Errata-P444text.pdf

(1)

Correction 1.1 Table 1.7 should read

Quark Bound States

BARYONS: qqq (a 3-quark bound state) MESONS: q6=q (a quark-antiquark bound state)

HADRONS

(big bracket now in correct place)

Correction 2.1 Equation (2.15) should read

gµνx0µx0ν =gµν(Λµαx α_{) Λ}ν

βx β

= gµνΛµαΛ ν

β

xαxβ

gµνΛµαΛνβ=gαβ

(indices now fixed)

Rotations compared to Lorentz transformations

Rotations Lorentz Transformations

RikRjk=δij gµνΛµαΛνβ=gαβ

φ0(~x) =φ(R−1_~_x) _SCALAR _φ0_(xµ_{) =}_{φ( Λ}−1µ

νx ν₎

V_i0(~x) =R_ijVj(R−1_~_x) _VECTOR _V0α_(~_{x) = Λ}α

µVµ( Λ−1

µ

νx ν₎

T_ij0(~x) =R_ikR_ilTkl(R−1~x) TENSOR T0αβ(~x) = ΛαµΛβνTµν( Λ−1

µ

νx ν₎

(indices now fixed in right-hand column)

A0_µB0µ=gµνA0µB0ν =gµν(ΛµαA α_{) Λ}ν

βB β

= gµνΛµαΛ ν

β

(2)

p0 = mu0=γmc= _p mc 1−v2_/c2

= 1 c

mc2+1 2mv

2₊3

8m v4

c2 +· · ·

Correction 2.6 Eq (2.37) should read

(mc)2+ (mc)2+ 2

_E

1E2

c2 −~p1·~p2

= (M c)2

This is just a correction in the font size of a subscript.

Correction 2.7 The first line of question 8 on page 42 should read

A pion travelling at speedv decays into a lepton of massm`and its corresponding antineutrino.

Part (a) of this question should read (a) Find an expression for the angle that the lepton is emitted relative to the original direction of motion.

R=





 



1 0 0

0 cosθx sinθx 0 −sinθx cosθx



, 



cosθy 0 −sinθy

0 1 0

sinθy 0 cosθy



, 



cosθz sinθz 0

−sinθz cosθz 0

0 0 1



 





Correction 3.2 Equation 3.12 should read

~

A·B~ =δjkAjBk A~×B~ i

=ijkAjBk

Correction 3.3 Question 3.7 should read

Consider a set of three objects{i, j, k} with the following properties

ij=k jk=i ki=j

and wherei2₌_j2₌_k2₌₋_{1. Show that the set}_{± {}_{1, i, j, k}_}_{forms a group under multiplication using}

these rules.

Correction 4.1 Pg 67, chapter 4, 2nd last paragraph last line should read

“... based on an action principle.”

Show that the operatorsP~ ·P~ andP~·~Lcommute with all elements of the algebra in question #4.

Consider the operatorU = exp−iθ

}nˆ·

~

Lwhere ~L=~x×(−i_}∇~) is the angular momentum operator

(3)

Correction 5.1 The last line of eq (5.18) should read

=−ψa

Correction 5.2 Question 5.6 (a) should read

(a) exphi~θ·~σi=P∞_n₌₀(i~θ·~σ)

n

n! = cosθ+iθb·~σsinθ

Correction 6.1 Page 96, just before eq (6.11) should read

are the spherical harmonics, given in appendix F.

Correction 6.2 Page 102, just before eq (6.23) should read

Using the Clebsch-Gordon tables in appendix F we have

Correction 6.3 The equations in Question 6.2 should read

(a) ¯n−→p¯+e++νe (b) γ+Z0−→νe+π0

(c) µ++τ− −→γ+γ+γ (d) Z0−→νµ¯ +νµ

(e) D0−→K−+ρ+ (f) e−+ ¯νe−→¯_t₊_b

Correction 6.4 Question 6.7 (b) should read

(b) Given thatT~J=−~JTwhere~Jis the angular momentum operator, show that

|χ1i ≡ −T

1 2,−

1 2

and |χ2i ≡T

1 2,

1 2

form a spin-1/2 doublet in the time-reversed system.

Correction 7.1 Equation (7.5) and the sentence before it should read

Because the charge of the electron is 1.60217733×10−19 _Coulombs _×₍₃_×₁₀9_{)esu/Coulomb, we find}

from eq.(7.3)

Br=pc q =

1.42×106_×₁₀7

3×109 = 4.73×10

3_Gauss-cm

_dE

dx

ionization

=−4πNA(ze)

2_e2

mec2_β2

_ρZ

A ln

_2m

ec2β2

I γ

2

−β2

(factor of densityρincluded)

The two sentences after this equation should read

(4)

_dE

dx

ionization

→ −4πNA(ze)

2_e2

mec2β2

_ρZ

A ln

_2mec2_β2

I

(factor of densityρincluded)

_dE

dx

ionization

' −2ρ MeV cm2/g

(spacing and units corrected)

The rate of energy loss of protons traveling a material of densityρtypically follows a power law

dE

dx =−Kρ

E E0

−p

where the constantsKandpare characteristic of the material andE0 is a reference calibration energy

that we can take to be 1 MeV.

(a) Find an expression for the thickness of material that will reduce the energy of a proton by half the value it has upon entering the material.

(b) For Carbon,K= 212 MeV cm2_{/g, and}_p_{= 0.757; for Lead,}_K_{= 89.5 MeV cm}2_{/g, and} _p_{= 0.694.}

How much thicker must a slab of Carbon be than a slab of Lead in order to reduce a proton of incident energy 150 MeV by half?

Correction 9.1 The bottom of page 175 to the top of page 176 should read

Independent measurements indicate that the proton spin is 1₂ and the deuteron spin is 1, so for|~pπ|2=

|~pp|2

2σ(p+p−→π+₊

D)

3 (2sπ+ 1) =σ π

+₊

D−→p+p (9.45)

Measurements by Cartwright, Clark and Durbin [102] showed that

σ π++_D−→p+p

= 3.1±0.3 mb 2σ(p+p−→π

+₊

D)

3 (2sπ+ 1) = 3.0±1.0 mb (9.46)

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(p0₁−p1) 2

−m2_B = (p0₁)2+ (p1)2−2p01·p1−m2

= 0 +m2−2 (E0E− |p~||p~0|cosθ)−m2

= −2E(E− |~p|cosθ)

(middle line corrected)

Correction 10.2 The first paragraph after eq (10.16) should read

Let’s pause to note some things about this formula. At high energies|~pc| 'E,and_ddσ_Ω → (}c)2₍_cg₎4

E6(1−cos2θ)2 ∼

(}c)2(cg)4

E6sin4_θ . The cross-section. . .

Correction 10.3 In figure 10.12, one B-particle should be a C-particle. The correct diagram should be

Find spinorsuthat satisfy (γµ_pµ₋_m)_{u(p) = 0 that have positive energy and that are eigenstates} of the operator ˆp·S~ where ˆpis the unit vector of the 3-momentum of the spinor and

~ S=}

2 ~ Σ = }

2

~σ 0 0 ~σ

is the spin angular momentum operator.

ψ(↑)ψ(↑)

=u(↑)(p)u(↑)(p)

= 2mψ₀2 qE+m

2m ξ

(↑)† qE−m

2m ξ

(↑)† _p_ˆ_·_~_σ†

I 0

0 −I



 q

E+m

2m ξ

(↑)

q

E−m

2m (ˆp·~σ)ξ

(↑)



(6)

= 2mψ₀2 qE+m

2m ξ

(↑)† qE−m

2m ξ

(↑)† _p_ˆ_·_~_σ† 

 q

E+m

2m ξ

(↑)

−qE−m

2m (ˆp·~σ)ξ

(↑)





= 2mψ₀2

_E₊_m

2m

ξ(↑)†ξ(↑)−

_E

−m 2m

ξ(↑)† pˆ·~σ†

(ˆp·~σ)ξ(↑)

= 2mψ₀2

_E₊_m

2m

−

_E₋_m

2m

= 2mψ₀2

(correct brackets inserted into 3rd line)

Consider a theory of one complex scalar particle with wavefunction ϕand two spin-1₂ particlesψ and χ, each of which couples to the photon via the equations

(iγµDµ−mψ)ψ = gϕ∗χ

(iγµDµ−mχ)χ = gϕψ

DµDµϕ−m2_ϕϕ = gψχ¯

wherem2

ϕ, mψ, and mχ are the respective masses of theϕ,ψandχ particles.

(a) Find the most general local phase transformation of ϕ , ψ and χ that leaves this system gauge-covariant.

(b) What is the relationship between the charges of ϕ,ψ andχ?

(c) In Maxwell’s equations

∂µFµν =Jv

what is the currentJv for this theory?

−iM=−iM_left−−iM_right

=hu(i01)_(p0

1)γ

µ v(i02)_(p0

2)

i ig_e2

(p1+p2) 2

h

¯ v(i2)_(p

2)γµu(i1)(p1)

i

−hu(i01)_(p0

1)γ

µ_u(i1)_(p

1)

i ig2

e (p0₁−p1)

2

h

¯ v(i2)_(p

2)γµv(i 0

2)_(p0

2)

i

Correction 13.2 Equation (13.24) should read (note – only the last line of this equation has changed)

Trhγµ/p0₁+mγν/p₁+mi

= Trhγµ/p0₁γνp/₁i+m2Tr [γµγν]

= Trγµ p0₁_λγλγν(p1ηγη)

+ 4m2gµν

=p0₁_λp1ηTr

γµγλγνγη

+ 4m2gµν

= 4p0₁_λp1η gµλgνη−gµνgλη+gµηgνλ+ 4m2gµν

= 4 p0₁µpν₁+p₁0νpµ₁+gµν

m2−p0₁·p1

(7)

(definition ofLµν corrected to be consistent with chapter 17)

Trhγ_µ/p0₂+Mγν/p₂+Mi = 4 p0₂_µp2ν+p02νp2µ+gµνM2−p02·p2

= Lµν(p02, p2;M2)

Correction 13.4 Equation (13.27) should read (note – only the first line of this equation is changed)

M 2 = 1 4 e2

(p0₁−p1) 2

!2

Lµν(p0₁, p1;m2)Lµν(p02, p2;M2)

= 4 e

2

(p0

1−p1) 2

!2

p₁0µpν₁+p₁0νpµ₁+gµν

m2−p0₁·p1

× p0₂_µp2ν+p02νp2µ+gµν

M2−p0₂·p2

= 4 e

2

(p0₁−p1)2

!2

[2 (p0₁·p0₂) (p1·p2) + 2 (p01·p2) (p1·p02)

+2 (p0₂·p2)m2−p01·p1+ 2 (p01·p1)M2−p02·p2

+4

m2−p0₁·p1 M2−p02·p2

= 8

_e2

(p0

1−p1) 2

2

[(p01·p02) (p1·p2) + (p01·p2) (p1·p02)

−(p0₁·p1)M2−(p02·p2)m2+ 2m2M2

X

iA,iB=↑,↓

u(iB)_a (pB) Γ_I

abu

(iA)

b (pA)u

(iA)

c (pA) ΓIIcdu

(iB)

d (pB)

= X

iB=↑,↓

u(iB)

a (pB) ΓIab



 X

iA=↑,↓

u(iA)

b (pA)u

(iA)

c (pA)



 Γ_II

cdu

(iB)

d (pB)

= X

iB=↑,↓

u(iB)

a (pB) Γ_I

ab

/ p

A+mA

bc Γ_II

cdu

(iB)

d (pB)

(notation for spin-indices (iA,iB) has been corrected).

X

iB=↑,↓

u(iB)a (pB) Γ_I

ab

/

p_A+mA bc ΓII

cdu

(iB)

d (pB)

= X

iB=↑,↓

u(iB)_d (pB)u(iaB)(pB) Γ_I

ab

/ p

A+mA

bc Γ_II

cd

= /p B+mB

da Γ_I ab / p

A+mA

bc Γ_II

(8)

(notation for spin-indices ((iA,iB)) has been corrected).

Correction 14.1 Part (c) of problem 14.2, bottom of pg 269 should read

(c) Use this to obtain a differential cross-section that is averaged over initial photon polarizations and summed over final photon polarizations. Work in the lab-frame.

Correction 15.1 Page 276, just before eq. (15.21) should read

... a job easily done using the 1⊗1

2 Clebsch-Gordon tables in appendix F:

Correction 15.2 Page 280, sentence after eq. (15.37) should read

Clearly the antiparticlesK− and Σ+ _have_S₌₋_{1 and}_S_{= +1 respectively.}

Correction 15.3 Page 285, just before eq. (15.41) should read

To get the spin part of the wavefunction, we just use the Clebsch-Gordon tables in appendix F twice to get 1₂⊗1

2⊗ 1

2 = (1⊕0)⊗ 1

2 = 1⊗ 1 2

⊕ 0⊗1 2

= 3₂⊕1 2⊕

1 2:

Correction 16.1 Eq. (16.29) should read

∆EH.F.=hHdipoleidirections=

_8π

3

_4πα

mempc2 ~

Se·S~p|Ψn=0(0)| 2

(factor of_~2 _removed)

|M|2= 1 4

−e2

q2

2

Lµν(p01, p1;m2)Lµν(p02, p2;M2)

dσ= }

2_πM

2

q

(p1·p2) 2

−p2 1p22

_e2

q2

2 _{c d}3_p0

1

2E₁0(2π)3L µν_(p0

1, p1;m2)Wµν(p02, p2;M2)

dσ= π E

}e2 cq2

2_E0_(dE0_dΩ)

4 (2π)3 L µν_(p0

1, p1;m2)Wµν(p02, p2;M2)

dσ dE0_dΩ=

}α cq2

2 _E0

2EL µν_W

µν

(9)

LµνWµν = 8EE0

2W1 q2, q·p2sin2

θ

2 +W2 q

2_{, q}_·_p 2cos2

θ 2

W₁j q2, xj

= Q

2

j

2mjδ(xj−1) W j

2 q 2_{, xj}

=−2mjQ

2

j

q2 δ(xj−1)

Correction 18.1 The 3rd sentence in the last paragraph on page 337 should read:

These must be high energy (“hard”) gluons since they carry the total mass of theφmeson.

Correction 18.2 The last sentence in the paragraph at the top of page 345 should read: Figure 18.12 illustrates a typical top event.

Correction 18.3 The caption for figure 18.13 on page 346 should read:

Analysis of a top event. In fig. 18.12 (a), a proton and antiproton collide producing four distinct jets (b) and some other particles. These are sketched in figure 18.12 (b). Multiple jets and a positron signify the possible creation of a top quark. The energies and locations of the particles measured by the calorimeter surrounding the beam line are shown in figure ??, with the energies released by the particles indicated by the heights of the bars [?]. Copyrightc 1997 by Scientific American, Inc. All rights reserved.

Correction 18.4 The paragraph immediately after equation (18.17) should read The left-hand side of this equation for each gluon wavefunctionAa

ν is just like Maxwell’s equations. The first term on the right-hand side is the color-current due to the quarks (analogous to the electric current

¯

ψγµ_ψ _{in QED) – this gives rise to the expected quark-quark-gluon vertex.}

Correction 18.5 The last paragraph in the sentence after equation (18.25) should read

This additional effect contributes oppositely to the fermion loops, leading to the result in eq. (18.24).

FµνΦ = (ig)−1[Dµ,Dν] Φ

Show that the 2 constituents of a meson in the colour singlet state _√1 3

RR+BB+GG

experience a

potential V =−4 3 1

r.

(a) Given

FµνΦ = (ig)−1[DµΦ]

show that

Fµν=∂µAν−∂νAµ+ig[Aµ,Aν]

or alternatively

F_µνa =∂µAνa−∂νAaµ−gf a bcA

b µA

c ν

(b) Show that for a given representationR

(10)

Correction 19.1 The 4th sentence in the 2nd paragraph on page 378 should read

But not as short as it might have been – theB0-meson fortuitously has a relatively long lifetime of about 10−12 sec ...

You make contact with alien physicists through a wormhole into another part of the multiverse, and soon develop a common language of communication with them. They have developed methods of traveling through the wormhole in short times and are considering visiting Earth. However, before they visit, you want to be sure that they are not made of antimatter. Because of this lack of knowledge, it’s too dangerous to send objects through the wormhole, but you can ask them any questions you want about experiments they have performed, and you are able to communicate with them results of experiments performed here.

(a) Can you determine if they are made of antimatter if each of _C, _P, or _T are conserved in all interactions in their universe?

(b) Can you determine if they are made of antimatter ifCPis conserved byCandPare both violated in their universe?

(c) Can you determine if they are made of antimatter if_CP,_Cand_Pare each violated in their universe?

Correction 20.1 Question 20.4 should read (note: only part (d) has changed)

Consider muonium, a bound state ofµ+ withe−.

(a) What are the possible spins of the two lowest-energy states? Which has the lowest energy?

(b) There are two possible ways that muonium can decay. What are they?

(c) Draw a diagram for each decay mode in part (b).

(d) Only one decay mode is possible for one of the lowest energy states. Which one is it, and why?

(e) Compute the ratio of the decay rates for the state in which both decay modes are allowed. Which is the more likely decay?

Correction 21.1 The last line of equation (21.22) should read

(p·p01) =p0 2

1 + (p02·p01) =m 2

`+ 1 2 m

2

π−m

2

`

=1 2 m

2

π+m

2

`

K+−→π++νe+ ¯νe K+−→π0+νe+e+ (21.1)

BR K+−→π++νe+ ¯νe

= 1.7±1.1×10−10

BR K+−→π0+νe+e+

(11)

M = (−i)2cV

gZ

2

u(p0₁)γµ

₁

2− 1 2γ

5

u(p1)

gµν−qµqν/M2

Z q2₋_M2

Z+iMZΓZ

×

u(p0₂)γν 1−γ5u(p2)

' −c

e V 2

gZ 2MZ

2

u(p02)γµ 1−γ5

u(p2) u(p01)γµ 1−γ5

u(p1)

(primes are in the proper place and covariant indices have been corrected).

Neutral Current Candidates from the Gargamelle Experiment

ν-exposure ν¯-exposure

# of Neutral current candidates 102 64 # of Charged current candidates 428 148

σ= }cg

2

ZE

2

48π



   h

(ce V)

2

+ (ce A)

2i

cf_V

2

+cf_A

2

(2E)2−M2

Z

2

+ (MZΓZ)2



  

(22.3)

(erroneous exponent of 2 removed).

(12)

Correction 23.1 Figure 23.5 should not have a factor of √2 in the denominator. The correct figure is

F_µνa =∂µWνa−∂νWµa−gWW

c µW

b ν

a cb

gZjZ

ν = −

gYsinθW

2

Y_LLχL_LγνχL_L+Y_RLχL_RγνχL_R+Y_LQχQ_LγνχQ_L +Y_RQχQ_RγνχQ_R

+gWcosθWj 3

ν

= gWcosθWj 3

ν− sinθW

cosθW

[gejem

ν −gWsinθWj 3

ν]

= ge

sinθWcosθW

(j3₎µ₋_sin2_θW_(jem₎µ

(factor of 2 corrected in last term of first line)

Correction 23.4 Question 23.4 should read (note: only the last line has been changed)

Show that if

Φ =

0 v+h(x)

then

DµΦ =−igW

v 2

_W1

µ−iWµ2

− Zµ cosθW

(13)

and

igW

2

Φ†σaDµΦ−(DµΦ)†σaΦ = gWv 2

2

W_µ1, W_µ2, 1 cosθW

Zµ

+O(h)

igY

4

Φ†DνΦ−(DνΦ)†Φ = −v

2

2 g_Wg_Y

cosθW

Zµ+O(h)

(minus sign corrected in last line)

Correction 24.1 Figure 24.1 should be

P(νµ−→ντ, t) = sin2(2ϑ23) sin2

_(E

2−E3)t

2}

Correction 25.2 Figure 25.6 should be (variables appropriately repositioned)

!

µ

!

"

k

ig

_d

_#

5 $

=

k

p

(14)

(15)

(16)

Correction 0.2 On page 553, the Z-vertex factor should not have a factor of√2 in the denominator. The corrected diagram is

Correction 0.3 On page 542 the entry for the weak mixing angle should read