aainstitutetext: Leinweber Center for Theoretical Physics,
Randall Laboratory of Physics, Department of Physics,
University of Michigan, Ann Arbor, MI 48109, U.S.Abbinstitutetext: Joseph
Henry Laboratories, Princeton University
Princeton, NJ 08544, U.S.A

# Entanglement, Renormalization and Effective Field Theories

Bingzheng Han a and Ratindranath Akhoury<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract

We develop the idea that renormalization, decoupling of heavy particle effects
from low energy physics and the construction of effective field theories are
intimately linked to the momentum space entanglement of disparate modes of an
interacting quantum field theory. Using unitary transformations to decouple
these modes at the perturbative level, we show in a scalar field theoretical
model with light and heavy fields, how renormalization may be consistently
implemented and how the low energy effective field theory can be constructed.
We also obtain a renormalization group equation in this framework and apply it
to the scalar field theoretical model.

##  1 Introduction

Entanglement is ubiquitous in any quantum theory. In a free field theory the
different momentum modes are not entangled. However, the introduction of
interactions causes the entanglement, in particular, of the low momentum modes
with the inaccessible high energy ones. In experiments only the low energy or
larger wavelength modes are accessible and renormalization can be thought of
as a procedure to disentangle the high energy modes from those of low energy
at the same time incorporating the effects of the former in a modified
effective theory at low energies. In the usual Wilsonian approach
(Wilson:1971bg, )Wilson:1971dh Polchinski:1983gv , the high energy modes are
integrated out and in this way we arrive at a low energy effective action. An
alternative viewpoint, that we discuss here, is to directly address the
entanglement and by a series of unitary transformations decouple the low and
high energy modes. The effective low energy Hamiltonian is then obtained by
projecting onto the “high energy vacuum", i.e., the low energy subspace where
there are no modes of heavy masses or of momenta larger than some cut-off
scale which can appear as external states. In this paper, we discuss
renormalization, decoupling of heavy mass states Appelquist:1974tg  and the
construction of effective field theories, Georgi:1994qn Manohar:1996cq
Rothstein:2003mp , all from this perspective. Not only do we set up the
general formalism but also exemplify the procedure by explicit examples up to
the one-loop level and to order $1/M^{2}$, where $M$ is the heavy mass scale.
Our results are in agreement with those obtained by the standard methods (see
for example, Rothstein:2003mp ) showing that such a program can be
successfully implemented,and thereby providing another way to construct low
energy effective theories. In terms of an extended program, this paper is a
first step in exploring how entanglement measures may be used in general to
address problems in quantum field theory like the correlations between
different momentum scales and in this way provide another window into
renormalization and related phenomenon.

Envisioning the renormalization process as removing the entanglement between
the low and high momentum modes through unitary transformations has a clear
physical significance in the Hamiltonian or Schrodinger framework which we
adopt in this paper. This framework is not manifestly covariant in the
intermediate stages and has seldom been used for practical calculations,
though it has been studied at a formal level, for example, in Symanzik:1981wd
, Pi:1987df , Minic:1994ff .In particular, in spite of the great successes of
the effective field theory approach, rarely has work been done in a
Hamiltonian framework. Ours is a straightforward and direct attempt where we
formally connect the Hamiltonian approach to the standard one by mapping the
calculations in this framework to the usual Feynman diagram calculations of S
matrix elements and extend dimensional regularization techniques to perform
loop calculations. Methods to evaluate certain unusual integrals encountered
in this approach are discussed in appendices.

The study of renormalization through similarity transformations has a long
history, though most of the previous works do not reference the crucial
connection to momentum space entanglement. In Glazek:1994qc  similarity
transformations were introduced to control divergences in light-front field
theory. An exact renormalization group equation similar in appearance to the
one in section (5) of this paper was obtained there. In Gubankova:1997mq , a
similar set of Hamiltonian flow equations were obtained but it was tailored to
address problems in many body theory where no divergent renormalization is
necessary. The motivation of these two references was also very different from
the one pursued here. The analysis presented in sections, (2) and (3) are
closest in spirit to the perturbative Hamiltonian renormalization of a scalar
theory with purely quartic interactions discussed in Alexanian:1998wu . In
this paper our emphasis is on renormalization from the perspective of momentum
space entanglement and we also similarly study the decoupling of heavy
particle effects and the construction of effective field theories in the
context of a model involving both heavy and light fields. In addition, we also
address the problem of renormalization group flow in this approach. In a more
recent paper, Balasubramanian:2011wt  momentum space entanglement and
renormalization in a quantum field theory has been specifically addressed.
There, a relationship is obtained between the Wilsonian effective action and
the density matrix (with an entanglement entropy) describing the infrared
degrees of freedom of the theory. Though there are points of contact, our
approach is more direct and there is little connection with the techniques or
the results of Balasubramanian:2011wt  .

The paper is organized as follows. In section (2) we discuss our general
approach to a perturbative realization of the disentangling high and low
momentum modes by means of unitary transformations on the states. We discuss
the projection of the Hamiltonian of a theory onto the low energy subspace and
the procedure followed in the rest of the paper for renormalization and
construction of effective field theories . In section (3) we consider a scalar
field theoretical model with heavy and light fields and explicitly construct
the unitary transformation that shows clearly how renormalization and
decoupling works. In particular, we construct the light particle two and four
point functions up to order $\frac{1}{M^{2}}$ in the heavy mass expansion and
at one loop order. In section (4) we extend the previous construction to
obtain an effective field theory of the light fields alone and make connection
with previous work based on conventional methods. In section (5) we check the
consistency of our approach by setting up a renormalization group equation and
discuss an evaluation of the $\beta$ function in a scalar theory with only
quartic interactions. We conclude with a discussion of these results in
section (6). Certain technical details of the calculations, particularly those
encountered in section (3) are relegated to appendices.

##  2 Perturbative Decoupling, Renormalization and Matching of Hamiltonian
Operators

###  2.1 Decoupling with Unitary Transformations

The subject of decoupling in Effective Field Theory has been studied
extensively for the past many decades. The decoupling theorem states that if
the low energy effective theory is renormalizable, and a physical
renormalization scheme has been applied, then all effects due to heavy
particles will appear as changes to couplings or are suppressed as
$\frac{1}{M^{n}}$, where M is the mass of the heavy particle. As we discussed
in the introduction, an alternative way to consider the decoupling is to
introduce a series of unitary transformations to decouple high energy and low
momentum modes and then look at the low energy part of the spectrum. The
Hamiltonian framework is best suited to study decoupling and renormalization
from this perspective. In this fixed time approach, consider the action of a
unitary transformation $\omega$ on the states of a theory,

$\ket{\Psi(\mu)}=\omega^{\dagger}(\mu)\ket{\Psi},$ (1)

under which the Hamiltonian transforms as
$H^{\prime}=\omega^{\dagger}H\omega$. We would like to use this unitary
transformation to disentangle the momentum states above a scale $\mu$ from the
low energy ones. However, this exact diagonalization procedure, in general, is
impossible at present and we have to be content with a less ambitious,
perturbative approach, where we identify a low energy subspace by projecting
the unitarily transformed Hamiltonian on to a state which acts as a vacuum for
high energy particles. Let us label this state as $\ket{0_{H}}$ which
satisfies,

$\displaystyle a_{H}\ket{0_{H}}=0,$ (2)

where, $a_{H}$ is the annihilation operator of high energy modes. Thus in this
approach the job of the unitary transformation, order by order is to remove
the terms in the full Hamiltonian which will perturb the high energy vacuum
structure. These are terms containing only high energy creation operators.
Since Hamiltonian operator is Hermitian, $\omega$ will inevitably cancel terms
containing only high energy annihilation operators as well. This condition
allows us to identify the unitary transformations perturbatively and the
transformed Hamiltonian with this $\omega$ when projected on to the high
energy vacuum will give us what we refer to as the decoupled Hamiltonian at
low energy. The decoupled Hamiltonian encodes the effects of the high momentum
modes on the low energy physics. The precise way this is accomplished will now
be discussed. A similar method was used in a related context in
Alexanian:1998wu .

Let’s consider the Hamiltonian $H$ of a full theory, and denote $H_{eff}(\mu)$
as the effective Hamiltonian defined at an energy scale $\mu$. $H_{eff}(\mu)$
will generate the same physical results i.e. S-matrix elements for all the
physical processes that do not involve momenta greater than $\mu$. We can view
$H_{eff}(\mu)$ as the projection of the full theory onto the low energy
subspace:

$\displaystyle H_{eff}(\mu)=P(\mu)HP(\mu),$ (3)

where P($\mu$) is the projection operator at energy scale $\mu$.
As discussed, at least perturbatively we can decouple low energy modes from
high energy ones using a series of unitary transformations and thus construct
the high energy vacuum and obtain the low energy subspace. Let $H_{decoupled}$
denote the decoupled Hamiltonian at low energy:

$\displaystyle H_{decoupled}=\bra{0_{H}}\omega^{\dagger}H\omega\ket{0_{H}},$
(4)

In general, the $\omega$ here is a product series of unitary transformations.
Furthermore, we will normal order with respect to high energy vacuum. It’s
worth noting here that the $H_{decoupled}$ so calculated is an intermediate
step towards the calculation of the physical effective Hamiltonian $H_{eff}$.
However, as we will see later its components have an important physical
meaning regarding renormalization and further it will also be involved in the
matching process to get the physical effective Hamiltonian $H_{eff}$.
Let’s break $\omega$ into a product series:

$\displaystyle\omega=\omega_{0}\omega_{1}\omega_{2}...\omega_{n}...$ (5)

Each $\omega_{i}$ partially diagonalizes the Hamiltonian to a given order
$\sim\frac{1}{\mu}$, $\mu$ is the cut-off energy scale. We can decompose the
full Hamiltonian as:

$\displaystyle H=H_{1}+H_{2}+H_{A}+H_{B},$ (6)

where $H_{1}$ only contains low energy modes, $H_{2}$ is the free part for
high energy modes, $H_{A}$ contains terms that only have high energy
annihilation or creation operators and $H_{B}$ is whatever left. For
simplicity, we can set $H_{1}$ to be $O(1)$ in energy , and the other three
terms of $O(\mu)$.
Let’s consider the following:

$\displaystyle\omega_{0}^{\dagger}\left(H_{1}+H_{2}+H_{A}+H_{B}\right)\omega_{0}$
(7) $\displaystyle=$ $\displaystyle
e^{-i\Omega_{0}}\left(H_{1}+H_{2}+H_{A}+H_{B}\right)e^{i\Omega_{0}}$
$\displaystyle=$ $\displaystyle
H_{1}+H_{2}+H_{A}+H_{B}+i[H_{1},\Omega_{0}]+i[H_{2},\Omega_{0}]+i[H_{A},\Omega_{0}]+i[H_{B},\Omega_{0}]...$

We want to eliminate $H_{A}$ by choosing $\Omega_{0}$ such that

$\displaystyle i[H_{2},\Omega_{0}]+H_{A}=0.$ (8)

This is our decoupling condition at $O(\mu)$, and since both $H_{2}$ and
$H_{A}$ are of $O(\mu)$, we can deduce that $\Omega_{0}\sim O(1)$. Although we
cancel out $H_{A}$, we create a new term $i[H_{1},\Omega_{0}]$ of order $\sim
O(1)$ that contains only annihilation or creation operators and in order to
eliminate this new term, we need to introduce the next unitary operator
$\omega_{1}=e^{i\Omega_{1}}$ at $O(\frac{1}{\mu})$. Then the Hamiltonian
becomes

$\displaystyle
e^{-i\Omega_{1}}e^{-i\Omega_{0}}\left(H_{1}+H_{2}+H_{A}+H_{B}\right)e^{i\Omega_{0}}e^{i\Omega_{1}}$
(9) $\displaystyle=$ $\displaystyle
H_{1}+H_{2}+H_{B}+i[H_{1},\Omega_{0}]+i[H_{A},\Omega_{0}]+i[H_{B},\Omega_{0}]+i[H_{2},\Omega_{1}]+...$

We now choose $\Omega_{1}$ such that

$\displaystyle i[H_{1},\Omega_{0}]+i[H_{2},\Omega_{1}]=0,$ (10)

and it is obvious that $\Omega_{1}$ is of $O(\frac{1}{\mu})$. In general our
decoupling condition will become:

$\displaystyle i[H_{1},\Omega_{n}]+i[H_{2},\Omega_{n+1}]=0,$ (11)

with $\Omega_{n+1}\sim\frac{1}{\mu}\Omega_{n}$. Thus, we see that decoupling
can be consistently carried out iteratively in a perturbative fashion.

###  2.2 Decoupled Hamiltonian and S-matrix Elements

Putting together the expansions given above and using the decoupling
conditions, it is straightforward to calculate $H_{decoupled}$:

$\displaystyle H_{decoupled}=$
$\displaystyle\bra{0_{H}}H_{1}+H_{2}+H_{B}+\frac{i}{2}[H_{A},\Omega_{0}]+i[H_{B},\Omega_{0}]-\frac{1}{3}[[H_{A},\Omega_{0}],\Omega_{0}]$
(12)
$\displaystyle-\frac{1}{2}[[H_{B},\Omega_{0}],\Omega_{0}]-\frac{1}{2}[[H_{1},\Omega_{0}],\Omega_{0}]+i[H_{B},\Omega_{1}]+O(\frac{1}{\mu})\ket{0_{H}}.$

As we pointed out in section 2.1, this $H_{decoupled}$ is not the physical
effective Hamiltonian operator at low energies. The path to obtain the
effective Hamiltonian will be discussed in the next subsection. Here we wish
to point out a useful connection of the various terms in Eq.(12) with
corresponding ones in Feynman diagram calculations. Explicit evaluation of the
various contributions using mode expansions will be done in section (3).

There are two parts in $H_{decoupled}$. The first part is
$\bra{0_{H}}H_{1}\ket{0_{H}}=H_{1}$ which is simply the low energy part in the
original Hamiltonian. The second part is due to the contribution of
commutators and normal ordering of $H_{B}$ in the expansion. As shown later in
the scalar field theory example, each element in the second part can be
understood as an S-matrix element in the full theory but expanded in terms of
$\frac{1}{\mu}$. For instance, suppose we have the tree level scattering
process represented in Figure 1, the S-matrix element in the full theory
including all channels is $-3\lambda^{2}\frac{i}{p^{2}-M^{2}}$, and the
corresponding term in $H_{decoupled}$ will be
$(-\frac{3\lambda^{2}}{M^{2}}+O(\frac{1}{M^{4}}))\frac{\Phi_{L}^{4}}{4!}$.
This correspondence remains true for the one loop scattering process as well.
For example, consider the diagram shown in Figure 1,at order
$O(\frac{1}{M^{2}})$. A traditional calculation will give
$-\frac{3\lambda_{1}^{2}\lambda_{0}}{16\pi^{2}M^{2}}\left(\frac{1}{\overline{\epsilon}}-\ln\frac{m^{2}}{\mu^{2}}\right)$,
and the corresponding contributing terms in $H_{decoupled}$ are a combination
of $-\frac{1}{3}[[H_{A},\Omega_{0}],\Omega_{0}]$ and
$\frac{1}{2}[[H_{B},\Omega_{0}],\Omega_{0}]$. We will show in section
(3.2.2)that this latter contribution is also
$-\frac{3\lambda_{1}^{2}\lambda_{0}}{16\pi^{2}M^{2}}\left(\frac{1}{\overline{\epsilon}}-\ln\frac{m^{2}}{\mu^{2}}\right)\frac{\phi^{4}(x)}{4!}$,
where the factor 3 appears again because of the summation of contributions
from all three channels. This correspondence will continue to hold for all
terms in $H_{decoupled}$ as we will see in Section 3. In this sense, we are
able to make a term by term correspondence between the traditional Feynman
diagram calculations and the contributions in the Hamiltonian formulation of
this paper of the terms in Eq.(12). Due to this connection between the
decoupled Hamiltonian and S-matrix elements, Feynman diagrams in the full
theory provide useful guidance in organizing practical calculations, as we
will see in section 3.

Figure 1:  Two examples of two-to-two scattering process in a scalar field
theory with solid lines representing light fields $\Phi_{L}$ and the dashed
line representing a heavy particle $\Phi_{H}$. The coupling coefficient at the
vertex is $\lambda$. The heavy particle mass M is much larger than the
momentum p on the propagator. Hamiltonian infull
theory$H^{tree}_{decoupled}$$\bra{0_{H}}\omega^{\dagger}H\omega\ket{0_{H}}_{tree}$$H^{tree}_{eff}$projection$H^{oneloop}_{decoupled}$$H^{oneloop\
eff}_{decoupled}$$H^{oneloop}_{eff}$subtraction$\bra{0_{H}}\omega^{\dagger}H\omega\ket{0_{H}}_{oneloop}$$\bra{0_{H}}\omega^{{}^{\prime}\dagger}H^{tree}_{eff}\omega^{{}^{\prime}}\ket{0_{H}}_{oneloop}$
Figure 2:  A flow chart showing how to do matching in our theory and obtain
the effective Hamiltonian at one loop order.

###  2.3 Renormalization and Matching

Since the second part in $H_{decoupled}$ produces terms similar to the
expansion of S-matrix elements, inevitably, there are UV divergences emerging
from the loop calculations. Because only low energy modes can appear in
$H_{decoupled}$, all the UV divergences should be canceled by the
renormalization of $H_{1}$. In this way, we can determine the renormalization
Z-factor of light field, light field mass and coupling constants of purely
light interactions. It is shown in section 3.3 that our results obtained from
$H_{decoupled}$ indeed agree with those obtained using traditional
renormalization in the Lagrangian framework.
The effective Hamiltonian is obtained by matching order by order as we show in
Figure 2. First we decouple the full theory at tree level by doing tree level
calculation using the unitary transformation
$H_{decoupled}^{tree}=\bra{0_{H}}\omega^{\dagger}H\omega\ket{0_{H}}_{tree}$
and match it onto the low energy regime to get the tree level effective
Hamiltonian. This step is simple, we can just denote
$H_{eff}^{tree}=H_{decoupled}^{tree}$. Then we obtain the decoupled
Hamiltonian at one loop in both full and effective theories by calculating
$\bra{0_{H}}\omega^{\dagger}H\omega\ket{0_{H}}_{oneloop}$ and
$\bra{0_{H}}\omega^{{}^{\prime}\,\dagger}H_{eff}^{tree}\omega^{{}^{\prime}}\ket{0_{H}}_{oneloop}$
at one loop order, where $\omega^{{}^{\prime}}$ is the decoupling unitary
transformation for the effective theory. After renormalizing both full and
effective theories, we will get decoupled Hamiltonians
$H_{decoupled}^{oneloop}$ and $H_{decoupled}^{oneloop,eff}$ which include
interaction terms that account for the one loop corrections. Similar to the
traditional EFT, we can do the matching by subtracting
$H_{decoupled}^{oneloop,eff}$ from $H_{decoupled}^{oneloop}$ to get new
interaction terms to be added in the effective Hamiltonian at one loop order
which now does not contain any large logarithmic contributions. We can proceed
to higher orders iteratively in this fashion.

##  3 Decoupling and Renormalization of A Scalar Field Theory

In this section we will work in the weak coupling regime of a scalar field
theory with both heavy and light fields. Because we will be discussing
renormalization in the Hamiltonian framework, we will consider mode expansions
at a fixed time or effectively, we will be working in the Schrodinger picture.
Also, for the purpose of decoupling, we set the cut-off energy scale $\mu$ to
be the heavy mass M.

###  3.1 Preliminaries

Our subsequent analysis will apply to a scalar field theory with heavy and
light fields ($\Phi_{H}$ and $\Phi_{L}$ respectively) with dynamics given by
the following Hamiltonian:

$\displaystyle H=$ $\displaystyle\int
d^{3}x\left(\frac{1}{2}\left(\Pi_{L}^{2}(x)+(\nabla\Phi_{L}(x))^{2}+m^{2}\Phi_{L}^{2}(x)\right)+\frac{1}{2}\left(\Pi_{H}^{2}(x)+(\nabla\Phi_{H}(x))^{2}+M^{2}\Phi_{H}^{2}(x)\right)\right.$
(13)
$\displaystyle\left.+\frac{\lambda_{0}}{4!}\Phi_{L}^{4}(x)+\frac{\lambda_{1}}{2}\Phi_{H}(x)\Phi_{L}^{2}(x)+\frac{\lambda_{2}}{4}\Phi_{L}^{2}(x)\Phi_{H}^{2}(x)+\frac{\lambda_{3}}{4!}\Phi_{H}^{4}(x)\right).$

These fields have the usual mode expansions, however, we will need to consider
light fields carefully. This is because the light fields contain two parts,
one is the low frequency mode $\phi(x)$ and the other is the high frequency
mode $\chi(x)$. In order to correctly project onto the low energy subspace, we
want only low frequency fields $\phi(x)$ to appear in external lines. This can
be taken into account in the usual expansion of all the fields (in Schrodinger
picture) in the following manner:

$\displaystyle\phi(x)=\sum_{p<M}\frac{1}{\sqrt{2\epsilon_{p}}}\left(b_{p}e^{i\boldsymbol{p}\boldsymbol{x}}+b_{p}^{\dagger}e^{-i\boldsymbol{p}\boldsymbol{x}}\right),$
(14a)
$\displaystyle\chi(x)=\sum_{M<p}\frac{1}{\sqrt{2\epsilon_{p}}}\left(b_{p}e^{i\boldsymbol{p}\boldsymbol{x}}+b_{p}^{\dagger}e^{-i\boldsymbol{p}\boldsymbol{x}}\right),$
(14b)
$\displaystyle\Phi_{L}(x)=\sum_{p}\frac{1}{\sqrt{2\epsilon_{p}}}\left(b_{p}e^{i\boldsymbol{p}\boldsymbol{x}}+b_{p}^{\dagger}e^{-i\boldsymbol{p}\boldsymbol{x}}\right),$
(14c)
$\displaystyle\Phi_{H}(x)=\sum_{k}\frac{1}{\sqrt{2\omega_{k}}}\left(a_{k}e^{i\boldsymbol{k}\boldsymbol{x}}+a_{k}^{\dagger}e^{-i\boldsymbol{k}\boldsymbol{x}}\right).$
(14d)

In the above, and for the rest of this paper, we adopt the notation that
$\omega_{k}$ denotes the energy of the heavy particle and $\epsilon_{p}$ that
of the light one. Thus, for example, $\omega_{k}=\sqrt{\vec{k}^{2}+M^{2}}$ and
$\epsilon_{p}=\sqrt{\vec{p}^{2}+m^{2}}$. From the expansion, we see that the
$\phi$ and the $\chi$ fields are orthogonal, i.e., $\int
d^{3}x\,\phi(x)\chi(x)=0$. In the following we will not use the mode expansion
of $\phi(x)$ . Finally, as a notational convenience, in going from the
discrete momentum sum to the continuum we will use
$\sum_{k}\rightarrow\int\frac{d^{3}k}{(2\pi)^{3}}$,and omit all factors of the
volume $V$ since these will eventually cancel out.
As discussed earlier, we want to split the total Hamiltonian into four parts:
$H_{1}$ contains only low frequency modes of light particles; $H_{2}$ contains
the free parts of both heavy particles and high frequency modes of light
particles; $H_{A}$ contains only creation or annihilation operators, e.g.,
$aab$, $a^{\dagger}a^{\dagger}b^{\dagger}$, etc; and $H_{B}$ contains
combinations of creation and annihilation operators, e.g.,
$b^{\dagger}a\phi(x)$, $a^{\dagger}abb$, etc. Thus,

$\displaystyle H=H_{1}+H_{2}+H_{A}+H_{B}.$ (15)

For our case,

$\displaystyle H_{1}$ $\displaystyle=\int
d^{3}x\left(\frac{1}{2}\left(\Pi^{2}+(\nabla\phi)^{2}+m^{2}\phi^{2}\right)+\frac{\lambda_{0}}{4!}\phi^{4}(x)\right),$
(16a) $\displaystyle H_{2}$
$\displaystyle=\sum_{k}\omega_{k}a_{k}^{\dagger}a_{k}+\sum_{M<p}\epsilon_{p}b_{p}^{\dagger}b_{p}.$
(16b)

However, $H_{A}$ and $H_{B}$ are rather involved and will not be explicitly
displayed here. As we proceed with the calculation, we will simply pick out
relevant terms by analyzing the coupling coefficient and the number of low
energy light particles in external legs.
We argued earlier that
$H_{decoupled}=\bra{0_{H}}\omega^{\dagger}H\omega\ket{0_{H}}$, where $\omega$
denotes a series of unitary transformations,
$\omega=\omega_{0}\omega_{1}\dots\omega_{n}\dots$ Our calculation will be
limited to the first loop order and for this purpose, we only need the first
two terms in the unitary transformations:

$\displaystyle\omega^{\dagger}H\omega=e^{-i\Omega_{1}}e^{-i\Omega_{0}}\left(H_{1}+H_{2}+H_{A}+H_{B}\right)e^{i\Omega_{0}}e^{i\Omega_{1}}.$
(17)

The right hand side of the above equation can be simplified to

$\displaystyle
H_{1}+H_{2}+H_{B}+\frac{i}{2}[H_{A},\Omega_{0}]+i[H_{B},\Omega_{0}]-\frac{1}{3}[[H_{A},\Omega_{0}],\Omega_{0}]$
(18)
$\displaystyle-\frac{1}{2}[[H_{B},\Omega_{0}],\Omega_{0}]-\frac{1}{2}[[H_{1},\Omega_{0}],\Omega_{0}]+i[H_{B},\Omega_{1}]+O(\frac{1}{M}),$

where we have set the cut-off energy scale to the heavy mass M and used the
condition $i\left[H_{2},\Omega_{0}\right]+H_{A}=0$,
$i\left[H_{1},\Omega_{0}\right]+i\left[H_{2},\Omega_{1}\right]=0$.

In the next two sections we will study decoupling and renormalization in this
scalar field theory by calculating the decoupled Hamiltonian up to order
$O(\frac{1}{M^{2}})$ at one loop level for the two and four point functions.
The techniques involved in the calculation are different from the traditional
Feynman diagram methods, however, as we have noticed earlier, Feynman diagrams
can provide a good indication of which term in Eq.(18) contributes to the
process of interest. Thus in the following, even though we are not using the
usual Feynman-Dyson perturbative expansion, we will still refer to the
corresponding Feynman diagrams in guiding the choice of the relevant terms in
Eq.(18).

###  3.2 Decoupling

In this section we will explicitly calculate the decoupled Hamiltonian from
Eq.(18), for the light particle two and four point functions at the one loop
level. To simplify the notations we will take it as understood that all
operator expressions must be sandwiched between the high energy vacuum state
to obtain the decoupled Hamiltonian from Eq.(18).

####  3.2.1 Two Point Function of Light Fields

Figure 3:  One loop two point functions. Dashed lines represent heavy fields
and solid lines represent light fields.

There are four contributions to the two point function at one loop order which
we choose to specify by ordinary Feynman diagrams. Let’s consider the Figure 3
which comes from the $\frac{\lambda_{0}}{4!}\Phi_{L}^{4}(x)$ term in the total
Hamiltonian and make the mode expansion for the $\phi^{2}\chi^{2}$ piece:

$\displaystyle 6\frac{\lambda_{0}}{4!}\int
d^{3}x\,\sum_{M<k}\sum_{M<p}\frac{1}{\sqrt{2\epsilon_{k}}}\frac{1}{\sqrt{2\epsilon_{p}}}e^{i\left(\boldsymbol{k}-\boldsymbol{p}\right)\boldsymbol{x}}b_{k}b_{p}^{\dagger}\phi^{2}(x).$
(19)

Normal ordering this term gives

$\displaystyle 6\frac{\lambda_{0}}{4!}\int
d^{3}x\,\sum_{M<k}\sum_{M<p}\frac{1}{\sqrt{2\epsilon_{k}}}\frac{1}{\sqrt{2\epsilon_{p}}}e^{i\left(\boldsymbol{k}-\boldsymbol{p}\right)\boldsymbol{x}}\left(\left[b_{k},b_{p}^{\dagger}\right]+b_{p}^{\dagger}b_{k}\right)\phi^{2}(x).$
(20)

Keeping the commutator piece since only this survives upon sandwiching the
above expression between high energy vacuum states, we get

$\displaystyle\frac{\lambda_{0}}{8}\int
d^{3}x\int\frac{d^{3}k}{\left(2\pi\right)^{3}}\,\frac{\phi^{2}(x)}{\sqrt{k^{2}+m^{2}}}-\frac{\lambda_{0}}{8}\int
d^{3}x\int_{k<M}\frac{d^{3}k}{(2\pi)^{3}}\,\frac{\phi^{2}(x)}{\sqrt{k^{2}+m^{2}}}.$
(21)

We use dimensional regularization in $d=3-2\epsilon$ dimensions, where usual
UV divergence appears as a pole in Hamiltonian framework at $d=3$. The result
is

$\displaystyle-\int
d^{3}x\,\frac{\phi^{2}(x)}{2}\frac{\lambda_{0}m^{2}}{32\pi^{2}}\left(\frac{1}{\overline{\epsilon}}-\ln{\frac{m^{2}}{\mu^{2}}}+1\right)+\frac{\lambda_{0}}{4}C\int
d^{3}x\,\frac{\phi^{2}(x)}{2},$ (22)
$\displaystyle\frac{1}{\overline{\epsilon}}=\frac{1}{\epsilon}-\gamma+\ln{4\pi},$
$\displaystyle
C=-\int_{k<M}\frac{d^{3}k}{(2\pi)^{3}}\,\frac{1}{\sqrt{k^{2}+m^{2}}}.$

The term proportional to C arises from the restriction imposed on the momentum
of the high frequency part of the light field, $\chi(x)$, namely $p>M$. In
other words, in order to use dimensional regularization we extend the momentum
integral to the full range and subtract the infrared region, whose
contribution is $C$. This procedure will be followed for all such integrals
and we will show later that finite infrared contributions like $C$ will cancel
out when we do the matching to construct the effective Hamiltonian.
Similarly, the Figure 3 arises from the term

$\displaystyle\frac{\lambda_{2}}{4}\int
d^{3}x\,\sum_{k}\sum_{p}\frac{1}{\sqrt{2\omega_{k}}}\frac{1}{\sqrt{2\omega_{p}}}e^{i\left(\boldsymbol{k}-\boldsymbol{p}\right)\boldsymbol{x}}a_{k}a_{p}^{\dagger}\phi^{2}(x).$
(23)

Normal ordering gives,

$\displaystyle\frac{\lambda_{2}}{4}\int
d^{3}x\,\sum_{k}\sum_{p}\frac{1}{\sqrt{2\omega_{k}}}\frac{1}{\sqrt{2\omega_{p}}}e^{i\left(\boldsymbol{k}-\boldsymbol{p}\right)\boldsymbol{x}}\left(\left[a_{k},a_{p}^{\dagger}\right]+a_{p}^{\dagger}a_{k}\right)\phi^{2}(x).$
(24)

Using dimensional regularization, we can get as the contribution to the
decoupled Hamiltonian,

$\displaystyle-\int
d^{3}x\,\frac{\phi^{2}(x)}{2}\frac{\lambda_{2}M^{2}}{32\pi^{2}}\left(\frac{1}{\overline{\epsilon}}-\ln{\frac{M^{2}}{\mu^{2}}}+1\right).$
(25)

Figure 3 is proportional to $\lambda_{1}^{2}$ and $\phi^{2}$. Since it is
second order in coupling constant, it must come from the term
$\frac{i}{2}\left[H_{A},\Omega_{0}\right]$ in Eq.(18). Let us denote the
corresponding $H_{A}$ in this case by $H_{A}^{2,1}$ and consider the mode
expansion of $\frac{\lambda_{1}}{2}\Phi_{H}\Phi_{L}^{2}$

$\displaystyle\frac{\lambda_{1}}{2}\Phi_{H}\Phi_{L}^{2}$
$\displaystyle=\frac{\lambda_{1}}{2}\int
d^{3}x\,\sum_{k}\sum_{M<p}\sum_{M<q}\frac{1}{2^{\frac{3}{2}}\sqrt{\omega_{k}\epsilon_{p}\epsilon_{q}}}\left(a_{k}e^{i\boldsymbol{k}\boldsymbol{x}}+a_{k}^{\dagger}e^{-i\boldsymbol{k}\boldsymbol{x}}\right)$
(26)
$\displaystyle\left(b_{p}e^{i\boldsymbol{p}\boldsymbol{x}}+b_{p}^{\dagger}e^{-i\boldsymbol{p}\boldsymbol{x}}\right)\left(b_{q}e^{i\boldsymbol{q}\boldsymbol{x}}+b_{q}^{\dagger}e^{-i\boldsymbol{q}\boldsymbol{x}}\right)+\frac{\lambda_{1}}{2}\int
d^{3}x\,\sum_{k}\frac{1}{\sqrt{2\omega_{k}}}$
$\displaystyle\phi^{2}(x)\left(a_{k}e^{i\boldsymbol{k}\boldsymbol{x}}+a_{k}^{\dagger}e^{-i\boldsymbol{k}\boldsymbol{x}}\right)+...$

where the dots represent terms proportional to $\phi(x)$.
This expansion has a piece proportional to $\frac{\lambda_{1}}{2}\int
d^{3}x\,\sum_{k}\frac{1}{\sqrt{2\omega_{k}}}\left(a_{k}e^{i\boldsymbol{k}\boldsymbol{x}}+a_{k}^{\dagger}e^{-i\boldsymbol{k}\boldsymbol{x}}\right)\phi^{2}(x)$
which contributes to $H_{A}^{2,1}$. In addition, there’s another term, which
is in the form of $H_{B}$:

$\displaystyle\frac{\lambda_{1}}{2}\int
d^{3}x\,\sum_{k}\sum_{M<p}\sum_{M<q}\frac{1}{2^{\frac{3}{2}}\sqrt{\omega_{k}\epsilon_{p}\epsilon_{q}}}\left(a_{k}e^{i\boldsymbol{k}\boldsymbol{x}}+a_{k}^{\dagger}e^{-i\boldsymbol{k}\boldsymbol{x}}\right)b_{p}e^{i\boldsymbol{p}x}b_{q}^{\dagger}e^{-i\boldsymbol{q}\boldsymbol{x}}.$
(27)

Normal ordering this gives,

$\displaystyle\frac{\lambda_{1}}{2}\int
d^{3}x\,\sum_{k}\sum_{M<p}\sum_{M<q}\frac{1}{2^{\frac{3}{2}}\sqrt{\omega_{k}\epsilon_{p}\epsilon_{q}}}\left(a_{k}e^{i\boldsymbol{k}\boldsymbol{x}}+a_{k}^{\dagger}e^{-i\boldsymbol{k}\boldsymbol{x}}\right)e^{i\left(\boldsymbol{k}-\boldsymbol{p}\right)\boldsymbol{x}}\left(\left[b_{p},b_{q}^{\dagger}\right]+b_{q}^{\dagger}b_{p}\right).$
(28)

Evaluating this we get the net contribution for $H_{A}^{2,1}$ to be

$\displaystyle H_{A}^{2,1}=\int
d^{3}x\,\frac{\lambda_{1}}{2}\sum_{k}\frac{1}{\sqrt{2\omega_{k}}}\left(a_{k}e^{i\boldsymbol{k}\boldsymbol{x}}+a_{k}^{\dagger}e^{-i\boldsymbol{k}\boldsymbol{x}}\right)\left(-\frac{m^{2}}{16\pi^{2}}\left(\frac{1}{\overline{\epsilon}}-\ln{\frac{m^{2}}{\mu^{2}}}+1\right)+\frac{C}{2}+\phi^{2}(x)\right),$
(29)

where again
$C=-\int_{k<M}\frac{d^{3}k}{(2\pi)^{3}}\,\frac{1}{\sqrt{k^{2}+m^{2}}}$ as
before. We next evaluate $\Omega_{0}$ for this case by which we denote by
$\Omega_{0}^{2,1}$. This is obtained from from the condition
$i\left[H_{2},\Omega_{0}^{2,1}\right]+H_{A}^{2,1}=0$ which gives,

$\displaystyle\Omega_{0}^{2,1}=\int
d^{3}y\,\frac{\lambda_{1}}{2}\sum_{p}\frac{-i}{\sqrt{2\omega_{p}}\omega_{p}}\left(a_{p}e^{i\boldsymbol{p}\boldsymbol{y}}-a_{p}^{\dagger}e^{-i\boldsymbol{p}\boldsymbol{y}}\right)\left(-\frac{m^{2}}{16\pi^{2}}\left(\frac{1}{\overline{\epsilon}}-\ln{\frac{m^{2}}{\mu^{2}}}+1\right)+\frac{C}{2}+\phi^{2}(y)\right).$
(30)

From this we obtain,

$\displaystyle\frac{i}{2}\left[H_{A}^{2,1},\Omega_{0}^{2,1}\right]$
$\displaystyle=-\frac{i}{2}\int\int
d^{3}xd^{3}y\,\frac{\lambda_{1}^{2}}{4}\sum_{k}\sum_{p}\frac{-i}{2\sqrt{\omega_{k}\omega_{p}}\omega_{p}}\left[a_{k},-a_{p}^{\dagger}\right]e^{i\boldsymbol{k}\boldsymbol{x}}e^{-i\boldsymbol{p}\boldsymbol{y}}$
(31)
$\displaystyle\left(\frac{m^{2}}{16\pi^{2}}\left(\frac{1}{\overline{\epsilon}}-\ln{\frac{m^{2}}{\mu^{2}}}+1\right)-\frac{C}{2}\right)\phi^{2}(y)+\dots$
$\displaystyle=\int\int
d^{3}xd^{3}y~{}\frac{\lambda_{1}^{2}}{16}\int\frac{d^{3}k}{{(2\pi)}^{3}}\frac{1}{\omega_{k}^{2}}e^{i\boldsymbol{k}\left(\boldsymbol{x}-\boldsymbol{y}\right)}\left(\frac{m^{2}}{16\pi^{2}}\left(\frac{1}{\overline{\epsilon}}-\ln{\frac{m^{2}}{\mu^{2}}}+1\right)-\frac{C}{2}\right)\phi^{2}(y)+\dots,$

where the dots denote three other similar terms arising from the commutator.
The integral over $x$ forces the momentum k to zero, and
$\omega_{k}^{2}\rightarrow M^{2}$. Note that in terms of the associated
Feynman diagram, $k$ is the (zero) external momenta flowing into the vacuum
bubble of the tadpole. This association with the external momentum flow in
Feynman diagrams is a general feature of the terms in $H_{decoupled}$, as we
will see. Including all these contributions we get for this contribution to
the decoupled Hamiltonian,

$\displaystyle\frac{i}{2}\left[H_{A}^{2,1},\Omega_{0}^{2,1}\right]$
$\displaystyle=\int
d^{3}y\frac{4\lambda_{1}^{2}}{16M^{2}}\left(\frac{m^{2}}{16\pi^{2}}\left(\frac{1}{\overline{\epsilon}}-\ln{\frac{m^{2}}{\mu^{2}}}+1\right)-\frac{C}{2}\right)\phi^{2}(y)$
(32) $\displaystyle=\int
d^{3}x\,\frac{\lambda_{1}^{2}m^{2}}{32\pi^{2}M^{2}}\left(\frac{1}{\overline{\epsilon}}-\ln{\frac{m^{2}}{\mu^{2}}}+1\right)\frac{\phi^{2}(x)}{2}-\frac{\lambda_{1}^{2}C}{4M^{2}}\int
d^{3}x\,\frac{\phi^{2}(x)}{2}.$

Next, consider Figure 3 which is the last contribution to the two point
function at one loop order. Figure 3 has both light particle and heavy
particle propagators. It also arises from the expansion of
$\frac{\lambda_{1}}{2}\Phi_{H}\Phi_{L}^{2}$, which is proportional to $\phi$,
and is not explicitly listed in Eq.(26). This contribution is

$\displaystyle H_{A}^{2,2}=2\int
d^{3}x\,\sum_{k}\sum_{M<p}\frac{\lambda_{1}}{2}\frac{\phi(x)}{2\sqrt{\omega_{k}\epsilon_{p}}}\left(a_{k}e^{i\boldsymbol{k}\boldsymbol{x}}b_{p}e^{i\boldsymbol{p}\boldsymbol{x}}+h.c.\right).$
(33)

Similarly, from $i\left[H_{2},\Omega_{0}^{2,2}\right]+H_{A}^{2,2}=0$, we can
get the $\Omega_{0}^{2,2}$ that corresponds to $H_{A}^{2,2}$,

$\displaystyle\Omega_{0}^{2,2}=2\int
d^{3}y\,\sum_{q}\sum_{M<r}\frac{\lambda_{1}}{2}\left(\frac{-i\phi(y)}{2\sqrt{\omega_{q}\epsilon_{r}}\left(\omega_{q}+\epsilon_{r}\right)}a_{q}e^{i\boldsymbol{q}\boldsymbol{y}}b_{r}e^{i\boldsymbol{r}\boldsymbol{y}}+h.c.\right).$
(34)

Then, we need to calculate
$\frac{i}{2}\left[H_{A}^{2,2},\Omega_{0}^{2,2}\right]$. After a simple
calculation, and normal ordering we get:

$\displaystyle\frac{i}{2}\left[H_{A}^{2,2},\Omega_{0}^{2,2}\right]$
$\displaystyle=-\int\int
d^{3}xd^{3}y\,\sum_{k}\sum_{M<p}\lambda_{1}^{2}\frac{\phi(x)\phi(y)}{4\omega_{k}\epsilon_{p}\left(\omega_{k}+\epsilon_{p}\right)}e^{i\left(\boldsymbol{k}+\boldsymbol{p}\right)\left(\boldsymbol{x}-\boldsymbol{y}\right)},$
(35) $\displaystyle=-\int\int
d^{3}xd^{3}y\,\sum_{k}\sum_{M<p}\lambda_{1}^{2}\frac{\phi(x)\phi(y)e^{i\left(\boldsymbol{k}+\boldsymbol{p}\right)\left(\boldsymbol{x}-\boldsymbol{y}\right)}}{4\sqrt{k^{2}+M^{2}}\sqrt{p^{2}+m^{2}}\left(\sqrt{k^{2}+M^{2}}+\sqrt{p^{2}+m^{2}}\right)}.$

This integral is evaluated in Appendix A.1. The net contribution from Figure 3
to the decoupled Hamiltonian is then found to be

$\displaystyle\frac{m^{2}\lambda_{1}^{2}}{16\pi^{2}M^{2}}\left(\frac{1}{\overline{\epsilon}}-\ln{\frac{m^{2}}{\mu^{2}}}+1\right)\int
d^{3}x\,\frac{\phi^{2}(x)}{2}-\frac{\lambda_{1}^{2}C}{2M^{2}}\int
d^{3}x\,\frac{\phi^{2}(x)}{2}$ (36)
$\displaystyle-\frac{\lambda_{1}^{2}}{16\pi^{2}}\left(\frac{1}{\overline{\epsilon}}-\ln{\frac{M^{2}}{\mu^{2}}}+1\right)\left(1+\frac{m^{2}}{M^{2}}\right)\int
d^{3}x\,\frac{\phi^{2}(x)}{2}+\frac{\lambda_{1}^{2}}{32\pi^{2}M^{2}}\int
d^{3}x\,\frac{1}{2}(\nabla\phi(x))^{2}.$

Note that there is a finite kinetic energy correction term in Eq.(36), which
implies that there should as well be a momentum term
$\frac{\lambda_{1}^{2}}{32\pi^{2}M^{2}}\int d^{3}x\,\frac{1}{2}\Pi^{2}(x)$
with the same coefficient. In fact, this term can also be calculated, based on
Figure 3. However, it arises from the commutator
$\bra{0_{H}}-\frac{1}{2}[[H_{1},\Omega_{0}],\Omega_{1}]\ket{0_{H}}$ in the
higher order expansion of $H_{decoupled}$, which we have omitted in Eq.(18).
The exact calculation of this term is given in Appendix A.2 and it confirms
the expectations above. Putting contributions from all the diagrams together
and taking $\mu\approx M$ to get rid of log terms
$\ln{\frac{M^{2}}{\mu^{2}}}$, we have the net result for two point functions
in $H_{decoupled}$ as

$\displaystyle-\frac{\lambda_{0}m^{2}}{32\pi^{2}}\left(\frac{1}{\overline{\epsilon}}-\ln{\frac{m^{2}}{M^{2}}}+1\right)\int
d^{3}x\,\frac{\phi^{2}(x)}{2}-\frac{\lambda_{2}M^{2}}{32\pi^{2}}\left(\frac{1}{\overline{\epsilon}}+1\right)\int
d^{3}x\,\frac{\phi^{2}(x)}{2}$ (37)
$\displaystyle+\frac{3\lambda_{1}^{2}m^{2}}{32\pi^{2}M^{2}}\left(\frac{1}{\overline{\epsilon}}-\ln{\frac{m^{2}}{M^{2}}}+1\right)\int
d^{3}x\,\frac{\phi^{2}(x)}{2}-\frac{\lambda_{1}^{2}}{16\pi^{2}}\left(\frac{1}{\overline{\epsilon}}+1\right)\left(1+\frac{m^{2}}{M^{2}}\right)\int
d^{3}x\,\frac{\phi^{2}(x)}{2}$
$\displaystyle+\frac{\lambda_{1}^{2}}{32\pi^{2}M^{2}}\int
d^{3}x\,\frac{1}{2}(\Pi^{2}(x)+(\nabla\phi(x))^{2})+\left(\frac{\lambda_{0}}{4}-\frac{3\lambda_{1}^{2}}{2M^{2}}\right)C\int
d^{3}x\,\frac{\phi^{2}(x)}{2}.$

There are several finite contributions proportional to
$C=-\int_{k<M}\frac{d^{3}k}{(2\pi)^{3}}\,\frac{1}{\sqrt{k^{2}+m^{2}}}$, which
will not affect the renormalization of $H_{1}$ as we will show in section 3.3.
In fact, we will show in section 4,that they will only appear in
$H_{decoupled}$ and not in the physical effective Hamiltonian $H_{eff}$ which
is constructed by a matching process. Another way to think about the effects
of these finite terms is that whenever we have a light field in the loop, our
calculation will produce these finite terms along with (in a linear way) the
"troublesome" large log terms $\ln{\frac{m^{2}}{M^{2}}}$. Therefore, as long
as the large log terms can be canceled during matching, we can convince
ourselves that these extra finite terms will also be canceled and therefore
not appear in $H_{eff}$. All this is essentially a reflection of the fact that
the infrared structure of the full and effective theory are the same.

####  3.2.2 Four Point Function of Light Fields

Figure 4:  Four point function at tree level.

We will begin with the tree level contribution at order $1/M^{2}$. In the
calculation of Figure 3 in section 3.2.1, there is one more term from the
commutator $\frac{i}{2}\left[H_{A}^{2,1},\Omega_{0}^{2,1}\right]$ left
unexplored that contributes to the four point function at tree level shown in
Figure 4:

$\displaystyle\frac{i}{2}\left[H_{A}^{2,1},\Omega_{0}^{2,1}\right]_{tree}$
$\displaystyle=i\int d^{3}x\int
d^{3}y\,\frac{\lambda_{1}^{2}}{4}\sum_{k}\sum_{p}\frac{-i}{2\sqrt{\omega_{k}\omega_{p}}\omega_{p}}\phi^{2}(x)\phi^{2}(y)[a_{k},-a_{p}^{\dagger}]e^{i\boldsymbol{kx}-i\boldsymbol{py}},$
(38) $\displaystyle=-\int d^{3}x\int
d^{3}y\int\frac{d^{3}k}{(2\pi)^{3}}\frac{\lambda_{1}^{2}}{8(k^{2}+M^{2})}\phi^{2}(x)\phi^{2}(y)e^{i\boldsymbol{k}(\boldsymbol{x}-\boldsymbol{y})},$
$\displaystyle\approx-\frac{3\lambda_{1}^{2}}{M^{2}}\int
d^{3}x\,\frac{\phi^{4}(x)}{4!}.$

As noted in the discussion of the tadpole contribution, similarly here, $k$
should be viewed as the sum of the incoming momenta at vertex $x$ and the
outgoing ones at $y$, and therefore $k\ll M$.Expanding in $k^{2}/M^{2}$ then
gives the leading contribution given above.

Figure 5:  Four point functions at one loop level.

In loop calculation using dimensional regularization, we will again encounter
extra finite terms proportional to
$C=-\int_{k<M}\frac{d^{3}k}{(2\pi)^{3}}\,\frac{1}{\sqrt{k^{2}+m^{2}}}$,
similar to those we encountered in section 3.2.1. However, in the interest of
making our expressions appear more compact, we will omit writing them down
explicitly.We will confirm in section 4 that all large log terms
$\ln{\frac{m^{2}}{M^{2}}}$ produced in the four point function calculation are
canceled out when constructing the effective Hamiltonian, which implies that
these extra finite terms like $C$ will be canceled as well. Furthermore, to
simplify calculations, we will set the total incoming and outgoing external
momenta to zero.
Let’s consider Figure 5. This diagram is of order $\lambda_{0}^{2}$, and it
arises from the term $\frac{i}{2}\left[H_{A}^{4,1},\Omega_{0}^{4,1}\right]$
where $H_{A}^{4,1}$ comes from the mode expansion of
$\frac{\lambda_{0}}{4!}\Phi_{L}^{4}(x)$. In this expansion we must pick a term
of form $\phi^{2}\chi^{2}$ which gives

$\displaystyle H_{A}^{4,1}=\int
d^{3}x\,\frac{\lambda_{0}}{4}\phi^{2}(x)\sum_{k}\sum_{p}\frac{1}{2\sqrt{\epsilon_{p}\epsilon_{k}}}\left(b_{p}b_{k}e^{i\left(\boldsymbol{p}+\boldsymbol{k}\right)\boldsymbol{x}}+b_{p}^{\dagger}b_{k}^{\dagger}e^{-i\left(\boldsymbol{p}+\boldsymbol{k}\right)\boldsymbol{x}}\right).$
(39)

From the equation $i\left[H_{2},\Omega_{0}^{4,1}\right]+H_{A}^{4,1}=0$, we can
get the corresponding $\Omega_{0}^{4,1}$ to be

$\displaystyle\Omega_{0}^{4,1}=\int
d^{3}y\,\frac{\lambda_{0}}{4}\phi^{2}(y)\sum_{k^{{}^{\prime}}}\sum_{p^{{}^{\prime}}}\left(\frac{-ie^{i\left(\boldsymbol{k}^{{}^{\prime}}+\boldsymbol{p}^{{}^{\prime}}\right)\boldsymbol{y}}}{2\sqrt{\epsilon_{k^{{}^{\prime}}}\epsilon_{p^{{}^{\prime}}}}\left(\epsilon_{k^{{}^{\prime}}}+\epsilon_{p^{{}^{\prime}}}\right)}b_{k^{{}^{\prime}}}b_{p^{{}^{\prime}}}+\frac{ie^{-i\left(\boldsymbol{k}^{{}^{\prime}}+\boldsymbol{p}^{{}^{\prime}}\right)\boldsymbol{y}}}{2\sqrt{\epsilon_{k^{{}^{\prime}}}\epsilon_{p^{{}^{\prime}}}}\left(\epsilon_{k^{{}^{\prime}}}+\epsilon_{p^{{}^{\prime}}}\right)}b_{k^{{}^{\prime}}}^{\dagger}b_{p^{{}^{\prime}}}^{\dagger}\right).$
(40)

Then $\frac{i}{2}\left[H_{A}^{4,1},\Omega_{0}^{4,1}\right]$ yields

$\displaystyle-\frac{3\lambda_{0}^{2}}{32\pi^{2}}\left(\frac{1}{\overline{\epsilon}}-\ln\frac{m^{2}}{\mu^{2}}\right)\int
d^{3}x\,\frac{\phi^{4}(x)}{4!}.$ (41)

In Eq.(39) and Eq.(40) and subsequently their commutator, we may view
$\boldsymbol{k}+\boldsymbol{p}$ as the total incoming momentum since that is
the momentum entering the vertex at $x$. In evaluating the commutator we have
set this to zero which then implies
$\epsilon_{k}=\epsilon_{p}=\sqrt{k^{2}+m^{2}}$.
The next contribution is from Figure 5. The calculation is similar to the
first one except the corresponding $H_{A}$ is different. In this case we have

$\displaystyle H_{A}^{4,2}=\int
d^{3}x\,\frac{\lambda_{2}}{4}\phi^{2}(x)\sum_{k}\sum_{p}\frac{1}{2\sqrt{\omega_{p}\omega_{k}}}\left(a_{p}a_{k}e^{i\left(\boldsymbol{p}+\boldsymbol{k}\right)\boldsymbol{x}}+a_{p}^{\dagger}a_{k}^{\dagger}e^{-i\left(\boldsymbol{p}+\boldsymbol{k}\right)\boldsymbol{x}}\right),$
(42a) $\displaystyle\Omega_{0}^{4,2}=\int
d^{3}y\,\frac{\lambda_{2}}{4}\phi^{2}(y)\sum_{k^{{}^{\prime}}}\sum_{p^{{}^{\prime}}}\left(\frac{-ie^{i\left(\boldsymbol{k}^{{}^{\prime}}+\boldsymbol{p}^{{}^{\prime}}\right)\boldsymbol{y}}}{2\sqrt{\omega_{k^{{}^{\prime}}}\omega_{p^{{}^{\prime}}}}\left(\omega_{k^{{}^{\prime}}}+\omega_{p^{{}^{\prime}}}\right)}a_{k^{{}^{\prime}}}a_{p^{{}^{\prime}}}+\frac{ie^{-i\left(\boldsymbol{k}^{{}^{\prime}}+\boldsymbol{p}^{{}^{\prime}}\right)\boldsymbol{y}}}{2\sqrt{\omega_{k^{{}^{\prime}}}\omega_{p^{{}^{\prime}}}}\left(\omega_{k^{{}^{\prime}}}+\omega_{p^{{}^{\prime}}}\right)}a_{k^{{}^{\prime}}}^{\dagger}a_{p^{{}^{\prime}}}^{\dagger}\right).$
(42b)

The commutator is then evaluated as:

$\displaystyle\frac{i}{2}\left[H_{A}^{4,2},\Omega_{0}^{4,2}\right]=-\frac{3\lambda_{2}^{2}}{32\pi^{2}}\left(\frac{1}{\overline{\epsilon}}-\ln\frac{M^{2}}{\mu^{2}}\right)\int
d^{3}x\,\frac{\phi^{4}(x)}{4!}.$ (43)

$H_{A}^{4,3}$ | $\Omega_{0}^{4,3,1}$ | $\Omega_{0}^{4,3,2}$
---|---|---
First combination (a) |  |
$\lambda_{1}a\phi^{2}$ | $\lambda_{1}a^{\dagger}b^{\dagger}b^{\dagger}$ | $\lambda_{0}bb\phi^{2}$
$\lambda_{1}a^{\dagger}\phi^{2}$ | $\lambda_{1}abb$ | $\lambda_{0}b^{\dagger}b^{\dagger}\phi^{2}$
Second combination (b) |  |
$\lambda_{1}abb$ | $\lambda_{1}a^{\dagger}\phi^{2}$ | $\lambda_{0}b^{\dagger}b^{\dagger}\phi^{2}$
$\lambda_{1}a^{\dagger}b^{\dagger}b^{\dagger}$ | $\lambda_{1}a\phi^{2}$ | $\lambda_{0}bb\phi^{2}$
Third combination (c) |  |
$\lambda_{1}abb$ | $\lambda_{0}b^{\dagger}b^{\dagger}\phi^{2}$ | $\lambda_{1}a^{\dagger}\phi^{2}$
$\lambda_{1}a^{\dagger}b^{\dagger}b^{\dagger}$ | $\lambda_{0}bb\phi^{2}$ | $\lambda_{1}a\phi^{2}$
Fourth combination (d) |  |
$\lambda_{0}bb\phi^{2}$ | $\lambda_{1}a^{\dagger}b^{\dagger}b^{\dagger}$ | $\lambda_{1}a\phi^{2}$
$\lambda_{0}b^{\dagger}b^{\dagger}\phi^{2}$ | $\lambda_{1}abb$ | $\lambda_{1}a^{\dagger}\phi^{2}$
Table 1: Combinations of
$-\frac{1}{3}\left[\left[H_{A}^{4,3},\Omega_{0}^{4,3,1}\right],\Omega_{0}^{4,3,2}\right]$
that contribute to Figure 5. Various entries are identified only by coupling
constant and operator structure.

Next we calculate the contributions that correspond to Figure 5. Notice that,
in this diagram, we have the product of coupling constants as
$\lambda_{0}\lambda_{1}^{2}$, which means we need both
$-\frac{1}{3}\left[\left[H_{A},\Omega_{0}\right],\Omega_{0}\right]$ and
$-\frac{1}{2}\left[\left[H_{B},\Omega_{0}\right],\Omega_{0}\right]$ from
Eq.(18). For consistency of notation we label $H_{A}$ in the first commutator
as $H_{A}^{4,3}$, the first $\Omega_{0}$ next to $H_{A}^{4,3}$ as
$\Omega_{0}^{4,3,1}$, the second $\Omega_{0}$ as $\Omega_{0}^{4,3,2}$ and
$H_{B}$ as $H_{B}^{4,3}$. There are many ways to pick $H_{A}^{4,3}$,
$\Omega_{0}^{4,3,1}$, and $\Omega_{0}^{4,3,2}$ in
$-\frac{1}{3}\left[\left[H_{A}^{4,3},\Omega_{0}^{4,3,1}\right],\Omega_{0}^{4,3,2}\right]$,
and we split them into four kinds of combinations shown in Table 1. We will
explicitly show the calculation of the first combination in Table 1, i.e., (a)
while a detailed discussion of contributions of the others is relegated to
appendix A.3.
Consider the following terms from the first combination in Table 1, i.e., (a):

$\displaystyle H_{A}^{4,3}$ $\displaystyle=\frac{\lambda_{1}}{2}\int
d^{3}x\,\sum_{k}\frac{\phi^{2}(x)}{\sqrt{2\omega_{k}}}a_{k}e^{i\boldsymbol{k}\boldsymbol{x}},$
(44a) $\displaystyle\Omega_{0}^{4,3,1}$
$\displaystyle=\frac{\lambda_{1}}{2}\int
d^{3}y\,\sum_{k^{{}^{\prime}}}\sum_{p}\sum_{q}\frac{ie^{-i\left(\boldsymbol{k}^{{}^{\prime}}+\boldsymbol{p}+\boldsymbol{q}\right)\boldsymbol{y}}}{2^{\frac{3}{2}}\sqrt{\omega_{k^{{}^{\prime}}}\epsilon_{p}\epsilon_{q}}\left(\omega_{k^{{}^{\prime}}}+\epsilon_{p}+\epsilon_{q}\right)}a_{k^{{}^{\prime}}}^{\dagger}b_{p}^{\dagger}b_{q}^{\dagger},$
(44b) $\displaystyle\Omega_{0}^{4,3,2}$
$\displaystyle=\frac{\lambda_{0}}{4}\int
d^{3}z\,\sum_{p^{{}^{\prime}}}\sum_{q^{{}^{\prime}}}\phi^{2}(z)\frac{-ie^{i\left(\boldsymbol{p}^{{}^{\prime}}+\boldsymbol{q}^{{}^{\prime}}\right)\boldsymbol{z}}}{2\sqrt{\epsilon_{p^{{}^{\prime}}}\epsilon_{q^{{}^{\prime}}}}\left(\epsilon_{p^{{}^{\prime}}}+\epsilon_{q^{{}^{\prime}}}\right)}b_{p^{{}^{\prime}}}b_{q^{{}^{\prime}}}.$
(44c)

We first calculate $\left[H_{A}^{4,3},\Omega_{0}^{4,3,1}\right]$

$\displaystyle\left[H_{A}^{4.3},\Omega_{0}^{4,3,1}\right]$
$\displaystyle=\frac{\lambda_{1}^{2}}{4}\int d^{3}x\int
d^{3}y\,\phi^{2}(x)\sum_{k}\sum_{k^{{}^{\prime}}}\sum_{p}\sum_{q}\frac{ie^{i\boldsymbol{k}\boldsymbol{x}}e^{-i\left(\boldsymbol{k}^{{}^{\prime}}+\boldsymbol{p}+\boldsymbol{q}\right)\boldsymbol{y}}}{2^{2}\sqrt{\omega_{k}\omega_{k^{{}^{\prime}}}\epsilon_{p}\epsilon_{q}}\left(\omega_{k^{{}^{\prime}}}+\epsilon_{p}+\epsilon_{q}\right)}$
(45)
$\displaystyle\left[a_{k},a_{k^{{}^{\prime}}}^{\dagger}b_{p}^{\dagger}b_{q}^{\dagger}\right],$
$\displaystyle=\frac{\lambda_{1}^{2}}{4}\int d^{3}x\int
d^{3}y\,\phi^{2}(x)\sum_{k}\sum_{p}\sum_{q}\frac{ie^{i\boldsymbol{k}\left(\boldsymbol{x}-\boldsymbol{y}\right)}e^{-i\left(\boldsymbol{p}+\boldsymbol{q}\right)\boldsymbol{y}}}{2^{2}\omega_{k}\sqrt{\epsilon_{p}\epsilon_{q}}\left(\omega_{k}+\epsilon_{p}+\epsilon_{q}\right)}b_{p}^{\dagger}b_{q}^{\dagger},$
$\displaystyle=\frac{i\lambda_{1}^{2}}{16}\int
d^{3}x\,\phi^{2}(x)\sum_{p}\sum_{q}\frac{e^{-i\left(\boldsymbol{p}+\boldsymbol{q}\right)\boldsymbol{x}}}{M\sqrt{\epsilon_{p}\epsilon_{q}}\left(M+\epsilon_{p}+\epsilon_{q}\right)}b^{\dagger}_{p}b^{\dagger}_{q}.$

In the above, we have put the total external momentum to zero, which implies
the momentum $k$ associated with heavy particle is therefore 0 and $w_{k}=M$.
Next we have

$\displaystyle-\frac{1}{3}\left[\left[H_{A}^{4,3},\Omega_{0}^{4,3,1}\right],\Omega_{0}^{4,3,2}\right]$
$\displaystyle=-\frac{\lambda_{1}^{2}\lambda_{0}}{192}\int d^{3}x\int
d^{3}z\,\sum_{p}\sum_{q}\sum_{p^{{}^{\prime}}}\sum_{q^{{}^{\prime}}}\phi^{2}(x)\phi^{2}(z)$
(46)
$\displaystyle\frac{e^{i\left(\boldsymbol{p}^{{}^{\prime}}+\boldsymbol{q}^{{}^{\prime}}\right)\boldsymbol{z}-i\left(\boldsymbol{p}+\boldsymbol{q}\right)\boldsymbol{x}}}{2^{2}M\sqrt{\epsilon_{p}\epsilon_{q}\epsilon_{p^{{}^{\prime}}}\epsilon_{q^{{}^{\prime}}}}\left(\epsilon_{p^{{}^{\prime}}}+\epsilon_{q^{{}^{\prime}}}\right)\left(M+\epsilon_{p}+\epsilon_{q}\right)}\left[b_{p}^{\dagger}b_{q}^{\dagger},b_{p^{{}^{\prime}}}b_{q^{{}^{\prime}}}\right],$
$\displaystyle=\frac{\lambda_{1}^{2}\lambda_{0}}{192}\int d^{3}x\int
d^{3}z\,\sum_{p}\sum_{q}\phi^{2}(x)\phi^{2}(z)\frac{e^{i\left(\boldsymbol{p}+\boldsymbol{q}\right)\left(\boldsymbol{x}-\boldsymbol{z}\right)}}{2^{2}M\epsilon_{p}^{3}\left(M+2\epsilon{p}\right)},$
$\displaystyle=\frac{\lambda_{1}^{2}\lambda_{0}}{192}\int
d^{3}x\,\phi^{4}(x)\int\frac{d^{3}p}{\left(2\pi\right)^{3}}\,\frac{1}{2^{2}M\epsilon_{p}^{3}\left(M+2\epsilon_{p}\right)},$

where we have used the fact that
$\boldsymbol{p}+\boldsymbol{q}=\boldsymbol{k}=0$, and thus
$\epsilon_{p}=\epsilon_{q}$. Including the Hermitian conjugate, we have the
result for combination Table 1(a) to be,

$\frac{\lambda_{1}^{2}\lambda_{0}}{96}\int
d^{3}x\,\phi^{4}(x)\int\frac{d^{3}p}{\left(2\pi\right)^{3}}\,\frac{1}{2^{2}M\epsilon_{p}^{3}\left(M+2\epsilon_{p}\right)}.$
(47)

The combinations Table 1(b) and Table 1(c) will yield the same result:

$\displaystyle\frac{\lambda_{1}^{2}\lambda_{0}}{96}\int
d^{3}x\,\phi^{4}(x)\int\frac{d^{3}p}{(2\pi)^{3}}\,\frac{1}{2M^{2}\epsilon_{p}^{3}}.$
(48)

and Table 1(d) gives,

$\displaystyle\frac{\lambda_{1}^{2}\lambda_{0}}{96}\int
d^{3}x\,\phi^{4}(x)\int\frac{d^{3}p}{(2\pi)^{3}}\,\frac{1}{M^{2}\epsilon_{p}^{2}(M+2\epsilon_{p})}.$
(49)

Adding all four kinds of combinations together, we will get

$\displaystyle\frac{\lambda_{1}^{2}\lambda_{0}}{96}\int
d^{3}x\,\phi^{4}(x)\int\frac{d^{3}p}{\left(2\pi\right)^{3}}\,\left(\frac{1}{M^{2}\epsilon_{p}^{2}\left(M+2\epsilon_{p}\right)}+\frac{1}{2\epsilon_{p}^{3}M\left(M+2\epsilon_{p}\right)}+\frac{1}{M^{2}\epsilon_{p}^{3}}\right),$
(50) $\displaystyle=\frac{\lambda_{1}^{2}\lambda_{0}}{64}\int
d^{3}x\phi^{4}(x)\int\frac{d^{3}p}{\left(2\pi\right)^{3}}\,\frac{1}{M^{2}\epsilon_{p}^{3}},$
$\displaystyle=\frac{3\lambda_{1}^{2}\lambda_{0}}{32\pi^{2}M^{2}}\int
d^{3}x\,\frac{\phi^{4}(x)}{4!}\left(\frac{1}{\overline{\epsilon}}-\ln\frac{m^{2}}{\mu^{2}}\right).$

$H_{B}^{4,3}$ | $\Omega_{0}^{4,3,1}$ | $\Omega_{0}^{4,3,2}$
---|---|---
$\lambda_{1}b^{\dagger}b^{\dagger}a$ | $\lambda_{0}bb\phi^{2}$ | $\lambda_{1}a^{\dagger}\phi^{2}$
$\lambda_{1}b^{\dagger}b^{\dagger}a$ | $\lambda_{1}a^{\dagger}\phi^{2}$ | $\lambda_{0}bb\phi^{2}$
$\lambda_{1}a^{\dagger}bb$ | $\lambda_{1}a\phi^{2}$ | $\lambda_{0}b^{\dagger}b^{\dagger}\phi^{2}$
$\lambda_{1}a^{\dagger}bb$ | $\lambda_{0}b^{\dagger}b^{\dagger}\phi^{2}$ | $\lambda_{1}a\phi^{2}$
Table 2: Combinations of
$-\frac{1}{2}\left[\left[H_{B}^{4,3},\Omega_{0}^{4,3,1}\right],\Omega_{0}^{4,3,2}\right]$
that contribute to Figure 5.

All possible combinations of
$-\frac{1}{2}\left[\left[H_{B}^{4,3},\Omega_{0}^{4,3,1}\right],\Omega_{0}^{4,3,2}\right]$
are shown in Table 2, and the result is

$\displaystyle\frac{3\lambda_{1}^{2}\lambda_{0}}{32\pi^{2}M^{2}}\int
d^{3}x\,\frac{\phi^{4}(x)}{4!}\left(\frac{1}{\overline{\epsilon}}-\ln\frac{m^{2}}{\mu^{2}}\right).$
(51)

Therefore, the total contribution from Figure 5 is,

$\displaystyle\frac{3\lambda_{1}^{2}\lambda_{0}}{16\pi^{2}M^{2}}\int
d^{3}x\,\frac{\phi^{4}(x)}{4!}\left(\frac{1}{\overline{\epsilon}}-\ln\frac{m^{2}}{\mu^{2}}\right).$
(52)

$H_{B}^{4,4}$ | $\Omega_{0}^{4,4,1}$ | $\Omega_{0}^{4,4,2}$
---|---|---
First combination (a) |  |
$\lambda_{0}b^{\dagger}b^{\dagger}\phi^{2}$ | $\lambda_{1}a^{\dagger}b^{\dagger}\phi$ | $\lambda_{1}ab\phi$
$\lambda_{0}b^{\dagger}b\phi^{2}$ | $\lambda_{1}ab\phi$ | $\lambda_{1}a^{\dagger}b^{\dagger}\phi$
Second combination (b) |  |
$\lambda_{1}b^{\dagger}a\phi$ | $\lambda_{1}a^{\dagger}b^{\dagger}\phi$ | $\lambda_{0}bb\phi^{2}$
$\lambda_{1}a^{\dagger}b\phi$ | $\lambda_{1}ab\phi$ | $\lambda_{0}b^{\dagger}b^{\dagger}\phi^{2}$
Third combination (c) |  |
$\lambda_{1}b^{\dagger}a\phi$ | $\lambda_{0}bb\phi^{2}$ | $\lambda_{1}a^{\dagger}b^{\dagger}\phi$
$\lambda_{1}a^{\dagger}b^{\dagger}\phi$ | $\lambda_{0}b^{\dagger}b^{\dagger}\phi^{2}$ | $\lambda_{1}ab\phi$
Table 3: Combinations of
$-\frac{1}{2}\left[\left[H_{B}^{4,4},\Omega_{0}^{4,4,1}\right],\Omega_{0}^{4,4,2}\right]$
that contribute to Figure 5.

As for Figure 5, only the term
$-\frac{1}{2}\left[\left[H_{B}^{4,4},\Omega_{0}^{4,4,1}\right],\Omega_{0}^{4,4,2}\right]$
will contribute. Similarly, we divide the whole commutator into several kinds
of combinations shown in Table 3 and calculate each separately. The final
results are given below and details may be found in appendix A.4.
Table 3(a) will give

$\displaystyle\frac{\lambda_{0}\lambda_{1}^{2}}{16}\int
d^{3}x\,\phi^{4}(x)\int\frac{d^{3}k}{\left(2\pi\right)^{3}}\frac{1}{\omega_{k}\epsilon_{k}^{2}\left(\omega_{k}+\epsilon_{k}\right)^{2}}.$
(53)

Since Table 3(b) and Table 3(c) only differ from an exchange of
$\Omega_{0}^{4,4,1}$ and $\Omega_{0}^{4,4,2}$, they will give the same result:

$\displaystyle\frac{\lambda_{1}^{2}\lambda_{0}}{32}\int
d^{3}x\,\phi^{4}(x)\int\frac{d^{3}p}{\left(2\pi\right)^{3}}\,\frac{1}{\omega_{p}\epsilon_{p}^{3}\left(\omega_{p}+\epsilon_{p}\right)}.$
(54)

Adding the contributions from three combinations we get,

$\displaystyle-\frac{1}{2}\left[\left[H_{B}^{4,4},\Omega_{0}^{4,4,1}\right],\Omega_{0}^{4,4,2}\right]=\frac{\lambda_{1}^{2}\lambda_{0}}{16}\int
d^{3}x\,\phi^{4}(x)\int\frac{d^{3}k}{\left(2\pi\right)^{3}}\,\left[\frac{1}{\omega_{k}\epsilon_{k}^{2}\left(\omega_{k}+\epsilon_{k}\right)^{2}}+\frac{1}{\omega_{k}\epsilon_{k}^{3}\left(\omega_{k}+\epsilon_{k}\right)}\right].$
(55)

Let’s consider
$\int\frac{d^{3}k}{\left(2\pi\right)^{3}}\frac{1}{\omega_{k}\epsilon_{k}^{3}\left(\omega_{k}+\epsilon_{k}\right)}$
first,

$\displaystyle\int\frac{d^{3}k}{\left(2\pi\right)^{3}}\,\frac{1}{\omega_{k}\epsilon_{k}^{3}\left(\omega_{k}+\epsilon_{k}\right)}$
$\displaystyle=\int\frac{d^{3}k}{\left(2\pi\right)^{3}}\,\left(\frac{1}{\epsilon^{3}_{k}\left(M^{2}-m^{2}\right)}-\frac{1}{\omega_{k}\epsilon_{k}^{2}\left(\omega_{k}+\epsilon_{k}\right)\left(\omega_{k}-\epsilon_{k}\right)}\right),$
(56)
$\displaystyle\approx\int\frac{d^{3}k}{\left(2\pi\right)^{3}}\,\left(\frac{1}{\epsilon^{3}_{k}M^{2}}-\frac{1}{\omega_{k}\epsilon_{k}^{2}\left(\omega_{k}+\epsilon_{k}\right)\left(\omega_{k}-\epsilon_{k}\right)}\right).$

Combining the last term with the first one in in Eq.(55), we get,

$\displaystyle\int\frac{d^{3}k}{\left(2\pi\right)^{3}}\,\left(\frac{1}{\omega_{k}\epsilon_{k}^{2}\left(\omega_{k}+\epsilon_{k}\right)^{2}}-\frac{1}{\omega_{k}\epsilon_{k}^{2}\left(\omega_{k}+\epsilon_{k}\right)\left(\omega_{k}-\epsilon_{k}\right)}\right),$
(57)
$\displaystyle=\int\frac{d^{3}k}{\left(2\pi\right)^{3}}\,\frac{-2\epsilon_{k}}{\omega_{k}\epsilon_{k}^{2}\left(\omega_{k}+\epsilon_{k}\right)^{2}\left(\omega_{k}-\epsilon_{k}\right)},$
$\displaystyle=\int\frac{d^{3}k}{\left(2\pi\right)^{3}}\,\frac{-2\epsilon_{k}\left(\omega_{k}-\epsilon_{k}\right)}{\omega_{k}\epsilon_{k}^{2}\left(\omega_{k}+\epsilon_{k}\right)^{2}\left(\omega_{k}-\epsilon_{k}\right)^{2}},$
$\displaystyle\approx\int\frac{d^{3}k}{\left(2\pi\right)^{3}}\,\left(\frac{-2}{\epsilon_{k}M^{4}}+\frac{2}{\omega_{k}M^{4}}\right).$

To the order $O(\frac{1}{M^{2}})$, we only need to keep
$\int\frac{d^{3}k}{\left(2\pi\right)^{3}}\frac{-2}{\omega_{k}M^{4}}$. The net
contribution from Figure 5 is then,

$\displaystyle\frac{\lambda_{1}^{2}\lambda_{0}}{16}\int
d^{3}x\,\phi^{4}(x)\int\frac{d^{3}k}{\left(2\pi\right)^{3}}\left(\frac{2}{\omega_{k}M^{4}}+\frac{1}{\epsilon_{k}^{3}M^{2}}\right),$
(58) $\displaystyle=-\frac{\lambda_{1}^{2}\lambda_{0}}{64\pi^{2}M^{2}}\int
d^{3}x\,\phi^{4}(x)\left(\left(\frac{1}{\overline{\epsilon}}+1-\ln\frac{M^{2}}{\mu^{2}}\right)-\left(\frac{1}{\overline{\epsilon}}-\ln\frac{m^{2}}{\mu^{2}}\right)\right)m,$
$\displaystyle=-\frac{3\lambda_{0}\lambda_{1}^{2}}{8\pi^{2}M^{2}}\left(1-\ln\frac{M^{2}}{\mu^{2}}+\ln\frac{m^{2}}{\mu^{2}}\right)\int
d^{3}x\,\frac{\phi^{4}(x)}{4!}.$

Taking $\mu\approx M$ and summing the contributions from the four diagrams
together, we get the final result of four point functions at one loop level,
up to order $O(\frac{1}{M^{2}})$:

$\displaystyle-\frac{3\lambda_{0}^{2}}{32\pi^{2}}\left(\frac{1}{\overline{\epsilon}}-\ln\frac{m^{2}}{M^{2}}\right)\int
d^{3}x\,\frac{\phi^{4}(x)}{4!}-\frac{3\lambda_{2}^{2}}{32\pi^{2}}\frac{1}{\overline{\epsilon}}\int
d^{3}x\,\frac{\phi^{4}(x)}{4!}$ (59)
$\displaystyle+\frac{3\lambda_{1}^{2}\lambda_{0}}{16\pi^{2}M^{2}}\left(\frac{1}{\overline{\epsilon}}-\ln\frac{m^{2}}{M^{2}}\right)\int
d^{3}x\,\frac{\phi^{4}(x)}{4!}-\frac{3\lambda_{1}^{2}\lambda_{0}}{8\pi^{2}M^{2}}\left(1+\ln\frac{m^{2}}{M^{2}}\right)\int
d^{3}x\,\frac{\phi^{4}(x)}{4!}.$

###  3.3 Renormalization

We argued in section 2.3 that the UV divergence in the calculation of
$H_{decoupled}$ contains information regarding the renormalization of $H_{1}$,
where $H_{1}$ is just the low energy part in the full Hamiltonian. Let’s first
write $H_{1}$ as

$\displaystyle H_{1}=\int
d^{3}x\,\frac{1}{2}(\Pi_{bare}^{2}(x)+(\nabla\phi_{bare}(x))^{2})+\frac{1}{2}m_{bare}^{2}\phi_{bare}^{2}(x)+\frac{\lambda_{0}^{bare}}{4!}\phi_{bare}^{4}(x),$
(60)

and then introduce the renormalization Z factor such that

$\displaystyle\phi_{bare}=\sqrt{Z_{\phi}}\phi,$ (61a)
$\displaystyle\Pi_{bare}=\frac{\Pi}{\sqrt{Z_{\phi}}},$ (61b) $\displaystyle
m_{bare}=\sqrt{Z_{m}}m,$ (61c)
$\displaystyle\lambda_{0}^{bare}=Z_{\lambda_{0}}\lambda_{0}.$ (61d)

Expanding these Z factors in terms of $\frac{1}{\overline{\epsilon}}$, we have

$\displaystyle
Z_{\phi}=1+\delta^{1}_{\phi}(\frac{1}{\overline{\epsilon}})+O(\frac{1}{\overline{\epsilon}^{2}}),$
(62a) $\displaystyle
Z_{m}=1+\delta^{1}_{m}(\frac{1}{\overline{\epsilon}})+O(\frac{1}{\overline{\epsilon}^{2}}),$
(62b) $\displaystyle
Z_{\lambda_{0}}=1+\delta^{1}_{\lambda_{0}}(\frac{1}{\overline{\epsilon}})+O(\frac{1}{\overline{\epsilon}^{2}}).$
(62c)

Implementing these expansion, we can rewrite $H_{1}$ as

$\displaystyle H_{1}=$ $\displaystyle\int
d^{3}x\,\frac{1}{2}(\Pi^{2}(x)+(\nabla\phi(x))^{2})+\frac{1}{2}m^{2}\phi^{2}(x)+\frac{\lambda_{0}}{4!}\phi^{2}(x)$
(63)
$\displaystyle\frac{1}{2}\delta_{\phi}^{1}(-\Pi^{2}(x)+(\nabla\phi(x))^{2})+\frac{1}{2}(\delta_{\phi}^{1}+\delta_{m}^{1})m^{2}\phi^{2}(x)+\frac{\lambda_{0}}{4!}(2\delta_{\phi}^{1}+\delta_{\lambda_{0}}^{1})\phi^{4}(x)+O(\frac{1}{\overline{\epsilon}^{2}}).$

Using $\overline{MS}$ scheme, we can cancel the UV divergence in
$H_{decoupled}$ by counterterms in $H_{1}$. From two point calculation in
section 3.2.1, we have divergent terms:

$\displaystyle\left(-\frac{\lambda_{0}m^{2}}{32\pi^{2}}-\frac{\lambda_{2}M^{2}}{32\pi^{2}}+\frac{\lambda_{1}^{2}m^{2}}{32\pi^{2}M^{2}}-\frac{\lambda_{1}^{2}}{16\pi^{2}}\right)\frac{1}{\overline{\epsilon}}\int
d^{3}x\,\frac{\phi^{2}(x)}{2}.$ (64)

From four point calculation in section 3.2.2, we have divergent terms:

$\displaystyle\left(-\frac{3\lambda_{0}^{2}}{32\pi^{2}}-\frac{3\lambda_{2}^{2}}{32\pi^{2}}+\frac{3\lambda_{1}^{2}\lambda_{0}}{16\pi^{2}M^{2}}\right)\frac{1}{\overline{\epsilon}}\int
d^{3}x\,\frac{\phi^{4}(x)}{4!}.$ (65)

Since, there is no term proportional to the momentum, we know
$\delta_{\phi}^{1}=0$. Then we can use the following renormalization
conditions:

$\displaystyle\left(\delta_{m}^{1}m^{2}+\left(-\frac{\lambda_{0}m^{2}}{32\pi^{2}}-\frac{\lambda_{2}M^{2}}{32\pi^{2}}+\frac{\lambda_{1}^{2}m^{2}}{32\pi^{2}M^{2}}-\frac{\lambda_{1}^{2}}{16\pi^{2}}\right)\frac{1}{\overline{\epsilon}}\right)\int
d^{3}x\,\frac{\phi^{2}(x)}{2}=0,$ (66a)
$\displaystyle\left(\delta_{\lambda_{0}}^{1}\lambda_{0}+\left(-\frac{3\lambda_{0}^{2}}{32\pi^{2}}-\frac{3\lambda_{2}^{2}}{32\pi^{2}}+\frac{3\lambda_{1}^{2}\lambda_{0}}{16\pi^{2}M^{2}}\right)\frac{1}{\overline{\epsilon}}\right)\int
d^{3}x\,\frac{\phi^{4}(x)}{4!}=0,$ (66b)

to obtain

$\displaystyle\delta_{\phi}^{1}=0,$ (67a)
$\displaystyle\delta_{m}^{1}=\frac{1}{m^{2}}\left(\frac{\lambda_{0}m^{2}}{32\pi^{2}}+\frac{\lambda_{2}M^{2}}{32\pi^{2}}-\frac{\lambda_{1}^{2}m^{2}}{32\pi^{2}M^{2}}+\frac{\lambda_{1}^{2}}{16\pi^{2}}\right)\frac{1}{\overline{\epsilon}},$
(67b)
$\displaystyle\delta_{\lambda_{0}}^{1}=\frac{1}{\lambda_{0}}\left(\frac{3\lambda_{0}^{2}}{32\pi^{2}}+\frac{3\lambda_{2}^{2}}{32\pi^{2}}-\frac{3\lambda_{1}^{2}\lambda_{0}}{16\pi^{2}M^{2}}\right)\frac{1}{\overline{\epsilon}},$
(67c)

which agree with the results from the traditional implementation of
renormalization in the Lagrangian framework. After renormalization, we get the
decoupled Hamiltonian at one loop order to be

$\displaystyle H_{decoupled}^{oneloop}=$ $\displaystyle\int
d^{3}x\,\frac{1}{2}(\Pi^{2}(x)+(\nabla\phi(x))^{2})+\frac{1}{2}m^{2}\phi^{2}(x)+\frac{\lambda_{0}}{4!}\phi^{2}(x)$
(68)
$\displaystyle(\frac{\lambda_{0}m^{2}}{32\pi^{2}}(\ln{\frac{m^{2}}{M^{2}}}-1)-\frac{\lambda_{2}M^{2}}{32\pi^{2}}+\frac{3\lambda_{1}^{2}m^{2}}{32\pi^{2}M^{2}}(-\ln{\frac{m^{2}}{M^{2}}}+1)$
$\displaystyle-\frac{\lambda_{1}^{2}}{16\pi^{2}}(1+\ln{\frac{m^{2}}{M^{2}}})+(\frac{\lambda_{0}}{4}-\frac{3\lambda_{1}^{2}}{2M^{2}})C)\frac{\phi^{2}(x)}{2}+\frac{\lambda_{1}^{2}}{32\pi^{2}M^{2}}\frac{1}{2}(\Pi^{2}(x)+(\nabla\phi(x))^{2})$
$\displaystyle+(\frac{3\lambda_{0}^{2}}{32\pi^{2}}\ln{\frac{m^{2}}{M^{2}}}-\frac{3\lambda_{1}^{2}\lambda_{0}}{16\pi^{2}M^{2}}\ln{\frac{m^{2}}{M^{2}}}-\frac{3\lambda_{1}^{2}\lambda_{0}}{8\pi^{2}M^{2}}(1+\ln{\frac{m^{2}}{M^{2}}}))\frac{\phi^{4}(x)}{4!},$

where we again omit the term proportional to
$C=-\int_{k<M}\frac{d^{3}k}{(2\pi)^{3}}\frac{1}{\sqrt{k^{2}+m^{2}}}$ in the
four point function calculation.

##  4 Construction of the Effective Field Theory

In this section we will use the results from section 3 to construct the one
loop matched effective field theory up to $O(\frac{1}{M^{2}})$. We will also
check that the matching corrections are analytic in low energy parameters.
This is essential for the overall consistency of the program and requires the
cancellation during matching of the infrared contributions in $H_{decoupled}$
like the $\ln{\frac{m^{2}}{M^{2}}}$ terms and those labelled by $C$.

Figure 6:  Tree level contributions in $H_{decoupled}$.

At tree level up to order $O(\frac{1}{M^{2}})$, we have

$H_{decoupled}^{tree}=\int
d^{3}x\,\frac{1}{2}(\Pi^{2}(x)+(\nabla\phi(x))^{2}+\frac{1}{2}m^{2}\phi^{2}(x)+(\lambda_{0}-\frac{3\lambda_{1}}{M^{2}})\frac{\phi^{4}(x)}{4!}+O(\frac{1}{M^{4}}),$
(69)

that corresponds to Figure 6 and Figure 6. First, projecting the decoupled
Hamiltonian at tree level onto the low energy subspace, we have

$\displaystyle H_{eff}^{tree}=H_{decoupled}^{tree}=\int
d^{3}x\,\frac{1}{2}(\Pi^{2}(x)+(\nabla\phi(x))^{2})+\frac{1}{2}m^{2}\phi^{2}(x)+(\lambda_{0}-\frac{3\lambda_{1}}{M^{2}})\frac{\phi^{4}(x)}{4!}+O(\frac{1}{M^{4}}).$
(70)

In order to obtain the physical effective Hamiltonian defined in the complete
light field Hilbert space and to include loop corrections, we need to switch
the low frequency modes $\phi$’s in Eq.(70) back to the full light field
$\Phi_{L}$. Then we can decouple the high frequency and low frequency modes by
unitary transformations:

$\displaystyle
H_{decoupled}^{{}^{\prime}}=\bra{0_{H}}\omega^{{}^{\prime}\dagger}H_{eff}^{tree}\omega^{{}^{\prime}}\ket{0_{H}},$
(71)

where $H_{decoupled}^{{}^{\prime}}$ is the decoupled Hamiltonian corresponding
to $H_{eff}^{tree}$ and $\omega^{{}^{\prime}}$ is the unitary transformation.
Similar to what we did in section 2.1, we decompose $\omega^{{}^{\prime}}$
into a series of unitary transformations

$\omega^{{}^{\prime}}=\omega_{0}^{{}^{\prime}}\omega_{1}^{{}^{\prime}}\omega_{2}^{{}^{\prime}}...,$
(72)

and each $\omega_{i}$ can be further written as
$\omega^{\prime}_{i}=e^{i\Omega^{\prime}_{i}}$. The function of this series of
unitary transformations $\omega^{{}^{\prime}}$ is to disentangle the high
energy states above scale $M$ from the low energy ones order by order.
Furthermore, we decompose the effective Hamiltonian at tree level into four
parts, analogous to the decomposition in section 2.1,

$H_{eff}^{tree}=H_{eff\,1}^{tree}+H_{eff\,2}^{tree}+H_{eff\,A}^{tree}+H_{eff\,B}^{tree},$
(73)

where $H_{eff\,1}^{tree}$ only contains the low frequency modes,
$H_{eff\,2}^{tree}$ is the free part for high frequency modes,
$H_{eff\,A}^{tree}$ contains terms that only have high frequency annihilation
or creation operators, and $H_{eff\,B}^{tree}$ denotes whatever is left over.
Starting from the zeroth order transformation $\omega_{0}^{{}^{\prime}}$, we
have

$\displaystyle\omega_{0}^{{}^{\prime}\dagger}\left(H_{eff\,1}^{tree}+H_{eff\,2}^{tree}+H_{eff\,A}^{tree}+H_{eff\,B}^{tree}\right)\omega_{0}^{{}^{\prime}}$
(74) $\displaystyle=$ $\displaystyle
e^{-i\Omega_{0}^{{}^{\prime}}}\left(H_{eff\,1}^{tree}+H_{eff\,2}^{tree}+H_{eff\,A}^{tree}+H_{eff\,B}^{tree}\right)e^{i\Omega_{0}^{{}^{\prime}}}$
$\displaystyle=$ $\displaystyle
H_{eff\,1}^{tree}+H_{eff\,2}^{tree}+H_{eff\,A}^{tree}+H_{eff\,B}^{tree}$
$\displaystyle+i[H_{eff\,1}^{tree},\Omega_{0}^{{}^{\prime}}]+i[H_{eff\,2}^{tree},\Omega_{0}^{{}^{\prime}}]+i[H_{eff\,A}^{tree},\Omega_{0}^{{}^{\prime}}]+i[H_{eff\,B}^{tree},\Omega_{0}^{{}^{\prime}}]...$

We can then eliminate $H_{eff\,A}^{tree}$ by imposing the decoupling condition
at zeroth order:

$i\left[H_{eff\,2}^{tree},\Omega^{{}^{\prime}}_{0}\right]+H_{eff\,A}^{tree}=0.$
(75)

Next, we go to first order and eliminate the new term
$i\left[H_{eff\,1}^{tree},\Omega^{{}^{\prime}}_{0}\right]$ generated in the
zeroth order decoupling that contains only annihilation or creation operators
by imposing

$i\left[H_{eff\,1}^{tree},\Omega^{{}^{\prime}}_{0}\right]+i\left[H_{eff\,2}^{tree},\Omega^{{}^{\prime}}_{1}\right]=0.$
(76)

Our result for the decoupled effective Hamiltonian at first order then is,

$\displaystyle H_{decoupled}^{{}^{\prime}}=$
$\displaystyle\bra{0_{H}}H_{eff\,1}^{tree}+H_{eff\,2}^{tree}+H_{eff\,B}^{tree}+\frac{i}{2}[H_{eff\,A}^{tree},\Omega^{{}^{\prime}}_{0}]+i[H_{eff\,B}^{tree},\Omega_{0}^{{}^{\prime}}]+...\ket{0_{H}},$
(77)

which is good enough for constructing the effective field theory up to
$O(\frac{1}{M^{2}})$.

Figure 7:  One loop two point function contribution in
$H_{decoupled}^{{}^{\prime}}$

Let us begin with the two point function shown in Figure 7, which arises from
the normal ordering of term
$(\lambda_{0}-\frac{3\lambda_{1}}{M^{2}})\frac{\Phi_{L}^{4}(x)}{4!}$ in
$H_{eff}^{tree}$. The calculation here is similar to the calculation of Figure
3, and we only need to substitute $\lambda_{0}$ with
$\lambda_{0}-\frac{3\lambda_{1}^{2}}{M^{2}}$. Therefore, the result is

$\displaystyle-\frac{m^{2}}{32\pi^{2}}(\lambda_{0}-\frac{3\lambda_{1}^{2}}{M^{2}})(\frac{1}{\overline{\epsilon}}-\ln{\frac{m^{2}}{\mu^{2}}}+1)\int
d^{3}x\,\frac{\phi^{2}(x)}{2}+\frac{\lambda_{0}-\frac{3\lambda_{1}^{2}}{M^{2}}}{4}C\int
d^{3}x\,\frac{\phi^{2}(x)}{2},$ (78)

where $C=-\int_{k<M}\frac{d^{3}k}{(2\pi)^{3}}\,\frac{1}{\sqrt{k^{2}+m^{2}}}$
is the same as in section 3.2.1. Then we do the renormalization in
$\overline{MS}$ scheme to cancel out the divergent part and take $\mu\approx
M$. The finite terms left are,

$\displaystyle-\frac{m^{2}}{32\pi^{2}}(\lambda_{0}-\frac{3\lambda_{1}^{2}}{M^{2}})(-\ln{\frac{m^{2}}{M^{2}}}+1)\int
d^{3}x\,\frac{\phi^{2}(x)}{2}+\frac{\lambda_{0}-\frac{3\lambda_{1}^{2}}{M^{2}}}{4}C\int
d^{3}x\,\frac{\phi^{2}(x)}{2}.$ (79)

Figure 8:  One loop four point function contribution in
$H_{decoupled}^{{}^{\prime}}$

We next consider the contribution from the four point function shown in Figure
8. This arises from the commutator
$\frac{i}{2}[H_{eff\,A}^{tree},\Omega^{{}^{\prime}}_{0}]$, where,

$\displaystyle H_{eff\,A}^{tree}$ $\displaystyle=\int
d^{3}x\,\frac{\lambda_{0}-\frac{3\lambda_{1}^{2}}{M^{2}}}{4}\phi^{2}(x)\sum_{k}\sum_{p}\frac{1}{2\sqrt{\epsilon_{p}\epsilon_{k}}}\left(b_{p}b_{k}e^{i\left(\boldsymbol{p}+\boldsymbol{k}\right)\boldsymbol{x}}+b_{p}^{\dagger}b_{k}^{\dagger}e^{-i\left(\boldsymbol{p}+\boldsymbol{k}\right)\boldsymbol{x}}\right),$
(80a) $\displaystyle\Omega_{0}^{{}^{\prime}}$ $\displaystyle=\int
d^{3}y\,\frac{\lambda_{0}-\frac{3\lambda_{1}^{2}}{M^{2}}}{4}\phi^{2}(y)\sum_{k^{{}^{\prime}}}\sum_{p^{{}^{\prime}}}\left(\frac{-ie^{i\left(\boldsymbol{k}^{{}^{\prime}}+\boldsymbol{p}^{{}^{\prime}}\right)\boldsymbol{y}}}{2\sqrt{\epsilon_{k^{{}^{\prime}}}\epsilon_{p^{{}^{\prime}}}}\left(\epsilon_{k^{{}^{\prime}}}+\epsilon_{p^{{}^{\prime}}}\right)}b_{k^{{}^{\prime}}}b_{p^{{}^{\prime}}}+\frac{ie^{-i\left(\boldsymbol{k}^{{}^{\prime}}+\boldsymbol{p}^{{}^{\prime}}\right)\boldsymbol{y}}}{2\sqrt{\epsilon_{k^{{}^{\prime}}}\epsilon_{p^{{}^{\prime}}}}\left(\epsilon_{k^{{}^{\prime}}}+\epsilon_{p^{{}^{\prime}}}\right)}b_{k^{{}^{\prime}}}^{\dagger}b_{p^{{}^{\prime}}}^{\dagger}\right).$
(80b)

Again the calculation is similar to the one of Figure 5, and we only need to
substitute $\lambda_{0}$ with $\lambda_{0}-\frac{3\lambda_{1}^{2}}{M^{2}}$.
Hence, we have

$\displaystyle-\frac{3\left(\lambda_{0}-\frac{3\lambda_{1}^{2}}{M^{2}}\right)^{2}}{32\pi^{2}}\left(\frac{1}{\overline{\epsilon}}-\ln{\frac{m^{2}}{M^{2}}}\right)\int
d^{3}x\,\frac{\phi^{4}(x)}{4!},$ (81)

where we have omitted the finite terms proportional to C in the four point
calculation as we did in section 3.2.2.
After renormalization, up to $O(\frac{1}{M^{2}})$, we are left with

$\displaystyle\left(\frac{3\lambda_{0}^{2}}{32\pi^{2}}-\frac{9\lambda_{1}^{2}\lambda_{0}}{16\pi^{2}M^{2}}\right)\ln{\frac{m^{2}}{M^{2}}}\int
d^{3}x\,\frac{\phi^{4}(x)}{4!}.$ (82)

Putting the results from two and four point calculation together, we get

$\displaystyle H_{decoupled}^{oneloop,eff}=$ $\displaystyle\int
d^{3}x\,\frac{1}{2}(\Pi^{2}(x)+(\nabla\phi(x))^{2})+\frac{1}{2}m^{2}\phi^{2}(x)$
(83)
$\displaystyle+(-\frac{m^{2}}{32\pi^{2}}(\lambda_{0}-\frac{3\lambda_{1}^{2}}{M^{2}})(-\ln{\frac{m^{2}}{M^{2}}}+1)+\frac{\lambda_{0}-\frac{3\lambda_{1}^{2}}{M^{2}}}{4}C)\frac{\phi^{2}(x)}{2}$
$\displaystyle+(\lambda_{0}-\frac{3\lambda_{1}}{M^{2}}+(\frac{3\lambda_{0}}{32\pi^{2}}-\frac{9\lambda_{1}^{2}\lambda_{0}}{16\pi^{2}M^{2}})\ln{\frac{m^{2}}{M^{2}}})\frac{\phi^{4}(x)}{4!}.$

To get the effective theory at one loop order, we simply subtract Eq.(83) from
Eq.(68) and then switch $\phi$ back to $\Phi_{L}$. Thus, to
$O(\frac{1}{M^{2}})$, the effective Hamiltonian at one loop so constructed in
this theory is:

$\displaystyle H_{eff}^{oneloop}=$ $\displaystyle\int
d^{3}x\,\frac{1}{2}(\Pi_{L}^{2}+(\nabla\Phi_{L}(x))^{2})+\frac{1}{2}m^{2}\Phi_{L}^{2}(x)-\left(\frac{\lambda_{2}M^{2}}{32\pi^{2}}+\frac{\lambda_{1}^{2}}{16\pi^{2}}(1+\frac{m^{2}}{M^{2}})\right)\frac{\Phi_{L}^{2}(x)}{2}$
(84)
$\displaystyle+\left(\lambda_{0}-\frac{3\lambda_{1}^{2}}{M^{2}}-\frac{3\lambda_{0}\lambda_{1}^{2}}{8\pi^{2}M^{2}}\right)\frac{\Phi_{L}^{4}(x)}{4!}+\frac{\lambda_{1}^{2}}{32\pi^{2}M^{2}}\frac{1}{2}(\Pi^{2}_{L}(x)+\nabla\Phi_{L}(x))^{2}).$

As promised, the extra finite terms proportional to
$C=-\int_{k<M}\frac{d^{3}k}{(2\pi)^{3}}\frac{1}{\sqrt{k^{2}+m^{2}}}$ have all
canceled out along with the large log terms proportional to
$\ln{\frac{m^{2}}{M^{2}}}$ during matching. The final result agrees with
Rothstein:2003mp  where starting from the same full theory an effective
lagrangian was obtained using standard methods.

##  5 The Renormalization Group Equations

In this section we first obtain the exact renormalization group equations in
the context of the renormalization scheme advocated here and apply it at the
perturbative level to the scalar field theory example. For simplicity of
presentation we will limit the application to the pure $\lambda\phi^{4}$
theory.As in the previous sections, we will be working in the Hamiltonian
framework at fixed time.

As we have emphasized in this paper and explicitly shown in the scalar field
theoretical example of the previous sections, the process of renormalization
and decoupling of heavy particle effects can be regarded as the result of
unitary transformations which decouple the entanglement between the low and
high momentum modes of an interacting field theory. The transformed
Hamiltonian then incorporates effects of the high energy modes on the low
energy physics.In this section, we will generically denote by $\mu$ the cut-
off scale separating the low momentum modes from those at high momenta. Let
the disentanglement of these low and high momentum modes in the Hilbert space
be implemented by the unitary transformation $\omega(\mu)$ whose action on the
states is given by:

$\ket{\Psi(\mu)}=\omega^{\dagger}(\mu)\ket{\Psi},$ (85)

and under a change in the cut-off scale,

${\partial\over\partial\mu}\ket{\Psi(\mu)}=G_{\mu}\ket{\Psi(\mu)},$ (86)

with the generator of scale transformations identified as
$G_{\mu}=({\partial\omega^{\dagger}\over\partial\mu})\omega$.As discussed
earlier, the corresponding change in the Hamiltonian when the unitary
transformations are time-independent is

$H^{\prime}(\mu)=\omega(\mu)^{\dagger}H\omega(\mu),$ (87)

where $H$ denotes the scale independent starting Hamiltonian. Taking the
derivative of both sides of Eq.(87) with respect to $\mu$, we obtain,

${\partial H^{\prime}\over\partial\mu}=[G_{\mu},H^{\prime}].$ (88)

The renormalization group equation follows from the observation that the
change in the Hamiltonian with scale is compensated by the corresponding
changes in the coupling parameters of the theory, i.e.,

${\partial m^{2}\over\partial\mu}{\partial H^{\prime}\over\partial
m^{2}}+{\partial\lambda\over\partial\mu}{\partial
H^{\prime}\over\partial\lambda}=-[G_{\mu},H^{\prime}]$ (89)

This is an exact equation which is not very useful in practical calculations.
The procedure, which is more appropriate for a perturbative expansion, that we
have followed in the previous sections is to project the Hamiltonian
$H^{\prime}$ on to the high energy vacuum and to expand the right hand side of
Eq.(89). We will implement this below at the one-loop level using the results
from the previous sections as needed. By expanding, $\omega=e^{i\Omega_{0}}$,
we get,

$G_{\mu}=-i{\partial\Omega_{0}\over\partial\mu}+\frac{1}{2}[{\partial\Omega_{0}\over\partial\mu},\Omega_{0}]+....$
(90)

This can now be used to evaluate the expansion of the right hand side of
Eq.(89). For $H^{\prime}$ we will use the expansion Eq.(18), and keeping only
the terms which are needed here we get,

$[G_{\mu},H^{\prime}]=\frac{i}{2}{\partial\over\partial\mu}[H_{A},\Omega_{0}]+...$
(91)

In obtaining these we have used the fact that to this order $H_{A}$ is
independent of $\mu$. Let us apply this next to the four point function and
obtain the $\beta$ function for the coupling constant $\lambda$.

Since we are working to the one loop order, the contribution on the left hand
side of Eq.(89) comes from the term ${\lambda\over 4!}\int d^{3}x\phi^{4}$ in
$H^{\prime}$ and on the right hand side we have the $\mu$ dependent
contribution from Eq.(41)(after $\overline{MS}$ renormalization),

$\frac{i}{2}[H_{A}^{4,1},\Omega_{0}^{4,1}]={-3\lambda^{2}\over
32\pi^{2}}\ln{{\mu^{2}\over m^{2}}}\int d^{3}x{\phi^{4}\over 4!}.$ (92)

Putting this together, we finally get the well known result,

$\mu{\partial\lambda\over\partial\mu}=\beta(\lambda)={3\lambda^{2}\over
16\pi^{2}}.$ (93)

This provides a consistency check of our approach. The above procedure can, in
principle, be extended iteratively to higher orders, however,at the expense of
growing tedium. In order to get some new information, it would be interesting
to extend this analysis to explore the decoupling at different momentum
scales. Work in this direction is in progress.

##  6 Discussion

In this paper we have shown the consistency of a Hamiltonian renormalization
framework which emphasizes its basic origin as due to the momentum space
entanglement between the various modes of a quantum field theory. Using
unitary transformations on states to decouple the high energy modes from the
low energy ones and projecting the transformed Hamiltonian to the low energy
subspace, correctly accounts for renormalization effects and the property of
decoupling in quantum field theories. We have also shown how the same approach
can be consistently used in the construction of effective field theories.
Novel renormalization group equations were also obtained and shown to lead to
beta functions which are consistent with more conventional approaches. The
next step would be to understand how different measures of entanglement like
entanglement entropy and mutual information (for a review see Headrick:2018ctr
) may be used to analyze the properties of decoupling and to shed light on
another striking property of quantum field theories, namely the insensitivity
of the low energy physics to the details of the short distance structure.
Taking inspiration from the Ryu-Takayanagi formula Ryu:2006ef , another
related future direction could be to look for a possible role of geometry in
the renormalization program.

###### Acknowledgements.

Bingzheng Han gratefully acknowledges a summer fellowship awarded by the
Leinweber Center for Theoretical Physics.

##  Appendix A Appendix for Section 3

###  A.1 Explicit Calculation of Equation 35

We start from this integral:

$\displaystyle\frac{i}{2}\left[H_{A}^{2,2},\Omega_{0}^{2,2}\right]$
$\displaystyle=-\int\int
d^{3}xd^{3}y\,\sum_{k}\sum_{M<p}\lambda_{1}^{2}\frac{\phi(x)\phi(y)}{4\omega_{k}\epsilon_{p}\left(\omega_{k}+\epsilon_{p}\right)}e^{i\left(\boldsymbol{k}+\boldsymbol{p}\right)\left(\boldsymbol{x}-\boldsymbol{y}\right)},$
(94) $\displaystyle=-\int\int
d^{3}xd^{3}y\,\sum_{k}\sum_{M<p}\lambda_{1}^{2}\frac{\phi(x)\phi(y)e^{i\left(\boldsymbol{k}+\boldsymbol{p}\right)\left(\boldsymbol{x}-\boldsymbol{y}\right)}}{4\sqrt{k^{2}+M^{2}}\sqrt{p^{2}+m^{2}}\left(\sqrt{k^{2}+M^{2}}+\sqrt{p^{2}+m^{2}}\right)}.$

In order to calculate this complicated integral, we need to split the fraction
into two parts:

$\displaystyle\frac{1}{\sqrt{k^{2}+M^{2}}\sqrt{p^{2}+m^{2}}\left(\sqrt{k^{2}+M^{2}}+\sqrt{p^{2}+m^{2}}\right)},$
(95) $\displaystyle=$
$\displaystyle\frac{\sqrt{k^{2}+M^{2}}-\sqrt{p^{2}+m^{2}}}{\sqrt{k^{2}+M^{2}}\sqrt{p^{2}+m^{2}}\left(k^{2}+M^{2}-p^{2}-m^{2}\right)},$
$\displaystyle=$
$\displaystyle\frac{1}{\sqrt{p^{2}+m^{2}}\left(k^{2}+M^{2}-p^{2}-m^{2}\right)}-\frac{1}{\sqrt{k^{2}+M^{2}}\left(k^{2}+M^{2}-p^{2}-m^{2}\right)}.$

Let $\boldsymbol{k}+\boldsymbol{p}=\boldsymbol{r}$. We know $\boldsymbol{r}$
is total external momentum, thus r is much smaller than M. Also since
$\boldsymbol{k}-\boldsymbol{p}$ is of order $O(M)$, we conclude that
$\left(k^{2}-p^{2}\right)\ll M^{2}$. Hence we can write:

$\displaystyle\frac{1}{\sqrt{p^{2}+m^{2}}\left(k^{2}+M^{2}-p^{2}-m^{2}\right)},$
(96) $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{p^{2}+m^{2}}M^{2}\left(1-\frac{m^{2}}{M^{2}}+\frac{k^{2}-p^{2}}{M^{2}}\right)},$
$\displaystyle\simeq$
$\displaystyle\frac{1}{\sqrt{p^{2}+m^{2}}M^{2}}\left(1+\frac{m^{2}}{M^{2}}-\frac{k^{2}-p^{2}}{M^{2}}+\left(\frac{k^{2}-p^{2}}{M^{2}}\right)^{2}\right).$

To the order $\sim O(\frac{1}{M^{2}})$ this gives

$\displaystyle\int
d^{3}x\frac{m^{2}\lambda_{1}^{2}}{16\pi^{2}M^{2}}\left(\frac{1}{\overline{\epsilon}}-\ln{\frac{m^{2}}{\mu^{2}}}+1\right)\frac{\phi^{2}(x)}{2}-\frac{\lambda_{1}^{2}C}{2M^{2}}\int
d^{3}x\,\frac{\phi^{2}(x)}{2}.$ (97)

Similarly, the second term from Eq.(95) gives

$\displaystyle\frac{1}{\sqrt{k^{2}+M^{2}}M^{2}}\left(1+\frac{m^{2}}{M^{2}}-\frac{k^{2}-p^{2}}{M^{2}}+\left(\frac{k^{2}-p^{2}}{M^{2}}\right)^{2}\right).$
(98)

Straightforwardly, the first two terms in Eq.(98) give

$\displaystyle\int\int
d^{3}xd^{3}y\,\lambda_{1}^{2}\phi(x)\phi(y)\int\frac{d^{3}k}{\left(2\pi\right)^{3}}\int\frac{d^{3}r}{\left(2\pi\right)^{3}}\frac{1}{4\sqrt{k^{2}+M^{2}}M^{2}}e^{i\boldsymbol{r}\left(\boldsymbol{x}-\boldsymbol{y}\right)}\left(1+\frac{m^{2}}{M^{2}}\right),$
(99) $\displaystyle=-\int
d^{3}x\frac{\lambda_{1}^{2}}{16\pi^{2}}\left(\frac{1}{\overline{\epsilon}}-\ln{\frac{M^{2}}{\mu^{2}}}+1\right)\left(1+\frac{m^{2}}{M^{2}}\right)\frac{\phi^{2}(x)}{2}.$

The last two terms are a bit tricky to handle. First, we note that
$k^{2}-p^{2}=\left(\boldsymbol{k}+\boldsymbol{p}\right)\left(\boldsymbol{k}-\boldsymbol{p}\right)=2\boldsymbol{k}\boldsymbol{r}-r^{2}$.
Then we get from the $\frac{k^{2}-p^{2}}{M^{2}}$ term:

$\displaystyle\int\int
d^{3}xd^{3}y\,\lambda_{1}^{2}\phi(x)\phi(y)\int\frac{d^{3}k}{\left(2\pi\right)^{3}}\int\frac{d^{3}r}{\left(2\pi\right)^{3}}\frac{1}{4\sqrt{k^{2}+M^{2}}M^{2}}e^{i\boldsymbol{r}\left(\boldsymbol{x}-\boldsymbol{y}\right)}\left(-\frac{k^{2}-p^{2}}{M^{2}}\right),$
(100) $\displaystyle=\int\int
d^{3}xd^{3}y\,\lambda_{1}^{2}\phi(x)\phi(y)\int\frac{d^{3}k}{\left(2\pi\right)^{3}}\int\frac{d^{3}r}{\left(2\pi\right)^{3}}\frac{1}{4\sqrt{k^{2}+M^{2}}M^{2}}e^{i\boldsymbol{r}\left(\boldsymbol{x}-\boldsymbol{y}\right)}\left(\frac{-2\boldsymbol{k}\boldsymbol{r}+r^{2}}{M^{2}}\right),$
$\displaystyle=\int\int
d^{3}xd^{3}y\,\lambda_{1}^{2}\phi(x)\phi(y)\int\frac{d^{3}k}{\left(2\pi\right)^{3}}\int\frac{d^{3}r}{\left(2\pi\right)^{3}}\frac{1}{4\sqrt{k^{2}+M^{2}}M^{2}}e^{i\boldsymbol{r}\left(\boldsymbol{x}-\boldsymbol{y}\right)}\frac{r^{2}}{M^{2}},$
$\displaystyle=\int\int
d^{3}xd^{3}y\,\frac{\lambda_{1}^{2}}{4}\phi(x)\phi(y)\int\frac{d^{3}k}{\left(2\pi\right)^{3}}\frac{1}{M^{2}\sqrt{k^{2}+M^{2}}}\int\frac{d^{3}r}{\left(2\pi\right)^{3}}r^{2}e^{i\boldsymbol{r}\left(\boldsymbol{x}-\boldsymbol{y}\right)},$
$\displaystyle=-\int\int
d^{3}xd^{3}y\,\frac{\lambda_{1}^{2}}{4}\phi(x)\phi(y)\int\frac{d^{3}k}{\left(2\pi\right)^{3}}\frac{1}{M^{2}\sqrt{k^{2}+M^{2}}}\left(\nabla_{y}^{2}\delta^{3}(\boldsymbol{x}-\boldsymbol{y})\right),$
$\displaystyle=-\int
d^{3}x\,\frac{\lambda_{1}^{2}}{4}\phi(x)\nabla^{2}\phi(x)\int\frac{d^{3}k}{\left(2\pi\right)^{3}}\frac{1}{M^{2}\sqrt{k^{2}+M^{2}}},$
$\displaystyle=-\int
d^{3}x\,\frac{\lambda_{1}^{2}}{16\pi^{2}M^{2}}\left(\frac{1}{\overline{\epsilon}}-\ln{\frac{M^{2}}{\mu^{2}}}+1\right)\frac{1}{2}(\nabla\phi(x))^{2}.$

To the $O(\frac{1}{M^{2}})$, we also need to consider the
$\frac{4\left(\boldsymbol{k}\boldsymbol{r}\right)^{2}}{M^{4}}$ term from
$\left(\frac{k^{2}-p^{2}}{M^{2}}\right)^{2}$:

$\displaystyle\int\int
d^{3}xd^{3}y\,\lambda_{1}^{2}\phi(x)\phi(y)\int\frac{d^{3}k}{\left(2\pi\right)^{3}}\int\frac{d^{3}r}{\left(2\pi\right)^{3}}\frac{1}{4\sqrt{k^{2}+M^{2}}M^{2}}e^{i\boldsymbol{r}\left(\boldsymbol{x}-\boldsymbol{y}\right)}\frac{4\left(\boldsymbol{k}\boldsymbol{r}\right)^{2}}{M^{4}},$
(101) $\displaystyle=\int\int
d^{3}xd^{3}y\,\lambda_{1}^{2}\phi(x)\phi(y)\int\frac{d^{3}k}{\left(2\pi\right)^{3}}\frac{1}{\sqrt{k^{2}+M^{2}}M^{6}}\int\frac{d^{3}r}{\left(2\pi\right)^{3}}\left(\boldsymbol{k}\boldsymbol{r}\right)^{2}e^{i\boldsymbol{r}\left(\boldsymbol{x}-\boldsymbol{y}\right)},$
$\displaystyle=\int\int
d^{3}xd^{3}y\,\lambda_{1}^{2}\phi(x)\phi(y)\int\frac{d^{3}r}{\left(2\pi\right)^{3}}r^{2}e^{i\boldsymbol{r}\left(\boldsymbol{x}-\boldsymbol{y}\right)}\int\frac{d^{3}k}{\left(2\pi\right)^{3}}\frac{k^{2}\cos^{2}{\theta}}{\sqrt{k^{2}+M^{2}}M^{6}},$
$\displaystyle=\int\int
d^{3}xd^{3}y\,\lambda_{1}^{2}\phi(x)\phi(y)\int\frac{d^{3}r}{\left(2\pi\right)^{3}}r^{2}e^{i\boldsymbol{r}\left(\boldsymbol{x}-\boldsymbol{y}\right)}\frac{1}{32\pi^{2}M^{2}}\left(\frac{1}{\overline{\epsilon}}-\ln{\frac{M^{2}}{\mu^{2}}}+\frac{3}{2}\right),$
$\displaystyle=-\int\int
d^{3}xd^{3}y\,\lambda_{1}^{2}\phi(x)\delta^{3}(\boldsymbol{x}-\boldsymbol{y})\nabla_{y}^{2}\phi(y)\frac{1}{32\pi^{2}M^{2}}\left(\frac{1}{\overline{\epsilon}}-\ln{\frac{M^{2}}{\mu^{2}}}+\frac{3}{2}\right),$
$\displaystyle=\int
d^{3}x\,\frac{\lambda_{1}^{2}}{16\pi^{2}M^{2}}\left(\frac{1}{\overline{\epsilon}}-\ln{\frac{M^{2}}{\mu^{2}}}+\frac{3}{2}\right)\frac{1}{2}(\nabla\phi(x))^{2}.$

Putting together all the previous results, we can get

$\displaystyle\frac{m^{2}\lambda_{1}^{2}}{16\pi^{2}M^{2}}\left(\frac{1}{\overline{\epsilon}}-\ln{\frac{m^{2}}{\mu^{2}}}+1\right)\int
d^{3}x\,\frac{\phi^{2}(x)}{2}-\frac{\lambda_{1}^{2}C}{2M^{2}}\int
d^{3}x\,\frac{\phi^{2}(x)}{2}$ (102)
$\displaystyle-\frac{\lambda_{1}^{2}}{16\pi^{2}}\left(\frac{1}{\overline{\epsilon}}-\ln{\frac{M^{2}}{\mu^{2}}}+1\right)\left(1+\frac{m^{2}}{M^{2}}\right)\int
d^{3}x\,\frac{\phi^{2}(x)}{2}+\frac{\lambda_{1}^{2}}{32\pi^{2}M^{2}}\int
d^{3}x\,\frac{1}{2}(\nabla\phi(x))^{2}.$

###  A.2 Explicit Calculation of the Momentum Term in Equation LABEL:eq:3.25

As we mentioned in section 3.2.1 the momentum term arises from
$\bra{0_{H}}-\frac{1}{2}[[H_{1},\Omega_{0}],\Omega_{1}]\ket{0_{H}}$, we need
to first find the corresponding $\Omega_{1}$. Since the momentum term is
proportional to $\lambda_{1}^{2}\Pi^{2}$, we should determine $\Omega_{1}$
from

$\displaystyle i[H_{1},\Omega_{0}^{2,2}]+i[H_{2},\Omega_{1}]=0.$ (103)

It’s straightforward to calculate $[H_{1},\Omega_{0}^{2,2}]$ using
$[\phi(x),\Pi(y)]=i\delta^{3}(\boldsymbol{x}-\boldsymbol{y})$, and the result
is

$\displaystyle\left[H_{1},\Omega_{0}^{2,2}\right]$ $\displaystyle=2\int
d^{3}x\int
d^{3}y\int\sum_{k}\sum_{M<p}\left(\frac{\lambda_{1}}{2}\frac{-ia_{k}b_{p}e^{i(\boldsymbol{p}+\boldsymbol{k})\boldsymbol{y}}}{2\sqrt{\omega_{k}\epsilon_{p}}(\omega_{k}+\epsilon_{p})}[\frac{1}{2}\Pi^{2}(x),\phi(y)]+h.c.\right),$
(104) $\displaystyle=-\lambda_{1}\int
d^{3}x\int\frac{d^{3}k}{(2\pi)^{3}}\int\frac{d^{3}p}{(2\pi)^{3}}\left(\frac{a_{k}b_{p}e^{i(\boldsymbol{p}+\boldsymbol{k})\boldsymbol{x}}}{2\sqrt{\omega_{k}\epsilon_{p}}(\omega_{k}+\epsilon_{p})}\Pi(x)-h.c.\right).$

Then $\Omega_{1}$ is calculated to be

$\displaystyle\Omega_{1}=-\int d^{3}y\int d^{3}q\int
d^{3}r\left(\frac{\lambda_{1}a_{q}b_{r}e^{i(\boldsymbol{q}+\boldsymbol{r})\boldsymbol{y}}}{2\sqrt{\omega_{q}\epsilon_{r}}(\omega_{q}+\epsilon_{r})^{2}}\Pi(y)+h.c.\right).$
(105)

Using Eq.(104) and Eq.(105) we are able to calculate the commutator
$-\frac{1}{2}[[H_{1},\Omega_{0}],\Omega_{1}]$. We will focus on commutator
$[a_{k}b_{p},a^{\dagger}_{q}b^{\dagger}_{r}]$ since the other one will yield
the same result.

$\displaystyle-\frac{1}{2}[[H_{1},\Omega_{0}],\Omega_{1}]$
$\displaystyle=-\frac{\lambda_{1}^{2}}{2}\int d^{3}x\int
d^{3}y\int\frac{d^{3}k}{(2\pi)^{3}}\int\frac{d^{3}p}{(2\pi)^{3}}\int\frac{d^{3}q}{(2\pi)^{3}}\int\frac{d^{3}r}{(2\pi)^{3}}$
(106)
$\displaystyle\frac{\Pi(x)\Pi(y)e^{i(\boldsymbol{k}+\boldsymbol{p})\boldsymbol{x}-i(\boldsymbol{q}+\boldsymbol{r})\boldsymbol{y}}}{4\sqrt{\omega_{k}\epsilon_{p}\omega_{q}\epsilon_{r}}(\omega_{k}+\epsilon_{p})(\omega_{q}+\epsilon_{r})^{2}}[a_{k}b_{p},a^{\dagger}_{q}b^{\dagger}_{r}],$
$\displaystyle=-\frac{\lambda_{1}^{2}}{8}\int d^{3}x\int
d^{3}y\int\frac{d^{3}k}{(2\pi)^{3}}\int\frac{d^{3}p}{(2\pi)^{3}}\,\frac{e^{i(\boldsymbol{p}+\boldsymbol{k})(\boldsymbol{x}-\boldsymbol{y})}}{\omega_{k}\epsilon_{p}(\omega_{k}+\epsilon_{p})^{3}}\Pi(x)\Pi(y).$

Here we have done normal ordering to get the final result.
Similar to section A.1, we can decompose the integral into four parts by
multiplying both denominator and numerator by $(\omega_{k}-\epsilon_{p})^{3}$
to get

$\displaystyle\frac{(\omega_{k}-\epsilon_{p})^{3}}{\omega_{k}\epsilon_{p}(k^{2}+M^{2}-p^{2}-m^{2})}=$
$\displaystyle\frac{\omega_{k}^{2}}{\epsilon_{p}M^{6}(1-\frac{m^{2}}{M^{2}}+\frac{k^{2}-p^{2}}{M^{2}})^{3}}-\frac{\epsilon_{p}^{2}}{\omega_{k}M^{6}(1-\frac{m^{2}}{M^{2}}+\frac{k^{2}-p^{2}}{M^{2}})^{3}}$
(107)
$\displaystyle+\frac{3\epsilon_{p}}{M^{6}(1-\frac{m^{2}}{M^{2}}+\frac{k^{2}-p^{2}}{M^{2}})^{3}}-\frac{3\omega_{k}}{M^{6}(1-\frac{m^{2}}{M^{2}}+\frac{k^{2}-p^{2}}{M^{2}})^{3}}.$

Since we are only interested in terms proportional to momentum $\Pi$, and it’s
from section A.1 that $\frac{k^{2}-p^{2}}{M^{2}}$ will mainly contribute to
terms proportional to gradient, it’s reasonable to make the assumption that
$k\approx p$. Moreover up to order $O(M^{2})$, we can neglect the $m^{2}$ in
$\epsilon_{p}$ and only need to consider the following three terms:

$\displaystyle\frac{3\lambda_{1}^{2}}{8}\int
d^{3}x\,\Pi^{2}(x)\int\frac{d^{3}k}{(2\pi)^{3}}\frac{\sqrt{k^{2}+M^{2}}}{M^{6}},$
(108a) $\displaystyle\frac{\lambda_{1}^{2}}{8}\int
d^{3}x\,\Pi^{2}(x)\int\frac{d^{3}k}{(2\pi)^{3}}\frac{k^{2}}{\sqrt{k^{2}+M^{2}}M^{6}},$
(108b) $\displaystyle-\frac{\lambda_{1}^{2}}{8}\int
d^{3}x\,\Pi^{2}(x)\int\frac{d^{3}k}{(2\pi)^{3}}\frac{k^{2}+M^{2}}{kM^{6}}.$
(108c)

The first term is simply calculated using dimensional regularization and the
result is

$\displaystyle-\frac{3\lambda_{1}^{2}}{128M^{2}\pi^{2}}\left(\frac{1}{\overline{\epsilon}}-\ln\frac{M^{2}}{\mu^{2}}+\frac{3}{2}\right)\int
d^{3}x\,\frac{1}{2}\Pi^{2}(x).$ (109)

The second term can be calculated by splitting the integral into two terms:

$\displaystyle\frac{\lambda_{1}^{2}}{8}\int
d^{3}x\,\Pi^{2}(x)\int\frac{d^{3}k}{(2\pi)^{3}}\frac{\sqrt{k^{2}+M^{2}}}{M^{6}}-\frac{\lambda_{1}^{2}}{8}\int
d^{3}x\,\Pi^{2}(x)\int\frac{d^{3}k}{(2\pi)^{3}}\frac{1}{\sqrt{k^{2}+M^{2}}M^{4}}.$
(110)

Again, using dimensional regularization we get:

$\displaystyle-\frac{\lambda_{1}^{2}}{128M^{2}\pi^{2}}\left(\frac{1}{\overline{\epsilon}}-\ln\frac{M^{2}}{\mu^{2}}+\frac{3}{2}\right)\int
d^{3}x\,\frac{1}{2}\Pi^{2}(x)+\frac{\lambda_{1}^{2}}{32M^{2}\pi^{2}}\left(\frac{1}{\overline{\epsilon}}-\ln\frac{M^{2}}{\mu^{2}}+1\right)\int
d^{3}x\,\frac{1}{2}\Pi^{2}(x).$ (111)

The third term can be calculated as follows,

$\displaystyle-\frac{\lambda_{1}^{2}}{8}\int
d^{3}x\,\Pi^{2}(x)\int\frac{d^{3}k}{(2\pi)^{3}}\frac{k^{2}+M^{2}}{kM^{6}}$
(112) $\displaystyle=$
$\displaystyle-\frac{\lambda_{1}^{2}}{8M^{6}}\frac{2\pi^{\frac{3}{2}}}{\Gamma(\frac{3}{2})}\int\frac{dk}{(2\pi)^{3}}\,\frac{k}{(k^{2}+M^{2})^{-1}}\int
d^{3}x\,\Pi^{2}(x),$ $\displaystyle=$
$\displaystyle-\frac{\lambda_{1}^{2}}{8M^{6}}\frac{2\pi^{\frac{3}{2}}}{\Gamma(\frac{3}{2})}\frac{1}{\Gamma(-1+\delta)}\int\frac{dk}{(2\pi)^{3}}\,k\int
d\lambda\,\lambda^{-2+\delta}e^{-\lambda(k^{2}+M^{2})}\int
d^{3}x\,\Pi^{2}(x),$ $\displaystyle=$
$\displaystyle-\frac{\lambda_{1}^{2}}{8M^{6}}\frac{2\pi^{\frac{3}{2}}}{\Gamma(\frac{3}{2})\Gamma(-1+\delta)}\frac{1}{(2\pi)^{3}}\int
d\lambda\,\lambda^{-2+\delta}e^{-\lambda M^{2}}\int dk\,ke^{-\lambda
k^{2}}\int d^{3}x\,\Pi^{2}(x),$ $\displaystyle=$
$\displaystyle-\frac{\lambda_{1}^{2}}{16M^{6}}\frac{\pi^{\frac{3}{2}}}{\Gamma(\frac{3}{2})\Gamma(-1+\delta)}\frac{1}{(2\pi)^{3}}\int
d\lambda\,\lambda^{-2+\delta}e^{-\lambda M^{2}}\int dk^{2}\,e^{-\lambda
k^{2}}\int d^{3}x\,\Pi^{2}(x),$ $\displaystyle=$
$\displaystyle-\frac{\lambda_{1}^{2}}{16M^{6}}\frac{\pi^{\frac{3}{2}}}{\Gamma(\frac{3}{2})\Gamma(-1+\delta)}\frac{1}{(2\pi)^{3}}\int
d\lambda\,\lambda^{-3+\delta}e^{-\lambda M^{2}}\int d^{3}x\,\Pi^{2}(x),$
$\displaystyle=$
$\displaystyle-\frac{\lambda_{1}^{2}}{16M^{6}}\frac{\pi^{\frac{3}{2}}\Gamma(-2+\delta)}{\Gamma(\frac{3}{2})\Gamma(-1+\delta)}\frac{M^{4}}{(2\pi)^{3}}\int
d^{3}x\,\Pi^{2}(x),$ $\displaystyle=$
$\displaystyle\frac{\lambda_{1}^{2}}{32M^{2}\pi^{2}}\int
d^{3}x\,\frac{1}{2}\Pi^{2}(x).$

In the above, $\delta$ is a regulator which is taken to zero at the end of the
calculation, and we have used the standard representation:

${1\over x^{a}}=\frac{1}{\Gamma(a)}\int d\lambda\lambda^{a-1}e^{-\lambda x}.$

Putting results from all three terms together we get the contribution from the
commutator $[a_{k}b_{p},a^{\dagger}_{q}b^{\dagger}_{r}]$ to be:

$\frac{\lambda_{1}^{2}}{64M^{2}\pi^{2}}\int d^{3}x\,\frac{1}{2}\Pi^{2}(x).$
(113)

Multiplying this result by 2 to take into account the contribution from
commutator $[a_{k}^{\dagger}b_{p}^{\dagger},a_{q}b_{r}]$, we arrive at the
final result for the canonical momentum piece in the decoupled Hamiltonian:

$\frac{\lambda_{1}^{2}}{32M^{2}\pi^{2}}\int d^{3}x\,\frac{1}{2}\Pi^{2}(x).$
(114)

###  A.3 Explicit Calculation of Figure 5

Since combination Table 1(b) and Table 1(c) only differ from an exchange of
$\Omega_{0}^{4,3,1}$ and $\Omega_{0}^{4,3,2}$, they will yield the same
result. Let’s consider the following expansions for the quantities in Table
1(b):

$\displaystyle H_{A}^{4,3}=\frac{\lambda_{1}}{2}\int
d^{3}x\,\sum_{k}\sum_{p}\sum_{q}\frac{e^{i\left(\boldsymbol{k}+\boldsymbol{p}+\boldsymbol{q}\right)\boldsymbol{x}}}{2)^{\frac{3}{2}}\sqrt{\omega_{k}\epsilon_{p}\epsilon_{q}}}a_{k}b_{p}b_{q},$
(115a) $\displaystyle\Omega_{0}^{4,3,1}=\frac{\lambda_{1}}{2}\int
d^{3}y\,\sum_{k^{{}^{\prime}}}\frac{i\phi^{2}(y)}{\sqrt{2\omega_{k^{{}^{\prime}}}}\omega_{k^{{}^{\prime}}}}a_{k^{{}^{\prime}}}^{\dagger}e^{-i\boldsymbol{k}^{{}^{\prime}}\boldsymbol{y}},$
(115b) $\displaystyle\Omega_{0}^{4,3,2}=\frac{\lambda_{0}}{4}\int
d^{3}z\,\sum_{p^{{}^{\prime}}}\sum_{q^{{}^{\prime}}}\phi^{2}(z)\frac{ie^{-i\left(\boldsymbol{p}^{{}^{\prime}}+\boldsymbol{q}^{{}^{\prime}}\right)\boldsymbol{z}}}{2\sqrt{\epsilon_{p^{{}^{\prime}}}\epsilon_{q^{{}^{\prime}}}}\left(\epsilon_{p^{{}^{\prime}}}+\epsilon_{q^{{}^{\prime}}}\right)}b_{p^{{}^{\prime}}}^{\dagger}b_{q^{{}^{\prime}}}^{\dagger}.$
(115c)

The first commutator gives

$\displaystyle\left[H_{A}^{4,3},\Omega_{0}^{4,3,1}\right]$
$\displaystyle=\frac{i\lambda_{1}^{2}}{4}\int d^{3}x\int
d^{3}y\,\sum_{k}\sum_{k^{{}^{\prime}}}\sum_{p}\sum_{q}\frac{\phi^{2}(y)}{2^{2}}\frac{e^{i\left(\boldsymbol{k}+\boldsymbol{p}+\boldsymbol{q}\right)\boldsymbol{x}-i\boldsymbol{k}^{{}^{\prime}}\boldsymbol{y}}}{\sqrt{\omega_{k}\omega_{k^{{}^{\prime}}}\omega_{q}\omega_{p}}\omega_{k^{{}^{\prime}}}}\left[a_{k}b_{p}b_{q},a_{k^{{}^{\prime}}}^{\dagger}\right],$
(116) $\displaystyle=\frac{i\lambda_{1}^{2}}{16}\int d^{3}x\int
d^{3}y\,\sum_{p}\sum_{q}\frac{e^{i\left(\boldsymbol{p}+\boldsymbol{q}\right)\boldsymbol{x}}}{\sqrt{\epsilon_{p}}\epsilon_{q}M^{2}}\int\frac{d^{3}k}{\left(2\pi\right)^{3}}\,e^{i\boldsymbol{k}\left(\boldsymbol{x}-\boldsymbol{y}\right)}\phi^{2}(y)b_{p}b_{q},$
$\displaystyle=\frac{i\lambda_{1}^{2}}{16}\int
d^{3}x\,\phi^{2}(x)\sum_{p}\sum_{q}\frac{e^{i\left(\boldsymbol{p}+\boldsymbol{q}\right)\boldsymbol{x}}}{M^{2}\sqrt{\epsilon_{p}\epsilon_{q}}}b_{p}b_{q}.$

The second one is

$\displaystyle\left[\left[H_{A}^{4,3},\Omega_{0}^{4,3,1}\right],\Omega_{0}^{4,3,2}\right]$
$\displaystyle=-\frac{\lambda_{1}^{2}\lambda_{0}}{64}\int d^{3}x\int
d^{3}z\,\phi^{2}(x)\phi^{2}(z),$ (117)
$\displaystyle\sum_{p}\sum_{q}\sum_{p^{{}^{\prime}}}\sum_{q{{}^{\prime}}}\frac{e^{i\left(\boldsymbol{p}+\boldsymbol{q}\right)\boldsymbol{x}-i\left(\boldsymbol{p}^{{}^{\prime}}+\boldsymbol{q}^{{}^{\prime}}\right)\boldsymbol{z}}}{2M^{2}\sqrt{\epsilon_{p}\epsilon_{q}\epsilon_{p^{{}^{\prime}}}\epsilon_{q^{{}^{\prime}}}}\left(\epsilon_{p^{{}^{\prime}}}+\epsilon_{q^{{}^{\prime}}}\right)}\left[b_{p}b_{q},b_{p^{{}^{\prime}}}^{\dagger}b_{q^{{}^{\prime}}}^{\dagger}\right],$
$\displaystyle=-\frac{\lambda_{1}^{2}\lambda_{0}}{64}\int
d^{3}x\,\phi^{4}(x)\int\frac{d^{3}p}{\left(2\pi\right)^{3}}\frac{1}{2M^{2}\epsilon_{p}^{3}}.$

As a result,

$\displaystyle-\frac{1}{3}\left[\left[H_{A}^{4,3},\Omega_{0}^{4,3,1}\right],\Omega_{0}^{4,3,2}\right]=\frac{\lambda_{1}^{2}\lambda_{0}}{192}\int
d^{3}x\,\phi^{4}(x)\int\frac{d^{3}p}{\left(2\pi\right)^{3}}\frac{1}{2M^{2}\epsilon_{p}^{3}}.$
(118)

Including the Hermitian conjugate we have the final result for both Table 1(b)
and Table 1(c) as,

$\displaystyle\frac{\lambda_{1}^{2}\lambda_{0}}{96}\int
d^{3}x\,\phi^{4}(x)\int\frac{d^{3}p}{\left(2\pi\right)^{3}}\frac{1}{2M^{2}\epsilon_{p}^{3}}.$
(119)

For the combination in Table 1(d), we have,

$\displaystyle H_{A}^{4,3}=\int
d^{3}x\,\frac{\lambda_{0}}{4}\phi^{2}(x)\sum_{p}\sum_{q}\frac{e^{i\left(\boldsymbol{p}+\boldsymbol{q}\right)\boldsymbol{x}}}{2\sqrt{\epsilon_{p}\epsilon_{q}}}b_{p}b_{q},$
(120a) $\displaystyle\Omega_{0}^{4,3,1}=\int
d^{3}y\,\frac{\lambda_{1}}{2}\sum_{k}\sum_{p^{{}^{\prime}}}\sum_{q^{{}^{\prime}}}\frac{ie^{-i\left(\boldsymbol{k}+\boldsymbol{p}^{{}^{\prime}}+\boldsymbol{q}^{{}^{\prime}}\right)\boldsymbol{y}}}{2^{\frac{3}{2}}\sqrt{\omega_{k}\epsilon_{p^{{}^{\prime}}}\epsilon_{q^{{}^{\prime}}}}\left(\omega_{k}+\epsilon_{p^{{}^{\prime}}}+\epsilon_{q^{{}^{\prime}}}\right)}a^{\dagger}_{k}b^{\dagger}_{p^{{}^{\prime}}}b^{\dagger}_{q^{{}^{\prime}}},$
(120b) $\displaystyle\Omega_{0}^{4,3,2}=\int
d^{3}z\,\frac{\lambda_{1}}{2}\phi^{2}(z)\sum_{k^{{}^{\prime}}}\frac{-ie^{\boldsymbol{k}^{{}^{\prime}}\boldsymbol{z}}}{\sqrt{2\omega_{k^{{}^{\prime}}}}\omega_{k^{{}^{\prime}}}}a_{k^{{}^{\prime}}}.$
(120c)

Following similar steps as before, we first calculate the commutator:

$\displaystyle\left[H_{A}^{4,3},\Omega_{0}^{4,3,1}\right]$
$\displaystyle=\frac{i\lambda_{0}\lambda_{1}}{8}\int d^{3}x\int
d^{3}y\,\phi^{2}(x)\sum_{p}\sum_{q}\sum_{k}\sum_{p^{{}^{\prime}}}\sum_{q^{{}^{\prime}}}\frac{e^{i\left(\boldsymbol{p}+\boldsymbol{q}\right)\boldsymbol{x}-i\left(\boldsymbol{k}+\boldsymbol{p}^{{}^{\prime}}+\boldsymbol{q}^{{}^{\prime}}\right)\boldsymbol{y}}}{2^{\frac{5}{2}}\left(\omega_{K}+\epsilon_{p^{{}^{\prime}}}+\epsilon_{q^{{}^{\prime}}}\right)}$
(121)
$\displaystyle\left[b_{p}b_{q},a_{k}^{\dagger}b_{p^{{}^{\prime}}}^{\dagger}b_{q^{{}^{\prime}}}^{\dagger}\right],$
$\displaystyle=\frac{i\lambda_{1}\lambda_{0}}{8}\int d^{3}x\int
d^{3}y\,\phi^{2}(x)\sum_{k}\sum_{p}\sum_{q}\frac{2e^{i\left(\boldsymbol{p}+\boldsymbol{q}\right)\left(\boldsymbol{x}-\boldsymbol{y}\right)-i\boldsymbol{k}\boldsymbol{y}}}{2^{\frac{5}{2}}\sqrt{\omega_{k}}\epsilon_{p}\epsilon_{q}\left(\omega_{k}+\epsilon_{p}+\epsilon_{q}\right)}a^{\dagger}_{k},$
$\displaystyle=\frac{i\lambda_{1}\lambda_{0}}{8}\int d^{3}x\int
d^{3}y\,\phi^{2}(x)\sum_{k}\sum_{p}\frac{e^{-i\boldsymbol{k}\boldsymbol{y}}}{2^{\frac{3}{2}}\epsilon_{p}^{2}\sqrt{\omega_{k}}\left(\omega_{k}+2\epsilon_{p}\right)}a^{\dagger}_{k}$
$\displaystyle\int\frac{d^{3}\left(p+q\right)}{\left(2\pi\right)^{3}}e^{i\left(\boldsymbol{p}+\boldsymbol{q}\right)\left(\boldsymbol{x}-\boldsymbol{y}\right)},$
$\displaystyle=\frac{i\lambda_{1}\lambda_{0}}{8}\int
d^{3}x\,\phi^{2}(x)\sum_{k}\sum_{p}\frac{e^{-i\boldsymbol{k}\boldsymbol{x}}}{2^{\frac{3}{2}}\epsilon_{p}^{2}\sqrt{\omega_{k}}\left(\omega_{k}+2\epsilon_{p}\right)}a^{\dagger}_{k}.$

The total commutator then is

$\displaystyle\left[\left[H_{A}^{4,3},\Omega_{0}^{4,3,1}\right],\Omega_{0}^{4,3,2}\right]$
$\displaystyle=\frac{\lambda_{1}^{2}\lambda_{0}}{16}\int d^{3}x\int
d^{3}z\,\phi^{2}(z)\phi^{2}(x)\sum_{k}\sum_{p}\sum_{k^{{}^{\prime}}}\frac{e^{i\boldsymbol{k}^{{}^{\prime}}\boldsymbol{z}-i\boldsymbol{k}\boldsymbol{x}}}{2^{2}\epsilon_{p}^{2}\sqrt{\omega_{k}\omega_{k^{{}^{\prime}}}}\omega_{k^{{}^{\prime}}}\left(\omega_{k}+2\epsilon_{p}\right)}$
(122) $\displaystyle\left[a_{k}^{\dagger},a_{k^{{}^{\prime}}}\right],$
$\displaystyle=-\frac{\lambda_{1}^{2}\lambda_{0}}{64}\int
d^{3}x\,\phi^{4}(x)\int\frac{d^{3}p}{\left(2\pi\right)^{3}}\,\frac{1}{M^{2}\epsilon_{p}^{2}\left(M+2\epsilon_{p}\right)}.$

Hence,

$\displaystyle-\frac{1}{3}\left[\left[H_{A}^{4,3},\Omega_{0}^{4,3,1}\right],\Omega_{0}^{4,3,2}\right]=\frac{\lambda_{1}^{2}\lambda_{0}}{192}\int
d^{3}x\,\phi^{4}(x)\int\frac{d^{3}p}{\left(2\pi\right)^{3}}\,\frac{1}{M^{2}\epsilon_{p}^{2}\left(M+2\epsilon_{p}\right)}.$
(123)

Including the Hermitian conjugate, the final contribution from the combination
in Table 1(d) is

$\displaystyle\frac{\lambda_{1}^{2}\lambda_{0}}{96}\int
d^{3}x\,\phi^{4}(x)\int\frac{d^{3}p}{\left(2\pi\right)^{3}}\,\frac{1}{M^{2}\epsilon_{p}^{2}\left(M+2\epsilon_{p}\right)}.$
(124)

The calculation of the contributions from
$-\frac{1}{2}\left[\left[H_{B}^{4,3},\Omega_{0}^{4,3,1}\right],\Omega_{0}^{4,3,2}\right]$
is very similar to that of Table 1(b) and Table 1(c).

###  A.4 Explicit Calculation of Figure 5

Consider Table 3 and in particular, the combination labelled (a). We will need
the following:

$\displaystyle H_{B}^{4,4}=\int
d^{3}x\,\frac{\lambda_{0}}{2}\phi^{2}(x)\sum_{p}\sum_{q}\frac{e^{i\left(\boldsymbol{p}-\boldsymbol{q}\right)\boldsymbol{x}}}{2\sqrt{\epsilon_{p}\epsilon_{q}}}b_{p}^{\dagger}b_{q},$
(125a) $\displaystyle\Omega_{0}^{4,4,1}=\int
d^{3}y\,\lambda_{1}\phi(y)\sum_{k}\sum_{p^{{}^{\prime}}}\frac{ie^{-i\left(\boldsymbol{k}+\boldsymbol{p}^{{}^{\prime}}\right)\boldsymbol{y}}}{2\sqrt{\omega_{k}\epsilon_{p^{{}^{\prime}}}}\left(\omega_{k}+\epsilon_{p^{{}^{\prime}}}\right)}a_{k}^{\dagger}b_{p^{{}^{\prime}}}^{\dagger},$
(125b) $\displaystyle\Omega_{0}^{4,4,2}=\int
d^{3}z\,\lambda_{1}\phi(z)\sum_{k^{{}^{\prime}}}\sum_{q^{{}^{\prime}}}\frac{-ie^{i\left(\boldsymbol{k}^{{}^{\prime}}+\boldsymbol{q}^{{}^{\prime}}\right)\boldsymbol{z}}}{2\sqrt{\omega_{k^{{}^{\prime}}}\epsilon_{q^{{}^{\prime}}}}\left(\omega_{k^{{}^{\prime}}}+\epsilon_{q^{{}^{\prime}}}\right)}a_{k^{{}^{\prime}}}b_{q^{{}^{\prime}}}.$
(125c)

The first commutator is,

$\displaystyle\left[H_{B}^{4,4},\Omega_{0}^{4,4,1}\right]$
$\displaystyle=\frac{i\lambda_{0}\lambda_{1}}{2}\int d^{3}x\int
d^{3}y\,\phi^{2}(x)\phi(y)\sum_{p}\sum_{q}\sum_{k}\sum_{p^{{}^{\prime}}}\frac{e^{-i\left(\boldsymbol{p}-\boldsymbol{q}\right)\boldsymbol{x}-i\left(\boldsymbol{k}+\boldsymbol{p}^{{}^{\prime}}\right)\boldsymbol{y}}}{4\sqrt{\epsilon_{p}\epsilon_{q}\omega_{k}\epsilon_{p^{{}^{\prime}}}}\left(\omega_{k}+\epsilon_{p^{{}^{\prime}}}\right)}$
(126)
$\displaystyle\left[b_{p}^{\dagger}b_{q},a_{k}^{\dagger}b_{p^{{}^{\prime}}}^{\dagger}\right],$
$\displaystyle=\frac{i\lambda_{0}\lambda_{1}}{8}\int d^{3}x\int
d^{3}y\,\phi^{2}(x)\phi(y)\sum_{p}\sum_{q}\sum_{k}\frac{e^{i\boldsymbol{q}\left(\boldsymbol{x}-\boldsymbol{y}\right)-i\boldsymbol{p}\boldsymbol{x}-i\boldsymbol{k}\boldsymbol{y}}}{\sqrt{\epsilon_{p}\omega_{k}}\epsilon_{q}\left(\omega_{k}+\epsilon_{q}\right)}b_{p}^{\dagger}a_{k}^{\dagger},$
$\displaystyle=\frac{i\lambda_{0}\lambda_{1}}{8}\int d^{3}x\int
d^{3}y\,\phi^{2}(x)\phi(y)\sum_{p}\sum_{k}\frac{e^{-i\boldsymbol{p}\boldsymbol{x}-i\boldsymbol{k}\boldsymbol{y}}}{\sqrt{\epsilon_{p}\omega_{k}}\epsilon_{p}\left(\omega_{k}+\epsilon_{p}\right)}b_{p}^{\dagger}a_{k}^{\dagger}\int\frac{d^{3}q}{\left(2\pi\right)^{3}}\,e^{i\boldsymbol{q}\left(\boldsymbol{x}-\boldsymbol{y}\right)},$
$\displaystyle=\frac{i\lambda_{0}\lambda_{1}}{8}\int
d^{3}x\,\phi^{3}(x)\sum_{p}\sum_{k}\frac{e^{-i\boldsymbol{p}\boldsymbol{x}-i\boldsymbol{k}\boldsymbol{y}}}{\sqrt{\epsilon_{p}\omega_{k}}\epsilon_{p}\left(\omega_{k}+\epsilon_{p}\right)}b_{p}^{\dagger}a_{k}^{\dagger},$

where we have used the condition that external momenta are zero, and therefore
$\left|k\right|=\left|p\right|=\left|q\right|$, which gives
$\epsilon_{p}=\epsilon_{q}$. The second commutator is

$\displaystyle\left[\left[H_{B}^{4,4},\Omega_{0}^{4,4,1}\right],\Omega_{0}^{4,4,2}\right]$
$\displaystyle=\frac{\lambda_{0}\lambda_{1}^{2}}{16}\int d^{3}x\int
d^{3}z\,\phi^{3}(x)\phi(z)\sum_{p}\sum_{k}\sum_{k^{{}^{\prime}}}\sum_{q^{{}^{\prime}}}$
(127)
$\displaystyle\frac{e^{i\left(\boldsymbol{k}^{{}^{\prime}}+\boldsymbol{q}^{{}^{\prime}}\right)\boldsymbol{z}-i\left(\boldsymbol{k}+\boldsymbol{p}\right)\boldsymbol{x}}}{\sqrt{\omega_{k}\omega_{k^{{}^{\prime}}}\epsilon_{p}\epsilon_{q^{{}^{\prime}}}}\left(\omega_{k}+\epsilon_{p}\right)\left(\omega_{k^{{}^{\prime}}}+\epsilon_{q^{{}^{\prime}}}\right)\epsilon_{p}}\left[b_{p}^{\dagger}a_{k}^{\dagger},a_{k^{{}^{\prime}}}b_{q^{{}^{\prime}}}\right],$
$\displaystyle=\frac{-\lambda_{0}\lambda_{1}^{2}}{16}\int d^{3}x\int
d^{3}z\,\phi^{3}(x)\phi(z)\sum_{k}\sum_{q}\frac{e^{i\left(\boldsymbol{k}+\boldsymbol{p}\right)\left(\boldsymbol{z}-\boldsymbol{x}\right)}}{\omega_{k}\epsilon_{p}^{2}\left(\omega_{k}+\epsilon_{p}\right)^{2}},$
$\displaystyle=\frac{-\lambda_{0}\lambda_{1}^{2}}{16}\int
d^{3}x\,\phi^{4}(x)\int\frac{d^{3}k}{\left(2\pi\right)^{3}}\frac{1}{\omega_{k}\epsilon_{k}^{2}\left(\omega_{k}+\epsilon_{k}\right)^{2}}.$

Including the Hermitian conjugate, the final result for Table 3(a) is:

$\displaystyle\frac{\lambda_{0}\lambda_{1}^{2}}{16}\int
d^{3}x\,\phi^{4}(x)\int\frac{d^{3}k}{\left(2\pi\right)^{3}}\frac{1}{\omega_{k}\epsilon_{k}^{2}\left(\omega_{k}+\epsilon_{k}\right)^{2}}.$
(128)

Next we turn to the combinations (b) and (c) of Table 3. We will need the
expansions:

$\displaystyle H_{B}^{4,4}=\lambda_{1}\int
d^{3}x\,\phi(x)\sum_{k}\sum_{p}\frac{e^{i\left(\boldsymbol{k}-\boldsymbol{p}\right)\boldsymbol{x}}}{2\sqrt{\omega_{k}\epsilon_{p}}}b_{p}^{\dagger}a_{k},$
(129a) $\displaystyle\Omega_{0}^{4,4,1}=\lambda_{1}\int
d^{3}y\,\phi(y)\sum_{k^{{}^{\prime}}}\sum_{q}\frac{ie^{-i\left(\boldsymbol{k}^{{}^{\prime}}+\boldsymbol{q}\right)\boldsymbol{y}}}{2\sqrt{\omega_{k^{{}^{\prime}}}\epsilon_{q}}\left(\omega_{k^{{}^{\prime}}}+\epsilon_{q}\right)}a_{k^{{}^{\prime}}}^{\dagger}b_{q}^{\dagger},$
(129b) $\displaystyle\Omega_{0}^{4,4,2}=\frac{\lambda_{0}}{4}\int
d^{3}z\,\phi^{2}(z)\sum_{p^{{}^{\prime}}}\sum_{q^{{}^{\prime}}}\frac{-ie^{i\left(\boldsymbol{p}^{{}^{\prime}}+\boldsymbol{q}^{{}^{\prime}}\right)\boldsymbol{z}}}{2\sqrt{\epsilon_{p^{{}^{\prime}}}\epsilon_{q^{{}^{\prime}}}}\left(\epsilon_{p^{{}^{\prime}}}+\epsilon_{q^{{}^{\prime}}}\right)}b_{p^{{}^{\prime}}}b_{q^{{}^{\prime}}}.$
(129c)

First commutator is

$\displaystyle\left[H_{B}^{4,4},\Omega_{0}^{4,4,1}\right]$
$\displaystyle=i\lambda_{1}^{2}\int d^{3}x\int
d^{3}y\,\phi(x)\phi(y)\sum_{k}\sum_{p}\sum_{k^{{}^{\prime}}}\sum_{q}\frac{e^{i\left(\boldsymbol{k}-\boldsymbol{p}\right)\boldsymbol{x}-i\left(\boldsymbol{k}^{{}^{\prime}}+\boldsymbol{q}\right)\boldsymbol{y}}}{4\sqrt{\omega_{k}\omega_{k^{{}^{\prime}}}\epsilon_{p}\epsilon_{q}}\left(\omega_{k^{{}^{\prime}}}+\epsilon_{q}\right)}$
(130)
$\displaystyle\left[a_{k}b_{p}^{\dagger},a{k^{{}^{\prime}}}^{\dagger}b_{q}^{\dagger}\right],$
$\displaystyle=\frac{i\lambda_{1}^{2}}{4}\int d^{3}x\int
d^{3}y\,\phi(x)\phi(y)\sum_{k}\sum_{p}\sum_{q}\frac{e^{i\left(\boldsymbol{x}-\boldsymbol{y}\right)\boldsymbol{k}-i\boldsymbol{p}\boldsymbol{x}-i\boldsymbol{q}\boldsymbol{y}}}{\omega_{k}\sqrt{\epsilon_{p}\epsilon_{q}}\left(\omega_{k}+\epsilon_{q}\right)}b_{p}^{\dagger}b_{q}^{\dagger},$
$\displaystyle=\frac{i\lambda_{1}^{2}}{4}\int
d^{3}x\,\phi^{2}(x)\sum_{p}\sum_{q}\frac{e^{-i\left(\boldsymbol{p}+\boldsymbol{q}\right)\boldsymbol{x}}}{\omega_{p}\sqrt{\epsilon_{p}\epsilon_{q}}\left(\omega_{p}+\epsilon_{q}\right)}b_{p}^{\dagger}b_{q}^{\dagger}.$

Again, we use the condition that external momenta are zero, and rewrite
$\omega_{k}$ as $\omega_{p}$, since $\left|k\right|=\left|p\right|$. The
second commutator gives:

$\displaystyle\left[\left[H_{B}^{4,4},\Omega_{0}^{4,4,1}\right],\Omega_{0}^{4,4,2}\right]$
$\displaystyle=\frac{\lambda_{1}^{2}\lambda_{0}}{16}\int d^{3}x\int
d^{3}z\,\phi^{2}(x)\phi^{2}(z)\sum_{p}\sum_{q}\sum_{p^{{}^{\prime}}}\sum_{q^{{}^{\prime}}}$
(131)
$\displaystyle\frac{e^{-i\left(\boldsymbol{p}+\boldsymbol{q}\right)\boldsymbol{x}+i\left(\boldsymbol{p}^{{}^{\prime}}+\boldsymbol{q}^{{}^{\prime}}\right)\boldsymbol{z}}}{2\omega_{p}\sqrt{\epsilon_{p}\epsilon_{q}\epsilon_{p^{{}^{\prime}}}\epsilon_{q^{{}^{\prime}}}}\left(\epsilon_{p^{{}^{\prime}}}+\epsilon_{q^{{}^{\prime}}}\right)\left(\omega_{p}+\epsilon_{q}\right)}\left[b_{p}^{\dagger}b_{q}^{\dagger},b_{p^{{}^{\prime}}}b_{q^{{}^{\prime}}}\right],$
$\displaystyle=\frac{-\lambda_{1}^{2}\lambda_{0}}{16}\int d^{3}x\int
d^{3}z\,\phi^{2}(x)\phi^{2}(z)\sum_{p}\sum_{q}\frac{e^{i\left(\boldsymbol{p}+\boldsymbol{q}\right)\left(\boldsymbol{z}-\boldsymbol{x}\right)}}{\omega_{p}\epsilon_{p}\epsilon_{q}\left(\epsilon_{p}+\epsilon_{q}\right)\left(\omega_{p}+\epsilon_{q}\right)},$
$\displaystyle=\frac{-\lambda_{1}^{2}\lambda_{0}}{32}\int
d^{3}x\,\phi^{4}(x)\int\frac{d^{3}p}{\left(2\pi\right)^{3}}\,\frac{1}{\omega_{p}\epsilon_{p}^{3}\left(\omega_{p}+\epsilon_{p}\right)}.$

Including the Hermitian conjugate, the final result for both Table 3(b) and
Table 3(c) is:

$\displaystyle\frac{\lambda_{1}^{2}\lambda_{0}}{32}\int
d^{3}x\,\phi^{4}(x)\int\frac{d^{3}p}{\left(2\pi\right)^{3}}\,\frac{1}{\omega_{p}\epsilon_{p}^{3}\left(\omega_{p}+\epsilon_{p}\right)}.$
(132)

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