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QUANTUM FIELD THEORY
Dr. W.L.G.A.M. van Neerven
Instituut-Lorentz
Universiteit Leiden
postbus 9506, 2300 RA Leiden
Nederland
1
Content Quantum Field Theory
1 page 5. Relativistic classical field theory.
1.1 page 5. The principle of special relativity.
1.2 page 7. The real scalar (spin zero) field.
1.3 page 9. The complex scalar (spin zero) field.
1.4 page 10. The complex Dirac (spin half) field.
1.5 page 17. The electromagnetic (massless spin one) field.
2 page 23. Symmetries in the classical field theory.
2.1 page 23. Lagrange en Hamilton formalism for classical fields.
2.2 page 28. Noether theorema for classical fields.
2.3 page 35. Global and local gauge symmetries.
3 page 47. Relativistic quantum field theory.
3.1 page 48. Quantisation of the real scalar field.
3.2 page 52. Quantisation of the complex scalar field.
3.3 page 54. Quantisation of the complex Dirac field.
3.9 page 59. Quantisation of the elektromagnetic field.
4 page 62. Interacting fields.
4.1 page 62. Greens functions for the scalar field.
4.2 page 65. Perturbation theory.
4.3 page 88. Ultraviolet singularities and renomalization.
5 page 97. Spontaneous symmetry breaking
5.1 page 97. Nambu-Goldstone mechanism.
5.2 page 102. Higgs mechanism.
5.3 page 106. Landau-Ginzburg theory.
5.4 page 115. The ε expansion and the renormalisation group.
5.5 page 126. The 1/N expansion.
6 page 132. Quantum electrodynamics
6.1 page 132. Green’s function for the Dirac field.
2
6.2 page 135. Green’s function for the electromagnetic field.
6.3 page 138. Feynman rules for QED.
6.4 page 139. Lowest order processes in QED.
3
Recommended literature
1 A. Bailin and A. Love, ”Introduction to Gauge Field Theory”, Adam Hilger,ISBN 0-85274-817-5.
2 L. Ryder, ”Quantum Field Theory”, Cambridge Univ. Press, ISBN 0 521 33859X.
3 R.J. Rivers, ”Path Integral Methods in Quantum Field Theory”, CambridgeUniv. Press, ISBN 0 521 25979 7.
4 ”Quantum Field Theory, A Modern Introduction”, M. Kaku, Oxford UniversityPress, ISBN 0-19-509158-2 (pbk). hoofdstukken 1 t/m 4.
5 Ta-Pei Cheng and Ling-Fong Li, ”Gauge theory of Elementary Particle Physics”,Oxford Univ. Press, ISBN 0-19-851956-7.
6 D.J. Amit, ”Field Theory, The Renormalization Group and Critical Phenomena”,World Scientific, ISBN 9971-966-10-7.
7 M. Le Bellac, ”Quantum and Statistical Field Theory”, Clarendon Press Oxford,ISBN 0-19-853929-0.
4
1 Relativistic classical field theory
1.1 The principle of special relativity
In the classical theory t is shown that the speed of light is not the same in all inertialsystems i.e. c′ 6= c (c′ = c+ v0 follows from x′ = x+ v0t). Further Galilei transforma-tions do not leave the Maxwell equations, which describe electromagnetism, covariant(Larmoor, H.A. Lorentz 1904, H. Poincare 1905).However the Michelson-Morley experiment (1887) reveals that the speed of light in allinertial systems is the same and finite. From this we conclude that we need anothersymmetry group. It turns out that the Maxwell equations and the invariance of thespeed of light are maintained if we assume the Lorentz group as the symmetry groupof nature. (or her extension the Poincare group). This follows from the Principle ofSpecial Relativity as formulated by A. Einstein in 1905. This states
1 Relativity principle: The laws of nature are the same in all inertial systems.(analogue of the classical relativity).
2 Principle of the invariant speed of light: The speed of light is the same in allinertial systems en and it is the largest speed at all.
A light signal is emitted from the origin in the coordinate system S (ct, x, y, z) andS ′ (c′t′, x′, y′, z′). The origins coincide at t = t′ = 0. The radius of the spherical waveis given by |~x| = ct en |~x′| = c′t′ respectively. According to Einstein is c′ = c. Next wehave the equation
c2t2 − |~x|2 = c2t′2 − |~x′|2 = 0 (1.1)
The above equation can be rewritten using the metric tensor. For that purpose weintroduce a four dimensional space, the Minkowski space. This unifies the three di-mensional space with time. Choose a basis in Minkowski space which is given by
e0, e1, e2, e3 → eµ µ = 0, 1, 2, 3 (1.2)
where e0 the time vector. A four vector (denoted without an arrow) can then bedenoted as x = xµ eµ (watch the index). This can be also written in coordinate formx = (x0, x1, x2, x3) = (ct, ~x). From mathematics it is known that each vector spaceV of dimension n has a dual vector space V ∗ with the same dimension. The latterconsists of all linear functions (mappings) f for which holds
f : V → ℜ met x ∈ V f ∈ V ∗ f(x) ∈ ℜ (1.3)
Here ℜ is the set of real numbers. The basis in V ∗ is denoted by
e0, e1, e2, e3 → eµ µ = 0, 1, 2, 3
with eµ eν = δµν (1.4)
5
Here δ represents the Kronecker delta tensor in four dimensions with the properties
δµν = 1 µ = ν δµ
ν = 0 µ 6= ν (1.5)
If x ∈ V ∗ then it can be written as x = xµ eµ or x = (x0, x1, x2, x3). In the formal-
ism above eµ and eµ are called covariant and contra variant basis vectors respectivelywhereas xµ en xµ are the covariant and contra variant components.Remind: µ is lowerindex → µ is co − variant, µ is upperindex → µ is contra −variant. We introduce a metric tensor g with components gµν . It has the properties
g00 = 1 gij = −1 i = j = 1, 2, 3 gµν = 0 µ 6= ν
0 1 2 3
0 1 0 0 01 0 −1 0 02 0 0 −1 03 0 0 0 −1
(1.6)
Notice that 0 is the time index and i, j are the spatial indices. The Minkowski metricis indefinite this is in contrast to the Euclidian metric which is represented by the
Kronecker delta tensor~~δ. That is why g does not define a norm in contrast to
~~δ.Equation (1.1) can now be cast in the form
c2t2 − |~x|2 = c2t′2 − |~x′|2 = 0 = gµν xµ xν (1.7)
Besides gµν which is a covariant tensor we have also a contra variant tensor gµν withexactly the same properties as in (1.6).
gµν gνλ = δλ
µ (1.8)
gµν gνλ = δλ
µ (1.9)
Further we have the relations
eµ · eν = gµν eµ · eν = gµν
eµ = gµν eν eµ = gµν eν
xµ = gµν xν xµ = gµν xν
x · y = xµyνeµeν = xµyνg
µν = gµνgµλxλyν = δν
λxλyν
x · y = xµyνeµeν = xµyνgµν = gµνgµλxλy
ν = δλνxλy
ν
x · y = x0y0 − x1y1 − x2y2 − x3y3 x · y = x0y0 − x1y1 − x2y2 − x3y3
6
x · y = x0y0 + x1y
1 + x2y2 + x3y
3 = x0y0 + x1y1 + x2y2 + x3y3
x · y = x0y0 − ~x · ~y = x0y0 − xiyi = x0y0 − xiyi = x0y0 + xiyi = x0y0 + xiy
i
(1.10)
Let A be a transformation which connects the components x and the basis vectors eof systems S and S ′ then the transformations read
e′µ = Aµν e
ν x′µ = (A−1)νµ xν
e′µ = (A−1)νµ eν x′µ = Aµ
ν xν
(AT )µν ≡ Aν
µ Aµν (A−1)ν
λ = δµλ A =
A00 A1
0 A20 A3
0
A01 A1
1 A21 A3
1
A02 A1
2 A22 A3
2
A03 A1
3 A23 A3
3
(1.11)
1.2 The real scalar (spin zero) field
A relativistic classical field, notation ϕr(~x, t) ≡ ϕr(x), is a function defined on spaceand time. This field satisfies a covariant differential equation. Further ϕr(x) is a rep-resentation of a certain symmetry group where r is the representation index. In theselecture the symmetry group is given by the Lorentz group or Poincare group. A clas-sical field can be considered as an object which describes a system of infinite degreesof freedom. In classical mechanics can a system with N particles be described by nindependent generalised coordinates qi(t) (i = 1, 2, · · · , n). If one takes the index icontinuous i.e. qi(t) → q(i, t) is a function of i and t. When we take for i the spatialcoordinates (x1, x2, x3) = (x, y, z) then q(~x, t) becomes a classical field.
The scalar free field satisfies the Klein-Gordon equation(
2 +m2c2
h2
)
φ(x) = 0 met 2 ≡ ∂µ∂µ (1.12)
The symbol 2 is called the d’Alembertian. This is a scalar quantity under the Lorentzgroup. It can be written as
2 =∂2
c2∂ t2−
3∑
i=1
∂2
∂ x2i
~∇ · ~∇ ≡ ∆ =3∑
i=1
∂2
∂ x2i
(1.13)
Here ∆ is the Laplacian. Further is h = h/2π where h is Planck’s constant. This isintroduced here on the classical level because later the field will be quantised. This fieldis also called the Klein-Gordon field. The most general solution of the Klein-Gordonequation looks like
φ(x) =∫
d3~p[
a(~p) fp(x) + a∗(~p) f ∗p (x)]
(1.14)
7
met(
2 +m2c2
h2
)
fp(x) = 0
(
2 +m2c2
h2
)
f ∗p (x) = 0 (1.15)
The solutions are
fp(x) = Ae−i px/h f ∗p (x) = Aei px/h
p = (E
c, ~p) E =
√
m2c4 + |~p|2c2 px
h=
1
h(E t− ~p · ~x) (1.16)
We have written the field φ(x) as a Fourier expansion in the plane wave functions fp,f ∗p . The quantities a(~p), a∗(~p) are the Fourier coefficients. A is a real normalisationconstant which has still to be determined. This depends on the convention which hasto be introduced. Define the inner product
(f, g) =i
hc
∫
d3~x f ∗(x)↔∂0 g(x)
f ∗(x)↔∂µ g(x) = f ∗(x)(∂µg(x)) − (∂µf
∗(x))g(x) (1.17)
Notice that this product does not lead to a norm. Calculate (fp, fp′)
(fp, fp′) = iA2
hc
∫
d3~x e−i ~p·~x/h ei ~p′·~x/h(
ei Et/h ∂
c∂ te−i E′t/h − e−i E′t/h ∂
c∂ tei Et/h
)
= A2 ei (E−E′)t/hE + E ′
(hc)2
∫
d3~x e−i (~p−~p′)·~x/h
= A2 ei (E−E′)t/hE + E ′
(hc)2(2π)3 δ(3)
(
(~p− ~p′)
h
)
= A2 2E
(hc)2(2πh)3 δ(3)(~p− ~p′) (1.18)
Notice that from ~p = ~p′ follows E = E ′. The properties of the delta function are
δ(n)(x) =1
(2π)n
∫
dnk e±i kx n = 1, 2, · · ·
δ(n)(x) = δ(x1)δ(x2) · · · δ(xn)
δ(n)(ax) =1
anδ(n)(x) (1.19)
8
Choose A such that (fp, fp′) = δ(3)(~p− ~p′). One obtains
A =1
(2πh)3/2
hc√2E
(1.20)
Further one can show
(f ∗p , f∗p′) = −δ(3)(~p− ~p′) (f ∗p , fp′) = (fp, f
∗p′) = 0
a(~p) = (fp, φ) a∗(~p) = −(f ∗p , φ) (1.21)
Notice that (f, g) does not define a norm as is the case in non relativistic relativisticquantum mechanics. In the latter case each wave function has to satisfy
|ψ|2 = (ψ, ψ) ≥ 0 (1.22)
Here the norm is definite positive! In field theory this is not the case e.g. (f ∗p , f∗p′) <
0. Hence φ(x) is field and no wave function as one originally thought. After thenormalisation choice we can write for the Klein-Gordon field
φ(x) =∫
d3~p
(2πh)3/2
hc√2E
[
a(~p) e−i p·x/h + a∗(~p) ei p·x/h]
(1.23)
Watch that in the most field theory books is written h = c = 1.
1.3 The complex scalar (spin zero) field
This field is defined by
φ(x) =1√2(φ1(x) + i φ2(x)) (1.24)
where φ1(x) and φ2(x) are real fields which satisfy Klein-Gordon equations(
2 +m2c2
h2
)
φ(x) = 0
(
2 +m2c2
h2
)
φ∗(x) = 0 (1.25)
Hence follow the solutions
φ(x) =∫ d3~p
(2πh)3/2
hc√2E
[
a+(~p) e−i p·x/h + a∗−(~p) ei p·x/h]
φ∗(x) =∫
d3~p
(2πh)3/2
hc√2E
[
a−(~p) e−i p·x/h + a∗+(~p) ei p·x/h]
(1.26)
From the definition of φ(x) the next relations follow
a+(~p) =1√2(a1(~p) + i a2(~p)) a∗−(~p) =
1√2(a∗1(~p) + i a∗2(~p))
a∗+(~p) =1√2(a∗1(~p) − i a∗2(~p)) a−(~p) =
1√2(a1(~p) − i a2(~p)) (1.27)
where ai(~p), a∗i (~p) are the Fourier coefficients of φi(x).
9
1.4 The complex Dirac (spin half) field
To describe spin half field we follow a procedure which is also applied in non relativisticquantum mechanics to describe spin half wave functions. To that purpose we introducespinors and the Pauli matrices.
σx = σ1 =
(
0 11 0
)
σy = σ2 =
(
0 −ii 0
)
σz = σ3 =
(
1 00 −1
)
(1.28)
with the properties
[σk, σl] = 2 i ǫklm σm σk, σl = 2 δkl σ†k = σk
⇒ σk σl = δkl + i ǫklm σm (1.29)
The Pauli spinors are
χ(1/2) =
(
10
)
χ(−1/2) =
(
01
)
σz χ(s) = 2 s χ(s) (1.30)
Here χ(s) is a eigen state of σz . In the above representation the quantisation axis ischosen to be in the z direction. If an arbitrary direction ~n is chosen with |~n = 1 thenthe spinors are eigen states of ~σ · ~n
~σ · ~nχ(s)n = 2 s χ(s)
n met ~n = (sin θ cosϕ, sin θ sinϕ, cos θ)
~σ · ~n =
(
cos θ sin θ e−i ϕ
sin θ ei ϕ − cos θ
)
χ(1/2)n =
(
e−i ϕ/2 cos θ/2ei ϕ/2 sin θ/2
)
χ(−1/2)n =
(
−e−i ϕ/2 sin θ/2ei ϕ/2 cos θ/2
)
(1.31)
From (1.29) follows
~σ · ~a~σ ·~b = ~a ·~b1 + i ~σ · (~a×~b) 1 =
(
1 00 1
)
→ (~σ · ~a)2 = |~a|2 1 (1.32)
Using the above we can write the Klein-Gordon equation as follows
1
(
− ∂2
∂ (x0)2+ ~∇ · ~∇
)
φ(x) =
(
−1∂2
∂ (x0)2+ (~σ · ~∇)2
)
φ(x) = 1m2c2
h2 φ(x) (1.33)
or(
i h1∂
∂ x0+ i h ~σ · ~∇
)(
i h1∂
∂ x0− i h ~σ · ~∇
)
φ(x) = 1m2c2 φ(x) (1.34)
10
Call φ(x) ≡ φL(x) then we can define
φR(x) =1
mc
(
i h1∂
∂ x0− i h ~σ · ~∇
)
φL(x) (1.35)
where φR(x) and φL(x) are two component vectors. We get two equations
i h
(
1∂
∂ x0− ~σ · ~∇
)
φL(x) = 1mcφR(x)
i h
(
1∂
∂ x0+ ~σ · ~∇
)
φR(x) = 1mcφL(x) (1.36)
Take the sum and difference of the equations in (1.36)
i h
(
1∂
∂ x0
(
φR(x) + φL(x))
+ ~σ · ~∇(
φR(x) − φL(x))
)
= 1mc(
φR(x) + φL(x))
−i h(
1∂
∂ x0
(
φR(x) − φL(x))
+ ~σ · ~∇(
φR(x) + φL(x))
)
= 1mc(
φR(x) − φL(x))
(1.37)
We introduce a four component spinor which is called Dirac spinor. It is given by
ψ(x) =
(
ψA(x)ψB(x)
)
=1√2
(
φR(x) + φL(x)φR(x) − φL(x)
)
=
ψ1(x)ψ2(x)ψ3(x)ψ4(x)
φR(x) + φL(x) =√
2
(
ψ1(x)ψ2(x)
)
φR(x) − φL(x) =√
2
(
ψ3(x)ψ4(x)
)
(1.38)
One can rewrite the equation in (1.37)
i h
(
1 ∂∂ x0 ~σ · ~∇
−~σ · ~∇ −1 ∂∂ x0
) (
ψA(x)ψB(x)
)
= mc
(
ψA(x)ψB(x)
)
(1.39)
Define the four by four γ-matrices (Dirac-representation)
γ0 =
(
1 /0/0 −1
)
γk =
(
/0 σk
−σk /0
)
k = 1, 2, 3 of k = x, y, z (1.40)
Further we have the conventions
~∇ = (∂
∂ x,∂
∂ y,∂
∂ z) = (
∂
∂ x1,∂
∂ x2,∂
∂ x3) = (∂1, ∂2, ∂3) ∂0 =
∂
∂ x0(1.41)
11
In this notation equation (1.39) looks like
i h
(
1∂0 σk∂k
−σk∂k −1∂0
) (
ψA(x)ψB(x)
)
= mc
(
ψA(x)ψB(x)
)
(1.42)
The equation above can be written in a shorter way
(i h γµ ∂µ −mc) ψ(x) = 0 Dirac-vergelijking (1.43)
Because φR(x), φL(x) satisfy the Klein-Gordon equation (1.1), ψ(x) satisfies this equa-tion too. From this one can derive a general relation between the γ-matrices
h2 γν γµ ∂ν ∂µψ(x) = −i hmc γν ∂ν ψ(x) = −m2c2 ψ(x)
h2 γµ γν ∂µ ∂νψ(x) = −m2c2 ψ(x) (1.44)
Addition gives
h2(
γµ γν + γν γµ)
∂µ ∂νψ(x) = −2m2c2ψ(x) = 2 h22ψ(x) =
2 h2 gµν ∂µ ∂νψ(x) ⇒ γµ, γν = 2 gµν 1 (1.45)
The four by four unit matrix 1 is often omitted in the literature but it is present!! Theanti-commutator is defined as
a, b = a b+ b a (1.46)
Further one can derive the following relations
(γ0)† = γ0 (γi)† = −γi (γ0)2 = 1 (γi)2 = −1
Sp γµ = 0 Sp 1 = 4 γµ = gµνγν → γ0 = γ0 γi = −γi (1.47)
where 1 is a four by four unit matrix which very often is omitted. We introduce thefollowing notation
/a = aµγµ (1.48)
We can now derive the following relations
γµ /a γµ = −2 /a γµ /a /b γ
µ = 4 a · b γµ, /a /b /c γµ = −2/c /b /a
Sp /a /b = 4a · b Sp /a1 /a2 · · ·/an = 0 if n is odd
Sp /a /b /c /d = 4[(a · b)(c · d) − (a · c)(b · d) + (a · d)(b · c)] (1.49)
12
Definition adjoint Dirac spinor
ψ(x) = ψ†(x) γ0 ψ†(x) = (ψT (x))∗
ψT (x) = (ψ1(x), ψ2(x), ψ3(x), ψ4(x))
ψ†(x) = (ψ∗1(x), ψ∗2(x), ψ
∗3(x), ψ
∗4(x))
ψ†(x) γ0 = (ψ∗1(x), ψ∗2(x),−ψ∗3(x),−ψ∗4(x)) (1.50)
The adjoint spinor satisfies the Dirac equation. This follows from hermitian conjugation
(
i h γ0 ∂0 + i h γk ∂k −mc)
ψ(x) = 0 → ψ†(x)
−i h γ0←∂0 +i h γk
←∂k −mc
= 0
ψ†(x)
−i h γ0←∂0 +i h γk
←∂k −mc
γ0 = 0
→ ψ(x)
−i h γ0←∂0 −i h γk
←∂k −mc
= 0 (1.51)
or
ψ(x)
i h γµ←∂µ +mc
= 0 (1.52)
Theorem: Each four by four matrix can be expressed in the complete set of matricesΓk which contains 16 elements. They are
ΓS = 1 ΓµV = γµ Γµν
T = σµν ≡ i
2[γµ, γν ]
ΓP = γ5 = γ5 ≡ i γ0 γ1 γ2 γ3 =
(
/0 11 /0
)
ΓµA = γ5 γµ (1.53)
Further holds
γµ, γ5 = 0 Sp γ5 γµ γν γλ γσ = 4 i ǫµνλσ (1.54)
We have to find the solutions of the Dirac equation. There are two solutions i.e. apositive energy solution and a negative energy solution
1 Positive energy solution
ψ1(x) =
(
ψ1A(x)ψ1B(x)
)
=∑
s
∫
d3~pA
(
u1A(~p, s)u1B(~p, s)
)
e−i px/h s = ±1
2(1.55)
13
where A is a normalisation factor which depends on the energy. Substitute thissolution in the Dirac equation
(
E
cγ0 − ~γ · ~p−mc
)
(
u1A(~p, s)u1B(~p, s)
)
= 0
(
1Ec
−~σ · ~p~σ · ~p −1E
c
)(
u1A(~p, s)u1B(~p, s)
)
= mc
(
u1A(~p, s)u1B(~p, s)
)
(1.56)
The solution is
u1B(~p, s) =c
E +mc2~σ · ~p u1A(~p, s) met u1A(~p, s) = χ(s)
n Pauli-spinor
(1.57)
The solution
u1A(~p, s) =c
E −mc2~σ · ~p u1B(~p, s) (1.58)
can be dropped because the limit E → mc2 or ~p→ 0 does exist.
2 Negative energy solution
ψ2(x) =
(
ψ2A(x)ψ2B(x)
)
=∑
s
∫
d3~pA
(
u2A(~p, s)u2B(~p, s)
)
ei px/h s = ±1
2(1.59)
Substitute this solution in the Dirac equation
(
−Ecγ0 + ~γ · ~p−mc
)
(
u2A(~p, s)u2B(~p, s)
)
= 0
(
−1Ec
~σ · ~p−~σ · ~p 1E
c
)(
u2A(~p, s)u2B(~p, s)
)
= mc
(
u2A(~p, s)u2B(~p, s)
)
(1.60)
The solution is
u2A(~p, s) =c
E +mc2~σ · ~p u2B(~p, s) met u2B(~p, s) = χ(−s)
n Pauli-spinor
(1.61)
The −s which occurs in the Pauli spinor is a convention which becomes clearwhen we quantise the Dirac field. The solution
u2B(~p, s) =c
E −mc2~σ · ~p u2A(~p, s) (1.62)
can be dropped because the limit E → mc2 or ~p→ 0 does not exist.
14
From the above we can derive the following solutions. The first Dirac spinor corre-sponds to the fermion
u(~p, s) =
√
E +mc2
2mc2
(
χ(s)n
c~σ·~pE+mc2
χ(s)n
)
(1.63)
The second Dirac spinor belongs to the anti-fermion anti-fermion
v(~p, s) =
√
E +mc2
2mc2
(
c ~σ·~pE+mc2
χ(−s)n
χ(−s)n
)
(1.64)
The square roots are chosen in such a way that Dirac spinors satisfy the followingorthogonality relations
u(~p, s)† u(~p, s′) =E
mc2δss′ v(~p, s)† v(~p, s′) =
E
mc2δss′ (1.65)
However in the literature other conventions are also accepted√
E +mc2
2mc2→ =
√
E +mc2
2E(1.66)
In this case we have the orthonormal relations
u(~p, s)† u(~p, s′) = δss′ v(~p, s)† v(~p, s′) = δss′ (1.67)
The advantage of this convention is that we can immediately take the limit m → 0.In these lectures the convention (1.63)- (1.65) is assumed. There are other relations.They can be derived from (1.63) and (1.64)
u(~p, s) u(~p, s′) = δss′ v(~p, s) v(~p, s′) = −δss′
u(~p, s) v(~p, s′) = 0 u(−~p, s)† v(~p, s′) = 0
v(~p, s) u(~p, s′) = 0 v(−~p, s)† u(~p, s′) = 0
u(~p, s) = u(~p, s)† γ0 = u(−~p, s)† v(~p, s) = v(~p, s)† γ0 = −v(−~p, s)† (1.68)
The Dirac equation for spinors read
(/p−mc)u(~p, s) = 0 u(~p, s)(/p−mc) = 0
(/p+mc)v(~p, s) = 0 v(~p, s)(/p+mc) = 0 (1.69)
Further one can define the projection operators
Λ+ =∑
s
u(~p, s) u(~p, s) Λ− = −∑
s
v(~p, s) v(~p, s)
Λ+ u(~p, s) = u(~p, s) Λ+ v(~p, s) = 0
Λ− u(~p, s) = 0 Λ− v(~p, s) = v(~p, s) (1.70)
15
From which follows
Λ+ =/p+mc
2mcΛ− =
−/p +mc
2mc(1.71)
The most general solution of the Dirac equation follows from (1.57) and (1.61)
ψ(x) =∑
s
∫
d3~p[b(~p, s)ψ1,~p,s(x) + d∗(~p, s)ψ2,~p,s(x)]
ψ1,~p,s(x) = Au(~p, s) e−i px/h ψ2,~p,s(x) = Av(~p, s) ei px/h (1.72)
Here b(~p, s) and d(~p, s) are complex Fourier coefficients. We can define again a innerproduct
(ψi,~p,s, ψj,~p′,s′) =∫
d3~xψ†i,~p,s(x)ψj,~p′,s′(x) = δij δ(3)(~p− ~p′) δss′ (1.73)
Hence it follows
A =1
(2πh)3/2
√
mc2
E(1.74)
Notice that the inner product above defines a real norm in contrast to the inner productfor scalar fields in (1.17). The classical Dirac field gets the following notation
ψ(x) =∑
s
∫ d3~p
(2πh)3/2
√
mc2
E[b(~p, s) u(~p, s) e−i px/h + d∗(~p, s) v(~p, s) ei px/h]
(1.75)
From which follows
ψ†(x) =∑
s
∫
d3~p
(2πh)3/2
√
mc2
E[b∗(~p, s) u†(~p, s) ei px/h + d(~p, s) v†(~p, s) e−i px/h]
ψ(x) =∑
s
∫ d3~p
(2πh)3/2
√
mc2
E[b∗(~p, s) u(~p, s) ei px/h + d(~p, s) v(~p, s) e−i px/h]
(1.76)
Remarks
~n =~p
|~p| → χ(s)n = χ(s)
p met~σ · ~p|~p| χ
(s)p = 2s χ(s)
p (1.77)
In this case the Dirac spinors are eigen vectors of
~Σ · ~p|~p| with ~Σ = (Σ1,Σ2,Σ3) Σk =
(
σk /0/0 σk
)
~Σ · ~p|~p| u(~p, s) = 2s u(~p, s)
~Σ · ~p|~p| v(~p, s) = −2s v(~p, s) s = ±1
2(1.78)
The quantity h = 2s is called the helicity.
16
1.5 The electromagnetic (massless spin one) field
This field is given by Maxwell’s equations
~∇ · ~E = ρ ~∇× ~B − ∂ ~E
c∂ t=
~J
c
~∇ · ~B = 0 ~∇× ~E +∂ ~B
c∂ t= 0 (1.79)
where the electrostatic field and the magnetic field (magnetic induction) are given by~E and ~B respectively. Furthermore ρ and ~J represent the charge distribution and thecurrent density. One can now express ~E and ~B in the electrostatic potential φ and themagnetic potential ~A
~E = −~∇φ− ∂ ~A
c∂ t~B = ~∇× ~A (1.80)
We will now represent the Maxwell equations in a Lorentz covariant form. Define theelectromagnetic field tensor (second rank tensor)
Fµν(x) = ∂µAν − ∂ν Aµ Fµν = −Fνµ
A(x) = (φ(x), ~A(x)) = (A0(x), A1(x), A2(x), A3(x)) (1.81)
Using
∂ = (∂0, ~∂) = (∂0, ∂1, ∂2, ∂3) =
(
∂
c∂ t,∂
∂ x1,∂
∂ x2,∂
∂ x3,
)
=
(
∂
c∂ t,− ∂
∂ x1,− ∂
∂ x2,− ∂
∂ x3,
)
=
(
∂
c∂ t,−~∇
)
(1.82)
we can derive
Ei = Fi0 → ~E = −~∇A0 −∂ ~A
c∂ t
Bi = −1
2ǫijk Fjk Fij = −ǫijk Bk → ~B = ~∇× ~A
B1 = −F23 B2 = −F31 B3 = −F12 (1.83)
The current density (four vector) is given by
J = (c ρ, ~J) = (J0, J1, J2, J3) (1.84)
17
where ρ represents the charge density. One can now write Maxwell equations in arelativistic covariant manner
∂µFµν =Jν
cof 2Aν − ∂ν∂
µAµ =Jν
c
~∇ · ~E = ρ ~∇× ~B − ∂ ~E
c∂ t=
~J
c(1.85)
The other two equations in (1.79) follow from the Bianchi identity
∂µ Fνλ + ∂ν Fλµ + ∂λ Fµν = 0
⇒ ~∇ · ~B = 0 ~∇× ~E +∂ ~B
c∂ t= 0 (1.86)
Further we have current conservation
∂µ∂νFµν =1
c∂νJν = 0 (1.87)
because Fµν is antisymmetric. therefore Jν is a conserved current. The solution for Aµ
is determined up to a gauge-transformation in other words if Aµ is a solution then isthe gauge transformed potential A′µ = Aµ + ∂µΛ also a solution
A′µ(x) = Aµ(x) + ∂µΛ(x) → F ′µν = ∂µ A′ν − ∂ν A
′µ
= Fµν + ∂µ∂νΛ(x) − ∂ν∂µΛ(x) = Fµν (1.88)
Hence it follows that ~E and ~B are independent of the choice for Λ. this has to be for~E and ~B are observables. To find an expression for Aµ we have to make a choice forΛ. This is called a gauge choice. Let us limit ourselves to the free electromagnetic fieldwhich implies Jµ = 0
2Aν − ∂ν∂µAµ = 0 (1.89)
The equation above becomes the Klein-Gordon equation when we choose
1 Lorentz gauge
∂µAµ = 0 (1.90)
Suppose ∂µAµ 6= 0, choose a Λ for which holds
∂µA′µ = 0 → ∂µAµ = −2Λ (1.91)
such that
2A′ν = 0. (1.92)
The Lorentz gauge is a Lorentz covariant gauge.
18
2 radiation gauge
φ = 0 ~∇ · ~A = 0 → ∂µAµ = 0 (1.93)
The disadvantage of this gauge choice is that it is not Lorentz covariant. Forexample in a system S it appears that φ = 0 and ~∇ · ~A = 0. One can now makea Lorentz transformation to a system S ′ for which holds φ′ 6= 0 and φ′ 6= 0 en~∇′ · ~A′ 6= 0. However ∂
′µA′µ = 0 is still valid in all Lorentz frames. The advantageof the radiation gauge is that it only contains the physical components of theradiation field so that for the time being we shall choose the radiation gauge.
In the radiation gauge ~A satisfies the Klein-Gordon equation 2 ~A = 0. The solution is
~A(x) =∑
α
∫
d3 ~k
(2π)3/2c
√
h
2ω
[
a(~k, α)~ǫ(α)(~k) e−i kx + a∗(~k, α)~ǫ(α)∗(~k) ei kx]
(1.94)
with k2 = 0.
~∇ · ~A = 0 → ~k · ~ǫ(α)(~k) = 0 (1.95)
Here ~ǫ(α)(~k) represents a polarisation vector. From the condition above it follows thatthere are two independent polarisations α for the electromagnetic field. These physicalpolarisation vectors are perpendicular on the direction of propagation (transversalityrequirement). Notice that the Klein-Gordon equation for the electromagnetic field hasno mass term hence k2 = 0. If we compare with the real scalar field equation (1.12)we have made the following substitutions
~p → h~k E → hω a(~p) → h−3/2 a(~k, α)
k =(
ω
c,~k)
k2 = 0 → ω = c|~k| (1.96)
We can also write equation (1.94)
~A(x) =∑
α
∫
d3~k[
a(~k, α) ~f(α)k (x) + a∗(~k, α) ~f
(α)∗k (x)
]
with ~f(α)k (x) =
c
(2π)3/2
√
h
2ω~ǫ(α)(~k) e−i kx (1.97)
From the requirement:(
f(α)k , f
(α′)k′
)
= δαα′ δ(3)(~k − ~k′) (1.98)
where
(
f(α)k , f
(α′)k′
)
=i
hc
∫
d3~x f(α)∗k (x)
↔∂0 f
(α′)k′ (x) (1.99)
19
follows that the polarisation vectors are mutually perpendicular.
ǫ(α)∗(~k) · ~ǫ(α′)(~k) = δαα′ (1.100)
Proof:
(
f(α)k , f
(α′)k′
)
=i
(2π)3
c
2√ωω′
∫
d3~x ǫ(α)∗(~k) · ~ǫ(α′)(~k′) e−i~k·~x ei~k′·~x
ei ωt ∂
c∂ te−i ω′t
−e−i ω′t ∂
c∂ tei ωt
=1
(2π)3
c
2√ωω′
ω + ω′
cǫ(α)∗(~k) · ~ǫ(α′)(~k′)
∫
d3~x ei (~k′−~k)·~x ei (ω−ω′)t
=ω + ω′
2√ωω′
ǫ(α)∗(~k) · ~ǫ(α′)(~k′) δ(3)(~k − ~k′) = ǫ(α)∗(~k) · ~ǫ(α′)(~k′) δ(3)(~k − ~k′)
= δαα′δ(3)(~k − ~k′) (1.101)
Watch that from ~k = ~k′ follows ω = ω′. The Fourier coefficients can be expressed inthe inner product
a(~k, α) =(
~f(α)k , ~A
)
a∗(~k, α) = −(
~f(α)∗k , ~A
)
(1.102)
Further two different polarisation vectors are chosen in the literature
1 Linear polarisation vectors
~k = |~k|(sin θ cosϕ, sin θ sinϕ, cos θ)
ǫ(1)(~k) = (cos θ cosϕ, cos θ sinϕ,− sin θ)
ǫ(2)(~k) = (− sinϕ, cosϕ, 0) (1.103)
so that ~ǫ(α) is real. Further we have the properties
~ǫ(1)(~k) ×~ǫ(2)(~k) =~k met
~k =
~k
|~k|
~ǫ(1)(−~k) = −~ǫ(1)(~k) ~ǫ(2)(−~k) = ~ǫ(2)(~k)
→ ~ǫ(1)(−~k) ×~ǫ(2)(−~k) = −~k (1.104)
From the above it appears that ~ǫ(1), ~ǫ(2) and ~k form a right handed system. If wesum over the polarisations then we get
2∑
α=1
ǫ(α)i (~k) ǫ
(α)j (~k) = Tij(~k) = δij −
ki kj
|~k|2(1.105)
20
This follows from
Tij(~k) = a δij + bki kj
|~k|2met ki Tij(~k) = Tij(~k) kj = 0 Tii = 2 (1.106)
2 Circular polarisations vectorsWe can now choose a left handed α = L and a right handed α = R polarisation.They are given by
~ǫ(R)(~k) = −~ǫ(1)(~k) + i~ǫ(2)(~k)√
2~ǫ(L)(~k) =
~ǫ(1)(~k) − i~ǫ(2)(~k)√2
(1.107)
in other words the polarisation vectors are complex. The Fourier coefficientsbecome
a(~k, R) = −a(~k, 1) − i a(~k, 2)√
2a(~k, L) =
a(~k, 1) + i a(~k, 2)√2
(1.108)
The properties of these polarisation vectors are
~ǫ(R)∗(~k) = −~ǫ(L)(~k) ~ǫ(R)(−~k) = ~ǫ(L)(~k) ~ǫ(L)(−~k) = ~ǫ(R)(~k)
~ǫ(α)∗(~k) · ~ǫ(α′)(~k) = δαα′ → ~ǫ(α)(~k) · ~ǫ(α)(~k) = 0 α = R,L (1.109)
Finally one can derive from equation (1.107)
2∑
α=1
ǫ(α)i (~k) ǫ
(α)∗j (~k) = δij −
ki kj
|~k|2(1.110)
One can now determine from ~A in (1.94) the field ~E and ~B. From (1.80) with φ = 0it follows
~E(x) = i∑
α
∫ d3 ~k
(2π)3/2ω
√
h
2ω
[
a(~k, α)~ǫ(α)(~k) e−i kx − a∗(~k, α)~ǫ(α)∗(~k) ei kx]
~B(x) = i c∑
α
∫
d3 ~k
(2π)3/2
√
h
2ω~k ×
[
a(~k, α)~ǫ(α)(~k) e−i kx − a∗(~k, α)~ǫ(α)∗(~k) ei kx]
(1.111)
In stead of the radiation gauge one could also choose the Lorentz gauge This has thedisadvantage that the electromagnetic field Aµ has unphysical unphysical polarisations.The field becomes
Aµ(x) =∑
α
∫
d3 ~k
(2π)3/2c
√
h
2ω
[
a(~k, α) ǫ(α)µ (~k) e−i kx + a∗(~k, α) ǫ(α)∗
µ (~k) ei kx]
∑
α
kµ ǫ(α)µ = 0 en k2 = 0 (1.112)
21
We have besides the two transverse polarisation ǫ(α)µ α = 1, 2 or α = R,L a scalar
polarisation ǫ(0)µ and a longitudinal polarisation ǫ(3)µ . The latter two are unphysical andsatisfy the following relation
kµ ǫ(0)µ = −kµ ǫ(3)µ (1.113)
This follows from (1.90). For k = (|~k|, ~k) the polarisation vectors are
ǫ(0) = (1,~0) ǫ(3) = (0,~k) ǫ(1) = (0,~ǫ(1)) ǫ(2) = (0,~ǫ(2))
∑
α
ǫ(α)µ (~k) ǫ(α)∗
ν (~k)
ǫ(α)(~k) ǫ(α)∗(~k)= gµν ǫ(α)(~k) ǫ(β)∗(~k) = gαβ (1.114)
One can assume that the polarisation indices are Lorentz indices.
22
2 Symmetries in classical field theory
2.1 Lagrange en Hamilton formalism for classical fields
The Lagrangian for classical fields is defined by
L[ϕr, ϕr, t] =∫
d3~xL(ϕr(x), ∂µ ϕr, xµ) (2.1)
This is a functional! The action is then given by
S[ϕr, ϕr] = c∫ t2
t1dt L[ϕr, ϕr, t] =
∫
Rd4xL(ϕr(x), ∂µ ϕr, xµ) (2.2)
where L is the Lagrange density. In the subsequent part we omit the explicit depen-dence of L on xµ. R is a four dimensional space time volume element such that
∫
R d4x
is a Lorentz-invariant.
∫
d4x′ =∫
d4x∣
∣
∣
∂ x′µ∂ xν
∣
∣
∣ =∫
d4x | det Λ| =∫
d4x with x′µ = Λλµ xλ
→ ∂ x′µ∂ xν
= Λλµ δ
νλ = Λν
µ |Λνµ| ≡ | detΛ| (2.3)
The boundary of the space R is denoted by ∂R. The upper and lower boundary ofthe integral
∫
d4x is determined by the hyper surfaces which arise by the intersectionst = t1 and t = t2 which form a part ∂R.
Principle of the minimal action (Principle of Hamilton for fields)
The equation of motion for the field ϕr(x) is given by that value for the field ϕr(x) forwhich the action is minimal. To find the minimum we vary the field in such a way thatϕr(x) = const. if x ∈ ∂R.
ϕr(x) → ϕr(x) + δϕr(x) with δϕr(x) = 0 if x ∈ ∂R (2.4)
The action varies like
δS =∫
Rd4x
L(ϕr(x) + δϕr(x), ∂λϕr(x) + δ∂λϕr(x)) −L(ϕr(x), ∂λϕr(x))
=∫
Rd4x
δϕr(x)∂
∂ ϕr(x)L(ϕr(x), ∂λϕr(x))
+δ∂µϕr(x)∂
∂ ∂µϕr(x)L(ϕr(x), ∂λϕr(x))
(2.5)
23
Using ∂µδϕr(x) = δ∂µϕr(x) we can write
δS =∫
Rd4x δϕr(x)
∂
∂ ϕr(x)L(ϕr(x), ∂λϕr(x)) − ∂µ
∂
∂ ∂µϕr(x)L(ϕr(x), ∂λϕr(x))
+∫
Rd4x ∂µ
δϕr(x)∂
∂ ∂µϕr(x)L(ϕr(x), ∂λϕr(x))
=∫
Rd4x δϕr(x)
∂
∂ ϕr(x)L(ϕr(x), ∂λϕr(x)) − ∂µ
∂
∂ ∂µϕr(x)L(ϕr(x), ∂λϕr(x))
+∫
∂Rdσµ
δϕr(x)∂
∂ ∂µϕr(x)L(ϕr(x), ∂λϕr(x))
(2.6)
In the last step use Gauss theorem∫
Rd4x ∂µ V
µ(x) =∫
∂Rdσµ V
µ(x) (2.7)
which converts a four dimensional integral into a three dimensional integral∫
∂R dσµ.The surface integral is zero because δϕ(x) = 0 for x ∈ ∂R. From this we derive theEuler Lagrange equation.
δS = 0 ⇒∂
∂ ϕr(x)L(ϕr(x), ∂λϕr(x)) = ∂µ
∂
∂ ∂µϕr(x)L(ϕr(x), ∂λϕr(x)) (2.8)
Define the canonical momentum
Πr(x) =∂
∂ ϕr(x)L(ϕr(x), ∂λϕr(x)) with ϕr(x) =
∂ ϕr(x)
∂ t(2.9)
Definition Hamilton density
H(Πr(x), ϕr(x)) = Πr(x) ϕr(x) − L(ϕr(x), ∂µϕr(x)) (2.10)
The Hamilton equations read:
− ∂ H∂ ϕr
=∂ L∂ ϕr
= ∂µ∂ L
∂ ∂µ ϕr
= Πr + ∂k∂ L
∂ ∂k ϕr
because ∂0∂ L
∂ ∂0 ϕr
=∂
∂ t
∂ L∂ ϕr
=∂ Πr
∂ t= Πr
∂ H∂ Πr
= ϕr (2.11)
Applications
24
A Scalar field
The Lagrange density is given by
L(φ, ∂µφ) =1
2∂µφ ∂µφ− 1
2
m2c2
h2 φ2 (2.12)
Application of the Euler Lagrange equation yields
∂ L∂ ∂λφ
=∂
∂ ∂λφ
[1
2∂µφ ∂µφ− 1
2
m2c2
h2 φ2]
=1
2gµσ δλ
σ ∂µφ
+1
2δλµ ∂
µφ = ∂λφ
∂ L∂ φ
= −m2c2
h2 φ
⇒(
2 +m2c2
h2
)
φ(x) = 0 (2.13)
The canonical momentum is
Π =1
c
∂ L∂ ∂0φ
=1
c∂0φ =
1
c2φ (2.14)
The Hamilton density is equal to
H =1
c2(φ)2 − 1
2c2(φ)2 +
1
2~∇φ · ~∇φ+
1
2
m2c2
h2 φ2
=1
2
(
1
c2(φ)2 + ~∇φ · ~∇φ+
m2c2
h2 φ2
)
(2.15)
For the complex scalar field the Lagrangian density becomes
L(φ, φ∗, ∂φ, ∂φ∗) = ∂µφ∗ ∂µφ− m2c2
h2 φ∗ φ
∂ L∂ ∂λφ
= ∂λφ∗∂ L∂ ∂λφ∗
= ∂λφ
∂ L∂ φ
= −m2c2
h2 φ∗∂ L∂ φ∗
= −m2c2
h2 φ (2.16)
From which follows(
2 +m2c2
h2
)
φ(x) = 0
(
2 +m2c2
h2
)
φ∗(x) = 0
25
Π =1
c2φ∗ Π∗ =
1
c2φ
H =1
c2φ∗ φ+ ~∇φ∗ · ~∇φ+
m2c2
h2 φ∗ φ (2.17)
B Dirac-veld
The Lagrange density is given by
L(ψ, ψ, ∂µψ, ∂µψ) = i c h ψ γµ ∂µψ −MC2 ψ ψ
∂ L∂ ∂µψ
= i c h ψ γµ ∂ L∂ ∂µψ
= 0
∂ L∂ ψ
= −mc2 ψ ∂ L∂ ψ
= i c h γµ ∂µψ −mc2 ψ (2.18)
The equations of motion become equal to
i h ∂µψ γµ +mc ψ = 0 i h γµ ∂µψ −mcψ = 0 (2.19)
The canonical momentum becomes equal to
Π =1
c
∂ L∂ ∂0ψ
= i h ψ γ0 = i h ψ† Π† =1
c
∂ L∂ ∂0ψ
= 0 (2.20)
One can also derive the Dirac equation from
L′(ψ, ψ, ∂µψ, ∂µψ) = −i c h (∂µ ψ) γµ ψ −mc2 ψ ψ
∆L = L′ −L = −i c h ∂µ
(
ψ γµ ψ)
(2.21)
In other words the Lagrange density is determined up to a total derivative. Thisis a general theorem which can be derived from the action.
S ′ − S =∫
Rd4x ∂µ V
µ(x) =∫
∂Rdσµ V
µ(x) (2.22)
From which follows
S minimal←−→ S ′ minimal (2.23)
The Hamilton density becomes
H = i h ψ† ψ − i h ψ† ψ − i c h ψ γk ∂kψ +mc2 ψ ψ
= −i c h ψ γk ∂kψ +mc2 ψ ψ (2.24)
26
C Electromagnetic field
The Lagrange density is given by
L(Aµ, ∂νAµ) = −1
4Fµν F
µν with Fµν = ∂µAν − ∂νAµ (2.25)
Apply the Euler Lagrange equations
∂ Fµν
∂ ∂λAσ= δλ
µ δσν − δλ
ν δσµ
∂ F µν
∂ ∂λAσ= gµα gνβ ∂ Fαβ
∂ ∂λAσ= gµλ gνσ − gµσ gνλ
⇒ ∂ L∂ ∂λAσ
= −1
4
∂ Fµν
∂ ∂λAσF µν − 1
4F µν ∂ F µν
∂ ∂λAσ
= −1
4
(
δλµ δ
σν − δλ
ν δσµ
)
F µν − 1
4Fµν
(
gµλ gνσ − gµσ gνλ)
= −1
4F λσ +
1
4F σλ − 1
4F λσ +
1
4F σλ = −F λσ
∂ L∂ Aσ
= 0 (2.26)
From this follows the equation of motion
∂λ Fλσ = 0 of 2Aσ − ∂σ∂λA
λ = 0 (2.27)
The canonical momentum equals
Πσ =1
c
∂ L∂ ∂0Aσ
= −1
cF 0σ =
1
cF σ0
→ Π0 = 0 Πk =1
cEk Electrostatic field (2.28)
If the field interacts with an external source then the Lagrange density becomes
L(Aµ, ∂νAµ) = −1
4Fµν F
µν − e
hcAµ J
µ (2.29)
with e as elementary unit charge. The Euler Lagrange equation yields
∂ L∂ ∂λAσ
= −F λσ ∂ L∂ Aσ
= − e
hcJσ
→ ∂λ Fλσ =
e
hcJσ (2.30)
27
Expressed in the E and B fields the Lagrange density becomes
L = −1
4Fµν F
µν =1
4Fµν F
νµ =1
4Sp(F FD) =
1
2
(
| ~E|2 − | ~B|2)
(2.31)
Hint use the notation
F =
0 −E1 −E2 −E3
E1 0 −B3 B2
E2 B3 0 −B1
E3 −B2 B1 0
→
FD =
0 −E1 −E2 −E3
E1 0 B3 −B2
E2 −B3 0 B1
E3 B2 −B1 0
=
0 E1 E2 E3
−E1 0 −B3 B2
−E2 B3 0 −B1
−E3 −B2 B1 0
(2.32)
The Hamilton density becomes equal to
H = Πσ Aσ −L = Ek ∂ Ak
c∂ t− 1
2
(
| ~E|2 − | ~B|2)
Ek = Fk0 = ∂k A0 −∂ Ak
c∂ t→ Ek = ∂k A0 +
∂ Ak
c∂ t
→ H =1
2
(
| ~E|2 + | ~B|2)
− Ek ∂kA0 =1
2
(
| ~E|2 + | ~B|2)
− ~E · ~∇φ
with A0 = φ electrostatic potential (2.33)
In the radiation gauge φ = 0 so that
H =1
2
(
| ~E|2 + | ~B|2)
(2.34)
2.2 Noether theorem for classical fields
If the action S is invariant under a continuous transformation then there exists aconserved current.
This conserved current implies a conserved charge.
To prove this theorem we take all transformations infinitesimal.
xµ → x′µ = xµ + ∆xµ with ∆xµ = Xµi (x) ∆ωi(x) (2.35)
28
The fields transform as
ϕr(x) → ϕ′r(x′) = ϕr(x) + ∆ϕr(x)
∂µϕr(x) → ∂′µϕ′r(x′) = ∂µϕr(x) + ∆ ∂µϕr(x)
with ∆ϕr(x) = Φri(x) ∆ωi(x) (2.36)
Further we define the infinitesimal transformations
ϕr(x) → ϕ′r(x) = ϕr(x) + δϕr(x)
∂µϕr(x) → ∂µϕ′r(x) = ∂µϕr(x) + δ ∂µϕr(x) (2.37)
Watch that the following holds
δ ∂µϕr(x) = ∂µ δϕr(x) but ∆ ∂µϕr(x) 6= ∂µ ∆ϕr(x) (2.38)
because δ, in contrast to ∆, does not transform xµ. Notice that ∆ transforms x as wellas ϕr(x) whereas δ only changes ϕr(x). One can derive the following relations betweenboth infinitesimal transformations ∆ and δ.
ϕ′r(x′) = ϕ′r(x
µ +Xµi (x)∆ωi(x)) = ϕ′r(x) +Xµ
i (x) ∆ωi(x) ∂µϕ′r(x)
= ϕ′r(x) +Xµi (x) ∆ωi(x) ∂µϕr(x) +Xµ
i (x) ∆ωi(x) ∂µ δϕr(x) (2.39)
The last term can be omitted since this is of order ∆ δ. Further we have
δϕr(x) = ϕ′r(x) − ϕr(x) = ϕ′r(x′) − ϕr(x) −Xµ
i (x) ∆ωi(x) ∂µϕr(x)
= ∆ϕr(x) −Xµi (x) ∆ωi(x) ∂µϕr(x)
→ δϕr(x) =(
Φri(x) −Xµi (x) ∂µϕr(x)
)
∆ωi(x) (2.40)
Also the following statement holds
δ ∂µϕr(x) = ∆∂µϕr(x) −Xνi (x) ∆ωi(x) ∂ν ∂µϕr(x) (2.41)
Under the transformations above the action changes into
δS =∫
R′
d4x′ L(ϕ′r(x′), ∂′λϕ
′r(x′)) −
∫
Rd4xL(ϕr(x), ∂λϕr(x)) (2.42)
The Jacobian of the transformation is
∂ x′µ
∂ xν= δµ
ν + ∂ν
(
Xµi (x) ∆ωi(x)
)
→ d4x′ =
∣
∣
∣
∣
∣
∂ x′µ
∂ xν
∣
∣
∣
∣
∣
d4x
=[
1 + ∂µ
(
Xµi (x) ∆ωi(x)
)]
(2.43)
29
Expand L in ∆ω
L(ϕ′r(x′), ∂′λϕ
′r(x′)) = L(ϕr(x) + ∆ϕr(x), ∂λϕr(x) + ∆ ∂λϕr(x))
= L(ϕr(x), ∂λϕr(x)) + ∆ϕr(x)∂
∂ ϕr(x)L(ϕr(x), ∂λϕr(x))
+∆ ∂µϕr(x)∂
∂ ∂µϕr(x)L(ϕr(x), ∂λϕr(x)) (2.44)
δS is then equal to
δS =∫
Rd4x
[
∂µ
Xµi (x) ∆ωi(x)
L(ϕr(x), ∂λϕr(x))
+∆ϕr(x)∂
∂ ϕr(x)L(ϕr(x), ∂λϕr(x))
+∆ ∂µϕr(x)ϕr(x)∂
∂ ∂µϕr(x)L(ϕr(x), ∂λϕr(x))
]
=∫
Rd4x
[
∂µ
Xµi (x) ∆ωi(x)L(ϕr(x), ∂λϕr(x))
−Xµi (x) ∆ωi(x) ∂µL(ϕr(x), ∂λϕr(x)) +
δϕr(x)
+Xµi (x) ∆ωi(x) ∂µϕr(x)
∂
∂ ϕr(x)L(ϕr(x), ∂λϕr(x)) +
δ ∂µϕr(x)
+Xνi (x) ∆ωi(x) ∂ν ∂µϕr(x)
∂
∂ ∂µϕr(x)L(ϕr(x), ∂λϕr(x))
]
(2.45)
Exchange in the last term µ and ν. Subsequently we have
∂µL(ϕr(x), ∂λϕr(x)) = ∂µϕr(x)∂
∂ ϕr(x)L(ϕr(x), ∂λϕr(x))
+∂µ ∂νϕr(x)∂
∂ ∂νϕr(x)L(ϕr(x), ∂λϕr(x)) (2.46)
Using (2.46) the second, the fourth and the last term in equation (2.45) cancel againsteach other and we can write
δS =∫
Rd4x
[
∂µ
Xµi (x) ∆ωi(x)L(ϕr(x), ∂λϕr(x))
30
+δϕr(x)∂
∂ ϕr(x)L(ϕr(x), ∂λϕr(x))
+∂µ δϕr(x)∂
∂ ∂µϕr(x)L(ϕr(x), ∂λϕr(x))
]
=∫
Rd4x
[
∂µ
Xµi (x) ∆ωi(x)L(ϕr(x), ∂λϕr(x))
+δϕr(x)∂
∂ ∂µϕr(x)L(ϕr(x), ∂λϕr(x))
+δϕr(x) ∂
∂ ϕr(x)L(ϕr(x), ∂λϕr(x))
−∂µ∂
∂ ∂µϕr(x)L(ϕr(x), ∂λϕr(x))
]
(2.47)
The last two terms vanish because ϕr(x) satisfies the i Euler Lagrange equation. Sub-stitute δϕr(x) (2.40) in the formula above then can the change in the action be writtenas
δS = −∫
Rd4x ∂µ
[
T µν (x)Xν
i (x)
−Φri(x)∂
∂ ∂µϕr(x)L(ϕr(x), ∂λϕr(x))
∆ωi(x)
]
(2.48)
where the stress energy momentum tensor is defined as
T µν (x) = ∂νϕr(x)
∂
∂ ∂µϕr(x)L(ϕr(x), ∂λϕr(x)) − δµ
ν L(ϕr(x), ∂λϕr(x)) (2.49)
We define now the Noether current
jµi (x) = T µ
ν (x)Xνi (x) − Φri(x)
∂
∂ ∂µϕr(x)L(ϕr(x), ∂λϕr(x)) (2.50)
so that the change in the action can be written as
δS = −∫
Rd4x ∂µ
(
jµi (x) ∆ωi(x)
)
= −∫
∂Rdσµ
(
jµi (x) ∆ωi(x)
)
(2.51)
If S is left invariant under the transformation and ∆ωi(x) is independent of x then
δS = −∆ωi∫
Rd4x ∂µj
µi (x) = −∆ωi
∫
∂Rdσµ j
µi (x) = 0 (2.52)
31
This holds for an arbitrary space R and surface ∂R bounded by t = t1 and t = t2.Therefore the following theorem holds
∂µjµi (x) = 0 (2.53)
In other words jµi (x) is a conserved current. Hence it follows that the charge defined
by
Qi =1
c
∫
Vd3~x j0
i (~x, t) → d Qi
d t= 0 (2.54)
is conserved.Proof:
d Qi
d t=
1
c
∫
Vd3~x ∂0j
0i (~x, t) =
1
c
∫
Vd3~x
(
∂µjµi (~x, t) − ∂kj
ki (~x, t)
)
=1
c
∫
Vd3~x ~∇ ·~ji(~x, t) =
1
c
∫
∂Vd~S ·~ji(~x, t) = 0
Gauss theorem (2.55)
Here we have assumed that
~ji(~x, t) = 0 for x ∈ ∂V (2.56)
In general one assumes |~x| → ∞.Notice: In general one can show that
Qi =1
c
∫
σdσµ j
µi (~x, t) (2.57)
is conserved where σµ is a hyper surface. In the proof above we have chosen t =constant so that dσµ = d3~x δ0
µ.The Noether current is not unique. One can always add two types of terms withoutchanging the Noether charge. They are
1
∂µ jµi = 0 → ∂µ j
′µi = 0 with j
′µi = jµ
i + V µi (2.58)
provided V µi = 0 using the Euler Lagrange equations to which all fields have to
satisfy.
2
∂µ jµi = 0 → ∂µ j
′µi = 0 with j
′µi = jµ
i + ∂νAµνi
with the condition Aµνi = −Aνµ
i so that ∂µ ∂ν Aµνi = 0 (2.59)
32
This is given in the theory of relativity by the Poincare group We limit ourselves totranslations T4. Take these infinitesimal aµ ≪ 1.
x′µ = xµ + aµ → ∆xµ = aµ
∆xµ = Xµν (x) ∆ων with ∆ων = aν
→ Xµν (x) = δµ
ν or Xµν(x) = gµν
ϕ′r(x+ a) = ϕr(x) → ∆ϕr(x) = 0
∆ϕr(x) = Φrν(x) ∆ων → Φrν(x) = 0 (2.60)
The Noether current for translations is given by the stress energy momentum tensor(second rank tensor)
T µν (x) = jµ
ν (x) or T µν(x) = jµν(x) with ∂µTµν(x) = 0 (2.61)
The Noether charges are
P ν =1
c
∫
d3~x T 0ν d P ν
d t= 0
H = T 00 Hamilton density (2.62)
P ν is a constant of motion and represents de four momentum of the field ϕr(x). Westudy the Noether current and the Noether charge for the translations in the case ofscalar, Dirac en electromagnetic field
1 Scalar veld
T µν = ∂µφ ∂νφ− 1
2gµν
(
∂λφ ∂λφ− m2c2
h2 φ2)
H = T 00 =1
2
( 1
c2(φ)2 + ~∇φ · ~∇φ+
m2c2
h2 φ2)
=1
2
(
c2 Π2 + ~∇φ · ~∇φ+m2c2
h2 φ2)
T 0i = ∂0φ ∂iφ (2.63)
Substitute the solution of the free field (1.23) then we can express the four mo-mentum into its Fourier coefficients
P ν =1
c
∫
d3~x T 0ν =1
2
∫
d3~p pν[
a(~p) a∗(~p) + a∗(~p) a(~p)]
(2.64)
33
Keeping an eye on second quantisation, where the Fourier coefficients representoperators, we have assumed that the latter do not commute. For the classicalfield these coefficients represent complex numbers which of course commute. Forthe scalar field we also have
P 0 =1
c
∫
d3~xH P k =∫
d3~xΠ ∂k φ (2.65)
Notice that for P 0 one is using the equation of motion for the scalar field.
2 Dirac-veld
T µν = i c h ψ γµ ∂νψ − gµν(
i c h ψ γλ ∂λψ −mc2 ψ ψ)
H = T 00 = −i c h ψ γk ∂kψ +mc2 ψ ψ
T 0i = i c h ψ γ0 ∂iψ = i c h ψ† ∂iψ (2.66)
Because ψ(x) satisfies the equation of motion (here the Dirac equation) disap-pears the term which is multiplied by the tensor gµν of T µν .
T µν = i c h ψ γµ ∂νψ → T 0ν = i c h ψ† ∂νψ
P ν =1
c
∫
d3~x T 0ν(x) = i h∫
d3~xψ†(x) ∂νψ(x)
=∫
d3~xΠ(x) ∂νψ(x) (2.67)
Only P ν can be expressed in a simple way in the Fourier coefficients of the Diracfield. The results is equal to
P ν =∑
s
∫
d3~p pν[
b∗(~p, s) b(~p, s) − d(~p, s) d∗(~p, s)]
(2.68)
3 Electromagnetic field
The Noether currents are
T µν = −F µλ ∂νAλ − gµν L = −gλσ Fµλ ∂νAσ + gµν 1
4Fλσ F
λσ
T 0ν = −F 0λ ∂νAλ − g0ν L = cΠλ ∂νAλ − g0ν L
T 00 = H = Πλ Aλ − L =1
2
(
| ~E|2 + | ~B|2)
− ~E · ~∇φ
T 0i = cΠλ ∂iAλ = cΠk ∂iAk = Ek ∂iAk = −Ek∂ Ak
∂ xi
= −Ek∂ Ai
∂ xk− ( ~E × ~B)i = Ek
∂ Ai
∂ xk+ ( ~E × ~B)i (2.69)
34
The Noether charges are
P 0 = H =1
c
∫
d3~x T 00 =1
2c
∫
d3~x(
| ~E|2 + | ~B|2)
E.M. energy density
P i =1
c
∫
d3~x T 0i =1
c
∫
d3~x ( ~E × ~B)i Poynting-vector
or P i =1
c
∫
d3~xΠλ ∂iAλ (2.70)
Notice that
∫
d3~xEk∂ Ai
∂ xk= −
∫
d3~xAi ~∇ · ~E = 0 because ∇ · ~E = 0 (2.71)
In the case of the radiation gauge the momentum of the field can be expressedin the Fourier coefficients as follows
P ν =1
2
∑
α
∫
d3~k h kν[
a(~k, α) a∗(~k, α) + a∗(~k, α) a(~k, α)]
α = 1, 2 or α = R,L (2.72)
In the case of the Lorentz gauge we have also two unphysical polarisations
P ν = −1
2
∑
α
∫
d3~k h kν gλσ[
a(~k, λ) a∗(~k, σ) + a∗(~k, λ) a(~k, σ)]
(2.73)
2.3 Global en Local gauge symmetries
There are two kinds of symmetries in Physics.
a External symmetriesThey concern space and time. For example in special relativity these symmetriesare given by the Poincare group. For classical mechanics and non-relativisticquantum mechanics by the Galilei group.
b Internal symmetriesThe concern the internal degrees of freedom. Here the space time coordinatesare not changed. The consequence is that the action is not changed only but alsothe Lagrange density is left invariant. Examples of these symmetries are:
1 Degree of freedom: charge. Symmetry: group U(1). Theory: electrodynam-ics.
35
2 Degree of freedom: isospin. Symmetry: group SU(2). Theory: stronginteractions.
3 Degree of freedom: colour. Symmetry: group SU(3). Theory: quantumChromodynamics.
The internal symmetry transformations, provided they are continuous, are called gaugetransformations. The latter are represented by Lie groups.
Group theory
Symmetry transformations are linear transformations which change the coordinates ofone or more particles into other coordinates in such a way that the physical system isleft invariant. In these lectures we shall limit ourselves to group transformations.Definition: A group G is a set of finite elements g1, g2, · · · , gn and satisfies the followingproperties
1. Multiplication: gi.gj = gk.
2. unit element e: e.gi = gi.e = gi.
3. Inverse of gi is g−1i : gi.g
−1i = g−1
i .gi = e.
4. Associativity: gi.(gj.gk) = (gi.gj).gk
In the special case that the following relation is satisfied
5. Commutativity: gi.gj = gj.gi.
G is called an Abelian group.
Infinite group:
An infinite group G has elements g(t1, t2, · · · , tn) which are functions of the continu-ous variables t1, t2, · · · , tn. As infinite group we shall only consider Lie groups. Theset t1, t2, · · · , tn constitute a manifold. Further is g(t1, t2, · · · , tn) differentiable w.r.t.t1, t2, · · · , tn around 0, 0, · · · , 0 with g(0, 0, · · · , 0) ≡ e = 1. Each element element canthen be written as
g(t1, t2, · · · , tn) = exp
(
n∑
i=1
ai ti
)
ai =d g
d ti|(t1,t2,···,tn)=(0,0,···,0) (2.74)
The elements ai are the generators of the Lie group G. They constitute a Lie algebra ofG (notation L(G)). The dimension of the Group manifold is denoted by n. Notice thegi and ai can be represented by matrices but also other representations are possible.Properties of a Lie algebra L(G) of G. A Lie algebra is a linear vector space in whichis defined a product [, ] which is anticommutative
1
A,B ∈ L(G) → [A,B] = −[B,A] ∈ L(G) (2.75)
36
2 The elements of the Lie algebra satisfy the Jacobi identity
A,B,C ∈ L(G) → [[A,B], C] + [[B,C], A] + [[C,A], B] = 0 (2.76)
For matrices the product looks like
A,B ∈ L(G) → [A,B] = AB − BA ∈ L(G) (2.77)
External symmetry group
An example of a symmetry group is the rotation group in three dimensions. Rotationsare transformations in a vector space which leave the inner product invariant. Let usstart with rotations in one plane. The Lie group is represented by SO(2). For a generalelement R(θ) we can be write
R(θ) = exp (θ I12) with I12 =
(
0 −11 0
)
(2.78)
Here I12 is the generator of SO(2). From detR = 1 follows Tr I12 = 0 (Tr = trace).From the above it follows
In12 = E n = 0 mod 4
In12 = I12 n = 1 mod 4
In12 = −E n = 2 mod 4
In12 = −I12 n = 3 mod 4 (2.79)
so that
R(θ) =∞∑
k=0
θk
k!Ik12 = E cos θ + I12 sin θ
=
(
cos θ − sin θsin θ cos θ
)
(2.80)
Remark: the generator one can always be found by differentiating the element of thegroup to θ at θ = 0
⇒ I12 =d R(θ)
d θ|θ=0 (2.81)
The Lie algebra of SO(2) (notation L(SO(2)) contains one element only i.e. I12 andit is Abelian.
Orthogonal group in three dimensions SO(3).
37
If detR = 1 then R ∈ SO(3). There are three elementary rotations in SO(3) whichare given by rotations around x-, y- and z-axis. 1.
Rx(θ1) = exp (θ1I23) =
1 0 00 cos θ1 − sin θ10 sin θ1 cos θ1
I23 =
0 0 00 0 −10 1 0
(2.82)
Ry(θ2) = exp (θ2I31) =
cos θ2 0 sin θ20 1 0
− sin θ2 0 cos θ2
I31 =
0 0 10 0 0−1 0 0
(2.83)
Rz(θ3) = exp (θ3I12) =
cos θ3 − sin θ3 0sin θ3 cos θ3 0
0 0 1
I12 =
0 −1 01 0 00 0 0
(2.84)
Notice: all rotations are right handed. Define
K1 ≡ I23 K2 ≡ I31 K3 ≡ I12
⇒ [K1, K2] = K3 cyclic Ki ∈ L(SO(3)) (2.85)
Notice that instead of I23, I31 and I12 one can also choose I32, I13 and I21 providedIij = −Iji (There are three independent generators in SO(3)). The group SO(3) is non-Abelian for R(θ1)R(θ2) 6= R(θ2)R(θ1) (non commutative). Rotations leave quantitiesas kinetic energy T = 1/2mv2 = 1/2m~v · ~v and potentials V (r) with r = |~r| invariant.Newtons Law
~F = m~a = md2~r(t)
d t(2.86)
is left covariant under rotations.
~F ′ = R ~F ~a′ = R~a → ~F ′ = m~a′ (2.87)
Internal symmetry group
We consider Lie groups with the following Lie algebra which is given by the generatorsTi satisfying the commutation relations
[Ta, Tb] = i fabc Tc (2.88)
The structure constant fabc are antisymmetric in a, b, c. A special representation is theadjoint representation
(Ta)bc = −i fabc (2.89)
1For the components x, y, z has been also used the notation x1, x2, x3.
38
Let us limit ourselves to gauge transformations on fields. They look like
ϕr,k(x) → ϕ′r,k(x) =[
exp(
− i ~α(x) · ~T)]
klϕr,l(x)
~α(x) · ~T = αi(x)Ti =n∑
i=1
αi(x)Ti (2.90)
Here αi(x) are the n parameters of the Group manifold and are real functions of spaceand time. The indices k, l correspond to the internal symmetry group G whereas rrefers to the external group in this case the Lorentz group SO(3, 1). If the symmetrygroup is unitary then the following holds
[
exp(
− i ~α(x) · ~T)]†
=[
exp(
i ~α(x) · ~T †)]
=[
exp(
i ~α(x) · ~T)]
⇒ T †a = Ta Ta is hermitian ~T = (T1, T2, · · ·Tn) (2.91)
Examples
Electromagnetism
In the case of electromagnetism is the gauge group given by U(1). The elements consistsof a set of phase transformations given by
U(1) = ei α → T = 1 (2.92)
In the case of the real charge α is also multiplied by the elementary unit charge e =1, 6.10−19 C.
Isospin
Here is the gauge group represented by SU(2) which is non-Abelian. The structureconstants are fabc = ǫabc (Levi-Civita tensor components a, b, c = 1, 2, 3).
The fundamental representation is in the two dimensional space of iso-spinors. It isgiven by:
T1 =1
2
(
0 11 0
)
T2 =1
2
(
0 −ii 0
)
T3 =1
2
(
1 00 −1
)
(2.93)
In the adjoint representation(
Ta
)
mn= −i ǫamn, which is in three dimensional space,
one obtains the following matrices:
T1 = i
0 0 00 0 −10 1 0
T2 = i
0 0 10 0 0−1 0 0
T3 = i
0 −1 01 0 00 0 0
(2.94)
One distinguishes two kinds of gauge transformations.
39
1 Global gauge transformations. Here αi(x) are independent of x. They are alsocalled gauge transformations of the first kind.
2 Local gauge transformations. Here αi(x) is dependent on x. The are also calledgauge transformations of the second kind.
Global gauge transformations
For a global gauge transformation exist a Noether current and a conserved charge.Take the gauge transformation infinitesimal
ϕ′r,k(x) =(
1 − i ~α · ~T)
klϕr,l(x) =
[
δkl − i αa
(
Ta
)
kl
]
ϕr,l(x) (2.95)
Hence it follows from (2.35) and (2.36)
∆xµ = 0 → Xµk (x) = 0
∆ϕr,k(x) = Φka(x) ∆ωa with ∆ωa = −αa and
Φr,ka(x) = i(
Ta
)
klϕr,l(x) Φ∗r,ka(x) = −i
(
T ∗a)
klϕ∗r,l(x) (2.96)
Notice that the space time coordinates do not change under gauge transformations.The Noether current in (2.50) becomes equal to
jµa (x) = −i ∂ L
∂ ∂µϕr,k(x)
(
Ta
)
klϕr,l(x) + i
∂ L∂ ∂µϕ
∗r,k(x)
(
T ∗a)
klϕ∗r,l(x) (2.97)
with the Noether charge
Qa =1
hc
∫
d3~x j0a(x) with
d Qa
d t= 0 (2.98)
Examples
A Complex scalar field
φ′k(x) =[
exp(
− i ~α · ~T)]
klφl(x)
φ′∗k (x) =
[
exp(
i ~α · ~T ∗)]
klφ∗l (x) = φ∗l (x)
[
exp(
i ~α · ~T)]
lk(2.99)
The Lagrangian density
LBmat = ∂µφ∗k ∂µφk −
m2c2
h2 φ∗k φk (2.100)
40
is invariant under this gauge transformation. The dependence of the fields on xis omitted for simplicity. The Noether current becomes equal to
jµa (x) = −i ∂ LB
mat
∂ ∂µφk
(
Ta
)
klφl + i
∂ LBmat
∂ ∂µφ∗k
(
T ∗a)
klφ∗l
= i(
∂µφk
(
T ∗a)
klφ∗l − ∂µφ∗k
(
Ta
)
klφl
)
= i(
φ∗l(
Ta
)
lk∂µφk − ∂µφ∗k
(
Ta
)
klφl
)
= i(
φ∗k(
Ta
)
kl∂µφl − ∂µφ∗k
(
Ta
)
klφl
)
= i φ∗k(
Ta
)
kl
↔∂µ φl (2.101)
From which follows that the Noether current is real
j∗µa =
(
i φ∗k(
Ta
)
kl
↔∂µ φl
)∗= −i φk
(
T ∗a)
kl
↔∂µ φ∗l
= −i φk
(
Ta
)
lk
↔∂µ φ∗l = i φ∗l
(
Ta
)
lk
↔∂µ φk = jµ
a
→ Q∗a = Qa (2.102)
Therefore the charge is also real. If we express the field in its Fourier coefficientsthen the charge can be written as follows
Qa =i
h
(
Ta
)
kl
∫
d3~x[
φ∗k Π∗l − Πk φl
]
Qa =∫
d3~p[
a+,k(~p)(
Ta
)
kla∗+,l(~p) − a∗−,k(~p)
(
Ta
)
kla−,l(~p)
]
(2.103)
B Complex Dirac field
ψ′r,k(x) =[
exp(
− i ~α · ~T)]
klψr,l(x)
ψ′r,k(x) =[
exp(
i ~α · ~T ∗)]
klψr,l(x) = ψr,l(x)
[
exp(
i ~α · ~T)]
lk(2.104)
The Lagrange density
LFmat = i c h ψr,k
(
γµ)
rs∂µψs,k −mc2 ψr,k ψr,k (2.105)
is invariant under this gauge transformation. The Noether current becomes
jµa (x) = −i ∂ LF
mat
∂ ∂µψr,k
(
Ta
)
klψr,l + i
∂ LFmat
∂ ∂µψr,k
(
T ∗a)
klψr,l
= c h ψr,k
(
γµ)
rs
(
Ta
)
klψs,l (2.106)
41
Hence it follows again that the Noether current is real in other words
j∗µa = jµa → Q∗a = Qa (2.107)
The charge is also real. If we express the field in its Fourier coefficients then thecharge can be written as follows
Qa = − i
h
(
Ta
)
kl
∫
d3~xΠr,k ψr,l
Qa =∑
s
∫
d3~p[
b∗k(~p, s)(
Ta
)
klbl(~p, s) + dk(~p, s)
(
Ta
)
kld∗l (~p, s)
]
(2.108)
Local gauge transformations
For local gauge transformations one can show that after application of the equationof motion the Noether current can be written as ∂µT
µν where T µν = −T νµ. In thiscase the Noether charge is equal to zero (see also the remark in (2.59)) and one has nogenuine Noether current. Lagrange densities which are global invariant are very oftennot local gauge invariant. Examples are the Lagrange densities in (2.12) and (2.18).For simplicity we shall take the gauge transformations infinitesimal and limit ourselvesto U(1)
φ′(x) =[
1 − i α(x)]
φ(x)
φ′∗(x) = φ∗(x)
[
1 + i α(x)]
(2.109)
∂µ φ′(x) = −i ∂µα(x)φ(x) +
[
1 − i α(x)]
∂µφ(x)
∂µ φ′∗(x) = i φ∗(x) ∂µ α(x) + ∂µφ
∗(x)[
1 + i α(x)]
(2.110)
The Lagrange density of the complex scalar field in (2.16) transforms as
δLBmat = LB
mat
(
φ′(x), ∂λφ′(x)
)
− LBmat
(
φ(x), ∂λφ(x))
(2.111)
with
δLBmat = i φ∗(x) ∂µ α(x) ∂µ φk(x) − i ∂µφ∗(x) ∂µ α(x)φ(x)
= ∂µ α(x) jµ(x) (2.112)
For the Dirac field one obtains after local gauge transformations
ψ′r(x) =[
1 − i α(x)]
ψr(x)
ψ′r(x) = ψr(x)[
1 + i α(x)]
(2.113)
42
∂µ ψ′r(x) = −i ∂µ α(x)ψr(x) +
[
1 − i α(x)]
∂µ ψr(x)
∂µ ψ′r(x) = i ψr(x) ∂µ α(x) + ∂µ ψr(x)
[
1 + i α(x)]
(2.114)
The Lagrange density for the Dirac field in (2.18) transforms as
δLFmat = h c ψr(x)
(
γµ)
rsψs(x) ∂µ α(x) = ∂µ α(x) jµ(x) (2.115)
One observes that in both cases the Lagrange densities are global (∂µα(x) = 0) but notlocal gauge invariant. To make the Lagrange densities local gauge invariant we haveto introduce a gauge field. Let us first remark that the gauge invariance breaking termis proportional to the Noether current which occurred at global gauge invariance. Forthe scalar fields the transformation proceeds as
j′µ(x) = −i
(
∂µ φ′∗(x) φ′(x) − φ
′∗(x) ∂µφ′(x))
= jµ(x)
+∂µ α(x)(
φ∗(x)φ(x) + φ∗(x) φ(x))
(2.116)
j′µ(x) = jµ(x) + 2 ∂µ α(x)φ∗(x)φ(x) (2.117)
For the Dirac field the transformation proceeds as
j′µ(x) = jµ(x) (2.118)
To make the Lagrange densities in (2.105) and(2.110) local gauge invariant we introducea new term in Lmat. Because this is simpler for the Dirac field we start with the latter.Introduce the following interaction Lagrange density
LFint = − g
hcjµ(x)Aµ(x) (2.119)
Here g is the gauge coupling constant which indicates the strength with which thegauge field Aµ(x) couples to the current. The gauge field transforms locally as follows
A′µ(x) = Aµ(x) +
hc
g∂µα(x) (2.120)
Under local gauge transformations transforms the interaction Lagrange density as
L′Fint = − g
hcj′µ(x)A
′µ(x)
= LFint − ∂µ α(x) jµ(x)
→ δLFint = L′F
int − LFint = −∂µ α(x) jµ(x) (2.121)
43
and the combination
LF = LFmat + LF
int (2.122)
is therefore local gauge invariant because
δLFmat = −δLF
int (2.123)
In the case of scalar fields is the term Lint not sufficient to render the Lagrange densitylocal gauge invariant. In this case we get after transformation
L′int = Lint − ∂µ α(x) jµ(x) − 2g
hc∂µα(x)φ∗(x)φ(x)Aµ(x) (2.124)
We add to Lint the next term
g2
(hc)2φ∗(x)φ(x)Aµ(x)Aµ(x) (2.125)
Under local gauge transformations this term becomes
g2
(hc)2φ
′∗(x)φ′(x)A′µ(x)A′µ(x)
=g2
(hc)2φ∗(x)φ(x)Aµ(x)Aµ(x)
+2g
hc∂µα(x)φ∗(x)φ(x)Aµ(x) (2.126)
Define for the scalar fields a new interaction Lagrange density
LBint = − g
hcjµ(x)Aµ(x) +
g2
(hc)2φ∗(x)φ(x)Aµ(x)Aµ(x) (2.127)
Under local gauge transformation this term becomes equal to
L′Bint = LB
int − ∂µ α(x) jµ(x) (2.128)
so that here also
LB = LBmat + LB
int (2.129)
is local gauge invariant. To complete the theory we need besides derivatives of thescalar field and the Dirac field also derivatives of the gauge field. For the gauge fieldthis is achieved by
Lijk = −1
4Fµν F
µν (2.130)
44
with the gauge field tensor
F µν = ∂µ Aν(x) − ∂ν Aµ(x) (2.131)
Under local gauge transformations remains the tensor unchanged
F′µν = F µν (2.132)
From the above it follows that Lijk is local gauge invariant. The full Lagrange densityreads:
L = Lijk + Lmat + Lint (2.133)
The index mat refers to matter fields. These are all the fields except for the gaugefields.
The notion covariant derivative
We have seen that under local gauge transformations ∂µφ(x) and ∂µψ(x) transformdifferently than φ(x) and ψ(x) as shown in (2.104) and (2.109) respectively. Therefore∂µφ(x) and ∂µψ(x) are the wrong vectors of the group. We define a new kind ofderivative namely the covariant derivative:
Dµ = ∂µ + ig
hcAµ(x) (2.134)
We now study the change in the covariant derivative under local gauge transformations.This we show for the scalar field
Dµ φ(x) =(
∂µ + ig
hcAµ(x)
)
φ(x)
Dµ φ∗(x) =(
∂µ − ig
hcAµ(x)
)
φ∗(x) (2.135)
Apply the local gauge transformation
D′µ φ′(x) =
(
∂µ + ig
hcA
′µ(x)) (
1 − i α(x))
φ(x)
=(
∂µ + ig
hc
Aµ(x) +hc
g∂µα(x)
)(
1 − i α(x))
φ(x)
= ∂µφ(x) − i ∂µα(x)φ(x) − i α(x) ∂µφ(x) + ig
hcAµ(x)φ(x)
+i ∂µ α(x)φ(x) +g
hcα(x) Aµ(x)φ(x)
=(
1 − i α(x))
∂µφ(x) + ig
hcAµ(x)φ(x) +
g
hcα(x)Aµ(x)φ(x)
=(
1 − i α(x)) (
∂µ + ig
hcAµ(x)
)
φ(x)
=(
1 − i α(x))
Dµ φ(x) (2.136)
45
The derivation for the Dirac field goes in an analogous way. The conclusion is thatDµφ(x) and Dµψ(x) transform in the same way as φ(x) and ψ(x). They are goodvectors in the representation space of the group. We can write the Lagrange densityin (2.138) in a manifest local gauge invariant way. For the scalar fields this becomes
LB = −1
4Fµν F
µν +Dµ φ∗Dµ φ− m2c2
h2 φ∗ φ (2.137)
and for the Dirac field
LF = −1
4Fµν F
µν + i c h ψr
(
γµ)
rsDµ ψs −mc2 ψr ψr (2.138)
After application of the Euler Lagrange equation one gets again the equations of mo-tion. For the scalar fields they are
∂µ Fµν =g
hcjν jν = i φ∗
↔Dν φ
(
DµDµ +m2c2
h2
)
φ = 0 φ∗(
←D∗µ
←D∗µ +
m2c2
h2
)
= 0 (2.139)
and for the Dirac fields
∂µ Fµν =g
hcjν jν = c h ψr
(
γν
)
rsψs
(
i h γµDµ −mc)
ψ = 0 ψ(
i h γµ←D∗µ +mc
)
= 0 (2.140)
46
3 Relativistic quantum field theory
The formalism to quantise relativistic fields is analogous to the procedure in non-relativistic quantum mechanics. In the latter we had to deal with a system of a finitedegrees of freedom qi(t), pi(t). In the former we had to deal with a system of infinitedegrees of freedom given by ϕ(~x, t),Π(~x, t).
ANALOGUE
first quantisation second quantisation
non-relativistic quantum mechanics relativistic quantum field theory
generalised coordinates: qi(t) generalised coordinates: ϕ(~x, t)
canonical momentum: pi(t) =∂ L∂ qi(t)
canonical momentum: Π(~x, t) =∂ L
∂ ϕ(~x, t)
H =N∑
i=1
p2i
2m+ V (q1, · · · qn) H =
1
2
[
c2 Π2(~x, t) + ~∇ϕ(~x, t) · ~∇ϕ(~x, t)
+m2c2
h2 ϕ2(~x, t)]
[qk(t), pl(t)] = i h δkl [ϕ(~x, t), Π(~x′, t)] = i h δ(3)(~x− ~x′)
[qk(t), ql(t)] = 0 [ϕ(~x, t), ϕ(~x′, t)] = 0
[pk(t), pl(t)] = 0 [Π(~x, t), Π(~x′, t)] = 0
The commutation relations for the fields are called equal time commutation (ETC)relations. Because of these relations the classical fields become quantum operatorswhich act in a Hilbert space. This is analogous to qi, pi in non-relativistic quantummechanics. This Hilbert space consist out of particles which are characterised bytheir momentum and spin. The above ETC relations hold only for integer spin fieldsas the scalar field (spin zero) and the electromagnetic field (spin one). In the caseof half numbered spin fields, like the Dirac field (spin half), the commutators arereplaced by anti commutators. This procedure follows from the spin statistics theorem.From this follows that integer spin field are quantised by commutation relations andspin half numbered fields by anti commutation relations. If this does not happenthe Einstein causality is violated. This principle implies that the commutator of two
47
physical operators A(x) and B(y) equals zero if x and y are spacelike
[A(x), B(y)] = 0 if (x− y)2 < 0 (3.1)
An example is the electromagnetic current jµ(x)
[jµ(x), jµ(y)] = 0 if (x− y)2 < 0
with jµ(x) = i : φ†(x)↔∂µ φ(x) : scalar field
jµ(x) = c h :¯ψ(x) γµ ψ(x) : Dirac field (3.2)
The explanation of : follows later on.
3.1 Quantisation of the real scalar field
φ(x) =∫ d3~p
(2πh)3/2
hc√2E
[
a(~p) e−i px/h + a†(~p) ei px/h]
(3.3)
Compared with the classical field we replace the complex conjugated ∗ by hermitianconjugated † because a(~p) and a†(~p) become operators by second quantisation. Noticethat φ(x) is hermitian is. The annihilation en creation operators can be expressed inthe inner products (see (1.21))
a(~p) = (fp, φ) =i
hc
∫
d3~x f ∗p (x)↔∂0 φ(x)
=i
hc
∫
d3~x[
c f ∗p (x) Π(x) − φ(x) ∂0f∗p (x)
]
a†(~p) = −(f ∗p , φ) = − i
hc
∫
d3~x fp(x)↔∂0 φ(x)
= − i
hc
∫
d3~x[
c fp(x) Π(x) − φ(x) ∂0fp(x)]
(3.4)
Hence it follows
[a(~p), a†(~p′)] =1
h2c2
∫
d3~x∫
d3~x′
− c f ∗p (~x, t) ∂0fp′(~x′, t) [Π(~x, t), φ(~x′, t)]
−c fp′(~x′, t) ∂0f
∗p (~x, t) [φ(~x, t), Π(~x′, t)]
=i
hc
∫
d3~x f ∗p (~x, t)↔∂0 fp′(~x, t) = (fp, fp′) = δ(3)(~p− ~p′) (3.5)
48
likewise one can show
[a(~p), a(~p′)] = − i
hc
∫
d3~x f ∗p (~x, t)↔∂0 f
∗p′(~x, t) = −(fp, f
∗p′) = 0
[a†(~p), a†(~p′)] = − i
hc
∫
d3~x fp(~x, t)↔∂0 fp′(~x, t) = −(f ∗p , fp′) = 0 (3.6)
The commutation relations above are the same as those derived for the harmonicoscillator in non-relativistic quantum mechanics. However now we have an infinitenumber of degrees of freedom (~p is continuous). Therefore the replacement becomesequal to
ai → a(~p)
[ai, a†j] = δij → [a(~p), a†(~p′)] = δ(3)(~p− ~p′) (3.7)
By second quantisation the momentum P µ and the angular momentum Jµν also op-erators in the Hilbert space because they are expressed in field. Analogous to theharmonic oscillator for a finite degrees of freedom (non relativistic quantum mechan-ics) there exists a ground state with the lowest energy in the Hilbert space H. This iscalled the vacuum, notation |0〉 ∈ H. this state does not contain any particles. Thenth exited state of the harmonic oscillator corresponds with the n-particle state whichcontains n different momenta, notation |~p1; ~p2; · · · ; ~pn〉 ∈ H. Introduce the numberdensity operator
N(~p) = a†(~p) a(~p) (3.8)
so that the number operator looks like
N =∫
d3~p N(~p) met N |~p1; ~p2; · · · ; ~pn〉 = n |~p1; ~p1; · · · ; ~pn〉 (3.9)
From the commutation relations for the creation and the annihilation operators follows
[N(~p), a(~p′)] = −δ(3)(~p− ~p′) a(~p) [N(~p), a†(~p′)] = δ(3)(~p− ~p′) a†(~p)
⇒ [N , a(~p)] = −a(~p) [N , a†(~p)] = a†(~p) (3.10)
One can now derive the following
a†(~p) |~p1; ~p2; · · · ; ~pn〉 = c+ |~p; ~p1; · · · ; ~pn〉
N a†(~p) |~p1; ~p2; · · · ; ~pn〉 = a†(~p) (N + 1) |~p1; ~p2; · · · ; ~pn〉
= (n+ 1) a†(~p)|~p1; ~p2; · · · ; ~pn〉 (3.11)
49
or
a†(~p) |~p1; ~p2; · · · ; ~pn〉 = c+ |~p; ~p1; ~p2; · · · ; ~pn〉 met ~p 6= ~pi (3.12)
From the above it follows that a†(~p) a creation operator which creates a particle withmomentum ~p in the Hilbert space. Because of normalisation reason we shall adoptc+ =
√n+ 1. An n-particle state can be written as
|~p1; ~p2; · · · ; ~pn〉 =1√n!a†(~p1) a
†(~p2) · · · a†(~pn) |0〉 (3.13)
a(~p) is an annihilation operator which destroys a particle in the Hilbert space
N a(~p) |~p1〉 = a(~p) (N − 1) |~p1〉 = 0 → a(~p) |~p1〉 = c−|0〉 (3.14)
Because only for the vacuum state holds that N |0〉 = 0. Choose c− = 1 for normalisa-tion reasons. Let a(~p) act on general state
a(~p) |~p1; ~p2; · · · ; ~pn〉 =1√n!a(~p)a†(~p1) a
†(~p2) · · · a†(~pn) |0〉
=1√n
[
δ(3)(~p− ~p1)|~p2; · · · ; ~pn〉 + δ(3)(~p− ~p2)|~p1; ~p3; · · · ; ~pn〉 + · · ·
+δ(3)(~p− ~pn)|~p1; ~p2; · · · ; ~pn−1〉]
(3.15)
Since the creation operators commute with each other is the state|~p1; ~p2; · · · ; ~pn〉 symmetric under interchange of the momenta. Because of this oneobtains the Bose-Einstein statistics for particles with spin zero.
N(~p) |~p1; ~p2; · · · ; ~pn〉 =1√n!a†(~p) a(~p) a†(~p1) a
†(~p2) · · · a†(~pn) |0〉
=1√n
δ(3)(~p− ~p1)√n |~p; ~p2; · · · ; ~pn〉 + δ(3)(~p− ~p2)
√n |~p1; ~p; · · · ; ~pn〉
+ · · ·+ δ(3)(~p− ~pn)√n |~p1; ~p2; · · · ; ~p〉
=
n∑
i=1
δ(3)(~p− ~pi)
|~p1; ~p2; · · · ; ~pn〉 (3.16)
Further we have the property
〈~p1; ~p2; · · · ; ~pn| = |~p1; ~p2; · · · ; ~pn〉† = 〈0| 1√n!a(~pn) · · · a(~p2) a(~p1) (3.17)
50
The inner product is equal to
〈~p1; ~p2; · · · ; ~pm|~p′1; ~p′2; · · · ; ~p′n〉 = 0 for m 6= n
〈~p1; ~p2; · · · ; ~pn|~p′1; ~p′2; · · · ; ~p′n〉 =1
n!
∑
P (p′1,p′
2,···,p′n)
δ(3)(~p1 − ~p′1)
δ(3)(~p2 − ~p′2) · · · δ(3)(~pn − ~p′n) for m = n (3.18)
P (p′1, p′2, · · · , p′n) represents the permutation of the momenta p′1, p
′2, · · · , p′n. One ob-
serves that for a continuous presentation of the state vectors in Hilbert space the sameproblem occur as for free waves in non-relativistic quantum mechanics namely the ap-pearance of delta functions. This has as consequence that the momenta in the multiparticle states are not equal to each other. For scattering process this is not neces-sary. The four momentum of the quantum field can now be expressed in creation andannihilation operators. Hence from equation (2.64) follows
P µ =1
2
∫
d3~p pµ[
a(~p) a†(~p) + a†(~p) a(~p)]
=∫
d3~p pµ[
N(~p) +1
2δ(3)(0)
]
(3.19)
The last term is very ugly and for the following reasons
a The distribution δ(3)(0) is not defined.
b P µ|0〉 6= 0. This means that the vacuum (no particles !!) has not the eigen valuezero for the energy en momentum-operator.
To remedy this fault we introduce normal ordered operators. It boils down that weomit the term δ(3)(0) everywhere where it appears.
Normal Ordering
Definition
: a(~p) a†(~p) :≡ a†(~p) a(~p) (3.20)
This operation implies that for any arbitrary product of creation and annihilationoperators after normal ordering the creation operators are to left of the annihilationoperators. For example
: a†(~p1) a(~p2) a(~p3) a†(~p4) := a†(~p1) a
†(~p4) a(~p2) a(~p3) (3.21)
Notice that the mutual order between the creation operators and between the annihi-lation operators does not matter. One can after normal ordering write
: P µ :=∫
d3~p pµ N(~p) met : P µ : |0〉 = 0
: P µ : |~p1; ~p2; · · · ; ~pn〉 =
n∑
i=1
pµi
|~p1; ~p2; · · · ; ~pn〉 (3.22)
51
3.2 Quantisation van the complex scalar field
The fields are given by
φ(x) =1√2
(
φ1(x) + i φ2(x))
φ†(x) =1√2
(
φ1(x) − i φ2(x))
(3.23)
where φi(x) (i = 1, 2) are hermitian (real) fields. From the latter it follows
[φk(~x, t), Πl(~x′, t)] = i h δkl δ
(3)(~x− ~x′)
[φk(~x, t), φl(~x′, t)] = 0
[Πk(~x, t), Πl(~x′, t)] = 0 (3.24)
In the case of the complex scalar field read the ETC commutation relations as follows
[φ(~x, t), Π(~x′, t)] = i h δ(3)(~x− ~x′) [φ†(~x, t), Π†(~x′, t)] = i h δ(3)(~x− ~x′)
[φ(~x, t), φ(~x′, t)] = 0 [φ†(~x, t), φ†(~x′, t)] = 0
[φ(~x, t), φ†(~x′, t)] = 0 [Π(~x, t), Π†(~x′, t)] = 0
[φ(~x, t), Π†(~x′, t)] = 0 [φ†(~x, t), Π(~x′, t)] = 0 (3.25)
The two quantum fields get the following representation
φ(x) =∫
d3~p
(2πh)3/2
hc√2E
[
a+(~p) e−i px/h + a†−(~p) ei px/h]
φ†(x) =∫
d3~p
(2πh)3/2
hc√2E
[
a−(~p) e−i px/h + a†+(~p) ei px/h]
(3.26)
The creation en annihilation operators can be expresses in those of the real fields φi
a+(~p) =1√2
(
a1(~p) + i a2(~p))
a†−(~p) =1√2
(
a†1(~p) + i a†2(~p))
a†+(~p) =1√2
(
a†1(~p) − i a†2(~p))
a−(~p) =1√2
(
a1(~p) − i a2(~p))
(3.27)
From the commutation relations
[ak(~p), a†l (~p′)] = δkl δ
(3)(~p− ~p′)
[ak(~p), al(~p′)] = 0
[a†k(~p), a†l (~p′)] = 0 (3.28)
52
we infer
[a±(~p), a†±(~p′)] = δ(3)(~p− ~p′) [a±(~p), a†∓(~p′)] = 0
[a±(~p), a±(~p′)] = 0 [a±(~p), a∓(~p′)] = 0
[a†±(~p), a†±(~p′)] = 0 [a†±(~p), a†∓(~p′)] = 0 (3.29)
The interpretation is that a†+(~p) and a+(~p) are the creation and annihilation operators
of the positive charged particles whereas a†−(~p) and a(−~p) represent the creation en
annihilation operators of the negative charged particles. The Hilbert spaces of thepositive and negative charges particles are denoted by H(+) and H(−) respectively. Thestate vectors are
|~p1; ~p2; · · · ; ~pm〉 ∈ H(+) |~q1; ~q2; · · · ; ~qn〉 ∈ H(−) (3.30)
From the above one can construct the following product space H = H(+) ⊗ H(−) withas state vector
|~p1; ~p2; · · · ; ~pm; ~q1; ~q2; · · · ; ~qn〉 ∈ H (3.31)
This can be created out of the vacuum
|0〉 = |0〉+ |0〉− (3.32)
via
|~p1; ~p2; · · · ; ~pm; ~q1; ~q2; · · · ; ~qn〉 =1√n!m!
a†+(~p1) a†+(~p2) · · · a†+(~pm)
a†−(~q1) a†−(~q2) · · · a†−(~qn) |0〉 (3.33)
Define again the number density operator
N±(~p) = a†±(~p)a±(~p) number density operator
N± =∫
d3~p N±(~p) number operator (3.34)
We can again after normal ordering write the momentum operator : P µ : and the charge: Q : as
: P µ :=∫
d3~p pµ(
N+(~p) + N−(~p))
: Q := e∫
d3~p(
N+(~p) − N−(~p))
(3.35)
53
with
: P µ : |~p1; ~p2; · · · ; ~pm; ~q1; ~q2; · · · ; ~qn〉 =
m∑
i=1
pµi +
n∑
i=1
qµi
|~p1; ~p2; · · · ; ~pm;
~q1; ~q2; · · · ; ~qn〉
: Q : |~p1; ~p2; · · · ; ~pm; ~q1; ~q2; · · · ; ~qn〉 = e (m− n) |~p1; ~p2; · · · ; ~pm;
~q1; ~q2; · · · ; ~qn〉(3.36)
Further we have the property
[: Q :, : P µ :] = 0 (3.37)
This commutator follows from
[N±(~p), N±(~p)] = 0 en [N∓(~p), N±(~p)] = 0 (3.38)
3.3 quantisation of the complex Dirac field
As mentioned in the beginning of this chapter a half integer spin field and his canonicalmomentum have to satisfy anti commutation relations. In the case of the Dirac fieldthey are given by
ψα(~x, t), Πβ(~x′, t) = i h δαβ δ(3)(~x− ~x′)
ψα(~x, t), ψβ(~x′, t) = 0
Πα(~x, t), Πβ(~x′, t) = 0
met Πα(~x, t) = i h ψ†α(~x, t) α, β = 1, 2, 3, 4 (3.39)
The quantum fields are
ψα(x) =∑
s
∫ d3~p
(2πh)3/2
√
mc2
E
[
b(~p, s) uα(~p, s) e−i px/h + d†(~p, s) vα(~p, s) ei px/h]
ψ†α(x) =∑
s
∫
d3~p
(2πh)3/2
√
mc2
E
[
b†(~p, s) u†α(~p, s) ei px/h + d(~p, s) v†α(~p, s) e−i px/h]
(3.40)
An abbreviation is (see (1.72))
ψ(x) =∑
s
∫
d3~p[
b(~p, s)ψ1,~p,s(x) + d†(~p, s)ψ2,~p,s(x)]
(3.41)
54
With the definitions (see (1.72)-(1.74)) the creation and annihilation operators can bewritten as
b(~p, s) =(
ψ1,~p,s, ψ)
=∫
d3~xψ†1,~p,s(x) ψ(x)
b†(~p, s) =(
ψ†1,~p,s, ψ†)
=1
ih
∫
d3~xψ1,~p,s(x) Π(x)
d(~p, s) =(
ψ†2,~p,s, ψ†)
=1
ih
∫
d3~xψ2,~p,s(x) Π(x)
d†(~p, s) =(
ψ2,~p,s, ψ)
=∫
d3~xψ†2,~p,s(x) ψ(x) (3.42)
Hence it follows
b(~p, s), b†(~p′, s′) =1
i h
∫
d3~x∫
d3~x′ ψ†1,~p,s(x)ψ1,~p′,s′(x′) ψ(~x, t),Π(~x′, t)
=∫
d3~xψ†1,~p,s(x)ψ1,~p′,s′(x) = δss′ δ(3)(~p− ~p′)
d(~p, s), d†(~p′, s′) = δss′ δ(3)(~p− ~p′)
b(~p, s), b(~p′, s′) = 0 d(~p, s), d(~p′, s′) = 0
b†(~p, s), b†(~p′, s′) = 0 d†(~p, s), d†(~p′, s′) = 0
b(~p, s), d(~p′, s′) = 0 b(~p, s), d†(~p′, s′) = 0
b†(~p, s), d(~p′, s′) = 0 b†(~p, s), d†(~p′, s′) = 0 (3.43)
b†(~p, s) and b(~p, s) are the creation and annihilation operators of the negative chargedspin half particle (e−, µ−,τ− etc.).d†(~p, s) and d(~p, s) are the creation and annihilation operators of the positive chargedspin half particle (e+, µ+,τ+ etc.).The Hilbert spaces of the negative charged particles and positive charged anti-particlesare given by H(−) and H(+) respectively. The state vectors are
|~p1, s1; ~p2, s2; · · · ; ~pm, sm〉 ∈ H(−) |~q1, t1; ~q2, t2; · · · ; ~qn, tn〉 ∈ H(+) (3.44)
with
|~p1, s1; ~p2, s2; · · · ; ~pm, sm〉 =1√m!
b†(~p1, s1) b†(~p2, s2) · · · b†(~pm, sm)|0〉−
|~q1, t1; ~q2, t2; · · · ; ~qn, tn〉 =1√n!d†(~q1, t1) d
†(~q2, t2) · · · d†(~qn, tn)|0〉+ (3.45)
55
Because the creation operators anti commute, the multi particle state is anti-symmetricunder interchange of momenta and spins.
|~p1, s1; · · · ~pi, si; · · · ~pj, sj; · · · ~pm, sm〉 = −|~p1, s1; · · · ~pj, sj; · · · ~pi, si; · · · ~pm, sm〉
|~q1, t1; · · ·~qi, ti; · · ·~qj, tj ; · · ·~qn, tn〉 = −|~q1, t1; · · ·~qj , tj; · · ·~qi, ti; · · ·~qn, tn〉 (3.46)
Hence it follows the Fermi-Dirac-statistics of the spin half particles. One can alsoconstruct the product space H = H(−) ⊗ H(+) with the state vector
|~p1, s1; ~p2, s2; · · · ; ~pm, sm; ~q1, t1; ~q2, t2; · · · ; ~qn, tn〉 = |~p1, s1; ~p2, s2; · · · ; ~pm, sm〉
|~q1, t1; ~q2, t2; · · · ; ~qn, tn〉 ∈ H (3.47)
This state can be created out of the vacuum
|~p1, s1; ~p2, s2; · · · ; ~pm, sm; ~q1, t1; ~q2, t2; · · · ; ~qn, tn〉
=1√n!m!
b†(~p1, s1) b†(~p2, s2) · · · b†(~pm, sm)d†(~q1, t1) d
†(~q2, t2) · · · d†(~qn, tn) |0〉
⇒ 〈~p1, s1; ~p2, s2; · · · ; ~pm, sm; ~q1, t1; ~q2, t2; · · · ; ~qn, tn|
=1√n!m!
〈0|d(~qn, tn) · · · d(~q2, t2) d(~q1, t1) b~pm, sm) · · · b(~p2, s2) b(~p1, s1) (3.48)
The normalisation is
〈~p′1, s′1; · · · ; ~p′m′ , s′m′ ; ~q′1, t′1; · · · ; ~q′n′, t′n′|~p1, s1; · · · ; ~pm, sm; ~q1, t1; · · · ; ~qn, tn〉
=δmm′δnn′
m!n!det |δsis′j
δ(3)(~pi − ~p′j)| det |δtkt′lδ(3)(~qk − ~q′l)| (3.49)
The creation operators act on a multiple particle state as follows
b†(~p, s) |~p1, s1; · · · ; ~pm, sm; ~q1, t1; · · · ; ~qn, tn〉 = (−1)i+1√m+ 1
|~p1, s1; · · · ~pi, si; · · · ; ~pm, sm; ~q1, t1; · · · ; ~qn, tn〉 ~pi = ~p si = s
d†(~q, t) |~p1, s1; · · · ; ~pm, sm; ~q1, t1; · · · ; ~qn, tn〉 = (−1)m+j+1√n + 1
|~p1, s1; · · · ; ~pm, sm; ~q1, t1; · · ·~qj , tj; · · ·~qn, tn〉 ~qj = ~q tj = t (3.50)
and for the annihilation operators
b(~p, s) |~p1, s1; · · · ; ~pm, sm; ~q1, t1; · · · ; ~qn, tn〉 =1√m
m∑
i=1
(−1)i+1
56
δssiδ(3)(~p− ~pi) |~p1, s1; · · · ~pi−1, si−1; ~pi+1, si+1; · · · ; ~pm, sm; ~q1, t1; · · · ; ~qn, tn〉
d(~q, t) |~p1, s1; · · · ; ~pm, sm; ~q1, t1; · · · ; ~qn, tn〉 =1√n
n∑
j=1
(−1)m+j+1
δttj δ(3)(~q − ~qi) |~p1, s1; · · · ; ~pm, sm; ~q1, t1; · · ·~qj−1, tj−1; ~qj+1, tj+1; · · ·~qn, tn〉
(3.51)
One can again define a number density operator for particles and anti-particles
N−(~p, s) = b†(~p, s) b(~p, s) N+(~p, s) = d†(~p, s) d(~p, s) (3.52)
from which follows a number operator
N± =∑
s
∫
d3~p N±(~p, s) (3.53)
The commutation relations read
[N−(~p, s), b(~p′, s′)] = −δss′ δ(3)(~p− ~p′) b(~p, s)
[N−(~p, s), b†(~p′, s′)] = δss′ δ(3)(~p− ~p′) b†(~p, s)
[N+(~p, s), d(~p′, s′)] = −δss′ δ(3)(~p− ~p′) d(~p, s)
[N+(~p, s), d†(~p′, s′)] = δss′ δ(3)(~p− ~p′) d†(~p, s) (3.54)
with as property for the number density operator
N−(~p, s) |~p1, s1; ~p2, s2; · · · ; ~pm, sm; ~q1, t1; ~q2, t2; · · · ; ~qn, tn〉 =
m∑
i=1
δssiδ(3)(~p− ~pi)
|~p1, s1; ~p2, s2; · · · ; ~pm, sm; ~q1, t1; ~q2, t2; · · · ; ~qn, tn〉
N+(~q, t) |~p1, s1; ~p2, s2; · · · ; ~pm, sm; ~q1, t1; ~q2, t2; · · · ; ~qn, tn〉 =
n∑
j=1
δttj δ(3)(~q − ~qj)
|~p1, s1; ~p2, s2; · · · ; ~pm, sm; ~q1, t1; ~q2, t2; · · · ; ~qn, tn〉
(3.55)
In the states all momenta and spins occur only once. This follows from the Fermi-Dirac statistics. This is in contrast to the Bose-Einstein statistics where an infinitenumber of particles with the same quantum numbers occur. For this we could notfind a continuous representation. The four momenta and the charge can be written as
57
follows.
P µ =∑
s
∫
d3~p pµ(
b†(~p, s) b(~p, s) − d(~p, s) d†(~p, s))
Q = e∑
s
∫
d3~p(
b†(~p, s) b(~p, s) + d(~p, s) d†(~p, s))
e < 0 (3.56)
This can be rewritten in
P µ =∑
s
∫
d3~p pµ(
N−(~p, s) + N+(~p, s) − δ(3)(0))
Q = e∑
s
∫
d3~p(
N−(~p, s) − N+(~p, s) + δ(3)(0))
(3.57)
Like in the case of the bosonic operators we have to normal order for the fermioniccase. Define
: b(~p, s) b†(~p, s) := −b†(~p, s) b(~p, s)
: d(~p, s) d†(~p, s) := −d†(~p, s) d(~p, s) (3.58)
The minus sign is coming from the anti commutator. Normal ordering eliminates theterm δ(3)(0). For example
b(~p, s) b†(~p, s) = b(~p, s), b†(~p, s) − b†(~p, s) b(~p, s)
= δ(3)(0) − b†(~p, s) b(~p, s) (3.59)
The same holds for d(~p, s) d†(~p, s). In general it means that for an arbitrary productof creation and annihilation operators the creation operators has to be left of theannihilation operator. Moreover one has to take into account that a minus sign appearsafter an odd number of interchanges.Example:
: b1 d2 d†3 b†4 := −b†4 d†3 b1 d2 = b†4 d
†3 d2 b1 (3.60)
After normal ordering we get equations (3.57) in the following form
: P µ :=∑
s
∫
d3~p pµ(
N−(~p, s) + N+(~p, s))
: Q := e∑
s
∫
d3~p(
N−(~p, s) − N+(~p, s))
= e(
N− − N+
)
(3.61)
Further we have the property
[: Q :, : P µ :] = 0 (3.62)
58
These operators act on the states as follows
: P µ : |~p1, s1; ~p2, s2; · · · ; ~pm, sm; ~q1, t1; ~q2, t2; · · · ; ~qn, tn〉
=
m∑
i=1
pµi +
n∑
j=1
qµj
|~p1, s1; ~p2, s2; · · · ; ~pm, sm; ~q1, t1; ~q2, t2; · · · ; ~qn, tn〉
: Q : |~p1, s1; ~p2, s2; · · · ; ~pm, sm; ~q1, t1; ~q2, t2; · · · ; ~qn, tn〉
= e (m− n) |~p1, s1; ~p2, s2; · · · ; ~pm, sm; ~q1, t1; ~q2, t2; · · · ; ~qn, tn〉 (3.63)
Finally we want to show the connexion between spin and statistics of fermions. Supposethat ψ and Π satisfy commutation relations instead of anti commutation relations.b, b†, d, d† satisfy commutation relations. In this case the energy operator : H : looksafter normal ordering
: H := c : P 0 :=∑
s
∫
d3~pE(
b†(~p, s) b(~p, s) − d†(~p, s) d(~p, s))
(3.64)
The eigenvalue for the anti-fermions then becomes
: H : |~q1, t1; ~q2, t2; · · · ; ~qn, tn〉 = −n∑
i=1
Ei |~q1, t1; ~q2, t2; · · · ; ~qn, tn〉 (3.65)
Because∑n
i=1Ei → ∞ for n→ ∞ are the eigenvalues for the energy operator : H : notbounded from below. This is unacceptable for a quantum theory where we have therequirement that the ground state (here the vacuum) has the lowest eigenvalue whichhas to be finite. We conclude that the spin half character of the Dirac field is relevantfor the fact that we have to take anti commutation relations to quantise the field. Thesame holds for integer spin fields like the scalar field or the electromagnetic field. Hereone has to take commutation relations to quantise the field for the same reason namelyto avoid that the energy spectrum is unbounded from below. This observation is one ofimportant successes of quantum field theory namely the connexion between spin andstatistics.
3.4 quantisation of the electromagnetic field
Except for the polarisation vectors ǫµ(~k) is there large similarity between the electro-magnetic field and the real scalar field φ(x). We shall only consider the quantisation ofthe electromagnetic field in the radiation gauge. In this case only physical polarisationsoccur. The quantised field becomes
~A(~x, t) =
∑
α
∫
d3~k c
√
h
2ω
[
a(~k, α)~ǫα(~k) e−i kx + a†(~k, α)~ǫ∗α(~k) ei kx]
59
α = 1, 2 or α = R,L (3.66)
where a†(~k, α) and a(~k, α) are creation and annihilation operators which satisfy therelations
[a(~k, α), a†(~k′, α′)] = δαα′ δ(3)(~k − ~k′)
[a†(~k, α), a†(~k′, α′)] = 0
[a(~k, α), a(~k′, α′)] = 0 (3.67)
The connexion between the creation and annihilation operators of the linear (α = 1, 2)and circular polarisation (α = R,L) is given by
a(~k, R) = − 1√2
(
a(~k, 1) − i a(~k, 2))
a(~k, L) =1√2
(
a(~k, 1) + i a(~k, 2))
(3.68)
The canonical commutation relations are equal to
[Ai(~x, t), Πj(~x′, t)] =
1
c[Ai(~x, t), Ej(~x
′, t)] =
−i h∫
d3~k
(2π)3e−i~k·(~x−~x′)
(
δij −kikj
|~k|2
)
[Πi(~x, t), Πj(~x′, t)] = 0
[Ai(~x, t), Aj(~x′, t)] = 0 (3.69)
Notice that the commutator is consistent with the radiation gauge condition
∂kAk = ~∇ · ~A = 0 (3.70)
Because the radiation gauge leads to physical polarisations of the electromagnetic fieldonly physical states (here called photons) arise in the multi particle space so that itbecomes a Hilbert space. This state is denoted by
|~k1, α1;~k2, α2; · · · ;~kn, αn〉 ∈ H met αi = R,L (3.71)
This can be created out of the vacuum
|~k1, α1;~k2, α2; · · · ;~kn, αn〉 =1√n!a†(~k1, α1) a
†(~k2, α2) · · ·a†(~kn, αn) |0〉
⇒ 〈~k1, α1;~k2, α2; · · · ;~kn, αn| =1√n!
〈0| a(~kn, αn) · · ·a(~k2, α2) a(~k1, α1)
(3.72)
60
Because the creation and annihilation operators commutate among each other thestates are again symmetric under interchange of momenta and polarisations. Thecreation and annihilation act on the photon states as follows
a†(~k, α) |~k1, α1; · · · ;~kn, αn〉 =√n+ 1 |~k, α;~k1, α1; · · · ;~kn, αn〉
a(~k, α) |~k1, α1; · · · ;~kn, αn〉 =1√n
n∑
i=1
∑
α
δααiδ(3)(~k − ~ki)
×|~k1, α1; · · · ;~ki−1, αi−1;~ki+1, αi+1; · · · ;~kn, αn〉 (3.73)
Next we define the number density operator
N(~k, α) = a†(~k, α) a(~k, α) (3.74)
with the number operator
N =∑
α
∫
d3~k N(~k, α) (3.75)
the former acts on the multi photon state as follows
N(~k, α) |~k1, α1; · · · ;~kn, αn〉 =n∑
i=1
∑
α
δααiδ(3)(~k − ~ki) |~k1, α1; · · · ;~kn, αn〉 (3.76)
The normalisation becomes
〈~k1, α1; · · · ;~km, αm|~k1, α1; · · · ;~kn, αn〉 =δmn
n!
∑
P (k′
1,α′
1···k′
n,α′
n)
δα1α′
1δ(3)(~k1 − ~k′1) · · · δαnα′
nδ(3)(~kn − ~k′n) (3.77)
After normal ordering we can express : P µ : again in creation and annihilation operators
: P µ := h∑
α=R,L
∫
d3~k kµ N(~k, α) (3.78)
It acts on the multi photon state as follows
: P µ : |~k1, α1; · · · ;~kn, αn〉 = h
n∑
i=1
kµi
|~k1, α1; · · · ;~kn, αn〉 (3.79)
61
4 Interacting fields
4.1 Green functions for the scalar field
From now on we shall put h = c = 1. Green functions play an important role inQuantum Field Theory. They occur in the perturbative expansion of the S-matrix fromwhich one can derive cross sections and decay widths of particles. They also appearin Statistical Mechanics and Condensed Matter where they represent the correlationfunctions from which one can derive quantities like susceptibility, magnetisation orspecific heat. Since we are mainly interested in scalar fields we will derive the Greenfunction together with other distributions for that case only. From Eq. (3.3) one canderive various distributions like the causal commutator
i∆(x− y) ≡[
φ(x), φ(y)]
(4.1)
or the Green function
i∆F (x− y) ≡ 〈0|T(
φ(x) φ(y))
|0〉 (4.2)
where the time ordered product for Boson fields is defined by
T(
φ(x) φ(y))
≡ θ(x0 − y0) φ(x) φ(y) + θ(y0 − x0) φ(y) φ(x) (4.3)
For convenience one can also define other distributions like
∆+(x− y) ≡ 〈0|φ(x) φ(y)|0〉
∆−(x− y) ≡ 〈0|φ(y) φ(x)|0〉
∆1(x− y) ≡ 〈0|
φ(x), φ(y)
|0〉 (4.4)
Let us first derive an expression for the causal commutator in Eq. (4.1) Insertion ofthe free fields into the commutator yields
i∆(x− y) =
1
(2π)3
∫
d3~p∫
d3~p′1
√
4EpEp′
(
ei (p′·y−p·x)[
a(~p), a†(~p′)]
+ ei (p·x−p′·y)[
a†(~p), a(~p′)])
=1
(2π)3
∫
d3~p∫
d3~p′1
√
4EpEp′
δ(3)(~p− ~p′)(
ei (p′·y−p·x) − ei (p·x−p′·y))
=1
(2π)3
∫
d3~p1
2Ep
(
ei p·(y−x) − ei p·(x−y))
(4.5)
62
Note that ~p′ = ~p→ Ep′ = Ep. Replace in the last exponent ~p→ −~p
i∆(x− y) = −i 1
(2π)3
∫
d3~p
Epei p ·(x−y) sinEp (x0 − y0) (4.6)
The latter we can write in a Lorentz invariant way
i∆(x− y) =1
(2π)3
∫
d3~p
2Epei ~p ·(~x−~y)
(
e−i Ep (x0−y0) − e−i (−Ep) (x0−y0))
=1
(2π)3
∫
d4p δ(p2 −m2) ǫ(p0) e−i p·(x−y) (4.7)
with
δ(p2 −m2) =1
2Ep
(
δ(p0 −Ep) + δ(p0 + Ep))
ǫ(p0) = θ(p0) − θ(−p0) (4.8)
Therefore we can split up δ(p2 −m2) into a positive and negative energy part. FromEq. (4.9) we can derive that for x0 = y0 the causal commutator ∆(x−y) vanishes. Forthose values x− y is spacelike. Because of the Lorentz invariance of ∆(x− y) (see Eq.(4.7)) the latter vanishes for all spacelike distances. Further we can derive
∂
∂y0i∆(x− y)|y0=x0
= i δ(3)(~x− ~y) (4.9)
Hence we obtain the canonical commutation relation
[
φ(~x, x0), Π(~y, x0)]
=∂
∂y0
[
φ(~x, x0), φ(~y, y0)]∣
∣
∣
y0=x0
=∂
∂y0i∆(x− y)|y0=x0
= i δ(3)(~x− ~y) (4.10)
where Π(~y, y0) is the canonical momentum operator given by ∂∂y0
φ(~y, y0). From Eq.
(4.7) it also follows that the causal commutator satisfies the Klein Gordon equation(
2x +m2)
i∆(x− y) = 0 (4.11)
Similar expressions as the one presented in Eq. (4.7) can be given for the distributionsin Eq. (4.4)
∆+(x− y) =1
(2π)3
∫
d4p δ(p2 −m2) θ(p0) e−i p·(x−y)
∆−(x− y) =1
(2π)3
∫
d4p δ(p2 −m2) θ(−p0) e−i p·(x−y)
∆1(x− y) =1
(2π)3
∫
d4p δ(p2 −m2) e−i p·(x−y) (4.12)
63
All expressions above satisfy the Klein Gordon equation as given in Eq. (4.11). Finallywe compute this equation when it acts on the time-ordered product in Eq. (4.3). Firstwe look at the effect on this product (see Eq. (4.3)) when it is differentiated twicew.r.t. the time coordinate x0. Differentiating once w.r.t. x0 yields
∂
∂x0
(
θ(x0 − y0) φ(x) φ(y) + θ(y0 − x0) φ(y) φ(x))
=
δ(x0 − y0)[
φ(x), φ(y)]
+
(
θ(x0 − y0)∂φ(x)
∂x0φ(y) + θ(y0 − x0) φ(y)
∂φ(x)
∂x0
)
=
(4.13)
The first term represented by the commutator equals zero because of[
φ(~x, x0), φ(~y, x0)]
= 0 (4.14)
(see the beginning of section 3). Differentiating the second term above again w.r.t. x0
gives
∂
∂x0
(
θ(x0 − y0)∂φ(x)
∂x0φ(y) + θ(y0 − x0) φ(y)
∂φ(x)
∂x0
)
=
δ(x0 − y0)[
Π(x), φ(y)]
+
(
θ(x0 − y0)∂2φ(x)
∂2x0
φ(y) + θ(y0 − x0) φ(y)∂2φ(x)
∂2x0
)
(4.15)
Using the equal time commutator in Eq. (4.10) we obtain
2x
(
θ(x0 − y0) φ(x) φ(y) + θ(y0 − x0) φ(y) φ(x))
= −i δ(4)(x− y)
+(
θ(x0 − y0) 2xφ(x) φ(y) + θ(y0 − x0) φ(y) 2xφ(x))
(4.16)
Using the equation of motion in Eq. (4.11) and sandwiching the expression betweenthe vacuum state we finally obtain
(
2x +m2)
i∆F (x− y) = −i δ(4)(x− y) (4.17)
Because of the δ-function on the right-hand side ∆F (x−y) becomes a Green’s function.The equation above admits the following solution
i∆F (x− y) = i limǫ→ 0
∫
d4p
(2π)4e−i p·(x−y) 1
p2 −m2 + i ǫǫ ≥ 0 (4.18)
The ǫ occurs in order to regularize the pole at p2 = m2. Very often one omits the limitsymbol which is implicitly understood. The sign of ǫ is determined in such a way thatit reproduces the answer which is obtained by substituting the free field in Eq. (4.2).
64
4.2 Perturbation theory
As an example we choose φ4-theory. The Lagrangian density is given by
L =1
2∂µ φ ∂
µ φ− 1
2µ2 φ2 − λ
4!φ4 (4.19)
The Hamiltonian density is equal to
H = Π ∂0 φ− L =1
2∂0 φ ∂
0 φ+1
2~∇φ ~∇φ+
1
2µ2 φ2 +
λ
4!φ4 (4.20)
From which follows
H0 =1
2∂0 φ ∂
0 φ+1
2~∇φ ~∇φ+
1
2µ2 φ2
V =λ
4!φ4 (4.21)
Suppose λ is a small parameter. In that case we can apply perturbation theory. Inquantum field theory this is an assumption which in practice is not always correct.From the above we obtain the Lagrangian and Hamiltonian
L =∫
d3xL(x) H =∫
d3xH(x) (4.22)
The Hamiltonian is split up as follows
H = H0 + V (4.23)
The solutions of H0 are known but the solutions of H are not. In the Interactionpicture the state vector is given by
|ΨI(t)〉 = ei H0 t |ΨS(t)〉 = ei H0 t e−i H t |ΨS(0)〉 (4.24)
|ΨS(t)〉 is a time dependent state vector in the Schrodinger picture. The time depen-dence in the interaction picture is given by the dynamics
AI(t) = ei H0 tAS(t) e−i H0 t (4.25)
AS(t) is the operator in the Schrodinger picture and is time independent unless itexplicitly depends on time. In that case it implies that the time dependence doesnot originate by the dynamics c.q. the Hamiltonian. The equations of motion in theinteraction picture are
id
d t|ΨI(t)〉 = HI(t) |ΨI(t)〉
d AI(t)
d t=
∂ AI(t)
∂ t+ i
[
H0, AI(t)]
(4.26)
65
HI = ei H0 t V e−i H0 t (4.27)
The time dependence of |ΨI(t)〉 is given by HI whereas the time dependence of AI(t)is given by H0. Define the time evolution operator U(t, t0) by
|ΨI(t)〉 = U(t, t0) |ΨI(t0)〉 (4.28)
From (4.26) we derive
id U(t, t0)
d t= HI(t)U(t, t0) met U(t0, t0) = 1 (4.29)
The solution of the above equation is
U(t, t0) = 1 − i∫ t
t0dt′HI(t
′)U(t′, t0) (4.30)
Dyson series
U(t, t0) =∞∑
n=0
U (n)(t, t0) U (0)(t, t0) = 1 (4.31)
U (n)(t, t0) =(−i)n
n!
∫ t
t0dt1
∫ t
t0dt2 · · ·
∫ t
t0dtn T
(
HI(t1)HI(t2) · · ·HI(tn))
(4.32)
or in shorthand notation
U(t, t0) = T
e−i∫ t
t0dt′ HI(t′)
(4.33)
The definition of the time ordered product is
T
(
HI(t1)HI(t2) · · ·HI(tn)
)
= HI(ti)HI(tj) · · ·HI(tk) if ti > tj > · · · > tk (4.34)
This can be rewritten in using θ-functions
T
(
HI(t1)HI(t2)
)
= θ(t1 − t2)HI(t1)HI(t2) + θ(t2 − t1)HI(t2)HI(t1)
2! terms (4.35)
T
(
HI(t1)HI(t2)HI(t3)
)
= θ(t1 − t2) θ(t2 − t3)HI(t1)HI(t2)HI(t3)
+θ(t2 − t3) θ(t3 − t1)HI(t2)HI(t3)HI(t1)
66
+θ(t3 − t1) θ(t1 − t2)HI(t3)HI(t1)HI(t2)
+θ(t2 − t1) θ(t1 − t3)HI(t2)HI(t1)HI(t3)
+θ(t3 − t2) θ(t2 − t1)HI(t3)HI(t2)HI(t1)
+θ(t1 − t3) θ(t3 − t2)HI(t1)HI(t3)HI(t2)
3! terms (4.36)
T
(
HI(t1)HI(t2) · · ·HI(tn)
)
=⇒ n! terms (4.37)
Scattering matrix S
Physical picture: At t = −∞ we have free particles which do not interact. Theyare described by free fields. For −∞ < t < ∞ the particles interact mutually. Thisinteraction is described by HI . For t = ∞ we have again free particles which donot interact. The latter are again described by free fields. The S-operator maps theincoming state (t = −∞) on the outgoing state (t = ∞)
|ΨI(∞)〉 = S |ΨI(−∞)〉
|i〉 = |ΨI(−∞)〉 incoming state |f〉 = |ΨI(∞)〉 outgoing state (4.38)
The S-operator is then given by
S = limT → ∞U
(
T
2,−T
2
)
= T
exp
(
− i∫ ∞
−∞dtHI(t)
)
(4.39)
This leads to the definition of the transition operator T
S = 1 − 2 π i δ(Ei − Ef)T (4.40)
S is a unitary operator
S S† = S† S = 1 (4.41)
This follows from
H†I = HI =⇒ U U † = U † U = 1 (4.42)
From this follows the unitary relation for the transition operator
i
(
Tfi − T †fi
)
= 2 π∑
n
δ(Ef − En)T †fn Tni with Tfi = 〈f |T |i〉 (4.43)
67
p1
p2
p3
p4
Figure 1: Scattering process for a scalar field theory 1 + 2 → 3 + 4.
Let us take for example φ4-theory and calculate the S-matrix. Consider the quantumfield operator
HI =∫
d3x
(
λ
4!φ4(x)
)
I
(4.44)
wherein the interaction picture φ(x) becomes a free field operator. The latter describesa scalar particle. We describe a process where two scalar particles come in and twoscalar particles of the same kind go out.
|i〉 = |~p1, ~p2〉 |f〉 = |~p3, ~p4〉 (4.45)
The S-matrix becomes
Sfi = 〈f |S|i〉 = 〈f |U(−∞,∞)|i〉 =∞∑
n=0
〈f |U (n)(−∞,∞)|i〉 = 〈f |i〉
+〈f |U (1)(−∞,∞)|i〉 + 〈f |U (2)(−∞,∞)|i〉 + · · · (4.46)
The first term equals
〈f |i〉 = 〈~p3, ~p4|~p1, ~p2〉 = δ(3)(~p1 − ~p3) δ(3)(~p2 − ~p4) + δ(3)(~p1 − ~p4) δ
(3)(~p2 − ~p3) (4.47)
In lowest order of perturbation theory in λ we have
〈f |U (1)(−∞,∞)|i〉 = −i∫ ∞
−∞dt 〈f |HI(t)|i〉
= −i λ4!
∫ ∞
−∞dt∫
d3x 〈~p3, ~p4|φ4(x)|~p1, ~p2〉 (4.48)
Substitution of the free field
φ(x) =∫
d3k
(2π)3/2
1√2ωk
[
a(~k) e−i k·x + a†(~k) ei k·x]
(4.49)
68
leads to
〈f |U (1)(−∞,∞)|i〉 = −i λ4!
∫
d4x∫
d3k1
(2π)3/2
1√2ωk1
∫
d3k2
(2π)3/2
1√2ωk2
∫
d3k3
(2π)3/2
1√2ωk3
∫
d3k4
(2π)3/2
1√2ωk4
〈~p3, ~p4|
a(~k1) e−i k1·x
+a†(~k1) ei k1·x
a(~k2) e−i k2·x + a†(~k2) e
i k2·x
a(~k3) e−i k3·x + a†(~k3) e
i k3·x
a(~k4) e−i k4·x
+a†(~k4) ei k4·x
|~p1, ~p2〉 (4.50)
Remember ωki=√
|~ki|2 + µ2 = ki,0. In the products of a and a† only contribute thefollowing matrix elements
1
〈~p3, ~p4|a(~k1) a(~k2) a†(~k3) a
†(~k4)|~p1, ~p2〉 (4.51)
2
〈~p3, ~p4|a(~k1) a†(~k2) a(~k3) a
†(~k4)|~p1, ~p2〉 (4.52)
3
〈~p3, ~p4|a(~k1) a†(~k2) a
†(~k3) a(~k4)|~p1, ~p2〉 (4.53)
4
〈~p3, ~p4|a†(~k1) a(~k2) a(~k3) a†(~k4)|~p1, ~p2〉 (4.54)
5
〈~p3, ~p4|a†(~k1) a(~k2) a†(~k3) a(~k4)|~p1, ~p2〉 (4.55)
6
〈~p3, ~p4|a†(~k1) a†(~k2) a(~k3) a(~k4)|~p1, ~p2〉 (4.56)
69
We have to rewrite equations 1-6 otherwise we cannot calculate the expectation values
1
〈~p3, ~p4||a†(~k3) a†(~k4) a(~k1) a(~k2) + δ(3)(~k2 − ~k3) a
†(~k4) a(~k1)
+δ(3)(~k2 − ~k4) a†(~k3) a(~k1) + δ(3)(~k1 − ~k4) a
†(~k3) a(~k2)
+δ(3)(~k1 − ~k3) a†(~k4) a(~k2) + δ(3)(~k2 − ~k3) δ
(3)(~k1 − ~k4)
+δ(3)(~k1 − ~k3) δ(3)(~k2 − ~k4) |~p1, ~p2〉 (4.57)
2
〈~p3, ~p4||a†(~k2) a†(~k4) a(~k1) a(~k3) + δ(3)(~k3 − ~k4) a
†(~k2) a(~k1)
+δ(3)(~k1 − ~k4) a†(~k2) a(~k3) + δ(3)(~k1 − ~k2) a
†(~k4) a(~k3)
+δ(3)(~k1 − ~k2) δ(3)(~k3 − ~k4) |~p1, ~p2〉 (4.58)
3
〈~p3, ~p4||a†(~k2) a†(~k3) a(~k1) a(~k4) + δ(3)(~k1 − ~k2) a
†(~k3) a(~k4)
+δ(3)(~k1 − ~k3) a†(~k2) a(~k4) |~p1, ~p2〉 (4.59)
4
〈~p3, ~p4||a†(~k1) a†(~k4) a(~k2) a(~k3) + δ(3)(~k3 − ~k4) a
†(~k1) a(~k2)
+δ(3)(~k2 − ~k4) a†(~k1) a(~k3) |~p1, ~p2〉 (4.60)
5
〈~p3, ~p4||a†(~k1) a†(~k3) a(~k2) a(~k4) + δ(3)(~k2 − ~k3) a
†(~k1) a(~k4) |~p1, ~p2〉(4.61)
We can now, by interchanging of the integration variables, collect a great number ofterms
〈f |U (1)(−∞,∞)|i〉 =∫
d4x∫
d4x∫ d3k1
(2π)3/2
1√2ωk1
∫ d3k2
(2π)3/2
1√2ωk2
∫
d3k3
(2π)3/2
1√2ωk3
∫
d3k4
(2π)3/2
1√2ωk4
ei x (k3+k4−k1−k2)
70
13
Figure 2: Feynman diagram
1
2
3
4
Figure 3: Vertex diagram in φ4-theory.
[(
−i λ4
)
〈~p3, ~p4||a†(~k3) a†(~k4) a(~k1) a(~k2) |~p1, ~p2〉
+
(
−i λ2
)
δ(3)(~k2 − ~k3) 〈~p3, ~p4|a†(~k4) a(~k1) |~p1, ~p2〉
+
(
−i λ8
)
δ(3)(~k2 − ~k3) δ(3)(~k1 − ~k4) 〈~p3, ~p4||~p1, ~p2〉
]
(4.62)
Use the following identities
a(~k1) a(~k2) |~p1, ~p2〉 = a(~k1) a(~k2) a†(~p1) a
†(~p2) |0〉 =
δ(3)(k1 − p1) δ(3)(k2 − p2)
+δ(3)(k1 − p2) δ(3)(k2 − p1)
|0〉 (4.63)
likewise
〈~p3, ~p4| a†(~k3) a†(~k4) = 〈0|
δ(3)(k3 − p3) δ(3)(k4 − p4) + δ(3)(k3 − p4) δ
(3)(k4 − p3)
(4.64)
a(~k1) |~p1, ~p2〉 = δ(3)(k1 − p1) |~p2〉 + δ(3)(k1 − p2) |~p1〉 (4.65)
71
p1 p3
k
Figure 4: Tadpole diagram
k1 k2
Figure 5: Vacuum bubble diagram
〈~p3, ~p4| a†(~k4) = 〈~p3|δ(3)(k4 − p4) + 〈~p4|δ(3)(k4 − p3) (4.66)
Finally we get
〈f |U (1)(−∞,∞)|i〉 =(2π)4
(2π)6
1√16E1E2 E3E4
δ(4)(p1 + p2 − p3 − p4) (−i λ)
+(2π)4
(2π)6
1
2E1
δ(4)(p1 − p4) 〈~p3|~p2〉 +1
2E1
δ(4)(p1 − p3) 〈~p4|~p2〉
+1
2E2
δ(4)(p2 − p4) 〈~p3|~p1〉 +1
2E2
δ(4)(p2 − p3) 〈~p4|~p1〉
(
−i λ2
)
∫
d3k2
2ω2
+(2π)4
(2π)6δ(0)(0) 〈~p3, ~p4|~p1, ~p2〉
(
−i λ8
)
×∫
d3k1
2ω1
∫
d3k2
2ω2
(4.67)
In the language of Feynman diagrams the expresion
〈~p3|~p1〉 = δ(3)(~p1 − ~p3) (4.68)
72
is represented by the graph in Fig. 2. The first term in Eq. (4.67) i.e.
δ(4)(p1 + p2 − p3 − p4) (−i λ) (4.69)
can be found in Fig. 3. The next term represented by
1
2δ(4)(p1 − p3)
(
−i λ
(2π)3
)
∫
d3k
2ω(4.70)
is depicted in Fig. 4. Finally the term
1
(2π)6δ(4)(0)
(
−i λ8
)
∫
d3k1
2ω1
∫
d3k2
2ω2
(4.71)
can be found in Fig. 5. Vacuum bubble diagrams arise even when |i〉 = |0〉 and|f〉 = |0〉 in other words 〈0|φ4(x)|0〉. The S-matrix gets the form depicted in Fig. 6.Diagrams multiplied by vacuum bubbles are disconnected and therefore unacceptablefor a physical S-matrix. Moreover the vacuum bubbles give a infinite contribution (thediagrams diverge). The latter also hold for the tadpole diagrams. However in this casethe divergences can be removed via the renormalisation procedure. To obtain a physicalS-matrix we have to redefine the naive S-matrix above. This goes via normal ordering.Take for example an arbitrary product of creation and annihilation operators. Normalordering implies that one puts all creation operators left of the annihilation operators.In the case of fermions one has to watch out. One has to add an extra minus sign ifthe normal ordered product is obtained via an odd number of interchanges from theordinary product.
: b d d† b† := −b† d† b d : b d d† b† := d† b† b d (4.72)
The consequence is that the expectation value of normal ordered operators sandwichedbetween the vacuum state is zero.
〈0| : φ2(x) : |0〉 = 0 〈0| : φ4(x) : |0〉 = 0 (4.73)
Watch that it is not necessary to apply normal ordering to products like
φ(x)φ(y) or : φ4(x) : : φ4(y) : (4.74)
as long x 6= y We have to give a new definition to the quantum Lagrangian densityand Hamiltonian density. For φ4-theory this means that
: L(x) : =1
2: ∂µ φ(x) ∂µ φ(x) : −1
2µ2 : φ2(x) : − λ
4!: φ4(x) :
: H(x) : =1
2: ∂0 φ(x) ∂0 φ(x) :
1
2: ~∇φ(x) · ~∇φ(x) : +
1
2µ2 : φ2(x) :
+λ
4!: φ4(x) : (4.75)
73
S = + +
1
2
3
4
1
2
3
4
1
2
4
3
1
2
3
4
+
1
2
3
4
1
2
4
3
×
1
2
3
4
1
2
4
3
1
2
3
4
1
2
4
3
Figure 6: The lowest order diagrams of the S-matrix in φ4-theory.
74
In lowest order perturbation theory the S-matrix becomes equal to
〈~p3, ~p4| : U (1)(∞,−∞) : |~p1, ~p2〉 = −i λ4!
∫
d4x 〈~p3, ~p4| : φ4(x) : |~p1, ~p2〉 (4.76)
We bring all creation operators to the left and the annihilation operators to the right.This implies that all terms proportional with [a(~ki), a
†(~kj)] = δ(3)(~ki − ~kj) disappear(see page 63 1-5). The infinite contributions, represented by the bubble graphs, do notoccur. However the tadpole graphs also disappear. The result in our example is
Sfi = 〈~p3, ~p4|~p1, ~p2〉 − i λ(2π)4
(2π)6δ(4)(p1 + p2 − p3 − p4)
1√16E1E2 E3E4
(4.77)
The transition operator becomes
S = 1 − 2 π i δ(E1 + E2 − E3 −E4)
Tfi =λ
(2π)3δ(3)(~p1 + ~p2 − ~p3 − ~p4)
1√16E1E2 E3E4
(4.78)
For our convenience it is easier to define a Lorentz covariant transition operator M
S = 1 + (2π)4 δ(4)(p1 + p2 − p3 − p4)1
(2π)6
1√16E1E2 E3E4
M (4.79)
Up to lowest order in λ
M = −i λ (4.80)
We now calculate the differential cross section of the process
p1 + p2 → p3 + p4 (4.81)
The transition amplitude per unit volume per unit time is defined by
Wfi =|〈f |S − 1|i〉|2
V T
=1
V T
1
(2π)4
1
16E1E2E3E4
δ(4)(0) δ(4)(p1 + p2 − p3 − p4) |Mfi|2
(4.82)
We use the identities
δ(4)(p) =1
(2π)4
∫
d4x ei p x δ(4)(0) =1
(2π)4V T
∫
d4x =∫ T/2
−T/2dt
∫
d3x|~x| ≤ V
= T V |δ(4)(p)|2 = δ(4)(0) δ(4)(p) (4.83)
75
Wfi =1
(2π)8
1
16E1E2E3E4δ(4)(p1 + p2 − p3 − p4) |Mfi|2 (4.84)
De differential effective cross section is defined by
d σ =V ol
| ~Ji|dNf Wfi (4.85)
where | ~Ji| is the flux of the incoming particles (e.g. ~pi). V ol−1 is the number of particles
per volume unit in the target (e.g. ~p2). The number of particles which per time unitcollide through a plane on each other is given by
| ~Ji| =|~v1 − ~v2|V ol
(4.86)
for collinear process (~v1 ‖ ~v2). The V ol is determined by the normalisation of the planewaves which appear in φ(x)
φ(x) =∫
d3k
(2π)3/2
1√2ωk
[
a(~k) e−i k x + a†(~k) ei k x]
(4.87)
plane wave → 1
(2π)3/2e±i k x → V ol = (2π)3 (4.88)
dNf is the number of final states with momentum ~p3 in the volume element d3~p3 andmomentum ~p4 in the volume element d3~p4 therefore dNf = d3~p3 d
3~p4. Hence it follows
d σ =1
4E1E2 |~v1 − ~v2|d3~p3
2E3 (2π)3
d3~p4
2E4 (2π)3(2π)4 δ(4)(p1 + p2 − p3 − p4)
|Mfi|2 (4.89)
If ~v1 ‖ ~v2 (collinear collision) then is E1 E2 |~v1 −~v2| a Lorentz invariant. We shall provethis. We can put the direction of the collision along the z-axis
p1 = (E1, 0, 0, p1,z) p2 = (E2, 0, 0, p2,z)
~v1 =(
0, 0,p1,z
E1
)
~v2 =(
0, 0,p2,z
E2
)
(4.90)
E1E2 |~v1 − ~v2| = E2 p1,z − E1 p2,z
(p1 · p2)2 −m2
1m22 = E21 E
22 − 2 p1,z p2,z E1E2 + p2
1,z p22,z −m2
1m22
p21,z = E2
1 −m21 p2
2,z = E22 −m2
2
76
(p1 · p2)2 −m2
1m22 = E2
1 (p22,z +m2
2) − 2 p1,z p2,z E1E2 + p21,z (E2
2 −m22)
−m21 m
22
= E21 p
22,z − 2 p1,z p2,z E1E2 + p2
1,z E22 +m2
2 (E21 − p2
1,z −m21)
= (E1 p2,z − E2 p1,z)2
E1E2 |~v1 − ~v2| =√
(p1 · p2)2 −m21m
22 (4.91)
The last statement holds for any arbitrary frame. Further we have∫
d3~p
2E=∫
d4p θ(p0) δ(p2 −m2) (4.92)
so that the cross section becomes∫
d σ =1
4√
(p1 · p2)2 −m21m
22
∫
d4p3
(2π)3θ(p3,0) δ(p
23 −m2
3)∫
d4p4
(2π)3θ(p4,0)
δ(p24 −m2
4) (2π)4 δ(4)(p1 + p2 − p3 − p4) |Mfi|2 (4.93)
For example let us calculate the effective cross section for Mfi = −i λ. The centre ofmass energy is equal to
√s where the Mandelstam variable is defined as
s = (p1 + p2)2 (4.94)
Proof: The centre of mass is given by
~p1 + ~p2 = 0 → p1 + p2 = (ECM , 0, 0, 0)
ECM = E1 + E2 =√
|~p1|2 +m21| +
√
|~p2|2 +m22|
s = E2CM → √
s = ECM = E1 + E2 (4.95)
Further we can rewrite
p1 · p2 =1
2(s−m2
1 −m22) →
√
(p1 · p2)2 −m21, m
22 =
1
2λ1/2(s,m2
1, m22) (4.96)
The Kallen function is equal to
λ(x, y, z) = x2 + y2 + z2 − 2 x y − 2 x z − 2 y z (4.97)
Integrate∫
d σ over p4. Using the delta-function
∫
d σ =1
2 λ1/2(s,m21, m
22)
λ2
(2π)2
∫
d4p3 θ(p3,0) θ(p1,0 + p2,0 − p3,0) δ(p23 −m2
3)
δ(
(p1 + p2 − p3)2 −m2
4) (4.98)
77
We calculate the effective cross section in the centre of mass system (the z-axis is thedirection of the collision)
p1 = (E1, 0, 0, p1,z) p2 = (E2, 0, 0,−p1,z)
p3 = (E3, |~p3| sin θ cos φ, |~p3| sin θ sinφ, |~p3| cos θ)
p4 = (E4,−|~p3| sin θ cosφ,−|~p3| sin θ sinφ,−|~p3| cos θ) (4.99)
p21,z = E2
1 −m21 = E2
2 −m22 = (E1 −
√s)2 −m2
2
E1 =s+m2
1 −m22
2√s
E2 =s+m2
2 −m21
2√s
→ |~p1| = |~p2| =λ1/2(s,m2
1, m22)
2√s
(4.100)
The measure can be written as
d4p3 = |~p3|2 d |~p3| d cos θ d φ d p3,0 (4.101)
so that the cross section becomes
∫
d σ =λ2
8 π2 λ1/2(s,m21, m
22)
∫
d |~p3| d cos θ d φ d p3,0 |~p3|2 θ(p3,0) θ(√s− p3,0)
δ(p23,0 − |~p3|2 −m2
3) δ(s− 2√s p3,0 +m2
3 −m24)
=λ2
16 π2 λ1/2(s,m21, m
22)
∫
d φ d cos θ d |~p3| |~p3|2 θ(
s+m23 −m2
4
2√s
)
θ
(
s+m24 −m2
3
2√s
)
δ
(
|~p3|2 −λ(s,m2
3, m24)
4 s
)
(4.102)
Use a property of the delta-functions
δ(f(x)) =n∑
i=1
δ(x− ai)
|f ′(ai)|with f(ai) = 0 (4.103)
δ
(
|~p3|2 −λ(s,m2
3, m24)
4 s
)
=
√s
λ1/2(s,m23, m
24)
δ
(
|~p3| −λ1/2(s,m2
3, m24)
2√s
)
+δ
(
|~p3| +λ1/2(s,m2
3, m24)
2√s
)
(4.104)
78
The last term vanishes. Integration over |~p3| gives
∫
d σ =λ2
64 π2
λ1/2(s,m23, m
24)
λ1/2(s,m21, m
22)
1
sθ(s+m2
3 −m24) θ(s+m2
4 −m23)∫
d cos θ d φ
(4.105)
From the theta-functions we can conclude s ≥ 0. From the lambda-functions follows
λ(s,m21, m
22) =
s− (m1 +m2)2
s− (m1 −m2)2
λ(s,m23, m
24) =
s− (m3 +m4)2
s− (m3 −m4)2
(4.106)
s ≥ (m1 +m2)2 or s ≤ (m1 −m2)
2
s ≥ (m3 +m4)2 or s ≤ (m3 −m4)
2 (4.107)
Because of
√s =
√
|~p1|2 +m21 +
√
|~p1|2 +m22 > 0 ~p1 = −~p2 =⇒ √
s ≥ m1 +m2 (4.108)
In the same way because ~p1 + ~p2 = ~p3 + ~p4 = ~0 it also holds
√s =
√
|~p3|2 +m23 +
√
|~p4|2 +m24 > 0 =⇒ √
s ≥ m3 +m4 (4.109)
We have now the condition√s ≥ max(m1 +m2, m3 +m4) otherwise d σ = 0
d σ
dΩ=
λ2
64 π2 s
λ1/2(s,m23, m
24)
λ1/2(s,m21, m
22)
with dΩ = d cos θ d φ (4.110)
In the case of φ4-theory we have to deal with identical scalar bosons which have thesame mass µ. The differential cross section becomes
d σ
dΩ=
λ2
64 π2 s(4.111)
and the total effective cross section is equal to
σtot =∫
dΩd σ
dΩ=∫ 1
−1d cos θ
∫ 2π
0d φ
λ2
64 π2 s=
λ2
16 π s(4.112)
Identical particle effect
If the outgoing particles 3 and 4 are identical one cannot distinguish which particleappears in the detector provided one integrates over all angles. The measured totalcross section is then two times smaller than the calculated one.
σtot =λ2
32 π s(4.113)
79
1 3
2 4
1 4
2 3
Figure 7: The bubble graph contribution to the S-matrix in φ4-theory.
In general: Suppose we have a 2 → n process with ni identical particles of species iand
∑Ni=1 ni = n the the total cross section becomes
σtot =N∏
i=1
1
ni!
1√
(p1 · p2)2 −m21 m
22
n∏
j=1
∫
d3~kj
2Ej (2π)3(2π)4 δ(4)(p1 + p2 −
n∑
i=1
ki)
|Mfi|2 (4.114)
Besides the effective cross section there exists another interesting quantity i.e. thedecay width Γ. Suppose we have a decay process 1 → 3 + 4. This process can onlyhappen if m1 ≥ m3 +m4. The average life time τ1 of particle 1 is given by τ1 = 1/Γ1.The S-matrix for the decay 1 → 3 + 4 is given by
S = 1 + (2π)4 δ(4)(p1 − p3 − p4)1
(2π)9/2
1
8E1E3E4Mfi (4.115)
The transition probability is the equal to
Wfi =|〈f |S − 1|i〉|2
V T=
1
V T
1
2π
1
8E1E3E4δ(4)(0) δ(4)(p1 − p3 − p4) |Mfi|2
=1
(2π)5
1
8E1E3 E4δ(4)(p1 − p3 − p4) |Mfi|2 (4.116)
The decay width is defined by (V ol = (2π)3)
dΓ = V ol dNf Wfi =1
2E1
∫
d3~p3
2E3 (2π)3
∫
d3~p4
2E4 (2π)3(2π)4 δ(4)(p1 − p3 − p4)
|Mfi|2 (4.117)
In the case that there are ni identical particles in the final state we get again the extrafactor
∏Ni=1 1/ni!. Further the decay width will be always calculated in the rest frame
80
k1
k2
k3
k4
Figure 8: The bubble graph in φ4-theory.
of the decaying particle. From the latter follows E1 = m1. The generalisation of dΓfor 1 → n is trivial.
In the example above (φ4(x)-theory) we have calculated the S-matrix up to orderλ. The normal ordering procedure eliminates all contributions which arise by inter-changing creation and annihilation operators in a product of fields which were definedin the same space time point. For example φ4(x), φ2(x). However this not the end ofthe problems encountered in a relativistic quantum field theory. This arises if we go tosecond order in λ. The second order term in the S-matrix is
〈f |U (2)(∞,−∞)|i〉 = − 1
2!
∫ ∞
−∞dt1
∫ ∞
−∞dt2 〈f |T
(
: HI(t1) : : HI(t2) :)
|i〉
= − 1
2!
(
λ
4!
)2∫
d4x∫
d4y 〈f |T(
: φ4(x) : : φ4(y) :)
|i〉
(4.118)
〈f |T(
: φ4(x) : : φ4(y) :)
|i〉 = θ(x0 − y0) 〈f | : φ4(x) : : φ4(y) : |i〉
+ θ(y0 − xo) 〈f | : φ4(y) : : φ4(x) : |i〉 (4.119)
A long calculation shows that the following terms appear in the S-matrix The firstterm in 〈f |U (2)(∞,−∞)|i〉 is represent by the bubble graph in Fig. 7
−λ2
24(2π)4 δ(4)(0)
∫ d4k1
(2π)4
∫ d4k2
(2π)4
∫ d4k3
(2π)4
i
k1 − µ2 + i ε
i
k2 − µ2 + i ε
i
k3 − µ2 + i ε
i
(k1 + k2 + k3)2 − µ2 + i ε〈p3, p4|p1, p2〉 (4.120)
81
1 3
2 4
1 3
2 4
1 4
2 3
1 4
2 3
Figure 9: The self energy contribution to the S-matrix in φ4-theory.
This contribution is represented by the graph in Fig. 8
(−i λ)2
4!
4∏
i=1
∫
d4ki
(2π)4(2π)4 δ(4)(
4∑
j=1
kj)
(
i
k2l − µ2 + i ε
)
(4.121)
We see that even after normal ordering vacuum bubbles appear which are ultra-violetdivergent. This can be attributed to contractions of creation and annihilation operatorsbetween : φ4(x) : and : φ4(y) :. They also occur in
〈0|T(
: φ4(x) : : φ4(y) :)
|0〉 = −λ2
24(2π)4 δ(4)(0)
4∏
i=1
∫
d4ki
(2π)4δ(4)(
4∑
j=1
kj)
(
1
k2i − µ2 + i ε
)
(4.122)
Moreover this vacuum expectation value is infinite. The second term in 〈f |U (2)(∞,−∞)|i〉is a correction to the self energy. It occurs 4 times (see Fig. 9)
iλ2
6
2π
2E1δ(4)(p1 − p3) 〈p4|p2〉
∫
d4k1
(2π)4
∫
d4k2
(2π)4
1
k1 − µ2 + i ε
1
k2 − µ2 + i ε
1
(k1 + k2 + p3) − µ2 + i ε
+iλ2
6
2π
2E1
δ(4)(p1 − p3) 〈p3|p2〉∫
d4k1
(2π)4
∫
d4k2
(2π)4
1
k1 − µ2 + i ε
1
k2 − µ2 + i ε
1
(k1 + k2 + p4) − µ2 + i ε
+iλ2
6
2π
2E2δ(4)(p1 − p4) 〈p4|p1〉
∫ d4k1
(2π)4
∫ d4k2
(2π)4
1
k1 − µ2 + i ε
1
k2 − µ2 + i ε
82
k1
k2
k3
p3 p3
Figure 10: The self energy in φ4-theory.
1
(k1 + k2 + p3) − µ2 + i ε
+iλ2
6
2π
2E2
δ(4)(p2 − p3) 〈p3|p1〉∫
d4k1
(2π)4
∫
d4k2
(2π)4
1
k1 − µ2 + i ε
1
k2 − µ2 + i ε
1
(k1 + k2 + p4) − µ2 + i ε(4.123)
The expression for the self energy equals (see Fig. 10)
(−i λ)2
3!
3∏
l=1
∫
d4kl
(2π)4(2π)4 δ(4)
3∑
j=1
kj + p3
(
i
k2l − µ2 + i ε
)
(4.124)
these self energy graphs are divergent in the ultra-violet region. However they do notappear in 〈0|T (| : φ4(x) : : φ4(y) : |0〉 Watch the similarity between the contributionabove and the tadpole graph in Fig. 4. The tadpole graph disappeared after normalordering. This procedure does not work for self energy graphs. The last contributionto 〈f |U (2)(∞,−∞)|i〉 can be found in Fig. 11
1
2
λ2
(2π)2
1√16E1E2E3E4
δ(4)(p1 + p2 − p3 − p4)
[
∫
d4k1
(2π)4
i
k1 − µ2 + i ε
i
(k1 + p3 + p4)2 − µ2 + i ε
+∫
d4k1
(2π)4
i
k1 − µ2 + i ε
i
(k1 + p3 − p1)2 − µ2 + i ε
+∫ d4k1
(2π)4
i
k1 − µ2 + i ε
i
(k1 + p4 − p1)2 − µ2 + i ε
]
(4.125)
83
1 3
2 4
1 3
2 4
1 4
2 3a b c
Figure 11: The four point function contribution to the S-matrix in φ4-theory.
The graph in Fig. 11a is represented in Fig. 12.
−(−i λ)2
2!
2∏
l=1
∫
d4kl
(2π)4(2π)4 δ(4)
2∑
j=1
kj + p3 + p4
(
i
k2l − µ2 + i ε
)
(4.126)
These graphs are also ultraviolet divergent. The latter are the only ones which con-tribute toMfi. All integrals above are represented in momentum space. They are calledFeynman integrals. Before determining Mfi we will first deal with the proper defini-tion of the S-matrix wherein all vacuum bubbles vanish. Definition of the S-matrix:Definition I.
〈f |S|i〉I =〈f |Texp(−i ∫∞−∞ dt : HI(t) :)|i〉〈0|Texp(−i ∫∞−∞ dt : HI(t) :)|0〉 (4.127)
〈f |S|i〉I
=〈f |i〉 − i 〈f | ∫ dt1 : HI(t1) : i〉 − 1/2 〈f | ∫ dt1
∫
dt2 T ((: HI(t1) : : HI(t1) :)|i〉〈0|0〉 − i 〈0| ∫ dt1 : HI(t1) : |0〉 − 1/2 〈0| ∫ dt1
∫
dt2 T (: HI(t1) : : HI(t1) :)|0〉
=δfi + λ c1 + λ2 (a2 δfi + b2 + c2)
1 + λ2 a2= δfi + λ c1 + λ2 (b2 + c2) (4.128)
Notice that 〈0| ∫ dt1 : HI(t1) : |0〉 = 0.This follows from the assumption that the coupling constant λ is small. The various
contributions c1, a2, b2, c2 are shown in Fig. 13.
δfi + λ c1 + λ2(
a2 δfi + b2 + c2)
1 − a2 λ2
= δfi + λ c1 + λ2(
b2 + c2)
+ · · ·O(λ3) (4.129)
One can prove that in all orders the vacuum bubbles in the new S-matrix vanish.
84
p1 p3
p2 p4
k1
k2
Figure 12: The four point function contribution in φ4-theory.
c1= a2= b2= c2=
Figure 13: The various contributions to the S-matrix in φ4-theory.
85
1 3
2 4
1 4
2 3
1 3
2 4
1 3
2 4
1 4
2 3
1 4
2 3
1 3
2 4
1 3
2 4
1 4
2 3
1 3
2 4
1 3
2 4
1 4
2 3
1 4
2 3
1 3
2 4
1 3
2 4
1 4
2 3
1 4
2 3
1 3
2 4
1 3
2 4
1 4
2 3
1 4
2 3
1 3
2 4
1 3
2 4
1 3
2 4
1 3
2 4
1 3
2 4
1 3
2 4
1 4
2 3
Figure 14: The contributions to 〈f |S|i〉II up to order λ2 in φ4-theory.
86
1 3
2 4
1 3
2 4
1 3
2 4
1 3
2 4
1 3
2 4
1 3
2 4
1 4
2 3
Figure 15: Connected graphs in order λ2 contributing to Mfi in φ4-theory.
There are people who prefer no normal ordering. This is related to the renormal-isation procedure. We shall talk about this later on. Suppose we do not perform thenormal ordering procedure that there is a second definition for the S-matrix:Definition II
〈f |S|i〉II =〈f |Texp(−i ∫∞−∞ dtHI(t))|i〉〈0|Texp(−i ∫∞−∞ dtHI(t))|0〉
(4.130)
In the latter case all vacuum bubbles disappear but the tadpole graphs appear again.
〈f |S|i〉II
=〈f |i〉 − i 〈f | ∫ dt1HI(t1)i〉 − 1/2 〈f | ∫ dt1
∫
dt2 T ((HI(t1)HI(t1))|i〉〈0|0〉 − i 〈0| ∫ dt1HI(t1)|0〉 − 1/2 〈0| ∫ dt1
∫
dt2 T (HI(t1) : HI(t1))|0〉
(4.131)
but now −i 〈0| ∫ dt1HI(t1)|0〉 is equal to Fig. 5. Definition II has the following con-sequences for 〈f |S|i〉. How 〈f |S|i〉II looks like can be inferred up to order λ2 fromFig. 14. Definition I of the S-matrix (with normal ordering) follows from definitionII by omitting all tadpole graphs in Fig. 14. Diagrams with a closed line are calledloop diagrams. They represent the quantum fluctuations of the theory. The diagramswithout loops are the tree graphs. In the non relativistic limit the latter become equalto the classical result.The covariant transition matrix is given by all graphs which have the factor
87
pi∆F (p) = i
p2−µ2+i ε
p1
p3p2
p4
−i λ (2π)4 δ(4)(p1 + p2 + p3 + p4)
Figure 16: The Feynman rules for φ4-theory.
(2π)4
(2π)6
1√16E1E2 E3E4
δ(4)(p1 + p2 − p3 − p4) (4.132)
In lowest order in λ this given by the graph in Fig. 3. In next order in λ this is givenby the Feynman diagrams in Fig. 15. Notice that with normal ordering de diagramswith tadpoles disappear. The Feynman rules in φ4 theory for Mfi read (see Fig. 16)
1 Draw at a given order in λ all topological inequivalent connected diagrams.
2 Each internal line is represented by the factor i ∆F (p)
3 Each vertex where four lines with the momenta p1, p2, p3, p4 come together isrepresented by a factor −i λ. Here is imposed the condition that
∑4i=1 pi = 0
4 Integrate over every internal momentum k. The measure is equal to d4k/(2π)4
5 A combinatorial factor 1/n! if n internal lines start on one vertex and end onanother vertex.
In this way one obtains the expression for (2π)4 δ(4)(∑m
i=1 pi)Mfi.
4.3 Ultraviolet singularities and renormalization
While considering the transition amplitude Mfi we still encounter beyond lowest orderultraviolet divergences (see Fig. 15). Let us analyse where these divergences come
88
from and to remove them in a consistent way. Notice that the latter is only possible inso called renormalizable field theories where φ4-theory is an example. Let us thereforestudy the four point function in Fig. 12. In momentum space this function reads
Γ(4)(p1, · · · , p4) (2π)4 δ(4)
4∑
j=1
pj
=1
2λ2 (2π)4 δ(4)
4∑
j=1
pj
∫ d4k
(2π)4
1
k2 −m2 + i ǫ
1
(k + p1 + p2)2 −m2 + i ǫ(4.133)
We perform again a Wick rotation k0 = i k4, pj,0 = i pj,4 (j = 1 · · · 4) so that the fourpoint-function becomes
Γ(4)(p1, · · · , p4) = iλ2
2
∫
d4kE
(2π)4
1
k2E +m2
1
(kE + p1,E + p2,E)2 +m2(4.134)
Approximate the integrand for kE → ∞ then the integral looks like
−i λ2
16π2
∫ ∞
0
dkE
kE
(4.135)
which is logarithmic divergent at kE = ∞. The divergence at kE = 0 is artificialbecause we neglected masses. If there would be no masses then we also have aninfrared divergences at kE = 0. From the observations above we infer the following.The ultraviolet divergence corresponds with kE = ∞ (large momentum) whereas theinfrared divergence corresponds with kE = 0 (small momentum). Since we want tocompute the loop graphs we have to find a method to regularise these divergences.The only problem is that this procedure is arbitrary. This arbitrariness is reflectedby the introduction of a mass scale in the integral as we will show below. Amongthe methods to regularise the integrals we adopt n-dimensional regularization. It boilsdown to the reformulation of the field theory in n dimensions. The consequence is thatthe coupling constant acquires a dimension which is the scale mentioned above. Let usfirst determine the dimension of all quantities.
dim m = µ → dim x = µ−1 dim dnx = µ−n
dim S[φ] =∫
dnxL = 1 → dim L = µn
dim ∂µ = µ
dim µ2 φ2(x) = µn → dim φ(x) = µ(n−2)/2
dim λn φ4(x) = µn → dim λn = µ4−n λn = λµ4−n (4.136)
Because the Feynman propagator for massive theories is much more simple in momen-tum space than in normal space we will carry out all integrals in momentum space.The Feynman rules in below Eq. (4.132) have to be modified as follows
89
3. For every vertex in Fig. 16 we have to assign the factor
−i λ (2π)4 δ(4)(p1 + p2 + p3 + p4) → −i λn (2π)n δ(n)(p1 + p2 + p3 + p4)(4.137)
4. Integrate over each internal momentum k which has the measure
d4k
(2π)4→ dnk
(2π)n(4.138)
which together with the other rules mentioned in 1, 2, 5 at the end of the previoussection determines the scattering amplitude
Mfi (2π)n δ(n)
n∑
j=1
pj
(4.139)
As an example we compute the scattering amplitude Mfi for the reaction
p1 + p2 → p3 + p4 (4.140)
where the incoming scalar particles indicated by the momenta p1 and p2 belong to theinitial state |i〉 and the outgoing particles, indicated by the momenta p3 and p4 belongto the initial state 〈f |. Ignoring the tadpole graphs in the first row of Fig. 15 we haveup to order λ2 the scattering amplitude Mfi represented by the graphs in Figs. 3 and11 (definition I). It is given by (see Eq. (4.133))
Mfi = −i λn + J(s) + J(t) + J(u) (4.141)
with
J(s) =λ2
n
2
∫
dnk
(2π)n
1
k2 −m2 + i ǫ
1
(k + p1 + p2)2 −m2 + i ǫ(4.142)
If n < 4 this integral is defined. We will now calculate this expression by introducingFeynman parameters. This trick reads
1
aibj=
Γ(i+ j)
Γ(i) Γ(i)
∫ 1
0dx
xi−1 (1 − x)j−1
[a x+ b (1 − x)]i+j(4.143)
The Gamma function is defined as
Γ(x) =∫ ∞
0dt tx−1 e−t Γ(x+ 1) = xΓ(x)
Γ(n+ 1) = n! with n ∈ N (4.144)
Further Γ(x) has poles for x = −n with n ≥ 0. Using the Feynman parameter trickthe expression in Eq. (4.142) equals
J(s) =λ2
n
2
∫ dnk
(2π)n
∫ 1
0dx
1
[k2 + 2 x k · (p1 + p2) + x.(p1 + p2)2 −m2 + i ǫ]2(4.145)
90
Define
k′ = k + x(p1 + p2) s = (p1 + p2)2 (4.146)
J(s) =λ2
n
2
∫
dnk′
(2π)n
∫ 1
0dx
1
[k′2 + x (1 − x) s−m2 + i ǫ]2(4.147)
If a = m2 − x (1 − x) s > 0 we can drop i ǫ. Let us calculate integrals of the type
I1 =∫
dnk
(2π)n
1
k2 − awith a > 0 (4.148)
Perform a Wick rotation: k0 = i kE,n so that dnk = i dnkE and k2 = −k2E
I1 = −i∫ dnk
(2π)n
1
k2E + a
with dnkE =2 πn/2
Γ(n/2)kn−1
E dkE (4.149)
I1 = −i 2 πn/2
(2π)n Γ(n/2)
∫ ∞
0dkE
kn−1E
k2E + a
(4.150)
Substitute kE =√ax (a > 0)
I1 = −i πn/2
(2π)n Γ(n/2)an/2−1
∫ ∞
0dx xn/2−1 (1 + x)−1 (4.151)
Substitute x = y/(1− y)
I1 = −i πn/2
(2π)n Γ(n/2)an/2−1
∫ 1
0dy yn/2−1 (1 − y)−n/2 = −i π
n/2
(2π)nΓ(1 − n/2) an/2−1
(4.152)
Differentiate the expression above l − 1 times w.r.t. a we get
Il =∫
dnk
(2π)n
1
(k2 − a)l= (−1)l i
πn/2
(2π)n
Γ(l − n/2)
Γ(l)an/2−l (4.153)
For a theory with spin we also need the following integrals.∫ dnk
(2π)n
kµ
(k2 − a)l= 0
∫
dnk
(2π)n
kµ
(k2 + 2 k · p− a)l= (−1)l+1 i
πn/2
(2π)n
Γ(l − n/2)
Γ(l)(a+ p2)n/2−l pµ
∫
dnk
(2π)n
kµ kν
(k2 + 2 k · p− a)l= (−1)l i
πn/2
(2π)n
Γ(l − n/2 − 1)
Γ(l)
×
pµ pν (l − n/2 − 1) − 1
2gµν (a+ p2)
(a+ p2)n/2−l (4.154)
91
Notice that gµν gµν = n. For the case a = m2 − x (1 − x) s < 0 we have to include theeffect of i ǫ. The integral I1 in Eq. (4.150) becomes
I1 = −i 2 πn/2
(2π)n Γ(n/2)
∫ ∞
0dkE
kn−1E
k2E + a− i ǫ
=
−i 2 πn/2
(2π)n Γ(n/2)
P∫ ∞
0dkE
kn−1E
k2E + a
+ i π∫ ∞
0dkE k
n−1E δ(k2
E + a)
(4.155)
The first integral we already evaluated. The second integral becomes equal to∫ ∞
0dkE k
n−1E δ(k2
E + a) =1
2
∫ ∞
0dk2
E (k2E)n/2−1 δ(k2
E + a) =1
2(−a)n/2−1 (4.156)
Hence for general a we can write
I1 = −i πn/2
(2π)n
Γ(1 − n/2) |a|n/2−1 + iπ
Γ(n/2)(−a)n/2−1
∣
∣
∣
a<0
(4.157)
Differentiation w.r.t. a yields
Il = iπn/2
(2π)n(−1)l
Γ(l − n/2) |a|n/2−l + iπ
Γ(n/2 − l)(−a)n/2−l
∣
∣
∣
a<0
(4.158)
Let us first calculate J(s) in Eq. (4.147) for s < 0 (a > 0)
JI(s) = iλ2
n
2
πn/2
(2π)nΓ(2 − n/2)
∫ 1
0dx (m2 − x (1 − x) s)n/2−2 (4.159)
Since we are interested in the limit n→ 4 we expand de Gamma function around n−4
Γ(1 + x) = Γ(1) + xΓ′(1) +x2
2Γ
′′
(1) + · · · = 1 − γE x+1
2
(
γ2E +
π2
6
)
x2 + · · ·
= e−γE x
(
1 +π2
6x2 + · · ·
)
(4.160)
where γE = 0.5772156649 · · · is the Euler Mascheroni constant. Define ω = n− 4 thenλn = λ−ω and the integral equals
JI(s) = iλ2
32π2µ−ω
[
− 2
ω− γE + ln 4π −
∫ 1
0dx ln
(
m2 − x (1 − x) s
µ2
) ]
(4.161)
The calculation of the last integral is straightforward and we obtain
JI(s) = iλ2
32π2µ−ω
[
− 2
ω− γE + ln 4π − ln
m2
µ2+ 2
+√
1 − 4m2/s ln
√
1 − 4m2/s− 1√
1 − 4m2/s+ 1
]
(4.162)
92
For 0 < s < 4m2 we have to analytically continuate J(s)
√
1 − 4m2/s = i√
4m2/s− 1 (4.163)
JII(s) = iλ2
32π2µ−ω
[
− 2
ω− γE + ln 4π − ln
m2
µ2+ 2
−2√
4m2/s− 1 arctan
1√
4m2/s− 1
]
JI(o) = JII(0) (4.164)
For s > 4m2 (III) we have two options.a) We continuate JII(s) analytically to s > 4m2. This means that we take into accountthe i ǫ in J(s)
−2
√
4 (m2 − i ǫ)
s− 1 arctan
1√
4 (m2−i ǫ)s
− 1
=
√
1 − 4m2/s ln
1 −√
1 − 4m2/s
1 +√
1 − 4m2/s
− π i θ(s− 4m2)√
1 − 4m2/s (4.165)
b) We take into account the second term in formula (4.158)
i µ−2ω λ2
32π2[−i π
∫ 1
0dx θ(x (1 − x) s−m2)] =
i µ−2ω λ2
32π2
[
− i π θ(s− 4m2)∫ x2
x1
dx]
=i µ−2 ω λ2
32π2
[
− i π θ(s− 4m2)√
1 − 4m2/s]
(4.166)
In this way the whole expression for JIII(s) becomes
JIII(s) =i µ−ω λ2
32π2
[
− 2
ω− γE + ln 4 pi− ln
m2
µ2+ 2 +
√
1 − 4m2/s
ln
1 −√
1 − 4m2/s
1 +√
1 − 4m2/s
− i π θ(s− 4m2)√
1 − 4m2/s
]
(4.167)
Resuming
1. We have found a way to define a Feynman integral. This procedure is calledregularization.
2. It Appears that the Feynman integral is an analytical function which has cuts.
93
3. In the Feynman integral is a term that diverges for n → 4 (ω → 0). This termcannot contribute to physical quantities like effective cross sections.
Let us take for example de effective cross section of the process p1 +p2 → p3 +p4 calcu-lated in higher order in λ. We take definition 1 for the S-matrix because here no tadpoleoccur. In lowest order in λ we have for Mfi the diagram in Fig. 3. This diagram isenhanced in order λ2 by the the diagrams in Fig. 11. Definition: Mandelstam variables
s = (p1 + p2)2 t = (p1 − p3)
3 u = (p1 − p4)2 s+ t+ u = 4m2 (4.168)
|~p1| = |~p2| = |~p3| = |~p4| =1
2sqrts− 4m2 (4.169)
In the physical region
s = 4 (|~p1|2 +m2) ≥ 4m2 t = −2 |~p1|2 (1 − cos θ) ≤ 0
u = −2 |~p1|2 (1 + cos θ) ≤ 0 (4.170)
Mfi = −i λ µ−ε
[
1 +3 λ
32π2
2
ω+ γE − ln 4π + ln
m2
µ2− 1
3FIII(s) −
1
3FI(t)
−1
3FI(u)
]
(4.171)
Watch that FIII(s) has a real as well as an imaginary part because s ≥ 4m2. HoweverFI(t) and FI(u) have no imaginary parts since t < 0 and u < 0.
|Mfi|2 = λ2 µ−2 ε
[
1 +3 λ
16 π2
2
ω+ γE − ln 4 π + ln
m2
µ2− 1
3ℜFIII(s) −
1
3FI(t)
−1
3FI(u)
]
(4.172)
Therefore the cross section becomes
d σ =1
4
1√
(p1.p2)2 −m21m
22
∫ d4p3
(2π)3θ(p3,0) δ(p
23 −m2
3)∫ d3p4
(2π)3θ(p4,0) δ(p
24 −m2
4)
(2π)4 δ(4)(p1 + p2 − p3 − p4) |Mfi|2 (4.173)
d σ
dΩ=
λ2 µ−2 ε
64π2s
[
1 +3λ
16π2
2
ω+ γE − ln 4π + ln
m2
µ2− 1
3ℜFIII(s) −
1
3FI(t)
−1
3FI(u)
]
(4.174)
94
If we assume that this effective cross section describes the experimental data can λbe no physical parameter because 2/ω → ∞ when n → 4. In order to observe aphysical coupling constant we have to define a renormalized coupling constant. Thisis called renormalisation of the coupling constant. This is done as follows Define therenormalized coupling constant λR(i) by
λ = λR(i)
[
1 − 3λr(i)
32π2
2
ω+ Ci
+O(λ2R(i))
]
(4.175)
The renormalized cross section becomes now
d σR
dΩ=
λ2R(i)
64π2s
[
1 +3λR(i)
16π2
γE − ln 4π − Ci + lnm2
µ2− 1
3
(
ℜFIII(s) + FI(t)
+FI(u))
+O(λ2R(i))
]
(4.176)
The arbitrary constant Ci indicates that the perturbation series for d σR is renormal-ization dependent. This dependence is indicated by the parameter i. However natureis not sensitive to the renormalisation choice. Remind that the coupling constant isnot given by the theory. We have to determine it experimentally. The cross section issymmetric in t and u or equivalently cos θ ⇔ − cos θ. If we fit the cross section to thedata then we get a whole list of values for λR(i) that depends on Ci. We have to watchthe following.
a. The Ci cannot be chosen so that λR(i) becomes too large (condition λR(i) ≪ 1)otherwise the perturbation series diverges.
b. It can happen that the form of the curve disagrees with the measured values. Inthis case one has to take higher powers in λR(i) into account.
c. It is also possible that the form never agrees with the experimental data. Higherorder terms cannot bring agreement. One has to take λR(i) so large that one canget a conflict with a.. The conclusion is then that the theory is wrong.
d. However even if the cross section is in agreement with the data we are not there.We have to show that for any arbitrary other process the data can be accuratelypredicted with the same renormalisation scheme i.
If we use the same renormalization scheme as in the previous process then we can usethe value for the coupling constant λR(i) in the new process. In this we can give anabsolute prediction for the latter process.
RemarkWe can absorb the constant Ci in µ2 by defining µ2 = µ2
i exp(−Ci). In the literature one
95
sees instead of λR(i) the coupling constant λR(µ2i ) or simply λR(µ2) (µ2 is a continuous
parameter). We can now renormalize in various ways
Ci = lnµ2
i
µ2
λ = λR(µ2i )
[
1 − 3λR(µ2i )
32π2
2
ω+ ln
µ2i
µ2
]
= λR(µ2j)
[
1 − 3λR(µ2j)
32π2
2
ω+ ln
µ2j
µ2
]
(4.177)
so that we can express λR(µ2j) into λR(µ2
i ) by
λR(µ2j) =
λR(µ2i )
1 − (3λR(µ2i )/32π2) (lnµ2
j/µ2i )
(4.178)
Take µi fixed and vary µj. Let µ2j → 0 then λR(µ2
j) → 0. The zero is IR stable point.Besides coupling constant renormalization there exists also mass renormalization. Inlowest order this given by the diagrams in the first row of Fig. 15. Finally we want tomake another remark. Perturbation theory will not finally determine coupling constantrenormalization. Therefore the coupling constant is unknown. More information isobtained by non-perturbative arguments for instance lattice calculations.
λ = λR(µ)Zλ
(
Λ
µ
)
(4.179)
Λ is a cut off or an inverse lattice distance in lattice theories. Suppose λ is finite theneither
limΛ → ∞ Zλ = finite for λR 6= 0
(4.180)
or
limΛ → ∞ Zλ = infinite for λR = 0
(4.181)
Non-perturbative arguments in φ4-theory lead to the result that λR = 0.
96
5 Spontaneous symmetry breaking
5.1 Nambu-Goldstone mechanism
Goldstone theorem
If the Lagrange density L is invariant under the global symmetry group G (n-parametergroup) whereas the Hilbert space H is invariant under the subgroup H ⊂ G (H is am-parameter group) then there are n-m massless (Goldstone) spin zero bosons in thespectrum of H.
Suppose we have the following Lagrange density which is globally invariant under thegroup SU(2).
L = i ψi(x) γµ ∂µ ψi(x) −M ψi(x)ψi(x) +
1
2∂φa(x) ∂
µφa(x) −1
2µ2 |φa(x)|2
+g ψi(x) (τa)ij ψj(x)φa(x) −λ
4|φa(x)|2 |φb(x)|2 (5.1)
The Dirac fields ψi are in the fundamental representation i = 1, 2 and the scalar fieldsφa are in the adjoint representation a = ±, 0.
~ψ =
(
ψ1
ψ2
)
~φ =
φ1
φ2
φ3
=
1√2(φ+ + φ−)
i√2(φ− − φ+)
φ0
(5.2)
~φ+ =1√2
φi φ0
~φ− =1√2
φ−i φ
0
~φ0 =
00φ
(5.3)
~ψ is a doublet and ~φ is a triplet.
φ0 = φ3 φ± =1√2
(φ1 ± i φ2)
|φa|2 = φ21 + φ2
2 + φ23 = 2φ+ φ− + φ2
0 (5.4)
Under the group the fields transform as
ψ′k(x) =[
exp(−i αa τa)]
klψl(x) φ′k(x) =
[
exp(−i αa Ta)]
klφl(x) (5.5)
with the matrices
τ+ =1
2(τ1 + i τ2) τ− =
1
2(τ1 − i τ2) τ0 = τ3
T+ =1√2
(T1 + i T2) T− =1√2
(T1 − i T2) T0 = T3 (5.6)
97
τ+ =
(
0 10 0
)
τ− =
(
0 01 0
)
τ0 =
(
1 00 −1
)
(5.7)
T+ =1√2
0 0 −10 0 −i1 i 0
T− =1√2
0 0 10 0 −i−1 i 0
T0 = i
0 −1 01 0 00 0 0
(5.8)
Further we have
T+~φ+ = 0 T+
~φ0 = −~φ+ T+~φ− = ~φ0
T0~φ+ = ~φ+ T0
~φ0 = 0 T0~φ− = −~φ−
T− ~φ+ = −~φ0 T− ~φ0 = ~φ− T− ~φ− = 0 (5.9)
and
τa φa = τ1 φ1 + τ2 φ2 + τ3 φ3 = τ0 φ0 + τ+ φ− + τ− φ+ (5.10)
The Noether current equals (i, j = 1, 2, a, b, c = 1, 2, 3)
jµa (x) = ψ(x)i γ
µ(
τa)
ijψj(x) + i φb(x)
†↔∂µ
(
Ta)bc φc(x) (5.11)
The Lagrangian above has the global symmetry group SU(2). Now we have to constructa Hilbert space H such that the fields are invariant under U(1). Let us consider thepotential
V (|~φ|) =1
2µ2 |~φ|2 +
λ
4|~φ|4 (5.12)
with λ > 0 (V (|φ|) has be bounded from below). There are now two possibilities
A µ2 > 0: All scalar fields are massive. Moreover H is symmetric under G = SU(2).
Further 〈0|~Φ|0〉 = 0
B µ2 < 0: One field is massive and two fields are massless. Moreover H is symmetric
under G = U(1). Further 〈0|~Φ|0〉 =√
−µ2/λ.
The minimum of the potential in case B is given by ∂V (|~φ|)/∂|~φ| = 0. From which
follows |~φ| =√
−µ2/λ. Notice that all solutions ~φ(x) for which |~φ| =√
−µ2/λ = v
lead to a minimum of V (|~φ|). From case B we can derive the following properties ofthe vacuum |0〉
a |〈0|~Φ|0〉| =√
−µ2/λ
98
|~φ|
µ2 > 0
V (|~φ|)
Figure 17: Case µ2 > 0.
|~φ|
µ2 < 0
V (|~φ|)
√
−µ2/λ
−µ4/4λ
Figure 18: Case µ2 < 0.
99
b |0〉 is not anymore invariant under G but under the subgroup H . The latter iscalled the stabiliser of |0〉.
U(g)|0〉 6= |0〉 g ∈ G g 6∈ H but U(h)|0〉 = |0〉 h ∈ H (5.13)
c From b it follows that there an infinite number of vacuums |0〉i with U(g)|0〉i =|0〉j (i 6= j) but with U(h)|0〉i = |0〉i
d i〈0|0〉i = ∞
Proof:
a:This is evident. The vacuum is always given by the expectation value of Φ for whichV (|〈0|~Φ|0〉|) is a minimum.b:Define 〈0|~Φ(x)|0〉 = ~v with |~v| =
√
−µ2/λ. The vector ~v is invariant under H but not
invariant under g ∈ G with g 6∈ H . Hence it follows that the ground state |0〉 is notinvariant under U(g) with g 6∈ H but invariant under U(h) with h ∈ H . This meansU(g)|0〉 6= |0〉 with g 6∈ H but it is invariant under U(h)|0〉 = |0〉 with h ∈ H . Thiscan be shown in the following way. The Wigner theorem states
U(g)† ~Φ(x) U(g) = e−i αa Ta ~Φ(x) (5.14)
with U(g) is unitary. Further is ~Φ no singlet under the transformation e−i αa Ta in otherwords
U(
e−i αa Ta
)† ~Φ(x) U(
e−i αa Ta
)
= e−i αa Ta ~Φ(x) 6= ~Φ(x) (5.15)
From Eq. (5.14) it follows that
〈U(g) 0|~Φ(x)|U(g) 0〉 = e−i αa Ta 〈0|~Φ(x)|0〉 (5.16)
Suppose U(g)|0〉 = |0〉. In this case we have
〈0|~Φ(x)|0〉 = e−i αa Ta 〈0|~Φ(x)|0〉 e−i αa Ta = g ∈ G and g 6∈ H
=⇒ 〈0|~Φ(x)|0〉 = 0 (5.17)
because 〈0|~Φ(x)|0〉 is no singlet under the group. But that contradicts 〈0|~Φ(x)|0〉 =~v 6= 0. Therefore U(g)|0〉 = |0〉 cannot be true. Conclusion
U(g)|0〉 = |0〉′ 6= |0〉 (5.18)
However
〈0|~Φ(x)|0〉 = e−i αa Ta 〈0|~Φ(x)|0〉 if e−i αa Ta = h ∈ H =⇒ U(h)|0〉 = |0〉 (5.19)
100
c:Because there exist an infinite number of transformations g = exp(−i αa Ta) there mustalso exist an infinite number of vacuums |0〉i which are transformed in each other. Inother words i in
i〈0|~Φ|0〉i = ~v (5.20)
is infinited:
U(g) = U(
e−i αa Ta
)
= e−i αa D(Ta) D(Ta)† = D(Ta) because U−1 = U † (5.21)
D(Ta) is hermitian and a representation of the generator Ta in the Hilbert space.Further is
D(Ta) =∫
d3~x J0a (x) (5.22)
Jµa is a quantum Noether current in Hilbert space. We show that i〈0|0〉i = ∞. FirstD(Ta)|0〉i = |0〉j for arbitrary a (notice if Ta ∈ L(H) then D(Ta)|0〉i = 0).
j〈0|0〉j =i 〈0|D(Ta) D(Ta)|0〉i =∫
d3~x i〈0|J0a(x) D(Ta)|0〉i (5.23)
translational invariance
J0a (x) = e−i P x J0
a(0) ei P x (5.24)
Further[
P , D(Ta)]
= 0 because[
P , U(g)]
= 0 (5.25)
j〈0|0〉j =∫
d3~x i〈0|J0a(0) D(Ta)|0〉i = ∞ (5.26)
Conclusion: the space characterised by i〈0|~Φ|0〉i 6= 0 is not a good Hilbert space (Foran unbroken symmetry is D(Ta)|0〉 = 0 therefore the above argument does not hold).We construct a good Hilbert space which is invariant under H. As vacuum we chooseone of the old vacuums |0〉i. Which i we take does not matter.Further we define a new scalar quantum field
~Φ(x) = ~Φ(x) − ~v 1 (5.27)
such that
i〈0|~Φ|0〉i = 0 U(h)|0〉i = |0〉i (5.28)
101
The classical field~φ(x) which corresponds to ~Φ(x) is equal to
~φ(x) = ~φ(x) − ~v. Ex-
pressed in the new fields ~φ(x) the Lagrange density becomes
L = i ψk γµ ∂µ ψk −M ψk ψk +
1
2∂µ~φ · ∂µ~φ+ g ψi (τa)ij ψj va + g ψi(τa)ijψj φa
−1
2µ2 |~φ|2 − µ2 ~φ · ~v − 1
2µ2 |~v|2 − λ
4
|~φ|2 + 2~v · ~φ+ |~v|22 |µ2=−λ |~v2|
= i ψk γµ ∂µ ψk − ψi
(
M δij − g (~τ · ~v)ij
)
ψj +1
2∂µ~φ · ∂µ~φ− λ
(
~v · ~φ)2
−λ4|~φ|4 − λ~v · ~φ |~φ|2 +
λ
4|~v|4 (5.29)
The minimum of the potential V (|~φ|) is now −λ/4 |~v|4 for |~φ| = 0. Hence it follows
that 〈0|~Φ|0〉 = 0 (there is one vacuum |0〉i = |0〉 for a specific i). In our example was
the field ~φ0 invariant under H = U(1) with T0 ∈ L(U(1)).
~v =
00v
~φ0 =
00
φ
~φ =
1√2(φ+ + φ−)
i√2(φ− − φ+)
φ0
(5.30)
From which follows T0 ~v = 0 and exp(−i α0 T0)~v = ~v. Further exp(−i α0 T0)~φ0 =
~φ0.
The consequence is that ~v · ~φ = v φ0 and φ0 has a mass 2 λ v2 = −2µ2 > 0. φ+ and φ−are the massless fields and they represent the Goldstone bosons. For the fermions wehave the mass matrix
M =
(
M − g v 00 M + g v
)
~ψ =
(
ψ1
ψ2
)
(5.31)
ψ1 and ψ2 have mass M − g v and M + g v respectively. Therefore the degeneracy islifted.
Conclusions:
a For case II the Hilbert space H is not invariant under the group G (here SU(2))but under its subgroup H (here U(1)). If dim G= n and dim H= m then thenumber of Goldstone bosons is equal to dim G/H= n−m.
b The Lagrange density L expressed in ~φ is still invariant under G!! The replace-
ment of ~φ by~φ+~v does not break the symmetry. We have still conserved Noether
currents jµa for all a.
102
c The fields which are left invariant by the subgroup H are those fields which de-scribe massive particles. The fields which belong to the co-set G/H are massless(Goldstone bosons). Notice that G/H is not a group unless H is a normal sub-group and the Goldstone bosons are therefore not representations of G/H .
The Goldstone theorem is only correct for global symmetries. In the case of local sym-metries the Goldstone bosons vanish and they appear as the longitudinal componentof the massive gauge boson. The massive scalar field describes then the Higgs boson.
5.2 Higgs mechanism
If the symmetry group G is local then the Noether currents disappear and the argu-ments which leads to the Goldstone mechanism does hold anymore. That this happensis shown by P. Higgs. We illustrate this via a simple example. Choose the local gaugegroup U(1). we take the Lagrangian density which is invariant under U(1)
L = −1
4Fµν(x)F
µν(x) +(
Dµ φ(x))∗Dµ φ(x) − µ2 φ∗(x)φ(x) − λ
(
φ∗(x)φ(x))2
(5.32)
with
Dµ φ(x) =(
∂µ + i eAµ(x))
φ(x) (Dµ φ(x))∗ =(
∂µ − i eAµ(x))
φ∗(x) (5.33)
Further the Lagrangian density is invariant under the transformations
φ′(x) = ei e α(x) φ(x) φ′∗(x) = e−i e α(x) φ(x)∗ (5.34)
Let us express the complex fields into real fields
φ(x) =1√2
(φ1(x) + i φ2(x)) φ(x)∗ =1√2
(φ1(x) − i φ2(x)) (5.35)
Under gauge transformations the real fields transform like
φ′1(x) = cos(e α(x))φ1(x) − sin(e α(x))φ2(x)
φ′2(x) = sin(e α(x))φ1(x) + cos(e α(x))φ2(x) (5.36)
For λ > 0 and µ2 < 0 we get again spontaneous symmetry breaking. The minimum ofthe potential is at
|φ(x)| =
√
−µ2
2λ=
v√2
|φ(x)| =1√2
√
φ1(x)2 + φ2(x)2 (5.37)
There are again an infinite number of degenerated vacuum which are characterised by
〈0|Φ(x)|0〉 =v√2ei δ (5.38)
103
for arbitrary δ. Let us choose a specific vacuum en build around that vacuum a newquantum theory. Since the symmetry group is U(1) is totally broken it does not matterwhich vacuum is chosen. There remains one Goldstone boson. Choose δ = 0 then
〈0|Φ1(x)|0〉 = v and 〈0|Φ2(x)|0〉 = 0
φ1(x) = φ1(x) − v φ2(x) = φ2(x) (5.39)
φ2 becomes the Goldstone boson field. the Lagrange density expressed in the new fieldsreads
L = −1
4Fµν F
µν +1
2∂µ φ1 ∂
µ φ1 +1
2∂µ φ2 ∂
µ φ2 +1
2e2Aµ A
µ (φ21 + φ2
2)
+eAµ (φ1 ∂µ φ2 − φ2 ∂µ φ1) −µ2
2(φ2
1 + φ22) −
λ
4(φ2
1 + φ22)
2
= −1
4Fµν F
µν +1
2∂µ φ1 ∂
µ φ1 +1
2∂µ φ2 ∂
µ φ2 +1
2e2 v2AµA
µ
+1
2e2AµA
µ (φ21 + φ2
2) + e2 v AµAµ φ1 + e v Aµ ∂µ φ2 + eAµ (φ1 ∂µ φ2
−φ2 ∂µ φ1) −λ
4(φ2
1 + φ22)
2 − λ v φ1 (φ21 + φ2
2) −µ2
2(φ2
1 + φ22) − µ2 v φ1
−1
2µ2 v2 − λ v2 φ2
1 −λ
4v4 − λ
2v2 (φ2
1 + φ22) − λ v3 φ1 (5.40)
After substitution of µ2 = −λ v2 all terms linear in φ1 disappear and we have
L = −1
4Fµν F
µν +1
2∂µ φ1 ∂
µ φ1 +1
2e2 v2 (Aµ +
1
ev∂µ φ2) (Aµ +
1
ev∂µ φ2)
+1
2e2 AµA
µ (φ21 + φ2
2) + e2 v Aµ Aµ φ1 + eAµ (φ1 ∂µ φ2 − φ2 ∂µ φ1) + µ2 φ2
1
+λ
4v4 − λ
4(φ2
1 + φ22)
2 − λ v φ1 (φ21 + φ2
2) (5.41)
For Aµ = 0 we have a field φ1 with mass −2µ2 (µ2 < 0) and a massless Goldstone fieldφ2.For Aµ 6= 0 we have two possibilities
1 A massless gauge field Aµ, a massless Goldstone field φ2 and massive field φ1.This is called the Goldstone mode.
2 A massive gauge field Bµ = Aµ + 1/ev ∂µ φ2 with mass e v and a massive scalarfield φ1 with mass −2µ2. This is called the Higgs mode. The massive scalar field
104
is called the Higgs field. The vanished massless scalar field φ2 is called the wouldbe Goldstone field.
The most people adopt the second point of view. In the Higgs mode is there a cancel-lation between the scalar polarisation of the gauge field and the would be Goldstonefield. In this case only the three polarisations (two transversal and one longitudinal)of the massive gauge field contribute. We shall now rewrite the Lagrangian density.Define the massive vector field
Bµ = Aµ +1
e v∂µ φ2 ∂µ Bν − ∂ν Bµ = ∂µ Aν − ∂ν Aµ = Fµν (5.42)
The Lagrangian density becomes
L = −1
4Fµν F
µν +1
2∂µ φ1 ∂
µ φ1 +1
2e2 v2BµB
µ +1
2e2Bµ B
µ (φ21 + φ2
2)
−evBµ ∂
µ φ2 (φ21 + φ2
2) + e2 v Bµ Bµ φ1 − 2 eBµ ∂
µ φ2 φ1 + eBµ (φ1 ∂µ φ2
−φ2 ∂µ φ1) +1
2 v2∂µ φ2 ∂
µ φ2 (φ21 + φ2
2) + µ2 φ21 +
λ
4v4 − λ
4(φ2
1 + φ22)
2
−λ v φ1 (φ21 + φ2
2) +1
vφ2 ∂µ φ2 ∂
µ φ1 (5.43)
with mB = e v th mass of the vector boson and m2H = −2µ2 the mass of the Higgs
boson. In this case the field φ2 has become unphysical. Notice that the symmetry isstill present in L it is only hidden. Only the Hilbert space is not symmetric anymore.It is now convenient to rewrite the Lagrange density such that only physical fieldsoccur. There is one to one correspondence between the classical and the quantumfields. Because we can choose a gauge such that that φ2 in the Lagrange densitydisappears. This is achieved in the following way. Define polar fields
ρ(x) =√
φ21(x) + φ2
2(x) θ(x) = arctanφ2(x)
φ1(x)(5.44)
φ(x) =1√2ρ(x) ei θ(x) φ∗(x) =
1√2ρ(x) e−i θ(x) (5.45)
or
φ1(x) = ρ(x) cos θ(x) φ2(x) = ρ(x) sin θ(x) (5.46)
The choice
〈0|Φ1(x)|0〉 = v and 〈0|Φ2(x)|0〉 = 0 (5.47)
105
corresponds with
〈0|P(x)|0〉 = v and〈0|Θ(x)|0〉 = 0 (5.48)
Therefore P(x) describes the Higgs field and Θ(x) the would be Goldstone field. Writenow
φ(x) =1√2
(
ρ(x) + v)
ei θ(x) φ∗(x) =1√2
(
ρ(x) + v)
e−i θ(x) (5.49)
and the Lagrange density expressed into the new field becomes
L = −1
4Fµν F
µν +1
2∂µ ρ ∂µ ρ+
e2
2
(
ρ+ v)2 (
Aµ +1
e∂µ θ
) (
Aµ +1
e∂µ θ
)
+ µ2 ρ2
−λ4ρ4 − λ v ρ3 +
1
4λ v4 (5.50)
Define again
Bµ(x) = Aµ(x) +1
e∂µ θ(x) (5.51)
In this way we remove θ(x) from the Lagrangian density. This can be also achieved bythe following gauge transformation
φ′(x) = e−i θ(x) φ(x) → ρ′(x) = e−i θ(x)(
ρ(x) ei θ(x))
= e−i θ(x)(
ρ(x) + v)
ei θ(x)
= ρ(x) + v ρ(x) = ρ(x) + v (5.52)
A′µ(x) = Aµ(x) = Aµ(x) +1
e∂µ θ(x) (5.53)
Notice that this gauge transformation is not arbitrary because α(x) = θ(x). Hence itis a gauge choice. It makes Aµ(x) massive. Finally the Lagrange density becomes inthis gauge
L = −1
4Fµν F
µν +1
2m2
A Aµ Aµ +
1
2∂µ ρ ∂µ ρ+
e2
2ρ2 Aµ A
µ + v e2 ρ Aµ Aµ − 1
2m2
H ρ2
−λ4ρ4 − λ v ρ3 +
1
4λ v4 (5.54)
The masses of the vector boson and the Higgs boson are
mA = e v m2H = −2µ2 = 2 λ v2 (5.55)
respectively.
106
5.3 Landau-Ginzburg theory
We will derive the Landau-Gimzburg Hamiltonian using the partition function for anferromagnet. In order to define the model we consider a lattice, in any number ofdimensions n, and on each site we place a two-valued variable si = ±1. The latticespacing is a. To a given configuration of the spins, of which there are 2N (where N isthe number of sites) we ascribe the energy
Esi = −∑
i,j
1
2Vij si sj −
∑
i
hi si (5.56)
For a ferromagnet Vij > 0 over the range a. Therefore we get lower energies to con-figurations with parallel spins (equal sign). The variable hi is the value of an externalmagnetic field at the site i. Hence the partition function is given by
Z[Hi] =∑
si
expβ
2
∑
i,j
Vij si sj +∑
i
Hi si
(5.57)
where β = 1/kT and Hi = β hi. In this discrete model Z[Hi] generates all the correla-tion functions. The average magnetisation at site i is
Mi = 〈si〉Hi=0 = Z−1 ∂ Z
∂ Hi|Hi=0 (5.58)
UsuallyMi = 0 atHi = 0 because of the symmetry in the energy Esi under si → −si.Unless one encounters spontaneous symmetry breaking. The susceptibility is given by
χ =∂ Mi
∂ Hi
|Hi=0=∑
i,j
(
〈si sj〉 − 〈si〉 〈sj〉)
=∑
i,j
〈(
si − 〈si〉) (
sj − 〈sj〉)
〉 (5.59)
The mean energy is equal to
〈Ei〉 = 〈−1
2Vij si sj〉 = −∂ lnZ[0]
∂ β(5.60)
and the specific heat
Cv =∂ 〈Ei〉∂ T
= −k β2 ∂ 〈Ei〉∂ β
= k β2 ∂2 lnZ[0]
∂ β2(5.61)
Insertion of the partition function (see Eq. (5.56)) provides us with the result
Cv = k β2
[
〈(1
2Vij si sj
)(1
2Vkl sk sl
)
〉 − 〈12Vij si sj〉2
]
= k β2
[
∑
ik
〈(
Ei − 〈Ei〉) (
Ek − 〈Ek〉)
〉]
(5.62)
107
Put Hi = 0 in Eq. (5.56) and use the following identity
N∏
i=1
∫ ∞
−∞dφi exp−1
2
∑
i,j
φi (β Vij)−1 φj +
∑
i
φi si
= (2π)N/2 (det βV )1/2 expβ2
∑
i,j
Vij si sj (5.63)
Hence
Z[0] =∑
si
expβ2
∑
i,j
Vij si sj = (2π)−N/2 (det βV )−1/2∑
si
N∏
i=1
∫ ∞
−∞dφi
exp−1
2
∑
i,j
φi (β Vij)−1 φj +
∑
i
φi si (5.64)
Further we have∑
si
exp∑
i
φi si =∏
i
(
expφi + exp−φi)
=∏
i
2 coshφi
= 2N exp∑
i
ln(cosh φi) (5.65)
Hence
Z[0] = 2N (2π)−N/2 (det βV )−1/2
×N∏
i=1
∫ ∞
−∞dφi exp−1
2
∑
i,j
φi (β Vij)−1 φj +
∑
i
ln(cosh φi) (5.66)
Expand
ln coshφi =1
2φ2
i −1
12φ4
i + · · · =1
2φ2
i −∞∑
k=2
M2k
(2k)!φ2k
i (5.67)
The partition function becomes
Z[0] = 2N (2π)−N/2 (det βV )−1/2
×N∏
i=1
∫ ∞
−∞dφi exp
− 1
2
∑
i,j
φi (β Vij)−1 φj +
1
2
∑
i
φ2i −
∞∑
k=2
M2k
(2k)!φ2k
i
(5.68)
We now go in the limit of long wavelength approximation. In this limit it is legitimateto replace
∑
i
=∫
dnx
anφ(x) = a−(n−2)/2 φi λ2k = an(k−1)−2k M2k (5.69)
108
so that Z[0] becomes
Z[0] = C∫
Dφ(x) exp
∫
dnx[
− a−2
2φ(x) (β V (x))−1 φ(x) +
a−2
2φ2(x)
−∞∑
k=2
λ2k
(2k)!φ2k(x)
]
(5.70)
where C is an arbitrary constant. To see this we go from position space to momentumspace
Z[0] = C∫
Dφ(q) exp
∫
dnq[
− a−2
2φ(q) (β V (q))−1 φ(q) +
a−2
2φ2(q)
−∞∑
k=2
λ2k
(2k)!φ2k(q)
]
(5.71)
Since V (q) > 0 over a range a we have maxV (q) = V (0) and for qa ≪ 1 we mayexpand according to V (q) = V (0)(1− (qa)2 + · · ·). Insertion in (5.70) provides us withthe form
Z[0] = C∫
Dφ(q) exp
∫
dnq[
− a−2
2φ(q)
1 − βV (0) + βV (0) (qa)2
βV (0)φ(q)
−∞∑
k=2
λ2k
(2k)!φ2k(q)
]
(5.72)
where we neglected higher powers in qa in numerator and denominator. Define
µ2 =1 − βV (0)
βV (0)a−2 (5.73)
We get
Z[0] = C∫
Dφ(q) exp
∫
dnq[
− 1
2φ(q) q2 φ(q) − µ2
2φ2(q) −
∞∑
k=2
λ2k
(2k)!φ2k(q)
]
(5.74)
or in position space
Z[0] = C∫
Dφ(x) exp
∫
dnx[
− 1
2∇φ(x) · ∇φ(x) − µ2
2φ2(x) −
∞∑
k=2
λ2k
(2k)!φ2k(x)
]
=∫
Dφ(x) exp
−∫
dnxL(φ,∇φ)
(5.75)
109
So the above expression can be used to calculate fluctuations of wave number q suchthat qa ≪ 1. It is assumed that the singular part of G ∼ − lnZ[0] (which is the freeenergy) comes from the low q part and this is the region where the second order phasetransition occurs. The argument of the exponent is the Lagrangian but since we arein Euclidian space it is also called Hamiltonian (see Eq. (4.21)). It was original pro-posed by Landau without the 1
2∇φ(x) · ∇φ(x) term. However if we want to calculate
correlation functions we also need the latter term to account for fluctuations. This wasproposed by Ginzburg. Hence the Landau-Ginzburg Hamiltonian. Before we proceedsome comments have to be made. The first comment concerns the number of dimen-sions. In one dimension a phase transition cannot occur but the above Hamiltonianleads to phase transition. Therefore we conclude that the procedure outlined above iswrong in one dimension. In two dimensions phase transitions do occur with Eq. (5.56)the specific heat is logarithmic i.e. ∼ ln(T −Tc). This quantities with the Hamiltonianabove behave power like as (T − Tc)
−b. Therefore in this case the Landau-Ginzburgtheory leads to a wrong prediction too. In three and larger dimension the latter theoryis presumably right and from now on we shall apply it for n ≥ 3.
Add an external magnetic field H0 to the Hamiltonian H as follows
Z[H0] = C∫
Dφ(x) exp∫
dnx(
H(φ,∇φ) +H0(x)φ(x))
(5.76)
Further we make the Landau approximation and put φ(x) independent of x. The freeenergy reads
G =µ2
2φ2 +
λ
4!φ4 −H0 φ (5.77)
We assume that the fluctuations in φ are small so that we can neglect higher powersin φ. Because of (5.58) φ can be interpreted as the magnetisation. We determine φ byminimising G in the case that H0 = 0.
µ2 > 0∂G
∂φ= 0 → φ = 0 for T > Tc
µ2 < 0∂G
∂φ= 0 → φ = ±
√
−6µ2
λfor T < Tc (5.78)
This implies that in the paramagnetic case φ = 0 whereas in the ferromagnetic caseφ 6= 0 even if H0 = 0. We make now the following assumptions
µ2 = a (T − Tc) → φ =
√
6a
λ(Tc − T )1/2 ∼ (Tc − T )β
β =1
2Landau value (5.79)
110
φ
TTc
Figure 19: The magnetisation φ as a function of T .
The quantity β is called a critical exponent. The magnetic field becomes at H0 = 0and T = Tc
H =∂G
∂φ=λ
6φ3 → φ ∼ H1/3 = H1/δ δ = 3 Landau value (5.80)
and δ is a critical exponent. We shall now calculate the susceptibility. Differentiate∂G/∂φ with respect to H but T 6= Tc
H =∂ G
∂ φ= −H0 + µ2 φ+
λ
6φ3
1 = µ2 ∂φ
∂H+
1
2φ2 ∂φ
∂H(5.81)
We get now
T > Tc χ =∂φ
∂H=
1
µ2=
1
a (T − Tc)∼ (T − Tc)
−γ γ = 1 Landau value
T < Tc χ =∂φ
∂H=
1
µ2=
1
a (T − Tc)∼ (T − Tc)
−γ γ = 1 Landau value
(5.82)
where we have neglected higher powers in φ in the limit T → Tc. The free energybecomes for H0 = 0
T > Tc φ = 0 → G = 0
T < Tc φ = ±√
−6µ2
λ→ G = −3 a2
2 λ(T − Tc)
2 (5.83)
111
Cv
TTc
Figure 20: The specific heat Cv as a function of T .
The specific heat Cv is
T > Tc → Cv = −2 k T∂ G
∂ T|T=Tc
− k T 2 ∂2 G
∂ T 2|T=Tc
= 0
T < Tc → Cv = −2 k T∂ G
∂ T|T=Tc
− k T 2 ∂2 G
∂ T 2|T=Tc
= 3 k T 2c
a2
λ
∼ (T − Tc)−α α = 0 Landau value (5.84)
The Landau theory neglects the fluctuations of φ. Hence it fails when these fluctuationsare large compared to φ. To account for fluctuations let us add the kinetic term12∇φ(x) · ∇φ(x) to the Hamiltonian (see (5.75)).
Z[H0] = C∫
Dφ(x) exp
∫
dnx[
− 1
2∇φ(x) · ∇φ(x) − µ2
2φ2(x)
− λ
4!φ4(x) +H0(x)φ(x)
]
(5.85)
Let us first put λ = 0 (free field theory). The path integral becomes
Z[H0] = C∫
Dφ exp
[
∫
dnx1
2φ(x)x φ(x) − µ2
2φ2(x) +H0(x)φ(x)
]
(5.86)
Define
G−10 (x, y) = G−1
0 (x− y) = −(
x − µ2)
δ(n)(x− y) (5.87)
The path integral can be rewritten as
Z[H0] = C∫
Dφ exp
[
− 1
2
∫
dnx∫
dny1
2φ(x)G0(x, y)φ(y)
112
+∫
dnxH0(x)φ(x)
]
(5.88)
Using Eq. (5.63)
Z[H0] =C
det1/2G−10
exp
[
1
2
∫
dnx∫
dny H0(x)G0(x, y)H0(y)
]
(5.89)
Choose C in such a way that Z[0] = 1.
G−10 (x, y) =
∫ dnp
(2π)ne−i p·(x−y)
(
p2 + µ2)
(5.90)
or
G0(x, y) =∫
dnp
(2π)ne−i p·(x−y) 1
p2 + µ2(5.91)
The correlation function becomes
〈φ(x)φ(y)〉 = Z[H0]−1 δ2 Z[H0]
δ H0(x) δ H0(y)
∣
∣
∣
∣
∣
H0(xi)=0
= G0(x, y) (5.92)
In momentum space (inverse Fourier transform) this equals
G0(k) =∫
dnx ei k (x−y)G0(x, y) =1
k2 + µ2=
k ≫ µk−2 = k−2+η
η = 0 Landau value (5.93)
The quantity η is a critical exponent which in zeroth order in λ is zero. The suscepti-bility χ is
G0(0) = µ−2 = (T − Tc)−γ γ = 1 Landau value (5.94)
Notice that the correlation length ξ = µ−1 it is defined by
ξ2 =
∫
dx x2 G0(x)∫
dxG0(x)=
−∂ G0(k)/∂ k2
G0(k)|k=0 = µ−2
G(x, y) =∫
dnk
(2π)ne−i k·(x−y) 1
k2 + µ2∼ e−µ |x−y|
|x− y|n−2with µ = ξ−1 (5.95)
From the behaviour of µ with respect to T we infer
ξ = (T − Tc)−ν = (T − Tc)
−1/2 ν =1
2Landau value (5.96)
Inserting the correlation length k = ξ−1 into Eq. (5.93) we obtain
113
k
p
p+ k
k
k
G0 Γφ2,φ2 Γφ2
Figure 21: Lowest order contribution to the Green’s function G0, Γφ2,φ2 and Γφ2 .
(T − Tc)ν (η−2) (5.97)
Equating this with Eq. (5.93) we get the fluctuation dissipation theorem
γ = ν (2 − η) (5.98)
The specific heat is derived from Cv = Γφ2,φ2 with
Γφ2,φ2 = −∂2 ln Z[H0]
∂(µ2)2= −
[
〈12
∫
dnxφ2(x)1
2
∫
dny φ2(y)〉 − 〈12
∫
dnxφ(x)〉2]
(5.99)
where µ2 = a (T − Tc). We can also write this as
Γφ2,φ2(k, µ) =∫ dnx
(2π)nei k·x 〈φ2(x)φ2(0)〉 (5.100)
This expression is equal to
Γφ2,φ2(k, µ) =∫
dnp
(2π)n
1(
p2 + µ2) (
(p+ k)2 + µ2)
= i Sn Γ(2 − n/2)∫ 1
0dx [k2 x (1 − x) + µ2]n/2−2 (5.101)
substitute x = 1/2 − y/2
Γφ2,φ2(k, µ) = i Sn Γ(2 − n/2)∫ 1
0dy
[
k2
4(1 − y2) + µ2
]n/2−2
(5.102)
Substitute y =√t then we can cast the integral into a hypergeometric function
F1,2(a, b, c; z)
Γφ2,φ2(k, µ)
114
=i
2Sn Γ(2 − n/2)
(
k2
4+ µ2
)n/2−2∫ 1
0dt t−1/2
[
1 − k2
k2 + 4µ2t
]n/2−2
= i Sn Γ(2 − n/2)
(
k2
4+ µ2
)n/2−2
F1,2
(
2 − n/2, 1, 3/2;k2
k2 + 4µ2
)
= i Sn Γ(2 − n/2) (µ2)n/2−2 F1,2
(
2 − n/2, 1, 3/2;− k2
4µ2
)
∼ µn−4 (5.103)
for k = 0. In the limit µ → 0 we have to make a Kummer transformation on thehypergeometric function
Γφ2,φ2(k, µ)
= i Sn Γ(2 − n/2) (µ2)n/2−2 F1,2
(
2 − n/2, 1, 3/2;− k2
4µ2
)
=
= i SnΓ(3/2) Γ(2 − n/2) Γ(n/2 − 1)
Γ(n/2 − 1/2)
(
k2
4
)n/2−2
×F1,2
(
2 − n/2, 3/2 − n/2, 2 − n/2;−4µ2
k2
)
+i SnΓ(3/2) Γ(2− n/2) Γ(n/2 − 1)
Γ(1/2) Γ(n/2− 1/2)
(
4µ2
k2
)
(µ2)n/2−2
×F1,2
(
1, 1/2, n/2;−4µ2
k2
)
∼ kn−4 if µ = 0 (5.104)
We can now derive the exponents
Γφ2,φ2(k, 0) = kn−4 = kλ λ = n− 4 Landau value (5.105)
and
Γφ2,φ2(0, µ) = µn−4 = (T − Tc)n/2−2 = (T − Tc)
−α α = 2 − n/2 (5.106)
Finally we have the Greens function
Γφ2(k, µ) =∫
dnx ei k·x∫
dny ei k·y 〈φ(x)φ2(y)φ(0)〉 = kµ (5.107)
In zeroth order perturbation theory we have
Γφ2(k, µ) = 1 → µ = 0 Landau value (5.108)
115
There exist a relation between γ, η and µ which is called hyper scaling.
1
γ= 1 − µ
2 − η(5.109)
This remains satisfied if we turn the coupling constant λ unequal to zero. Also another relation remains preserved
α = 2 − ν n (5.110)
5.4 The ε expansion and the renormalisation group.
Until now we have only calculated the critical exponents for N = 1. It would notbe difficult to generalize them to an order parameter having dimension N and thesymmetry group is in this case O(N). Further we take λ 6= 0 and the Ginzburg-Landaupartition function reads
Z[H0] = C∫
Dφ(x) exp
∫
dnx[
− 1
2
N∑
i=1
∇φi(x) · ∇φi(x) −µ2
2
N∑
i=1
φ2i (x)
− λ
4!(
N∑
i=1
φ2i (x))
2 +N∑
i=1
Hi(x)φi(x)]
(5.111)
We see that for µ2 = a (T−Tc) and T < Tc we get again a second order phase transition.The symmetry group O(N) breaks down to a smaller group. In other words we getmore order. Hence the name order parameter for φi. If T → Tc the mass µ approacheszero and the correlation length ξ = µ−1 becomes infinite. Therefore at the critical pointwe are at a maximally ordered state. At this critical temperature the thermodynamicfunctions show scaling in the form
C = (T − Tc)−b (5.112)
and b is called the critical exponents. Examples are for n = 3:
N = 0 Self avoiding walkN = 1 Ising Ferro magnetismN = 2 λ point of Helium Super fluidityN = 3 Heisenberg modelN = ∞ Spherical model
The Feynman rules of the partition function in Eq. (5.111) (generating functional forthe connected graphs) are:
1 Draw at a given order in λ all topological inequivalent connected diagrams.
116
pa b G0,ab(p) = δab
p2+µ2
p1
p3p2
p4
a
cb
d
−λn Iabcd (2π)4 δ(4)(p1 + p2 + p3 + p4)
Figure 22: The Feynman rules for Euclidian φ4-theory.
2 Each internal line with momentum p is represented by the factor G0,ab(p)
3 Each vertex where four lines with the momenta p1, p2, p3, p4 come together isrepresented by a factor −λn Iabcd with Iabcd = 1
3
(
δab δcd + δac δbd + δad δbc)
. Here
is imposed the condition that∑4
i=1 pi = 0
4 Integrate over every internal momentum k. The measure is equal to dnk/(2π)n
5 A combinatorial factor 1/n! if n internal lines start on one vertex and end onanother vertex.
We will now calculate the first order corrections to the critical exponents. We startwith the vertex correction and set the mass µ = 0 (the ultra violet divergence will notdepend on µ2). In lowest order the vertex is given by
Λ(0)abcd = −λn Iabcd (5.113)
The first order correction to Λ(0) is exhibited in Fig. 22. The result for the first graphin Fig. 23 is equal to
Λ(1)abcd(p) =
1
2λ2
n Iablm Ilmcd
∫
dnk
(2π)n
1
k2 (k + p)2(5.114)
Let us first calculate Iablm Ilmcd. The result is
Iablm Ilmcd =1
9
[
(N + 4) δab δcd + 2 δac δbd + 2 δad δbc]
(5.115)
117
k + p
k
l
m
a c
b d
a c
b d
a d
b c
Figure 23: The first order contribution to Γ(4)abcd.
Then we compute the momentum dependence of Λ(1). Using Feynman parameters
Λ(1)abcd(p) =
1
2Iablm Ilmcd
λ2n
(2π)n
∫
dnk∫ 1
0dx
1[
k2 + 2 x k · p+ x p2]2
=1
2Iablm Ilmcd λ
2n
πn/2
(2π)nΓ(2 − n/2)
∫ 1
0dx(
x (1 − x))n/2−2
(p2)n/2−2
=1
2Iablm Ilmcd λ
2n
πn/2
(2π)n
Γ(2 − n/2) Γ2(n/2 − 1)
Γ(n− 2)(p2)n/2−2 (5.116)
Define
Cn =πn/2
(2π)n
Γ(3 − n/2) Γ2(n/2 − 1)
Γ(n− 2)(5.117)
and
ε = 4 − n λn = λMε (5.118)
We get
Λ(1)abcd(p) =
1
2Iablm Ilmcd λn (Cn λ)
[
2
ε− ln
p2
M2
]
(5.119)
The other two graphs in Fig. 23 give the same except for the group theoretical factor.The second graph yields
Iackl Iklbd =1
9
[
(N + 4) δac δbd + 2 δab δcd + 2 δad δbc]
(5.120)
and the third graph gives
Iadkl Iklbc =1
9
[
(N + 4) δad δbc + 2 δac δbd + 2 δab δcd]
(5.121)
118
k
q − k
−p− q
p pc da b
i
j
m
Figure 24: The first order contribution to the self energy Γ(2)ab .
Adding all three contributions from Fig. 23 we get for Λ(1)abcd
Λ(1)abcd =
N + 8
6Iabcd λn (Cn λ)
[
2
ε− ln
p2
M2
]
(5.122)
The first order correction to Γ(2)ab (p) is given by Fig. 24. Notice that the tadpole graph
in Fig. 4 yields zero for µ = 0 The expression for this graph is
G1,ab(p) = G0,ac(p) Πcd(p)G0,db(p)
Πcd(p) =1
3!Icijm Idijm λ
2n
∫
dnk
(2π)n
∫
dnq
(2π)n
1
(q + p)2 (k − q)2 k2(5.123)
The group factor is
Icijm Idijm =1
9(3N + 6) δcd (5.124)
The contribution to the two point function becomes
Πcd(p) =N + 2
18δcd
∫
dnq
(2π)n
1
(q + p)2Λ(q) (5.125)
with
Λ(q) = λn (Cn λn)2
ε(q2)n/2−2 (5.126)
Therefore
Πcd(p) =N + 2
18δcd λn (Cn λn)
2
ε
∫
dnq
(2π)n
1
(q + p)2 (q2)2−n/2(5.127)
119
a
b
c
c
k
k + p
⊕
Figure 25: The first order contribution to the Γab,φ2 .
Let us work out the integral.
∫
dnq
(2π)n
1
(q + p)2 (q2)2−n/2=
Γ(3 − n/2)
Γ(2 − n/2)
∫
dnq
(2π)n
∫ 1
0dy
(1 − y)1−n/2
(q2 + 2 y q · p+ y p2)3−n/2
=πn/2
(2π)n
Γ(3 − n)
Γ(2 − n/2)(p2)n−3
∫ 1
0dy yn−3 (1 − y)n/2−2
=πn/2
(2π)n
Γ(3 − n) Γ(n− 2) Γ(n/2 − 1)
Γ(2 − n/2) Γ(3n/2 − 3)(p2)n−3
= −1
2Cn
Γ(5 − n) Γ2(n− 3)
Γ(3n/2 − 5) Γ(n/2 − 1) Γ2(3 − n/2)
n− 3
(3n/2 − 4) (3n/2 − 5)(p2)n−3
= −1
4Cn (1 + 5/4 ε+ · · ·) (p2)n−3 (5.128)
Therefore Πcd becomes
Πcd(p) =N + 2
18δcd(
Cn λ)2p2
[
− 1
2 ε+
1
2ln
p2
M2− 5
8
]
(5.129)
Finally we compute Γφ2 (Eq. (5.106)) up to first order
Γ(1)ab,φ2 =
1
2Iccab λn
∫
dnk
(2π)n
1
k2 (k + p)2(5.130)
The group theoretical factor becomes
Iccab =N + 2
3δab (5.131)
and the two point function Γ(1)ab,φ2 is equal to
Γ(1)ab,φ2 =
N + 2
6Cn λ
[
2
ε− ln
p2
M2
]
(5.132)
120
Collecting all terms we can write the correlation functions corrected up to one looporder
Γ(4)abcd(p) = Λ
(0)abcd(p) + Λ
(1)abcd(p) + · · · = −λn Iabcd
[
1 − N + 8
6Cn λ
2
ε− ln
p2
M2
+ · · ·]
Γ(2)ab (p) = G0,ab(p) +G0,ac(p) Πcd(p)G0,db(p) + · · · =
1
p2δab
[
1 − N + 2
18C2
n λ2
×
1
2 ε− 1
2ln
p2
M2+
5
8
+ · · ·]
Γab,φ2(p) = Γ(0)ab,φ2(p) + Γ
(1)ab,φ2(p) + · · · = δab
[
1 − N + 2
6Cn λ
2
ε− ln
p2
M2
+ · · ·]
(5.133)
We have now to renormalize to get rid of the infinity exhibited by the pole term 1/ε
Γ(4),Rabcd (p) = Z1 Γ
(4)abcd(p)
Γ(2),Rab (p) = Z−1
3 Γ(2)ab (p)
ΓRab,φ2(p) = Zφ2 Γab,φ2(p) (5.134)
Up to first order
Z1 = 1 + Cn λRN + 8
3
1
ε
Z3 = 1 − C2n λ
2R
N + 2
36
1
ε
Zφ2 = 1 + Cn λRN + 2
3
1
ε(5.135)
en the finite Green’s function equal
Γ(4),Rabcd (p) = −Mn−4 λR Iabcd
[
1 +N + 8
6Cn λR ln
p2
M2+ · · ·
]
Γ(2),Rab (p) =
1
p2δab
[
1 +N + 2
36C2
n λ2R
1
2ln
p2
M2− 5
8
+ · · ·]
121
ΓRab,φ2(p) = δab
[
1 +N + 2
6Cn λR ln
p2
M2+ · · ·
]
(5.136)
The Greens functions satisfy a renormalisation group equation (Callan-Symanzik equa-tion) which we will now derive. We have defined the renormalization group constantsZ1, Z3 and Zλ as follows
L =∫
dnx[1
2
N∑
i=1
∇φi(x) · ∇φi(x) +µ2
2
N∑
i=1
φ2i (x) +
λ
4!(
N∑
i=1
φ2i (x))
2
+Jφ2
N∑
i=1
φ2i (x) +
N∑
i=1
Hi(x)φi(x)]
(5.137)
We make the substitution
φi(x) = Z3(M)1/2 φi,R(x) (5.138)
Note that Z3 is independent of i. The Lagrangian becomes
L =∫
dnx[1
2Z3
N∑
i=1
∇φi,R(x) · ∇φi,R(x) +µ2
2Z3
N∑
i=1
φ2i,R(x)
+λ
4!Z2
3 (N∑
i=1
φ2i,R(x))2 + Jφ2 Zφ2
N∑
i=1
φ2i,R(x)
+Z1/23
N∑
i=1
Hi(x)φi,R(x)]
(5.139)
On the other hand
L =∫
dnx[1
2Z3
N∑
i=1
∇φi,R(x) · ∇φi,R(x) +µ2
R
2Zµ
N∑
i=1
φ2i,R(x)
+λR
4!M4−n Z1 (
N∑
i=1
φ2i,R(x))2 + Jφ2 Zφ2
N∑
i=1
φ2i,R(x)
+Z1/23
N∑
i=1
Hi(x)φi,R(x)]
(5.140)
We obtain the relations
λ = M4−n Z1(λR)
Z3(λR)2λR = Mε Zλ(M)λR (5.141)
µ2 =Zµ
Z3µ2
R (5.142)
122
φ2i (x) = Zφ2
iφ2
i,R(x) (5.143)
The bare coupling constant λ is independent of the renormalization scale M .
Mdλ
dM= 0 =⇒
0 = εMε λR(M)Zλ(M) +M1+ε d λR(M)Zλ(M)
dM
0 = ε λR(M)Zλ(M) +∂ λR(M)Zλ(M)
∂ λRM
∂ λR
∂M(5.144)
Define the beta-function
β (λR) = M∂ λR(M)
∂ M= − ε Zλ(M)λR
∂ λR Zλ(M)/∂ λR(5.145)
One can show that this function is finite in all orders in perturbation theory for ε→ 0.Following the same arguments for the unrenormalized Green function one can derive
Γ(n)i1,i2,···in(x1, x2, · · ·xn) = 〈φi1(x1)φi2(x2) · · ·φin(xn)〉
Γ(n),Ri1,i2,···in(x1, x2, · · ·xn) = 〈φi1,R(x1)φi2,R(x2) · · ·φin,R(xn)〉
Γ(n)i1,i2,···in(x1, x2, · · ·xn) = Z
n/23 Γ
(n),Ri1,i2,···in(x1, x2, · · ·xn)
Γ(n)i1,i2,···in(p1, p2, · · ·pn) = Z
n/23 Γ
(n),Ri1,i2,···in(p1, p2, · · ·pn) (5.146)
For massless theories we have
MdΓ
(n)i1,i2,···in(p1, p2, · · ·pn)
dM= 0 −→
[
M∂
∂M+ β (λR)
∂
∂ λR
+n
2M
∂ lnZ3(M)
∂ M
]
Γ(n),Ri1,i2,···in(p1, p2, · · ·pn) = 0
(5.147)
Define the anomalous dimension of the field φi,R(x) by
γφ (λR) =1
2M
∂ lnZ3(M)
∂M=
1
2β(λR)
∂ lnZ3(M)
∂ λR(5.148)
which is finite in all orders in the limit ε→ 0. The result is[
M∂
∂M+ β (λR)
∂
∂ λR+n
2γφ (λR)
]
Γ(n),Ri1,i2,···in(p1, p2, · · ·pn) = 0 (5.149)
123
For simplicity we shall only consider two-point functions which are only dependent onone momentum (n = 2). Equation (5.149) becomes
(
− ∂
∂ t+ β(λR)
∂
∂ λR
+ γφ(λR)
)
ΓRij(λR, t) = 0 t =
1
2ln
p2
M2(5.150)
The equation above admits the following solution
ΓRij
(
λR(M),1
2ln
p2
M2
)
= ΓRij(λ(t), 0) exp
∫ λ(t)
λ(0)dx
γφ(x)
β(x)
(5.151)
Here λ(t) is called the running coupling constant which satisfies the equation
(
− ∂
∂ t+ β(λR)
∂
∂ λR
)
λ(t) = 0 with λ(0) = λR(M) ≡ λR (5.152)
Eq. (5.152) can be also written as an integral equation
∫ λ(t)
λR
dx1
β(x)= t (5.153)
Taking the partial derivative of the equation above
∂
∂ t
∫ λ(t)
λR(M)dx
1
β(x)= 1
=⇒ ∂ λ(t)
∂ t
1
β(λ(t)= 1 (5.154)
Hence it follows that
∂ λ(t)
∂ t= β(λ(t)) (5.155)
To show that one can obtain Eq. (5.153) from Eq. (5.152) we differentiate the formerw.r.t. λR
∂ λ(t)
∂ λR
1
β(λ(t))− 1
β(λR)= 0
=⇒ β(λR)∂ λ(t)
∂ λR− β(λ(t)) = 0
=⇒ β(λR)∂ λ(t)
∂ λR− ∂λ(t)
∂ t= 0 (5.156)
124
p p p
k
a b a b
c c
Figure 26: Corrections to the propagator G(p).
Suppose the beta-function has a zero for small momenta (p2 → 0 then t→ −∞). Thezero shows up in the following way
∂ λ(t)
∂ t= β(λ(t)) = β0 (λ(t) − λ0) (5.157)
The solution is
λ(t) = λ0 + eβ0 t → β0 > 0 (5.158)
So β0 must be positive which is the case for φ4-theory. The solution of the Callan-Symanzik equation is
ΓRij
(
λR(M),1
2ln
p2
M2
)
∼ ΓRij(λ0, 0)
(
p
M
)γ0 λ2
0
(5.159)
where we have chosen
γφ(λR) = γ0 λ2R + · · · (5.160)
This is called scaling. So in second order phase transitions if we are at the criticaltemperature the correlation length ξ = (T − Tc)
−ν → ∞. In other words the momentap go to zero. At that point the correlation functions show scaling. The coefficient ofthis scaling functions i.e. γ0 λ
20 are the critical exponents. If we compare Eq. (5.93)
with (5.148) we conclude
η = 2 γφ(λ0) (5.161)
is two times the anomalous dimension of the φ field. According to Eq. (5.145) thebeta-function up to one-loop order is
β(λR) = −ε λR + λ2RCn
N + 8
3=Cn (N + 8)λR
3(λR − λ0)
β0 =Cn (N + 8)λR
3> 0 (5.162)
125
The zero is
λ0 =3 ε
Cn (N + 8)(5.163)
The critical exponents are
η = β(λR)∂ lnZ3(M)
∂ λR=(
− ε λR + λ2RCn
N + 8
3
) (
− λRC2n
N + 2
36
1
ε
)
=(N + 2)
2 (N + 8)2ε2 (5.164)
The critical exponent µ (Eq. (5.106)) is equal to the anomalous dimension of theoperator φ2
i,R
µ = −β(λR)∂ ln(Zφ2 Z−1
3 )
∂ λR
= −(
− ε λR + λ2RCn
N + 8
3
) (
CnN + 2
3
1
ε
)
=N + 2
N + 8ε (5.165)
From the hyper scaling equation Eq. (5.109) we get (up to order ε we can neglect η)
γ = 1 +µ
2+ · · · = 1 +
N + 2
2(N + 8)ε (5.166)
The γ can be also calculated from corrections to the two-point function Γ(2)(p) (seeFig. 26)
Γ(2)(0) = µ−2
[
1 − λRCnN + 2
6
(2
ε− ln
µ2
M2− 1
)
]
(5.167)
From which we can derive
Γ(2),R(0) = µ−2
[
1 + λR CnN + 2
6
(
lnµ2
M2+ 1
)
]
(5.168)
Zµ = 1 + λRCnN + 2
3
1
ε(5.169)
γµ(λR) = β(λR)∂ ln(Zm Z
−13 )
∂ λR= −Cn
N + 2
3λR (5.170)
From which we derive (see (5.94))
γ = 1 − 1
2γm = 1 +
N + 2
2 (N + 8)ε (5.171)
126
Which confirms hyper scaling. The critical exponent α is derived from the hyperscalingrelation γ = ν (2 − η). In next to leading order this implies (η = O(ε2))
ν =1
2+
N + 2
4 (N + 8)ε (5.172)
The critical exponent of the specific heat is α = 2 − n ν
α = 2 − n
2− n (N + 2)
4 (N + 8)ε =
ε
2− N + 2
N + 8ε =
4 −N
2 (N + 8)ε (5.173)
If we put the dimension of the physical system n = 3, therefore ε = 4 − n = 1, weobtain
order parameter ε-expansion high temperature series
N = 1 η = 0.019 η = 0.041 ± 0.010γ = 1.167 γ = 1.250 ± 0.003α = 0.167 α = 0.110 ± 0.005
N = 2 η = 0.020γ = 1, 200
N = 3 η = 0.021 η = 0.043 ± 0.014γ = 1.227 γ = 1.375 ± 0.040
Comparison of the ε expansion with the high temperature series.
5.5 The 1/N expansion.
Besides the ε expansion we can make another one i.e. the 1/N expansion. In this casewe take N → ∞. In that case the vertex correction behaves as follows see Fig. 27.The coupling constant in the large N limit is defined as
Λ(0)abcd = −λn δab δcd (5.174)
We get now a geometric series. All other graphs in a given order are suppressed bypowers in N . Only the bubble graphs dominate the perturbation series (see Fig. 28).The bubble sum in Fig. 27 provides us with the effective vertex
Λabcd(p, µ2)= − λn
1 + λnN Π(p, µ2)δab δcd with
Π(p, µ2) =1
2
∫
dnq
(2π)n
1
(q2 + µ2)(
(q + p)2 + µ2) (5.175)
127
=
+ · · · =Λabcd(p, µ2)
Figure 27: Correction to the vertex in the limit N → ∞.
The effective vertex will now be the variable in which we perturb. As an example wewill calculate the critical exponent η. The latter is inferred from the graph in Fig. 29.First we put µ = 0 in Eq. (5.101). The effective vertex Λ(p, 0) becomes
Λabcd(p, 0) = − λn
1 + λnN Π(p, 0)δab δcd with
Π(p, 0) =1
2
∫
dnq
(2π)n
1
q2 (q + p)2= S−1
n pn−4 and
Sn = 2n+1 πn/2 Γ(n− 2)
Γ(2 − n/2) Γ2(n/2 − 1)(5.176)
The graph in Fig. 29 is equal to
Σab(k) = −δab
∫
dnp
(2π)n
λn
1 + λnN Π(p, 0)
1
(p+ k)2(5.177)
We are interested in the infrared behaviour k → 0. Therefore we approximate theintegral by providing a cut off
Σab(k) = −δab
∫ λn dnp
(2π)n
1
N Π(p, 0)
1
(p+ k)2(5.178)
We can also remove the cut off and apply m-dimensional regularization
Σab(k) = −δab (λn)n−m∫ dmp
(2π)m
1
N Π(p, 0)
1
(p+ k)2(5.179)
128
a b
Figure 28: Graph a is of the order N2 whereas graph b is of the order of N .
The integration is straightforward
Σab(k) = −δab
NSn (λn)n−m
∫
dmp
(2π)m
1
(p2)n/2−2 (p+ k)2(5.180)
Using Feynman parameters
Σab(k) = −δab
NSn (λn)n−m Γ(n/2 − 1)
Γ(n/2 − 2)
∫ dmp
(2π)m
∫ 1
0dx
(1 − x)n/2−3
(p2 + 2 x p k + x k2)n/2−1
(5.181)
We are allowed to translate the momentum
Σab(k) = −δab
NSn (λn)n−m Γ(n/2 − 1)
Γ(n/2 − 2)
∫ dmp
(2π)m
∫ 1
0dx
(1 − x)n/2−3
(p2 + x (1 − x) k2)n/2−1
(5.182)
Σab(k) = −δab
NSn
πm/2
(2π)m
Γ(n/2 −m/2 − 1)
Γ(n/2 − 2)(λn)
n−m∫ 1
0dx xm/2−n/2+1 (1 − x)m/2−2
×(k2)m/2−n/2+1 (5.183)
Σab(k) = −δab
NSn
πm/2
(2π)m(λn)n−m Γ(n/2 −m/2 − 1)
Γ(n/2 − 2)
Γ(m/2 − 1) Γ(m/2 + 2 − n/2)
Γ(m+ 1 − n/2)
×(k2)m/2−n/2+1 (5.184)
Σab(k) = −δab
NSn
πm/2
(2π)m(λn)n−m Γ(n/2 −m/2 − 1)
Γ(n/2 − 2)
Γ(m/2 − 1) Γ(m/2 + 2 − n/2)
Γ(m+ 1 − n/2)
×(k2)m/2−n/2+1 (5.185)
129
Figure 29: Self energy diagram to order 1/N .
Put m = n− ε
Σab(k) =δab
NSn
πn/2−ε/2
(2π)n−ελε
n
2 Γ(1 + ε/2)
(1 − ε/2) ε
Γ(n/2 − 1 − ε/2) Γ(2 − ε/2)
Γ(n/2 + 1 − ε) Γ(n/2 − 2)
×(k2)1−ε/2 (5.186)
Since we are only interested in the logarithmic term we put ε = 0.
Σab(k) =δab
N
4
ε
Γ(n− 2)
Γ(2 − n/2) Γ(n/2 − 1) Γ(n/2 + 1) Γ(n/2 − 2)k2[
1 +ε
2lnλ2
n
k2
]
(5.187)
We can now remove the pole term because it is fictitious (it is not a real ultra-violetsingularity). Including the lowest order term we get (see Eq. (5.93))
Gab(k) = δab1
k2
[
1 +2 Γ(n− 2)
N Γ(2 − n/2) Γ(n/2 − 1) Γ(n/2 + 1) Γ(n/2 − 2)lnλ2
n
k2
]
= δab1
k2
[
1 − η
2lnλ2
n
k2
]
(5.188)
Hence
η = − 4
N
Γ(n− 2)
Γ(2 − n/2) Γ(n/2 − 1) Γ(n/2 + 1) Γ(n/2 − 2)(5.189)
Substitute n = 4 − ε then
η =1
2
ε2
N(5.190)
If we take the ε expansion for η in Eq. (5.164) and the limit N → ∞ we get exactlythe same formula.
130
6 Quantum electrodynamics
6.1 Green’s function for the Dirac field
The Dirac field is given by
ψα(x) =∑
s
∫ d3~p
(2π)3/2
1√
2Ep
[
b(~p, s) uα(~p, s) e−i p·x + d†(~p, s) vα(~p, s) ei p·x]
(6.1)
with the anti-commutation relations
b(~p, s), b†(~p′, s′)
= δss′ δ(3)(~p− ~p′)
d(~p, s), d†(~p′, s′)
= δss′ δ(3)(~p− ~p′) (6.2)
The other anti-commutation relations are zero. We compute the following distributions
−i S(x− x′) =
ψ(x), ˆψ(x′)
=1
(2π)3
∑
s,s′
∫
d3p∫
d3p′1
√
4EpEp′
[
ei (p′·x′−p·x)
×u(~p, s) u(~p′, s′)
b(~p, s), b†(~p′, s′)
+ e−i (p′·x′−p·x) v(~p, s) v(~p′, s′)
×
d†(~p, s), d(~p′, s′)
]
(6.3)
We normalise the spinors in a different way and choose Eq. (1.66). Therefore theprojection operators become
∑
s
u(~p, s) u(~p, s) = /p+m∑
s
v(~p, s) v(~p, s) = /p−m (6.4)
The distribution becomes
S(x− x′) =i
(2π)3
∫
d3~p
2Ep
(/p+m) ei p·(x′−x) + (/p−m) e−i p·(x′−x)
(6.5)
Notice that S is a four by four matrix. Substitute in the last term ~p by −~p then
S(x− x′) =i
(2π)3
∫
d3~p
2Ep
(/p+m) ei p0·(x′
0−x0) −
(
γ0(−p0) − ~γ · ~p+m)
×ei (−p0)·(x′
0−x0)
e−i ~p·(~x′−~x)
=i
(2π)3
∫
d4p ǫ(p0) δ(p2 −m2) (/p+m) ei p·(x′−x) (6.6)
131
with p0 = Ep. The distribution satisfies the Dirac equation(
i γµ ∂µ −m)S(x− x′) = 0 (6.7)
Next we define the positive and negative energy contributions to the Feynman propa-gator
S+(x− x′) = 〈0|ψ(x) ˆψ(x′)|0〉 =1
(2π)3
∫
d4p θ(p0) δ(p2 −m2) (/p+m) e−i p·(x−x′)
S−(x− x′) = 〈0| ˆψ(x′) ψ(x)|0〉 = − 1
(2π)3
∫
d4p θ(−p0) δ(p2 −m2) (/p+m) e−i p·(x−x′)
=1
(2π)3
∫
d4p θ(p0) δ(p2 −m2) (/p−m) e−i p·(x′−x) (6.8)
Both satisfy the Dirac equation(
i γµ ∂µ −m)S±(x− x′) = 0 (6.9)
S can be expressed into S± as follows
−i S(x− x′) = S+(x− x′) + S−(x− x′) (6.10)
The time ordered product for the fermion field
i SF (x− x′) = 〈0|T(
ψ(x) ˆψ(x))
|0〉 (6.11)
or Feynman propagator is a Green’s function because it satisfies(
i γµ ∂µ −m)
i SF (x− x′) = i δ(4)(x− x′) (6.12)
Proof:
T(
ψ(x) ˆψ(x′))
= θ(x0 − x′0) ψ(x) ˆψ(x′) − θ(x′0 − x0)ˆψ(x′) ψ(x) (6.13)
or
i SF (x− x′) = θ(x0 − x′0)S+(x− x′) − θ(x′0 − x0)S−(x− x′) (6.14)
We now use
∂
∂ x0
θ(x0 − x′0) = δ(x0 − x′0)∂
∂ x0
θ(x′0 − x0) = −δ(x0 − x′0) (6.15)
so that
∂
∂ x0i SF (x− x′) = −δ(x0 − x′0) i S(x− x′) + θ(x0 − x′0)
∂
∂ x0S+(x− x′)
−θ(x′0 − x0)∂
∂ x0S−(x− x′) (6.16)
132
Further is
∂i θ(x0 − x′0) = 0 (6.17)
We obtain(
i γµ ∂µ −m
)
i SF (x− x′) = δ(x0 − x′0) i γ0
(
− i S(x− x′))
+ θ(x0 − x′0)
×(
i γµ ∂µ −m
)
S+(x− x′) − θ(x′0 − x0)(
i γµ ∂µ −m
)
×S−(x− x′) = γ0 δ(x0 − x′0)S(x− x′) (6.18)
Using the expression for S(x− x′)
(
i γµ ∂µ −m
)
i SF (x− x′) =1
(2π)3
∫ d3~p
2Ep
[
γ0
(
γ0 p0 + γi p
i +m)
ei p0 (x′
0−x0)
+γ0
(
γ0 p0 − γi p
i −m)
e−i p0 (x′
0−x0)
]
ei ~p·(~x−~x′)
×δ(x0 − x′0) =i
(2π)3
∫
d3~p ei ~p·(~x−~x′) δ(x0 − x′0)
= i δ(4)(x− x′) (6.19)
The canonical anti-commutation relation is derived from
−i S(x− x′) γ0|x0=x′
0=
ψ(x), ψ†(x′)
= i δ(3)(~x− ~x′) (6.20)
Furthermore we can write for SF (x− x′)
i SF (x− x′) = i∫ d4p
(2π)4e−i p·(x−x′) /p+m
p2 −m2 + i ε(6.21)
This satisfies the aforementioned differential equation Eq. (6.19). Finally we have thedistribution
−S1(x− x′) = 〈0|[
ψ(x), ˆψ(x′)]
|0〉 = S+(x− x′) − S−(x− x′)
=1
(2π)3
∫
d4p δ(p2 −m2)
[
θ(p0) (/p+m) e−i p·(x−x′)
+θ(−p0) (/p+m) e−i p·(x−x′)
]
=1
(2π)3
∫
d4p e−i p·(x−x′) δ(p2 −m2) (/p+m) (6.22)
133
The relations between the the scalar and fermion Green’s functions are
S±(x− x′) = ± (i γµ ∂µ +m) ∆±(x− x′)
−i S(x− x′) = (i γµ ∂µ +m) i∆(x− x′)
−i S1(x− x′) = (i γµ ∂µ +m) i∆1(x− x′)
i SF (x− x′) = (i γµ ∂µ +m) i∆F (x− x′)
(6.23)
Hence it follows that
−i S(x− x′) =
ψ(x), ˆψ(x′)
= 0 when (x− x′)2 < 0
because S(x− x′) = 0 when (x− x′)2 < 0 (6.24)
micro causality
6.2 Green’s function for the electromagnetic field
In defining the Feynman propagator for the electromagnetic field one encounters aproblem that is related to gauge invariance Define
iDµνF = 〈0|T
(
Aµ(x)Aν(x′))
|0〉 (6.25)
Because of Bose Einstein statistics
T(
Aµ(x)Aν(x′))
= θ(x0 − x′0)Aµ(x)Aν(x′) + θ(x′0 − x0)A
ν(x′)Aµ(x′) (6.26)
Further we assume that in analogy with the scalar field and the fermion field iDµνF is
a Green’s function that satisfies the following differential equation
Lµν(x) iDµνF (x− x′) = i δλ
µ δ(4)(x− x′) (6.27)
with the equation of motion
Lµν(x)Aν(x) = 0 (6.28)
Lµν follows from the Euler Lagrange equation
L = −1
4Fµν F
µν with F µν = ∂µ Aν(x) − ∂ν Aµ(x)
∂ν ∂ L∂ ∂ν Aµ
=∂ L∂ Aµ
→ ∂ν Fµν = 0 (6.29)
134
or(
gµν 2 − ∂µ ∂ν
)
Aν(x) = 0 =⇒ Lµν(x) = gµν 2 − ∂µ ∂ν (6.30)
We have to invert Lµν in other words
DνλF = i
(
L−1)νµ
δλµ δ
(4)(x− x′) (6.31)
Unfortunately L−1 does not exist this is because L is locally gauge invariant. F µν isinvariant under the local gauge transformation
Aµ(x) → Aµ(x) + ∂µ Λ(x) =⇒ Fµν(x) → Fµν(x) (6.32)
This implies that if
Lµν(x)Aν(x) = 0 (6.33)
holds it is also true that
Lµν(x)(
Aν(x) + ∂ν Λ(x))
= 0 (6.34)
in other words
Lµν(x) ∂ν Λ(x) = 0 (6.35)
This implies that Lµν has an eigenvalue which is zero. Therefore Lµν is not invertible.To find a solution for Dµν
F we have to break the gauge invariance. This implies that wehave to make a gauge choice for Aµ. Let us make a covariant gauge choice
L′ = L − 1
2α
(
∂µAµ(x)
)2(6.36)
Application of the Euler-Lagrange equations yields
Lµν(x) = gµν 2 −(
1 − 1
α
)
∂µ ∂ν (6.37)
Because L′ is not invariant under local gauge transformations ∂ν Λ is not an eigenvectoranymore with eigenvalue zero. We can now invert Lµν
(
gµν 2 −(
1 − 1
α
)
∂µ ∂ν
)
iDνλF (x) = i δλ
µ δ(4)(x− x′) (6.38)
Take the following general solution
DνλF =
1
(2π)4
∫
d4k e−i k·(x−x′) 1
k2 + i ε
(
a gνλ + bkν kλ
k2
)
(6.39)
135
Substitution in Eq. (6.38) yields
1
(2π)4
∫
d4k e−i k·(x−x′) 1
k2 + i ε
(
− gµν k2 +
(
1 − 1
α
)
kµ kν
) (
a gνλ + bkν kλ
k2
)
=1
(2π)4
∫
d4k e−i k·(x−x′) δλµ (6.40)
Hence we have the solution
a = −1 a(
1 − 1
α
)
− b
α= 0 → b = 1 − α (6.41)
The Feynman propagator becomes
DνλF (x− x′) = − 1
(2π)4
∫
d4k e−i k·(x−x′) 1
k2 + i ε
(
gνλ − (1 − α)kν kλ
k2
)
(6.42)
Notice that the photon is massless. Two gauge choices are very popular. The Feynmangauge is given by α = 1 and the Landau gauge is given by α = 0. We could also choosethe radiation gauge. However this is not covariant. In this gauge the electromagneticfield is equal to
~A(~x, t) =∑
α=1,2
∫
d3 ~k
(2π)3/2
1√2ω
[
a(~k, α)~ǫ(α)(~k) e−i k·x + a†(~k, α)~ǫ(α)(~k) ei k·x]
(6.43)
Define
ǫ(α),µ(~k) =(
0,~ǫ(α)(~k))
(6.44)
so that we can write Aµ in a pseudo covariant way
Aµ(x) =∑
α=1,2
∫
d3 ~k
(2π)3/2
1√2ω
[
a(~k, α) ǫ(α),µ(~k) e−i k·x + a†(~k, α) ǫ(α),µ(~k) ei k·x]
(6.45)
The sum over the transverse polarisations is∑
α
ǫ(α),i(~k) ǫ(α),j(~k) = δij −ki kj
|~k|2(6.46)
Define
n = (1, 0, 0, 0) |~k|2 = (k · n)2 − k2 (6.47)
then
P µν(k) =∑
α
ǫ(α),µ(k) ǫ(α),ν(k) = −gµν − n2 kµ kν − k · n (kµ nν + nµ kν) + k2 nµ nν
(k · n)2 − k2
(6.48)
Therefore the Feynman propagator in this gauge looks like
iDµνF (x− x′) =
1
(2π)4
∫
d4k e−i k·(x−x′) i
k2 + i εP µν(k) (6.49)
In the subsequent sections we will choose the covariant gauge. Notice that S-matrixelements are independent of any gauge choice but Green’s functions are not.
136
6.3 Feynman rules for QED
The Lagrangian density in the case of a spin half field (electron, muon) is given by
LQED(x) = ψ(x) γµ ∂µ ψ(x) −mψ(x)ψ(x) − e ψ(x) γµ ψ(x)Aµ(x)
−1
4Fµν(x)F
µν(x) (6.50)
with
Fµν(x) = ∂µ Aν(x) − ∂ν Aµ(x) (6.51)
L(x) is invariant under local gauge transformations
Aµ(x) → A′µ(x) = Aµ(x) − 1
e∂µ α(x)
ψ(x) → ψ′(x) = ei α(x) ψ(x) (6.52)
The Feynman rules are given by
1 Draw at a given order in e all topological inequivalent connected graphs.
2 Each internal photon line is represented by the propagator i DµνF (k)
i DµνF (k) = − i
k2 + i ε
(
gµν − (1 − α)kµ kν
k2
)
(6.53)
3 Each internal fermion line is represented by the propagator i SF (p)
i SF (p) =i (/p+m)
p2 −m2 + i ε=
i
/p−m(6.54)
4 Each fermion-fermion-photon vertex with incoming momenta p1, p2, p3 is givenby
−i e γµ (2π)4 δ(4)(p1 + p2 + p3) (6.55)
5 Each incoming fermion line gets the spinor u(~p, s). Each outgoing fermion linegets the spinor u(~p, s). Each incoming anti-fermion line gets the anti-spinorv(~p, s). Each outgoing anti-fermion line gets the anti-spinor v(~p, s).
6 Each incoming or outgoing photon gets the polarisation vector ǫ(α)µ (k).
7 Integrate over each internal loop momentum k with the measure
∫ d4k
(2π)4(6.56)
137
8 A factor −1 is needed for each internal fermion loop.
These rules provide us with the amplitude Mfi (2π)4 δ(4)(∑
i pi). There are some ad-ditional comments. We get a relative minus sign if one Feynman diagram is obtainedfrom another one by exchanging two identical fermions. Finally we get an extra minussign if an incoming (outgoing) fermion is exchanged with an outgoing (incoming) anti-fermion. The latter is called crossing. Finally the anti-electron is called the positron.
6.4 Lowest order processes in QED
Our first application will be electron-electron scattering (Moller scattering)
e−(p1, s1) + e−(p2, s2) → e−(p3, s3) + e−(p4, s4) (6.57)
The amplitude Mfi is equal to
Mfi = i (−i e)2
[
u(p3, s3) γµ u(p1, s1) DµνF (p1 − p3) u(p4, s4) γν u(p2, s2)
−u(p3, s3) γµ u(p2, s2) DµνF (p2 − p3) u(p4, s4) γν u(p1, s1)
]
(6.58)
Notice that apparently Mfi depends on the gauge choice parameter α. However itturns out that this is not the case. This we will show below. Substitute i Dµν
F
Mfi = i e2[
u(p3, s3) γµ u(p1, s1)
u(p4, s4) γµ u(p2, s2)
(p1 − p3)2 + i ε
−
u(p3, s3) γµ u(p2, s2)
u(p4, s4) γµ u(p1, s1)
(p2 − p3)2 + i ε
−(1 − α)
u(p3, s3) (/p3 − /p1) u(p1, s1)
u(p4, s4) (/p3 − /p1) u(p2, s2)
(p1 − p3)2[
(p1 − p3)2 + i ε]
+(1 − α)
u(p3, s3) (/p3 − /p2) u(p2, s2)
u(p4, s4) (/p3 − /p2) u(p1, s1)
(p2 − p3)2[
(p2 − p3)2 + i ε]
]
(6.59)
Substitute at the end in the last two terms p3 − p1 = p2 − p4 and p3 − p2 = p1 − p4.Because of current conservation the last two terms in Mfi are equal to zero. Proof:
(p3 − p1)µ u(p3, s3) γµ u(p1, s1) = u(p3, s3) (/p3 − /p1) u(p1, s1)
= u(p3, s3)[
(/p3 −m) − (/p1 −m)]
u(p1, s1) = 0 (6.60)
138
idem
u(p4, s4) (/p2 − /p4) u(p2, s2) = 0 (6.61)
etc. In other words Mfi is independent of the gauge choice. Therefore only the term−i gµν/k
2 in DµνF (k) contributes. In general Feynman graphs depend on α. However
S-matrix elements or transition amplitudes Mfi never depend on α. On the other handGreen’s function depend on α. The calculation of the effective cross section for Mollerscattering proceeds as follows. Call the Mandelstam variables
s = (p1 + p2)2 = (p3 + p4)
2 t = (p1 − p3)2 = (p2 − p4)
2
u = (p2 − p3)2 = (p1 − p4)
2 (6.62)
The amplitude squared reads
|Mfi|2 = e4[
u(p3, s3) u(p3, s3) γµ u(p1, s1) u(p1, s1) γ
ν
u(p4, s4) u(p4, s4) γµ
u(p2, s2) u(p2, s2) γν
1
t2+
u(p3, s3) u(p3, s3) γµ u(p2, s2) u(p2, s2) γ
ν
u(p4, s4) u(p4, s4) γµ u(p1, s1) u(p1, s1) γν
1
u2−
u(p3, s3) u(p3, s3) γµ
u(p1, s1) u(p1, s1) γν u(p4, s4) u(p4, s4) γµ u(p2, s2) u(p2, s2) γν 1
t u
−
u(p3, s3) u(p3, s3) γν u(p2, s2) u(p2, s2) γµ u(p4, s4) u(p4, s4) γν
u(p1, s1) u(p1, s1) γµ 1
t u
]
(6.63)
Since t < 0 and u < 0 we can omit i ε. We consider the cross section where we sum overall spins in the final state 3 and 4. Therefore we look at the production of electronswhere each spin is allowed. Furthermore the incoming electrons unpolarised. Thisimplies that we also sum over the initial spins 1 and 2 provided we average over theinitial spins. The average factor for each incoming particle with spin si is 1/(2 si + 1).Notice that
∑
s
u(p, s) u(p, s) = /p+m and∑
s
v(p, s) v(p, s) = /p−m (6.64)
Using the above we have
|Mfi|2 =1
(2 s1 + 1) (2 s2 + 1)
∑
s1,s2,s3,s4
|Mfi|2 =e4
4
[
Tr
(/p3 +m) γµ (/p1 +m) γν
139
×Tr
(/p4 +m) γµ (/p2 +m) γν
1
t2+ Tr
(/p3 +m) γµ (/p2 +m) γν
×Tr
(/p4 +m) γµ (/p1 +m) γν
1
u2− Tr
(/p3 +m) γµ (/p1 +m) γν (/p4 +m)
×γµ (/p2 +m) γν 1
t u− Tr
(/p3 +m) γν (/p2 +m) γµ (/p4 +m) γν
×(/p2 +m) γµ 1
t u
]
(6.65)
We can calculate these traces. Use the following relations in n dimensions.
γµ, γν
= 2 gµν γµ γµ = δµ
µ = n γµ γα γµ = (2 − n) γα
γµ γα γβ γµ = 4 gαβ + (n− 4) γα γβ
γµ γα γβ γλ γµ = −2 γλ γβ γα − (n− 4) γα γβ γλ
γ5, γµ
= 0 (6.66)
The traces are
Tr(odd number of γ-matrices) = 0 Tr 1 = n Tr γα γβ = 4 gαβ
Tr γα γβ γλ γρ = 4
gαβ gλρ − gαλ gβρ + gαρ gβλ
Tr γ5 = 0 Tr γ5 γα γβ = 0 (6.67)
The generalisation to n dimensions is only necessary if we calculate loop integrals whichcontain ultraviolet divergences (n dimensional regularisation) otherwise we put n = 4.Let us calculate the traces which occur above
Tr (/p3 +m) γµ (/p1 +m) γν = Tr /p3 γµ /p1 γ
ν +m2 Tr γµ γν
= 4[
pµ3 p
ν1 + pν
3 pµ1 + gµν (m2 − p1 · p3)
]
Tr (/p4 +m) γµ (/p2 +m) γν = 4[
p4,µ p2,ν + p4,ν p2,µ + gµν (m2 − p2 · p4)]
Tr (/p3 +m) γµ (/p2 +m) γν = 4[
pµ3 p
ν2 + pν
3 pµ2 + gµν (m2 − p2 · p3)
]
Tr (/p4 +m) γµ (/p1 +m) γν = 4[
p4,µ p1,ν + p4,ν p1,µ + gµν (m2 − p1 · p4)]
(6.68)
Tr (/p3 +m) γµ (/p1 +m) γν (/p4 +m) γµ (/p2 +m) γν = Tr
/p3 γµ /p1 γν +mγµ /p1 γν
140
+m/p3 γµ γν +m2 γµ γν
/p4 γµ /p2 γν +mγµ /p2 γ
ν +m/p4 γµ γν +m2 γµ γ
ν
=
Tr /p3 γµ /p1 γν /p4 γµ /p2 γ
ν +m2 Tr /p3 γµ /p1 γν γµ γ
ν +m2 Tr γµ /p1 γν γµ /p2 γν
+m2 Tr γµ /p1 γν /p4 γµ γν +m2 Tr /p3 γ
µ γν γµ /p2 γν +m2 Tr /p3 γ
µ γν /p4 γµ γν
+m2 Tr γµ γν /p4 γµ /p2 γν +m4 Tr γµ γν γµ γ
ν = −2 Tr /p3 γµ /p1 /p2 γµ /p4
+4m2 p1,ν Tr /p3 γν + 4m2 p2,µ Tr γ
µ /p1 + 4m2 p4,µ Tr γµ /p1 + 4m2 p2,µ Tr /p3 γ
µ
+4m2 p4,µ Tr /p3 γµ + 4m2 p4,µ Tr /p2 γ
ν − 2m4 Tr γµ γµ
= −32 (p1 · p2) (p3 · p4) + 16m2(
p1 · p3 + p1 · p2 + p1 · p4 + p2 · p3 + p3 · p4
+p2 · p4
)
− 32m4 = −8 (s− 2m2)2 + 32m2 (s−m2) − 32m4
= −8 (s− 2m2) (s− 6m2) (6.69)
Tr (/p3 +m) γν (/p2 +m) γµ (/p4 +m) γν (/p1 +m) γµ = −8 (s− 2m2)2
+32m2 (s−m2) − 32m4 = −8 (s− 2m2) (s− 6m2) (6.70)
Tr
(/p3 +m) γµ (/p1 +m) γν
Tr
(/p4 +m) γµ (/p2 +m) γν
= 16
[
2 (p1 · p2) (p3 · p4)
+2 (p1 · p4) (p2 · p3) +m2(
4m2 − 2 (p1 · p3) − 2 (p2 · p4))
]
= 8 (s− 4m2)2
+8 (u− 4m2)2 − 64m4 (6.71)
Tr
(/p3 +m) γµ (/p2 +m) γν
Tr
(/p4 +m) γµ (/p1 +m) γν
= 8 (s− 4m2)2
+8 (t− 4m2)2 − 64m4 (6.72)
The spin average of the amplitude squared is then equal to
|Mfi|2 = 2 e4[
(s− 4m2)2 + (u− 4m2)2 − 8m4
t2+
(s− 4m2)2 + (t− 4m2)2 − 8m4
u2
+2 (s− 2m2) (s− 6m2)
t u
]
= 16 e4[
(s/2 −m2) (s/4 −m2)
t2 u2
−s2/2 − sm2 −m4
t u+
1
4
]
(6.73)
141
Here we have used s + t+ u = 4m2. Because the kinematics is the same as for scalarparticle scalar particle scattering (see Eq. (4.111)) we can write the effective crosssection (replace on page 78 in Eq. (4.111) λ2 by |Mfi|2).
d σ
dΩ=
1
64 π2 s|Mfi|2 (6.74)
Define the fine structure constant α = e2/4 π the cross section becomes equal to
d σ
dΩ=
α2
2 s
[
(s− 4m2)2 + (u− 4m2)2 − 8m4
t2+
(s− 4m2)2 + (t− 4m2)2 − 8m4
u2
+2 (s− 2m2) (s− 6m2)
t u
]
(6.75)
On page 77 we have calculated the energies and momenta of the incoming and outgoingparticles. Here we have for all i the value mi = m. This implies
Ei =1
2
√s |~pi| =
1
2
√s− 4m2 ≡ |~p| (6.76)
where |~p| is the centre of mass momentum and Ei the centre of mass energy for eachparticle i. The Mandelstam variables become
s = 4(
|~p|2 +m2)
t = −2 |~p|2 (1 − cos θ) = −4 |~p|2 sin2 θ
2
u = −2 |~p|2 (1 + cos θ) = −4 |~p|2 cos2 θ
2(6.77)
In the ultra relativistic limit |~p| ≫ m when s ∼ 4 |~p|2 the cross section becomes
(
d σ
dΩ
)UR
=α2
8 |~p|2[
1 + cos4 θ/2
sin4 θ/2+
1 + sin4 θ/2
cos4 θ/2+
2
sin2 θ/2 cos2 θ/2
]
(6.78)
Notice that in the forward direction θ = 0 and in the backward direction θ = π thecross section becomes infinite. This is because in the t-channel and in the u-channela massless particle (here the photon) is exchanged. In the non-relativistic limit where|~p| ≪ m and s ∼ 4m2, −t ≪ m2, −u ≪ m2 the cross section becomes
(
d σ
dΩ
)NR
=α2m2
16 |~p|4[
1
sin4 θ/2+
1
cos4 θ/2− 1
sin2 θ/2 cos2 θ/2
]
(6.79)
Because of the divergences which occur at θ = 0, π we cannot compute the total crosssection. The cross section for the process
e+(p′1, s′1) + e+(p′2, s
′2) → e+(p′3, s
′3) + e+(p′4, s
′4) (6.80)
142
is identical to that for Moller scattering.
The electron-positron annihilation process (Bhabha scattering)
e−(p1, s1) + e+(p′2, s′2) → e+(p′3, s
′3) + e−(p4, s4) (6.81)
can be obtained from the scattering amplitude for electron-electron scattering via cross-ing p2 → −p′3 and p3 → −p2. The Mandelstam variables s′, t′, u′ are
s = (p1 + p2)2 → t′ = (p1 − p′3)
2
t = (p1 − p3)2 → s′ = (p1 + p′2)
2
u = (p1 − p4)2 → u′ = (p1 − p4)
2 (6.82)
Under crossing the amplitude becomes
|Mfi|2 = 2 e4[
(t′ − 4m2)2 + (u′ − 4m2)2 − 8m4
s′2
+(t′ − 4m2)2 + (s′ − 4m2)2 − 8m4
u′2+
2 (s′ − 2m2) (s′ − 6m2)
s′ u′
]
(6.83)
The cross section becomes
d σ
dΩ=
1
64 π2 s′|Mfi|2 (6.84)
In the centre of mass system we have
s′ = 4(
|~p|2 +m2)
t′ = −2 |~p|2 (1 + cos θ) = −4 |~p|2 cos2 θ
2
u′ = −2 |~p|2 (1 − cos θ) = −4 |~p|2 sin2 θ
2(6.85)
In the ultra relativistic limit |~p| ≫ m
(
d σ
dΩ
)UR
=α2
8 |~p|2[
1 + cos4 θ2
sin4 θ2
− 2 cos4 θ2
sin2 θ2
+1 + cos2 θ
2
]
(6.86)
Notice that also in this case the cross section diverges if θ = 0. Therefore the totalcross section does not exist.
The photon-electron scattering (Compton scattering) process is given by
γ(p1, i) + e−(p2, s2) → γ(p3, j) + e−(p4, s4) (6.87)
143
The transition amplitude becomes
Mfi = i (−i e)2 u(p4, s4) ǫνj (p3)
[
γν (/p1 + /p2 +m) γµ
(p1 + p2)2 −m2 + i ε+
γµ (/p2 − /p3 +m) γν
(p2 − p3)2 −m2 + i ε
]
×u(p2, s2) ǫµi (p1) = ǫνj (p3)Tµν ǫ
µi (p1) (6.88)
with
Tµν = −i e2 u(p4, s4)
[
γν (/p1 + /p2 +m) γµ
(p1 + p2)2 −m2 + i ε+
γµ (/p2 − /p3 +m) γν
(p2 − p3)2 −m2 + i ε
]
u(p2, s2) (6.89)
The transverse polarisations are given i, j = 1, 2. Further we have the Ward identity
pµ1 Tµν = 0 pν
3 Tµν = 0 (6.90)
The proof of the Ward identity involves the equations of motion
(/p2 −m) u(p2, s2) = 0 u(p4, s4) (/p4 −m) (6.91)
The ward identity implies that we can make a gauge transformation
Aµ → A′µ = Aµ − 1
e∂µ α(x) ǫµi (k) → ǫ
′µi (k) = ǫµi (k) + λ kµ (6.92)
Under this transformation Mfi is invariant
ǫ′νj (p3)Tµν ǫ
′µi (p1) =
(
ǫνj (p3) + λ pν3
)
Tµν
(
ǫµi (p1) + λ pµ1
)
= ǫνj (p3)Tµν ǫµi (p1) (6.93)
Because kµ ǫµi (k) = 0 (i = 1, 2) we can choose an arbitrary vector nµ for which
nµ ǫµi (k) = 0. This is equivalent with a gauge choice n · A = 0. We have in a given
Lorentz frame four mutual perpendicular vectors i.e. kµ, nµ, ǫµ1 (k), ǫµ2(k). For on-shell
photons we have the relation
P µν(k) =∑
i
ǫµi (k) ǫνi (k) = −gµν +kµ nν + nµ kν
n · k − n2 kµ kν
(n · k)2(6.94)
If we now sum again over all polarisation of the in- and outgoing photons and averageover the incoming polarisations then
|Mfi|2 =1
2P νσ(p3)Tµν Tλσ P
µλ(p1) (6.95)
From de Ward identity follows
|Mfi|2 =1
2(−gνσ)Tµν Tλσ (−gµλ) =
1
2Tµν T
µν (6.96)
144
If Tµν satisfies the Ward identity then we are allowed to replace P µν by −gµν . Summingalso over the spins of the incoming and outgoing electron and average over the initialspins we get
|Mfi|2 =e4
4
[
Tr (/p4 +m) γν (/p1 + /p2 +m) γµ (/p2 +m) γµ (/p1 + /p2 +m) γν
(s−m2)2
+Tr (/p4 +m) γµ (/p2 − /p3 +m) γν (/p2 +m) γν (/p2 − /p3 +m) γµ
(u−m2)2
+Tr (/p4 +m) γν (/p1 + /p2 +m) γµ (/p2 +m) γν (/p2 − /p3 +m) γµ
(s−m2) (u−m2)
+Tr (/p4 +m) γµ (/p2 − /p3 +m) γν (/p2 +m) γµ (/p1 + /p2 +m) γν
(s−m2) (u−m2)
]
(6.97)
The Mandelstam variables are
s = (p1 + p2)2 = m2 + 2 p1 · p2 = (p3 + p4)
2 = m2 + 2 p3 · p4
t = (p1 − p3)2 = −2 p1 · p3 = (p2 − p4)
2 = 2m2 − 2 p2 · p4
u = (p1 − p4)2 = m2 − 2 p1 · p4 = (p2 − p3)
2 = m2 − 2 p2 · p3
s+ t+ u = 2m2 (6.98)
The traces are
Tr (/p4 +m) γν (/p1 + /p2 +m) γµ (/p2 +m) γµ (/p1 + /p2 +m) γν
= 8(
− s u+ 3 sm2 + um2 +m4)
Tr (/p4 +m) γµ (/p2 − /p3 +m) γν (/p2 +m) γν (/p2 − /p3 +m) γµ
= 8(
− s u+ 3 um2 + sm2 +m4)
Tr (/p4 +m) γν (/p1 + /p2 +m) γµ (/p2 +m) γν (/p2 − /p3 +m) γµ
= 8m2(
s+ u+ 2m2)
Tr (/p4 +m) γµ (/p2 − /p3 +m) γν (/p2 +m) γµ (/p1 + /p2 +m) γν
= 8m2(
s+ u+ 2m2)
(6.99)
145
The amplitude squared becomes
|Mfi|2 = 2 e4[
−s u+ 3 sm2 + um2 +m4
(s−m2)2+
−s u+ 3 um2 + sm2 +m4
(u−m2)2
+2m2 (s+ u+ 2m2)
(s−m2) (u−m2)
]
(6.100)
In the centre of mass frame the energies and momenta get the following form
E1 =s−m2
2√s
E2 =s+m2
2√s
E3 =s−m2
2√s
E4 =s+m2
2√s
|~pi| =s−m2
2√s
(6.101)
The two Mandelstam variables become
t =(s−m2)2
2 s(1 − cos θ) u =
m4
s− (s−m2)2
2 s(1 + cos θ) (6.102)
The differential cross section in the centre of mass frame is equal to (replace λ2 by|Mfi|2 in Eq. (4.111) )
d σ
dΩ=
1
64 π2 s|Mfi|2
=α2
2 s
[
−s u+ 3 sm2 + um2 +m4
(s−m2)2+
−s u+ 3 um2 + sm2 +m4
(u−m2)2
+2m2 (s+ u+ 2m2)
(s−m2) (u−m2)
]
(6.103)
We can now calculate the total cross section because the particle in the u-channel ismassive
σtot =∫
dΩd σ
dΩ=∫ 1
−1d cos θ
∫ 2π
0d φ
d σ
dΩ
=π α2
s
[
− 16m6
(s−m2)3− 24m4
(s−m2)2− 6m2
s−m2+ 2
lns
m2+
16m4
(s−m2)2
+16m2
s−m2+m2
s+ 1
]
(6.104)
146
In the non-relativistic limit (s ∼ m2)
σNRtot =
8 π α2
3m2Thomson cross section (6.105)
r0 = α/m is the classical radius of the electron (r0 = 2.82.10−15m). In the ultrarelativistic limit (s≫ m2) we have
σURtot =
2 π α2
s
[
lns
m2+
1
2
]
(6.106)
From Compton scattering we can derive the amplitude of the two other processes viacrossing. They are given by fermion pair creation
γ(p1, i) + γ(p′2, j) → e+(p′3, s′3) + e−(p4, s4) (6.107)
and fermion pair annihilation
e+(p′1, s′1) + e−(p2, s2) → γ(p3, i) + γ(p′4, j) (6.108)
If we start with fermion pair creation we have to substitute for the process in Eq. (6.87)
p3 → −p′2 p2 → −p′3 =⇒ |Mfi|2 → −|Mfi|2 (6.109)
The Mandelstam variables become
s = (p1 + p2)2 → t′ = (p1 − p′3)
2
t = (p1 − p3)2 → s′ = (p1 + p′2)
2
u = (p1 − p4)2 → u′ = (p1 − p4)
2 (6.110)
In the centre of mass system the energies and momenta are
E1 = E2′ =1
2
√s′ E3′ = E4 =
1
2
√s′
|~p1| = |~p2′ | =1
2
√s′ |~p3′ | = |~p4| =
1
2
√s′ − 4m2 (6.111)
and the Mandelstam variables become
t′ = −s′
2+m2 +
1
2
√
s′(s′ − 4m2) cos θ
u′ = −s′
2+m2 − 1
2
√
s′(s′ − 4m2) cos θ (6.112)
147
The differential cross section becomes (see Eq. (4.110))
d σ
dΩ=
1
64 π2 s′
√
s′(s′ − 4m2)
s′|Mfi|2
=α2
2 s′
√
1 − 4m2
s′
[
t′ u′ − 3 t′m2 − u′m2 −m4
(t′ −m2)2+t′ u′ − 3 u′m2 − tm2 −m4
(u′ −m2)2
+2m2 (s′ −m2)
(t′ −m2) (u′ −m2)
]
(6.113)
The total cross section is equal to
σtot =4 π α2
s′
[
−(
1 +4m2
s′
)
√
1 − 4m2
s′+
(
1 +4m2
s′− 8m4
s′2
)
× ln1 +
√
1 − 4m2/s′
1 −√
1 − 4m2/s′
]
(6.114)
Notice that fermion pair creation is a typical relativistic process which has no counterpart in the non-relativistic limit. Near threshold it behaves like
σtot =s′ ∼ 4m2
π α2
m2
√
1 − 4m2
s′(6.115)
and in the ultra relativistic limit the total cross section behaves like
σURtot =s′ ≫ m2
4 π α2
s′
[
lns′
m2− 1
]
(6.116)
The amplitude for fermion pair annihilation Eq. (6.108) is obtained from the processin Eq. (6.87) by substituting
p4 → −p′1 p1 → −p′4 =⇒ |Mfi|2 → −|Mfi|2 (6.117)
The Mandelstam variables become
s = (p1 + p2)2 → t′ = (p2 − p′4)
2
t = (p1 − p3)2 → s′ = (p3 + p′4)
2
u = (p2 − p3)2 → u′ = (p2 − p3)
2 (6.118)
148
In the centre of mass system the energies and momenta are
E1′ = E2 =1
2
√s′ E3 = E4′ =
1
2
√s′
|~p1′ | = |~p2| =1
2
√s′ − 4m2 |~p3| = |~p4′| =
1
2
√s′ (6.119)
and the Mandelstam variables become
t′ = −s′
2+m2 +
1
2
√
s′(s′ − 4m2) cos θ
u′ = −s′
2+m2 − 1
2
√
s′(s′ − 4m2) cos θ (6.120)
The differential cross section reads
d σ
dΩ=
1
64 π2
1√
s′(s′ − 4m2)|Mfi|2
=α2
2 s′1
√
1 − 4m2/s′
[
t′ u′ − 3 t′m2 − u′m2 −m4
(t′ −m2)2+t′ u′ − 3 u′m2 − t′m2 −m4
(u′ −m2)2
+2m2 (s′ −m2)
(t′ −m2) (u′ −m2)
]
(6.121)
The total cross section is equal to
σtot =1
2
∫
dΩd σ
dΩ
=2 π α2
s′ − 4m2
[
−(
1 +4m2
s′
)
√
1 − 4m2
s′+
(
1 +4m2
s′− 8m4
s′2
)
× ln1 +
√
1 − 4m2/s′
1 −√
1 − 4m2/s′
]
(6.122)
The factor 1/2 is due to the fact that we have two identical particles (here photons)in the final state. Like in the previous case this process is relativistic and it has nonon-relativistic counter part. Near threshold it behaves like
σtot =s′ ∼ 4m2
π α2
2m2
1√
1 − 4m2/s′(6.123)
and in the ultra relativistic limit the total cross section equals
σURtot =s′ ≫ m2
2 π α2
s′
[
lns′
m2− 1
]
(6.124)
149