qft

1. Workouts #1 in Basic QFT
One Scalar Particle Scattering Into One Scalar Particle
Roa, F. J.P.
Let us suppose that a particle propagates from a spacetime region 𝑥 to some other spacetime
region𝑥′. (Cautionary remark: In this draft I use the word region to mean a point for the basic
reason that we perform Fourier integrations at those points that allow us to have integral
definitions of Dirac-delta functions.) As the particle enters the latter spacetime region it carries a
spatial momentum𝑘⃗ , then scatters as a scalar particle carrying a new spatial momentum𝑘⃗ ′. This
process is given with the following scattering matrix
(1)
⟨𝑘⃗ ′|𝑈( 𝑇𝑜𝑢𝑡, 𝑇𝑖𝑛 )|𝑘⃗ ⟩ =
√2𝜔( 𝑘⃗ ′)
√ℏ
∫
𝑑3
𝑥′
√(2𝜋)3
𝑒−𝑖𝑘⃗ ′∙𝑥′
√2𝜔 ( 𝑘⃗ )
√ℏ
∫
𝑑3
𝑥
√(2𝜋)3
𝑒 𝑖𝑘⃗ ∙𝑥
(−
ℏ
𝑖
)
𝛿
𝛿𝐽( 𝑥′)
(
ℏ
𝑖
)
𝛿
𝛿𝐽(𝑥)
⟨0| 𝑈(𝑇𝑜𝑢𝑡, 𝑇𝑖𝑛 )|0⟩
This involves second order in the derivative operations with respect to sources J’s on the
vacuum-to-vacuum matrix
(2)
𝛿
𝛿𝐽( 𝑥′)
𝛿
𝛿𝐽(𝑥)
⟨0| 𝑈(𝑇𝑜𝑢𝑡, 𝑇𝑖𝑛 )|0⟩
where already in its factored form, this said matrix has the form
(3)
⟨0| 𝑈(𝑇𝑜𝑢𝑡 , 𝑇𝑖𝑛 )|0⟩ = 𝑒𝑥𝑝(−
𝑖
ℏ
( 𝑇𝑜𝑢𝑡 − 𝑇𝑖𝑛) 𝐸0
0
) 𝑒𝑖 𝑆 𝑐 / ℏ
The form of the matrix (1) is a consequence following from path integration. However, we shall
no longer tackle the details of this path integration leading to the right-hand-side (rhs) of (1). It is
to be noticed in (3) that we have not yet normalized the resulting matrix (1) so the vacuum-to-
vacuum matrix (3) still carries the factor that involves the ground state energy 𝐸0
0
.
We can conveniently write out the vacuum-to-vacuum matrix explicitly as a Taylor/Maclaurin
expansion with
(4)
𝑒 𝑖 𝑆 𝑐 /ℏ = 1 +
𝑖
ℏ
𝑆 𝑐 + ∑
1
𝑛!
(
𝑖
ℏ
)
𝑛
𝑆 𝐶
𝑛
∞
𝑛=2

2. As a basic recollection recall that in the case for a boson such as a scalar field in this exercise, we
can raise a one-particle state of certain spatial momentum 𝑘⃗ from the vacuum state with the
application of bosonic creation operator 𝑎†
and such creation operation is given by
(5.1)
|𝑘⃗ ⟩ = 𝑎†
(𝑘⃗ )|0 ⟩
with its Hermitian adjoint
(5.2)
⟨𝑘⃗ | = (|𝑘⃗ ⟩ )
†
= ⟨0| 𝑎(𝑘⃗ )
We may think of (5.1) as the one-particle state at the initial time 𝑇𝑖𝑛 and evolve such state into
some other state at 𝑇𝑜𝑢𝑡 with the application of the time evolution operator 𝑈(𝑇𝑜𝑢𝑡, 𝑇𝑖𝑛 ). The
projection of this evolved one-particle state on some other one-particle state |𝑘⃗ ′⟩ will give the
left-hand-side (lhs) of the scattering matrix (1), which bears a quantum field theory interpretation
of being associated with a probability that the one-particle state of momentum 𝑘⃗ at an initial time
𝑇𝑖𝑛 can be found as a one-particle state of spatial momentum 𝑘⃗ ′ at a later time 𝑇𝑜𝑢𝑡 . Note in here
that 𝑇𝑖𝑛 is in the initial spacetime region 𝑥, while 𝑇𝑜𝑢𝑡 is in the latter spacetime region 𝑥′. This
matrix is then thought of as a scattering matrix that can be evaluated via path integration
resulting in (1).
In (1) we take note that we have two different spatial Fourier integrations, one over the spatial
region represented by 𝑥, while the other one with the spatial region of 𝑥′. Each of these
integrations defines a Dirac-delta function at the spatial region of integration. That is, for
example
(6)
𝛿3
(𝑘⃗ ± 𝑘⃗ (𝑗)
)
𝑥
= ∫
𝑑3
𝑥
(2𝜋)3
𝑒±𝑖(𝑘⃗ ± 𝑘⃗ (𝑗)
)∙𝑥
Ofcourse, such delta functions assume symmetric integral limits in those space regions where
these integrations are performed.
Note as to be explicit we have for the initial spacetime region 𝑥 = (𝑥0
, 𝑥) and for the latter
spacetime region x′ = (𝑥′0
, 𝑥′).
Of prior note also is the connected two-point function for scalars not a two-point function. This
is connected in the sense that it connects two sources J’s, each of which belongs to the two
different spacetime regions that act as end regions for the propagating scalar particle.
Such connected two-point function will simply be given by the scalar classical action as
expressed in the functional of the sources with a scalar Green’s function that plays the role of
propagator.

3. (7)
𝑆 𝑐 = −
1
2
1
(2𝜋)2
∫ 𝑑4 𝑦 𝑑4 𝑦 ′ 𝐽( 𝑦) 𝐺( 𝑦 − 𝑦 ′) 𝐽(𝑦′) = − 〈 𝐽 𝑦 𝐺𝑦𝑦′ 𝐽 𝑦′〉
For convenience we specify in notation that
(8)
𝛿𝐽(𝑥) =
𝛿
𝛿𝐽(𝑥)
𝛿𝐽(𝑥′) 𝛿𝐽(𝑥) =
𝛿2
𝛿𝐽(𝑥′)𝛿𝐽(𝑥)
Then for (2) we write
(9)
𝛿𝐽(𝑥′) 𝛿𝐽(𝑥) 𝑒 𝑖 𝑆 𝑐/ ℏ =
𝑖
ℏ
𝛿𝐽(𝑥′) 𝛿𝐽(𝑥) 𝑆𝑐 +
1
2
(
𝑖
ℏ
)
2
𝛿𝐽(𝑥′) 𝛿𝐽(𝑥) 𝑆 𝐶
2
+ ∑
1
𝑛!
(
𝑖
ℏ
)
𝑛
𝛿𝐽(𝑥′) 𝛿𝐽(𝑥) 𝑆 𝐶
𝑛
∞
𝑛=3
where to the first power of the connected two-point function we have (7), while to the second
power of this function we write as
(10.1)
𝑆 𝑐 = −
1
2
1
(2𝜋)2
∫ 𝑑4 𝑦 𝑑4 𝑦 ′ 𝐽( 𝑦) 𝐺( 𝑦 − 𝑦 ′) 𝐽(𝑦′) = − 〈 𝐽 𝑦 𝐺𝑦𝑦′ 𝐽 𝑦′〉
with
(10.2)
𝑆 𝐶
2
= (−1)(−1) 〈 𝐽 𝑦 𝐺𝑦𝑦′ 𝐽𝑦′〉1 〈 𝐽 𝑦 𝐺𝑦𝑦′ 𝐽 𝑦′〉
and
(10.3)
𝐺( 𝑥′ − 𝑥) =
−1
(2𝜋)2
∫ 𝑑4 𝑘
𝑒 𝑖𝑘 𝜎(𝑥′ 𝜎− 𝑥 𝜎)
−𝑘 𝜇 𝑘 𝜇 + 𝑀2 + 𝑖𝜖
The derivative operation via functional derivative in 3 + 1 spacetime
(10.4)
𝛿4( 𝑥 − 𝑦) =
𝛿𝐽(𝑥)
𝛿𝐽(𝑦)
The first order differentiation of (7) yields

4. (11.1)
𝛿𝐽(𝑥) 𝑆𝑐 = − 𝛿 𝐽(𝑥)〈 𝐽 𝑦 𝐺𝑦𝑦′ 𝐽 𝑦′〉 = − (〈 𝐽 𝑦 𝐺𝑦𝑥〉+ 〈 𝐺 𝑥𝑦′ 𝐽 𝑦′〉)
= −
1
2
1
(2𝜋)2
(∫ 𝑑4 𝑦 𝐽( 𝑦) 𝐺( 𝑦 − 𝑥) + ∫ 𝑑4 𝑦′ 𝐺( 𝑥 − 𝑦′) 𝐽(𝑦′) )
and with the setting of y = y’, this becomes
(11.2)
𝛿𝐽(𝑥) 𝑆𝑐|
𝑦 = 𝑦′
= −(2)〈 𝐺𝑥𝑦 𝐽 𝑦〉
In (11.2) we note
(11.3)
〈 𝐺𝑥𝑦 𝐽 𝑦〉 =
1
2
1
(2𝜋)2
∫ 𝑑4 𝑦 𝐽( 𝑦) 𝐺( 𝑦 − 𝑥)
and it is important to take note that the number inside the parenthesis (2) means the number of
terms originally involved in the first differentiation. Consequently, from (11.2) we have the
second order differentiation resulting as
(11.4)
𝛿𝐽(𝑥′) 𝛿𝐽(𝑥) 𝑆𝑐 = − (2)𝐺𝑥′𝑥
After carrying out the indicated differentiation in (9) and only up to second order in i/hbar, we
write (9) explicitly as
(12.1)
𝛿𝐽(𝑥′) 𝛿𝐽(𝑥) 𝑒 𝑖 𝑆 𝑐 /ℏ
=
𝑖
ℏ
(− (2) 𝐺 𝑥′ 𝑥) +
1
2
(
𝑖
ℏ
)
2
(2)(2)〈 𝐽 𝑦 𝐺 𝑦𝑦′ 𝐽 𝑦′〉 𝐺 𝑥′ 𝑥
+
1
2
(
𝑖
ℏ
)
2
(2)(2)(2)〈 𝐺𝑥′𝑦′ 𝐽 𝑦′〉〈 𝐺𝑥𝑦 𝐽 𝑦〉
The first major term of this consists two terms, the second major term four terms and the third
major term has eight terms. So (12.1) has a total of fourteen terms.
Taking note from (1) we perform the Fourier integrations involving the first major term in (12.1)
and these integrations are given by

5. (12.2)
∫
𝑑3
𝑥′
√(2𝜋)3
𝑒−𝑖𝑘⃗ ′∙𝑥′
∫
𝑑3
𝑥
√(2𝜋)3
𝑒 𝑖𝑘⃗ ∙𝑥
𝐺( 𝑥′ − 𝑥)
= −
1
(2𝜋)2
∫ 𝑑𝑘(2)
0
∫ 𝑑3
𝑘⃗
(2) 𝑒
𝑖𝑘(2)
0 ( 𝑥′ 0
− 𝑥0)
−𝑘(2)
2
(𝑘⃗ (2)) + 𝑀2
+ 𝑖𝜖
(2𝜋)3
× 𝛿3
(𝑘⃗ ′ + 𝑘⃗ (2))
𝑥′
𝛿3
(𝑘⃗ + 𝑘⃗ (2))
𝑥
where it is specified that to the space region 𝑥 ′ a vertex with the Dirac-delta function
(12.3)
𝛿3
(𝑘⃗ ′ + 𝑘⃗ (2))
𝑥′
and to the space region 𝑥 a vertex with its own Dirac-delta function
(12.4)
𝛿3
(𝑘⃗ + 𝑘⃗ (2))
𝑥
Clearly in (12.2) we initially see two vertices, one as already mentioned at the space region of 𝑥,
where we sum up two space momenta 𝑘⃗ and 𝑘⃗ (2) and the vertex at the space region of 𝑥 ′ where
𝑘⃗ ′ and 𝑘⃗ (2) are summed up. Since it is indicated that we are to perform integration ove the 𝑘⃗ (2)
vec momentum variable, the vertex at the initial space region will disappear as there will be
picking of 𝑘⃗ (2) = − 𝑘⃗ . Given such picking, we have
(12.5)
𝑘(2)
2
( 𝑘⃗⃗ (2)) → 𝑘(2)
2
(−𝑘⃗⃗ ) = ( 𝑘(2)
0 )
2
− 𝑘⃗⃗ ∙ 𝑘⃗⃗
and then just relabel 𝑘(2)
0
to 𝑘0
so that
(12.6)
𝑘(2)
2
(−𝑘⃗⃗ ) = 𝑘 𝜇 𝑘 𝜇 = ( 𝑘
0
)
2
− 𝑘⃗⃗ ∙ 𝑘⃗⃗
Thus, (12.2) further results to
(12.7)
∫
𝑑3
𝑥′
√(2𝜋)3
𝑒−𝑖𝑘⃗ ′∙𝑥′
∫
𝑑3
𝑥
√(2𝜋)3
𝑒 𝑖𝑘⃗ ∙𝑥
𝐺 𝑥′
𝑥
=
1
2(2𝜋)4
∫ 𝑑 𝑘0
𝑒
𝑖𝑘0 ( 𝑥′ 0
− 𝑥0)
( 𝑘0)2 − ( 𝑘⃗ ∙ 𝑘⃗ + 𝑀2
+ 𝑖𝜖)
𝛿3
(𝑘⃗ ′ − 𝑘⃗ )
𝑥′

6. We have one more integration to perform and this is a contour integration on a complex z-plane.
In our convenience we will only choose the upper half-contour that encloses the complex pole
𝑧0 = 𝑏′ where
(12.8)
𝑏′
= √ 𝑘⃗ ∙ 𝑘⃗ + 𝑀2
+ 𝑖𝜖 ≈ √ 𝑘⃗ ∙ 𝑘⃗ + 𝑀2
+
𝑖𝜖
2√ 𝑘⃗ ∙ 𝑘⃗ + 𝑀2
We will no longer dig into the details of such integration and simply write here the result. The
result in the limit as 𝜖 → 0 is given by
(12.9)
𝑙𝑖𝑚 𝜖 → 0 ∫ 𝑑 𝑘0
∞
−∞
𝑒
𝑖𝑘0( 𝑥′ 0
− 𝑥0)
( 𝑘0)2 − ( 𝑘⃗ ∙ 𝑘⃗ + 𝑀2
+ 𝑖𝜖)
=
𝑖𝜋
𝜔(𝑘⃗ )
𝑒
𝑖𝑘0 ( 𝑥′ 0
− 𝑥0)
at
(12.10)
𝑘0
= 𝜔(𝑘⃗ ) = √ 𝑘⃗ ∙ 𝑘⃗ + 𝑀2
Proceeding, we may write the scattering matrix up to a major third term only and having to note
that only the first major term is relevant upon the setting of all sources J’s to zero.
(13.1)
⟨𝑘⃗ ′|𝑈( 𝑇𝑜𝑢𝑡, 𝑇𝑖𝑛 )|𝑘⃗ ⟩ = 1𝑠𝑡 + 2𝑛𝑑 + 3𝑟𝑑 + ⋯
Given (12.9), we may write
(13.2)
1𝑠𝑡 = 𝑒𝑥𝑝(−
𝑖
ℏ
( 𝑇𝑜𝑢𝑡 − 𝑇𝑖𝑛) 𝐸0
0
) 𝑒
𝑖𝑘0( 𝑥′ 0
− 𝑥0)
𝛿3
(𝑘⃗ ′ − 𝑘⃗ )
𝑥′
This is associated with the Feynman diagram at the space region of 𝑥′.

7. (Fig.1)
While for the other major terms in (13.1) we have
(13.3)
2𝑛𝑑 = −
𝑖
ℏ
〈 𝐽 𝑦
𝐺 𝑦𝑦′ 𝐽 𝑦′〉 × 1𝑠𝑡
and
(13.4)
3𝑟𝑑 = −
1
ℏ
√ 𝜔(𝑘⃗ ′)√ 𝜔(𝑘⃗ ) 23
∫
𝑑3
𝑥′
√(2𝜋)3
𝑒−𝑖 𝑘⃗ ′∙𝑥′ 〈 𝐺 𝑥′
𝑦′ 𝐽 𝑦′〉 ∫
𝑑3
𝑥
√(2𝜋)3
𝑒 𝑖𝑘⃗ ∙𝑥 〈 𝐺 𝑥𝑦 𝐽 𝑦
〉
References
[1]Baal, P., A COURSE IN FIELD THEORY
[2]Cardy, J., Introduction to Quantum Field Theory
[3]Gaberdiel, M., Gehrmann-De Ridder, A., Quantum Field Theory