Chapter 1 IntroductionNecesscity of field theory in relativistic systemSchrodingerSchrodinger equation ⇒ conservation of particle number.Hψ = i h ∂ψ
∂t ⇒ ddt
∫d 3xψ†ψ = 0 →
∫d 3x (ψ†ψ) indep of time
If H is hermitian, H = H †. Then number of particles is conserved and no particle creation orannihilation.
Canonical commutation relation gives uncertainty relation,
[x , p] = −i h, ⇒ 4x4p > h
Fromp2c2 +m2c4 = E 2
get
4E = p4pE
c2 > p hc2
E4x or 4x > pcE(hc4E )
To avoid new particle creation we require 4E 6 mc2. Then we get a lower bound on 4x
4x > pcE
hmc
= (vc)(
hmc)
(Institute) Chapter 1 Introduction 2 / 37
For relativistic particlevc≈ 1, then
4x > ( hmc) Compton wavelength
⇒ Particle can not be confined to a interval smaller than its Compton wavelength
(Institute) Chapter 1 Introduction 3 / 37
Klein paradoxTo illustrate this feature we will study Klein’s paradox in the context of the Klein-Gordonequation given by (
∂2
∂t2−∇2 −m2
)ψ (x , t) = 0
Let us consider a square potential with height V0 > 0 as shown in the figure,
(Institute) Chapter 1 Introduction 4 / 37
A solution to the wave equation in regions I and II is given by
ψI (x , t) = e−iEt−ip1x + R e−iEt+ip1x
ψII (x , t) = Te−iEt−ip2x
where
p1 =√E 2 −m2, p2 =
√(E − V0)2 −m2
The constants R(reflection) and T(transmission) are computed by matching the twosolutions across the boundary x = 0. The conditions ψI (t , 0) = ψII (t , 0) and∂xψI (t , 0) = ∂xψII (t , 0) give
1+ R = T , (1− R ) p1 = Tp2
Solve for R and T
T =2p1
p1 + p2, R =
p1 − p2p1 + p2
if E > V0 +m both p1 and p2 are real and there are both transmitted and reflected wave.If E < V0 +m and E > m, then p2 is imaginary and p1 real, we get a reflected wave,transmitted wave being exponentially damped within Compton wavelength inside thebarrier and there is total reflection.when V0 > 2m and m < V0 − E then both p1 and p2 are real and there are both reflectedand transmitted waves . This implies that there is a nonvanishing probability of findingthe particle at any point across the barrier with negative kinetic energy (E −m − V0 < 0)!This result is known as Klein’s paradox. This result can only be understood in terms ofparticle creation at sudden potential step.
(Institute) Chapter 1 Introduction 5 / 37
Gauge Theory—Quantum Field Theory with Local SymmetryGauge principleAll fundamental Interactions are descibed in terms of gauge theories;
1 Strong Interaction-QCD;gauge theory based on SU(3) symmetry
2 Electromagnetic and Weak interaction-gauge theory based on SU (2)× U (1) symmetry
3 Gravitational interaction-Einstein’s theory-gauge theory of local coordinate transformation.
(Institute) Chapter 1 Introduction 6 / 37
Nature UnitsIn high energy physics, convienent to us the natural unit
h = c = 1
=⇒ many formulae simplified. Recall that in MKS units
h = 1.055× 10−34J sec
Thus } = 1 implies that energy has same dimension as (time)−1. Also
c = 2.99× 108m/sec
so c = 1 =⇒ time and length have the same dimension. In this unit, at the end of thecalculation one needs put back the factors of h and c to get the right dimension for the physicalquantities in the problem. For example, the quantity me can have following different meaningsdepending on the contexts;
1 Reciprocal length
me =1hme c
=1
3.86× 10−11cm
(Institute) Chapter 1 Introduction 7 / 37
2 Reciprocal time
me =1h
me c2
=1
1.29× 10−21 sec
3 energyme = me c2 = 0.511 Mev
4 momentumme = me c = 0.511 Mev/c
The following conversion relations
h = 6.58× 10−22Mev − sec hc = 1.973× 10−11Mev − cm
are quite useful in getting the physical quantities in the right units.Example:
(Institute) Chapter 1 Introduction 8 / 37
1 Thomson cross section
σ =8πα2
3m2e
Cross section has dimension of (length)2 and the only quantity here with dimension is meFirst compute this in natural unit,
σ =8πα2
3m2e=
8× 3.143× (0.5Mev )2
×(1137
)2=(1.78× 10−3
)(Mev )−2
Then we multiply this by(hc = 1.973× 10−11Mev − cm
)2 to convert this into cm2σ =
(1.78× 10−3
)(Mev )−2
(1.973× 10−11Mev − cm
)2= 6.95× 10−25cm2
(Institute) Chapter 1 Introduction 9 / 37
2 Decay rate of W−bosonIn Standard Model the decay rate for W − → eν is given by
Γ(W − → eν) =GF√2
M 3W6π
where MW = 80.4Gev/c2 is the mass of W−boson and GF = 1.166× 10−5Gev−2 is theweak coupling constant. We will first compute the rate in Gev and then convert to(sec)−1 , the unit for decay.
Γ(W − → eν) =GF√2
M 3W6π
=
(1.166× 10−5Gev−2
)× (80.4Gev )3√
26π= 0.227Gev
Then we divide by } to get the right unit
Γ(W − → eν) =0.228Gev
6.58× 10−22Mev − sec = 3.5× 1023/ sec
(Institute) Chapter 1 Introduction 10 / 37
3 Neutrino cross sectionFor quasi-elastic neutrino scattering νµ + e → µ+ νe , the cross section at low energies isof form
σ = 2G 2FmeE
where E is the energy of the neutrino. We want to find σ for E = 10Gev . Again, innatural unit.
σ = 2×(1.166× 10−5Gev−2
)2 × (.5Mev ) (10Gev ) = 1.34× 10−12Gev−2
Now we use hc = 1.973× 10−11Mev − cm to convert
σ = 1.34× 10−12Gev−2(1.973× 10−11Mev − cm
)2= 5.2× 10−40cm2
Note that this is a very small cross section which means that neutrino hardly interactswhen encounter matter with many electrons. Another feature is that in these lowenergies, this cross section increases with energies.
(Institute) Chapter 1 Introduction 11 / 37
4 Conversion of Newton constant
GN = 6.67× 10−11m3kg−1sec−2
to some energy scale, Planck scale. Use
hc = 3.16× 10−26Jm
we can write
GN = 6.67× 10−11(m2kgs2
)mkg 2
= 6.67× 10−11J mkg 2
Then we get for the combination
(hcGN) = 3.16× 10−26Jm × 1
6.67× 10−11Jm kg2 = 4.73× 10−16(kg )2
Use √hcGN
= 2.176× 10−8(kg )
and
1kg c2 = 1kg × (3× 108m/sec )2 = 9× 1016m2kgs2
= 9× 1016J or 1kg = 9× 1016J/c2
(Institute) Chapter 1 Introduction 12 / 37
we have
=⇒√hcGN
= 1.96× 109J/c2
Use the conversion factor
1Gev = 1.6× 10−10J =⇒ 1J =11.6× 1010Gev
we finally get
mp ≡√hcGN
=1.96× 109
1.6× 1010Gev = 1.225× 1019Gev/c2
This is the usual statement that the energy associated with gravity (Planck scale) is~1019Gev . Another way to express the result is
GN = 6.07× 10−39( hc )(Gev/c2)−2
(Institute) Chapter 1 Introduction 13 / 37
Review of Special RelativityBasic principles of special relativity :
1 The speed of light : same in all inertial frames.
2 Physical laws: same forms in all inertial frames.
Lorentz transformation—relate coordinates in different inertial frame
x ′ =x − vt√1− v 2
y ′ = y , z ′ = z , t ′ =t − vx√1− v 2
⇒t2 − x 2 − y 2 − z 2 = t ′2 − x ′2 − y ′2 − z ′2
Proper time τ2 = t2 −−→r 2 invariant under Lorentz transfomation.Particle moves from
→r1(t1) to
→r2(t2) .The speed is
|−→v | = 1|t2 − t1 |
√(x1 − x2)2 + (y1 − y2)2 + (z1 − z2)2
For |−→v | = 1,(t1 − t2)2 = |−→r1 −−→r2 |2
this is invariant under Lorentz transformation ⇒ speed of light same in all inertialframes .
(Institute) Chapter 1 Introduction 14 / 37
Another form of the Lorentz transformation
x ′ = coshω x − sinhω t , y ′ = y , z ′ = z , t ′ = sinhω x − coshω t
wheretanhω = v
For infinitesmal interval (dt , dx , dy , dz), proper time is
(dτ)2 = (dt)2 − (dx )2 − (dy )2 − (dz )2
(Institute) Chapter 1 Introduction 15 / 37
Minkowski space,xµ = (t , x , y , z ) = (x 0, x 1, x 2, x 3), 4− vector
Lorentz invariant product can be written as
x 2 = (x0)2 − (x1)2 − (x2)2 − (x3)2 = xµx νgµν
where
gµν =
1 0 0 00 −1 0 00 0 −1 00 0 0 −1
Define another 4-vector
xµ = gµνx ν = (t ,−x 1,−x 2,−x 3) = (t ,−−→r )
so thatx 2 = xµxµ
For infinitesmal coordinates
(dx )2 = (dxµ)(dxµ) = dxµdx νgµν = (dx 0)2 − (d−→x )2
(Institute) Chapter 1 Introduction 16 / 37
Write the Lorentz transformation as
xµ → x ′µ = Λµνx
ν
For example for Lorentz transformation in the x−direction, we have
Λµν =
1√1−β2
−β√1−β2
0 0−β√1−β2
1√1−β2
0 0
0 0 1 00 0 0 1
Write
x ′2 = x ′µx ′νgµν = Λµα Λν
β gµνx αx β
then x 2 = x ′2 impliesΛµ
α Λνβ gµν = gαβ
and is called pseudo-orthogonality relation.
(Institute) Chapter 1 Introduction 17 / 37
Energy and MomentumStart from
dxµ = (dx 0, dx 1, dx 2, dx 3)
Proper time is Lorentz invariant and has the form,
(dτ)2 = dxµdxµ = (dt)2 − (d−→xdt)2(dt)2 = (1−→v
2)(dt)2
4− velocity ,
uµ =dxµ
dτ= (
dx 0
dτ,d−→xdτ
)
there is a constraint
uµuµ =dxµ
dτ
dxµ
dτ= 1
Note that−→u = d−→x
dτ=d−→xdt(dtdτ) =
1√1− v 2
−→v ≈ −→v , for v � 1
4− velocity =⇒ 4−momentum
pµ = muµ = (m√1− v 2
,m−→v√1− v 2
)
(Institute) Chapter 1 Introduction 18 / 37
For v � 1,
p0 =m√1− v 2
= m(1+12v 2 + ...) = m +
m2v 2 + ..., energy
−→p = m−→v 1√1− v 2
= m−→v + ... momentum
pµ = (E ,−→p )
Note that
p2 = E 2 −−→p 2 = m2
1− v 2 [1− v2 ] = m2
(Institute) Chapter 1 Introduction 19 / 37
Tensor analysisPhysical laws take the same forms in all inertial frames, if we write them in terms of tensors inMinkowski space.Basically, tensors are
tensors ∼ product of vectors
2 different types of vectors,x ′µ = Λµ
νxν, x ′µ = Λ ν
µ xν
multiply these vectors to get 2nd rank tensors,
T ′µν = ΛµαΛν
βTαβ, T ′µν = Λ α
µ Λ βν Tαβ, T ′µν = Λµ
αΛ βν T
αβ
In general,
T′µ1 ···µnν1 ···νm = Λµ1
α1 · · ·ΛµnαnΛ β1
ν1 · · ·Λβm
νm Tα1 ···αnβ1 ···βm
transformation of tensor components is linear and homogeneous.
(Institute) Chapter 1 Introduction 20 / 37
Tensor operations; operation which preserves the tensor property
1 Multiplication by a constant, (cT ) has the same tensor properties as T2 Addition of tensor of same rank3 Multiplication of two tensors
4 Contraction of tensor indices. For example,T µαβγµ is a tensor of rank 3 while T µαβγ
ν is atensor or rank 5. This follows from the psudo-orthogonality relation
5 Symmetrization or anti-symmetrization of indices. This can be seen as follows. SupposeT µν is a second rank tensor,
T ′µν = ΛµαΛν
βTαβ
interchanging the indices
T ′νµ = ΛναΛµ
βTαβ = Λν
βΛµαT
βα
ThenT ′µν + T ′νµ = Λµ
αΛνβ
(T αβ + T βα
)symmetric tensor transforms into symmetric tensor. Similarly, the anti-symmetric tensortransforms into antisymmetic one.
6 gµν, and εαβγδ have the property
Λµα Λν
β gµν = gαβ, εαβγδ det (Λ) = ΛαµΛβ
ν Λγρ Λδ
σεµνρσ
gµν, and εαβγδ transform in the same way as tensors if det (Λ) = 1.
Example: M µν = xµpν − x νpµ, F µν = ∂µAν − ∂νAµ 2nd rank antisymmetric tensor.
(Institute) Chapter 1 Introduction 21 / 37
Note that if all components of a tensor vanish in one inertial frame they vanish in all inertialframe. Suppose
f µ = maµ
Definetµ = f µ −maµ
then tµ = 0 in this inertial frame. In another inertial frame,
t ′µ = f µ′ −ma′µ = 0
we get
f µ′ = ma′µ
Thus physical laws in tensor form are same in all inertial frames .
(Institute) Chapter 1 Introduction 22 / 37
Action principle: actual trajectory of a partilce minimizes the actionParticle mechanicsA particle moves from x1 at t1 to x2 at t2. Write the action as
S =∫ t2
t1L(x , x ) dt L : Lagrangian
For the least action, make a small change x (t),
x (t)→ x ′(t) = x (t) + δx (t)
with end points fixed
i .e . δx (t1) = δx (t2) = 0 initial conditions
Then
δS =∫ t2
t1[∂L∂x
δx +∂L∂x
δ(x )] dt
Note that
δx = x ′(t)− x (t) = ddt[δ(x )]
Integrate by parts and used the initial conditions
δS =∫ t2
t1[∂L∂x
δx +∂L∂x
ddt(δx )] dt =
∫ t2
t1[∂L∂x− ddt(
∂L∂x)]δx dt
(Institute) Chapter 1 Introduction 23 / 37
Since δx (t) is arbitrary, δS = 0 implies
∂L∂x− ddt(
∂L∂x) = 0 Euler-Lagrange equation
Conjugate momentum is
p ≡ ∂L∂x
Hamiltonian is ,H (p, q) = px − L(x , x )
Consider the simple case
md 2xdt2
= − ∂V∂x
Suppose
L =m2(dxdt)2 − V (x )
then∂L∂x=ddt(
∂L∂x), ⇒ − ∂V
∂x= m
d 2xdt2
Hamiltonian
H = px − L = m2(x )2 + V (x ) where p =
∂L∂x= mx
is just the total energy.
(Institute) Chapter 1 Introduction 24 / 37
Generalizationx (t)→ qi (t), i = 1, 2, ..., n
S =∫ t2
t1L(qi , qi ) dt
Euler-Lagrange equationsddt(
∂L∂qi)− ∂L
∂qi= 0 i = 1, 2, ..., n
pi =∂L∂qi
, H = ∑ipi qi − L
Example: harmonic oscillator in 3-dimensionsLagrangian
L = T − V = m2(x12 + x22 + x32)−
mw 2
2(x 21 + x
22 + x
23 )
=m2
(d→xdτ
)2− mw
2
2
(→x)2
and∂L∂xi
= −mw 2xi ,∂L∂xi
= mxi
(Institute) Chapter 1 Introduction 25 / 37
Euler-Langarange equationm
..x i = −mw 2xi
same as Newton’s second law.Remarks:
We need action principle for quantization
In action principle formulation, the discussion of symmetry is simpler
Can take into account the constraints in the coordinates in terms of Lagrange multiplers
(Institute) Chapter 1 Introduction 26 / 37
Field TheoryField theory ∼ limiting case where number of degrees of freedom is infinite. qi (t)→ φ(−→x , t).Action
S =∫L(φ, ∂µφ) d 3xdt L : Lagrangian density
Variation of action
δS =∫[∂L∂φ
δφ+∂L
∂(∂µφ)δ(∂µφ)] dx 4 =
∫[∂L∂φ− ∂µ
∂L∂(∂µφ)
]δφ dx 4
Use δ(∂µφ) = ∂µ(δφ) and do the integration by part. then δS = 0 implies
=⇒ ∂L∂φ= ∂µ(
∂L∂(∂µφ)
) Euler-Lagrange equation
Conjugate momentum density
π(−→x , t) =∂L
∂(∂0φ)
and Hamiltonian densityH = πφ− L
Generalization to more than one field
φ(−→x , t)→ φi (−→x , t), i = 1, 2, ..., n
(Institute) Chapter 1 Introduction 27 / 37
Equations of motion are∂L∂φi
= ∂µ(∂L
∂(∂µφi )) i = 1, 2, ..., n
and conjugate momentum
πi (−→x , t) =
∂L∂(∂0φi )
Hamiltonian density isH = ∑
iπi φi − L
(Institute) Chapter 1 Introduction 28 / 37
Symmetry and Noether’s TheoremContinuous symmetry =⇒ conservation law, e.g. invariance under time translation
t → t + a, a is arbitrary constant
gives energy conservation. Newton’s equation for a force derived from a potential V (→x , t) is,
md 2→x
dt2= −
→∇V (→x , t)
Suppose V (→x , t) =V (
→x ), then invariant under time translation and
md−→xdt·(d 2−→xdt2
)= −
(d−→xdt
)· −→∇V = − d
dt[V (−→x )]
Orddt[12m(d−→xdt)2 + V (−→x )] = 0, energy conservation
Similarity, invariance under spatial translation
−→x → −→x +−→a
gives momentum conservation and invariance under rotations gives angular momentumconservation. Noether’s theorem : unified treatment of symmetries in the Lagrangian formalism.
(Institute) Chapter 1 Introduction 29 / 37
Particle mechanicsAction in classical mech
S =∫L(qi , qi ) dt
Suppose S is invariant under a continuous symmetry transformation,
qi → q ′i = fi (qj ) ,
For infinitesmal changeqi → q ′i ' qi + δqi
The change of S
δS =∫[
∂L∂qi
δqi +∂L∂qi
δqi ] dt where δqi →ddt(δqi )
Using the equation of motion,∂L∂qi
=ddt(
∂L∂qi)
we can write δS as
δS =∫[ddt(
∂L∂qi)δqi +
∂L∂qi
ddt(δqi )] dt =
∫[ddt(
∂L∂qi
δqi )] dt
Thus δS = 0 ⇒ddt(
∂L∂qi
δqi ) = 0
(Institute) Chapter 1 Introduction 30 / 37
This can be written as
ordAdt
= 0, A =∂L∂qi
δqj
A is the conserved charge.
Note if δL 6= 0 but changes by a total time derivative δL =ddtK ,we still get the conservation
law in the form,ddt(
∂L∂qi
δqi −K ) = 0
because the action is still invariant. For example, for translation in time, t −→ t + ε,
q (t + ε) = q (t) + εdqdt
, =⇒ δq = εdqdt
Similarly,
δL =dLdt
The conservation law is thenddt(
∂L∂qi
δqi − L) = 0
OrdHdt
= 0, with H =∂L∂qi
qi − L
(Institute) Chapter 1 Introduction 31 / 37
Example: rotational symmetry in 3-dimensionaction
S =∫L(xi , xi ) dt
Suppose S is invariant under rotation,
xi → x ′i = Rij xj , RRT = RT R = 1 or RijRik = δjk
For example, a rotation around z−axis,
Rz (θ) =
cos θ sin θ 0− sin θ cos θ 00 0 1
For infinitesmal rotations, we write
Rij = δij + εij , |εij | � 1
Orthogonality requires,
(δij + εij )(δik + εik ) = δjk =⇒ εjk + εkj = 0 i , e , εjk is antisymmetric
For example, for θ � 1,
Rz (θ)→ 1+
0 θ 0−θ 0 00 0 0
(Institute) Chapter 1 Introduction 32 / 37
and ε12 = θ corresponds to infinitesmal rotation about z -axis. Similarly ε23 corresponds torotation about x -axis and ε31 rotation about y -axis.For the general case, we can compute the conserved quantities as
J =∂L∂xi
εij xj = εijpi xj
More explicitly,
J = ε12 (p1x2 − p2x1) + ε23 (p2x3 − p3x2) + ε13 (p1x3 − p1x3)
If we writeε12 = −θ3, ε23 = −θ1, ε31 = −θ2
orεij = −εijk θk
thenJ = −θk εijkpi xj = θk Jk Jk = εijk xipj
More explicitly,
J1 = (x2p3 − x3p2) , J2 = (x3p1 − x1p3) , J3 = (x1p2 − x2p1) ,
Hence Jk can be identified with k -th component of the usual angular momentum.If we write εij = −εijk θk
J = −θk εijkpi xj = −θk Jk Jk = εijk xipj
Jk k-th component of angular momentum.
(Institute) Chapter 1 Introduction 33 / 37
Field TheoryStart from the action
S =∫L(φ, ∂µφ) d 4x
Symmetry transformation,φ(x )→ φ′(x ′),
which includes the change of coordinates,
xµ → x ′µ 6= xµ
Infinitesmal transformation
δφ = φ′(x ′)− φ (x ) , δx ′µ = x ′µ − xµ
need to include the change in the volume element
d 4x ′ = Jd 4x where J =
∣∣∣∣ ∂(x ′0, x ′1, x ′2, x ′3)∂(x0, x1, x2, x3)
∣∣∣∣J :Jacobian for the coordinate transformation. For infinitesmal transformation,
J = | ∂x′µ
∂x ν| ≈ |g µ
ν +∂(δxµ)
∂x ν| ≈ 1+ ∂µ(δxµ)
(Institute) Chapter 1 Introduction 34 / 37
we have used the relation
det(1+ ε) ≈ 1+ Tr (ε) for |ε| � 1
Thend 4x ′ = d 4x (1+ ∂µ(δxµ))
change in the action is
δS =∫[∂L∂φ
δφ+∂L
∂(∂µφ)δ(∂µφ) + L∂µ(δxµ)] dx 4
Define the change of φ for fixed xµ,
δφ(x ) = φ′(x )− φ(x ) = φ′(x )− φ′(x ′) + φ′(x ′)− φ(x ) = −∂µφ′δxµ + δφ
or δφ = δφ+ (∂µφ)δxµ
Similarly,δ(∂µφ) = δ(∂µφ) + ∂ν(∂µφ)δx ν
Operator δ commutes with the derivative operator ∂µ,
δ(∂µφ) = ∂µ(δφ)
(Institute) Chapter 1 Introduction 35 / 37
Then
δS =∫[∂L∂φ(δφ+ (∂µφ)δxµ) +
∂L∂(∂µφ)
(δ(∂µφ) + ∂ν(∂µφ)δx ν) + L∂µ(δxµ)] dx 4
Use equation of motion∂L∂φ= ∂µ(
∂L∂(∂µφ)
)
we get∂L∂φ
δφ+∂L
∂(∂µφ)δ(∂µφ) = ∂µ ∂L
∂(∂µφ)δφ+
∂L∂(∂µφ)
∂µ(δφ) = ∂µ[∂L
∂(∂µφ)δφ]
Combine other terms as
[∂L∂φ(∂νφ) +
∂L∂(∂µφ)
∂ν(∂µφ)]δx ν + L∂ν(δx ν) = (∂νL)δx ν + L∂ν(δx ν)
= ∂ν(Lδx ν)
Then
δS =∫dx 4∂µ[
∂L∂(∂µφ)
δφ+ Lδxµ]
and if δS=0 under the symmetry ransformation, then
∂µJµ = ∂µ[∂L
∂(∂µφ)δφ+ Lδxµ] = 0 current conservation
(Institute) Chapter 1 Introduction 36 / 37
Simple case: space-time translationHere the coordinate transformation is,
xµ → x ′µ = xµ + aµ =⇒ φ′(x + a) = φ(x )
thenδφ = −aµ∂µφ
and the conservation laws take the form
∂µ[∂L
∂(∂µφ)(−aν∂νφ) + Laµ] = −∂µ(Tµνaν) = 0
where
Tµν =∂L
∂(∂µφ)∂νφ− gµνL energy momentum tensor
In particular,
T0i =∂L
∂(∂0φ)∂iφ
andPi =
∫dx 3T0i momentum of the fields
(Institute) Chapter 1 Introduction 37 / 37