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(1)

The Nucleon Mass in Chiral Perturbation Theory

Beyond one Loop

September 12, 2017

Institute of Theoretical Physics II

Hadron and Particle Physics

(2)

Contents

1

Introduction

2

The Lagrangian and Feynman Rules

3

Basics in QFT

4

Diagrams

5

Calculate the Diagrams

Contact Interaction

Preface

apply Feynman Rules

add Zeros

scalar integrals

Master Integrals

6

Extended-on-mass-shell Renormalization

7

Conclusion

(3)

Introduction

Quantum Chromodynamics (QCD)

ˆ

= strong interaction

→ nucleon mass

Lattice Calculation ˆ

= high energies

Chiral Pertubation Theory

ˆ

= low energies

calculations in baryon ChPT

Lattice calculation, Abdhel-Rehim et. al, 2015

BChPT power counting

7

HBChPT power counting

X

, Lorentz covariance

7

IR power counting

X

, Lorentz covariance

X

→ problems due to analytic properties of the loop integrals

(4)

Introduction

nucleon mass up to Order O(q

3

) (Scherer and Schindler 2012)

in terms of one loop integrals:

mN=m − 4c1M2+ 3g2 A 8F2(p · p)

n

/p m2− (p · p)



+ −/p (p · p) + m2



− 2m (p · p)



TN + /p2m2+ M2



(p · p) + (p · p)2+ m4− m2M2



− 2mM2(p · p)



TπN

o

renormalization of the integrals leads to:

m

N

= m − 4c

1

r

M

2

+

3g

2

A

r

M

2

32π

2

F

2

r

m −

3g

2

A

r

M

3

32πF

2

r

+ O(q

4

)

nucleon mass up to order O(q

6

) with IR by Schindler 2007

goal: calculate nucleon mass up to order O(q

6

) with EOMS

(5)

The Lagrangian and Feynman Rules

The Lagragian

chiral Lagrangian up to chiral order O(q

4

) in SU(2) case

L = L

(2)

π

+ L

(4)

π

+ L

(1)

πN

+ L

(2)

πN

+ L

(3)

πN

+ L

(4)

πN

L

(1)

πN

= ¯

Ψ



i /

D − m +

g

A

2

/

5



Ψ

L

(2)

π

=

F

2

4

Tr



µ

U (∂

µ

U)



M = O(q

1

)

µ

U = O(q

1

)

µ

Ψ = O(q

1

)

with nucleon and pion field:

Ψ = (p, n)

t

U =

1 +

i

F

~

τ ~

π −

1

2F

2

~

π

2

− iα

1

F

3

~

π

2

~

τ ~

π + (8α − 1)

1

8F

4

~

π

4

u =

12

+

i

2F

~

τ ~

π −

8

F

2

(~

τ ~

π)

2

+

i (8α − 1)

16F

3

(~

τ ~

π)

3

+

32α − 5

128F

4

(~

τ · ~

π)

4

(6)

The Lagrangian and Feynman Rules

The Lagrangian

expand the Lagrangian in pion fields (using for example trace rules)

L

(1)

πN

= − ¯

ΨmΨ + i ¯

Ψ /

∂Ψ

+

g

A

2F

¯

Ψγ

5

~

τ · /

∂~

πΨ +

1

4F

2

Ψ~

¯

π · ~

τ × /

∂~

π

 Ψ

g

A

4F

3

Ψγ

¯

5

2α(~

π · ~

π)(~

τ · /

∂~

π) + (4α − 1)(~

π · ~

τ )(~

π · /

∂~

π)

 Ψ

1

16F

4

Ψ(8α − 1)(~

¯

π · ~

π)~

π · ~

τ × /

∂~

π

 Ψ + O(π

5

)

(7)

The Lagrangian and Feynman Rules

Feynman Rules

L

(2)

π

:

2

L

(4)

π

:

4

L

(1)

πN

:

1

1

1

1

L

(2)

πN

:

2

2

2

L

(3)

πN

:

3

3

3

L

(4)

πN

:

4

4

L

(6)

πN

:

6

(8)

The Lagrangian and Feynman Rules

Feynman Rules

q

=

ˆ

i

q

2

− M

2

+ i ε

p

=

ˆ

i

/

p − m + i ε

=

i /

p + m



p · p − m

2

+ i ε

p

1

p

2

q

1

, a

q

2

, b

1

ˆ

=(2π)

4

δ

4

(p

1

− p

2

+ q

1

+ q

2

)

1

4F

2

(/

q

1

− /q

2

abc

τ

c

(9)

Basics in QFT

Nucleon Mass

p

h0|T {Ψ

H

(x ) ¯

Ψ

H

(y )}|0i

p

=

any

inter-action

=

+

1PI

+

1PI

1PI

+...

=

i

/

p − m + i ε

+

i

/

p − m + i ε

(−i Σ)

i

/

p − m + i ε

+ ...

=

i

/

p − m + i ε

1 +

Σ

/

p − m + i ε

+

Σ

/

p − m + i ε

!2

+ ...

=

i

/

p − m − Σ + i ε

=

i



/

p + m + Σ



p · p − (m + Σ)

2

+ i ε

where the self energy Σ is defined

−iΣ :=

(10)

Basics in QFT

Nucleon Mass

Definition

The nucleon mass is a pole in the two-point

nucleon function.

h

p · p − (m + Σ)

2

+ i ε

i

p·p=m

2 N

= 0

⇒ m

N

= m + Σ

(11)

Basics in QFT

Power Counting

q

m

< 1

and

q

m

N

< 1

M = O(q)

|~p| = O(q)

For integrals over pion and nucleon propagators one can show

integrals in d dimensions count as O(q

d

)

pion propagators count as O(q

−2

)

nucleon propagators count as O(q

−1

)

for example

m

Z

d

d

l

(2π)

d

1

(l · l − M

2

+ i ε) ((l − p) · (l − p) − m

2

+ i ε)

= O(q

d −3

)

(12)

Basics in QFT

Power Counting

for diagrams one can show that vertices of chiral order k count like

δ

d

(q)q

k

→ t

k−d

δ

d

(q)q

k

So a diagram with N

I

π

internal pion lines N

I

N

internal nucleon lines and

N

V

k

vertices of chiral order k count in d -dimensions as

FD → t

D

FD

D = d + (d − 2)N

I

π

+ (d − 1)N

I

N

+

X

k

N

V

k

(k − d )

with

N

L

− 1 = N

I

− N

V

k

⇒ D = dN

L

− 2N

π

I

− N

I

N

+

X

k

kN

V

k

(13)

Basics in QFT

Power Counting

For example (in four dimensions):

1

1

2

is of order

(14)

Diagrams

contact terms

2

O(q

2

)

4

O(q

4

)

6

O(q

6

)

(15)

Diagrams

one loop

1

1

(a) O(q

3

)

1

3

(b)O(q

5

)

3

1

(c) O(q

5

)

1

1

4

(d) O(q

5

)

2

(e) O(q

4

)

4

(f) O(q

6

)

2

4

(g) O(q

6

)

(16)

Diagrams

two loop

1

1

1

1

(a) O(q

5

)

1

1

1

1

(b) O(q

5

)

1

1

(c) O(q

5

)

1

2

1

(j) O(q

6

)

(17)

Diagrams

two loop

1

1

1

(d) O(q

5

)

1

1

1

(e) O(q

5

)

1

1

2

(f) O(q

6

)

2

1

1

(g) O(q

6

)

1

1

1

(h) O(q

5

)

1

2

1

(i) O(q

6

)

(18)

Diagrams

two loop

1

1

(k) O(q

5

)

1

1

(l) O(q

5

)

2

(m) O(q

6

)

2

2

(n) O(q

6

)

1

1

2

(o) O(q

5

)

(19)

Diagrams

two loop further

1

2

[f] O(q

6

)

2

1

(20)

Calculate the Diagrams

Contact Interaction

Σ

c

= −4c

1

M

2

− 2 e

115

+ e

116

+ 8e

38



M

4

+ ˆ

g

1

M

6

.

For the simplification of the calculations make a shift

m → m + Σ

c

⇒ m

N

− m = Σ

l

= O(q

3

).

This implies

lim

p→m

N

p · p = m

2

+ m

2

N

− m

2

= m

2

+ O(mq

3

)

lim

p→m

N

¯

u(~

p)/

pu(~

p) = ¯

u(~

p)



/

p − m

N

+ m + m

N

− m



u(~

p) = ¯

u(~

p) m u(~

p) + O(q

3

)

For diagrams, which are of minimal order O(q

4

) terms the replacments

/

p → m and p · p → m

2

(21)

Calculate the Diagrams

Preface

1.

The Feynman Rules are applied to get the matematical expression.

2.

Suitable zeros are added to reduce the tensor rank.

3.

All integrals are reduced to scalar integrals in d + ... dimensions

(Davydychev 1991, Tarasov 1996).

4.

The integrals are expressed with a short set of “Master Integrals”

(using the program TARCER based on Tarasov’s algorithm).

example:

l

1

, a

l

2

, b

−iΣ

(2)

c

=

p

p − l

1

− l

2

p

1

1

(22)

Calculate the Diagrams

apply Feynman Rules

apply FR

use {γ

µ

, γ

5

} = 0, γ

5

γ

5

=

1

d

, τ

a

τ

b

= δ

ab

12

+ i ε

abc

τ

c

to simplify the

expressions

simplify and order the expression using

µ

, γ

ν

} = 2g

µν

1

d

(23)

Calculate the Diagrams

apply Feynman Rules – Example

−2iΣ(2)c =



/l1⊗ (τcεabc) 4F2 − /l2⊗ (τcεabc) 4F2





i1 d⊗12 (l1· l1) + i ε − M2 i1d⊗12 (l2· l2) + i ε − M2 i (m1d⊗12+ (−/l1− /l2+ /p) ⊗12) ((−l1− l2+ p) · (−l1− l2+ p)) + i ε − m2





−/l1⊗ (τdεabd) 4F2 − −/l2⊗ (τdεabd) 4F2



−2iΣ(2)c = 3i 8F4 l1· l1− M2+ i ε



l2· l2− M2+ i ε



(l1+ l2− p) · (l1+ l2− p) − m2+ i ε





1dm(−2 (l1· l2) + (l1· l1) + (l2· l2)) − /p(−2 (l1· l2) + (l1· l1) + (l2· l2)) −/l1(2 (l1· l2) − 2 (l1· p) + (l1· l1) + 2 (l2· p) − 3 (l2· l2)) +/l2(−2 (l1· l2) − 2 (l1· p) + 3 (l1· l1) + 2 (l2· p) − (l2· l2))



⊗12

(24)

Calculate the Diagrams

add Zeros

idea: reduce tensor rank by adding suitable zeros

l · l

l · l − M

2

+ i ε

=

l · l − M

2

+ M

2

l · l − M

2

+ i ε

= 1 +

M

2

l · l − M

2

+ i ε

(in the case of one loop diagrams) rewrite in the denominator

q · q − m

2

+ i 0 → P[q, m]

and in the numerator

l · p → −

1

2

[(l − p) · (l − p) − (l · l + p · p)]

(l − p) · (l − p) → P[l − p, M] + m

2

l · l → P[l , M] + M

2

individual procedure – for example:

(25)

Calculate the Diagrams

add Zeros

further simplification:

scale-less integrals vanish in dimensional renormalization

Z

d

d

l

(2π)

d

l

ν1

l

ν2

→ 0

use T -notation

cancel odd integrals

Z

d

d

l

(2π)

d

l

ν

(p − l ) · (p − l ) − m

2

+ i ε

= 0

(26)

Calculate the Diagrams

add Zeros – Notation

T

(1),ν

1

...ν

n

p,M,m

(d ; α

1

, α

2

) =

=

Z

d

d

l

(2π)

d

l

ν

1

...l

ν

n

(l · l − M

2

+ i ε)

α

1

((l − p) · (l − p) − m

2

+ i ε)

α

2

T

(2),ν

1 1

...ν

n11

ν

21

...ν

2n2

p,m

1

,m

2

,m

3

,m

4

,m

5

(d ; β

1

, β

2

, β

3

, β

4

, β

5

) =

=

Z

d

d

l

1

(2π)

d

Z

d

d

l

2

(2π)

d

l

ν

1 1

1

...l

ν

1 n1

2

l

ν

21

2

...l

ν

2 n2

2

(l

1

· l

1

− m

2

1

+ i ε)

β

1

(l

2

· l

2

− m

2

2

+ i ε)

β

2

1

((l

1

− p) · (l

1

− p) − m

2

3

+ i ε)

β

3

((l

2

− p) · (l

2

− p) − m

2

4

+ i ε)

β

4

1

((l

1

+ l

2

− p) · (l

1

+ l

2

− p) − m

2

5

+ i ε)

β

5

(27)

Calculate the Diagrams

add Zeros – Example

−2iΣ(2)c = − 3igµ1ν1 pµ1(1dm − /p) − 2M2γµ1



T(2),ν11 p,M,M,m,m,m(1, 1, 0, 0, 1) 4F4 − 3igµ1ν1 pµ1(1dm − /p) − 2M2γµ1



T(2),ν21 p,M,M,m,m,m(1, 1, 0, 0, 1) 4F4 + 3iM2(1 dm − /p)Tp,M,M,m,m,m(2) (1, 1, 0, 0, 1) 2F4 + 3i (1dm − /p)Tp,M,M,m,m,m(2) (0, 1, 0, 0, 1) 4F4 + 3i (1dm − /p)Tp,M,M,m,m,m(2) (1, 0, 0, 0, 1) 4F4 − 3i (1dm − /p)Tp,M,M,m,m,m(2) (1, 1, 0, 0, 0) 8F4 − 3i γµ1pµ2 gµ1ν2gµ2ν1+ gµ1ν1gµ2ν2



T(2),ν11ν22 p,M,M,m,m,m(1, 1, 0, 0, 1) 2F4 + 3i γµ1gµ1ν1T(2),ν21 p,M,M,m,m,m(0, 1, 0, 0, 1) 2F4 + 3i γµ1gµ1ν1T(2),ν11 p,M,M,m,m,m(1, 0, 0, 0, 1) 2F4 − 3i γµ1gµ1ν1T(2),ν11 p,M,M,m,m,m(1, 1, 0, 0, 0) 8F4 − 3i γµ1gµ1ν1T(2),ν21 p,M,M,m,m,m(1, 1, 0, 0, 0) 8F4

(28)

Calculate the Diagrams

scalar integrals

factorize into one loop integrals for example

T

ν

1 1

...ν

1 n1

ν

2 1

...ν

2 n2

p,{m

k

}

5k=1

(d ; α

1

, α

2

, 0, 0, 0) = T

ν

1

...ν

n1

p,m

1

,0

(d ; α

1

, 0) T

ν

1

...ν

n2

p,m

2

,0

(d ; α

2

, 0)

T

ν

1 1

...ν

1n1

ν

12

...ν

n22

p,{m

k

}

5k=1

(d ; 0, α

2

, 0, 0, α

5

) =

Z

d

d

l

1

(2π)

d

Z

d

d

l

2

(2π)

d

1

(l

1

· l

1

− m

2

5

+ i ε)

α

5

(l

2

· l

2

− m

2

2

+ i ε)

α

2



l

ν

11

1

...l

ν

n11

1

l

ν

2 1

2

...l

ν

2n2

2



l

1

→l

1

−l

2

+p

reduce tensor integrals to integrals in d + ...-dimensions

1.

Rewrite the tensor structure by using derivatives.

2.

Rewrite the propagators with the Schwinger trick/ in λ-representation

3.

Evaluate the integral(s) over the loop momentum/ momenta per Gaussian

integration.

4.

Construct a operator and reverse the Gaussian integration and the

λ-parametrization for the scalar integral.

(29)

Calculate the Diagrams

scalar integrals

Tp,m1,m2(1),ν1...νn(d ; α1, α2) =

Z

ddl (2π)d lν1...lνn (l · l ) − m2 1+ i ε



α1 (l − p) · (l − p) − m2 2+ i ε



α2

first two steps yield

(−i )

n

n

Y

k=1

∂a

ν

k

a=0

(−i )

α

1

2

Γ (α

1

) Γ (α

2

)

Z

0

Z

0

1

2

λ

α

1

1

−1

λ

α

2

−1

2

G

(1)

Gauss integration

G

(1)

=

1

(2π)

d

i



π

i



d2

1

(D(λ))

d2

exp



i

Q(λ, a)

D(λ)



exp



i λ

1

−m

2

1

+ i ε + iλ

2

p

2

− m

2

2

+ i ε − i

λ

2

2

p

2

λ

1

+ λ

2



D(λ) = λ

1

+ λ

2

Q(λ, a, b) = (p · a) Q

1

+ a

2

Q

11

Q

1

= λ

2

Q

11

= −

1

4

(30)

Calculate the Diagrams

scalar integrals

G

(1)

=

1

(2π)

d

i



π

i



d2

1

(D(λ))

d2

exp



i

Q(λ, a)

D(λ)



exp



i λ

1

−m

2

1

+ i ε + iλ

2

p

2

− m

2

2

+ i ε − i

λ

2

2

p

2

λ

1

+ λ

2



1

D(λ)

1

(2π)

d



π

i



d2

1

(D(λ))

d2

= (2π)

2

i

π

1

(2π)

d +2



π

i



d +22

1

(D(λ))

d +22

.

1

D(λ)

=i 4π

ˆ

2

d

+

λ

i

=i

ˆ

∂m

2

i

(numerator)

(31)

Calculate the Diagrams

scalar integrals

one can build a operator

T

ν

1

...ν

n

p,m

1

,m

2

(d ; α

1

, α

2

) =T

ν

1

...ν

n

p, {∂/∂m

2

}, d

+

 T

p,m

1

,m

2

(d ; α

1

, α

2

)

T

ν

1

...ν

n

p, {∂/∂m

2

}, d

+

 = (−i)

n

n

Y

k=1

∂a

ν

k

exp (iQ (λ, a) ρ)

a=0

λ

j

=i

∂m2 j

ρ=i 4πd

+

similar in the two loop case

T

ν

1 1

...ν

n11

ν

12

...ν

n22

p,{m

k

}

5k=1

(d ; α

1

, α

2

, α

3

, α

4

, α

5

)

=T

ν

11

...ν

n11

ν

12

...ν

n22

p, {∂/∂m

2

}, d

+

 T

p,{m

k

}

5k=1

(d ; α

1

, α

2

, α

3

, α

4

, α

5

)

T

ν

11

...ν

1 n1

ν

2 1

...ν

2 n2

p, {∂/∂m

2

}, d

+



= (−i )

n

1

+n

2

n

1

Y

k

1

=1

n

2

Y

k

2

=1

∂a

ν

k1

∂b

ν

k2

exp (iQ (λ, a, b) ρ)

a=b=0

λ

j

=i

∂m2 j

ρ=−16π

2

d

+

(32)

Calculate the Diagrams

scalar integrals – Example

−2iΣ(2)c = 12i π2 /p (p · p) + 2M2



−1dm (p · p)



Tp,m1,m2,m3,m4,m5(2) (d + 2; 1, 2, 0, 0, 2) F4 + 12i π2 / p (p · p) + 2M2



−1dm (p · p)



Tp,m1,m2,m3,m4,m5(2) (d + 2; 2, 1, 0, 0, 2) F4 + 3iM2(1dm − /p)Tp,m1,m2,m3,m4,m5(2) (d ; 1, 1, 0, 0, 1) 2F4 − 3iT (m5, 0)(d ; 1, 0)(/p(Tm1,0(1) (d ; 1, 0) + T (m2, 0)(d ; 1, 0)) −1dmT (1) m2,0(d ; 1, 0)) 4F4 − 3iT(1) m1,0(d ; 1, 0)((1dm − /p)Tm2,0(1) (d ; 1, 0) − 21dmTm5,0(1) (d ; 1, 0)) 8F4 −1536i π 4 / p (p · p) Tp,m1,m2,m3,m4,m5(2) (d + 4; 2, 2, 0, 0, 3) F4 + 24i π2/pTp,m1,m2,m3,m4,m5(2) (d + 2; 1, 1, 0, 0, 2) F4

(33)

Calculate the Diagrams

Master Integrals – Integration by Parts

0 =

Z

d

d

l

1

(2π)

d

Z

d

d

l

2

(2π)

d

∂l

1

µ

l

µ

1

P

α

1

l

1

,m

1

P

α

2

l

2

,m

2

P

α

5

l

1

−l

2

,m

5

=

Z

d

d

l

1

(2π)

d

Z

d

d

l

2

(2π)

d

dP

α

1

l

1

,m

1

P

α

2

l

2

,m

2

P

α

5

l

1

+l

2

−p,m

5

− 2α

1

(l

1

· l

1

) P

α

1

+1

l

1

,m

1

P

α

2

l

2

,m

2

P

α

5

l

1

−l

2

,m

5

−2α

2

(l

1

· l

2

) P

l

α

1

,m

1 1

P

α

2

+1

l

2

,m

2

P

α

5

l

1

−l

2

,m

5

− 2α

5

((l

1

· l

1

) − (l

1

· l

2

)) P

α

1

l

1

,m

1

P

α

2

l

2

,m

2

P

α

5

+1

l

1

−l

2

,m

5

one can cancel the scalar products to obtain



d − 2α

1

1 + m

2

1

1

+

 − 2α

2

1 + m

2

2

2

+



−2α

5

5

+



1

+ m

2

1

+

1

2

5

+ m

2

5

− 1

− m

2

1

− 2

− m

2

2



 

¯

T

p,{m

(2)

k

}

5k=1

(d ; α

1

, α

2

, 0, 0, α

5

) = 0

(34)

Calculate the Diagrams

Master Integrals – Dimension Reduction

T

p,m1,m2

(1)

(d ; α1

, α

2)

=

(−i )

α1

2

Γ (α1

) Γ (α2)

2

Y

j=1

Z

0

j

λ

α

j

−1

j

1

(2π)

d

i



π

i



d2

1

(D(λ))

d2

exp

i λ

j



ˆl

j

− m

2

j

+ i ε



− i

λ

2

2

p

2

D(λ)

!

apply on both sides D(λ) = λ1

+ λ2

in operator form

D(λ) = i ∂

m

2

1

+ i ∂

m

22

gives

i 4πT

p,m1,m2

(1)

(d − 2; α

1

, α

2

)

(35)

Calculate the Diagrams

Master Integrals

relations between integrals in same dimension

integration by parts

relations between integrals in different dimensions

Schwinger representation and Gauss integration

⇒ recurrence relations to reduce integrals to a set of

(36)

Calculate the Diagrams

Master Integrals – Example

−iΣ

(2)

c

=

1

2

i

2(d − 2)(3d − 4)F

4

m

n

2

h

2 2d

2

− 7d + 5 m

4

+

−7d

2

+ 23d − 16 m

2

M

2

+ d

2

− 5d + 6 M

4

T

(2)

M,m,0,0,M

(d ; 1, 1, 0, 0, 1)

−4M

2

M

2

− m

2



(d − 2)M

2

− 2(d − 1)m

2

 T

(2)

M,m,0,0,M

(d ; 2, 1, 0, 0, 1)

i

+(d − 2) (d − 1)m

2

+ (d − 2)M

2

 (T

M,0

(1)

(d ; 1, 0))

2

+(2 − d ) 4(d − 1)m

2

+ (d − 2)M

2

 T

(1)

m,0

(d ; 1, 0)T

(1)

M,0

(d ; 1, 0)

o

(37)

Calculate the Diagrams

Master Integrals – one loop

−iΣ(1)a = 3g2 A 8F2(p · p)

n

/ p m2− (p · p)



T (M, 0)({d }, 1, 0) + −/p (p · p) + m2



− 2m (p · p)



T (m, 0)({d }, 1, 0) + /p2m2+ M2



(p · p) + (p · p)2+ m4− m2M2



− 2mM2(p · p)



T (M, m)({d }, 1, 1)

o

−iΣ(1) b = − i Σ (1) c = 3gAmM2 F2

n

M2(d18 − 2d16)T (M, m)({d }, 1, 1) + (d18 − 2d16)T (m, 0)({d }, 1, 0)

o

−iΣ(1) d = 3g2 A 2F4 4m2− M2



n

− 2mM4

h

l3 2(d − 1)m2− (d − 2)M2



+ l4 M2− 4m2



i

T (M, m)({d }, 1, 1) + 2m (d − 2)l3M4+ l4 4m2M2− M4



T (m, 0)({d }, 1, 0) − (d − 2)l3mM4T (M, 0)({d }, 1, 0)

o

(38)

Calculate the Diagrams

Master Integrals – one loop

−iΣ(1)e =1 2 6M2T (M, 0)({d }, 1, 0) dm2(2c1 − c3) − c2m2



dF2m2 −iΣ(1) f = − 1 2 24M4T (M, 0)({d }, 1, 0) d (d + 2)F2

n

d2(2e14+ 2e19− e36− 4e38) + 2d (2e14+ e15+ 2e19+ e20+ e35− e36− 4e38) + 4e15+ 6e16+ 4e20+ 4e35

o

−iΣ(1)g = −1 2 6M4(2c1d (2l4− (d − 2)l3) + (c2 + c3d )(dl3− 2l4))T (M, 0)({d }, 1, 0) dF4

(39)

Calculate the Diagrams

Master Integrals – two loop

−iΣ(2) k = − i Σ (2) l = 1 2 6i (10α − 1)gA2mT (M, 0)({d }, 1, 0) M2TM,m(1) (1, 1) + Tm,0(1)(1, 0)



4F4 −iΣ(2)m = −1 8 12iM2(TM,0(1)(1, 0))2 (2 − 20α)c2m2+ dm2(5(8α − 1)c1− 20αc3+ 2c3)



dF4m2 −iΣ(2)n =1 4 3iM2(TM,0(1)(1, 0))2 dm2(2c1(40α + d − 6) − c3(40α + d − 4)) − c2(40α + d − 4)m2



dF4m2 −iΣ(2)o = −1 2 6ig2 A 8F4 4m2− M2



mT(1) M,0(1, 0)

h

M2 2 (d − 3)m2+ m2(80α + d − 7) − M2(20α + d − 4)



TM,m(1) (1, 1) + (d − 2)TM,0(1)(1, 0)



+ 2 (80α − 8)m2− M2(20α + d − 4)



Tm,0(1)(1, 0)

i

(40)

Extended-on-mass-shell Renormalization

one loop results

T

π

= T

M,0

(1)

(d ; 1, 0)

T

N

= T

m,0

(1)

(d ; 1, 0)

T

πN

= T

M,m

(1)

(d ; 1, 1)

R =

2

d − 4

− [ln(4π) + γ

E

+ 1]

Ω =

p · p − m

2

1

− m

2

2

2m

1

m

2

F (Ω) =

p

2

− 1 arccos (−Ω)

− 1 ≤ Ω ≤ 1

Dimensional Regularization leads to

µ

4−d

T

π

= −i

M

2

16π

2



R + ln

 M

2

µ

2



+ O(4 − d )

µ

4−d

T

N

= −i

m

2

16π

2



R + ln

 m

2

µ

2



+ O(4 − d )

µ

4−d

T

πN

= −i

1

16π

2



R + ln

 m

2

µ

2



− 1 +

p · p − m

2

− M

2

p · p

ln

 M

m



+

2Mm

p · p

F (Ω)



(41)

Extended-on-mass-shell Renormalization

one loop results

Dimensional Regularization (µ = m, omit O(4 − d )) leads to

µ

4−d

T

π

= −i

M

2

16π

2



R + ln

 M

2

m

2



µ

4−d

T

N

= −i

m

2

16π

2

R

µ

4−d

T

πN

= −i

1

16π

2



R − 1 +

p · p − m

2

− M

2

p · p

ln

 M

m



+

2Mm

p · p

F (Ω)



using ˜

MS and let d → 4

T

π

= −i

M

2

16π

2

ln

 M

2

m

2



T

πN

= −i

1

16π

2



−1

+

p · p − m

2

− M

2

p · p

ln

 M

m



+

2Mm

p · p

F (Ω)



(42)

Extended-on-mass-shell Renormalization

one loop results

EOMS leads to

T

π

= −i

M

2

16π

2

ln

M

2

m

2

!

T

N

= 0

T

πN

= −i

1

16π

2

"

p · p − m

2

− M

2

p · p

ln



M

m



+

2Mm

p · p

F (Ω)



(43)

Extended-on-mass-shell Renormalization

two loop idea

l

1

, a

l

2

, b

p

p − l

1

p − l

1

− l

2

p − l

1

p

1

1

1

1

problematic cases

1.

l

1

= O(q) and l

2

>> q

2.

l

1

>> q and l

2

= O(q)

3.

l

1

>> q and l

2

>> q

4.

l

1

>> q and l

2

>> q

(44)

Conclusion and Outlook

Using a naive power counting scheme all self-energy diagrams up to

chiral order O(q

6

) are constructed.

The expressions of the diagrams contain tensor integrals.

The Tensor Integrals are reduced by (adding zeros and) going to higher

dimensions.

All Integrals are reduced to a set of Master Integrals in d -dimension.

The renormalized expression of diagrams are derived for one loop and

“factorizing” two loop diagrams.

outlook

renormalize all two loop integrals

get a numeric expression for the nucleon mass depending on the pion

mass

(45)

Thanks for your attention!

(46)

References

Lagrangian:

Nadia Fettes et al. “The Chiral Effective Pion-Nucleon Lagrangian of Order p4”.

In: Annals of Physics 283.2 (Aug. 1, 2000)

Reduction:

A.I. Davydychev. “A simple formula for reducing Feynman diagrams to scalar

integrals”. In: Physics Letters B 263.1 (July 1991)

O.V. Tarasov. “Generalized recurrence relations for two-loop propagator

integrals with arbitrary masses”. In: Nuclear Physics B 502.1 (Sept. 1997)

Programs:

R. Mertig and R. Scharf. “TARCER—A mathematica program for the reduction

of two-loop propagator integrals”. In: Computer Physics Communications 111.1

(June 1998)

Michele Re Fiorentin. “FaRe: A Mathematica package for tensor reduction of

Feynman integrals”. In: International Journal of Modern Physics C 27.3 (Mar.

2016)

(47)

References

Basics and Compare:

Stefan Scherer and Matthias R. Schindler. A primer for chiral perturbation

theory. Lecture notes in physics 830. OCLC: 724844640, 2012

Matthias R Schindler. “Higher-order calculations in manifestly Lorentz-invariant

baryon chiral perturbation theory”. PhD thesis. 2007

Matthias R. Schindler et al. “Infrared renormalization of two-loop integrals and

the chiral expansion of the nucleon mass”. Nuclear Physics A 803.1 (Apr. 15,

2008)

Lattice Simulation:

A. Abdel-Rehim et al., “Nucleon and pion structure with lattice QCD

simulations at physical value of the pion mass”, physical review D 92, 31

December 2015

Renormalization:

T. Fuchs et al. “Renormalization of relativistic baryon chiral perturbation theory

and power counting”. In: Physical Review D 68.5 (Sept. 24, 2003)

J.M.Alarcon et al. “Improved description of the πN-scattering phenomenology

in covariant baryon chiral Perturbation Theory”, 2012

References

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