1) A scalar particle travels from one spacetime region to another, carrying an initial momentum k and scattering into a final momentum k'. This scattering process is described by a scattering matrix. 2) The scattering matrix involves second order derivatives of the vacuum-to-vacuum matrix element with respect to sources. This vacuum-to-vacuum matrix can be written as a Taylor expansion involving the connected scalar classical action. 3) The left-hand side of the scattering matrix gives the probability that a one-particle state with momentum k at initial time Tin will be found with momentum k' at final time Tout, and can be evaluated via path integration.

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 momentum 𝑘⃗ , then scatters as a scalar particle carrying a new spatial momentum 𝑘⃗ ′. 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
𝑆𝑐 + ∑
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 (spin zero) 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)
𝑆𝑐 = −
1
2
1
(2𝜋)2
∫ 𝑑4
𝑦 𝑑4
𝑦 ′ 𝐽(𝑦)𝐺(𝑦 − 𝑦 ′)𝐽(𝑦′) = − 〈𝐽 𝑦 𝐺 𝑦𝑦′ 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)
𝑆𝑐 = −
1
2
1
(2𝜋)2
∫ 𝑑4
𝑦 𝑑4
𝑦 ′ 𝐽(𝑦)𝐺(𝑦 − 𝑦 ′)𝐽(𝑦′) = − 〈𝐽 𝑦 𝐺 𝑦𝑦′ 𝐽 𝑦′〉
with
(10.2)
𝑆 𝐶
2
= (−1)(−1) 〈𝐽 𝑦 𝐺 𝑦𝑦′ 𝐽 𝑦′〉1 〈𝐽 𝑦 𝐺 𝑦𝑦′ 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 = − 𝛿𝐽(𝑥)〈𝐽 𝑦 𝐺 𝑦𝑦′ 𝐽 𝑦′〉 = − (〈𝐽 𝑦 𝐺 𝑦𝑥〉 + 〈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)〈𝐺 𝑥𝑦 𝐽 𝑦〉
In (11.2) we note
(11.3)
〈𝐺 𝑥𝑦 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)𝐺 𝑥′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)〈𝐺 𝑥′𝑦′ 𝐽 𝑦′〉〈𝐺 𝑥𝑦 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
(12.2)

5. ∫
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
= ( 𝑘0
)
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)
∫ 𝑑 𝑘0
𝑒 𝑖 𝑘0( 𝑥′ 0 − 𝑥0)
(𝑘0)2 − ( 𝑘⃗ ∙ 𝑘⃗ + 𝑀2
+ 𝑖𝜖)
𝛿3
(𝑘⃗ ′ − 𝑘⃗ ) 𝑥′

6. We have on 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
𝑖 𝑘0( 𝑥′ 0
− 𝑥0)
(𝑘0)2 − ( 𝑘⃗ ∙ 𝑘⃗ + 𝑀2
+ 𝑖𝜖)
=
𝑖𝜋
𝜔(𝑘⃗ )
𝑒 𝑖 𝑘0( 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 𝑖 𝑘0( 𝑥′ 0 − 𝑥0)
𝛿3
(𝑘⃗ ′ − 𝑘⃗ ) 𝑥′
This is associated with the Feynman diagram at the space region of 𝑥′.
(Fig.1)

7. 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−𝑖𝑘⃗ ′∙𝑥′ 〈 𝐺 𝑥′ 𝑦′ 𝐽 𝑦′〉 ∫
𝑑3
𝑥
√(2𝜋)3
𝑒 𝑖𝑘⃗ ∙𝑥 〈 𝐺 𝑥𝑦 𝐽 𝑦
〉
As emphasized earlier, the only relevant terms (that are identical) in (13.1) are those that take the
form of the 1st
. They are already independent of the sources J’s so these terms don’t vanish upon
the setting of these sources to zero. That is, they are the only terms that contribute to the
scattering process illustrated in this exercise when sources are made to vanish.
How about the 2nd
and 3rd
major terms in (13.1)?
For the 2nd
major term, since this involves the product of the connected two-point function (7)
and (13.2), in coordinate space, it depicts simultaneously the propagation of the scalarfield
between two end spacetime points y and y’, where the propagator connects the two sources
located at these end spacetime points and the scattering process of the 1st
. These two
simultaneous processes are independent of each other. However, this term is dependent on
sources and vanishes upon the setting of these sources to zero.
Regarding the 3rd
term, in coordinate space we write one major component of it as
(13.5)
∫
𝑑3
𝑥
√(2𝜋)3
𝑒 𝑖𝑘⃗ ∙𝑥 〈 𝐺 𝑥𝑦 𝐽 𝑦
〉 =
√(2𝜋)3
2(2𝜋)2
∫ 𝑑4
𝑦 𝐽(𝑦) ∫
𝑑3
𝑥
(2𝜋)3
𝑒 𝑖𝑘⃗ ∙𝑥
𝐺(𝑦 − 𝑥)
The integration over the spacetime point/region of y involves the source J(y) and given the
Green’s function that acts as a propagator with a causality from spacetime y to spacetime x, the

8. said integration is just a Fourier transformation of the said source from coordinate space to
momentum space thus, obtaining the Fourier component of this source at the spacetime
point/region of y. Meanwhile, the other integration over the spatial region of 𝑥 implies a Fourier
integral definition of a Dirac-delta function 𝛿3
(𝑘⃗ + 𝑘⃗ (1))
𝑥
at the cited spatial region.
In momentum space, (13.5) is given by
(13.6)
∫
𝑑3
𝑥
√(2𝜋)3
𝑒 𝑖𝑘⃗ ∙𝑥 〈 𝐺 𝑥𝑦 𝐽 𝑦
〉 = −
√(2𝜋)3
2(2𝜋)2
∫
𝑑4
𝑘(1) 𝑒− 𝑖 𝑘(1)
0
𝑥0
𝐽̃(−𝑘(1)) 𝑦
−𝑘 𝜇
(1)
𝑘(1)
𝜇
+ 𝑀2
+ 𝑖𝜖
𝛿3
(𝑘⃗ + 𝑘⃗ (1))
𝑥
𝛿3
(𝑘⃗ + 𝑘⃗ (1))
𝑥
= ∫
𝑑3
𝑥
(2𝜋)3
𝑒 𝑖(𝑘⃗ + 𝑘⃗ (1)) ∙ 𝑥
The remaining major component of 3rd
term can be likewise written as
(13.7)
∫
𝑑3
𝑥 ′
√(2𝜋)3
𝑒− 𝑖𝑘⃗ ′∙ 𝑥 ′
〈 𝐺 𝑥 ′ 𝑦 ′ 𝐽 𝑦′
〉 =
√(2𝜋)3
2(2𝜋)2
∫ 𝑑4
𝑦′
𝐽(𝑦′
) ∫
𝑑3
𝑥′
(2𝜋)3
𝑒− 𝑖𝑘⃗ ′ ∙ 𝑥′
𝐺(𝑦′
− 𝑥′
)
The integration over the spacetime point/region of y’ involves the source J(y’) and given the
Green’s function that acts as a propagator with a causality from spacetime y’ to spacetime x’, the
said integration is just a Fourier transformation of the said source from coordinate space to
momentum space thus, obtaining the Fourier component of this source at the spacetime
point/region of y’. Meanwhile, the other integration over the spatial region of 𝑥′
implies a
Fourier integral definition of a Dirac-delta function at the cited spatial region.
In momentum space, (13.7) is given by
(13.8)
∫
𝑑3
𝑥 ′
√(2𝜋)3
𝑒− 𝑖𝑘⃗ ′∙ 𝑥 ′
〈 𝐺 𝑥 ′ 𝑦 ′ 𝐽 𝑦′
〉
= −
√(2𝜋)3
2(2𝜋)2
∫
𝑑4
𝑘 ′(1) 𝑒− 𝑖 𝑘 ′ (1)
0
𝑥 ′
0
𝐽̃(−𝑘 ′ (1)) 𝑦 ′
−𝑘′
𝜇
(1)
𝑘′
(1)
𝜇
+ 𝑀2
+ 𝑖𝜖
𝛿3
(𝑘⃗ ′
− 𝑘⃗ ′ (1))
𝑥′

9. 𝛿3
(𝑘⃗ ′
− 𝑘⃗ ′ (1))
𝑥′ = ∫
𝑑3
𝑥′
(2𝜋)3
𝑒− 𝑖(𝑘⃗ ′− 𝑘⃗ ′
(1)) ∙ 𝑥′
As a whole what does this 3rd
term signify? Taking (13.5) and (13.7) altogether, there are two
separate propagations of the scalar field starting at two different initial spacetime points and
ending up to scatter at two different spatial points. The scalar field propagating in (13.5) starts at
the spacetime point y, then propagates towards the spatial point of 𝑥, carrying the spatial
momentum 𝑘⃗ (1) and then it scatters at this point carrying the spatial momentum 𝑘⃗ . For this said
scalar field, the spatial point of 𝑥 is where the scattering vertex is. At this scattering vertex
spatial momenta 𝑘⃗ and 𝑘⃗ (1) are summed up to zero and in turn implies a picking 𝑘⃗ (1) = −𝑘⃗
over the integration variable 𝑘⃗ (1). Meanwhile, the scalar field propagating in (13.7) starts at the
spacetime point y’, then propagates towards the spatial point of 𝑥′
, carrying the spatial
momentum 𝑘⃗ ′
(1) and then it scatters at this point carrying the spatial momentum 𝑘⃗ ′
. For this said
scalar field, the spatial point of 𝑥′
is where the scattering vertex is. At this scattering vertex
spatial momenta 𝑘⃗ ′
and 𝑘⃗ ′
(1) are summed up to zero and in turn implies a picking 𝑘⃗ ′
(1) = 𝑘⃗ ′
over the integration variable 𝑘⃗ ′
(1). So in view of the 3rd
term, there are two different scattering
vertices (processes), one at the spatial point of 𝑥 and the other one at 𝑥′
and that the scatterings at
these vertices may not be simultaneous, one may happen earlier than the other. However, these
scatterings depend on the presence of their corresponding sources at two different initial
spacetime points.
Basic derivations
In this later portion of the draft let us attempt to dig into the basic details and derivations that
lead us to the form of the scattering matrix (1).
We start with the field operator given for the scalar field (spin 0 boson)
(14.1)
𝜑̂( 𝑥 ) =
1
√𝐿3
∑
√ℏ
√2𝜔(𝑘⃗ )
(𝑒 𝑖𝑘⃗ ⋅ 𝑥
𝑎( 𝑘⃗ ) + 𝑒− 𝑖𝑘⃗ ⋅ 𝑥
𝑎†
( 𝑘⃗ ))
𝑘⃗
In here we shall be reminded that spatial momentum vector 𝑝 is related to the wave number
vector 𝑘⃗ via de Broglie’s hypothesis, 𝑝 = ℏ𝑘⃗ , where |𝑘⃗ | = 2𝜋/𝜆. To make things momentarily
convenient, we resort to the Heaviside units (𝑐 = ℏ = 1 ) so that 𝑝 = 𝑘⃗ and reinsert the
appropriate values of the constants involved whenever required at the end of calculations
although I still retain ℏ in some expressions like in the above. (So for loose convenience we refer
to 𝑘⃗ as spatial momentum.) Also we must take note that in my own personal convenient notation
I usually write the equivalence
(14.2)

10. 1
√𝐿3
∑
𝑘⃗
≡ ∫
𝑑3
𝑘⃗
√(2𝜋)3
We also like to re-emphasize (5.1) here that we can raise a one-particle state |𝑘⃗ ⟩ of spatial
momentum 𝑘⃗ from the vacuum state by the application of the raising operator on the vacuum
state |0 ⟩
| 𝑘⃗ ⟩ = 𝑎†
( 𝑘⃗ )|0 ⟩
with its Hermitian adjoint given by (5.2)
⟨ 𝑘⃗⃗⃗ | = (| 𝑘⃗ ⟩ )
†
= ⟨0|𝑎( 𝑘⃗ )
In here, annihilating a vacuum state means 𝑎(𝑘⃗ )|0 ⟩ = 0. We can use these in the application of
the field operator (14.1) on the vacuum state to obtain
(14.3)
𝜑̂( 𝑥 )|0 ⟩ =
1
√𝐿3
∑
√ℏ
√2𝜔(𝑘⃗ )
𝑒− 𝑖𝑘⃗ ⋅ 𝑥
|𝑘⃗ ⟩
𝑘⃗
also along its Hermitian adjoint
(14.4)
⟨0|𝜑̂( 𝑥 ) = ⟨0| 𝜑̂†( 𝑥 ) =
1
√𝐿3
∑
√ℏ
√2𝜔(𝑘⃗ )
𝑒 𝑖𝑘⃗ ⋅ 𝑥
⟨ 𝑘⃗ |
𝑘⃗
In the above expression, we have a loose definition of a Hermitian field operator that depends on
the spatial coordinates 𝑥
(14.5)
𝜑̂( 𝑥 ) = 𝜑̂†( 𝑥 )
although this is already apparent in (14.1).
Proceeding, we evolve the vacuum state
(15.1)
|0 ⟩ → 𝑈(𝑇)|0 ⟩
and obtain the projection of this state vector on the state vector (14.3) in terms (14.4) to get
(15.2)

11. ⟨0|𝜑̂( 𝑥 )𝑈(𝑇)|0⟩ =
1
√𝐿3
∑
√ℏ
√2𝜔(𝑘⃗ )
𝑒 𝑖𝑘⃗ ⋅ 𝑥
⟨ 𝑘⃗ |𝑈(𝑇)|0⟩
𝑘⃗
From here we take note that the given field operator in the expression does not evolve with time
but we can perform a similarity transformation on this operator so that in the Heisenberg picture
this operator can have time-dependence.
(16.1)
𝜑̂(𝑥) = 𝜑̂(𝑇, 𝑥 ) = 𝑈†(𝑇) 𝜑̂ ( 𝑥 ) 𝑈(𝑇)
from which we also take note of
(16.2)
𝜑̂ ( 𝑥 ) 𝑈(𝑇) = 𝑈(𝑇)𝜑̂(𝑥)
whereby we make use of the following very important unitary property of the time evolution
operator 𝑈(𝑇)
(16.3)
𝑈†(𝑡) 𝑈(𝑡) = 𝑈(𝑡)𝑈†(𝑡) = 𝑈(𝑡, 𝑡) = 1
𝑈(𝑡𝑖, 𝑡𝑗) = 𝑈(𝑡𝑖, 𝑡 𝑘)𝑈(𝑡 𝑘, 𝑡𝑗)
As a consequence from all of these, (15.2) can also be given by
(16.4)
⟨0| 𝑈(𝑇)𝜑̂( 𝑥 )|0⟩ =
1
√𝐿3
∑
√ℏ
√2𝜔(𝑘⃗ )
𝑒 𝑖𝑘⃗ ⋅ 𝑥
⟨ 𝑘⃗ |𝑈(𝑇)|0⟩
𝑘⃗
where on the left-hand-side (lhs) it is already clear that the field operator 𝜑̂ ( 𝑥 ) in the former
expression now becomes a Heisenberg operator 𝜑̂(𝑥) = 𝜑̂(𝑇, 𝑥 ) in this latter expression, given
(16.2). With this latter expression we can proceed to take its inverse Fourier transform
(16.4.1)
∫
𝑑3
𝑥
√(2𝜋)3
𝑒− 𝑖𝑘⃗ ′ ⋅ 𝑥
while we also have my own personal notation for my convenience
1
√𝐿3
=
1
√(2𝜋)3
and also with the
Dirac-delta function
(16.5)

12. 𝛿3
( 𝑘⃗ ′
− 𝑘⃗ ) 𝑥
= ∫
𝑑3
𝑥
√(2𝜋)3 𝐿3
𝑒− 𝑖(𝑘⃗ ′− 𝑘⃗ ) ∙ 𝑥
→ 𝛿 𝑘⃗ ′ 𝑘⃗
3
where in (16.5) the Dirac-delta function can approach a discrete case. So we write
(16.6)
⟨ 𝑘⃗ |𝑈(𝑇)|0⟩ =
√2𝜔(𝑘⃗ )
√ℏ
∫
𝑑3
𝑥
√(2𝜋)3
𝑒−𝑖𝑘⃗ ∙ 𝑥 ⟨0| 𝑈(𝑇)𝜑̂( 𝑥 )|0⟩
The matrix involved in the right-hand side can be handled straightforwardly using path
integration, and we would no longer dig into the details associated with this integration and
simply quote the result here. This is given by
(16.7)
⟨0| 𝑈(𝑇)𝜑̂( 𝑥 )|0⟩ = 𝑖ℏ
𝛿
𝛿𝐽(𝑥)
⟨0| 𝑈(𝑇)|0⟩
noting that 𝑥 = (𝑇; 𝑥). It might be useful to remember that
(16.8)
𝜑̂†(𝑥) = 𝑈†(𝑇)𝜑̂†(𝑥)𝑈(𝑇) = 𝑈†(𝑇)𝜑̂(𝑥 )𝑈(𝑇) = 𝜑̂(𝑥 )
𝜑̂(𝑥 )𝑈†(𝑇) = 𝑈†(𝑇) 𝜑̂(𝑥 )
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