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Quantum Field Theory-II Solutions Set n. 2

UZH and ETH, FS-2020 Prof. G. Isidori

Assistants: C. Cornella, D. Faroughy, F. Kirk, J. Pagès, A. Rolandi www.physik.uzh.ch/en/teaching/PHY552

1 Saddle point evaluation of the Path Integral

Consider the Generating functional in Euclidean space ( x

2

= x

20

+ x

21

+ x

22

+ x

23

) of a generic scalar field theory (~ = 1 )

W

E

[J] = N Z

Dφ e

−SE[φ,J]

, (1)

where

S

E

[φ, J ] = Z

d

4

x 1

2 (∂

µ

φ)

2

+ 1

2 m

2

φ

2

+ V (φ) − J (x)φ(x)

. (2)

Define the classical field configuration φ

0

,

δS

E

[φ, J]

δφ

φ=φ0

= 0 , (3)

and expand S

E

[φ, J ] around φ

0

:

S

E

[φ, J ] = S

E

0

, J ] + Z

d

4

x ∆S

J(1)

(x)[φ(x) − φ

0

(x)]

+ 1 2

Z

d

4

xd

4

y ∆S

J(2)

(x, y)[φ(x) − φ

0

(x)][φ(y) − φ

0

(y)] + . . . (4)

I. Show that in the limit where we can neglect the dots in Eq. (4), W

E

[J] is given by W

E

[J ] ≈ N

0

e

−SE0,J]

Det ˆ K

−1/2

(5) where K ˆ is defined by

Kf ˆ = Z

d

4

x

−∂

µ2

+ m

2

+ V

00

0

)

f (x) . (6)

Put back the ~ factors and discuss the physical meaning of this approximation.

Solution Let us start by reintroducing ~ in the functional integral as W

E

[J ] = N

Z

Dφ e

1~SE[φ,J]

. (7)

Now we find the classical field configuration which satisfies the equation of motion in presence of the source J , that is a stationary point for the action:

δS

E

[φ, J ]

= −∂

2

φ (x) + m

2

φ (x) + V

0

(φ (x)) − J(x) = 0 . (8)

(2)

We now expand the action around this solution and stop at the quadratic order. The aim will be to compute explicitly the functional integral with this approximation. The terms ∆S

J(1)

(x) and

∆S

J(2)

(x, y) are given by

∆S

J(1)

(x) = δS

E

[φ, J]

δφ(x)

φ=φ0

= 0 ,

∆S

J(2)

(x, y) = δ

2

S

E

[φ, J]

δφ(x)δφ(y)

φ=φ0

= [−∂

2µ

+ m

2

+ V

00

0

)]

x

δ

4

(x − y) ≡ K ˆ

xy

.

(9)

Higher order terms in the expansion will depend only on the potential, since the rest of the action is at most quadratic in φ:

∆S

n>2(n)

= δ

n

S

E

[φ, J ] δφ(x

1

) . . . δφ(x

n

)

φ=φ0

= V

(n)

0

4

(x

1

− x

2

) . . . δ

4

(x

1

− x

n

) . (10) After defining δφ ≡ φ − φ

0

, and since Dφ = Dδφ (it is just a shift), the functional integral is given by

W

E

[J ] = N e

~1SE0,J]

Z

Dδφ exp (

− 1 2 ~

Z

d

4

xδφ(x) ˆ K

x

δφ(x) + X

p>2

Z

d

4

x δφ

p

p! V

(p)

0

) )

. (11) Let us now define δφ

0

≡ δφ/ √

~ such that W

E

[J] =

~ N e

1~SE0,J]

Z

Dδφ

0

exp (

− 1 2

Z

d

4

xδφ

0

(x) ˆ K

x

δφ

0

(x) + X

p>2

~

p 2−1

Z

d

4

x δφ

0p

p! V

(p)

0

) )

. We see from here that the higher-order terms are suppressed by powers of ~, therefore can be neglected (12) in the limit ~ → 0 . Once we neglect those terms, the remaining is a simple Gaussian integral, which is formally given by

1

W

E

[J] ≈N

0

e

1~SE0,J]

Det ˆ K

−1/2

=

=N

0

exp

− 1

~

S

E

0

, J] + ~

2 Tr [log ˆ K ]

,

(13)

from which we see that the second term represents the first quantum correction (1-loop diagrams).

1Kˆ is an operator which can be thought of as a (diagonal) matrix in position space:Kˆxδ4(x−y). Its determinant is defined as the product of its eigenvalues. This implies that detM = exp(TrlogM).

(3)

II. Assuming

V (φ) = λ

4! φ

4

, (14)

derive the expressions of the connected Green Functions of the theory in the classical limit (leading contribution for ~ → 0), in momentum space, up to O(λ

2

).

Suggestion: expand φ

0

in powers of λ .

Solution In the previous exercise we obtained that, at the classical level (therefore ignoring the second term in eq. (13)), the generating functional for connected Euclidean Green functions Z

E

[J] , defined by W

E

[J] = exp −Z

E

[J ] , is simply given by

Z

E

[J ] = 1

~

S

E

0

, J] − ln N

0

. (15)

From now on we will take ~ = 1 and neglect ln N

0

since it is a constant and we are only interested in the derivatives of Z

E

[J] . Using eq. (8) and after integrating by parts, we can write the classical action as

S

E

0

, J] = 1 2

Z d

4

x

2 − φ

0

d dφ

0

(V (φ

0

) − φ

0

J) = − 1 2

Z d

4

x

J φ

0

+ 2 λ 4! φ

40

. (16) Let us expand the classical solution in powers of λ , φ

0

= φ

[0]

+ λφ

[1]

+ λ

2

φ

[2]

+ . . . , plug this in the classical action above and keep only terms up to O(λ

2

):

S

E

0

, J] = Z

d

4

x

− 1

2 J φ

[0]

− λ

2 (J φ

[1]

+ 1

12 φ

[0]4

) − λ

2

2 (J φ

[2]

+ 1

3 φ

[1]

φ

[0]3

)

+ O(λ

3

) . (17) In order to obtain the J -dependence of φ

0

we have to solve eq. (8). This can be done independently for each power of λ . In particular we take the φ

0

(x) expansion and plug it in eq. (8). Let us also define the Euclidean two-point Green function as

(∂

2

− m

2

)

x

G(x, y) = −δ

4

(x − y) , (18) which is solved by

G(x, y) =

Z d

4

q

E

(2π)

4

e

iqE·(x−y)

q

2E

+ m

2

. (19)

The λ

0

term in eq. (8) is given by

−(∂

2

− m

2

)

x

φ

[0]

(x) = J(x) . (20) Let us multiply it by the two-point Green function and then integrate:

− Z

d

4

a G

xa

(∂

2

− m

2

)

a

φ

[0]

(a) = Z

d

4

a G

xa

J(a) , (21)

(4)

which gives (after integrating by part on the l.h.s.) φ

[0]

(x) =

Z

d

4

a G(x, a)J (a) ≡ G

xa

J

a

, (22) where we integrated over repeated indices.

The terms proportional to λ and λ

2

in eq. (8) are

−λ(∂

2

− m

2

)

x

φ

[1]

(x) = − λ 6 φ

[0]3

.

−λ

2

(∂

2

− m

2

)

x

φ

[2]

(x) = − λ

2

2 φ

[0]2

φ

[1]

.

(23)

Manipulating it as did above we get φ

[1]

(x) = − 1

6 G

xa

φ

[0]3a

= − 1

6 G

xa

G

ab

G

ac

G

ad

J

b

J

c

J

d

, φ

[2]

(x) = − 1

2 G

xa

φ

[0]2a

φ

[1]a

= − 1

2 φ

[1]a

G

xa

G

ab

G

ac

J

b

J

c

= 1

12 G

xa

G

ab

G

ac

G

ay

G

yd

G

ye

G

yf

J

b

J

c

J

d

J

e

J

f

. (24)

Plugging these solutions in the action, eq. (17), we get Z

E

[J] = S

E

[J] = − 1

2 J

x

G

xa

J

a

+ λ

4! J

x

G

xa

G

ab

G

ac

G

ad

J

b

J

c

J

d

+

− λ

2

72 G

ax

G

xb

G

xc

G

xy

G

yd

G

ye

G

yf

J

a

J

b

J

c

J

d

J

e

J

f

+ O(λ

3

)

(25)

The Euclidean connected Green functions are given by G

(n)E

(x

1

, . . . x

n

) = − δ

n

Z

E

δJ

1

. . . δJ

n

J=0

. (26)

Up to order O(λ

2

) they are given by (we set ~ = 1 ) G

(2)E

(x

1

, x

2

) = G(x

1

, x

2

) , G

(4)E

(x

1

, x

2

, x

3

, x

4

) = −λ

Z

d

4

y G(x

1

, y)G(x

2

, y)G(x

3

, y)G(x

4

, y) . (27) These are simply the propagator and the 4-point local interaction. By taking six functional derivatives we get all connected diagrams with six external lines (from a, b, c, d, e, f ) and two interaction vertices ( x, y ), connected by a propagator ( G

xy

):

G

(6)E

(x

1

, ..., x

6

) = λ

2

Z

d

4

xd

4

y G(x, y)P (x, y, x

1

, ..., x

6

) , (28) where

P (x, y, x

1

, ..., x

6

) = X

(ijk)∈A

G(x, x

i

)G(x, x

j

)G(x, x

k

)G(y, x

l

)G(y, x

m

)G(y, x

n

) (29)

(5)

with (lmn) being the complementary values of (ijk) (eg. (lmn) = (456) for (ijk) = (123) ) and A = {(123), (124), (125), (126), (135), (136), (145), (146), (156)} . Note that (ijk) runs only over half of its possible values since the whole expression of the connected Green function is symmetric under exchange of x and y .

The momentum space’s Green functions are defined by:

G ˜

(n)E

(p

1

, ..., p

n

)(2π)

4

δ(p

1

+ ... + p

n

) = Z

d

4

x

1

...d

4

x

n

e

i(x1p1+...+xnpn)

G

(n)E

(x

1

, ..., x

n

) . (30) Therefore it is easy to see that, in the classical limit

G ˜

(2)E

(p) = 1 p

2

+ m

2

=

p

, (31)

where we removed the second parameter since we have to impose p

1

= −p

2

,

G ˜

(4)E

(p

1

, p

2

, p

3

, p

4

) = −λ

4

Y

i=1

1

p

2i

+ m

2

= −λ p

1

p

2

p

4

p

3

, (32)

and

G ˜

(6)E

(p

1

, ..., p

6

) = λ

2

6

Y

i=1

1 p

2i

+ m

2

! X

(ijk)∈A

1

(p

i

+ p

j

+ p

k

)

2

+ m

2

(33)

= λ

2

 p

2

p

3

p

1

p1+p2+p3

p

4

p

5

p

6

+ permutations

. (34)

(6)

2 Scalar QED

Consider the theory of a complex scalar field φ interacting with the electromagnetic field A

µ

. The Lagrangian of the system is

L = − 1

4 F

µν2

+ (D

µ

φ)

(D

µ

φ) − m

2

φ

φ , (35) where D

µ

= ∂

µ

+ ieA

µ

is the gauge-covariant derivative.

I. Using functional methods, show that the propagator of the free complex scalar field, in momentum space, is

i

p

2

− m

2

+ i (36)

as in the case of a real scalar field.

Solution The generating functional of this theory is given by W [J, J

, J

µ

] =

Z

DφDA

µ

exp

i Z

d

4

x L + J

φ + J φ

+ J

µ

A

µ

, (37)

where we have to redefine m

2

→ m

2

− i in order for the integral to be convergent. Since for this question we are interested in the free scalar field, let us fix e = 0 and neglect the gauge field A

µ

. We can now complete the square in the action by shifting the scalar field:

φ

0

(x) ≡ φ(x) − i Z

D(x, y)J(y) , φ

0†

(x) ≡ φ

(x) − i Z

D(x, y)J

(y) , (38) where

D(x, y) =

Z d

4

q (2π)

4

i

q

2

− m

2

+ i e

iq·(x−y)

, (∂

2

+ m

2

− i)

x

D(x, y) = −iδ

4

(x − y) , (39) is the Feynman propagator. In fact, we can see that after this shift the exponent in the functional integral can be rewritten as

W [J, J

] = Z

0†

0

exp

i

− Z

d

4

0†

(∂

2

+ m

2

− i)φ

0

+ i Z

d

4

xd

4

yJ

(x)D(x, y)J(y)

= W

0

exp

− Z

d

4

xd

4

yJ

(x)D(x, y)J (y)

(40)

with W

0

= W [0, 0] . The first term is just a constant, which simplifies with the numerator when computing Green functions. Now we can compute the two point green functions and it is obvious that the only non-vanishing one is

h0|T φ

(x)φ(y)|0i = 1 W

0

−i δ

δJ (x) −i δ δJ

(y)

W [J, J

] = D(x, y) . (41)

Therefore the propagator for a complex scalar field is given by G(x, y) , as for a real scalar field.

(7)

II. Using functional methods, derive the expressions of all the connected Green Functions of the theory up to O(e

2

) , in the classical limit, in momentum space.

Solution From exercise 1 we know that, at the classical level, the generating functional for connected Green functions is given simply by the action evaluated at the classical field configuration, in the presence of an external current: Z [J, J

J

µ

] = S[φ

0

, φ

0

, A

µ0

, J, J

, J

µ

]. For this theory we have

S[φ

0

, φ

0

, A

µ0

, J, J

, J

µ

] =

= Z

d

4

x

− 1

4 F

µν2

+ (D

µ

φ

0

)

(D

µ

φ

0

) − (m

2

− i)φ

0

φ

0

− 1

2ξ (∂

µ

A

µ0

)

2

+ J

φ

0

+ J φ

0

+ J

µ

A

µ0

,

= Z

d

4

x 1

2 A

µ0

(g

µν

2

− (1 − 1

ξ )∂

µ

ν

)A

ν0

+ φ

0

(−∂

2

− m

2

+ i)φ

0

− ieA

µ0

0

µ

φ

0

) + e

2

A

2

φ

0

φ

0

+ J

φ

0

+ J φ

0

+ J

µ

A

µ0

.

(42)

where we included the i term in order to make the functional integral convergent and a Faddeev- Popov gauge fixing. Also, we defined (φ

0

µ

φ

0

) ≡ φ

0

µ

φ

0

− (∂

µ

φ

0

0

. To find the classical field configurations we have to solve 3 equations analogous to eq. (8), though it turns out that these equations are quite lengthy to solve because of the presence of derivatives in one of the interaction terms. To solve this problem we rewrite the action by taking the Fourier transform of the fields.

φ(x) =

Z d

4

p (2π)

4

φ(p)e ˜

ipx

, φ

(x) =

Z d

4

p (2π)

4

φ ˜

(p)e

−ipx

, A

µ

(x) =

Z d

4

p (2π)

4

A ˜

µ

(p)e

ipx

(43) By plugging these definitions into the action and integrating over x we get

S[φ

0

, φ

0

, A

µ0

, J, J

, J

µ

] =

− 1 2

A ˜

µ0

(−k)(g

µν

k

2

− (1 − 1

ξ )k

µ

k

ν

) ˜ A

ν0

(k)

k

+ D

φ ˜

0

(p)(p

2

− m

2

+ i) ˜ φ

0

(p) E

p

+ e D

(p + p

0

)

µ

A ˜

µ0

(−p + p

0

) ˜ φ

0

(p

0

) ˜ φ

0

(p) E

p,p0

+ e

2

D

A ˜

µ

(k) ˜ A

µ

(−k − p + p

0

) ˜ φ

0

(p

0

) ˜ φ

0

(p) E

k,p,p0

+ D

J ˜

(p) ˜ φ

0

(p) E

p

+ D

J(p ˜

0

) ˜ φ

0

(p

0

) E

p0

+ D

J ˜

µ

(−k) ˜ A

µ0

(k) E

k

,

(44)

where we used the short notation hf (p

1

, . . . , p

n

)i

p

1,...pn

= R

d4p1

(2π)4

. . .

(2π)d4pn4

f (p

1

, . . . , p

n

) . Now, to find

the classical field configurations we can notice when the action is minimized a small change in φ (or

equivalently φ

or A ) will leave the action unchanged (at first order), so a small change in φ ˜ will also

(8)

leave the action unchanged. We can use this to find the the defining equations for φ ˜

0

, φ ˜

0

and A

µ0

: 0 = δS

δ φ(p) ˜

˜

φ= ˜φ0

= (p

2

− m

2

+ i) ˜ φ

0

(p) + e D

(p + p

0

)

µ

A ˜

µ0

(−p + p

0

) ˜ φ

0

(p

0

) E

p0

+ e

2

D

A ˜

µ0

(k) ˜ A

µ,0

(−k − p + p

0

) ˜ φ

0

(p

0

) E

k,p0

+ ˜ J

(p) , 0 = δS

δ φ ˜

(p

0

)

˜

φ= ˜φ0

= (p

2

− m

2

+ i) ˜ φ

0

(p

0

) + e D

(p + p

0

)

µ

A ˜

µ0

(−p + p

0

) ˜ φ

0

(p) E

p

+ e

2

D

A ˜

µ0

(k) ˜ A

µ,0

(−k − p + p

0

) ˜ φ

0

(p) E

k,p

+ ˜ J(p

0

) , 0 = δS

δ A ˜

µ

(k)

˜

A= ˜A0

= − A ˜

ν0

(−k)(g

νµ

k

2

− (1 − 1

ξ )k

ν

k

µ

) + e D

(k + 2p)

µ

φ ˜

0

(k + p) ˜ φ

0

(p) E

p

+ 2e

2

D

A ˜

µ,0

(p

0

− k − p) ˜ φ

0

(p

0

) ˜ φ

0

(p) E

p,p0

+ ˜ J

µ

(−k) .

(45)

As did for the first exercise, let us expand the fields in powers of e and solve the equations above iteratively:

ψ

0

= ψ

[0]

+ eψ

[1]

+ e

2

ψ

[2]

+ . . . , ψ = ˜ φ, φ ˜

, A ˜

µ

(46) For the zero-th order, equations (45) become

0 = (p

2

− m

2

+ i) ˜ φ

[0]†

(p) + ˜ J

(p) , 0 = (p

2

− m

2

+ i) ˜ φ

[0]

(p

0

) + ˜ J(p

0

) , 0 = − A ˜

[0]µ

(k)(g

µν

k

2

− (1 − 1

ξ )k

µ

k

ν

) + ˜ J

µ

(k) .

(47)

And the solutions are simply given by φ ˜

[0]†

(p) = i

p

2

− m

2

+ i i J ˜

(p) ≡ P

S

(p)i J ˜

(p), φ ˜

[0]

(p) = i

p

2

− m

2

+ i i J ˜ (p) ≡ P

S

(p)i J(p), ˜ A ˜

[0]µ

(k) = −i

k

2

(g

µν

− (1 − ξ)k

µ

k

ν

/k

2

) i J ˜

ν

(k) ≡ P

Aµν

(k)i J ˜

ν

(k) ,

(48)

where you can recognise the scalar and gauge propagators. For the first order, equations (45) become:

−(p

2

− m

2

+ i) ˜ φ

[1]†

(p) = D

(p + p

0

)

µ

A ˜

[0]µ

(p

0

− p) ˜ φ

[0]†

(p

0

) E

p0

,

−(p

2

− m

2

+ i) ˜ φ

[1]

(p

0

) = D

(p + p

0

)

µ

A ˜

[0]µ

(p

0

− p) ˜ φ

[0]

(p) E

p

, A ˜

[1]µ

(k)(g

µν

k

2

− (1 − 1

ξ )k

µ

k

ν

) = D

(2p − k)

ν

φ ˜

[0]†

(p − k) ˜ φ

[0]

(p) E

p

.

(49)

(9)

Thus the solutions are

φ ˜

[1]†

(p) = P

S

(p)i D

(p + a)

µ

P

Aµν

(a − p)P

S

(a)i J ˜

ν

(a − p)i J ˜

(a) E

a

, φ ˜

[1]

(p) = P

S

(p)i D

(a + p)

µ

P

Aµν

(p − a)P

S

(a)i J ˜

ν

(p − a)i J(a) ˜ E

a

, A ˜

[1]µ

(k) = P

Aµν

(k)i D

(2a − k)

ν

P

S

(a − k)P

S

(a)i J ˜

(a − k)i J(a) ˜ E

a

,

(50)

Finally, for the second order, the equations we get are

−(p

2

− m

2

+ i) ˜ φ

[2]†

(p) = D

(p + p

0

)

µ

A ˜

[1]µ

(−p + p

0

) ˜ φ

[0]†

(p

0

) + A

[0]µ

(−p + p

0

) ˜ φ

[1]†

(p

0

) E

p0

+

D A ˜

[0]µ

(k) ˜ A

[0]µ

(−k − p + p

0

) ˜ φ

[0]†

(p

0

) E

k,p0

,

−(p

2

− m

2

+ i) ˜ φ

[2]

(p

0

) = D

(p + p

0

)

µ

A ˜

[1]µ

(−p + p

0

) ˜ φ

[0]

(p) + A

[0]µ

(−p + p

0

) ˜ φ

[1]

(p) E

p

+ D

A ˜

[0]µ

(k) ˜ A

[0]µ

(−k − p + p

0

) ˜ φ

[0]

(p) E

k,p

, A ˜

[2]µ

(k)(g

µν

k

2

− (1 − 1

ξ )k

µ

k

ν

) = D

(2p − k)

ν

φ ˜

[1]†

(p − k) ˜ φ

[0]

(p) + ˜ φ

[0]†

(p − k) ˜ φ

[1]

(p) E

p

+ 2 D

A ˜

[0]µ

(p

0

+ k − p) ˜ φ

[0]†

(p

0

) ˜ φ

[0]

(p) E

p,p0

.

(51)

Plugging in the previous results we get φ ˜

[2]†

(p) = P

S

(p)i

D

P

Aµν

(a)P

Aµρ

(b − a − p)P

S

(b)i J ˜

ν

(a)i J ˜

ρ

(b − a − p)i J ˜

(b) E

a,b

+ i D

(p + a)

µ

P

Aµν

(a − p)P

S

(a)(b + a)

ρ

P

Aρσ

(a − b)P

S

(b)i J ˜

ν

(a − p)i J ˜

σ

(a − b)i J ˜

(b) E

a,b

+ i D

(p + a)

µ

P

Aµν

(a − p)(p − a + 2b)

ν

P

S

(b + p − a)P

S

(b)P

S

(a)i J ˜

(b + p − a)i J ˜ (b)iJ

(a) E

a,b

, φ ˜

[2]

(p) = P

S

(p)i

D

P

Aµν

(a)P

Aµρ

(p − a − b)P

S

(p)i J ˜

ν

(a)i J ˜

ρ

(p − a − b)i J ˜

(b) E

a,b

+ i

D

(p + a)

µ

P

Aµν

(p − a)P

S

(a)(b + a)

ρ

P

Aρσ

(a − b)P

S

(b)i J ˜

ν

(p − a)i J ˜

σ

(a − b)i J ˜

(b) E

a,b

+ i D

(p + a)

µ

P

Aµν

(p − a)(a − p + 2b)

ν

P

S

(b + a − p)P

S

(b)P

S

(a)i J ˜

(b + a − p)i J ˜ (b)i J ˜

(a) E

a,b

, A ˜

[2]µ

(k)a = P

Aµν

(k)i

2 D

P

Aνρ

(b + k − a)P

S

(b)P

S

(a)i J ˜

ρ

(b + k − a)i J ˜

(b)i J(a) ˜ E

ab

+ i D

(2a − k)

µ

P

S

(a − k)(a + b − k)

ρ

P

Aρσ

(b + k − a)P

S

(b)P

S

(a)i J ˜

σ

(b + k − a)i J ˜

(b)i J ˜ (a) E

a,b

+ i D

(2a − k)

µ

P

S

(a − k)P

S

(a)(a + b)

ρ

P

Aρσ

(a − b)P

S

(b)i J ˜

(a − k)i J ˜

σ

(a − b)i J ˜ (b) E

a,b

.

(52)

(10)

Now we can substitute these expressions in the generating functional eq. (44) and keep only the terms up to O(e

2

) :

Z[J, J

J

µ

] =

− 1 2

A ˜

[0]µ

(−k)(g

µν

k

2

− (1 − 1

ξ )k

µ

k

ν

) ˜ A

[0]µ

(k)

k

+ D

φ ˜

[0]†

(p)(p

2

− m

2

+ i) ˜ φ

[0]

(p) E

p

+ D

J ˜

(p) ˜ φ

[0]

(p) E

p

+ D

J ˜ (p

0

) ˜ φ

[0]†

(p

0

) E

p0

+ D

J ˜

µ

(−k) ˜ A

[0]µ

(k) E

k

+ e − A ˜

[1]µ

(−k)(g

µν

k

2

− (1 − 1

ξ )k

µ

k

ν

) ˜ A

[0]µ

(k)

k

+ D

φ ˜

[1]†

(p)(p

2

− m

2

+ i) ˜ φ

[0]

(p) + ˜ φ

[0]†

(p)(p

2

− m

2

+ i) ˜ φ

[1]

(p) E

p

+ D

J ˜

(p) ˜ φ

[1]

(p) E

p

+ D

J ˜ (p

0

) ˜ φ

[1]†

(p

0

) E

p0

+ D

J ˜

µ

(−k) ˜ A

[1]µ

(k) E

k

+ D

(p + p

0

)

µ

A ˜

[0]µ

(−p + p

0

) ˜ φ

[0]†

(p

0

) ˜ φ

[0]

(p) E

p,p0

+ e

2

− A ˜

[2]µ

(−k)(g

µν

k

2

− (1 − 1

ξ )k

µ

k

ν

) ˜ A

[0]µ

(k) − 1 2

A ˜

[1]µ

(−k)(g

µν

k

2

− (1 − 1

ξ )k

µ

k

ν

) ˜ A

[1]µ

(k)

k

+ D

φ ˜

[2]†

(p)(p

2

− m

2

+ i) ˜ φ

[0]

(p) + φ

[1]†

(p)(p

2

− m

2

+ i) ˜ φ

[1]

(p) + φ

[0]†

(p)(p

2

− m

2

+ i) ˜ φ

[2]

(p) E

p

+ D

J ˜

(p) ˜ φ

[2]

(p) E

p

+ D

J ˜ (p

0

) ˜ φ

[2]†

(p

0

) E

p0

+ D

J ˜

µ

(−k) ˜ A

[2]µ

(k) E

k

+ D

(p + p

0

)

µ

A ˜

[1]µ

(−p + p

0

) ˜ φ

[0]†

(p

0

) ˜ φ

[0]

(p) E

p,p0

+ D

(p + p

0

)

µ

A ˜

[0]µ

(−p + p

0

) ˜ φ

[1]†

(p

0

) ˜ φ

[0]

(p) E

p,p0

+ D

(p + p

0

)

µ

A ˜

[0]µ

(−p + p

0

) ˜ φ

[0]†

(p

0

) ˜ φ

[1]

(p) E

p,p0

+

D A ˜

[0]µ

(k) ˜ A

[0]µ

(−k − p + p

0

) ˜ φ

[0]†

(p

0

) ˜ φ

[0]

(p) E

k,p,p0

+ O(e

3

)

(53) By plugging our previous results in this last expression we get

Z [J, J

J

µ

] = i

2

D J ˜

µ

(−k)P

Aµν

(k) ˜ J

ν

(k) E

k

+ i D

J ˜

(p)P

S

(p) ˜ J (p) E

p

+ + e

c

1

i D

(p + p

0

)

µ

P

Aµν

(p

0

− p)P

S

(p

0

)P

S

(p) ˜ J

ν

(p

0

− p) ˜ J

(p

0

) ˜ J(p) E

p,p0

+ + e

2

c

2

D

P

Aµν

(k)P

Aµρ

(p

0

− k − p)P

S

(p

0

)P

S

(p) ˜ J

ν

(k) ˜ J

σ

(p

0

− k − p) ˜ J

(p

0

) ˜ J(p) E

k,p,p0

+ + c

3

i D

P

S

(p)(p + a)

µ

P

Aµν

(a − p)P

S

(a)(b + a)

ρ

P

Aρσ

(a − b)P

S

(b)i J ˜ (p)i J ˜

σ

(a − b)i J ˜

ν

(a − p)i J ˜

(b) E

p,a,b

+ + c

4

i D

P

S

(p)(p + a)

µ

P

Aµν

(a − p)(p − a + 2b)

ν

P

S

(b + p − a)P

S

(b)P

S

(a)i J ˜ (p)i J ˜

(b + p − a)i J(b)i ˜ J ˜

(a) E

p,a,b

(54)

We can already see, by the number of source terms, that the first line will give the scalar and gauge

propagators, the second line to the 3-point connected green functions. And finally the e

2

terms give

(11)

rise to the 4-point connected green function. Since the generating functional is already expressed in momentum space we can directly derive the expression of the connected green functions in momentum space. So we need to derive the correct formula to get the result directly in momentum space. To do so let us consider generating functional (for a real scalar field) containing only one term with n sources, expressed in momentum space:

Z[J] =

Z d

4

p

1

. . . d

4

p

n−1

(2π)

4(n−1)

J(p ˜

1

) . . . J ˜ (p

n−1

) ˜ J(−p

1

− . . . − p

n−1

)f(p

1

, ..., p

n−1

) . (55) The terms in (54) can all be written in this form: n sources, n − 1 integrals and other terms which are represented by f (p

1

, . . . , p

n−1

) . Now we can notice that

J(p ˜

1

) . . . J(p ˜

n−1

) ˜ J(−p

1

− . . . − p

n−1

) = Z

d

4

p

n

δ(p

1

+ . . . + p

n

) ˜ J(p

1

) . . . J ˜ (p

n−1

) ˜ J(p

n

) . (56) And using the fact that we impose conservation of momentum we can rewrite f as a function of all momenta. Therefore we get

Z [J] = (2π)

4

Z d

4

p

1

. . . d

4

p

n

(2π)

4n

J ˜ (p

1

) . . . J(p ˜

n

)f (p

1

, ..., p

n

)δ(p

1

+ . . . + p

n

) . (57) Now we can use the definition of Fourier transform to express the generating functional in terms of sources in position space.

Z [J] = (2π)

4

Z d

4

p

1

d

4

x

1

. . . d

4

p

n

d

4

x

n

(2π)

4n

J (x

1

) . . . J (x

n

)f (p

1

, ..., p

n

)e

−i(p1x1+...pnxn)

δ(p

1

+ . . . + p

n

) . (58) From this we can compute the n-point connected green function in position space:

G

(n)c

(x

1

, ..., x

n

) =

−i δ δJ(x

1

)

. . .

−i δ δJ(x

n

)

iZ [J]

J=0

= (2π)

4

(−i)

n−1

X

σ∈Sn

Z d

4

p

1

. . . d

4

p

n

(2π)

4n

f (p

σ(1)

, ..., p

σ(n)

)e

−i(pσ(1)x1+...pσ(n)xn)

δ(p

1

+ . . . + p

n

) . (59) Now we can plug this result in the momentum space’s green function definition:

G ˜

(n)c

(p

1

, ..., p

n

)δ(p

1

+ ... + p

n

) = (2π)

−4

Z

d

4

x

1

...d

4

x

n

e

i(x1p1+...+xnpn)

G

(n)c

(x

1

, ..., x

n

)

= (−i)

n−1

Z d

4

x

1

d

4

q

1

. . . d

4

x

n

d

4

q

n

(2π)

4n

X

σ∈Sn

f (q

σ(1)

, ..., q

σ(n)

)e

i[x1(p1−qσ(1))+...+xn(pn−qσ(n))]

δ(q

1

+ . . . + q

n

)

= (−i)

n−1

X

σ∈Sn

Z

d

4

q

1

. . . d

4

q

n

f(q

σ(1)

, ..., q

σ(1)

)δ(p

1

− q

σ(1)

) . . . δ(p

n

− q

σ(n)

)δ(q

1

+ . . . + q

n

)

= (−i)

n−1

δ(p

1

+ . . . + p

n

) X

σ∈Sn

f(p

σ(1)

, ..., p

σ(n)

)

= (2π)

4(n−1)

−i δ δ J(p ˜

1

)

. . .

−i δ δ J(p ˜

n

)

iZ [J ]

˜

J=0

,

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