecture IV

Let us recall the path integral prescription to compute Greensfunctions:

i1(x1)

in(xn)

=

Ni=1

Di ei S[]i1(x1) in(xn)

Ni=1

Di ei S[]

where {i} are the set of fields in the theory. Splitting the actionS = S0 + SI one develops a diagramatic procedure for finding thecorrelation functions i1(x1) in(xn) upto any given order inperturbation theory. But this procedure fails for gauge theories.In order to see this, consider the action of a gauge theory

Sgauge = 12

d4k

(2)4Aa(k)abMa,bAb(k) + SI, (4.34)

7/28/2019 tyty hhtt

56/100


where, the matrix M is

Ma,b = ab(k2 kk). (4.35)

The propagator is the matrix inverse of M. However, since

(k2 kk)k = 0, (4.36)M has a zero eigenvalue, hence its inverse does not exist. In otherwords, the gauge field propagator cannot be calculated .

Now consider a more general set of objects:

0|T( O1 On)|0 =

n

=1

O

, (4.37)

where,

O is some combination of fields. For example, in QCD,

we may consider the elementary fields Aa(x), ks (x), Fa(x) or thecomposites ks (x)

kt (x), Fa(x)F

a(x), etc

1. The path integralprescription to compute the expectation value of such operatorsis given by

O

=

Ni=1

Di ei S[]

ON

i=1

Di ei S[]. (4.38)

The above is also ill-defined for reasons given above.

If two configuration {i(k)} and {i(k)} are related by a gaugetransform defined by

U(x) = exp (iga(x)a) ,

then {i} and {i} are said to be on the same gauge orbit (seeFig.4.1).

1One may ask as to why we are considering these more general operatorswhen we have not solved the problem for the elementary fields. The point isthat the composite operators are sometime simpler and often physically moremeaningful.

7/28/2019 tyty hhtt

57/100

4.4. Lecture IV 157

Gaugeorbit

Gauge

inequivalent

direction

Figure 4.1: The space of gauge field configurations.

Naively on would think that we can decompose the measureas

Ni=1

Di =Ni=1

Diineq

Ni=1

Diorbit

(4.39)

so as to integrate along the gauge orbits and along the directionsperpendicular to these. Using the fact that S[] is gauge invariant,we arrive at

O

=

Ni

Di|ineqei S[]N

i

Di|orbit

O

N

i D i|ineqei S[] N

i D i|orbit. (4.40)

We can integrate along the gauge orbits first, but this fails becausethere is no convergence factor from an action, indeed the actionfor this part of the integration is zero. This is the old problemof the non-invertible matrix in the quadratic term in a gauge the-ory action. However, the infinities

Di|orbit from the numeratorand the denominator will cancel if we could pull out O out ofthe integral. This happens if all the Os are gauge invariant.From now on we shall consider gauge invariant operators only (al-though we shall need to refer to gauge non-invariant ones later).

7/28/2019 tyty hhtt

58/100


Gauge

inequivalent

direction

Figure 4.2: Two inequivalent ways of picking representatives fromthe gauge orbits.

Of the examples above, only ks (x)kt (x) and F

a(x)F

a(x) are

allowed.In order to define the path integral corresponding to the cor-relators of gauge invariant operators, we do the following:

1. The first step involves gauge fixing, which essentially meanspicking one point from each gauge orbit (see Fig.4.2).

Since there are eight functions (x) parametrising thegauge orbits of SU(3), we need eight constraints to specifya point. Let us choose eight (gauge non-invariant) functionsof fields Ga(x) with constraints Ga = Ba(x), where Ba(x)are some fixed functions. For example we may choose

Ga = nAa(x),

or, Ga = Aa(x),

where n is an arbitrary vector. After Fourier transform tomomentum space, Ga = Ba(k). Roughly speaking, we want

i

Diineq

i

Dik,a

Ga Ba(k)

,

where the -function is at each point of the momentumspace. However, this is not quite correct as we shall also

7/28/2019 tyty hhtt

59/100

4.4. Lecture IV 159

have to make sure that any other way of gauge fixing willalso lead to the same result. In the following we shall learnto manipulate the integral so as to get something like the

above from the original expression for the correlators.2. Define i as the tranformation of i(x) by the gauge

transform U = eiga(x)Ta . Let i + i be the tran-

formation of i by an infinitesimal gauge transformation(1 iga(x)Ta). We denote Ga evaluated for i by Gaand similarly Ga + G

a and G

a have obvious meaning. In

general,

Ga = d4k(2)4 Kab,(k, k) b(k) (4.41)For example, for the choice Ga(x) =

Aa(x), we have

Ga

(k) =

ikAa

(k) and Ga

(k) =

ikAa

(k). From the

gauge transformation of Aa

Aa(x) = a(x) + gfabcb(x)Ac(x)

Aa(k) = ika(k) + gfabc

d4k

(2)4b(k)Ac(k k),

hence,

Kab,(k, k) = k2 (2)4ab 4(k k)

igfabckAc(k k). (4.42)At this point, we claim that

N Da(k)k,a

Ga(k) Ba(k)det Kab(k, k)=

(2)4

b4, (4.43)

(recall that b is the lattice size of the discrete momentumspace). The above follows from the analogous finite dimen-sional identityN

i=1

dui

Ni=1

(Fi(u)) det

Fi(u)

uj

= N, (4.44)

7/28/2019 tyty hhtt

60/100


where we have assumed that Fi(u), for all i, vanishes onlyat one point u0. The translation from the finite dimensionalcase to the functional identity (4.43) is provided by the dic-

tionary:

i (a, k),ui a(k),

Fi(u) Ga(k) Ba(k), b4/(2)4,

Fiuj

Kab(k, k).

The numerator of the path integral, by inserting an 1, canbe rewritten as

Ni

Di ei S[]

O() N1

Da(k)k,a

Ga(k, ) Ba(k)

det Kab(k, k, ).The first (original) part of the above integrand is indepen-dent of the gauge parameters a. We can therefore changethe order of integration to write

N1

a Da

iD

i

ei S[]

O()k,a

Ga(k, ) Ba(k)

det Kab(k, k, ).

Now, using the fact that the action S[], the operators O()and the measure D are all gauge invariant2, we can replacethese by S[], O() and D respectively. Now all de-pendence on is through the dummy variable of integration

2The functional measure is gauge invariant in QCD, but this is not neces-sarily the case for all gauge theories.

7/28/2019 tyty hhtt

61/100

4.5. Lecture V 161

, which can be redefined to be . So we make a change ofvariable from to and get

N1

aD

a i

Di

ei S[]

O()

k,a

Ga(k, ) Ba(k)

det Kab(k, k, ).

as the final expression for the numerator.

4.5 Lecture V

In the previous lecture we have obtained an expression for the cor-relators of gauge invariant operators in a gauge theory. Actuallywe manipulated only the numerator. However, a similar manipu-lation can be done for the denominator as well. It is now easy tosee that the integral

a

Da

in the numerator and the denominator cancel each other. Paren-thetically let us notice that only the absolute value of det Kab(k, k

)matters as its sign can be absorbed in N1 which cancels outin the end. We shall now express

k,a

Ga(k) Ba(k)

and

det Kab(k, k) in a way so that we can use standard perturbation

theory. In order to do this we use

a

DBa(k) exp

i2

d4k(2)4

Ba(k)Ba(k)

= N(),(4.45)

where N() is independent of Ba(k). Inserting N1() times theLHS above in the correlator we get:

O()

=1

N()

a

DBa(k)ei2

d4k(2)4

Ba(k)Ba(k)

1N

a

Da

i

DieiS[]

7/28/2019 tyty hhtt

62/100


O()k,a

Ga(k) Ba(k)

det Kab(k, k)

=N0 i Di exp

i

2 d4k

(2)4Ga(

k)Ga(k)

eiS[]

O()det Kab(k, k), (4.46)

where N0 is a new constant.We shall now rewrite det Kab(k, k

). To this end, let us recallthe results of gaussian integrationN

i=1

dxiNj=1

dyj exp

i

2xiAijy

j

=1

det A, (4.47)

for commuting variables x, y andNi=1

diNj=1

dj exp

i

2iAij

j

, = det A (4.48)

for anti-commuting (Grassmann) variables , . We may write

det Kab(k, k) =

DaDa exp

i

d4k

(2)4d4k

(2)4

a(k)Kab(k, k)b(k)

,

where a

and a are two independent spin-zero anti-commutatingfields. They do not carry any spinor index (but only gauge index),

yet they anti-commute and, hence they violate the spin-statisticstheorem. Due to this peculiar property, these fields, which wehave introduced for our convenience, are known as ghost fields.

Finally we have the desired expression for the correlator:

O()

=

i

Dia

Daa

Da eiStot

O (4.49)

where,

Stot = S[] 12

d4k

(2)4Ga(k)Ga(k)

7/28/2019 tyty hhtt

63/100

4.5. Lecture V 163

+

d4k

(2)4d4k

(2)4a

(k)Kab(k, k)b(k). (4.50)

Notice that Stot depends on as well as on Ga although the orig-

inal action we started with was independent of these. It turns outthat the Greens functions of gauge invariant operators actuallyare independent of and Ga. However, this is not true of Greensfunctions of operators that are not gauge invariant but appear inthe intermediate steps of calculations. This independence providesa nice check on the calculations.

Let us now make an explicit gauge choice, called the Lorenzgauge

Ga(k) = ikA(k), (4.51)(which in coordinate space is Ga(x) = A

a(x)). For this choice,we have

12

d4k(2)4

Ga(k)Ga(k) = 12

d4k(2)4

Aa(k)kkAb(k)ab,(4.52)

which is quadratic in gauge fields. The complete quadratic termfor A in Stot is

1

2

d4k

(2)4Aa(k)ab

k2 + kk 1

kk

Ab(k), (4.53)

where the first two terms (independent of ) come from the origi-nal kinetic term from S[] and the last term is from gauge fixing.It should be noted that the modified kernelk2 + kk 1

kk

(4.54)

no longer has a zero eigenvalue. Hence it is is invertible and onecan define the propagator of the gauge fields without any diffi-culty. Indeed, that was the purpose of this exercise. Since theparameter is arbitrary, we can choose a specific value to sim-plify computation. A particularly simple choice is = 1, for whichthe quadratic term in the action is

S(2) =1

2

d4k

(2)4Aa(k)

abk2

Ab(k). (4.55)

7/28/2019 tyty hhtt

64/100


Therefore, the gauge field propagator is

i ab 1k2

. (4.56)

The gauge fixing condition Ga(x) = Aa(k) results in

Kab(k, k) = ab(2)4(k k) k2 igkfabc Ac(k k). (4.57)

Hence, the third term in the Stot, which we shall call Sghost, be-comes

Sghost =

d4k

(2)4a

(k)k2 abb(k)

ig

d4k

(2)4d4k

(2)4fabck

a(

k)b(k)Ac(k

k). (4.58)

The Feynman rules with the modified action involves two ad-ditional diagrams:

The ghost propagator:

k ka b

= ik2

ab

Ghost-gauge boson coupling:

k1 k2a b

Ac

= gfabck1

The contribution from these diagrams will also have to betaken into account in the computation of amplitudes.

7/28/2019 tyty hhtt

65/100

4.6. Lecture VI 165

4.6 Lecture VI

We have obtained the Feynman rules of QCD in the path inte-

gral formalism. One can now proceed to calculate amplitude forphysical processes exactly as in the canonical method. We do,however, encounter divergences at the loop level. These requireregularization and renormalization before we can extract physi-cally meaningful expressions. An example of a divergent graph isthe following.

This graph contributesd4

(2)41

p21p22

2 ( +p1)2V(3)()V(3)( +p1), (4.59)

where the vertices contain one momentum factor each and the mo-mentum conservation condition (2)4(4)(p1 + p2) has been sup-pressed. For large values of internal momentum , the integralgoes as

d4(2)4

1

2 1

2 , (4.60)

which diverges quadratically.Regularization and renormalization are two steps in the pro-

cess of renormalization which makes sense of these divergent inte-grals. In the first step of regularization, the integrals are changedin such a way that they become finite. (It should be pointed outthat these are not the same integrals anymore.) There are variousways to do this, for example, with momentum cut-off one puts anupper limit on the value of the internal momenta || < , where

7/28/2019 tyty hhtt

66/100


is some fixed mass parameter. One takes at the end ofthe renormalization procedure. However, momentum cut-off is notvery convenient in QCD. We shall use a different regularization

called dimensional regularization.In dimensional regularization, one pretends to work in dimen-

sion D < 4, where D is chosen such that the integrals are notdivergent any longer. Only at the end of renormalization, we takethe limit D 4. We face a problem here since taking the limitis a continuous process. The solution is to write the formulas asfunctions of D and treat the dimension D (formally) as a contin-uous variable. In other words, we derive the relevant expressionsfor general D, without

tyty hhtt

Documents

Transcript of tyty hhtt