Post on 30-Apr-2019
Fundamental experimental objects
1 1 2( ... )na b b bΓ → Decay width = 1/lifetime
1 2 1 2( ... )na a b b bσ → Cross section
Cross section = Transition rate x Number of final states
Initial flux
3
31 (2 )Vd pn
i π=Π
a1v 1V V×
1a 1av
# particles passing through unit area in unit time
# target particles per unit volume
2a
(Lab frame)
The transition rate 4 * ( ) ( ) ( ) ...fi f iT d x x V x xφ φ= − +∫
φi, f → f p
± = e(− ,+ )ip.x 1
2 p0V≡ N
Ve(− ,+ )ip.x
A
B
C
D
Transition rate per unit volume 2
fiTfi TVW =
24 4(2 ) ( )A B C DN N N N
fi C D A BVT p p p pπ δ= − + − − fiM
e.g.
φ f ,i = e(− ,+ )ip.x
244
4
( ) 1 1 1 1(2 )2 2 2 2
C D A Bfi
A B C D
p p p pW
V E E E Eδ
π+ − − ⎛ ⎞⎛ ⎞⎛ ⎞ ⎛ ⎞
= ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠
M
Cross section = Transition rate x Number of final states
Initial flux
42 4
32 32
64
1 (2 ) ( )2 (22 ) 2 2C D A B
A B C D
C Dd p p p pE E
dE E
pVV
p d Vπ δπ
σ = + − −Av
M
2
d dQF
σ =M
3 34 4
3 3(2 ) ( )(2 ) 2 (2 ) 2
C DC D A B
C D
d p d pdQ p p p pE E
π δπ π
= + − −Lorentz Invariant Phase space
2 2 2 1/ 2
2 2
4(( . ) )A B
A B A B
F E E
p p m m
=
= −Av
The cross section
The decay rate
212 A
d dQE
Γ = M
1
1
1
334 4
3 3(2 ) ( ... ) ...(2 ) 2 (2 ) 2
n
n
n
BBA B B
B B
d pd pdQ p p p
E Eπ δ
π π= − −
Compton scattering of a π meson
,k λ ', 'k λ
p 'pp k+ p 'pp k−
,k λ ', 'k λ
,k λ ', 'k λ
p 'p
Feynman rules 2
2 2( )
( ) V
V ie A A A
m
eµ µµ
µµ
µ
ψ ψ∂ ∂ +
∂
−
= − + −
=
∂
Klein Gordon
( ' )ie p pλ λ− +
p 'p
,k λ
2 2
ip m− External photon λε
p
2ie
,k λ ', 'k λ
p 'p
γπ γπ→
Compton scattering of a π meson
,k λ ', 'k λ
p 'pp k+ p 'p'p k−
,k λ ', 'k λ ,k λ ', 'k λ
p 'p
22 2
2 2
( ) [ .(2 ) '.(2 ' ')( )
.(2 ' ) '.(2 ') 2 . ']( ')
fiiie p k p k
p k mip k p k i
p k m
ε ε
ε ε ε ε
= − + ++ −
+ − − −− −
iM
3 32 42 4 2
4 6
1 (2 ) ( )2 2 (2 ) 2 2
C DC D A B
A B C D
d p d pVd p p p p VE E V E E
πσ δπ
= + − −Av
M
Compton scattering of a π meson
,k λ ', 'k λ
p 'pp k+ p 'pp k−
,k λ ', 'k λ ,k λ ', 'k λ
p 'p
1
e2M fi = ε.(2 p + k)
i( p + k)2 − m2 ε '.(2 p '+ k ')
+ ε.(2 p '− k ')i
( p '− k ')2 − m2 ε '.(2 p '− k ') − 2iε.ε '
[ ]2 2
22
( . ')1 (1 cos )k
lab m
dd mσ α ε ε
θ⎛ ⎞ =⎜ ⎟Ω⎝ ⎠ + −
22 2 2
0 2
8| . 8.10 3.103total k
d d GeV mbd mπ
σ πασ − − −= = Ω = =
Ω∫
2
/ 12|total k m mkπασ >>
ε.p = ε '.p = 0 Transverse polarisation
Causality?
3
3
'
(2 ) 2( ' )
( ) ( )
p
p
i t t id pF x x i e
D x y D y x
ω
π ω
− − −Δ − = −
= − − −∫ p.(x'-x)
2When ( ) 0, we can perform a Lorentz transformation taking ( ) ( )x y x y x y− < − → − −-m|r| | |...causality preserved (e )m re−−
2No (continuous) transformation possible for (x-y) 0 >
•
•
y
y−
x
0x
x
-imt...and amplitude nonvanishing (e )imte−
2 2-m ( ' ) ( ' )QM : ( '- ) e t tU x x − − −∝ x x
Field theory :
×See Peskin & Schroeder “Quantum Field Theory” p28
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∝
•
•
Lagrangian invariant under all the symmetries of nature
(Nonrelativistic mechanics)
-makes it easy to construct viable theories
3 lagrangian densi, tyL d x= ∫L L
Klein Gordon field ( )xφ
( )† 2 †( ) ( ) ( ) ( )x x m x xµµφ φ φ φ∂ ∂ −L= } }
T V
Manifestly Lorentz invariant
Lagrangian formulation of the Klein Gordon equation
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∂φ
δφ +∂L
∂(∂µφ)δ (δ ∂µφ)
=
∂L∂φ
− ∂µ ∂L∂(∂µφ)
⎛⎝⎜
⎞⎠⎟
⎡
⎣⎢⎢
⎤
⎦⎥⎥δφ + ∂µ ∂L
∂(∂µφ)δφ
⎛⎝⎜
⎞⎠⎟
surface integral in S..vanishes
δSδφ
= 0
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→
Noether current
( )† 2 †( ) ( ) ( ) ( )x x m x xµµφ φ φ φ∂ ∂ −L=
Is invariant under ( ) ( )ix e xαφ φ→ …an Abelian (U(1)) gauge symmetry
†0 ( ) ( )( )
µµδ δφ δ φ φ φ
φ φ∂ ∂= + ∂ + ↔∂ ∂ ∂L L
L=
iαφ i µα φ∂
†( )( ) ( )
i iµ µµ µα φ α φ φ φ
φ φ φ⎡ ⎤⎛ ⎞ ⎛ ⎞∂ ∂ ∂= −∂ + ∂ − ↔⎢ ⎥⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎣ ⎦
L L L
0 (Euler lagrange eqs.)
††0,
2 ( ) ( )iej jµ
µ µ µ µφ φφ φ
⎛ ⎞∂ ∂∂ = = −⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠
L L Noether current
The Klein Gordon current
( )† 2 †( ) ( ) ( ) ( )x x m x xµµφ φ φ φ∂ ∂ −L=
Is invariant under ( ) ( )ix e xαψ ψ→ …an Abelian (U(1)) gauge symmetry
††0,
2 ( ) ( )iej jµ
µ µ µ µφ φφ φ
⎛ ⎞∂ ∂∂ = = −⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠
L L
( )* *KGj ieµ µ µφ φ φ φ= − ∂ − ∂
This is of the form of the electromagnetic current we used for the KG field
The Klein Gordon 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,
2 ( ) ( )iej jµ
µ µ µ µφ φφ φ
⎛ ⎞∂ ∂∂ = = −⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠
L L
( )* *KGj ieµ µ µφ φ φ φ= − ∂ − ∂
This is of the form of the electromagnetic current we used for the KG field
3 0Q d x j= ∫ is the associated conserved charge
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)
L = ∂µφ(x)( )†
∂µφ(x) − m2φ(x)†φ(x) + λ φ4+
λ 'M 2 φ
6+ ...
Additional terms
}
Renormalisable D ≤ 4
If M 103GeV , "Effective" Field theory approximately renormalisable
Terms allowed by U(1) symmetry
U(1) local gauge invariance and QED
( )† 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 → eiα ( x )Q Dµφ
U(1) local gauge invariance and QED
( )† 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 i xx QQ i x Qe e iQe xiQA iQA iQe xαµ µ
α α αµ µ µ µµφ φ φ φ φαφ φ α∂ → ∂ = ∂− − − ∂+ ∂
To obtain invariant Lagrangian look for a modified derivative transforming covariantly
Dµφ → eiα ( x )Q Dµφ
D iQAµ µ µ= ∂ −
A Aµ µ µα→ + ∂Need to introduce a new vector field
( )( ) ( )iQ xx e xαφ φ→
Dµφ → eiα ( 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 p p eAµ µ µ→ +
universal coupling of electromagnetism follows from local gauge invariance
2 2 2( ) ( )m V where V ie A A e Aµ µ µµ µ µψ ψ∂ ∂ + = − = − ∂ + ∂ −
The Euler lagrange equation give the KG equation:
( )( ) ( )iQ xx e xαφ φ→
Dµφ → eiα ( 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 p p eAµ µ µ→ +
universal coupling of electromagnetism follows from local gauge invariance
( )†KG 2 † 2( ) ( ) ( ) ( ) ( )KGx x m x x j A O eµ µµ µφ φ φ φ∂ ∂ − − +i.e. L = L =
The electromagnetic Lagrangian
F A Aµν µ ν ν µ= ∂ − ∂
, F F A Aµν µν µ µ µα→ → + ∂
14
EM j AF F µνµν
µµ= − −L
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( )
The photon propagator
( )F A A jµν µ ν ν µ νµ µ µ∂ = ∂ ∂ − ∂ ∂ =
• The propagators determined by terms quadratic in the fields, using the Euler Lagrange equations.
2 2 2( ) ( )m V where V ie A A e Aµ µ µµ µ µψ ψ∂ ∂ + = − = − ∂ + ∂ −
The Klein Gordon propagator (reminder)
2 4( ) ( ' ) ( ' )Fm x x x xµµ δ∂ ∂ + Δ − = −
In momentum space:
Δ 'F ( p) = i
− p2 +m2 ± iε
With normalisation convention used in Feynman rules = inverse of momentum space operator multiplied by -i
The photon propagator
( )F A A jµν µ ν ν µ νµ µ µ∂ = ∂ ∂ − ∂ ∂ =
Gauge ambiguity
A Aµ µ µα→ + ∂ 2A Aµ µµ µ α∂ → ∂ + ∂
i.e. with suitable “gauge” choice of α (“ξ” gauge) want to solve
21 1(1 ) ( ) ( (1 ) )A A g A jµ ν ν µ νµ ν µ νµ µ µξ ξ
∂ ∂ − − ∂ ∂ ≡ ∂ − − ∂ ∂ =
In momentum space the photon propagator is
12
2 2
1(1 ) (1 )p pii g p p p g
p pµ νµν µ ν
µν ξξ
− ⎛ ⎞⎛ ⎞− − − = − + −⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠
(‘t Hooft Feynman gauge ξ=1)
Choose as1
(gauge fixing)
Aµ µξ− ∂
• The propagators determined by terms quadratic in the fields, using the Euler Lagrange equations.