This document provides an overview of field theory concepts, including: 1) Constructing a toy Standard Model using an SU(2)xU(1) symmetry, which unifies scalar and spinor fields into a single Lagrangian. 2) Describing the transformations of fields like the left-handed spinor doublet under the full and diagonal subgroups. 3) Explaining how the U(1) and SU(2) gauge fields can be rotated into a massive Z field and massless photon through mixing angles. 4) Defining covariant derivative operators that incorporate the distinguishing hypercharges of fields.

1. Field Theory
Highlights 2015
By: Roa, F. J. P.
Topics:

1 SU(2)XU(1) Construction in toy Standard Model

2 Basic Practice Calculations Involving Scalar Fields

1 SU(2)XU(1) Construction in toy Standard Model
We can initially unify a scalar doublet  with a left-handed spinor doublet L
 and a
right-handed spinor singlet R
2
 under the SU(2)XU(1) symmetry. These distinct fields
can be incorporated into a single Lagrangian )
,
,
( 2 

 R
L
L that observes the said
symmetry
(1)
  )
(
2
1
)
,
,
(
2
2
2
2
)
(
2
)
(
2






















V
D
y
D
i
D
i
L
R
L
L
dagger
R
R
R
R
L
L
L
R
L






Aside from spin, each of the fields is also distinct by its assigned hypercharge that this
field carries in a given SU(2)XU(1) transformation. For the case of the left-handed spinor
doublet, its transformation
(2)
L
q
L
L
Q
i
Y
i 



 )
exp(
)
exp(







is distinctly characterized by the hypercharge L
Y . The full SU(2) part in the
transformation is represented by the matrix )
(
exp 





 Q
i and is common in all
transformations of these distinct fields.
To observe the full L
XU
SU )
1
(
)
2
( symmetry for the case of the left-handed spinor
doublet, the covariant derivative operator for the spinor doublet is also assigned with this
distinguishing hypercharge L
Y ,
(3)







3
1
)
(
)
(
l
l
l
L
L W
Q
i
B
Y
Q
i
D 


 

2. Basically, this includes the U(1) gauge field 
B and the three components of SU(2)
gauge field, )
,
,
( )
3
(
)
2
(
)
1
( 

 W
W
W .
Under the full L
XU
SU )
1
(
)
2
( as also in the other )
1
(
)
2
( XU
SU cases, )
1
(
U gauge field
must transform as
(4.1)
q
Q
B
B 


 

 1
and to 1st
order in Q, the SU(2) gauge field 
W

transforms as
(4.2)




 
 W
Q
W
W
W













 2
To accommodate for the diagonal subgroup of symmetry say for example, L
XU
SU )
1
(
)
2
(
diagonal subgroup
(5.1)
)
(
exp
)
(
exp 3 q
q
L i
Y
i 

 

we replace 
B field and the 
)
3
(
W component of 
W

field with a massive 
Z field and a
massless electromagnetic field em
A . This is implemented through the linear combinations
(5.2)

 

 cos
sin )
3
(
W
B
Z 

(5.3)

 

 sin
cos )
3
(
W
B
Aem


The covariant derivative operator for the spinor doublet would then be written as
(5.4)
   
  












sin
cos
sin
cos
3
2
3
2
)
2
(
2
)
1
(
1
)
(
m
e
L
L
L
A
Y
Q
i
Z
Y
Q
i
W
W
Q
i
D











Under this particular diagonal subgroup (5.1), the electromagnetic field transforms as

3. (5.5)
q
m
e
m
e
e
A
A 


 

 1
2
Here,
(5.6)
2
cos
sin
e
Q
Q 

 

where  is the mixing angle. The massive vector field 
Z is neutral )
( 
 Z
Z  as the
left-handed spinor doublet transforms under this diagonal L
XU
SU )
1
(
)
2
( subgroup.
The transformation of the right-handed spinor singlet is characterized by its own
hypercharge R
Y as it transforms under a unimodular symmetry
(5.7)
R
q
R
R
Y
i 2
2 )
(
exp 

 

In the diagonal subgroup (5.1), the designated 3rd
component 
)
3
(
W of the SU(2)gauge
field 
W

transforms as
(5.8)
q
Q
W
W 


 


 1
)
3
(
)
3
(
while the other components of this gauge field take SO(2)-like rotations
(5.9)
q
q W
W
W
W 
 


 2
sin
2
cos )
2
(
)
1
(
)
1
(
)
1
( 



(5.10)
q
q W
W
W
W 
 


 2
cos
2
sin )
2
(
)
1
(
)
2
(
)
2
( 




Under this unimodular transformation of the given right-handed spinor singlet, the
covariant derivative operator in turn is also characterized by the hypercharge R
Y although
with the same U(1) gauge field 
B as with the case for the left-handed spinor doublet.
(6.1)


 B
Y
Q
i
D R
R 


)
(
Using (5.2) and (5.3), we can also write this operator in its usually useful form,

4. (6.2)





 sin
cos
sin2
)
(
m
e
R
R
R A
Y
Q
i
Z
Y
Q
i
D 





With regards to parity operator, I stick to Baal’s and ‘tHooft’s notation so in here, I write
a right-handed spinor singlet as
(7.1)
  2
5
2 1
2
1


 

R
and the left-handed component as
(7.2)
  2
5
2 1
2
1


 

L
Note here that the given fifth Dirac-Gamma matrix 5
 is with a superscript not a
subscript as we shall note also that in the metric signature -2, we have
(7.3)
5
5

 

So in this way, we may write a combinatory between the right-handed and left-handed
component as
(7.4)
2
2
2
2
2
2 




 
 R
L
L
R
where  is an adjoint spinor defined by
(7.5)
0


 dagger

( T
dagger
)
( 
 
 )
We may think of the scalar doublet as having a given hypercharge, 1

Y , under its own
SU(2) X U(1) transformation,
(8.1)




 )
exp(
)
exp(






 Q
i
i q
So we write this scalar doublet’s covariant derivative operator in the following form

5. (8.2)







3
1
)
(
l
l
l W
Q
i
B
Q
i
D 


 
as distinguished from that of the spinor doublet’s by the particular hypercharge that goes
with the U(1) gauge field 
B .
To accommodate the transformation of  under the diagonal subgroup
(8.3)




 )
exp(
)
exp( 3 q
q Q
i
i 



we also switch to the alternative form of (8.2)
(8.4)
   
  












sin
1
cos
sin
cos
3
2
3
2
)
2
(
2
)
1
(
1
m
e
A
Q
i
Z
Q
i
W
W
Q
i
D











which we can write, again using (5.2) and (5.3).
Technically we think of (5.2) and (5.3) as simultaneous rotations of the U(1) gauge field
and the third component of the SU(2) gauge field into one massive field 
Z and one
massless electromagnetic field m
e
A under the Special Orthogonal Group SO(2) given for
the mixing angle alpha defined by (5.6)
(9)















 

















B
W
A
Z
m
e
)
3
(
cos
sin
sin
cos
Concerning Lagrangian (1), this is not yet the full electroweak theory under the SU(2) X
U(1) gauge group. This Lagrangian includes a left-handed electron L
2
 and its associated
left-handed electron neutrino L
1
 , which is rendered massless in the Yukawa coupling
terms. Together L
1
 and L
2
 are the two left-handed components of the left-handed
spinor doublet L
 . A right-handed electron R
2
 is also contained in (1) and with the
left-handed counterpart L
2
 is made massive by the Yukawa coupling.
Left-handed Spinor Doublet

6. (10.1)








 L
L
L
2
1



Note that each component in the doublet (10.1) is in itself a spinor applied with a parity
operator like that of (7.2). For the case of right-handed electron, it is a singlet spinor
given by (7.1). Then we can combine these two electrons via (7.4) to yield a single
electron that goes with its adjoint spinor.
What is also included in (1) is the Higgs field  , which is a scalar doublet
(10.2)




















2
1
0
0






with the constant part 0
 that can be thought as the non-zero value of the Higgs field in
the ground state. In this doublet, there are three Goldstone bosons: ]
[
Re 1
 , ]
[
Im 1
 and

 
]
[
Im 2 . The real part of 2
 is the Higgs boson
(10.3)
]
[
Re 2

 
That is, the Higgs boson is the real part of a complex scalar field 2
 whose imaginary
part  disappears by a gauge condition imposed on the Goldstone bosons of which this
imaginary part of 2
 is a part.
(10.4)
0
]
Im[
]
Im[
]
Re[ 2
1
1 

 


We still need to explore further on the Higgs potential )
(
V but for the present purposes
the following form of this potential may suffice.
(11.1)
 2
2
2
1
4
)
( 
 


V
This potential increases without bound as the absolute magnitude  of the scalar
doublet  increases to infinity. However, there is a set of values of this magnitude where

7. 0


d
V
d
and one of these values is where the potential is minimum
(11.2)
  0

n
i
m
V 
at
(11.3)

 
n
i
m
This minimal value of  represents the stable ground state or the true vacuum state and
the minima of the component scalars of the scalar doublet
(11.4)









2
1



revolve along the circle defined with a radius of  ,
(11.5)
2
2
2
2
1 

 

n
i
m
n
i
m
Each of the components in the doublet is complex so it shall be noted that the scalar
doublet has four actual components and we can pick one of these scalars to assume the
minimal values 
 thereby, enforcing on the remaining three the zero values on their
minima
(11.6)
2
1
1
1
1 

 i


2
2
1
2
2 

 i


The other value 0

 , at which 0


d
V
d
, where the potential assumes the value
(11.7)
 
4
0
4
2


 

V

8. represents a false (unstable) vacuum state. This is so since this value of the potential is
above the true vacuum which the Higgs potential must roll down to.
As a quick drill one may verify that
(11.8)
0
2 2
2
2
2

 


d
V
d
at the value in (11.3). So indeed, by simple minimal test in elementary calculus, the value
presented by (11.3) is a minimal value that satisfies (11.8).
So, say by convenience, we pick
(11.9)

 
)
(
1
2 n
i
m .
This forces the other components to assume the minimum values
(11.10)
0
)
(
2
2
)
(
2
1
)
(
1
1 

 n
i
m
n
i
m
n
i
m 


That is, one component has a non-zero minimum value, enforcing the zero value for the
minima of the other remaining components. These remaining components that have the
zero for their minima are the would-be Goldstone bosons, while we select the field with
the non-zero minimum as a field perturbed from the minimum.
(11.11)


 

1
2
As identified earlier in (10.3),  is the Higgs boson, while the three Goldstone bosons
necessarily follow from selecting which components have the zero minima.
(11.12)


 
 ]
[
Im 2
2
2
2
1
1
1
1
1 


 i



The Higgs boson is the only component field in the doublet that acquires mass in the
expansion of the given Higgs potential in terms of the component scalars.

9. (11.13)





 







 






 



2
1
2
2
2
2
1
2
2
2
2
2
1
2
2
4
2
2
2
2
2
1
4
1
4
1
)
(



















V
The mass term in this expansion comes with a single Higgs boson raised to the power of
two and from this, we single out the Higgs boson mass as
(11.14)


 2

m
The gauge condition (10.4) is then imposed to reduce expansion (11.13) into
(11.15)
3
2
4
2
2
2
4
1
2
1
)
( 





  

 m
V
In the reduction, we find that the Higgs boson theory following from the given gauge
condition contains both the cubic and quartic self-interactions with all the Goldstone
bosons vanished. One quick note is that the mass term in (11.15) comes with a right sign
but if we take the absolute magnitude  of the Higgs field as a field itself, then this
field’s own mass term is tachyonic, having a negative sign.
In the case for the Yukawa coupling terms, say under all diagonal subgroups, one term
transforms as
(12.1)
)
exp(
)
)
1
(
exp(
2
2 q
R
q
L
R
L
R
L
Y
i
Y
i 






 



or with the other term as
(12.2)
)
)
1
(
exp(
)
exp(
*
* 2
2 q
L
q
R
L
T
R
L
T
R
Y
i
Y
i 






 

From these we see the relation of the hypercharges, if we are to invoke invariance of
these terms under the transformations. That is, the invariance requires
(12.3)
1

 L
R Y
Y
Let us note that under diagonal subgroup (8.3) for the scalar doublet, the constant part 0

of the Higgs field is invariant

10. (13.1)
0
0
0 

 


and we have phase transformation for 1

(13.2)
1
1
1 )
2
exp( 


 q
i




while the second component is also invariant
(13.3)
2
2
2 

 


One thing to notice immediately in writing the Higgs potential in terms of the four actual
components
(14.1)
2
2
1
2
1
2
2
2
4
)
( 








 
 
i j
j
i
j
i
V 



is that this potential is symmetric under reflection of the actual component scalars. That
is,
(14.2)
)
(
)
( j
i
j
i V
V 
 

under reflection
(14.3)
j
i
j
i 
 

By selecting one of these components to assume a non-zero minimum (ground state) and
express such scalar component as a field perturbed from that ground state (see (11.11)),
the potential becomes non-symmetric under reflection in terms of the new field
component  , which becomes massive and is earlier identified as the Higgs boson. That
is, by the following reflection of 
(14.4)

 

the symmetry of the potential is broken

11. (14.5)
)
(
)
( 
 
 V
V
Such result is noticeable in (11.13).
The negative one value given to the hypercharge of the spinor doublet ( 1


L
Y ) tends
to decouple the left-handed electron neutrino from the electromagnetic field. This is
because
(15.1)
  









 L
L
L
Y
2
3
2
0



That leaves the left-handed electron the only entry in the column vector, while the left-
handed electron neutrino disappears.
In the case for the scalar doublet with the hypercharge 1

Y , only the Goldstone bosons
contained in the complex scalar 1
 can interact with the electromagnetic field, while
those remaining components in 2
 decouple
(15.2)
  m
e
m
e
A
A 



 









0
2
1 1
3
Also, the hypercharge given to the scalar doublet renders the electromagnetic field
massless as we can notice from the term
(15.3)
  m
e
m
e
A
A 


 









0
0
1 0
3
which shows decoupling of m
e
A from the non-zero ground state of the Higgs field. The
other )
1
(
)
2
( U
SU  gauge bosons pick up masses via the term
(15.4)








 Z
Z
Q
W
W
Q
D 2
2
2
)
(
)
(
2
2
2
0
cos
2
2
1 


 

from which we underline the W and Z masses respectively,

12. (15.5.1)

Q
MW

 2
and
(15.5.2)

cos
2
1 W
Z
M
M 
Here, the )
( 

W gauge bosons are constructed as complex gauge fields
(15.6)
 


 )
2
(
)
1
(
)
(
2
1
W
i
W
W 


To Lagrangian (1) we add the free Lagrangians of the W gauge bosons and of the U(1)
gauge field 
B , and keeping in mind that Lagrangian (1) does not yet include the
additional Fermions needed in the full SU(2) X U(1) Electro-weak theory.
For the weak W bosons we have the quite familiar form
(16.1)







3
1
)
(
)
(
4
1
4
1
l
l
l
W F
F
F
F
L 









The anti-symmetric tensor 

F

is a short for
(16.2)







 W
W
Q
W
W
F











 2
To first order in Q, the gauge fields transform as
(16.3)



 
 W
Q
W
W










 2
This prompts the anti-symmetric tensor (16.2) to take the following transformation (also
to first order in Q),
(16.4)





  F
Q
F
F







 2

13. So, the corresponding transformation of Lagrangian (16.1) involves
(16.5)
)
(
)
2
(
2 










  F
F
Q
F
F
F
F














Again, to first order in Q.
Effectively, Lagrangian (16.1) is invariant under the gauge transformation (16.3)
because the second major term on the right-hand-side (rhs) of (16.5) vanishes by cyclic
permutation,
(16.6)
0
)
(
)
( 




 






 
 F
F
F
F






as then noted that
(16.7)
0

 



F
F


In some convenient notations, we shall write each component of the anti-symmetric
tensor 

F

in the following:
(16.8)
   
   



















 Z
W
Z
W
e
i
Z
W
Z
W
e
i
W
D
W
D
W
D
W
D
F m
e
m
e
m
e
m
e
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
1
(
cot
2
cot
2
2
1
2
1




















(16.9)
   











  Z
W
Z
W
e
i
W
D
W
D
F
i
F m
e
m
e )
(
)
(
)
(
)
(
)
(
)
(
)
2
(
)
1
(
cot
2
2 




 




(16.10)
m
e
m
e
A
e
i
D 

 


)
(
In these expressions, we have to note the SO(2) rotations in (9) and the complex
construction (15.6).
The remaining third component is written as

14. (16.11)
   
)
(
)
(
)
(
)
(
)
3
(
sin
2
cos
sin 









 













 W
W
W
W
e
i
Z
Z
F
F m
e
with
(16.12)
m
e
m
e
m
e
A
A
F 




 



For the U(1) gauge field 
B , we have for its free Lagrangian
(17.1)
 2
4
1



 B
B
LB 




and in combination with (16.11), upon noting (9) and (15.6), we can conveniently
construct
(17.2)
       


















W
Z
e
i
W
e
i
Z
W
F
e
i
F
F
L m
e
m
e
B
cot
4
sin
2
4
1
4
1
4
4
1
4
1 2
2
2
2
2
)
3
(













where
(17.3)
)
(
)
(
)
(
)
( 




 




 W
W
W
W
W
and
(17.4)





 Z
Z
Z 



Then for the combined Lagrangian of (16.1) and (17.1) we have,
(17.5)
    
   2
2
2
2
)
)
(
)
)
(
)
(
)
(
sin
2
4
1
cot
4
4
1
4
4
1
4
1























 W
e
i
W
Z
e
i
Z
W
F
e
i
F
N
M
N
M
L
L m
e
m
e
B
W















 



with

15. (17.6)
   
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
cot
2 











 










  W
Z
W
Z
e
i
W
D
W
D
N
M m
e
m
e
and
(17.7)
   
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
cot
2 











 










  W
Z
W
Z
e
i
W
D
W
D
N
M m
e
m
e
So for our SU(2)XU(1) theory to be partially accomplished, we add (17.5) to (1) though
as earlier stated, Lagrangian (1) still lacks the additional Fermions needed in the complete
SU(2)XU(1) electroweak theory of the Standard Model (SM). We can then outline the
partial Higgs Boson theory in (1) as
(18.1)
 
3
2
4
2
2
2
)
(
)
(
2
2
2
4
1
cos
2
1
)
(
2
1
)
(
)
(
2
1










 
















 

Z
Z
W
W
Q
x
J
m
L T
(18.2)
2
2
2
)
(
)
(
2
cos
2
1
2
)
( 


 


 y
Z
Z
W
W
Q
x
JT 











 

The other piece of interest that we can also outline from (1) is an initial theory involving
the electron
(18.3)
 
)
(
1
2
)
(
2
1
1
2
2
1
2
2
2
2
2
2
2
2
2
2
2
)
sin
2
(
)
cos
sin
2
(
tan
)
(






















































W
W
e
Z
e
Z
e
y
A
e
y
i
L
L
L
L
L
L
L
m
e
Immediately, we can spot out the Quantum Electrodynamics (QED) pieces from that
(18.4)
m
e
D
E
Q A
e
y
i
L 












 2
2
2
2
2
2
2 )
( 



while excluding the free Lagrangian for the electromagnetic field m
e
A . Such Lagrangian
can be extracted out from (17.5), the term that involves the electromagnetic tensor m
e
F 

contracted with itself.

16. 
2 Basic Practice Calculations Involving Scalar Fields
Let us do some basic practice calculations here. Conveniently, we take the scalar field as
subject for the said calculations in relevance to quantum field theories involving vacuum-
to-vacuum matrices whose forms can be given by
(19.1)
⟨0|𝑈1(𝑇)|0⟩
The basic quantum field interpretation of these matrices is that they give probability
amplitudes that the particles (initially) in the vacuum state |0⟩ at an initial time (say,
𝑡 = 0) will still be (found) in the vacuum state at a later time 𝑇 > 0.
The time-evolution operator (teo) here 𝑈1(𝑇) comes with a system Hamiltonian 𝐻[ 𝐽(𝑡) ]
containing time-dependent source 𝐽(𝑡). (System Hamiltonian as obtained from the system
Lagrangian through Legendre transformation. We shall present the basic review details of
this transformation in the concluding portions of this draft. )
(19.2)
𝑈1(𝑇) = 𝑈1(𝑇, 0) = 𝑒𝑥𝑝 (−
𝑖
ℏ
∫ 𝑑𝑡 𝐻(𝑡)
𝑇
0
)
As an emphasis, we are starting with a system (Lagrangian) that does not contain yet self-
interaction terms. These terms are generated with the application of potential operator
𝑉(𝜑
̂). (Note that (19.2) is an operator although we haven’t put a hat on the Hamiltonian
operator, 𝐻(𝑡). We will only put a hat on operators in situations where we are compelled
to do so in order to avoid confusion. )
In the absence of the said self-interaction terms, the (19.1) matrices can be evaluated via
Path Integrations (PI’s) although the evaluation of these matrices can be quite easily
facilitated by putting them in their factored form
(19.3)
⟨0|𝑈1(𝑇)|0⟩ = 𝐶 [ 𝐽 = 0 ] 𝑒𝑖 𝑆𝑐 / ℏ
where 𝐶[𝐽 = 0] is considered as a constant factor that can be evaluated directly
(19.4)
𝐶[𝐽 = 0] = ⟨0|𝑒𝑥𝑝 (−
𝑖
ℏ
𝑇𝐻[𝐽 = 0]) |0⟩ = 𝑒𝑥𝑝 (−
𝑖
ℏ
𝑇 𝐸0
0
)
As can be seen in the result this constant matrix comes with the ground state energy, 𝐸0
0
.
(We shall also give review details of this result in the later parts of this document.)

17. In the later portions of this draft, we shall also provide the necessary details of expressing
the classical action 𝑆𝑐 involved in (19.3) as a functional of the sources as this action is to
be given by
(19.5)
𝑆𝑐 =
1
2
1
(2𝜋)2
∫ 𝑑4
𝑥 𝑑4
𝑦 𝐽(𝑥)𝐺(𝑥 − 𝑦)𝐽(𝑦)
The classical action given by (19.5) as evaluated in terms of the sources 𝐽(𝑥), 𝐽(𝑦) and
the Green’s function 𝐺(𝑥 − 𝑦) which serves here as the propagator. This form follows
from the fact that the solution to the classical equation of motion in the presence of
sources can be expressed using the Green’s function and the source
(19.6)
𝜑(𝐽, 𝑥) =
1
(2𝜋)2
∫ 𝑑4
𝑦 𝐺(𝑥 − 𝑦)(− 𝐽(𝑦))
(19.7)
𝐺(𝑥 − 𝑦) =
1
(2𝜋)2
∫ 𝑑4
𝑘
𝑒𝑖𝑘𝜎(𝑥𝜎− 𝑦𝜎)
−𝑘𝜇𝑘𝜇 + 𝑀2 + 𝑖𝜖
(We shall also give basic calculation details of this in the later continuing drafts.)
When the system (with its given Lagrangian) does indeed involve self-interaction terms
the exponential potential operator
(20.1)
𝑒𝑥𝑝 (−
𝑖
ℏ
∫ 𝑑4
𝑦 𝑉(𝜑
̂)
𝐵
𝐴
)
can be inserted into the vacuum-to-vacuum matrix thus writing this matrix as
(20.2)
⟨0|𝑈1(𝑇)𝑒𝑥𝑝 (−
𝑖
ℏ
∫ 𝑑4
𝑦 𝑉(𝜑
̂)
𝐵
𝐴
) |0⟩
(It is important to show the details of arriving at (20.2) but we do this in the later portions
of this document.)
We can easily evaluate (20.2) using Path Integral so as to express this matrix in terms of
the derivative operator with respect to the given source

18. (20.3)
⟨0|𝑈1(𝑇)𝑒𝑥𝑝 (−
𝑖
ℏ
∫ 𝑑4
𝑦 𝑉(𝜑
̂)
𝐵
𝐴
) |0⟩ = 𝑒𝑥𝑝 (−
𝑖
ℏ
∫ 𝑑4
𝑦 𝑉 (𝑖ℏ
𝛿
𝛿𝐽(𝑦)
)
𝐵
𝐴
) ⟨0|𝑈1(𝑇)|0⟩
Taking note of (19.3), we can employ the Taylor/Maclaurin expansion
(20.4)
𝑒𝑥𝑝 (−
𝑖
ℏ
∫ 𝑑4
𝑦 𝑉 (𝑖ℏ
𝛿
𝛿𝐽(𝑦)
)
𝐵
𝐴
) 𝑒𝑖 𝑆𝑐 / ℏ
= 1 + ∑
1
𝑚!
(
𝑖
ℏ
)
𝑚
𝑆𝐶
𝑚
[𝐽]
∞
𝑚=1
+ ∑
(−1)𝑛
𝑛!
(
𝑖
ℏ
)
𝑛
(∫ 𝑑4
𝑦 𝑉[𝜑(𝐽)]
𝐵
𝐴
)
𝑛
∞
𝑛=1
+ ∑
(−1)𝑛
𝑛!
(
𝑖
ℏ
)
𝑛
∞
𝑛=1
∑
1
𝑚!
∞
𝑚=1
(∫ 𝑑4
𝑦 𝑉 [𝑖ℏ
𝛿
𝛿𝐽(𝑦)
]
𝐵
𝐴
)
𝑛
(
𝑖𝑆𝑐[ 𝐽 ]
ℏ
)
𝑚
In my personal (convenient) notation, I write the integration in 𝑆𝑐[ 𝐽 ] as a bracket
(20.5)
𝑆𝑐[ 𝐽 ] = 〈𝐽𝑥𝐺𝑥𝑧𝐽𝑧〉 =
1
2
1
(2𝜋)2
∬ 𝑑4
𝑥 𝑑4
𝑧 𝐽(𝑥)𝐺(𝑥 − 𝑧)𝐽(𝑧)
Here, the bracket 〈𝐽𝑥𝐺𝑥𝑧𝐽𝑧〉 means double four-spacetime integrations involving two
sources 𝐽(𝑥) and 𝐽(𝑧), and a propagator 𝐺(𝑥 − 𝑧) in coordinate spacetime.
As a specific case in this draft let us take the cubic self-interactions
(21.1)
𝑉[𝜑(𝑥, 𝐽)] =
1
3!
𝑔(3)𝜑3(𝑥, 𝐽)
In my notation, I write
(21.2)
𝑆𝐶
𝑛[ 𝐽 ] = ∏〈 𝐽𝑥𝐺𝑥𝑧𝐽𝑧 〉𝑖
𝑛−1
𝑖=0
where 𝑖 is specified on both x and z. Meaning,

19. (21.3)
〈 𝐽𝑥𝐺𝑥𝑧𝐽𝑧 〉𝑖 = 〈𝐽(𝑥(𝑖)
) 𝐺(𝑥(𝑖)
, 𝑧(𝑖)
) 𝐽(𝑧(𝑖)
) 〉
Note here that 𝑆𝐶
𝑛
involves 2𝑛 J’s. For 𝜑3(𝑥, 𝐽) I shall also write this as
(21.4)
𝜑3(𝑥, 𝐽) = ∏〈 𝐺𝑥𝑧𝐽𝑧 〉𝑗
2
𝑗=0
where 𝑗 is specified on z only.
When there is a need to take the derivative with respect to the source I shall specify the
replacement
(21.5)
𝜑(𝑥) → 𝑖ℏ
𝛿
𝛿𝐽(𝑥)
Now, for (21.1) we write this as
(21.6)
𝑉[𝜑(𝑥, 𝐽)] =
1
3!
𝑔(3) ∏〈 𝐺𝑥𝑧𝐽𝑧 〉𝑗
2
𝑗=0
and so it follows that
(21.7)
∫ 𝑑4
𝑥 𝑉[𝜑(𝑥, 𝐽)] =
1
3!
𝑔(3) ∫ 𝑑4
𝑥 ∏〈 𝐺𝑥𝑧𝐽𝑧 〉𝑗
2
𝑗=0
𝐵
𝐴
𝐵
𝐴
where 𝑗 is specified on z only. Then raising (21.7) to the power of 𝑛
(21.8)
(∫ 𝑑4
𝑥 𝑉[𝜑(𝑥, 𝐽)]
𝐵
𝐴
)
𝑛
=
1
(3!)𝑛
𝑔(3)
𝑛
∫ ∏ 𝑑4
𝑥(𝑘)
𝑛−1
𝑘=0
𝐵
𝐴
∏〈𝐺𝑥(𝑘)𝑧𝐽𝑧〉𝑗
3𝑛−1
𝑗=0
where 𝑘 is on all x, while 𝑗 on all z.

20. In situations where replacement (21.5) is indicated, contrasting (21.7) we write
(21.9)
∫ 𝑑4
𝑦 𝑉 (𝑖ℏ
𝛿
𝛿𝐽(𝑦)
)
𝐵
𝐴
=
1
3!
𝑔(3) (∏ 𝑖ℏ
𝛿
𝛿𝐽(𝑣(𝑗))
3−1
𝑗=1
) ∫ 𝑑4
𝑦 𝑖ℏ
𝛿
𝛿𝐽(𝑦)
𝐵
𝐴
and raised to the power of n, this becomes
(21.10)
(∫ 𝑑4
𝑦 𝑉 (𝑖ℏ
𝛿
𝛿𝐽(𝑦)
)
𝐵
𝐴
)
𝑛
=
1
(3!)𝑛
𝑔(3)
𝑛
( ∏ 𝑖ℏ
𝛿
𝛿𝐽(𝑣(𝑗))
(3−1)𝑛
𝑗=1
) ∫ (∏ 𝑑4
𝑦(𝑖)
𝑛−1
𝑖=0
)
𝐵
𝐴
∏ 𝑖ℏ
𝛿
𝛿𝐽(𝑦(𝑘))
𝑛−1
𝑘 = 0
where
(21.11)
∏ 𝑖ℏ
𝛿
𝛿𝐽(𝑦(𝑘))
𝑛−1
𝑘 = 0
indicates differentiation 𝑛 times, while
(21.12)
∏ 𝑖ℏ
𝛿
𝛿𝐽(𝑣(𝑗))
(3−1)𝑛
𝑗=1
indicates differentiation (3 − 1)𝑛 times. So that combining (21.11) and (21.12) in (21.10)
(21.13)
( ∏ 𝑖ℏ
𝛿
𝛿𝐽(𝑣(𝑗))
(3−1)𝑛
𝑗=1
) ⋯ ∏ 𝑖ℏ
𝛿
𝛿𝐽(𝑦(𝑘))
𝑛−1
𝑘 = 0
indicates differentiation (3 − 1)𝑛 + 𝑛 = 3𝑛 times.

21. Given (20.3), then we can proceed to evaluate this matrix upon the setting of all sources
to zero. That is,
(21.14)
⟨0|𝑈1(𝑇)𝑒𝑥𝑝 (−
𝑖
ℏ
∫ 𝑑4
𝑦 𝑉(𝜑
̂)
𝐵
𝐴
) |0⟩|
𝐽=0
Upon the setting of 𝐽 = 0, the terms in (20.4), where the J’s are explicitly expressed
vanish. Such vanishing terms involve
(21.15)
𝑆𝑐[ 𝐽 ]|𝐽 = 0 = 0 → 𝑆𝐶
𝑛[ 𝐽 ]|𝐽 = 0 = 0
𝑉[𝜑(𝑥, 𝐽)]|𝐽 = 0 = 0
∫ 𝑑4
𝑥 𝑉[𝜑(𝑥, 𝐽)]
𝐵
𝐴
|
𝐽 = 0
= 0
(∫ 𝑑4
𝑥 𝑉[𝜑(𝑥, 𝐽)]
𝐵
𝐴
)
𝑛
|
𝐽 = 0
= 0
In an expression where the replacement (21.5) is involved such as an operation on the
classical action in the first power, we have the vanishing result
(21.16)
∫ 𝑑4
𝑦 𝑉 (𝑖ℏ
𝛿
𝛿𝐽(𝑦)
)
𝑖
ℏ
𝑆𝑐[ 𝐽 ] = −
1
3!
(𝑖ℏ)2
𝑔(3)
𝛿
𝛿𝐽(𝑣")
∫ 𝑑4
𝑦 𝐺〈𝑣′,𝑦〉 = 0
𝐵
𝐴
𝐵
𝐴
where in notation we take note of
(21.17)
𝐺〈𝑣′,𝑦〉 =
1
2
1
(2𝜋)2
(𝐺(𝑣′
− 𝑦) + 𝐺(𝑦 − 𝑣′
))
It follows that

22. (21.18)
(∫ 𝑑4
𝑦 𝑉 (𝑖ℏ
𝛿
𝛿𝐽(𝑦)
)
𝐵
𝐴
)
𝑛
𝑖
ℏ
𝑆𝑐[ 𝐽 ] = 0
The terms involving those integral powers of m and n, where ∀3𝑛 > 2𝑚 vanish also.
That is,
(21.19)
(∫ 𝑑4
𝑦 𝑉 (𝑖ℏ
𝛿
𝛿𝐽(𝑦)
)
𝐵
𝐴
)
𝑛
(
𝑖
ℏ
𝑆𝑐[ 𝐽 ])
𝑚
= 0
This is so since in here (for all the terms at ∀3𝑛 > 2𝑚), the indicated differentiation is
3𝑛 times on 𝑆𝐶
𝑚[ 𝐽 ] that involves 2𝑚 J’s or sources. Meanwhile, those terms at ∀3𝑛 <
2𝑚 yields
(21.20)
(∫ 𝑑4
𝑦 𝑉 (𝑖ℏ
𝛿
𝛿𝐽(𝑦)
)
𝐵
𝐴
)
𝑛
(
𝑖
ℏ
𝑆𝑐[ 𝐽 ])
𝑚
|
𝐽 ≠ 0
≠ 0
So in this instance, the differentiation does not lead to a vanishing result. However, such
terms will ultimately vanish upon the setting of 𝐽 = 0 because they still contain sources
at the end of the indicated differentiations. Thus, these terms are not at all relevant in
(21.14).
Proceeding, the only relevant terms in (21.14) are those in the integral powers of n and m
that satisfy 3𝑛 = 2𝑚. The end results of differentiations here no longer involve J’s so
these terms do not vanish when these sources are set to zero.
(21.21)
(∫ 𝑑4
𝑦 𝑉 (𝑖ℏ
𝛿
𝛿𝐽(𝑦)
)
𝐵
𝐴
)
𝑛
(
𝑖
ℏ
𝑆𝑐[ 𝐽 ])
𝑚
≠ 0
(Given for ∀3𝑛 = 2𝑚, where end results of differentiations are already independent of
J’s.)
The term of lowest order here that satisfies 3𝑛 = 2𝑚 are those in the powers of 𝑚 = 3
and 𝑛 = 2.

23. (21.22)
𝑚 = 3, 𝑛 = 2:
(∫ 𝑑4
𝑦 𝑉 (𝑖ℏ
𝛿
𝛿𝐽(𝑦)
)
𝐵
𝐴
)
2
(
𝑖
ℏ
𝑆𝑐[ 𝐽 ])
3
=
1
(3!)2
𝑔(3)
2
(𝑖ℏ)5
∫ (∏ 𝑑4
𝑦(𝑖)
1
𝑖=0
) (∏
𝛿
𝛿𝐽(𝑦(𝑗))
5
𝑗=0
)
𝐵
𝐴
(∏〈 𝐽𝑥𝐺𝑥𝑧𝐽𝑧 〉𝑖
2
𝑖=0
)
In a much later while we think of (as arbitrarily selected) 𝑦(0)
= 𝑦 and 𝑦′ as the
integration variables.
The greater task now to follow from (21.22) is to carry out the indicated differentiations
(21.23)
(∏
𝛿
𝛿𝐽(𝑦(𝑗))
5
𝑗=0
) (∏〈 𝐽𝑥𝐺𝑥𝑧𝐽𝑧 〉𝑖
2
𝑖=0
)
For convenience we write each differentiation in short hand as
(22.1)
𝛿𝐽(𝑦) =
𝛿
𝛿𝐽(𝑦)
and this acts on (20.5) via the functional derivative defined in 3 + 1 dimensional
spacetime
(22.2)
𝛿4(𝑥 − 𝑦) =
𝛿𝐽(𝑥)
𝛿𝐽(𝑦)
To first order in the differentiation we have
(22.3)
𝛿𝐽(𝑦)〈𝐽𝑥𝐺𝑥𝑧𝐽𝑧〉 = 〈𝐽𝑥𝐺𝑥𝑦〉 + 〈𝐺𝑦𝑧𝐽𝑧〉
=
1
2
1
(2𝜋)2
(∫ 𝑑4
𝑥 𝐽(𝑥)𝐺(𝑥 − 𝑦) + ∫ 𝑑4
𝑧 𝐺(𝑦 − 𝑧)𝐽(𝑧) )

24. The first order differentiation generates two terms and if we are to invoke symmetry of
the Green’s function under integration and upon the setting of 𝑥 = 𝑧, we can write (22.3)
in a more concise way
(22.4)
𝛿𝐽(𝑦)〈𝐽𝑥𝐺𝑥𝑧𝐽𝑧〉|
𝑥 = 𝑧
= (2)〈𝐽𝑥𝐺𝑥𝑦〉
In this notation we are to be reminded that we have two terms before the setting of z
equals x and invoking symmetry of the Green’s function and this is denoted by the factor
2 inside the parenthesis.
So second order differentiation yields
(22.5)
𝛿𝐽(𝑦′)𝛿𝐽(𝑦)〈𝐽𝑥𝐺𝑥𝑧𝐽𝑧〉 = 𝛿𝐽(𝑦)
2
〈𝐽𝑥𝐺𝑥𝑧𝐽𝑧〉 = (2)𝐺𝑦′𝑦 = (2)
𝐺(𝑦′
− 𝑦)
2(2𝜋)2
Again, (2) specifies that there were two terms involved.
Proceeding let us take the first order differentiation, 𝛿𝐽(𝑦)𝑆𝐶
3[ 𝐽 ]. This we shall put as
(22.6)
𝐴(𝐽5) = 𝛿𝐽(𝑦)𝑆𝐶
3[ 𝐽 ]
= 〈𝐽𝑥𝐺𝑥𝑧𝐽𝑧〉2 〈𝐽𝑥𝐺𝑥𝑧𝐽𝑧〉1 𝛿𝐽(𝑦)〈𝐽𝑥𝐺𝑥𝑧𝐽𝑧〉
+ 〈𝐽𝑥𝐺𝑥𝑧𝐽𝑧〉2 (𝛿𝐽(𝑦)〈𝐽𝑥𝐺𝑥𝑧𝐽𝑧〉1)〈𝐽𝑥𝐺𝑥𝑧𝐽𝑧〉
+ (𝛿𝐽(𝑦)〈𝐽𝑥𝐺𝑥𝑧𝐽𝑧〉2)〈𝐽𝑥𝐺𝑥𝑧𝐽𝑧〉1〈𝐽𝑥𝐺𝑥𝑧𝐽𝑧〉
This generates three major terms and each major term in turn has two terms. We are to
think of these major terms as identical. So further we write this concisely in the form
(22.7)
𝐴(𝐽5) = 𝛿𝐽(𝑦)𝑆𝐶
3[ 𝐽 ] = (2)(3)〈𝐽𝑥𝐺𝑥𝑧𝐽𝑧〉2 〈𝐽𝑥𝐺𝑥𝑧𝐽𝑧〉1 〈𝐽𝑥𝐺𝑥𝑦〉
We note in this that (2)(3) specifies that there were (2)(3) = 6 terms involved initially,
where it is noted as identical
(22.8)
𝛿𝐽(𝑦)〈𝐽𝑥𝐺𝑥𝑧𝐽𝑧〉𝑖 = (2)〈𝐽𝑥𝐺𝑥𝑦〉𝑖 = 𝛿𝐽(𝑦)〈𝐽𝑥𝐺𝑥𝑧𝐽𝑧〉𝑗 = (2)〈𝐽𝑥𝐺𝑥𝑦〉𝑗
upon the setting 𝑥 = 𝑧 after the indicated differentiation.
Next, we have the second order differentiation

25. (22.9)
𝐴(𝐽4) = 𝛿𝐽(𝑦′)𝐴(𝐽5) = 𝛿𝐽(𝑦)
2
𝑆𝐶
3[ 𝐽 ] = 𝛿𝐽(𝑦′)𝛿𝐽(𝑦)𝑆𝐶
3[ 𝐽 ]
= (3)(2)〈𝐽𝑥𝐺𝑥𝑧𝐽𝑧〉2 〈𝐽𝑥𝐺𝑥𝑧𝐽𝑧〉1𝐺𝑦′𝑦
+ (3)(2)(2)(2)〈𝐽𝑥𝐺𝑥𝑧𝐽𝑧〉2 〈𝐽𝑥𝐺𝑥𝑦′〉1〈𝐽𝑥𝐺𝑥𝑦〉
In this second partial differentiation, there are (3)(2) + (3)(2)(2)(2) = 30 terms.
Proceeding, we write the third order partial differentiation as
(22.10)
𝐴(𝐽3) = 𝛿𝐽(𝑦′′)𝐴(𝐽4) = 𝛿𝐽(𝑦)
3
𝑆𝐶
3[ 𝐽 ] = (∏ 𝛿𝐽(𝑦(𝑞))
2
𝑞=0
) 𝑆𝐶
3[ 𝐽 ]
= 𝛿𝐽(𝑦′′)𝐴(𝐽4)|
1
+ 𝛿𝐽(𝑦′′)𝐴(𝐽4)|
2
It is only for our convenience that we express this third order partial differentiation in
terms of two major terms.
Here we take note of the result
(22.11.1)
𝛿𝐽(𝑦′′)𝐴(𝐽4)|
1
= (3)(2)3 〈𝐽𝑥𝐺𝑥𝑧𝐽𝑧〉2〈𝐽𝑥𝐺𝑥𝑦′′〉1𝐺𝑦′𝑦
while the other piece is longer
(22.11.2)
𝛿𝐽(𝑦′′)𝐴(𝐽4)|
2
= (3)(2)3 〈𝐽𝑥𝐺𝑥𝑧𝐽𝑧〉2〈𝐽𝑥𝐺𝑥𝑦′〉1𝐺𝑦′′𝑦
+ (3)(2)3 〈𝐽𝑥𝐺𝑥𝑧𝐽𝑧〉2 𝐺𝑦′′𝑦′ 〈𝐽𝑥𝐺𝑥𝑦〉
+ (3)(2)4 〈𝐽𝑥𝐺𝑥𝑦′′〉2 〈𝐽𝑥𝐺𝑥𝑦′〉1 〈𝐽𝑥𝐺𝑥𝑦〉
and by straightforward inspection we can ascertain that this resulting differentiation has
(22.11.3)
𝐴(𝐽3) = 𝛿𝐽(𝑦′′)𝐴(𝐽4) = 𝐴(𝐽3)|1 + 𝐴(𝐽3)|2 = 𝛿𝐽(𝑦)
3
𝑆𝐶
3[ 𝐽 ]
= (3)(2)3
+ (3)(2)3
+ (3)(2)3
+ (3)(2)3(2) = 120 𝑡𝑒𝑟𝑚𝑠
From this we go to the fourth order differentiation, which we shall write as

26. (22.12)
𝐴(𝐽2) = 𝛿𝐽(𝑦(3))𝐴(𝐽3) = 𝛿𝐽(𝑦(3))𝐴(𝐽3)|1 + 𝛿𝐽(𝑦(3))𝐴(𝐽3)|2
The first part of which we write with two major terms in turn
(22.13.1)
𝛿𝐽(𝑦(3))𝐴(𝐽3)|1 = 𝛿𝐽(𝑦(3))𝐴(𝐽3)|11 + 𝛿𝐽(𝑦(3))𝐴(𝐽3)|12
Carrying out the indicated differentiation we have for
(22.13.2)
𝛿𝐽(𝑦(3))𝐴(𝐽3)|11 = (3)(2)3 〈𝐽𝑥𝐺𝑥𝑧〉2 𝐺𝑦(3)𝑦′′𝐺𝑦′𝑦
which has
𝛿𝐽(𝑦(3))𝐴(𝐽3)|11 = (3)(2)3
= 24𝑡𝑒𝑟𝑚𝑠
The other part of (22.13.1) on the other hand we write
(22.13.3)
𝛿𝐽(𝑦(3))𝐴(𝐽3)|12 = (3)(2)4 〈𝐽𝑥𝐺𝑥𝑦(3)〉2 〈𝐽𝑥𝐺𝑥𝑦′′〉1𝐺𝑦′𝑦
and this has twice the number of terms of (22.13.2)
(22.13.4)
𝛿𝐽(𝑦(3))𝐴(𝐽3)|12 = (3)(2)4
𝑡𝑒𝑟𝑚𝑠
It follows that
(22.13.5)
𝛿𝐽(𝑦(3))𝐴(𝐽3)|1 = 24 + 48 = 72𝑡𝑒𝑟𝑚𝑠
Continuing on, we have from (22.13.2)
(22.13.6)
(∏ 𝛿𝐽(𝑦(𝑞))
5
𝑞=3
) 𝐴(𝐽3)|11 = (3)(2)3
(2) 𝐺𝑦(5)𝑦(4)𝐺𝑦(3)𝑦′′𝐺𝑦′𝑦

27. (∏ 𝛿𝐽(𝑦(𝑞))
5
𝑞=3
) 𝐴(𝐽3)|11 = (3)(2)3(2) = 48 𝑡𝑒𝑟𝑚𝑠
while from (22.13.3) we have
(22.13.7)
(∏ 𝛿𝐽(𝑦(𝑞))
5
𝑞=3
) 𝐴(𝐽3)|12
= (3)(2)4
𝐺𝑦(5)𝑦(3)𝐺𝑦(4)𝑦′′𝐺𝑦′𝑦 + (3)(2)4
𝐺𝑦(5)𝑦′′𝐺𝑦(4)𝑦(3)𝐺𝑦′𝑦
(∏ 𝛿𝐽(𝑦(𝑞))
5
𝑞=3
) 𝐴(𝐽3)|12 = (3)(2)4
+ (3)(2)4
= 96 𝑡𝑒𝑟𝑚𝑠
and we shall also proceed from (22.13.1) to obtain
(22.13.8)
(∏ 𝛿𝐽(𝑦(𝑞))
5
𝑞=3
) 𝐴(𝐽3)|1 = (∏ 𝛿𝐽(𝑦(𝑞))
5
𝑞=3
) 𝐴(𝐽3)|11 + (∏ 𝛿𝐽(𝑦(𝑞))
5
𝑞=3
) 𝐴(𝐽3)|12
into which we substitute (22.13.6) and (22.13.7).
For the other major part in (22.10), we proceed with (22.11.2) to write
(22.14)
𝐴(𝐽3)|2 = 𝛿𝐽(𝑦′′)𝐴(𝐽4)|
2
= 𝐴(𝐽3)|21 + 𝐴(𝐽3)|22 + 𝐴(𝐽3)|23
= (3)(2)3 〈𝐽𝑥𝐺𝑥𝑧𝐽𝑧〉2〈𝐽𝑥𝐺𝑥𝑦′〉1𝐺𝑦′′𝑦
+ (3)(2)3 〈𝐽𝑥𝐺𝑥𝑧𝐽𝑧〉2 𝐺𝑦′′𝑦′ 〈𝐽𝑥𝐺𝑥𝑦〉
+ (3)(2)4 〈𝐽𝑥𝐺𝑥𝑦′′〉2 〈𝐽𝑥𝐺𝑥𝑦′〉1 〈𝐽𝑥𝐺𝑥𝑦〉
From here we go directly to what is ultimately needed to finish the job and so we get

28. (22.14.1)
(∏ 𝛿𝐽(𝑦(𝑞))
5
𝑞=3
) 𝐴(𝐽3)|21
= (3)(2)4
𝐺𝑦(5)𝑦(4)𝐺𝑦(3)𝑦′𝐺𝑦′′𝑦 + (3)(2)4
𝐺𝑦(5)𝑦(3)𝐺𝑦(4)𝑦′𝐺𝑦′′𝑦
+ (3)(2)4
𝐺𝑦(5)𝑦′𝐺𝑦(4)𝑦(3)𝐺𝑦′′𝑦
along with
(22.14.2)
(∏ 𝛿𝐽(𝑦(𝑞))
5
𝑞=3
) 𝐴(𝐽3)|22
= (3)(2)4
𝐺𝑦(5)𝑦(4)𝐺𝑦(3)𝑦𝐺𝑦′′𝑦′ + (3)(2)4
𝐺𝑦(5)𝑦(3)𝐺𝑦(4)𝑦𝐺𝑦′′𝑦′
+ (3)(2)4
𝐺𝑦(5)𝑦𝐺𝑦(4)𝑦(3)𝐺𝑦′′𝑦′
and
(22.14.3)
(∏ 𝛿𝐽(𝑦(𝑞))
5
𝑞=3
) 𝐴(𝐽3)|23
= (∏ 𝛿𝐽(𝑦(𝑞))
5
𝑞=3
) 𝐴(𝐽3)|231 + (∏ 𝛿𝐽(𝑦(𝑞))
5
𝑞=3
) 𝐴(𝐽3)|232
+ (∏ 𝛿𝐽(𝑦(𝑞))
5
𝑞=3
) 𝐴(𝐽3)|233
Into (22.14.3), we have the following results to substitute
(22.14.3.1)
(∏ 𝛿𝐽(𝑦(𝑞))
5
𝑞=3
) 𝐴(𝐽3)|231
= (3)(2)4
𝐺𝑦(5)𝑦′′𝐺𝑦(4)𝑦′𝐺𝑦(3)𝑦 + (3)(2)4
𝐺𝑦(5)𝑦′𝐺𝑦(4)𝑦′′𝐺𝑦(3)𝑦

29. (22.14.3.2)
(∏ 𝛿𝐽(𝑦(𝑞))
5
𝑞=3
) 𝐴(𝐽3)|232
= (3)(2)4
𝐺𝑦(5)𝑦′′𝐺𝑦(4)𝑦𝐺𝑦(3)𝑦′ + (3)(2)4
𝐺𝑦(5)𝑦𝐺𝑦(4)𝑦′′𝐺𝑦(3)𝑦′
and
(22.14.3.3)
(∏ 𝛿𝐽(𝑦(𝑞))
5
𝑞=3
) 𝐴(𝐽3)|233
= (3)(2)4
𝐺𝑦(5)𝑦′𝐺𝑦(4)𝑦𝐺𝑦(3)𝑦′′ + (3)(2)4
𝐺𝑦(5)𝑦𝐺𝑦(4)𝑦′𝐺𝑦(3)𝑦′′
Given all these results, we can now write the ultimately required order of differentiation
of (22.14) in the following form
(22.14.4)
(∏ 𝛿𝐽(𝑦(𝑞))
5
𝑞=3
) 𝐴(𝐽3)|
2
= (∏ 𝛿𝐽(𝑦(𝑞))
5
𝑞=3
) 𝐴(𝐽3)|21 + (∏ 𝛿𝐽(𝑦(𝑞))
5
𝑞=3
) 𝐴(𝐽3)|22
+ (∏ 𝛿𝐽(𝑦(𝑞))
5
𝑞=3
) 𝐴(𝐽3)|23
To go on further let us take (22.10) and write its ultimate higher order differentiation as
(22.14.5)
(∏ 𝛿𝐽(𝑦(𝑞))
5
𝑞=0
) 𝑆𝐶
3[ 𝐽 ] = (∏ 𝛿𝐽(𝑦(𝑞))
5
𝑞=3
) 𝐴(𝐽3)
= (∏ 𝛿𝐽(𝑦(𝑞))
5
𝑞=3
) 𝐴(𝐽3)|1 + (∏ 𝛿𝐽(𝑦(𝑞))
5
𝑞=3
) 𝐴(𝐽3)|2

30. (22.14.6)
(∏ 𝛿𝐽(𝑦(𝑞))
5
𝑞=3
) 𝐴(𝐽3)|1
= (3)(2)4
𝐺𝑦(5)𝑦(4)𝐺𝑦(3)𝑦′′𝐺𝑦′𝑦 + (3)(2)4
𝐺𝑦(5)𝑦(3)𝐺𝑦(4)𝑦′′𝐺𝑦′𝑦
+ (3)(2)4
𝐺𝑦(5)𝑦′′𝐺𝑦(4)𝑦(3)𝐺𝑦′𝑦
This consists of
(22.14.7)
(∏ 𝛿𝐽(𝑦(𝑞))
5
𝑞=3
) 𝐴(𝐽3)|1 = (3)(2)4
+ (3)(2)4
+ (3)(2)4
= 144 𝑡𝑒𝑟𝑚𝑠
Looking back at (22.14.4) we have for
(22.14.8)
(∏ 𝛿𝐽(𝑦(𝑞))
5
𝑞=3
) 𝐴(𝐽3)|21 = (3)(3)(2)4
= 144 𝑡𝑒𝑟𝑚𝑠
(22.14.9)
(∏ 𝛿𝐽(𝑦(𝑞))
5
𝑞=3
) 𝐴(𝐽3)|22 = (3)(3)(2)4
= 144 𝑡𝑒𝑟𝑚𝑠
(22.14.10)
+ (∏ 𝛿𝐽(𝑦(𝑞))
5
𝑞=3
) 𝐴(𝐽3)|23 = (2)(3)(3)(2)4
= 288 𝑡𝑒𝑟𝑚𝑠
So for (22.14.4), we have
(22.14.11)
(∏ 𝛿𝐽(𝑦(𝑞))
5
𝑞=3
) 𝐴(𝐽3)|
2
= 144 + 144 + 288 = 576 𝑡𝑒𝑟𝑚𝑠

31. To go on further with our long story let us assume that at the end we are given with the
following integration variables
(22.15.1)
𝑦(0)
= 𝑦′′
= 𝑦(4)
= 𝑦
𝑦′
= 𝑦(3)
= 𝑦(5)
So from (22.14.5) we would have
(22.15.2)
(∏ 𝛿𝐽(𝑦(𝑞))
5
𝑞=0
) 𝑆𝐶
3[ 𝐽 ] = (3)2(2)5
𝐺𝑦′𝑦𝐺𝑦𝑦′𝐺𝑦′𝑦 + (3)3(2)4
𝐺𝑦′𝑦′𝐺𝑦′𝑦𝐺𝑦𝑦
(∏ 𝛿𝐽(𝑦(𝑞))
5
𝑞=0
) 𝑆𝐶
3[ 𝐽 ] = (3)2(2)5
+ (3)3(2)4
= 720 𝑡𝑒𝑟𝑚𝑠
For 𝐺𝑦′𝑦𝐺𝑦𝑦′𝐺𝑦′𝑦 = 288 𝑖𝑑𝑒𝑛𝑡𝑖𝑐𝑎𝑙 𝑡𝑒𝑟𝑚𝑠
For 𝐺𝑦′𝑦′𝐺𝑦′𝑦𝐺𝑦𝑦 = 432 𝑖𝑑𝑒𝑛𝑡𝑖𝑐𝑎𝑙 𝑡𝑒𝑟𝑚𝑠
From (21.22) we are now able to write
(22.15.3)
(∫ 𝑑4
𝑦 𝑉 (𝑖ℏ
𝛿
𝛿𝐽(𝑦)
)
𝐵
𝐴
)
2
(
𝑖
ℏ
𝑆𝑐[ 𝐽 ])
3
= 𝑔(3)
2
(𝑖ℏ)5
∫ 𝑑4
𝑦 ′ 𝑑4
𝑦
𝐺(𝑦′
− 𝑦)
(2𝜋)2
𝐺(𝑦 − 𝑦 ′ )
(2𝜋)2
𝐺(𝑦′
− 𝑦)
(2𝜋)2
𝐵
𝐴
+
3
2
𝑔(3)
2
(𝑖ℏ)5
∫ 𝑑4
𝑦 ′ 𝑑4
𝑦
𝐺(𝑦′
− 𝑦 ′ )
(2𝜋)2
𝐺(𝑦 ′ − 𝑦 )
(2𝜋)2
𝐺(𝑦 − 𝑦)
(2𝜋)2
𝐵
𝐴
and we are happy to write (20.4) as

32. (22.15.4)
𝑒𝑥𝑝 (−
𝑖
ℏ
∫ 𝑑4
𝑦 𝑉 (𝑖ℏ
𝛿
𝛿𝐽(𝑦)
)
𝐵
𝐴
) 𝑒𝑖 𝑆𝑐 / ℏ
|
𝐽 = 0
= 1
+
1
12
𝑔(3)
2
(𝑖ℏ)3
∫ 𝑑4
𝑦 ′ 𝑑4
𝑦
𝐺(𝑦′
− 𝑦)
(2𝜋)2
𝐺(𝑦 − 𝑦 ′ )
(2𝜋)2
𝐺(𝑦′
− 𝑦)
(2𝜋)2
𝐵
𝐴
+
1
8
𝑔(3)
2
(𝑖ℏ)3
∫ 𝑑4
𝑦 ′ 𝑑4
𝑦
𝐺(𝑦′
− 𝑦 ′ )
(2𝜋)2
𝐺(𝑦 ′ − 𝑦 )
(2𝜋)2
𝐺(𝑦 − 𝑦)
(2𝜋)2
𝐵
𝐴
+ ⋯
[To be continued…]
Very Brief Highlights On Some Key Details: Vacuum-To-Vacuum Matrix In the
Presence Of Cubic Self Interaction
Remarks: This part here is just a quick continuation of Basic Practice Calculations
Involving Scalar Fields. The numbering of equations does not follow the
preceding parts of this draft.
Say we start at an initial time 𝑡 = 0 for a quantum mechanical description of a scalar
particle in the vacuum state |0⟩. With the application of the time evolution operator (teo)
𝑈1(𝑇) on the given vacuum state we evolve this vacuum state into another state |𝜓⟩ that
is given at a later time 𝑡 = 𝑇 > 0.
(1)
|0⟩ → |𝜓⟩ = 𝑈1(𝑇)|0⟩
We take the projection of this evolved state on the vacuum state. Thus, getting the
vacuum-to-vacuum matrix
(2)
⟨0|𝑈1(𝑇)|0⟩
for the probability amplitude that a particle initially in the vacuum state at an initial time
will still be in the vacuum state at a later time.
The time evolution operator (TEO)
(3)
𝑈1(𝑇) = 𝑈1(𝑇, 0) = 𝑒𝑥𝑝 (−
𝑖
ℏ
∫ 𝑑𝑡 𝐻(𝑡)
𝑇
0
)

33. is given for a system Hamiltonian for scalar fields considered in this particular drill and
this Hamiltonian has implicit time dependence being a Hamiltonian H(J(t)) that is a
functional of time dependent source J(t). This Hamiltonian can be derived from the scalar
field system Lagrangian via Legendre transformation. However, we shall no longer give
the details of such transformation here.
In this prior set-up, we are not yet taking into account the self-interactions in our
Lagrangian and consequently, in our Hamiltonian so our initial matrix given by (2) is in
the absence of the said interactions. These self-interactions enter into the Lagrangian and
Hamiltonian in the form of potential functions and in my personal convenience, I don’t
include in these functions the scalar field mass terms as I put these mass terms already
explicitly in the Lagrangian and Hamiltonian.
Without going into the details of how these self-interactions enter into the vacuum-to-
vacuum (VTV) matrix we will only give this VTV matrix here that comes with the
presence of the said interactions via potential operators
(4)
⟨0|𝑈1(𝑇)𝑒𝑥𝑝 (−
𝑖
ℏ
∫ 𝑑4
𝑦 𝑉(𝜑
̂)
𝐵
𝐴
) |0⟩|
𝐽=0
= 𝑒𝑥𝑝 (−
𝑖
ℏ
∫ 𝑑4
𝑦 𝑉 (𝑖ℏ
𝛿
𝛿𝐽(𝑦)
)
𝐵
𝐴
) ⟨0|𝑈1(𝑇)|0⟩|
𝐽=0
and in this particular drill we are considering cubic self- interaction in the form given by
(5)
𝑉[𝜑(𝑥, 𝐽)] =
1
3!
𝑔(3)𝜑3(𝑥, 𝐽)
In this current presentation, I am also not going to dwell on the lengthy details to arrive at
the end results of this exercise but this more elaborate presentation will be done in future
drafts.
The VTV matrix that includes the cubic self-interaction (5) is obtained upon the setting of
all sources to zero, J = 0. This matrix is given by
(6)
⟨0|𝑈1(𝑇)𝑒𝑥𝑝 (−
𝑖
ℏ
∫ 𝑑4
𝑦 𝑉(𝜑
̂)
𝐵
𝐴
) |0⟩|
𝐽=0
= 𝐶 [ 𝐽 = 0] ×

34. (1 +
1
12
𝑔(3)
2
(𝑖ℏ)3
∫ 𝑑4
𝑦 ′𝑑4
𝑦
𝐺(𝑦 ′ − 𝑦)
(2𝜋)2
𝐺( 𝑦 − 𝑦 ′ )
(2𝜋)2
𝐺(𝑦 ′ − 𝑦)
(2𝜋)2
𝐵
𝐴
+
1
8
𝑔(3)
2
(𝑖ℏ)3
∫ 𝑑4
𝑦 ′𝑑4
𝑦
𝐺(𝑦 ′ − 𝑦 ′)
(2𝜋)2
𝐺( 𝑦 ′ − 𝑦 )
(2𝜋)2
𝐺(𝑦 − 𝑦)
(2𝜋)2
𝐵
𝐴
)
In my personal convenience, I am used to writing the Green’s function as (for example)
(7)
𝐺(𝑦 ′ − 𝑦) =
1
(2𝜋)2
∫ 𝑑4
𝑘
𝑒𝑖𝑘𝜎(𝑦 ′ 𝜎− 𝑦𝜎)
−𝑘𝜇𝑘𝜇 + 𝑀2 + 𝑖𝜖
where I have already inserted an imaginary part 𝑖𝜖 to shift the poles as contour integration
will be required afterwards. The Green’s functions in the matrix expression act as
propagators.
The resulting matrix (6), upon the setting of J = 0, consists of two groups (aside from the
term with numerical 1) of relevant identical terms. These terms are as written in
coordinate space.
Let us take the group of terms with the numerical factor
1
12
and write this in momentum
space
(8)
∫ 𝑑4
𝑦 ′𝑑4
𝑦
𝐺(𝑦 ′ − 𝑦)
(2𝜋)2
𝐺( 𝑦 − 𝑦 ′ )
(2𝜋)2
𝐺(𝑦 ′ − 𝑦)
(2𝜋)2
𝐵
𝐴
=
1
(2𝜋)4
∫ 𝑑4
𝑘 ∫ 𝑑4
𝑘 ′ ∫ 𝑑4
𝑘 ′′
1
−𝑘2 + 𝑀2 + 𝑖𝜖
1
−𝑘 ′2 + 𝑀2 + 𝑖𝜖
1
−𝑘 ′′2 + 𝑀2 + 𝑖𝜖
𝛿4(𝑘
− 𝑘 ′ + 𝑘 ′′ )𝑦 ′ 𝛿4(𝑘 − 𝑘 ′ + 𝑘 ′′ )𝑦
The Dirac-delta functions in this expression represent vertices at the respective space-
time points 𝑦 ′ and 𝑦. As indicated above there are three four-momentum integration
variables with two initial four-momentum vertices that we have just mentioned.
As written in coordinate space (8) can be depicted with the following Feynman graph

35. (Fig.1)
where at the spacetime point 𝑦 ′ is the Dirac-delta function 𝛿4(𝑘 − 𝑘 ′ + 𝑘 ′′ )𝑦 ′, while
at the spacetime point y the Dirac-delta function 𝛿4(𝑘 − 𝑘 ′ + 𝑘 ′′ )𝑦 . (Note the
symmetric property of the Dirac-delta function, 𝛿4(−𝑎) = 𝛿4(𝑎)).
We also have the Fourier components of the propagators
(8.1)
𝑔
̃(𝑘) =
1
−𝑘2 + 𝑀2 + 𝑖𝜖
(8.2)
𝑔
̃(𝑘 ′) =
1
−𝑘 ′2 + 𝑀2 + 𝑖𝜖
and
(8.3)
𝑔
̃(𝑘 ′′) =
1
−𝑘 ′′2 + 𝑀2 + 𝑖𝜖
We can integrate over 𝑘 ′′ at the space-time point 𝑦 with a picking 𝑘 ′′ = k ’ – k so that
we can reduce (8) into (noted (k ’ – k )2
= (k – k ′ )2
)
(9)
∫ 𝑑4
𝑦 ′𝑑4
𝑦
𝐺(𝑦 ′ − 𝑦)
(2𝜋)2
𝐺( 𝑦 − 𝑦 ′ )
(2𝜋)2
𝐺(𝑦 ′ − 𝑦)
(2𝜋)2
𝐵
𝐴
=
1
(2𝜋)4
∫ 𝑑4
𝑘 ′
1
−𝑘 ′2 + 𝑀2 + 𝑖𝜖
∫ 𝑑4
𝑘
1
−𝑘2 + 𝑀2 + 𝑖𝜖
1
− ( k – k ′ )2 + 𝑀2 + 𝑖𝜖
𝛿4(𝑘 ′
− 𝑘 + k – k ′ )𝑦 ′
In momentum space as given by (9), we have the corresponding Feynman graph

36. (Fig.2)
where
(9.1)
𝑔
̃(𝑘) =
1
−𝑘2 + 𝑀2 + 𝑖𝜖
(9.2)
𝑔
̃(𝑘 ′) =
1
−𝑘 ′2 + 𝑀2 + 𝑖𝜖
and
(9.3)
𝑔
̃(k – k ′ ) =
1
− (k – k ′ )2 + 𝑀2 + 𝑖𝜖
In (9), we have chosen the four-momentum k as internal four-momentum and k’ as the
external four-momentum and such recognition enables us to construct the Feynman graph
Fig.2.
In (9) we take note of the remaining Dirac-delta function that we write as (caution: with
symmetric integral limits)
(9.4)
𝛿4(𝑘 ′ − 𝑘 + k – k ′ )𝑦 ′ = 𝛿4(0 )𝑦 ′ =
1
(2𝜋)4
∫ 𝑑4
𝑦 ′
This might imply a singularity if we are to base this on the basic definition of a Dirac-
delta function with an argument that is vanishing, which is 𝛿4(0 )𝑦 ′ = ∞. However, for
the moment let us take (9.4) as a definite constant.

37. We continue with the other group of terms that collectively go with the numerical factor
1
8
as given by (6) in coordinate space. This we translate into momentum space
(10)
∫ 𝑑4
𝑦 ′𝑑4
𝑦
𝐺(𝑦 ′ − 𝑦 ′ )
(2𝜋)2
𝐺( 𝑦 ′ − 𝑦 )
(2𝜋)2
𝐺(𝑦 − 𝑦)
(2𝜋)2
𝐵
𝐴
=
1
(2𝜋)4
∫ 𝑑4
𝑘 ∫ 𝑑4
𝑘 ′ ∫ 𝑑4
𝑘 ′′
1
−𝑘2 + 𝑀2 + 𝑖𝜖
1
−𝑘 ′2 + 𝑀2 + 𝑖𝜖
1
−𝑘 ′′2 + 𝑀2 + 𝑖𝜖
𝛿4(𝑘
− 𝑘 " + 𝑘 ′′ )𝑦 ′ 𝛿4(𝑘 − 𝑘 ′ + 𝑘 ′ )𝑦
This is depicted in the following Feynman graph
(Fig.3)
Ref’s:
[1]W. Hollik, Quantum field theory and the Standard Model, arXiv:1012.3883v1 [hep-
ph]
[2]Baal, P., A COURSE IN FIELD THEORY,
http://www.lorentz.leidenuniv.nl/~vanbaal/FTcourse.html
[3]’t Hooft, G., THE CONCEPTUAL BASIS OF QUANTUM FIELD THEORY,
http://www.phys.uu.nl/~thooft/
[4]Siegel, W., FIELDS, arXiv:hep-th/9912205 v2
[5]Wells, J. D., Lectures on Higgs Boson Physics in the Standard Model and Beyond,
arXiv:0909.4541v1
[6]Cardy, J., Introduction to Quantum Field Theory
[7]Gaberdiel, M., Gehrmann-De Ridder, A., Quantum Field Theory