quantum field theory

1. Field Theory
Highlights 2015
Set B
By: Roa, F. J. P.
Topics:
1 SU(2)XU(1) Construction in toy Standard Model
2 Basic Practice Calculations Involving Scalar Fields
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(𝜑̂). (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

2. (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.)
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
𝑒 𝑖𝑘 𝜎(𝑥 𝜎
− 𝑦 𝜎
)
−𝑘 𝜇 𝑘 𝜇 + 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

3. (20.1)
𝑒𝑥𝑝(−
𝑖
ℏ
∫ 𝑑4
𝑦 𝑉(𝜑̂)
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(𝜑̂)
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
(20.3)
⟨0|𝑈1(𝑇)𝑒𝑥𝑝(−
𝑖
ℏ ∫ 𝑑4
𝑦 𝑉(𝜑̂)
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
) 𝑒 𝑖 𝑆𝑐 / ℏ
= 1 + ∑
1
𝑚!
(
𝑖
ℏ
)
𝑚
𝑆 𝐶
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 ] = 〈 𝐽𝑥 𝐺𝑥𝑧 𝐽𝑧〉 =
1
2
1
(2𝜋)2
∬ 𝑑4
𝑥 𝑑4
𝑧 𝐽( 𝑥) 𝐺( 𝑥 − 𝑧) 𝐽(𝑧)
Here, the bracket 〈 𝐽𝑥 𝐺𝑥𝑧 𝐽𝑧〉 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

4. (21.1)
𝑉[ 𝜑( 𝑥, 𝐽)] =
1
3!
𝑔(3) 𝜑3( 𝑥, 𝐽)
In my notation, I write
(21.2)
𝑆 𝐶
𝑛[ 𝐽 ] = ∏〈 𝐽𝑥 𝐺𝑥𝑧 𝐽𝑧 〉𝑖
𝑛−1
𝑖=0
where 𝑖 is specified on both x and z. Meaning,
(21.3)
〈 𝐽𝑥 𝐺𝑥𝑧 𝐽𝑧 〉𝑖 = 〈 𝐽(𝑥(𝑖)
) 𝐺(𝑥( 𝑖)
, 𝑧( 𝑖)
) 𝐽(𝑧(𝑖)
) 〉
Note here that 𝑆 𝐶
𝑛
involves 2𝑛 J’s. For 𝜑3( 𝑥, 𝐽) I shall also write this as
(21.4)
𝜑3( 𝑥, 𝐽) = ∏〈 𝐺𝑥𝑧 𝐽𝑧 〉 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) ∏〈 𝐺𝑥𝑧 𝐽𝑧 〉 𝑗
2
𝑗=0
and so it follows that

5. (21.7)
∫ 𝑑4
𝑥 𝑉[ 𝜑( 𝑥, 𝐽)] =
1
3!
𝑔(3) ∫ 𝑑4
𝑥 ∏〈 𝐺𝑥𝑧 𝐽𝑧 〉 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
∏ 〈 𝐺 𝑥(𝑘)
𝑧 𝐽𝑧〉 𝑗
3𝑛−1
𝑗 =0
where 𝑘 is on all x, while 𝑗 on all z.
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)

6. ∏ 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.
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(𝜑̂)
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 → 𝑆 𝐶
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)

7. ∫ 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
(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 𝑆 𝐶
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)

8. (∫ 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.
(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
(∏〈 𝐽𝑥 𝐺𝑥𝑧 𝐽𝑧 〉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
) (∏〈 𝐽𝑥 𝐺𝑥𝑧 𝐽𝑧 〉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)

9. 𝛿4( 𝑥 − 𝑦) =
𝛿𝐽(𝑥)
𝛿𝐽(𝑦)
To first order in the differentiation we have
(22.3)
𝛿𝐽(𝑦)〈 𝐽𝑥 𝐺𝑥𝑧 𝐽𝑧〉 = 〈 𝐽𝑥 𝐺𝑥𝑦〉+ 〈 𝐺 𝑦𝑧 𝐽𝑧 〉
=
1
2
1
(2𝜋)2
(∫ 𝑑4
𝑥 𝐽( 𝑥) 𝐺( 𝑥 − 𝑦) + ∫ 𝑑4
𝑧 𝐺( 𝑦 − 𝑧) 𝐽(𝑧) )
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)〈 𝐽𝑥 𝐺𝑥𝑧 𝐽𝑧〉|
𝑥 = 𝑧
= (2)〈 𝐽𝑥 𝐺𝑥𝑦〉
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)〈 𝐽𝑥 𝐺𝑥𝑧 𝐽𝑧〉 = 𝛿𝐽(𝑦)
2 〈 𝐽𝑥 𝐺𝑥𝑧 𝐽𝑧〉 = (2) 𝐺 𝑦′
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) 𝑆 𝐶
3[ 𝐽 ]. This we shall put as
(22.6)
𝐴( 𝐽5) = 𝛿𝐽( 𝑦) 𝑆 𝐶
3[ 𝐽 ]
= 〈 𝐽𝑥 𝐺𝑥𝑧 𝐽𝑧〉2 〈 𝐽𝑥 𝐺𝑥𝑧 𝐽𝑧〉1 𝛿𝐽( 𝑦)〈 𝐽𝑥 𝐺𝑥𝑧 𝐽𝑧〉
+ 〈 𝐽𝑥 𝐺𝑥𝑧 𝐽𝑧〉2 (𝛿𝐽( 𝑦)〈 𝐽𝑥 𝐺𝑥𝑧 𝐽𝑧〉1)〈 𝐽𝑥 𝐺𝑥𝑧 𝐽𝑧〉
+ (𝛿 𝐽( 𝑦)〈 𝐽𝑥 𝐺𝑥𝑧 𝐽𝑧〉2)〈𝐽𝑥 𝐺𝑥𝑧 𝐽𝑧〉1〈 𝐽𝑥 𝐺𝑥𝑧 𝐽𝑧〉
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)

10. 𝐴( 𝐽5) = 𝛿𝐽( 𝑦) 𝑆 𝐶
3[ 𝐽 ] = (2)(3)〈 𝐽𝑥 𝐺𝑥𝑧 𝐽𝑧〉2 〈 𝐽𝑥 𝐺𝑥𝑧 𝐽𝑧〉1 〈𝐽𝑥 𝐺𝑥𝑦〉
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)〈 𝐽𝑥 𝐺𝑥𝑧 𝐽𝑧〉𝑖 = (2)〈 𝐽𝑥 𝐺𝑥𝑦〉𝑖 = 𝛿𝐽( 𝑦)〈 𝐽𝑥 𝐺𝑥𝑧 𝐽𝑧〉 𝑗 = (2)〈𝐽𝑥 𝐺𝑥𝑦〉 𝑗
upon the setting 𝑥 = 𝑧 after the indicated differentiation.
Next, we have the second order differentiation
(22.9)
𝐴( 𝐽4) = 𝛿𝐽( 𝑦′) 𝐴( 𝐽5) = 𝛿𝐽(𝑦)
2
𝑆 𝐶
3[ 𝐽 ] = 𝛿𝐽( 𝑦′) 𝛿𝐽( 𝑦) 𝑆 𝐶
3[ 𝐽 ]
= (3)(2)〈 𝐽𝑥 𝐺𝑥𝑧 𝐽𝑧〉2 〈 𝐽𝑥 𝐺𝑥𝑧 𝐽𝑧〉1 𝐺 𝑦′
𝑦
+ (3)(2)(2)(2)〈 𝐽𝑥 𝐺𝑥𝑧 𝐽𝑧〉2 〈 𝐽𝑥 𝐺𝑥𝑦′〉1〈 𝐽𝑥 𝐺𝑥𝑦〉
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( 𝐽5) = 𝛿𝐽(𝑦)
3
𝑆 𝐶
3[ 𝐽 ] = (∏ 𝛿𝐽(𝑦(𝑞)
)
2
𝑞=0
) 𝑆 𝐶
3[ 𝐽 ]
= 𝛿𝐽( 𝑦′′) 𝐴( 𝐽4)|
1
+ 𝛿𝐽( 𝑦′′) 𝐴( 𝐽4)|
2
[To be continued…]
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