# QFT strikes back The gauge fields

Document Sample

```					            QFT strikes back:
The gauge elds

Jakub Jankowski
Institute of Theoretical Physics, University of Wrocªaw

December 2007
QED

•   We can write QED lagrangian in the form:
1
L=−      F F µν + ψ(D − m)ψ
/
4 µν
where Dµ = ∂µ − ieAµ and Fµν = ∂µ Aν − ∂ν Aµ
QED

•   We can write QED lagrangian in the form:
1
L=−      F F µν + ψ(D − m)ψ
/
4 µν
where Dµ = ∂µ − ieAµ and Fµν = ∂µ Aν − ∂ν Aµ
•   We nd the symmetry:
1
Aµ (x ) → Aµ (x ) −       ∂µ α(x )   ψ(x ) → e i α(x ) ψ(x )
e
Dµ ψ(x ) → e i α(x ) Dµ ψ(x )
QED

•   We can write QED lagrangian in the form:
1
L=−     F F µν + ψ(D − m)ψ
/
4 µν
where Dµ = ∂µ − ieAµ and Fµν = ∂µ Aν − ∂ν Aµ
•   We nd the symmetry:
1
Aµ (x ) → Aµ (x ) −       ∂µ α(x )   ψ(x ) → e i α(x ) ψ(x )
e
Dµ ψ(x ) → e i α(x ) Dµ ψ(x )

•   This is U (1) gauge symmetry !
¡
Weak Decays

• β−   decay n → p + e − + νe
¯
¡
Weak Decays

• β−   decay n → p + e − + νe
¯

• β+   decay p → n + e + + νe
The Standard Model
•   The elementary particles interactions (strong, weak,
electomagnetic) are described by the gauge group:
SU (3) × SU (2) × U (1)
Yang-Mills theory
1 a µν
L=−       F F + ψ(D − m)ψ
/
4 µν a
where Dµ = ∂µ − igAa t a and
µ

Fµν = ∂µ Aa − ∂ν Aa + gf abc Ab Ac
a
ν       µ           µ ν
Yang-Mills theory
1 a µν
L=−     F F + ψ(D − m)ψ
/
4 µν a
where Dµ = ∂µ − igAa t a and
µ

Fµν = ∂µ Aa − ∂ν Aa + gf abc Ab Ac
a
ν       µ           µ ν

•   Lie group G
[t a ; t b ] = if abc t c
each element is paremetrized U (α ; ...; αn ) ∈ G and for
1

innitezimal α we have
U (α) = 1 + i αa t a
Gauge symmetry

•   Transformation laws are given by:
i
Aa (x )tµ → V (x )(Aa (x )tµ +
a                  a
)∂ V † (x )
g µ
ψ(x ) → V (x )ψ(x )
where
V (x ) = exp(i αa (x )t a )
Gauge symmetry

•   Transformation laws are given by:
i
Aa (x )tµ → V (x )(Aa (x )tµ +
a                  a
)∂ V † (x )
g µ
ψ(x ) → V (x )ψ(x )
where
V (x ) = exp(i αa (x )t a )

•   Covariant derivative transforms like the eld:
Dµ ψ(x ) → V (x )Dµ ψ(x )
so like in QED we have gauge invariance.

•
I   =   DA exp(iS [A])

where S [A] = −   1
4
d xFµν Faµν
a
4

•
I   =   DA exp(iS [A])

where S [A] = − d xFµν Fa
1
4
4a   µν

•   Gauge xing condition G (A) = 0

•
I   =   DA exp(iS [A])

where S [A] = − d xFµν Fa
1
4
4a   µν

•   Gauge xing condition G (A) = 0
dg
1=    dg (x )δ(g (x )) =   dx det(      )δ(g (x ))
dx

•
I   =    DA exp(iS [A])

where S [A] = − d xFµν Fa
1
4
a
4   µν

•   Gauge xing condition G (A) = 0
dg
1=     dg (x )δ(g (x )) =       dx det(      )δ(g (x ))
dx
Now we can generalize this to get:
δ G (Au )
1=        Du det(             )δ(G (Au ))
δu

where Du = D(vu ) is ivariant Haar measure.
Theorem

•   Quantity
δ G (Au )
det(             )δ(G (Au ))
δu
is independent of u
Calculation

•   from our theorem and from the fact S [A] = S [Au ] and
DA = DAu it follows:

δ G (A)
I   =(   Du )   DA exp(iS [A]) det(           )δ(G (A))
δu
Calculation

•   from our theorem and from the fact S [A] = S [Au ] and
DA = DAu it follows:

δ G (A)
I   =(   Du )   DA exp(iS [A]) det(           )δ(G (A))
δu

•   Generalized Lorentz Gauge: G (A) = ∂ µ Aa − ω a
µ
Calculation

•   from our theorem and from the fact S [A] = S [Au ] and
DA = DAu it follows:

δ G (A)
I   =(       Du )   DA exp(iS [A]) det(           )δ(G (A))
δu

•   Generalized Lorentz Gauge: G (A) = ∂ µ Aa − ω a
µ

1
I       = |V |    DA exp(iS [A]) det( ∂ µ Dµ )δ(∂ µ Aa − ω a )
g              µ
Calculation

•   Taking proper normalization we obtain:

ω 2 (x )                        1
I   = N (ξ)   Dω exp(−i   d x
4
)   DA exp(iS [A]) det( ∂ µ Dµ )
2ξ                            g
δ(∂ µ Aa − ω a )
µ
Calculation

•   Taking proper normalization we obtain:

ω 2 (x )                              1
I   = N (ξ)   Dω exp(−i     d x
4
)   DA exp(iS [A]) det( ∂ µ Dµ )
2ξ                                  g
δ(∂ µ Aa − ω a )
µ

•   Using δ function:

(∂ µ Aa )2       1
I   =N   DA exp(iS [A] − i      d x
4         µ
) det( ∂ µ Dµ )
2ξ           g
Propagator

•   that gives us the gauge boson propagator:
i                   kµ kν
Dµν (k
ab      2
)=−        ηµν − (1 − ξ)           δ ab
k2
k2
Propagator

•   that gives us the gauge boson propagator:
i                   kµ kν
Dµν (k
ab      2
)=−        ηµν − (1 − ξ)           δ ab
k2
k2

for ξ = 1 Feynman gauge
Propagator

•   that gives us the gauge boson propagator:
i                   kµ kν
Dµν (k
ab      2
)=−        ηµν − (1 − ξ)           δ ab
k2
k2

for ξ = 1 Feynman gauge             for ξ = 0 Landau gauge
Ghosts
•   Recall that
DψDψ exp(i ψ M ψ) = det M
¯       ¯
Ghosts
•   Recall that
DψDψ exp(i ψ M ψ) = det M
¯       ¯

(∂ µ Aµ )2
I   ∝          c
DADc D¯ exp(iS [A] − i              + i c ∂ µ Dµ c )
¯
2ξ
c and c are called ghost elds
¯
Conclusion

•   To obtain Feynman rules in gauge theories we have to use
eective lagrangian:

1 a µν (∂ µ Aa )   2

L=−     F F −       µ
+ ψ(D − m)ψ + c ∂ µ Dµ c
/         ¯
4 µν a     2ξ
•   The naive rules have to be supplement by ghost loops
• Ghost elds do not obey spin-statistic theorem

```
DOCUMENT INFO
Shared By:
Categories:
Stats:
 views: 6 posted: 1/19/2010 language: English pages: 27