Advanced(Quantum(Field(Theory(–the(fundamental(forces((( · 2017. 1. 16. ·...
Transcript of Advanced(Quantum(Field(Theory(–the(fundamental(forces((( · 2017. 1. 16. ·...
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Advanced Quantum Field Theory – the fundamental forces
QED • Scalar electrodynamics Spinor electrodynamics
Nonabelian gauge theory • QCD Electroweak
Books Quantum Field Theory – Mark Srednicki IntroducEon to Quantum Field Theory – Peskin & Schroeder
The Quantum Field Theory of Fields – Weinberg
Field Theory: A Modern Primer – Pierre Ramond
Quantum Field Theory in a nutshell – Zee
} The Standard Model
Quantum Field Theory and the Standard Model -‐ Schwartz
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QED • Scalar electrodynamics Spinor electrodynamics
Nonabelian gauge theory • QCD Electroweak
Books Quantum Field Theory – Mark Srednicki hOp://www.physics.ucsb.edu/~mark/qT.html
Web
Advanced Quantum Field Theory – the fundamental forces
hOp://www2.physics.ox.ac.uk/contacts/people/ross
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Outline!!
• QED!Introduction• Photon!propagator• Scalar!QED• Radiative(corrections• Fermions• Fermion(path!integral• Feynman'rules'"!Yukawa&theory• Spinor!QED• Sample&calculations• Beta$functions• Nonabelian)gauge)theory• NAGT"Feynman"rules• BRST!symmetry• Gauge&fixing• Spontaneous*symmetry*breaking• Anomalies• The$Standard$Model
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Quantum Electrodynamics (QED)
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Construction of a relativistic field theory
Lagrangian L T V= −
Action
2
1
t
t
S L dt= ∫
Classical path … minimises action
Quantum mechanics … sum over all paths with amplitude /iSe∝ !
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Lagrangian invariant under all the symmetries of nature
(Nonrelativistic mechanics)
-makes it easy to construct viable theories
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L = L∫ d 3x, L lagrangian density - D=4
Klein Gordon field φ(x)
( )† 2 †( ) ( ) ( ) ( )x x m x xµµφ φ φ φ∂ ∂ −L= } } T V
Manifestly Lorentz invariant
Lagrangian formulation of the Klein Gordon equation
Dφ = 1
complex scalar field
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L = L∫ d 3x, L lagrangian density
Klein Gordon field ( )xφ
L = ∂µφ(x)( )† ∂µφ(x) − m2φ(x)†φ(x)
δS = 0 ⇒ ∂L
∂φ− ∂µ
∂L∂(∂µφ)
= 0
Manifestly Lorentz invariant
Euler Lagrange equation
Lagrangian formulation of the Klein Gordon equation
Classical path :
(∂µ∂
µ + m2 )φ = 0 Klein Gordon equation
δL = ∂L
∂φδφ + ∂L
∂(∂µφ)δ (∂µφ)
=
∂L∂φ
− ∂µ∂L
∂(∂µφ)⎛⎝⎜
⎞⎠⎟
⎡
⎣⎢⎢
⎤
⎦⎥⎥δφ + ∂µ ∂L
∂(∂µφ)δφ
⎛⎝⎜
⎞⎠⎟
surface integral in S..vanishes
δSδφ
= 0
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New symmetries
( )† 2 †( ) ( ) ( ) ( )x x m x xµµφ φ φ φ∂ ∂ −L=
Is invariant under
A symmetry implies a conserved current and charge.
Translation Momentum conservation
Rotation Angular momentum conservation
e.g.
What conservation law does the U(1) invariance imply?
…an Abelian (U(1)) gauge symmetry ( ) ( )ix e xαφ φ→
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Noether current
( )† 2 †( ) ( ) ( ) ( )x x m x xµµφ φ φ φ∂ ∂ −L=
Is invariant under φ(x)→ eiαφ(x) …an Abelian (U(1)) gauge symmetry
†0 ( ) ( )( )
µµδ δφ δ φ φ φφ φ
∂ ∂= + ∂ + ↔∂ ∂ ∂L L
L=
iαφ i µα φ∂
†( )( ) ( )
i iµ µµ µα φ α φ φ φφ φ φ⎡ ⎤⎛ ⎞ ⎛ ⎞∂ ∂ ∂= −∂ + ∂ − ↔⎢ ⎥⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎣ ⎦
L L L
0 (Euler lagrange eqs.)
∂µ jµ = 0, jµ = iα
∂L∂(∂µφ)
φ − ∂L∂(∂µφ† )
φ†⎛⎝⎜
⎞⎠⎟
Noether current
True locally
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The Klein Gordon current
( )† 2 †( ) ( ) ( ) ( )x x m x xµµφ φ φ φ∂ ∂ −L=
Is invariant under …an Abelian (U(1)) gauge symmetry
∂µ jµ = 0, jµ = iα
∂L∂(∂µφ)
φ − ∂L∂(∂µφ† )
φ†⎛⎝⎜
⎞⎠⎟
jµ
KG = −iα φ* ∂µφ −φ ∂µφ*( )
φ(x)→ eiαφ(x)
Classical electrodynamics: motion of particle of charge in EM potential Aµ = ( A0 ,A)
given by pµ → pµ +QAµ , i∂µ→ i∂µ +QAµ
⇒ jµ
KG ≡ jµEM
−Q
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The Klein Gordon current
( )† 2 †( ) ( ) ( ) ( )x x m x xµµφ φ φ φ∂ ∂ −L=
Is invariant under …an Abelian (U(1)) gauge symmetry
∂µ jµ = 0, jµ = iα
∂L∂(∂µφ)
φ − ∂L∂(∂µφ† )
φ†⎛⎝⎜
⎞⎠⎟
jµ
KG = −iα φ* ∂µφ −φ ∂µφ*( )
φ(x)→ eiαφ(x)
∂∂t
j0(x)+∇.J(x) = 0
3 0Q d x j= ∫ is the associated conserved charge (Gauss’ law)
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Suppose we have two fields with different U(1) charges :
1,21,2 1,2( ) ( )
i Qx e xαφ φ→
( )( )
† 2 †1 1 1 1
† 2 †2 2 2 2
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
x x m x x
x x m x x
µµ
µµ
φ φ φ φ
φ φ φ φ
∂ ∂ −
+ ∂ ∂ −
L=
..no cross terms possible (corresponding to charge conservation)
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( )† 2 †( ) ( ) ( ) ( )x x m x xµµφ φ φ φ∂ ∂ −L= not invariant due to derivatives
( )( ) ( )i x Qx e xαφ φ→
( ) ( ) ( ) ( )i x Q i x Q i x Qe e iQe xα α αµ µ µ µφ φ φ φ α∂ → ∂ = ∂ + ∂
To obtain invariant Lagrangian look for a modified derivative transforming covariantly
Dµφ → e
iα ( x )Q Dµφ
U(1) local gauge invariance and QED
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( )† 2 †( ) ( ) ( ) ( )x x m x xµµφ φ φ φ∂ ∂ −L= not invariant due to derivatives
( )( ) ( )i x Qx e xαφ φ→
∂µφ − iQAµφ → e
iα ( x )Q (∂µφ − iQAµφ)+ iQeiα ( x )Qφ ∂µα (x)− iQe
iα ( x )Qφ ∂µα (x)
To obtain invariant Lagrangian look for a modified derivative transforming covariantly
Dµφ → e
iα ( x )Q Dµφ
D iQAµ µ µ= ∂ −
A Aµ µ µα→ + ∂Need to introduce a new vector field
U(1) local gauge invariance and QED
Aµ ≡ (V ,A)( )
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= ∂µφ( )† ∂µφ − m2φ†φ − iα Aµ (φ ∂µφ* −φ* ∂µφ)+α 2 Aµ Aµ
J µ ≡ (ρ,J)( )
( )( ) ( )iQ xx e xαφ φ→
Dµφ → e
iα ( x )Q Dµφ
A Aµ µ µα→ + ∂
( )† 2 †( ) ( ) ( ) ( )D x D x m x xµµφ φ φ φ−L= is invariant under local U(1)
Dµ → eiαQ Dµe
− iαQ
jµ = iα
∂L∂(∂µφ)
φ − ∂L∂(∂µφ† )
φ†⎛⎝⎜
⎞⎠⎟
= iα (φ ∂µφ
* −φ* ∂µφ)− 2α2 Aµφ
*φNoether current in free theory
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Note : D iQAµ µ µ µ∂ → = ∂ − is equivalent to p p eAµ µ µ→ +
universal coupling of electromagnetism follows from local gauge invariance
i.e. L = LKG = ∂µφ(x)( )† ∂µφ(x)− m2φ(x)†φ(x)− jµKG Aµ +O(Q2 )
J µ ≡ (ρ,J)( )
( )( ) ( )iQ xx e xαφ φ→
Dµφ → e
iα ( x )Q Dµφ
A Aµ µ µα→ + ∂
( )† 2 †( ) ( ) ( ) ( )D x D x m x xµµφ φ φ φ−L= is invariant under local U(1)
Dµ → eiαQ Dµe
− iαQ
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( )( ) ( )iQ xx e xαφ φ→
Dµφ → e
iα ( x )Q Dµφ
A Aµ µ µα→ + ∂
( )† 2 †( ) ( ) ( ) ( )D x D x m x xµµφ φ φ φ−L= is invariant under local U(1)
Note : D iQAµ µ µ µ∂ → = ∂ − is equivalent to pµ → pµ +QAµ
universal coupling of electromagnetism follows from local gauge invariance
(∂µ∂
µ + m2 )φ = −Vφ where V = −iQ(∂µ Aµ + Aµ ∂µ )−Q
2 A2
The Euler lagrange equation give the KG equation:
Dµ → eiαQ Dµe
− iαQ
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The electromagnetic Lagrangian
F A Aµν µ ν ν µ= ∂ − ∂
, F F A Aµν µν µ µ µα→ → + ∂
− 14 Fµν F
µν
1 2 3
1 3 2
2 3 1
3 2 1
00
00
E E EE B BE B BE B B
− − −⎛ ⎞⎜ ⎟−⎜ ⎟⎜ ⎟−⎜ ⎟⎜ ⎟−⎝ ⎠
2M A Aµ µ Forbidden by gauge invariance
F µν = i
QDµ , Dν⎡⎣ ⎤⎦
⎛⎝⎜
⎞⎠⎟
Invariant:
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Scalar QED Lagrangian
L = (Dµφ)† Dµφ − m
2φ†φ − 14
F µν Fµν
Euler Lagrange equaEon
DµD
µφ − m2φ = 0
Noether current
jµ = −iQ[φ†Dµφ − (Dµφ)†φ]
L = − 14 Fµν F
µν + jµ Aµ + ∂µφ* ∂µφ − m
2φ*φ
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The electromagnetic Lagrangian
F A Aµν µ ν ν µ= ∂ − ∂
, F F A Aµν µν µ µ µα→ → + ∂
LEM = − 14 Fµν F
µν + jµ Aµ
1 2 3
1 3 2
2 3 1
3 2 1
00
00
E E EE B BE B BE B B
− − −⎛ ⎞⎜ ⎟−⎜ ⎟⎜ ⎟−⎜ ⎟⎜ ⎟−⎝ ⎠
2M A Aµ µ Forbidden by gauge invariance
F µν = i
QDµ , Dν⎡⎣ ⎤⎦
⎛⎝⎜
⎞⎠⎟
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The electromagnetic Lagrangian
F A Aµν µ ν ν µ= ∂ − ∂
, F F A Aµν µν µ µ µα→ → + ∂
LEM = − 14 Fµν F
µν + jµ Aµ
The Euler-Lagrange equations give Maxwell equations !
F jµν νµ∂ =. , 0
. 0,
tEt
ρ ∂∇ = ∇× + =∂∂∇ = ∇× − =∂
BE E
B B j≡
1 2 3
1 3 2
2 3 1
3 2 1
00
00
E E EE B BE B BE B B
− − −⎛ ⎞⎜ ⎟−⎜ ⎟⎜ ⎟−⎜ ⎟⎜ ⎟−⎝ ⎠
0)A A
µν µ ν
∂ ∂− ∂ =∂ ∂ ∂L L
(
2M A Aµ µ Forbidden by gauge invariance
EM dynamics follows from a local gauge symmetry!!
N.B. εµνρσ∂µFρσ = 0( )
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Degrees of freedom
Aµ (x) : 4 degrees of freedom?
Lorentz transformaEon: can remove one more degree of freedom
No Eme derivaEve of A0 …no conjugate momentum and no dynamics
Originates from gauge invariance …too many degrees of freedom
LEM = − 14 Fµν Fµν + jµ Aµ
= − 12 ∂µ Aν ∂µ Aν + 12 ∂
µ Aν ∂ν Aµ + jµ Aµ
(c.f. Chapter 8 Schwartz)
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No Eme derivaEve of A0 …no conjugate momentum and no dynamics
Originates from gauge invariance …too many degrees of freedom
Gauge fixing:
nµAµ = 0n2 > 0, axial gaugen2 = 0, lightcone gaugen2 < 0, temporal gauge{ •
∂µAµ = 0, Lorentz gauge•
A Aµ µ µα→ + ∂ 2A Aµ µµ µ α∂ → ∂ + ∂ Rξ − gauge: ∂
2α = − 1ξ∂µ Aµ•
Coulomb (radiaEon, transverse) gauge:
k.A k( ) = 0•
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e.g. Coulomb gauge L = − 14 Fµν F
µν
⇒
Gauge invariance:
can choose α to give
αα
⇒
Coulomb gauge preserved by gauge transformaEons with
i.e. have residual gauge symmetry to set A0 = 0
Other equs of moEon give
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Free photon field
∂2Aµ − ∂µ ∂ν Aν( ) = jµ
For a free field in Coulomb gauge ∂2A(x) = 0In analogy with the scalar case:
aλ (k) and aλ† (k) are annihilation and creation operators after canonical quantisation
See Srednicki chapter 3 and chapters 5, 55 (LSZ reducEon)
polarisaEon vectors
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i.e k.A k( ) = 0
PolarisaEon vectors -‐ Coulomb gauge
right-‐polarizaEon
leT-‐polarizaEon
= (0,1)
4-‐vector form
0 | Aµ | p,εj = εµ
je− ip.x
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Ward idenEty
Sample basis:
Lorentz transformaEon:
physical polarisaEon states
Unphysical ?
e.g.
Λνµ =
32 1 0 − 121 1 0 −10 0 1 012 1 0 12
⎛
⎝
⎜⎜⎜⎜
⎞
⎠
⎟⎟⎟⎟
, Λνµε1
ν = ε1µ + 1
Epµ , Λν
µ pν = pµ
Λ in liOle group, SO(3)
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Ward idenEty
Sample basis:
Lorentz transformaEon:
QED matrix elements:
M = ε µ Mµ → ε
µ +pµE
⎛
⎝⎜⎞
⎠⎟Mµ '
Unphysical polarisaEon must not contribute ⇒ pµ Mµ
' = 0
i.e. pµ Mµ ' = p '
µ Mµ ' = pµ Mµ = 0 Ward idenEty
physical polarisaEon states
Unphysical ?
(also required by gauge invariance)