Advanced Quantum Field Theory: Examples 3

Solution prepared by Peng Zhao

1. The superficial degree of divergence D of a diagram counts the power of momenta. For a diagram with L
loops, IB internal boson lines and IF internal fermion lines, D = dL 2IB IF in d-dimensional spacetime
1
1
because each scalar propagator tends to 2 , each fermionic propagator tends to and we integrate over
k
k
X
fr
nr IB IF +
each loop. Consider a theory with interaction vertices r br fr . Use Eulers formula
r
X
X
L = 1 and formula for the number of internal lines 2IB =
nr br EB , 2IF = 2
nr fr EF to rewrite
r

X
d2
d1
D =d
EB
EF +
nr r ,
2
2
r


r =


d2
br + (d 1)fr d .
2

X
3
d2
For d = 4, we find D = 4 EB EF +
nr r with r = br + 3fr 4. In d-dimensions, [] =
and
2
2
r
d1
d2
[] =
, so [r ] = d
br (d 1)fr = r and negative coupling dimension corresponds to a
2
2
non-renormalisable theory.
X
2. For a scalar field in d = 2, D = 2 2
nr so any polynomial interaction is renormalisable. The only
r

divergent connected 1PI diagrams in theory are those with only one vertex, i.e. and
. These are
subtracted upon renormalisation and is consistent with normal-ordering, which removes . For a fermionic
EF X
field in d = 2, D = 2
+
nr (fr 2) so only interactions with fr 2 is renormalisable.
2
r
Z
1
1
(n)
3. From power-counting, we see that Id =
dd k 2
diverges if d 2n 0. We use the
d
(4)
(k
+
m 2 )n
Z
Z


2
d/2
to write
identities x () =
d 1 ex and dd k ek =

0


Z
Z
Z
2
1
d
m2n+d
1
(n)
d
n d
n1 (k2 +m2 )
2 1 em =
Id =

d
k
d

e
=
.
d
d
(2)d (n)
2
(4) 2 (n) 0
(4) 2 (n)
0
4

d
= 0, 1, 2, . . ., which agrees with power-counting if d is even. Note that there is no
2
pole at odd d, so we have analytically continued the original integral.
4. With c, c0 > 0, consider
Z
Z
Z
Z
x1
x1
x1
y 1
0
I1 (c) =
dx
,
I2 (c, c ) =
dx
,
I
(c)
=
dx
dy
.
3
x+c
(x + c)(x + c0 )
x+c 0
(y + c)(x + y + c)
0
0
0
(n)

Id

has poles at n

From power-counting we see that I1 is convergent

Z if 1 <  < 2, I2 is convergent for 0 <  < 2, I3 is

convergent for 1/2 <  < 2. We use x () =

d 1 ex and (1 )() =
to write
sin

0
Z
Z
Z
1
I1 (c) =
dx x1
d e(x+c) = (2 )
d 2 ec = (2 )( 1)c1 =
c ,
sin 
0
0
0
 1

Z
1
x
x1
c1 c01
I2 (c, c0 ) = 0
dx

=
,
c c 0
x + c x + c0
sin 
c c0
Z
Z
Z
x1
y 1

x1 (x + c)1 c1

I3 (c) =
dx
dy
=
dx
.
x+c 0
(y + c)(x + y + c)
sin  0
x+c
x
0
The last integral can be written as

Z
Z
x1
 
1
dx (x + c) x c
dx
I3 (c) = (1 )()
(x + c)x
0
Z 0Z
= (1 )
dx
dy x y 1 ey(x+c) c12 (1 )2 ()2
0
0
Z



2
22 yc
12
2
= (1 )
dy y
e
c
() = (1 )2 (2 1) ()2 c12 .
0

I1 and I2 have simple poles at integer values of . I3 has double poles at integer values of  and at halfc
1
1
integers at 1/2, 1/2, 3/2, etc. Near  = 0, I1 + c log c and I2 . Since () E , (1 )



c
1
c
2c
12
1 + E , (2 1) + E 1 and c
c(1 2 log c), so I3 (c) 2
+
log c. Thus the log c
2

2

2
subdivergence may be subtracted by considering I3 I1 .

Z
d eA , rescaling = sx and using the identity (sx) = s1 (x), we find

5. Using A1 =

1
=
A1 A2 An

ds s

d
Rn
+

i i Ai

n1

ds s

d x
Rn
+

xi

es

xi A i

P
P
Z
i xi )
(1 i xi )
n (1
n
= (n)
= (n 1)!
.
d x P
d x P
( i x i Ai ) n
( i x i Ai ) n
x[0,1]n
x[0,1]n
Z

6. The bare Lagrangian is

1
1
1
1
1
1
LB = L + Lc.t = (1 + A)()2 (m2 + B 2 )2  (1 + D)4 = (B )2 m20 2B 0 4B ,
2
2
24
2
2
24

1+D 
m2 + B 2
are
and 0 =
1+A
(1 + A)2
"
!
"


#
#
2

2
m
2
1

2
5

2
2
2 2
3
2
2

3 ),
m0 = m +
+ m

+ O( ) = m 1 + +

1+
+ O(

2
2
12

2
12
!
"
"

#

#
2
2

3
1

9
17
3

2
4 ) = 16 2
+
3
4 ).
1+
+ 3

1+
+ O(
+

+ O(
 0 = 16 2

2

6

2
6

where the bare field B =

1 + A and the bare parameters m20 =

and m2 are determined by the simple poles of  0 and m0 , respectively.



32
173
d
32
173
f1 =
+

1
f1 =

+ O(4 ),

2
4
2
16
1536
d
16
768 4

52
d

52
b1 =

m2 = b1 =

+ O(3 ).
2
4
2
16
3072
d
16
1536 4




93
d
d
d
d
4
2
3

1 f2 =
+O( ) = f1 and b2 = 4 +O( ) = m2 +
The double poles satisfy
b1 ,
d
128 4
d
d
d
as expected.
7. To one loop, the 1PI diagrams that contribute to F2 (p, p) in 4 theory are
k
p

-p

-p

= p2 m2


2

dd k
1
.
(2)d k 2 + m2


(1)
Id from problem 3. In dimensional regularisation, d = 4 , we find
2




 m2 e 2 log m2  2
m2
m2
2
2
=

1
=

1
+

+O()
=

1
+

+
log
m

log
4
+O().


E
E
2

16 2

(4)2 2
(4)2 2

The integral is just

(1)

I4

In the M S scheme, we subtract the divergent and finite parts. So we are left with


m2
m2
F2 (p, p) = p2 m2 +
1

log
.
32 2
2
The 1PI diagrams that contribute to F4 (p1 , p2 , p3 , p4 ) are
p1

p3

p1

p3

+
p2

p4

p1

p2

p-k

p4

p3

p3

p4

p-k

p2

+
p2

p1

p-k

p4

With help of the Feynman parameter identity, we evaluate the integral

Z 1
Z 1
Z
Z
(1 )
1
2 2
dd k
2 2
dd k
d
d
.
=
d
2
2
2
2
d
2
2
2
(2) (k + m )((p k) + m )
2
(2) 0
[(k + m ) + ((p k)2 + m2 )]2
0
We can complete square in the denominator as
"
2

(k + m ) + ((p k) + m ) = ( + )

p
k
+

2

p2
+m +
( + )2

#
( + )(k 02 + m02 ).

(2)

The integral over k 0 is just Id from problem 3








1
2

2
(m02 ) 2   

1
(2)
02
02
=
1 + log 4 =
Id =
E
1 log m
E log m + log 4 +O().

2
16 2 
2
2
16 2 
(4)2 2
In the M S scheme, we find
Z
Z 1
Z 1
(1 ) (2)
m2 + (1 )p2
2
2 2 1
2
d
d
I
d
log

f (p2 ),
=

2
( + )2 d
32 2 0
2
32 2
0
0
The one-loop diagrams correspond to p2 = s, t, u, respectively. Including the tree level contribution,
4 (p1 , p2 , p3 , p4 ) =

2
(f (s) + f (t) + f (u)) ,
32 2

s = (p1 + p2 ),

t = (p1 + p3 ),

u = (p1 + p4 ).

It is straightforward to check that the Greens functions satisfy the Callan-Symanzik equations


2
2
2
2
32
m2

2 = 2m m m + O(2 ) = O(2 ),
+
+
F

16 2 16 2 m2
m2 3 32 2
16 2


32
m2
2 32
32

+
+
F4 = 2

+ O(3 ) = O(3 ).

2
2
2
2
16 16 m
32
16 2

 1
1
3
s
2

For s = t = u and s  m , then to second order, F4 (p1 , p2 , p3 , p4 ) =

log
. We check
32 2
2
that it satisfies the Callan-Symanzik equation to all orders




32
m2
3
2 2s
32 1

+
+
F4 =

F42 = 0.

16 2 16 2 m2
32 2 s 3
16 2 2
Suppose F4 (p, p, p, p) = 0 for p2 = m2 , then s = 4m2 , t = u = 0 and we can express 0 in terms of
0 = +

2
2
32
m2
2
f
(4m
)
+
2f
(0)

+
log
32 2
16 2 32 2
2

= 0 +

02
302
m2

log
+O(03 ).
16 2 32 2
2

Then we can express 4 in terms of 0

02
02
4 (p1 , p2 , p3 , p4 ) =
+
(g(s) + g(t) + g(u))+O(03 ),
16 2 32 2
0

Z
g(s)
0

h 
s i
d log 1 (1 ) 2 ,
m

which is independent of but diverges as m 0.

1 2 2
1 
m 2 g3 . The one point function is
2
3!
Z




2 g
dd k
1
2 g (1)
2 g m4

=
=

I
=

2
+

2
(2)d k 2 + m2
2 d
2 (4)3 2
2
2

m





log
2
g m4 e 2
1 3 E
gm4
1
m2
1
1 3 E
=
+

+
O()
=

log
+
log
4
+ O().

2 (4)3 2

4
2
128 3 
4
2
2
2
2

8. Consider 3 theory in d = 6  dimensions with V () =

k

1 gm4
. The two point function is identical to that in problem 7.
 128 3
Z
 g 2
dd k
1
 g 2 (2)
=
=
I
d
2
2
2
2
2
(2) (k + m )((p k) + m )
2 6




 g 2 (m02 )1 2  
m02
2
m02
=

1
=

1
+

+
log

log
4
+ O().


E
2 (4)3 2
2
128 3

2

Thus the divergent part is

k
p

-p

p- k



Z 1


1 g2
1 2
1 g2
2
2
2
d m + (1 )p =
m + p . The three point function
The divergent part is
 64 3 0
 64 3
6
can be computed again using Feynmans parametric identity
p2
p1

1
(k 2 + m2 )((p1 k)2 + m2 )(p2 k)2 + m2 )
Z
Z 1 Z 1 Z 1
3
(1 )
= 2 2 g 3 dd k d d
d
3.
2
2
[(k + m ) + ((p1 k)2 + m2 ) + ((p2 k)2 + m2 )]
0
0
0
3

= 2 g 3

p 2- k
p3

p 1- k

dd k

Completing the square in the denominator as usual, we find



( + )p21 + ( + )p22 2p1 p2
2
2
2
2
2
2
02
2
(k +m )+((p1 k) +m )+((p2 k) +m ) = (++) k + m +
,
( + + )2
where k 0 = k

p1 + 3 p2
(3)
. The integral over k 0 is Id
++


3
2

(3)
g 3 I6


(m02 ) 2
g3
= 2 g
=


2
128 3
(4)3 2 (3)
3
2


2
m02
E log 2 + log 4 + O().


1 g3
. These divergences can be cancelled by adding the counter term
 64 3





1
1
1
1
1 2
1 2
2
 00
3
2

2
4
2 2 2
3 3
2
g () + V () =
Lc.t. =
+ 3 gm + 3g m + g () + g .
 3!(4)3 2
 3!(4)3
2


1 2
1 1 2
1 1
12 
2
4

g m + p
to
gm to each 1-point vertex, A + B =
which effectively adds E =
 128 3
 64 3
6
1
1 1

each 2-point vertex and C =
2  g 3 to each 3-vertex. Since [] = 2 and [m] = 1. It follows that
 (4)3
2
[Lc.t. ] = 6 , as expected.
9. We work out the bare parameters, beta function and anomalous dimension as we did for for problem 6





m2 + B
1 g2
1 g2
1 5g 2
2
2
4
2
m0 =
=m 1
1+
+ O(g ) = m 1 +
+ O(g 4 ),
1+A
 64 3
 384 3
 384 3








3 1 g2
1+C
1 g2
1 3g 2
2g
2g
g0 = 2 g
1
+
=

+ O(g 5 ),
=

 64 3
2  384 3
 256 3
(1 + A)3/2


d
3g 3
d
5g 2
g = g
1 f1 =
,
m2 = g b1 =
.
3
dg
256
dg
192 3
The divergent part is

Since g is decreasing, we can have asymptotic freedom in this theory.

10. For a theory with multiple couplings g i , a change of variable g g 0 (g) changes the beta function as
g 0i dg j
g 0i j
dg 0i
0i (g 0 ) =
= j
=
(g). For (g) = b1 g 3 + b2 g 5 + O(g 7 ) and g 0 (g) = g + a1 g 3 + O(g 5 ),
d
g d
g j
0 (g 0 ) = (1 + 3a1 g 02 + O(g 04 ))(b1 (g 0 a1 g 03 )3 + b2 g 05 ) + O(g 07 ) = b1 g 03 + b2 g 05 + O(g 07 ).
So the first two terms in the beta function are invariant. Consider the higher order terms
(g) = b1 g 3 + b2 g 5 + +b3 g 7 + O(g 8 ), g 0 = g + a1 g 3 + a2 g 4 + a3 g 5 + O(g 6 )




0 (g 0 ) = 1 + 3a1 g 2 + 4a2 g 3 + 5a3 g 4 b1 g 3 + b2 g 5 + b3 g 7 + O(g 8 )
= b1 g 3 + (b2 + 3a1 b1 )g 5 + 4a2 b1 g 6 + (b3 + 3a1 b2 + 5a3 b1 )g 7 + O(g 8 )
= b1 (g 0 a1 g 03 a2 g 04 )3 + (b2 + 3a1 b1 )(g 0 a1 g 03 )5 + 4a2 b1 g 06 + (b3 + 3a1 b2 + 5a3 b1 )g 07 + O(g 08 )
= b1 g 03 + b2 g 05 + (4a2 b1 3a2 )g 06 + (b3 + 3a1 b2 + 5a3 b1 5a1 (b2 + 3a2 b1 ) + 3a21 )g 07 + O(g 08 ).
We see that we can make the g 06 term vanish by choosing a2 = 0 and the g 07 term vanish by choosing
appropriate values for a1 and a3 . In this way, we can recursively define g 0 (g) such that the beta function
vanishes beyond two loops.
4