self-interacting scalar field theory

8/7/2019 Self-Interacting Scalar Field Theory

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Self-Interacting Scalar Field Theories

1



I. CANONICAL QUANTIZATION

The action for the self-interacting scalar field or ϕ4 theory is given by

A[ϕ(x)] =

d4

x L =

d4

x1

2 [∂ µ

ϕ(x)∂ µϕ(x) − m20ϕ

2

(x)] −λ0

4! ϕ4

(x)

. (1)

We will consider the interaction to be repulsive, i.e., the coupling describing the strength of

the interaction λ0 > 0. m0 represents the bare mass, i.e., the mass of a particle if there were

no interaction. Interactions change m0. Physically, we can never measure m0. This is true

even for isolated particles that are not interacting with other particles, since they may still

interact with themselves. There is no way of turning off this self-interaction. Once we have

added interactions, we can no longer give simple physical interpretations to the parameters

appearing in the action.

From the action we find the field momentum conjugate to ϕ(x),

π(x) =∂ L

∂ (∂ tϕ(x))= ∂ tϕ(x). (2)

The field Hamiltonian is

H =

d3x(π∂ tϕ − L)

=

d3x

12

[π2(x) + ϕ(x) · ϕ(x) + m20ϕ2(x)] + λ0

4!ϕ4(x)

. (3)

We may canonically quantize the ϕ4 theory in the same way as the free fields. Canonical

quantization implies that the equal-time commutators are

[ϕ(x, t), π(y, t)] = iδ3(x − y), (4)

[ϕ(x, t), ϕ(y, t)] = 0 = [π(x, t), π(y, t)].

The Hamiltonian (3) and commutators (4) reproduce Eq.(2) and the classical field equation,

∂ µ∂ µϕ(x) + m20ϕ(x) +

λ0

3!ϕ3(x) = 0, (5)

as operator equations of motion.

∂ tϕ(x) = i[H, ϕ(x)] =i

2

d3y[π2(y, t), ϕ(x, t)] = π(x) (6)

2



∂ 2t ϕ(x) = ∂ tπ(x) = i[H, π(x)]

= i

d3y[1

2ϕ(y, t) · ϕ(y, t) +

1

2m2

0ϕ2(y, t) +λ0

4!ϕ4(y, t), π(x, t)]

= 2ϕ(x) − m20ϕ(x) − λ0

3!ϕ3(x) (7)

However, we cannot exactly solve the nonlinear equation of motion - we no longer have a

simple particle interpretation. ϕ(x) will no longer just create single particles of a definite

position. To make any progress, we must make a few assumptions about the states.

1. There is a Lorentz invariant vacuum |0 that has zero energy (with respect to the

interacting field theory Hamiltonian), and is translationally invariant (no momentum).

2. There are no bound states.

Each one of these assumptions is true for free field theory. By making these assumptions for

the interating field theory, we are implicitly saying that the coupling or interactions are not

strong enough to radically change the spectrum of states from the noninteracting case. The

value of making these assumptions is that we can borrow the particle interpretation from

the free field theory and build a Fock space.

II. IN AND OUT FIELDS

In high energy experiments, two particles are accelerated towards each other, they collide

and interact, and the particles resulting from the collision fly away from the small region

of interaction. Before collision, when the particles are far from each other and the region

of interaction, they act like free particles. After the collision is over, the particles leaving

the interaction region again act like free particles once they get far away from the region

of interaction. In an experiment, the cross section for such a process is measured. We

should therefore compute cross sections using our interacting field theories to compare with

experiments.

The cross section for a process is directly related to the probability for the process to

occur, i.e., the square of the transition amplitude A for the process to occur. The transition

amplitude for a process that starts in initial state |αIN and ends up in a final state |β OUTis the overlap between the two states,

A =

β OUT

|αIN

. (8)

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In an accelerator, we prepare the |αIN state and we measure the kinematics of the |β OUTstate. Since the particles are essentially free in the |αIN and |β OUT states, we know how to

write these states down using the corresponding free field theory where we have a particle

interpretation. For the ϕ4 theory, this is the free scalar field.

If the collision occurs at time t = 0, then the initial state long before the collision,

t → −∞, is a state describing free particles. An initial n-particle state at t → −∞ is

|αIN = |k1 · · · kn; IN = a†IN(k1) · · · a†IN(kn)|0, (9)

where aIN(k)|0 = 0 and both aIN(k) and a†IN(k) can be found in the plane wave expansion

of the free scalar field operator

ϕIN(x) = d3 k

(2π)31

2ωk [e−ik·x

aIN(k) + eik·x

a†

IN(k)]. (10)

ϕIN(x) creates free particles of physical (experimentally measured) mass m. It satisfies

(∂ µ∂ µ + m2)ϕIN(x) = 0. (11)

Long after the collision, t → ∞, we again have free particles propagating at their physical

mass m. These particles are created and destroyed by the free scalar field operator ϕOUT(x)

that satisfies

(∂ µ

∂ µ + m2

)ϕOUT(x) = 0. (12)

A final m-particle state at t → ∞ is

|β OUT = | p1 · · · pm;OUT = a†OUT( p1) · · · a†OUT( pm)|0, (13)

where aOUT(k)|0 = 0 and both aOUT(k) and a†OUT(k) can be found in the plane wave

expansion of the free scalar field operator

ϕOUT(x) = d3 k

(2π)3

1

2ωk

[e−ik·xaOUT(k) + eik·xa†OUT(k)]. (14)

The states |αIN and |β OUT are to be properly normalized, but we will not show the

smearing test function explicitly since it is not crucial to the computation of the S -matrix.

III. THE S -MATRIX

There is really no difference between the IN and OUT states. Both the IN field theory

and the OUT field theory describe the same free scalar particle of mass m. Their spectra

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are identical. This means that the Fock space built from application of a†IN(k) to |0 should

be isomorphic to that built from a†OUT(k) on |0. That is, there should be a mapping

S |αOUT = |αIN. (15)

In terms of the field operators,

ϕOUT(x) = S −1ϕIN(x)S. (16)

S is Lorentz invariant since ϕIN(x) and ϕOUT(x) are Lorentz scalars.

It follows from the definition of S , Eq.(15), that S is unitary.

β OUT|S †S |αOUT = β IN|αIN = δαβ . (17)

The matrix elements of S are the transition amplitudes for scattering processes.

(S †|β IN)†|αIN = β IN|S |αIN= β OUT|αIN= β OUT|S |αOUT= S βα. (18)

Therefore, in order to calculatecross sections

, we need to compute the S -matrix elements.S naturally depends on the interacting field theory, which determines the details of the

reactions that go on inside the interacting region. The LSZ reduction formula relates the

S -matrix to vacuum expectation values of time-ordered products of the interacting field

operators.

Since the vacuum can’t scatter off itself, S takes the vacuum into itself up to a phase:

0|S |0 = 0|0 = 1 (19)

Here, we set the unmeasureable phase to zero.

Likewise, a single stable particle has nothing to scatter with. So,

0|aOUT(k)a†IN(k)|0 = 0|S †aIN(k)S a†IN(k)|0= 0|aIN(k)S a†IN(k)|0= 0|aIN(k)a†IN(k)|0= (2π)32ωkδ3( k

− k) (20)

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IV. THE LSZ REDUCTION FORMULA

Consider an arbitrary element of the S -matrix S βα where the IN state

|αIN = |k1 · · · kn; IN,

contains n particles with momenta kµ1 , · · · , kµ

n, and the OUT state

|β OUT = | p1 · · · pm;OUT,

contains m particles with momenta pµ1 , · · · , pµ

m. Therefore,

S βα = p1 · · · pm;OUT|k1 · · · kn; IN (21)

The idea behind deriving the reduction formula is to first pull out 1 particle at a time

from the IN and OUT states by writing its creation operator explicitly.

S βα = p1 · · · pm;OUT|a†IN(k1)|0k2 · · · kn; IN= p1 · · · pm;OUT|a†OUT(k1)|0k2 · · · kn; IN

+ p1 · · · pm;OUT|[a†IN(k1) − a†OUT(k1)]|0k2 · · · kn; IN (22)

The first term will be zero unless one of the pi’s is equal to k1. If one does match, this term

represents a disconnected process. Suppose p1 = k1, then

p1 · · · pm;OUT|a†OUT(k1)|0k2 · · · kn; IN= 0 p2 · · · pm;OUT|aOUT( p1)a†OUT(k1)|0k2 · · · kn; IN= (2π)32ωk1δ3( p1 − k1) p2 · · · pm;OUT|k2 · · · kn; IN (23)

A process that is not disconnected is connected . Clearly, only the connected terms are

nontrivial and we only need to calculate S βα for connected process.

S cβα = p1 · · · pm;OUT|[a†IN(k1) − a†OUT(k1)]|0k2 · · · kn; IN (24)

Next, we express the creation operators in terms of the IN and OUT field operators.

Recall that for the free scalar field,

ϕ(x) = d3 k

(2π)31

2ωk

[e−ik·xa(k) + eik·xa†(k)] ≡ ϕ(+)(x) + ϕ(−)(x),

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π(x) = ∂ tϕ(x) = −i d3 k

(2π)31

2ωk

ωk[e−ik·xa(k) − eik·xa†(k)],

and

a†

(k) =

d3

xe−ik·x

[ωkϕ(x) − iπ(x)]

=

d3x[(i∂ te−ik·x)ϕ(x) − ie−ik·x∂ tϕ(x)]

= −i

d3xe−ik·x∂ tϕ(x), (25)

which is independent of time. Here,

a∂ tb ≡ −(∂ ta)b + a(∂ tb). (26)

Therefore,

S cβα = i

d3x1e−ik1·x1∂ t p1 · · · pm;OUT|[ϕOUT(x1) − ϕIN(x1)]|0k2 · · · kn; IN

=i√Z

( limx01→∞

− limx01→−∞

)

d3x1e−ik1·x1∂ t p1 · · · pm;OUT|ϕ(x1)|0k2 · · · kn; IN

=i√Z

d4x1∂ t

e−ik1·x1∂ t p1 · · · pm;OUT|ϕ(x1)|0k2 · · · kn; IN

=i√Z

d4x1

−∂ 2t e−ik1·x1 p1 · · · pm;OUT|ϕ(x1)|0k2 · · · kn; IN

+i

√Z d4x1 e−ik1·x1∂ 2t

p1

· · · pm;OUT

|ϕ(x1)

|0k2

· · ·kn; IN

=

i√Z

d4x1

(−2 + m2)e−ik1·x1 p1 · · · pm;OUT|ϕ(x1)|0k2 · · · kn; IN

+i√Z

d4x1

e−ik1·x1∂ 2t p1 · · · pm;OUT|ϕ(x1)|0k2 · · · kn; IN

=i√Z

d4x1

e−ik1·x1(x1 + m2) p1 · · · pm;OUT|ϕ(x1)|0k2 · · · kn; IN

, (27)

where ≡ ∂ 2t −2. The factor of 1/√

Z is a normalization to indicate that ϕ(x) does more

than just create or destroy single particles. Explicitly,

αIN|ϕIN(x)|0 = 0 (28)

unless |αIN = |1 is a 1-particle IN state, whereas

αIN|ϕ(x)|0 = 0 (29)

for any n-particle IN state |αIN = |n, with n = 1, 2, 3, · · ·. Z is determined by

1

|ϕ(x)

|0

=

√Z

1

|ϕIN(x)

|0

. (30)

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Hence, we expect Z to be between 0 and 1.

We note that one cannot implement

limt→−∞

ϕ(x) =√

Z ϕIN(x),

limt→∞

ϕ(x) = √Z ϕOUT(x), (31)

as operator equations (see Sec.VII). In the above we can use the asymptotic conditions

because ϕIN(x) and ϕOUT(x) are sandwiched between normalizable states.

To continue, we pull out another particle - this time one from the OUT state.

p1 · · · pm;OUT|ϕ(x1)|0k2 · · · kn; IN=

0 p2

· · · pm;OUT

|aOUT( p1)ϕ(x1)

|0k2

· · ·kn; IN

= 0 p2 · · · pm;OUT|ϕ(x1)aIN( p1)|0k2 · · · kn; IN

+0 p2 · · · pm;OUT|[aOUT( p1)ϕ(x1) − ϕ(x1)aIN( p1)]|0k2 · · · kn; IN (32)

And,

0 p2 · · · pm;OUT|[aOUT( p1)ϕ(x1) − ϕ(x1)aIN( p1)]|0k2 · · · kn; IN= i

d3y1eip1·y1∂ t0 p2 · · · pm;OUT|ϕOUT(y1)ϕ(x1) − ϕ(x1)ϕIN(y1)|0k2 · · · kn; IN

= i√Z

( limy01→∞

− limy01→−∞

)

d3y1eip1·y1∂ t0 p2 · · · pm;OUT|T [ϕ(x1)ϕ(y1)]|0k2 · · · kn; IN

=i√Z

d4y1eip1·y1(y1 + m2)0 p2 · · · pm;OUT|T [ϕ(x1)ϕ(y1)]|0k2 · · · kn; IN (33)

Here, T is the time-ordering operator. Therefore,

S cβα =

i√Z

2 d4x1d4y1e−ik1·x1eip1·y1(x1 + m2)(y1 + m2)

×0 p2

· · · pm;OUT

|T [ϕ(x1)ϕ(y1)]

|0k2

· · ·kn; IN

(34)

We continue this process of emptying the IN and OUT states until we are left with the

vacuum. The full reduction formula is

S cβα =

i√Z

(m+n) d4x1 · · · d4xnd4y1 · · · d4yme−ik1·x1 · · · e−ikn·xneip1·y1 · · · eipm·ym

×(x1 + m2) · · · (xn + m2)(y1 + m2) · · · (ym + m2)

×0|T [ϕ(x1) · · · ϕ(xn)ϕ(y1) · · · ϕ(ym)]|0 (35)

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V. PERTURBATION THEORY

Consider the n-point function

G(x1, · · · , xn)

≡ 0|T [ϕ(x1) · · · ϕ(xn)]|0= 0|T [U †(t1)ϕIN(x1)U (t1)U †(t2)ϕIN(x2)U (t2) · · · U †(tn)ϕIN(xn)U (tn)]|0≡ 0|T [U †(t1)ϕIN(x1)U (t1, t2)ϕIN(x2)U (t2, t3) · · · U (tn−1, tn)ϕIN(xn)U (tn)]|0, (36)

We cannot exactly compute G(x, · · · , xn) because we do not know the vacuum expectation

values of the interacting field operators. However, we do know the vacuum expectation

values for the IN field operators. So, we can perturbatively compute G(x1, · · · , xn) in termsof those. Here, the unitary operator U (t) is formally defined by

ϕ(x, t) ≡ U †(t)ϕIN(x, t)U (t). (37)

In view of the observation that Eq.(31) cannot be implemented as operator equations, we

know that the same must be true for Eq.(37). It is only meant to make any sense when sand-

wiched between normalizable states. Since U (t) below always appears sandwiched between

normalizable states, it follows from Eq.(31) that

ϕ(x, −∞) =√

Z U (−∞)ϕ(x, −∞)U †(−∞) (38)

and hence

U (−∞) ∝ I . (39)

Together with Eq.(16), we have

ϕOUT(x) =ˆS

†

ϕIN(x)ˆS =

ˆS

† ˆU (t)ϕ(x)

ˆU

†

(t)ˆS, (40)

which in the limit t → ∞,

ϕ(x, ∞) =√

Z ϕOUT(x, ∞) =√

Z S †U (∞)ϕ(x, ∞)U †(∞)S (41)

Therefore,

S = U (∞)U †(−∞) (42)

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To find what is U (t) exactly, consider

∂ tϕIN(x, t)

= ∂ t[U (t)ϕ(x, t)U †(t)]

= ∂ tU (t)ϕ(x, t)U †(t) + U (t)∂ tϕ(x, t)U †(t) + U (t)ϕ(x, t)∂ tU †(t)

= ∂ tU (t)U †(t)U (t)ϕ(x, t)U †(t) + U (t)∂ tϕ(x, t)U †(t) + U (t)ϕ(x, t)U †(t)U (t)∂ tU †(t)

= ∂ tU (t)U †(t)ϕIN(x, t) + U (t)∂ tϕ(x, t)U †(t) − ϕIN(x, t)∂ tU (t)U †(t)

= [∂ tU (t)U †(t), ϕIN(x, t)] + U (t)∂ tϕ(x, t)U †(t)

= [∂ tU (t)U †(t), ϕIN(x, t)] + iU (t)[H (ϕ, π), ϕ(x, t)]U †(t)

= [∂ tU (t)U †(t), ϕIN(x, t)] + i[U (t)H (ϕ, π)U †(t), ϕIN(x, t)]

= [∂ tU (t)U †(t), ϕIN(x, t)] + i[H (ϕIN, πIN), ϕIN(x, t)]

= i[H IN(ϕIN, πIN), ϕIN(x, t)] (43)

since

U (t)H (ϕ, π)U †(t)

=

d3xU (t)[1

2π2(x) +

1

2ϕ(x) · ϕ(x) +

1

2m2

0ϕ2(x) +λ0

4!ϕ4(x)]U †(t)

=

d

3

x[

1

2 π

2

IN(x) +

1

2ϕIN(x) · ϕIN(x) +

1

2m

2

0ϕ

2

IN(x) +

λ0

4! ϕ

4

IN(x)]= H (ϕIN, πIN) (44)

Here,

H IN(ϕIN, πIN) =1

2

d3x[π2

IN(x) + ϕIN(x) · ϕIN(x) + m2ϕ2IN(x)] (45)

Therefore, U (t) satisfies

[∂ tU (t)U †(t) + iH I (ϕIN, πIN), ϕIN(x, t)] = 0, (46)

where the interacting part of the Hamiltonian

H I (ϕIN, πIN) ≡ H (ϕIN, πIN) − H IN(ϕIN, πIN)

=

d3x1

2(m2 − m2

0)ϕ2IN(x) +

λ0

4!ϕ4IN(x) (47)

If we had started with πIN(x, t) = U (t)π(x, t)U †(t) and followed the same steps, we would

find that

[∂ tU (t)U †(t) + iH I (ϕIN, πIN), πIN(x, t)] = 0. (48)

10



Therefore, ∂ tU (t)U †(t) + iH I (ϕIN, πIN) must be an ordinary number and not an operator.

∂ tU (t)U †(t) + iH I (ϕIN, πIN) = if (t) (49)

f (t) is an arbitrary function that will eventually be absorbed in the normalization.Define

H f I (t) ≡ H I (ϕIN, πIN) − f (t). (50)

Then,

∂ tU (t) = −iH f I (t)U (t) (51)

or

U (t) = U (t)

−i

t

t

H f I (t1)U (t1)dt1 (52)

It follows that

U (t) = U (t) − i tt

dt1H f I (t1)

U (t) − i t1t

dt2H f I (t2)U (t2)

=

1 − i tt

dt1H f I (t1)

U (t) + (−i)2 tt

dt1

t1t

dt2H f I (t1)H f I (t2)U (t2)

=

1 − i tt

dt1H f I (t1)

U (t)

+(

−i)2

t

tdt1

t1

tdt2H f I (t1)H f I (t2) U (t)

−i

t2

tdt3H f I (t3)U (t3)

=

1 − i tt

dt1H f I (t1) + (−i)2 tt

dt1

t1t

dt2H f I (t1)H f I (t2)

U (t)

+(−i)3 tt

dt1

t1t

dt2

t2t

dt3H f I (t1)H f I (t2)H f I (t3)U (t3)

=

1 +

∞n=1

(−i)n tt

dt1

t1t

dt2 · · · tn−1t

dtnH f I (t1)H f I (t2) · · · H f I (tn)

U (t)

=

1 +

∞n=1

(−i)n

n!

tt

dt1

tt

dt2 · · · tt

dtnT [H f I (t1)H f I (t2) · · · H f I (tn)]

U (t)

≡T exp −i

t

tdtH f I (t) U (t) (53)

Here, we have assumed that the series converges to U (t).

It is conventional to define

U (t, t) ≡ U (t)U †(t) = T exp−i

tt

dtH f I (t)

, (54)

and we note that U (t, t) = 1.

11



Returning to the n-point function

G(x1, · · · , xn)

=

0

|T [U †(t1)ϕIN(x1)U (t1, t2)ϕIN(x2)U (t2, t3)

· · ·U (tn−1, tn)ϕIN(xn)U (tn)]

|0

= 0|T [U †(t)U (t, t1)ϕIN(x1) · · · ϕIN(xn)U (tn, −t)U (−t)]|0= 0|U †(t)T [ϕIN(x1) · · · ϕIN(xn)U (t, t1) · · · U (tn, −t)]U (−t)|0= 0|U †(t)T [ϕIN(x1) · · · ϕIN(xn)U (t, −t)]U (−t)|0 (55)

Here, we take t large enough so that t > t1 and −t < tn. With this large t, we can pull U †(t)

and U (−t) out of the time-ordering. Therefore,

G(x1, · · · , xn) = 0|U †

(t)T [ϕIN(x1) · · · ϕIN(xn)exp−i

t−t dtH

f

I (t)

]U (−t)|0 (56)

To guarantee that we may pull U †(t) and U (−t) out from the time-ordering operator, we will

only consider the limit as t → ∞ in the above equation. Even though we do not explicitly

show it, H f I (t) is normal ordered.

To proceed, we must determine the effect of U †(t) and U (−t) on the vacuum as t → ∞.

It follows from Eq.(39) that

U (

−∞)

|0

= α

|0

. (57)

Next, the S -matrix maps the vacuum into itself up to a phase. So, according to Eq.(42)

U (∞)U †(−∞)|0 = S |0 = β |0. (58)

Therefore,

U (∞)|0 = βα|0. (59)

And,

G(x1, · · · , xn) = β ∗

α∗

α0|T [ϕIN(x1) · · · ϕIN(xn)exp−i

∞−∞ dtH

f I (t)

]|0 (60)

To determine β ∗α∗α, consider removing all of the incoming particles by setting ϕIN(xi) =

1, which means setting ϕ(xi) = 1 by Eq.(37). Then,

1 = β ∗α∗α0|T [exp−i

∞−∞

dtH f I (t)

]|0. (61)

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Therefore,

G(x1, · · · , xn)

=0|T [ϕIN(x1) · · · ϕIN(xn)exp

−i

∞−∞ dtH f I (t)

]|0

0|T [exp−i

∞−∞ dtH f I (t)

]|0

=0|T [ϕIN(x1) · · · ϕIN(xn)exp

−i

∞−∞ dtH I (t)

]|0

0|T [exp−i

∞−∞ dtH I (t)

]|0

=

0|T [ϕIN(x1) · · · ϕIN(xn)]|0+∞m=1

(−i)m

m!

∞−∞ d4y1 · · · ∞−∞ d4ym0|T [ϕIN(x1) · · · ϕIN(xn)HI (y1) · · · HI (ym)]|0

1 +∞

m=1(−i)m

m!

∞−∞ d4y1 · · · ∞−∞ d4ym0|T [HI (y1) · · · HI (ym)]|0 ,(62)

where ˆ

HI (y) is the normal ordered Hamiltonian density.

H I (t) =

d3xHI (x)

A. Wick’s Theorem

In order to compute any term in the expansion (62), we must calculate the vacuum

expectation of the time-ordered product of IN field operators. First, consider the product

of two IN field operators.

ϕIN(x1)ϕIN(x2)

= [ϕ(+)IN (x1) + ϕ

(−)IN (x1)][ϕ

(+)IN (x2) + ϕ

(−)IN (x2)]

= ϕ(+)IN (x1)ϕ

(+)IN (x2) + ϕ

(+)IN (x1)ϕ

(−)IN (x2) + ϕ

(−)IN (x1)ϕ

(+)IN (x2) + ϕ

(−)IN (x1)ϕ

(−)IN (x2)

= ϕ(+)IN (x1)ϕ(+)

IN (x2) + ϕ(−)IN (x2)ϕ(+)

IN (x1) + ϕ(−)IN (x1)ϕ(+)

IN (x2) + ϕ(−)IN (x1)ϕ(−)

IN (x2)

+ϕ(+)IN (x1)ϕ(−)

IN (x2) − ϕ(−)IN (x2)ϕ(+)

IN (x1)

= : ϕIN(x1)ϕIN(x2) : +[ϕ(+)IN (x1), ϕ(−)IN (x2)] (63)

Here,

ϕ(+)IN (x) ≡

d3 k

(2π)31

2ωk

e−ik·xaIN(k),

ϕ(−)IN (x) ≡

d3 k

(2π)31

2ωk

eik·xa†IN(k). (64)

13



It follows that

[ϕ(+)IN (x1), ϕ(−)

IN (x2)] = d3 k

(2π)31

2ωk

d3 k

(2π)31

2ωke−ik·x1+ik·x2 [aIN(k), a†IN(k)]

= d3 k

(2π)31

2ωk e−ik·(x1−x2)

(65)

Therefore,

T [ϕIN(x1)ϕIN(x2)] = : ϕIN(x1)ϕIN(x2) :

+ d3 k

(2π)31

2ωk

[e−ik·(x1−x2)θ(t1 − t2) + eik·(x1−x2)θ(t2 − t1)]

= : ϕIN(x1)ϕIN(x2) : +i∆F (x1 − x2) (66)

And, the contraction of ϕIN(x1) with ϕIN(x2),

0|T [ϕIN(x1)ϕIN(x2)]|0 = i∆F (x1 − x2) (67)

Next, consider the product of three IN field operators.

ϕIN(x1)ϕIN(x2)ϕIN(x3)

= : ϕIN(x1)ϕIN(x2) : ϕIN(x3) + [ϕ(+)IN (x1), ϕ

(−)IN (x2)]ϕIN(x3)

= : ϕIN(x1)ϕIN(x2) : [ϕ(+)IN (x3) + ϕ(−)

IN (x3)] + [ϕ(+)IN (x1), ϕ(−)

IN (x2)]ϕIN(x3)

= : ϕIN(x1)ϕIN(x2)ϕIN(x3) : +[ϕ(+)IN (x1), ϕ(−)

IN (x2)]ϕIN(x3)

+[ϕ(+)IN (x2), ϕ

(−)IN (x3)]ϕIN(x1) + [ϕ

(+)IN (x1), ϕ

(−)IN (x3)]ϕIN(x2) (68)

since

: ϕIN(x1)ϕIN(x2) : ϕ(−)IN (x3)

= [ϕ(+)IN (x1)ϕ

(+)IN (x2) + ϕ

(−)IN (x2)ϕ

(+)IN (x1) + ϕ

(−)IN (x1)ϕ

(+)IN (x2) + ϕ

(−)IN (x1)ϕ

(−)IN (x2)]ϕ

(−)IN (x3)

= ϕ

(+)

IN (x1)[ϕ

(−)

IN (x3)ϕ

(+)

IN (x2) + [ϕ

(+)

IN (x2), ϕ

(−)

IN (x3)]] + ϕ

(−)

IN (x1)ϕ

(−)

IN (x2)ϕ

(−)

IN (x3)+ϕ

(−)IN (x2)ϕ

(−)IN (x3)ϕ

(+)IN (x1) + ϕ

(−)IN (x2)[ϕ

(+)IN (x1), ϕ

(−)IN (x3)]

+ϕ(−)IN (x1)ϕ(−)

IN (x3)ϕ(+)IN (x2) + ϕ(−)

IN (x1)[ϕ(+)IN (x2), ϕ(−)

IN (x3)]

= ϕ(−)IN (x3)ϕ

(+)IN (x1)ϕ

(+)IN (x2) + [ϕ

(+)IN (x1), ϕ

(−)IN (x3)]ϕ

(+)IN (x2)

+[ϕ(+)IN (x2), ϕ

(−)IN (x3)]ϕ

(+)IN (x1) + ϕ

(−)IN (x1)ϕ

(−)IN (x2)ϕ

(−)IN (x3)

+ϕ(−)IN (x2)ϕ

(−)IN (x3)ϕ

(+)IN (x1) + [ϕ

(+)IN (x1), ϕ

(−)IN (x3)]ϕ

(−)IN (x2)

+ϕ(−)IN (x1)ϕ(−)

IN (x3)ϕ(+)IN (x2) + [ϕ(+)

IN (x2), ϕ(−)IN (x3)]ϕ(−)

IN (x1) (69)

14



Therefore,

T [ϕIN(x1)ϕIN(x2)ϕIN(x3)]

= : ϕIN(x1)ϕIN(x2)ϕIN(x3) :

+i∆F (x1 − x2)ϕIN(x3) + i∆F (x2 − x3)ϕIN(x1) + i∆F (x1 − x3)ϕIN(x2) (70)

and

0|T [ϕIN(x1)ϕIN(x2)ϕIN(x3)]|0 = 0 (71)

Last, consider the product of four IN field operators.

ϕIN(x1)ϕIN(x2)ϕIN(x3)ϕIN(x4)

= : ϕIN(x1)ϕIN(x2)ϕIN(x3) : ϕIN(x4) + [ϕ

(+)

IN (x1), ϕ

(−)

IN (x2)]ϕIN(x3)ϕIN(x4)+[ϕ

(+)IN (x2), ϕ

(−)IN (x3)]ϕIN(x1)ϕIN(x4) + [ϕ

(+)IN (x1), ϕ

(−)IN (x3)]ϕIN(x2)ϕIN(x4)

= : ϕIN(x1)ϕIN(x2)ϕIN(x3)ϕIN(x4) :

+[ϕ(+)IN (x1), ϕ

(−)IN (x2)] : ϕIN(x3)ϕIN(x4) : +[ϕ

(+)IN (x1), ϕ

(−)IN (x3)] : ϕIN(x2)ϕIN(x4) :

+[ϕ(+)IN (x1), ϕ

(−)IN (x4)] : ϕIN(x2)ϕIN(x3) : +[ϕ

(+)IN (x2), ϕ

(−)IN (x3)] : ϕIN(x1)ϕIN(x4) :

+[ϕ(+)IN (x2), ϕ

(−)IN (x4)] : ϕIN(x1)ϕIN(x3) : +[ϕ

(+)IN (x3), ϕ

(−)IN (x4)] : ϕIN(x1)ϕIN(x2) :

+[ϕ(+)IN (x1), ϕ(−)

IN (x2)][ϕ(+)IN (x3), ϕ(−)

IN (x4)] + [ϕ(+)IN (x2), ϕ(−)

IN (x3)][ϕ(+)IN (x1), ϕ(−)

IN (x4)]

+[ϕ(+)IN (x1), ϕ

(−)IN (x3)][ϕ

(+)IN (x2), ϕ

(−)IN (x4)] (72)

since

: ϕIN(x1)ϕIN(x2)ϕIN(x3) :

= : [ϕ(+)IN (x1) + ϕ

(−)IN (x1)][ϕ

(+)IN (x2) + ϕ

(−)IN (x2)][ϕ

(+)IN (x3) + ϕ

(−)IN (x3)] :

= ϕ(+)IN (x1)ϕ

(+)IN (x2)ϕ

(+)IN (x3) + ϕ

(−)IN (x1)ϕ

(−)IN (x2)ϕ

(−)IN (x3)

+ϕ(−)IN (x1)ϕ(+)IN (x2)ϕ(+)IN (x3) + ϕ(−)IN (x2)ϕ(+)IN (x3)ϕ(+)IN (x1)

+ϕ(−)IN (x3)ϕ

(+)IN (x1)ϕ

(+)IN (x2) + ϕ

(−)IN (x1)ϕ

(−)IN (x2)ϕ

(+)IN (x3)

+ϕ(−)IN (x2)ϕ

(−)IN (x3)ϕ

(+)IN (x1) + ϕ

(−)IN (x3)ϕ

(−)IN (x1)ϕ

(+)IN (x2) (73)

and

: ϕIN(x1)ϕIN(x2)ϕIN(x3) : ϕ(−)IN (x4)

= ϕ(+)IN (x1)ϕ(+)

IN (x2)ϕ(+)IN (x3)ϕ(−)

IN (x4) + ϕ(−)IN (x1)ϕ(−)

IN (x2)ϕ(−)IN (x3)ϕ(−)

IN (x4)

15



+ϕ(−)IN (x1)ϕ(+)

IN (x2)ϕ(+)IN (x3)ϕ(−)

IN (x4) + ϕ(−)IN (x2)ϕ(+)

IN (x3)ϕ(+)IN (x1)ϕ(−)

IN (x4)

+ϕ(−)IN (x3)ϕ

(+)IN (x1)ϕ

(+)IN (x2)ϕ

(−)IN (x4) + ϕ

(−)IN (x1)ϕ

(−)IN (x2)ϕ

(+)IN (x3)ϕ

(−)IN (x4)

+ϕ(−)IN (x2)ϕ

(−)IN (x3)ϕ

(+)IN (x1)ϕ

(−)IN (x4) + ϕ

(−)IN (x3)ϕ

(−)IN (x1)ϕ

(+)IN (x2)ϕ

(−)IN (x4)

= ϕ(−)IN (x4)ϕ(+)

IN (x1)ϕ(+)IN (x2)ϕ(+)

IN (x3) + ϕ(−)IN (x1)ϕ(−)

IN (x2)ϕ(−)IN (x3)ϕ(−)

IN (x4)

+ϕ(+)IN (x1)ϕ(+)

IN (x2)[ϕ(+)IN (x3), ϕ(−)

IN (x4)] + ϕ(+)IN (x1)ϕ(+)

IN (x3)[ϕ(+)IN (x2), ϕ(−)

IN (x4)]

+ϕ(+)IN (x2)ϕ(+)

IN (x3)[ϕ(+)IN (x1), ϕ(−)

IN (x4)] + ϕ(−)IN (x1)ϕ(−)

IN (x4)ϕ(+)IN (x2)ϕ(+)

IN (x3)

+ϕ(−)IN (x1)ϕ

(+)IN (x2)[ϕ

(+)IN (x3), ϕ

(−)IN (x4)] + ϕ

(−)IN (x1)ϕ

(+)IN (x3)[ϕ

(+)IN (x2), ϕ

(−)IN (x4)]

+ϕ(−)IN (x2)ϕ

(−)IN (x4)ϕ

(+)IN (x3)ϕ

(+)IN (x1) + ϕ

(−)IN (x2)ϕ

(+)IN (x3)[ϕ

(+)IN (x1), ϕ

(−)IN (x4)]

+ϕ(−)IN (x2)ϕ(+)

IN (x1)[ϕ(+)IN (x3), ϕ(−)

IN (x4)] + ϕ(−)IN (x3)ϕ(−)

IN (x4)ϕ(+)IN (x1)ϕ(+)

IN (x2)

+ϕ(−)IN (x3)ϕ

(+)IN (x1)[ϕ

(+)IN (x2), ϕ

(−)IN (x4)] + ϕ

(−)IN (x3)ϕ

(+)IN (x2)[ϕ

(+)IN (x1), ϕ

(−)IN (x4)]

+ϕ(−)IN (x1)ϕ

(−)IN (x2)ϕ

(−)IN (x4)ϕ

(+)IN (x3) + ϕ

(−)IN (x1)ϕ

(−)IN (x2)[ϕ

(+)IN (x3), ϕ

(−)IN (x4)]

+ϕ(−)IN (x2)ϕ

(−)IN (x3)ϕ

(−)IN (x4)ϕ

(+)IN (x1) + ϕ

(−)IN (x2)ϕ

(−)IN (x3)[ϕ

(+)IN (x1), ϕ

(−)IN (x4)]

+ϕ(−)IN (x3)ϕ

(−)IN (x1)ϕ

(−)IN (x4)ϕ

(+)IN (x2) + ϕ

(−)IN (x3)ϕ

(−)IN (x1)[ϕ

(+)IN (x2), ϕ

(−)IN (x4)] (74)

Therefore,

T [ϕIN(x1)ϕIN(x2)ϕIN(x3)ϕIN(x4)]

= : ϕIN(x1)ϕIN(x2)ϕIN(x3)ϕIN(x4) : +σ

: ϕIN(xσ(1))ϕIN(xσ(2)) : i∆F (xσ(3) − xσ(4))

+i2[∆F (x1 − x2)∆F (x3 − x4) + ∆F (x1 − x3)∆F (x2 − x4) + ∆F (x1 − x4)∆F (x2 − x3)]

(75)

and

0|T [ϕIN(x1)ϕIN(x2)ϕIN(x3)ϕIN(x4)]|0

= i2[∆F (x1 − x2)∆F (x3 − x4) + ∆F (x1 − x3)∆F (x2 − x4) + ∆F (x1 − x4)∆F (x2 − x3)]

(76)

The sum in Eq.(75) is carried out over all permutations σ of {1, 2, 3, 4} where σ(1) < σ(2)

and σ(3) < σ(4).

16



In general, we have Wick’s theorem .

T [ϕIN(x1) · · · ϕIN(xn)]

= : ϕIN(x1)

· · ·ϕIN(xn) : +

σ

i∆F (xσ(1)

−xσ(2)) : ϕIN(xσ(3))

· · ·ϕIN(xσ(n)) :

+σ

i∆F (xσ(1) − xσ(2))i∆F (xσ(3) − xσ(4)) : ϕIN(xσ(5)) · · · ϕIN(xσ(n)) : + · · ·

+

σ i∆F (xσ(1) − xσ(2)) · · · i∆F (xσ(n−2) − xσ(n−1))ϕIN(xσ(n)) n oddσ i∆F (xσ(1) − xσ(2)) · · · i∆F (xσ(n−1) − xσ(n)) n even

(77)

where the permutations σ are such that σ(1) < σ(2) < · · · < σ(n − 1) < σ(n). This general

result may be proven by induction. It follows that

0|T [ϕIN(x1) · · · ϕIN(xn)]|0=

0 n oddσ i∆F (xσ(1) − xσ(2)) · · · i∆F (xσ(n−1) − xσ(n)) n even

(78)

The vacuum expectation of a time-ordered product of an even number of IN field operators

is the product of vacuum expectations of time-ordered pairs of IN field operators (i.e.,

propagators) summed over all ordered permutations. Each field operator ϕIN(xi) finds itself

contracted once with each of the other (n − 1) field operators.

From Eqs.(68) and (70), we observe that

T [: ϕIN(x1)ϕIN(x2) : ϕIN(x3)] + i∆F (x1 − x2)ϕIN(x3)

= T [ϕIN(x1)ϕIN(x2)ϕIN(x3)]


+i∆F (x1 − x2)ϕIN(x3) + i∆F (x2 − x3)ϕIN(x1) + i∆F (x1 − x3)ϕIN(x2) (79)

Therefore,

T [: ϕIN(x1)ϕIN(x2) : ϕIN(x3)]


+i∆F (x2 − x3)ϕIN(x1) + i∆F (x1 − x3)ϕIN(x2) (80)

The contraction i∆F (x1− x2) does not appear on the right-hand side above because ϕIN(x1)

and ϕIN(x2) enter already normal ordered with respect to each other on the left-hand side.

17



Similarly, from Eqs.(63) and (75), we have

ϕIN(x1)ϕIN(x2)ϕIN(x3)ϕIN(x4)

= : ϕIN(x1)ϕIN(x2) :: ϕIN(x3)ϕIN(x4) : +i∆F (x1

−x2) : ϕIN(x3)ϕIN(x4) :

+i∆F (x3 − x4) : ϕIN(x1)ϕIN(x2) : +i2∆F (x1 − x2)∆F (x3 − x4) (81)

and

0|T [: ϕIN(x1)ϕIN(x2) :: ϕIN(x3)ϕIN(x4) :]|0 + i2∆F (x1 − x2)∆F (x3 − x4)

= 0|T [ϕIN(x1)ϕIN(x2)ϕIN(x3)ϕIN(x4)]|0= i2[∆F (x1 − x2)∆F (x3 − x4) + ∆F (x1 − x3)∆F (x2 − x4) + ∆F (x1 − x4)∆F (x2 − x3)]

(82)

Therefore,

0|T [: ϕIN(x1)ϕIN(x2) :: ϕIN(x3)ϕIN(x4) :]|0= i2[∆F (x1 − x3)∆F (x2 − x4) + ∆F (x1 − x4)∆F (x2 − x3)] (83)

Neither the contraction i∆F (x1 − x2) nor the contraction i∆F (x3 − x4) appears. In general,

no contractions of field operators will appear that already appear together inside one ::symbol.

In conclusion, each term in the perturbative expansion of the n-point function, Eq.(62),

can be reduced via the above process of normal ordering to a product of IN field 2-point

functions. Which contractions appear in each term is only limited by the factors of HI (ym)

present, since the IN field operators that come from HI (ym) are already normal ordered.

B. Feynman Diagrams I

Each term in the perturbative expansion of the n-point function, Eq.(62), can be rep-

resented by a diagram or graph. To obtain the diagrammatic representation for each term

in the expansion, we merely draw lines for each propagator that appears and identify the

common endpoints.

18



For example, the first term in the numerator for the 4-point function G(x1, x2, x3, x4) is

0|T [ϕIN(x1)ϕIN(x2)ϕIN(x3)ϕIN(x4)]|0= i2[∆F (x1

−x2)∆F (x3

−x4) + ∆F (x1

−x3)∆F (x2

−x4) + ∆F (x1

−x4)∆F (x2

−x3)]

(84)

The diagram representing this term is given in Figure 1. The interaction Hamiltonian is not

present in this term, so this is just the free field theory result. None of the propagators cross

each other so no interaction takes place.

The next term in the numerator for the 4-point function G(x1, x2, x3, x4) is

−i d4y

0|T [ϕIN(x1)ϕIN(x2)ϕIN(x3)ϕIN(x4) ˆ

HI (y)]

|0, (85)

where

HI (y) =1

2δm2 : ϕ2

IN(y) : +λ0

4!: ϕ4

IN(y) :, (86)

with δm2 ≡ m20 − m2. Consider first

−iλ0

4!

d4y0|T [ϕIN(x1)ϕIN(x2)ϕIN(x3)ϕIN(x4) : ϕ4

IN(y) :]|0

= −iλ0 d4y∆F (x1 − y)∆F (x2 − y)∆F (x3 − y)∆F (x4 − y) (87)

This term is represented by the diagram in Figure 2. From the diagram, we can see that

two particles meet, interact at y, and then part ways. The 4! has disappeared from the

right-hand side of the above equation because there are 4! equivalent ways of contracting

the four ϕIN(xi)’s with the four ϕIN(y)’s in HI (y). The point y where the propagators meet

is called an elementary vertex . Since the λ0 part of HI (y) is proportional to ϕ4IN(y), each λ0

vertex will always have four propagators or lines radiating from it. Consider next

−iδm2

2

d4y0|T [ϕIN(x1)ϕIN(x2)ϕIN(x3)ϕIN(x4) : ϕ2

IN(y) :]|0= −iδm2

d4y[i∆F (x1 − x2)i∆F (x3 − y)i∆F (x4 − y) + permutations] (88)

The first term is represented by the diagram in Figure 3. This second type of vertex, the

δm2 vertex, is represented by the × as in Figure 3. Once again, the factor of 1/2 on the

left-hand side of the above equation is cancelled because there are two equivalent ways to

contract the two ϕIN(xi)’s with the two ϕIN(y)’s in HI (y).

19



In summary, the net result of all of the factors of HI (ym) that appear in any of the terms

in the expansion is to create vertices. Thus, any diagram in the expansion is made up of

lines representing propagators connected at points of interactions, the vertices. There will

be one vertex for each factor of ˆHI (ym) appearing in the particular term, and each diagram

representing a term in the expansion of the n-point function will have n propagators with

one free end not connected to a vertex. The lines or propagators with free ends are called

external legs.

C. Feynman rules and Feynman Diagrams II

From the above terms and their associated diagrams that we have considered, we derive

the following Feynman rules.

1. To compute an n-point function (n even), draw all distinct diagrams with n external

legs, and sum their contributions. A diagram is distinct from another if they are

topologically inequivalent after we have labeled the endpoints of the external legs. The

labeling of the vertices (with yi) is unimportant. (See Figure 4, for example.)

2. To compute the value of a given diagram,

(a) associate a propagator, i∆F (zi − z j), to each line connecting zi to z j.

(b) associate a factor

−iλ0

d4yi

to each λ0 vertex at yi. Each λ0 vertex must have FOUR propagators emerging

from it.

(c) associate a factor

−iδm2

d4yi

to each δm2 vertex at yi. Each δm2 vertex must connect TWO propagators.

(d) Divide diagram by symmetry factor S . (See Figure 5, for example.)

In practice, to determine the S -matrix for a given process, instead of formally manip-

ulating Eq.(62), we draw the associated diagrams and evaluate them using the Feynman

rules.

20



D. Momentum space Feynman rules

The free-field propagator

i∆F (x

− x) = lim→0+ d4k

(2π)4i

k2 − m2 + i e−ik·(x−x)

= lim→0+

d4k

(2π)4i∆F (k)e−ik·(x−x), (89)

where the momentum space propagator

i∆F (k) ≡ i

k2 − m2 + i. (90)

It follows, for instance, that

−iλ0

d4y∆F (x1 − y)∆F (x2 − y)∆F (x3 − y)∆F (x4 − y)

= −iλ0

d4y

d4k1(2π)4

d4k2(2π)4

d4k3(2π)4

d4k4(2π)4

×∆F (k1)e−ik1·(x1−y)∆F (k2)e−ik2·(x2−y)∆F (k3)e−ik3·(x3−y)∆F (k4)e−ik4·(x4−y)

= d4k1

(2π)4

d4k2(2π)4

d4k3(2π)4

d4k4(2π)4

−iλ0(2π)4δ4

4

i=1

ki

×∆F (k1)∆F (k2)∆F (k3)∆F (k4) exp[−i(k1 · x1 + k2 · x2 + k3 · x3 + k4 · x4)] (91)

And,

λ20

2

d4y1

d4y2∆F (x1 − y1)∆F (x2 − y1)∆F (x3 − y2)∆F (x4 − y2)∆2

F (y1 − y2)

=λ20

2

d4y1

d4y2

d4k1(2π)4

d4k2(2π)4

d4k3(2π)4

d4k4(2π)4

d4 p1(2π)4

d4 p2(2π)4

×∆F (k1)e−ik1·(x1−y1)∆F (k2)e−ik2·(x2−y1)∆F (k3)e−ik3·(x3−y2)

×∆F (k4)e−ik4·(x4−y2)∆F ( p1)e−ip1·(y1−y2)∆F ( p2)e−ip2·(y1−y2)

=

λ20

2 d4k1

(2π)4 d4k2

(2π)4 d4k3

(2π)4 d4k4

(2π)4 d4 p1

(2π)4 d4 p2

(2π)4

×(2π)4δ4(k1 + k2 + p1 + p2)(2π)4δ4(k3 + k4 − p1 − p2)

×∆F ( p1)∆F ( p2)∆F (k1)∆F (k2)∆F (k3)∆F (k4)

× exp[−i(k1 · x1 + k2 · x2 + k3 · x3 + k4 · x4)] (92)

Often, it is easier to compute the Fourier transform, G(k1, · · · , kn), of the n-point function

G(x1, · · · , xn). The momentum space Feynman rules are:

21



1. Draw all distinct diagrams with n external legs. A diagram is distinct from another

if they are topologically inequivalent after we have labeled the external legs with

momenta k1, · · · , kn. By convention, all momenta point into the diagram.

2. Associate a propagator, i∆F (kl) to the lth external leg.

3. Associate a factor of

−iλ0(2π)4δ4

4i=1

qi

to the λ0 vertices where qi are the momenta flowing into the vertex. This expresses

momentum conservation at the λ0 vertex.

4. Associate a factor of

−iδm2(2π)4δ4 2i=1

qi

to the δm2 vertices where qi are the momenta flowing into the vertex. This expresses

momentum conservation at the δm2 vertex.

5. Label the direction and momentum of all internal propagators. Associate the integral

d4 pm

(2π)4i∆F ( pm)

to the mth internal line. (See Figure 5, for example.)

6. Divide the diagram by the symmetry factor S .

7. Multiply the overall result by the factor

(2π)4δ4

ni=1

ki

.

This expresses momentum conservation of the entire process.

22



VI. 2- AND 4-POINT FUNCTIONS

The denominator in Eq.(62) is the same for all n-point functions. Since it does not depend

on xi, the diagrams representing these terms will contain no external legs. The diagrams

representing these terms to second-order are given in Figure 6.

1 + (−i) ∞−∞

d4y10|T [HI (y1)]|0 +(−i)2

2!

∞−∞

d4y1

∞−∞

d4y20|T [HI (y1)HI (y2)]|0

= 1 +(−i)2

2!

δm2

2!

2 ∞−∞

d4y1

∞−∞

d4y20|T [: ϕ2IN(y1) :: ϕ2

IN(y2) :]|0

+(−i)2

2!

λ0

4!

2 ∞−∞

d4y1

∞−∞

d4y20|T [: ϕ4IN(y1) :: ϕ4

IN(y2) :]|0

= 1 +(−iδm2)2

2!2! ∞

−∞

d4y1 ∞

−∞

d4y2[i∆F (y1−

y2)]2

+(−iλ0)2

2!4!

∞−∞

d4y1

∞−∞

d4y2[i∆F (y1 − y2)]4 (93)

The diagrams represent disconnected vacuum processes in the interacting vacuum and are

called vacuum bubbles. These just represent numbers.

A. 2-Point Function

The numerator of the 2-point function with terms to second-order

0|T [ϕIN(x1)ϕIN(x2)]|0 + (−i) ∞−∞

d4y10|T [ϕIN(x1)ϕIN(x2)HI (y1)]|0

+(−i)2

2!

∞−∞

d4y1

∞−∞

d4y20|T [ϕIN(x1)ϕIN(x2)HI (y1)HI (y2)]|0

= 0|T [ϕIN(x1)ϕIN(x2)]|0 + (−i)δm2

2!

∞−∞

d4y10|T [ϕIN(x1)ϕIN(x2) : ϕ2IN(y1) :]|0

+(−i)2

2!

δm2

2!

2 ∞−∞

d4y1

∞−∞

d4y20|T [ϕIN(x1)ϕIN(x2) : ϕ2IN(y1) :: ϕ2

IN(y2) :]|0

+(−i)22!

λ0

4!

2 ∞

−∞d4y1

∞

−∞d4y20|T [ϕIN(x1)ϕIN(x2) : ϕ4

IN(y1) :: ϕ4IN(y2) :]|0

= i∆F (x1 − x2) + (−iδm2) ∞−∞

d4y1i∆F (x1 − y1)i∆F (x2 − y1)

+(−iδm2)2

2!

∞−∞

d4y1

∞−∞

d4y2i∆F (x1 − y1)i∆F (x2 − y2)i∆F (y1 − y2)

+i∆F (x1 − x2)(−iδm2)2

2!2!

∞−∞

d4y1

∞−∞

d4y2[i∆F (y1 − y2)]2

+(−iλ0)2

2!3! ∞

−∞d4y1

∞

−∞d4y2i∆F (x1 − y1)i∆F (x2 − y2)[i∆F (y1 − y2)]3

23



+i∆F (x1 − x2)(−iλ0)2

2!4!

∞−∞

d4y1

∞−∞

d4y2[i∆F (y1 − y2)]4

The diagrams representing these terms are given in Figure 7. We note that the ”tadpole”

diagram in Figure 8 is absent because we have normal ordered ˆ

HI . If we had not normal

ordered HI , we would have to include the tadpoles. These amount to a mass renormalization ,

so when all is said and done, we end up with the same results. (Compare Figure 9.)

In Figure 10, we regroup the diagrams by factoring out the vacuum bubbles. If we wrote

out the diagrams to higher order, we would find that the same vacuum bubble factor mul-

tiplies each diagram that doesn’t contain any vacuum bubbles. Thus, the purpose of the

denominator is to remove all diagrams containing disconnected vacuum bubbles - physically

disconnected vacuum processes should have no effect on the propagation of particles.

The same vacuum bubbles will show up in diagrams for any n-point function and will be

cancelled by the denominator in each case. Therefore, we can forget about the denominator

in Eq.(62) as long as we add a rule that says not to calculate diagrams with disconnected

bubbles. The removal of the vacuum bubbles amounts to the removal of the phase generated

by the S -matrix acting on the vacuum.

B. 4-Point Function

The numerator of the 4-point function with terms to second-order

0|T [ϕIN(x1)ϕIN(x2)ϕIN(x3)ϕIN(x4)]|0+(−i)

∞−∞

d4y10|T [ϕIN(x1)ϕIN(x2)ϕIN(x3)ϕIN(x4)HI (y1)]|0

+(−i)2

2!

∞−∞

d4y1

∞−∞

d4y20|T [ϕIN(x1)ϕIN(x2)ϕIN(x3)ϕIN(x4)HI (y1)HI (y2)]|0=

0

|T [ϕIN(x1)ϕIN(x2)ϕIN(x3)ϕIN(x4)]

|0

+(−i)

δm2

2!

∞

−∞d4y10|T [ϕIN(x1)ϕIN(x2)ϕIN(x3)ϕIN(x4) : ϕ2

IN(y1) :]|0

+(−i)λ0

4!

∞−∞

d4y10|T [ϕIN(x1)ϕIN(x2)ϕIN(x3)ϕIN(x4) : ϕ4IN(y1) :]|0

+(−i)2

2!

∞−∞

d4y1

∞−∞

d4y20|T [ϕIN(x1)ϕIN(x2)ϕIN(x3)ϕIN(x4)HI (y1)HI (y2)]|0 (94)

The diagrams representing these terms (modulo vacuum bubbles) are given in Figure 11.

24



VII. SPECTRAL DENSITY

Consider

i∆

(x − y) ≡ 0|[ϕ(x), ϕ(y)]|0= 0|ϕ(x)ϕ(y)|0 − 0|ϕ(y)ϕ(x)|0=

i

0|ϕ(x)|kiki|ϕ(y)|0 −i

0|ϕ(y)|kiki|ϕ(x)|0

=i

0| exp(ix · P )ϕ(0) exp(−ix · P )|kiki| exp(iy · P )ϕ(0) exp(−iy · P )|0

−i

0| exp(iy · P )ϕ(0) exp(−iy · P )|kiki| exp(ix · P )ϕ(0) exp(−ix · P )|0

=

i[e−iki·(x−y)0|ϕ(0)|kiki|ϕ(0)|0 − eiki·(x−y)0|ϕ(0)|kiki|ϕ(0)|0]

=i

[e−iki·(x−y) − eiki·(x−y)]|0|ϕ(0)|ki|2

=i

|0|ϕ(0)|ki|2

d4kδ4(k − ki)[e−ik·(x−y) − eik·(x−y)]

=

d4k

i

δ4(k − ki)|0|ϕ(0)|ki|2

[e−ik·(x−y) − eik·(x−y)]

≡ d4k

(2π)3σ(k)[e−ik·(x−y) − eik·(x−y)], (95)

where thespectral density

σ(k) ≡ (2π)3i

δ4(k − ki)|0|ϕ(0)|ki|2. (96)

Here,

|ki = a†IN(ki)|0 (97)

with k0i = ωki =

k2i + m2 > 0, and

P µ

|ki

= kµ

i |ki

, (98)

i

|kiki| = 1.

We observe the following properties of σ(k).

1. σ(k) is real valued and positive definite.

2. Since ϕ(x) transforms as a Lorentz scalar, σ(k) is Lorentz invariant.

25



3. If ϕ(x) were just a free field, for example ϕIN(x), then

σ(k) = (2π)3i

δ4(k − ki)|0|ϕIN(0)|ki|2

= (2π)3i

δ4(k−

ki)0|ϕIN(0)

|ki

ki

|ϕIN(0)

|0

(99)

Substituting

ϕIN(0) = d3 k

(2π)31

2ωk

[aIN(k) + a†IN(k)], (100)

we have

σ(k) = (2π)3i

δ4(k − ki) d3 k

(2π)31

2ωk

d3 k

(2π)31

2ωk

×0

|aIN(k)a†IN(ki)

|0

0

|aIN(ki)a†IN(k)

|0

= (2π)3

i

δ4(k − ki)

d3 kδ3( k − ki)

d3 kδ3( k − ki)

= δ(k2 − m2)θ(k0) (101)

It follows that

d4k

(2π)3σ(k)[e−ik·(x−y) − eik·(x−y)] =

d4k

(2π)42πδ(k2 − m2)θ(k0)[e−ik·(x−y) − eik·(x−y)]

= d3 k

(2π)3

1

2ωk

[e−ik·(x−y)

−eik·(x−y)]

≡ i∆(x − y) (102)

Let σ(k) = ρ(k2)θ(k0), then

i∆(x − y) = d4k

(2π)3σ(k)[e−ik·(x−y) − eik·(x−y)]

= d4k

(2π)3ρ(k2)θ(k0)[e−ik·(x−y) − eik·(x−y)]

=

d

4

k(2π)3 ∞0

dµ2ρ(µ2)δ(k2 − µ2)θ(k0)[e−ik·(x−y) − eik·(x−y)]

= ∞0

dµ2ρ(µ2) d4k

(2π)3δ(k2 − µ2)θ(k0)[e−ik·(x−y) − eik·(x−y)]

= ∞0

dµ2ρ(µ2)i∆(x − y; µ) (103)

Now,

∞

0dµ2ρ(µ2) = 1. (104)

26



This is because

limy0→t

∂

∂x0i∆(x − y) = 0|[∂ tϕ(x, t), ϕ(y, t)]|0

= −iδ3(x − y), (105)

and

limy0→t

∂

∂x0i∆(x − y; µ) = −i lim

y0→t

d4k

(2π)3δ(k2 − µ2)θ(k0)k0[e−ik·(x−y) + eik·(x−y)]

= −iδ3(x − y). (106)

For µ2 < m2, ρ(µ2) = 0, since ϕ(x) cannot create a particle with mass less than m -

the single-particle mass. Let mt be the mass where multiparticle production begins. Then

for m2 < µ2 < m2t , ρ(µ2) = 0, since ϕ(x) cannot create a particle with mass above the

single-particle mass but below the multiparticle threshold. Therefore, for µ2 < m2t ,

ρ(µ2) = Zδ(µ2 − m2), (107)

where

1 − Z = ∞m2t

ρ(µ2)dµ2. (108)

0 < Z ≤ 1, since ρ(µ2) is positive. In our case here with one distinct particle of physical

mass m, the multiparticle threshold mt = 2m.

The relation

limt→−∞

ϕ(x) =√

Z ϕIN(x) (109)

cannot be implemented as an operator equation. This is because, at equal times,

0|[∂ tϕ(x, t), ϕ(y, t)]|0 = −iδ3(x − y), (110)

which holds no matter what t is. Thus,

limt→−∞

0

|[∂ tϕ(x, t), ϕ(y, t)]

|0

=

−iδ3(x

−y). (111)

If we take Eq.(109) as an operator equation, then

limt→−∞

0|[∂ tϕ(x, t), ϕ(y, t)]|0 = Z limt→−∞

0|[∂ tϕIN(x, t), ϕIN(y, t)]|0. (112)

Since the equal-time commutators for ϕ and ϕIN are identical, the above result implies

Z = 1. But, Eq.(108) with Z = 1 implies that ρ(µ2) = 0 except at µ2 = m2, so ϕ = ϕIN.

Therefore, if we implement Eq.(109) as an operator equation, we end up with a free field

theory.

27

self-interacting scalar field theory

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