Lecture 12
8.324 Relativistic Quantum Field Theory II
Fall 2010
8.324 Relativistic Quantum Field Theory II
MIT OpenCourseWare Lecture Notes
Hong Liu, Fall 2010
Lecture 12
3: GENERAL ASPECTS OF QUANTUM ELECTRODYNAMICS
3.1: RENORMALIZED LAGRANGIAN
Consider the Lagrangian of quantum electrodynamics in terms of the bare quantities:
1 B µβ
L = − Fµβ
FB − iψ¯B (γ µ (∂µ − ieB AB
µ ) − mB )ψB .
4
(1)
{γ µ , γ β } = 2η µβ ,
(2)
We use the convention:
γ0† = −γ0 ,
γ02 = −1,
γi† = γi,
ψ̄ = ψ † γ0 , /{k} = k µ γµ ,
2
k/ = k 2 .
where, in four dimensions, (η µβ ) = diag(−1, 1, 1, 1). We expect to find the mass and field renormalizations.
Note: we will omit the “B” signifying bare quantities in what follows.
ψ
:
+
+ ...
Aµ
:
+
+ ...
(3)
along with vertex corrections
e:
+
+ ....
(4)
We will look at how to introduce renormalized quantities.
3.1.1: Fermion self-energy
We have that
S�� (x) = ⟨0| T (ψ� (x)ψ̄� (0) |0⟩ ,
(5)
and
1P I = −�
S�� (k)
=
�
�
+
�
�
�
α
+
= S0 (k) + S0 (−�)S0 + S0 (−�)S0 (−�)S0 + . . .
1
= S0
,
1 + �S0
1
�
�
+ ...
(6)
Lecture 12
8.324 Relativistic Quantum Field Theory II
Fall 2010
where we have omitted the spinor indices in the second and third lines: these are to be read as matrix equations.
Hence, we have that
S −1 = S0−1 + �.
(7)
Recall that
S0 =
−1
, −� =
ik/ + mB
and so
+ ...,
(8)
S −1 = −(ik/ + mB ) + �,
(9)
and we have for the fully interacting two-point function
−1
.
ik/ + mB − iϵ − �
S=
(10)
Note that � = �(k/), since it can only depend on k/ and k 2 . Even though it is a function of matrix, we can treat it
as an ordinary function. The physical mass is defined by k/ = im, so that
−m + mB − �(im) = 0.
(11)
We note that, again, � will be divergent. Near the pole, we have that
S≈
with Z2−1 = 1 + i
d� / /
dk
k=im
−Z2
ik/ + m − iϵ
(12)
. The relations between the bare and physical quantities are given by
mB = m + δm, ψB =
√
Z2 ψ
(13)
where δm = �(im) and ψ is the renormalized field.
3.1.2: Photon self-energy
Similarly, we have that
B
Dµβ (x) = ⟨0| T (AB
µ (x)Aβ (0) |0⟩ ,
(14)
and
1P I = i�
Dµβ (k) =
=
=
µ
β
+
µ
γ
δ
β
+
µ
β
+ ...
D0 (k) + D0 (i�)D0 + D0 (i�)D0 (i�)D0 + . . .
1
D0
1 − i�D0
Hence,
(iD)
Recall that
iD0µβ
=
=
−1
= (iD0 )
−1
(15)
− �.
(16)
[
]
1
kµ kβ
µβ
η
−
(1
−
ξ)
k 2 − iϵ
k2
1
[P µβ + ξPLµβ ] ,
k 2 − iϵ T
µ β
µ β
where we have defined the transverse projector PTµβ ≡ η µβ − k kk2 , and the longitudinal projector PLµβ ≡ k kk2 . Note
that it is not a coincidence that the propagator can be built from these two tensors, η µβ and k µ k β : they are the
only two two-tensors allowed by symmetry. These projectors satisfy the properties
T
L
L
PTµβ Pβα
= PTµβ , PLµβ Pβα
= PLµα , PTµβ Pβα
= 0.
2
(17)
Lecture 12
8.324 Relativistic Quantum Field Theory II
Fall 2010
Note: T and L are just labels here, and the placing of these indices does not carry meaning. Hence, we have that
[
]
1 µβ
µβ
−1
2
(iD0 ) = k PT + PL .
(18)
ξ
We may also expand �µβ as
�µβ
PTµβ fT (k 2 ) + PLµβ fL (k 2 )
kµ kβ
= η µβ fT + 2 (fL − fT ).
k
=
(19)
Therefore,
(iD)
−1
= PTµβ (k 2 − fT ) + PLµβ (
and we have for the full interacting photon two-point function,
[
1
D = −i PTµβ 2
+ PLµβ
k − fT
k2
− fL ),
ξ
k2
η
]
1
.
− fL
(20)
(21)
We observe that if fT,L (k 2 = 0) ̸= 0, a mass will be generated for the photon. Because �µβ comes from 1PI
diagrams, it should not be singular at k 2 = 0, and so fL − fT = O(k 2 ), as k −→ 0. We will show that gauge
invariance ensures that no mass is generated from the loop corrections.
3.1.3: Ward identities
Consider the path integral for the generating functional:
ˆ
Z [Jµ , η, η̄] = DAµ DψDψ̄eiS [Aµ ,λ,λ̄]
(22)
´
where S = SQED + d4 x Jµ AµB + η̄ψB +ψ̄B η, where we note explicitly these couplings are in terms of bare quantities.
1
1
LQED = − Fµβ F µβ − iψ̄(γ µ Dµ − m)ψ − (∂µ Aµ )2 .
4
2ξ
(23)
We define the generating functional for connected diagrams, W [Jµ , η, η̄], by
Z [Jµ , η, η̄] = eiW [Jµ ,�,�¯] .
(24)
For example,
⟨0| T (ψ� (x)ψ̄� (y)) |0⟩
=
B
⟨0| T (AB
µ (x)Aβ (y)) |0⟩
=
δ 2 W [Jµ , η, η̄] i
,
δη� (x)δη� (y) J =�=�¯=0
δ 2 W [Jµ , η, η̄] i
.
δJ µ (x)δJ β (y) J=�=�¯=0
¯ Consider a change of
Recall, for infinitesimal gauge transformations, δAµ = ∂µ λ, δψ = ieB λψ, and δ ψ¯ = −ieB λψ.
variables in the path integral:
Aµ
ψ
ψ̄
Then we have
ˆ
−→ A′µ = Aµ + δAµ ,
−→ ψ ′ = ψ + δψ,
¯
−→ ψ̄ ′ = ψ̄ + δ ψ.
′
′
′
DA′µ Dψ ′ Dψ̄ ′ eiS [Aµ ,λ ,λ̄ ]
ˆ
=
¯ iS [Aµ ,λ,λ̄] ,
DAµ DψDψe
(25)
as this is just of a change of the dummy integration variables. Note that the measure is unchanged by this shift:
¯
DA′µ Dψ ′ Dψ̄ ′ = DAµ DψDψ,
3
(26)
Lecture 12
8.324 Relativistic Quantum Field Theory II
and the action for the two sets of variables are related by
ˆ
ˆ
[ ′ ′ ′]
[ ′ ′ ′] 1
4
µ 2
¯
S Aµ , ψ , ψ̄ = S Aµ , ψ , ψ̄ −
d x ∂µ A ∂ λ + d4 x Jµ ∂ µ λ + ieB λ¯
ηψ − ieB λψη.
ξ
Hence, we must have
[
]
ˆ
ˆ
¯ iS [A,λ,λ̄] − 1 ∂ 2 ∂µ Aµ − ∂µ J µ + ieB (η̄ψ − ψη)
¯
d4 x λ(x) DAµ DψDψe
= 0.
ξ
Fall 2010
(27)
(28)
Since
δZ
δW
=Z µ
,
µ
δJ (x)
δJ (x)
δZ
δW
ψ(x) ∼ −i
=Z
,
δη̄(x)
δη̄(x)
δZ
δW
¯
ψ(x)
∼ −i
=Z
,
δη(x)
δη(x)
Aµ (x) ∼ −i
we have that
1 2 µ
δ2 W
∂ ∂
+ ∂β δ (4) (x − y) = 0,
µ
β
ξ
δJ (x)δJ (y) J=�=�¯=0
(29)
i 2 µ
∂ ∂ Dµβ (x − y) + ∂β δ (4) (x − y) = 0,
ξ
(30)
i
− k 2 k µ Dµβ (k) + kβ = 0.
ξ
(31)
T
L
Dµβ (k) = Pµβ
DT (k 2 ) + Pµβ
DL (k 2 ),
(32)
that is,
or, written in momentum-space,
If we now write
L
with k µ Pµβ
= kβ , the Ward identity reduces to
−i 2
k kβ DL (k 2 ) + kβ = 0,
ξ
(33)
and so
iξ
,
(34)
k2
and the longitudinal part of the two-point function is completely determined. Comparing this with (21), we find
that fL (k 2 ) = 0, and we thus conclude that �µβ is purely transverse. That is, from (19), we have that
(
)
kµ kβ
�µβ = η µβ − 2
fT (k 2 ).
(35)
k
DL (k 2 ) = −
For �µβ (k) to be non-singular at k = 0, we must have
fT (k 2 ) = k 2 �(k 2 ),
where �(0) is non-singular. Hence, for the two-point function in the interacting theory, we have
[
]
T
Pµβ
−i
L
Dµβ = 2
+ ξPµβ .
k − iϵ 1 − �(k 2 )
(36)
(37)
Remarks:
1.
The longitudinal part of Dµβ does not receive any loop corrections: it is completely determined by the
Ward identities. The physics should not depend on this part. For example, in the Landau gauge, ξ = 0,
Dµβ is purely transverse.
4
Lecture 12
8.324 Relativistic Quantum Field Theory II
Fall 2010
2.
Since �(k 2 ) is non-singular at k 2 = 0, the photon remains massless to all orders. There are exceptions
to this: it is not true in quantum electrodynamics in 1 + 1 dimensions, or in theories where an additional
Higgs field is introduced.
3.
The residue at the k 2 = 0 pole is given by Z3−1 = 1 − �(0), and we have that
Z3
P T
− iϵ µβ
√
near k 2 = 0. The renormalized field is given by AB
Z3 Aµ .
µ =
T
iDµβ
≈
5
k2
(38)
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8.324 Relativistic Quantum Field Theory II
Fall 2010 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.