# Beliaev damping in Bose gas

Jan Dereziński  Department of Mathematical Methods in Physics, Faculty of
Physics, University of Warsaw, Pasteura 5, 02-093 Warszawa, Poland
<EMAIL_ADDRESS>,  Ben Li  Department of Mathematical Methods in
Physics, Faculty of Physics, University of Warsaw, Pasteura 5, 02-093
Warszawa, Poland<EMAIL_ADDRESS>and  Marcin Napiórkowski  Department of
Mathematical Methods in Physics, Faculty of Physics, University of Warsaw,
Pasteura 5, 02-093 Warszawa, Poland<EMAIL_ADDRESS>
###### Abstract.

According to the Bogoliubov theory the low energy behaviour of the Bose gas at
zero temperature can be described by non-interacting bosonic quasiparticles
called phonons. In this work the damping rate of phonons at low momenta, the
so-called Beliaev damping, is explained and computed with simple arguments
involving the Fermi Golden Rule and Bogoliubov’s quasiparticles.

##  1\. Introduction

The Bose gas near the zero temperature has curious properties that can be
partly explained from the first principles by a beautiful argument that goes
back to Bogoliubov [5]. In Bogoliubov’s approach the Bose gas at zero
temperature can be approximately described by a gas of weakly interacting
quasiparticles. The dispersion relation of these quasiparticles, that is,
their energy in function of the momentum is described by a function
${\mathbf{k}}\mapsto e_{\mathbf{k}}$ with an interesting shape. At low momenta
these quasiparticles are called phonons and $e_{\mathbf{k}}\approx ck$, where
$c>0$ and $k:=|{\mathbf{k}}|$. Thus the low-energy dispersion relation is very
different from the non-interacting, quadratic one. It is responsible for
superfluidity of the Bose gas.

It is easy to see that phonons could be metastable, because the energy-
momentum conservation may not prohibit them to decay into two or more phonons.
This decay rate was first computed in perturbation theory by Beliaev [2],
hence the name Beliaev damping. According to his computation, the imaginary
part of the dispersion relation behaves for small momenta as
$-c_{\mathrm{Bel}}k^{5}$. This implies the exponential decay of phonons with
the decay rate $2c_{\mathrm{Bel}}k^{5}$. The Beliaev damping has been observed
in experiments, and appears to be consistent with its theoretical predictions
[25, 22].

In our paper we present a systematic derivation of Beliaev damping. Our
presentation differs in several points from similar accounts found in the
physics literature. We try to make all the arguments as transparent as
possible, without hiding some of less rigorous steps. We avoid using
diagrammatic techniques, in favor of a mathematically much clearer picture
involving a Bogoliubov transformation and the 2nd order perturbation
computation (the so-called Fermi Golden Rule) applied to what we call the
effective Friedrichs Hamiltonian. We use the grand-canonical picture instead
of the canonical one found in a part of the literature. This is a minor
difference; on this level both pictures should lead to the same final result.
We believe that the derivation of Beliaev damping is a beautiful illustration
of methods many-body quantum physics, which is quite convincing even if not
fully rigorous.

In the remaining part of the introduction we provide a brief sketch of the
main steps of Beliaev’s argument. In the main body of our article we discuss
these steps in more detail, indicating which parts can be easily made
rigorous.

Let $v$ be a real function satisfying $v(x)=v(-x)$ that decays fast at
infinity. (Later we will need more assumptions on $v$.) The homogeneous Bose
gas of $N$ particles interacting with the pair potential $v$ is described by
the Hamiltonian and the total momentum

$\displaystyle H_{N}$
$\displaystyle=-\sum_{i=1}^{N}\frac{1}{2m_{i}}\Delta_{i}+\sum_{1\leq i<j\leq
N}v(x_{i}-x_{j}),$ (1) $\displaystyle P_{N}$
$\displaystyle=\sum_{i=1}^{N}\frac{1}{\mathrm{i}}\partial_{x_{i}}.$ (2)

These operators act on $L_{\mathrm{s}}^{2}\big{(}(\mathbb{R}^{3})^{N}\big{)}$,
the space of functions symmetric in the positions of $N$ 3-dimensional
particles. Note that $H_{N}$ commutes with $P_{N}$, which expresses the
spatial homogeneity of the system.

We would like to describe Bose gas of positive density in infinite volume.
This is difficult to do in terms of the Hamiltonian acting on the whole space
$\mathbb{R}^{3}$. Therefore we replace (1) and (2) with a system enclosed in a
box of size $L$, and then we take thermodynamic limit. In order to preserve
translation symmetry we consider periodic boundary conditions. They are not
very physical, but it is believed that they should not affect the overall
picture in thermodynamic limit.

Thus $v$ is replaced by its periodized version adapted to the box of size $L$.
The new Hilbert space is
$L_{\mathrm{s}}^{2}\big{(}([-L/2,L/2]^{3})^{N}\big{)}$. We will use the same
symbols $H_{N},P_{N}$ to denote the Hamiltonian and total momentum in the box.
Note that they still commute with one another.

It is very convenient to consider at the same time all numbers of particles.
In order to control the density, that is $\frac{N}{L^{3}}$, we introduce the
chemical potential given by a positive number $\mu>0$, and we use the grand-
canonical formalism. It is also convenient to pass from the position to the
momentum representation. Thus we replace $H_{N},P_{N}$ with

$\displaystyle H$
$\displaystyle:=\mathop{\oplus}\limits_{N=0}^{\infty}(H_{N}-\mu N)=\int
a_{x}^{*}(-\Delta_{x}-\mu)a_{x}\,{\rm d}x+\frac{1}{2}\int\int\,{\rm d}x\,{\rm
d}yv(x-y)a_{x}^{*}a_{y}^{*}a_{y}a_{x}$
$\displaystyle=\sum_{\mathbf{p}}({\mathbf{p}}^{2}-\mu)a_{\mathbf{p}}^{*}a_{\mathbf{p}}\,{\rm
d}{\mathbf{p}}+\frac{1}{2L^{3}}\sum_{\mathbf{p}}\sum_{\mathbf{q}}\sum_{\mathbf{k}}\hat{v}({\mathbf{k}})a_{{\mathbf{p}}+{\mathbf{k}}}^{*}a_{{\mathbf{q}}-{\mathbf{k}}}^{*}a_{\mathbf{q}}a_{\mathbf{p}},$
(3) $\displaystyle P$
$\displaystyle:=\mathop{\oplus}\limits_{N=0}^{\infty}P_{N}=\int
a_{x}^{*}\frac{1}{\mathrm{i}}\partial_{x}a_{x}\,{\rm
d}x=\sum_{\mathbf{p}}{\mathbf{p}}a_{\mathbf{p}}^{*}a_{\mathbf{p}}.$ (4)

$a_{x}^{*}$ and $a_{x}$ are the usual creation/annihilation operators for
$x\in[-L/2,L/2]^{3}$ in the position representation, commuting to the Dirac
delta. $a_{\mathbf{p}}^{*},a_{\mathbf{p}}$ are the usual
creration/annihilation operators for ${\mathbf{p}}\in 2\pi\mathbb{Z}^{3}/L$ in
the momentum representation commuting to the Kronecker delta. $H,P$ act on the
bosonic Fock space with the one-particle space
$L^{2}\big{(}[-L/2,L/2]^{3}\big{)}$ in the position representation, and
$l^{2}\big{(}2\pi\mathbb{Z}^{3}/L\big{)}$ in the momentum representation. $H$
and $P$ still commute with one another.

Now there comes the main idea of the Bogoliubov approach. At zero temperature,
one expects complete Bose–Einstein condensation. This is expressed by assuming
that the zero mode is populated macroscopically and nonzero modes are only
very few. The zero mode is treated classically, and essentially removed from
the picture. One obtains an approximate Hamiltonian, which does not preserve
the number of particles. One argues that its most important component is the
quadratic part which involves operators of the form
$a_{\mathbf{k}}a_{-{\mathbf{k}}}$, $a_{\mathbf{k}}^{*}a_{-{\mathbf{k}}}^{*}$
and $a_{\mathbf{k}}^{*}a_{\mathbf{k}}$, ${\mathbf{k}}\neq 0$. It can be
diagonalized by a linear transformation which mixes creation and annihilation
operators, called since [5] a Bogoliubov transformation, and becomes

$\displaystyle\qquad H_{\mathrm{Bog}}$ $\displaystyle:=$
$\displaystyle\sum_{{\mathbf{k}}\neq
0}e_{\mathbf{k}}b_{\mathbf{k}}^{*}b_{\mathbf{k}},$ (5) $\displaystyle
e_{\mathbf{k}}$ $\displaystyle:=$
$\displaystyle\sqrt{\frac{1}{4}|{\mathbf{k}}|^{4}+\frac{\hat{v}({\mathbf{k}})}{\hat{v}(0)}\mu|{\mathbf{k}}|^{2}}.$
(6)

Thus, the Bogoliubov approximation states that

$H\approx E_{\mathrm{Bog}}+H_{\mathrm{Bog}}$ (7)

where $E_{\mathrm{Bog}}$ is a constant, which will not be relevant for our
analysis. The operator $b_{\mathbf{k}}^{*}$ is the creation operator of the
quasiparticle with momentum ${\mathbf{k}}$. It is a linear combination of
$a_{\mathbf{k}}^{*},a_{-{\mathbf{k}}}$. (5) is sometimes called a Bogoliubov
Hamiltonian. It describes independent quasiparticles with the dispersion
relation $e_{\mathbf{k}}$. The Bogoliubov vacuum, annihilated by
$b_{\mathbf{k}}$ and denoted $\Omega_{\mathrm{Bog}}$, is its ground state, and
can be treated as an approximate ground state of the many-body system. The
Bogoliubov Hamiltonian is still translation invariant: in fact, it commutes
with the total momentum, described (without any approximation) by

$P=\sum_{{\mathbf{k}}\neq 0}{\mathbf{k}}b_{\mathbf{k}}^{*}b_{\mathbf{k}}.$ (8)

It is easy to describe the thermodynamic limit of (5) and (8): we simply
replace the summation by integration, without changing the dispersion
relation:

$\displaystyle H_{\mathrm{Bog}}$ $\displaystyle=\int
e_{\mathbf{k}}b_{\mathbf{k}}^{*}b_{\mathbf{k}}\,{\rm d}{\mathbf{k}},$ (9)
$\displaystyle P$
$\displaystyle=\int{\mathbf{k}}b_{\mathbf{k}}^{*}b_{\mathbf{k}}\,{\rm
d}{\mathbf{k}}.$ (10)

It is interesting to visualize possible energy-momentum values of the system
or, in a more precise mathematical language, the joint spectrum of the total
momentum $P$ and the Bogoliubov Hamiltonian $H_{\mathrm{Bog}}$. On the
1-quasiparticle space this joint spectrum is given by the graph of the
function ${\mathbf{k}}\mapsto e_{\mathbf{k}}$. On fig. 1 we show a typical
form of the dispersion relation in the low momentum region, marked with the
black line. The green line denotes the bottom of the 2-quasiparticle spectrum,
that is the joint spectrum of $(H_{\mathrm{Bog}},P)$ in the 2-quasiparticle
sector. The bottom of the full joint spectrum of $(H_{\mathrm{Bog}},P)$ is
marked with a red dashed line.

Figure 1. Joint spectrum of $(H_{\mathrm{Bog}},P)$ for generic potentials

One can perform an additional step in the Bogoliubov approach. If the
potential $v$ has a very small support, one can argue that
$\frac{\hat{v}({\mathbf{k}})}{\hat{v}(0)}$ can be approximated by $1$. One
then usually says that the interaction is given by contact potentials, which
are presented in the position representation as $v(x)=a\delta(x)$, where $a$
is a constant, called the scattering length. This, however, strictly speaking
is not correct. The delta function needs a renormalization to become a well-
defined interaction in the two-body case; in the $N$-body case the situation
is even more problematic. Anyway, in this approximation, which is valid in the
dilute case, we obtain a simpler dispersion relation

$e_{\mathbf{k}}=\sqrt{\frac{1}{4}|{\mathbf{k}}|^{4}+\mu|{\mathbf{k}}|^{2}}.$
(11)

On fig. 2 we show the energy-momentum spectrum corresponding to (11).

Figure 2. Joint spectrum of $(H_{\mathrm{Bog}},P)$ for contact potentials

The Hamiltonian $H_{\mathrm{Bog}}$, both with the dispersion relation (6) and
(11) has remarkable physical consequences. Note first that the dispersion
relation ${\mathbf{k}}\mapsto e_{\mathbf{k}}$ has a linear cusp at the bottom.
It also has a positive critical velocity, that is,

$c_{\mathrm{crit}}:=\sup\\{c\ |\ e_{\mathbf{k}}\geq
ck,\quad{\mathbf{k}}\in\mathrm{R}^{3}\\}>0.$ (12)

In other words, the graph ${\mathbf{k}}\mapsto e_{\mathbf{k}}$ is above
${\mathbf{k}}\mapsto c_{\mathrm{crit}}k$. The full joint spectrum
$\sigma(P,H_{\mathrm{Bog}})$ is still above ${\mathbf{k}}\mapsto
c_{\mathrm{crit}}k$. This is interpreted as one of the most important
properties of superfluidity: a droplet of the Bose gas travelling with
velocity less than $c_{\mathrm{crit}}k$ has negligible friction (see e.g.
[11]).

Of course, $H_{\mathrm{Bog}}$ yields only an approximate description of the
Bose gas. In reality, one cannot treat the quasiparticles given by
$b_{\mathbf{k}}^{*},b_{\mathbf{k}}$ as fully independent. In the derivation of
the Bogoliubov Hamiltonian various terms were neglected. In particular, terms
of the third and fourth degree in $b_{\mathbf{k}}^{*},b_{\mathbf{k}}$ were
dropped. Replacing $v$ by $\kappa v$ we obtain an (artificial) coupling
constant, to be set to $1$ at the end. The third order terms are multiplied by
$\sqrt{\kappa}$ and the quartic terms by $\kappa$. We argue that the quartic
terms are of lower order and can be dropped. The third order terms have the
form

$\displaystyle\frac{1}{\sqrt{L^{3}}}\sum_{{\mathbf{k}},{\mathbf{p}},{\mathbf{k}}+{\mathbf{p}}\neq
0}u_{{\mathbf{k}},{\mathbf{p}}}b_{\mathbf{k}}^{*}b_{\mathbf{p}}^{*}b_{{\mathbf{k}}+{\mathbf{p}}}+\overline{u_{{\mathbf{k}},{\mathbf{p}}}}b_{{\mathbf{k}}+{\mathbf{p}}}b_{\mathbf{k}}^{*}b_{\mathbf{p}}^{*}$
(13) $\displaystyle+$
$\displaystyle\frac{1}{\sqrt{L^{3}}}\sum_{{\mathbf{k}},{\mathbf{p}},{\mathbf{k}}+{\mathbf{p}}\neq
0}w_{{\mathbf{k}},{\mathbf{p}}}b_{\mathbf{k}}^{*}b_{\mathbf{p}}^{*}b_{-{\mathbf{k}}-{\mathbf{p}}}^{*}+\overline{w_{{\mathbf{k}},{\mathbf{p}}}}b_{-{\mathbf{k}}-{\mathbf{p}}}b_{\mathbf{k}}b_{\mathbf{p}}.$
(14)

We will argue (see Section 6) that triple creation and triple annihilation
terms do not contribute to the decay of phonons. Thus we drop also (14).

Let us investigate what happens with the quasiparticle state
$b_{\mathbf{k}}^{*}\Omega_{\mathrm{Bog}}$ under the perturbation (13). The
state $b_{\mathbf{k}}^{*}\Omega_{\mathrm{Bog}}$ couples only to the
2-quasiparticle sector. By taking thermodynamic limit we can assume that the
variable ${\mathbf{k}}$ is continuous. Thus the perturbed quasiparticle can be
described by the space $\mathbb{C}\oplus L^{2}(\mathbb{R}^{3})$ with the
Hamiltonian

$\displaystyle H_{\rm{Fried}}({\mathbf{k}})$
$\displaystyle:=\begin{bmatrix}e_{\mathbf{k}}&(h_{\mathbf{k}}|\\\
|h_{\mathbf{k}})&e_{\mathbf{p}}+e_{{\mathbf{k}}-{\mathbf{p}}}\end{bmatrix},$
(15)

and $h_{\mathbf{k}}$ can be derived from (13). Hamiltonians similar to this
one are well understood. They are often used as toy models in quantum physics
and are sometimes called Friedrichs Hamiltonians.

It is important to notice that, if we set $\hat{v}=0$, so that the off-
diagonal terms in (15) disappear, the unperturbed quasiparticle energy
$e_{\mathbf{k}}$ lies inside the continuous spectrum of 2-quasiparticle
excitations $\\{e_{\mathbf{p}}+e_{{\mathbf{k}}-{\mathbf{p}}}\ |\
{\mathbf{p}}\in\mathbb{R}^{3}\\}$, at least for small momenta. (To be able to
say this we need thermodynamic limit which makes the momentum continuous.) To
see this, note that if ${\mathbf{k}}\mapsto e_{\mathbf{k}}$ is convex we have
a particularly simple expression (cf. Lemma 1) for the infimum of the
2-quasiparticle spectrum:

$\inf_{{\mathbf{p}}}\\{e_{\mathbf{p}}+e_{{\mathbf{k}}-{\mathbf{p}}}\\}=2e_{{\mathbf{k}}/2}.$
(16)

Now (11) is strictly convex, hence $e_{\mathbf{k}}$ lies inside the continuous
spectrum of 2-quasiparticle excitations. The generic dispersion relation (6)
is convex for small momenta, hence this property is true at least for small
momenta.

Because of that, one can expect that the position of the singularity of the
resolvent of (15) becomes complex—it describes a resonance and not a bound
state. This is interpreted as the unstability of the quasiparticle: its decay
rate is twice the imaginary part of the resonance.

The second order perturbation theory, often called the Fermi Golden Rule, says
that in order to compute the (complex) energy shift of an eigenvalue we need
to find the so-called self-energy $\Sigma_{\mathbf{k}}(z)$, which for
$z\not\in\mathbb{R}$ in our case is given by the integral

$\displaystyle\Sigma_{\mathbf{k}}(z)$
$\displaystyle=\frac{1}{(2\pi)^{3}}\int\frac{h_{\mathbf{k}}^{2}({\mathbf{p}})\,{\rm
d}{\mathbf{p}}}{(z-e_{\mathbf{p}}-e_{{\mathbf{k}}-{\mathbf{p}}})}.$ (17)

Then $\Sigma_{\mathbf{k}}(e_{\mathbf{k}}+\mathrm{i}0)$ should give the energy
shift of the eigenvalue $e_{\mathbf{k}}$.

The imaginary part of this shift is much easier to compute. In fact, applying
the Sochocki-Plemelj formula
$\frac{1}{x+\mathrm{i}0}=\mathcal{P}\frac{1}{x}-\mathrm{i}\pi\delta(x)$ we
obtain

$\displaystyle\mathrm{Im}\Sigma_{\mathbf{k}}(e_{\mathbf{k}}+\mathrm{i}0)$
$\displaystyle=\frac{1}{8\pi^{2}}\int
h^{2}({\mathbf{p}})\delta(e_{\mathbf{k}}-e_{\mathbf{p}}-e_{{\mathbf{k}}-{\mathbf{p}}})\,{\rm
d}{\mathbf{p}}.$ (18)

In Theorem 2 we prove that if $e_{\mathbf{k}}$ is given by (11), then

$\mathrm{Im}\Sigma_{{\mathbf{k}}}(e_{k}+\mathrm{i}0)=-c_{\mathrm{Bel}}k^{5}+O(k^{6})\qquad\text{as}\qquad
k\to 0,\qquad c_{\mathrm{Bel}}=\frac{3\hat{v}(0)}{640\pi^{2}\mu}k^{5}.$ (19)

In fact, our result could be also extended to the case of (6), but for the
sake of clarity of the presentation we present the proof only for (11).
Physically (19) means that quasiparticles are almost stable for small $k$ with
the lifetime proportional to $k^{-5}$. (19) is the main result of our paper.

We remark that our analysis is based on the grand-canonical approach where
$\mu$ is the chemical potential. One can go back to the canonical picture. To
this end one determines the chemical potential as a function of the density.
In the Bogoliubov approximation one obtains to leading order that
$\rho\approx\mu/\hat{v}(0)$. Furthermore, also $\rho\approx\rho_{0}$, where
$\rho_{0}$ is the condensate density, holds to leading order and thus the
proportionality constant can be written as

$c_{\mathrm{Bel}}=\frac{3}{640\pi^{2}\rho_{0}},$ (20)

which is the form of this result which is usually stated in the physics
literature ([36, 19, 28, 13]).

One could naively expect that the same method gives the correction to the real
part of the dispersion relation. Unfortunately,
$\mathrm{Re}\Sigma_{{\mathbf{k}}}(z)$ obtained from (17) is ill defined
because of the divergence of the integral at large momenta. One can impose a
cut-off and try to renormalize. For instance, one can replace
$h_{\mathbf{k}}({\mathbf{p}})$ by

$h_{\mathbf{k}}^{\Lambda}({\mathbf{p}}):=h_{\mathbf{k}}({\mathbf{p}})\theta(\Lambda-{\mathbf{p}}-|{\mathbf{k}}-{\mathbf{p}}|),$
(21)

where $\theta$ is the Heaviside function. (Note that the details of the cutoff
are not physically relevant; (21) is especially convenient for computations,
because it respects the natural symmetry of the problem). The cut-off self-
energy

$\displaystyle\Sigma_{\mathbf{k}}^{\Lambda}(z)$
$\displaystyle=\frac{1}{(2\pi)^{3}}\int\frac{(h_{\bf
k}^{\Lambda}({\mathbf{p}}))^{2}\,{\rm
d}{\mathbf{p}}}{(z-e_{\mathbf{p}}-e_{{\mathbf{k}}-{\mathbf{p}}})}$ (22)

is well defined.

Let us now try to remove the dependence of the self-energy on the cutoff. The
most satisfactory renormalization scenario would be to find a counterterm
$c^{\Lambda}$ independent of ${\mathbf{k}}$ so that

$\lim_{\Lambda\to\infty}\big{(}\Sigma_{\mathbf{k}}^{\Lambda}(z)-c^{\Lambda}\big{)}\quad\text{
exists. }$ (23)

An initial positive result suggests that one can hope for a removal of the
ultraviolet cutoff in the self-energy: there exists the limit

$\lim_{\Lambda\to\infty}\big{(}\Sigma_{\mathbf{k}}^{\Lambda}(z)-\Sigma_{\mathbf{k}}^{\Lambda}(0)\big{)}.$
(24)

Unfortunately, $\lim\limits_{{\mathbf{k}}\to
0}\Sigma_{\mathbf{k}}^{\Lambda}(0)=\infty$, which implies that finding a
$c^{\Lambda}$ such that (23) is true is impossible. This is the content of
Theorem 3. Thus the physical meaning of the quantity (24) is dubious, because
the counterterm $\Sigma_{\mathbf{k}}^{\Lambda}(0)$ depends on the momentum
${\mathbf{k}}$. We leave the interpretation of this result open.

One can conclude that perturbation theory around the Bogoliubov Hamiltonian
provides a reasonable method to find the second order imaginary correction to
the dispersion relation. However, by this method we seem not able to compute
its real part. This is not very surprising. It is a general property of
Friedrichs Hamiltonians with singular off-diagonal terms: the imaginary part
of the perturbed eigenvalue can be computed much more reliably than its real
part. We describe this briefly in Sections 2 and 3.

The above problem is an indication of the crudeness of the Bogoliubov
approximation. Throwing out the zero mode from the picture (or, which is
essentially the same, treating it as a classical quantity), as well as
throwing out higher order terms, is a very violent act and we should not be
surprised by a punishment. By the way, one expects that the true dispersion
relation of phonons goes to zero as ${\mathbf{k}}\to 0$. This is the content
of the so called “Hugenholtz-Pines Theorem” [24], which is a (non-rigorous)
argument based on the gauge invariance. Perturbation theory around the
Bogoliubov Hamiltonian is compatible with this theorem where it comes to the
imaginary part. For the real part it fails.

Let us now make a few remarks about the literature. The orginal paper of
Bogoliubov [5] was heuristic, however in recent years there have been many
rigorous papers justifying Bogoliubov’s approximation in several cases. The
first result justifying (7) has been obtained in the mean-field scaling by
Seiringer in [35] (see also [26, 17, 20, 32] for related results). Recently,
corresponding results have been obtained in the Gross-Pitaevskii regime [3,
10, 33] and even beyond [9]. A time-dependent version of Bogoliubov theory has
been successful in describing the dynamics of Bose-Einstein condensates and
excitations thereof (see [30, 34] for reviews).

As explained above, to describe damping one has to go beyond Bogoliubov
theory. In the mean-field regime this has been done for the ground state
energy expansion in [31, 8] and for the dynamics in [7]. Very recently, the
extension of [8] to singular interactions has been obtained in [6].

None of the above rigorous papers, with exception of [17], addressed the
energy-momentum spectrum. In fact, it is very difficult to study rigorously
the dispersion relation in thermodynamic limit—which is essentially necessary
to analyze phonon damping.

The quasiparticle picture of the Bose gas at low temperatures has been
confirmed in experiments. The dispersion relation of ${}^{4}\mathrm{He}$ can
be observed in neutron scattering experiments, and is remarkably sharp. It has
been measured within a large range of wave numbers covering not only phonons,
but also the so-called maxons and rotons, see e.g. [21]. In particular, one
can see that the dispersion relation is slightly higher than the
2-quasiparticle spectrum for low wave numbers. The quasiparticle picture has
also been confirmed by experiments on Bose Einstein condensates involving
alkali atoms. The Beliaev damping has been observed in experiments on Bose
Einstein condensates. The results are consistent with theoretical predictions
[25, 22]. Note, however, that the precise prediction (18) is difficult to
verify experimentally. Bose-Einstein condensates created in labs are not very
large, so it is difficult to probe the large wavelength region.

Let us mention that there exists another phenomenon found in Bose-Einstein
condensates, the so-called Landau damping, which involves instability of
quasiparticles due to thermal excitations. The Landau damping is absent at
zero temperature and becomes dominant at higher temperatures. The Beliaev
damping occurs at zero temperature, and for very small temperatures it is
still stronger than the Landau damping.

In the physics literature, the damping of phonons was first computed by
Beliaev [2]. Lanadu damping has been for the first time computed by Hohenberg
and Martin in [23] (see also [29]). Both these results have been reproduced in
[36], also using the formalism of Feynman diagrams and many-body Green’s
functions. In [28] the damping rate was derived starting from an effective
action in the spirit of Popov’s hydrodynamical approach. [19] repeated the
same computation in the time-dependent mean-field approach. In [13] the mean-
field and hydrodynamic approaches were applied to the 2D case. Our derivation
is consistent with the above works, however, in our opinion, avoids some
unnecessary elements obscuring the simple mechanism of the Beliaev damping.

The plan of the paper is as follows. Sections 2 and 3 concern general well-
known facts about about 2nd order perturbation theory of embedded eigenvalues.
In Section 4 we define the Bose gas Hamiltonian and describe the Bogoliubov
approach in the grand-canonical setting. In Section 5 we derive heuristically
the effective model that we consider. Then, in Section 6 we discuss the shape
of the energy-momentum spectrum and explain why the contribution from term
(14) is irrelevant for the damping rate computation, which is the main result
of the paper is proven in Section 8 as Theorem 2. The analysis why computing
the real part of the self-energy fails by the method of this paper is
described in Section 9.

Acknowledgements. The work of all authors was supported by the Polish-German
NCN-DFG grant Beethoven Classic 3 (project no. 2018/31/G/ST1/01166).

##  2\. Friedrichs Hamiltonian

Suppose that $\mathcal{H}$ is a Hilbert space with a self-adjoint operator
$H$. Let $\Psi\in\mathcal{H}$ be a normalized vector. We can write
$\mathcal{H}\simeq\mathbb{C}\oplus\mathcal{K}$, where
$\mathbb{C}\simeq\mathbb{C}\Psi$ and $\mathcal{K}:=\\{\Psi\\}^{\perp}$. First
assume that $\Psi$ belongs to the domain of $H$ and set

$h:=H\Psi,\quad E_{0}:=(\Psi|H\Psi).$ (25)

Let $K$ denote $H$ compressed to $\mathcal{K}$. That means, if
$I:\mathcal{K}\to\mathcal{H}$ is the embedding, then $K:=I^{*}HI$. Then in
terms of $\mathbb{C}\oplus\mathcal{K}$ we can write

$H=\begin{bmatrix}E_{0}&(h|\\\ |h)&K\end{bmatrix}.$ (26)

Operators of this form were studied by Friedrichs in [18]. Therefore,
sometimes they are referred to as Friedrichs Hamiltonians.

Let $z\in\mathbb{C}$. The following identity is a special case of the so-
called Feshbach-Schur formula:

$\displaystyle(\Psi|(H-z)^{-1}\Psi)$
$\displaystyle=\frac{1}{E_{0}+\Sigma(z)-z},$ (27) $\displaystyle\Sigma(z)$
$\displaystyle=-(h|(K-z)^{-1}h).$ (28)

Following a part of the physics literature, we will call $\Sigma(z)$ the self-
energy. For further reference let us rewrite (27) as

$\Sigma(z)=\frac{1}{(\Psi|(H-z)^{-1}\Psi)}+z-E_{0},$ (29)

and let us describe the full resolvent:

$\displaystyle(H-z)^{-1}=$ $\displaystyle\begin{bmatrix}0&0\\\
0&(K-z)^{-1}\end{bmatrix}$ (30) $\displaystyle+$
$\displaystyle\begin{bmatrix}1\\\
(K-z)^{-1}|h)\end{bmatrix}\frac{1}{E_{0}+\Sigma(z)-z}\begin{bmatrix}1&(h|(K-z)^{-1}\end{bmatrix}.$

We can apply the above formulas also if $\Psi$ does not belong to the domain
of $H$, but belongs to its form domain, so that $(\Psi|H\Psi)$ is well
defined. Note that $E_{0}$ and $\Sigma(z)$ are then uniquely defined by (25)
and (29)).

If $\Psi$ does not belong to the form domain of $H$, then strictly speaking
the self-energy is ill defined. In practice in such situations one often
introduces a cutoff Hamiltonian $H^{\Lambda}$, which in some sense
approximates $H$. Then, setting $h^{\Lambda}:=H^{\Lambda}\Psi$,
$E_{0}^{\Lambda}:=(\Psi|H^{\Lambda}\Psi)$, and denoting by $K^{\Lambda}$ the
operator $H^{\Lambda}$ compressed to $\mathcal{K}$, one can use the cutoff
version of the Feshbach-Schur formula:

$\displaystyle(\Psi|(H^{\Lambda}-z)^{-1}\Psi)$
$\displaystyle=\frac{1}{E_{0}^{\Lambda}+\Sigma^{\Lambda}(z)-z},$ (31)
$\displaystyle\Sigma^{\Lambda}(z)$
$\displaystyle=-(h^{\Lambda}|(K^{\Lambda}-z)^{-1}h^{\Lambda}).$ (32)

The resolvent of the original Hamiltonian $H$ can be retrieved [16] in the
limit $\Lambda\to\infty$:

$(H-z)^{-1}=\lim_{\Lambda\to\infty}(H^{\Lambda}-z)^{-1}.$ (33)

Note that $E_{0}^{\Lambda}$ is a sequence of real numbers, typically
converging to $\infty$. They can be treated as counterterms renormalizing the
self-energy $\Sigma^{\Lambda}(z)$.

##  3\. Fermi Golden Rule

The meaning of the self-energy is especially clear in perturbation theory.
Again, let $\Psi$ be a normalized vector in $\mathcal{H}$. Consider a family
of self-adjoint operators $H_{\lambda}=H_{0}+\lambda V$ such that
$H_{0}\Psi=E_{0}\Psi$ and $(\Psi|V\Psi)=0$. Let $h:=V\Psi$ and $K_{\lambda}$
be $H_{\lambda}$ compressed to $\mathcal{K}$. Thus we can rewrite (26) as

$H_{\lambda}=\begin{bmatrix}E_{0}&\lambda(h|\\\
\lambda|h)&K_{\lambda}\end{bmatrix}.$ (34)

We extract $\lambda^{2}$ from the definition of the self-energy, so that (27)
and (28) are rewritten as

$\displaystyle(\Psi|(H_{\lambda}-z)^{-1}\Psi)$
$\displaystyle=\big{(}E_{0}+\lambda^{2}\Sigma_{\lambda}(z)-z\big{)}^{-1},$
(35) $\displaystyle\Sigma_{\lambda}(z)$
$\displaystyle:=-(h|(K_{\lambda}-z)^{-1}h)=\Sigma_{0}(z)+O(\lambda).$ (36)

Now (35) has a pole at

$E_{0}+\lambda^{2}\Sigma_{0}(E_{0}+\mathrm{i}0)+O(\lambda^{3}).$ (37)

This is often formulated as the Fermi Golden Rule: the pole of the resolvent,
originally at an eigenvalue $E_{0}$, is shifted in the second order by
$\lambda^{2}\Sigma_{0}(E_{0}+\mathrm{i}0)$. This shift can have a negative
imaginary part, and then the eigenvalue disappears. A singularity of the
resolvent with a negative imaginary part is usually called a resonance.

Resonances describe metastable states. A rigorous meaning of a resonance is
provided by the following version of the weak coupling limit ([12], see also
[14, 15])

$\lim_{\lambda\to
0}\big{(}\Psi\big{|}\exp\big{(}-\mathrm{i}\tfrac{t}{\lambda^{2}}(H_{\lambda}-E_{0})\big{)}\big{|}\Psi\big{)}=\mathrm{e}^{-\mathrm{i}t\Sigma_{0}(E_{0}+i0)}.$
(38)

If the perturbation is singular, so that $\Psi$ does not belong to the domain
of $V$, then $\Sigma_{0}(z)$ is in general ill defined and (37) may lose its
meaning. Strictly speaking, one then needs to introduce a cutoff on the
perturbation and a counterterm, and only then to apply the appropriately
modified Fermi Golden Rule.

Note that it is enough to consider real counterterms. Therefore, if we know
that the renormalized energy is close to $E_{0}$, then we can still expect
that (37) gives a correct prediction for the imaginary part of the resonance.
In other words, the imaginary part of the singularity of the resolvent
$(H_{\lambda}-z)^{-1}$ is

$\lambda^{2}\mathrm{Im}\Sigma_{0}(E_{0}+\mathrm{i}0)+O(\lambda^{3}),$ (39)

where we do not need to cut off the perturbation.

In practice, we start from a singular expression of the form 34. To make it
well-defined we need to choose a cutoff and counterterms. These choices will
not affect the imaginary part of the resonance, however in principle, one can
add an arbitrary real constant to a counterterm, which will affect the real
part of the resonance. Therefore, for singular perturbations it may be more
difficult to predict the real part of the resonance.

##  4\. Bose gas and Bogoliubov ansatz

We consider a homogeneous Bose gas of $N$ particles with a two-body potential
described by a function $v:\mathbb{R}^{3}\to\mathbb{R}$ whose Fourier
transform
$\hat{v}({\mathbf{k}})=\int_{\mathbb{R}^{3}}v(x)\mathrm{e}^{-\mathrm{i}{\mathbf{k}}\cdot{\mathbf{x}}}\,{\rm
d}{\mathbf{x}}$ is non-negative and rotation invariant. In the grand canonical
setting and the momentum representation such a system is governed by the
(second quantized) Hamiltonian

$H=\int\left(\frac{{\mathbf{k}}^{2}}{2}-\mu\right)a_{\mathbf{k}}^{*}a_{\mathbf{k}}\,{\rm
d}{\mathbf{k}}+\frac{\kappa}{2(2\pi)^{3}}\int\,{\rm d}{\mathbf{p}}\int\,{\rm
d}{\mathbf{q}}\int\,{\rm
d}{\mathbf{k}}\hat{v}({\mathbf{k}})a_{{\mathbf{p}}-{\mathbf{k}}}^{*}a_{{\mathbf{q}}+{\mathbf{k}}}^{*}a_{{\mathbf{p}}}a_{{\mathbf{q}}},$
(40)

where $\mu\geq 0$ is the chemical potential and
$a_{\mathbf{k}}^{*}/a_{\mathbf{k}}$ the creation/annihilation operators for
particles of mode ${\mathbf{k}}$. It acts on the bosonic Fock space
$\mathcal{F}=\Gamma_{\mathrm{s}}\big{(}L^{2}(\mathbb{R}^{3})\big{)}$, and for
each $N$ it leaves invariant its $N$-particle sector
$L^{2}_{\mathrm{s}}\big{(}(\mathbb{R}^{3})^{N}\big{)}$. Recall that the
creation and annihilation operators satisfy the canonical commutation relation
(CCR):

$[a_{\mathbf{p}},a_{\mathbf{q}}]=0=[a_{\mathbf{p}}^{*},a_{\mathbf{q}}^{*}],\ \
[a_{\mathbf{p}},a^{*}_{\mathbf{q}}]=\delta({\mathbf{p}}-{\mathbf{q}}),$ (41)

where $[\ ,\ ]$ is the usual commutator. We introduce the coupling constant
$\kappa>0$ mostly for bookkeeping purposes; note that in the introduction we
set $\kappa=1$.

For the reasons explained in the introduction, we replace the infinite space
$\mathbb{R}^{3}$ by the torus $[-L/2,L/2]^{3}$ with periodic boundary
conditions. In the momentum representation the Hamiltonian becomes

$H=\sum_{{\mathbf{k}}\in
2\pi\mathbb{Z}^{3}/L}\left(\frac{{\mathbf{k}}^{2}}{2}-\mu\right)a_{\mathbf{k}}^{*}a_{\mathbf{k}}+\frac{\kappa}{2L^{3}}\sum_{{\mathbf{p}},{\mathbf{q}},{\mathbf{k}}\in
2\pi\mathbb{Z}^{3}/L}\hat{v}({\mathbf{k}})a_{{\mathbf{p}}-{\mathbf{k}}}^{*}a_{{\mathbf{q}}+{\mathbf{k}}}^{*}a_{{\mathbf{p}}}a_{{\mathbf{q}}}.$
(42)

Note that $\hat{v}$ is the same function as in (40), however it is now sampled
only on the lattice $2\pi\mathbb{Z}^{3}/L$. The commutation relation involve
now the Kronecker delta:

$[a_{\mathbf{p}},a_{\mathbf{q}}]=0=[a_{\mathbf{p}}^{*},a_{\mathbf{q}}^{*}],\ \
[a_{\mathbf{p}},a^{*}_{\mathbf{q}}]=\delta_{{\mathbf{p}},{\mathbf{q}}}.$ (43)

Let us now pass to the quasiparticle representation. To this end we follow the
well-known grand-canonical version of the Bogoliubov approach (see e.g. [11]).
It involves two unitary transformations.

The first one is a Weyl transformation that introduces a macroscopic
occupation of the zero-momentum mode, the Bose-Einstein condensate. (In the
canonical version Bogoliubov approach this corresponds to the c-number
substitution [27].) To this end, for $\alpha\in\mathbb{C}$, we introduce the
Weyl operator of the mode ${\mathbf{k}}=0$

$W_{\alpha}=\exp(-\alpha a_{0}^{*}+\bar{\alpha}a_{0}).$ (44)

Then

$W_{\alpha}^{*}a^{*}_{\mathbf{k}}W_{\alpha}=a^{*}_{\mathbf{k}}-\bar{\alpha}\delta_{{\mathbf{k}},0}=:\tilde{a}^{*}_{{\mathbf{k}}}.$

The new annihilation operators with tildes kill the “new vacuum”
$\Omega_{\alpha}=W^{*}_{\alpha}\Omega$. We express our Hamiltonian in terms of
$\tilde{a}_{\mathbf{k}}^{*},\tilde{a}_{\mathbf{k}}$. To simplify the notation,
in what follows we drop the tildes and we obtain

$\displaystyle H$ $\displaystyle=$
$\displaystyle-\mu|\alpha|^{2}+\frac{\kappa\hat{v}(0)}{2L^{3}}|\alpha|^{4}+\left(\frac{\kappa\hat{v}(0)}{L^{3}}|\alpha|^{2}-\mu\right)(\alpha
a_{0}^{*}+\bar{\alpha}a_{0})$ $\displaystyle+$
$\displaystyle\sum_{{\mathbf{k}}}\left(\frac{{\mathbf{k}}^{2}}{2}-\mu+\frac{\kappa(\hat{v}({\mathbf{k}})+\hat{v}(0))}{L^{3}}|\alpha|^{2}\right)a_{\mathbf{k}}^{*}a_{\mathbf{k}}+\sum_{{\mathbf{k}}}\frac{\kappa\hat{v}({\mathbf{k}})}{2L^{3}}\left(\alpha^{2}a_{\mathbf{k}}^{*}a_{-{\mathbf{k}}}^{*}+\bar{\alpha}^{2}a_{\mathbf{k}}a_{-{\mathbf{k}}}\right)$
$\displaystyle+$
$\displaystyle\frac{\kappa}{L^{3}}\sum_{{\mathbf{k}}_{1},{\mathbf{k}}_{2}}\hat{v}({\mathbf{k}}_{1})\left(\bar{\alpha}a_{{\mathbf{k}}_{1}+{\mathbf{k}}_{2}}^{*}a_{{\mathbf{k}}_{1}}a_{{\mathbf{k}}_{2}}+\alpha
a_{{\mathbf{k}}_{1}}^{*}a_{{\mathbf{k}}_{2}}^{*}a_{{\mathbf{k}}_{1}+{\mathbf{k}}_{2}}\right)$
$\displaystyle+$
$\displaystyle\frac{\kappa}{2L^{3}}\sum_{{\mathbf{k}}_{1},{\mathbf{k}}_{2},{\mathbf{k}}_{3},{\mathbf{k}}_{4}}\delta({\mathbf{k}}_{1}+{\mathbf{k}}_{2}-{\mathbf{k}}_{3}-{\mathbf{k}}_{4})\hat{v}({\mathbf{k}}_{2}-{\mathbf{k}}_{3})a_{{\mathbf{k}}_{1}}^{*}a_{{\mathbf{k}}_{2}}^{*}a_{{\mathbf{k}}_{3}}a_{{\mathbf{k}}_{4}}.$

Note that we have

$(\Omega_{\alpha}|H\Omega_{\alpha})=-\mu|\alpha|^{2}+\frac{\kappa\hat{v}(0)}{2L^{3}}|\alpha|^{4},$

and we choose $\alpha=\sqrt{\frac{\mu L^{3}}{\kappa\hat{v}(0)}}$, so that
$\Omega_{\alpha}$ minimizes this expectation value. This leads to

$\displaystyle H$
$\displaystyle=\kappa^{-1}H_{0}+H_{2}+\sqrt{\kappa}H_{3}+\kappa H_{4},$ (45)
$\displaystyle H_{0}$ $\displaystyle:=-\frac{\mu^{2}L^{3}}{2\hat{v}(0)},$
$\displaystyle H_{2}$
$\displaystyle:=\sum_{{\mathbf{k}}}\left(\frac{{\mathbf{k}}^{2}}{2}+\frac{\mu\hat{v}({\mathbf{k}})}{\hat{v}(0)}\right)a_{\mathbf{k}}^{*}a_{\mathbf{k}}+\sum_{{\mathbf{k}}}\frac{\mu\hat{v}({\mathbf{k}})}{2\hat{v}(0)}\left(a_{\mathbf{k}}^{*}a_{-{\mathbf{k}}}^{*}+a_{\mathbf{k}}a_{-{\mathbf{k}}}\right),$
$\displaystyle H_{3}$
$\displaystyle:=\frac{1}{L^{3/2}}\sum_{{\mathbf{k}}_{1},{\mathbf{k}}_{2}}\frac{\hat{v}({\mathbf{k}}_{1})\sqrt{\mu}}{\sqrt{\hat{v}(0)}}\left(a_{{\mathbf{k}}_{1}+{\mathbf{k}}_{2}}^{*}a_{{\mathbf{k}}_{1}}a_{{\mathbf{k}}_{2}}+a_{{\mathbf{k}}_{1}}^{*}a_{{\mathbf{k}}_{2}}^{*}a_{{\mathbf{k}}_{1}+{\mathbf{k}}_{2}}\right),$
$\displaystyle H_{4}$
$\displaystyle:=\frac{1}{2L^{3}}\sum_{{\mathbf{k}}_{1},{\mathbf{k}}_{2},{\mathbf{k}}_{3},{\mathbf{k}}_{4}}\delta({\mathbf{k}}_{1}+{\mathbf{k}}_{2}-{\mathbf{k}}_{3}-{\mathbf{k}}_{4})\hat{v}({\mathbf{k}}_{2}-{\mathbf{k}}_{3})a_{{\mathbf{k}}_{1}}^{*}a_{{\mathbf{k}}_{2}}^{*}a_{{\mathbf{k}}_{3}}a_{{\mathbf{k}}_{4}}.$

We extract from the above Hamiltonian all terms containing only non-zero
modes:

$\displaystyle H_{2}$
$\displaystyle=\frac{\mu}{2}(a_{0}^{*2}+a_{0}^{2}+2a_{0}^{*}a_{0})+H_{2}^{\rm
exc},$ $\displaystyle H_{2}^{\rm exc}$ $\displaystyle:=\sum_{{\mathbf{k}}\neq
0}\left(\frac{{\mathbf{k}}^{2}}{2}+\frac{\mu\hat{v}({\mathbf{k}})}{\hat{v}(0)}\right)a_{\mathbf{k}}^{*}a_{\mathbf{k}}+\sum_{{\mathbf{k}}\neq
0}\frac{\mu\hat{v}({\mathbf{k}})}{2\hat{v}(0)}\left(a_{\mathbf{k}}^{*}a_{-{\mathbf{k}}}^{*}+a_{\mathbf{k}}a_{-{\mathbf{k}}}\right);$
(46) $\displaystyle H_{3}$
$\displaystyle=\frac{1}{L^{3/2}}\sum_{{\mathbf{k}}}\sqrt{\mu\hat{v}(0)}(a_{0}^{*}+a_{0})a_{\mathbf{k}}^{*}a_{\mathbf{k}}$
$\displaystyle+\frac{1}{L^{3/2}}\sum_{{\mathbf{k}}\neq
0}\frac{\sqrt{\mu}\hat{v}({\mathbf{k}})}{\sqrt{\hat{v}(0)}}\big{(}(a_{0}^{*}+a_{0})a_{\mathbf{k}}^{*}a_{\mathbf{k}}+a_{0}a_{\mathbf{k}}^{*}a_{-{\mathbf{k}}}^{*}+a_{0}^{*}a_{\mathbf{k}}a_{-{\mathbf{k}}}\big{)}+H_{3}^{\rm
exc},$ $\displaystyle H_{3}^{\rm exc}$
$\displaystyle:=\frac{1}{L^{3/2}}\sum_{{\mathbf{k}}_{1},{\mathbf{k}}_{2},{\mathbf{k}}_{1}+{\mathbf{k}}_{2}\neq
0}\frac{\hat{v}({\mathbf{k}}_{1})\sqrt{\mu}}{\sqrt{\hat{v}(0)}}\left(a_{{\mathbf{k}}_{1}+{\mathbf{k}}_{2}}^{*}a_{{\mathbf{k}}_{1}}a_{{\mathbf{k}}_{2}}+a_{{\mathbf{k}}_{1}}^{*}a_{{\mathbf{k}}_{2}}^{*}a_{{\mathbf{k}}_{1}+{\mathbf{k}}_{2}}\right);$
(47) $\displaystyle H_{4}$
$\displaystyle=\frac{1}{2L^{3}}\hat{v}(0)\Big{(}a_{0}^{*}a_{0}^{*}a_{0}a_{0}+2\sum_{{\mathbf{k}}\neq
0}a_{0}^{*}a_{0}a_{\mathbf{k}}^{*}a_{\mathbf{k}}\Big{)}$
$\displaystyle+\frac{1}{2L^{3}}\sum_{{\mathbf{k}}\neq
0}\hat{v}({\mathbf{k}})(a_{0}^{*}a_{0}^{*}a_{\mathbf{k}}a_{-{\mathbf{k}}}+a_{0}a_{0}a_{\mathbf{k}}^{*}a_{-{\mathbf{k}}}^{*}+2a_{0}^{*}a_{0}a_{\mathbf{k}}^{*}a_{\mathbf{k}})$
$\displaystyle+\frac{1}{L^{3}}\sum_{{\mathbf{k}}_{1},{\mathbf{k}}_{2},{\mathbf{k}}_{1}+{\mathbf{k}}_{2}\neq
0}\hat{v}(k_{1})\big{(}a_{0}^{*}a_{{\mathbf{k}}_{1}+{\mathbf{k}}_{2}}^{*}a_{{\mathbf{k}}_{1}}a_{{\mathbf{k}}_{2}}+a_{0}a_{{\mathbf{k}}_{1}}^{*}a_{{\mathbf{k}}_{2}}^{*}a_{{\mathbf{k}}_{1}+{\mathbf{k}}_{2}}\big{)}+H_{4}^{\rm
exc},$ $\displaystyle H_{4}^{\rm exc}$
$\displaystyle:=\frac{1}{2L^{3}}\sum_{{\mathbf{k}}_{1},{\mathbf{k}}_{2},{\mathbf{k}}_{3},{\mathbf{k}}_{4}\neq
0}\delta({\mathbf{k}}_{1}+{\mathbf{k}}_{2}-{\mathbf{k}}_{3}-{\mathbf{k}}_{4})\hat{v}({\mathbf{k}}_{2}-{\mathbf{k}}_{3})a_{{\mathbf{k}}_{1}}^{*}a_{{\mathbf{k}}_{2}}^{*}a_{{\mathbf{k}}_{3}}a_{{\mathbf{k}}_{4}}.$
(48)

We are going to apply a Bogoliubov transformation

$U_{\mathrm{Bog}}:=\exp\Bigg{(}\sum_{{\mathbf{k}}\neq
0}\beta_{\mathbf{k}}(a_{\mathbf{k}}^{*}a_{-{\mathbf{k}}}^{*}-a_{\mathbf{k}}a_{-{\mathbf{k}}})\Bigg{)},$
(49)

which transforms non-zero mode operators $a_{\mathbf{k}}^{*},a_{\mathbf{k}}$
into quasi-particle operators $b_{\mathbf{k}}^{*},b_{\mathbf{k}}$:

$\displaystyle b_{\mathbf{k}}$ $\displaystyle:=$ $\displaystyle
U_{\mathrm{Bog}}a_{\mathbf{k}}U_{\mathrm{Bog}}^{*}=\sigma_{\mathbf{k}}a_{\mathbf{k}}+\gamma_{\mathbf{k}}a_{-{\mathbf{k}}}^{*},$
$\displaystyle b_{\mathbf{k}}^{*}$ $\displaystyle:=$ $\displaystyle
U_{\mathrm{Bog}}a_{\mathbf{k}}^{*}U_{\mathrm{Bog}}^{*}=\sigma_{\mathbf{k}}a_{\mathbf{k}}^{*}+\gamma_{\mathbf{k}}a_{-{\mathbf{k}}},$
(50)

where

$\displaystyle\sigma_{\mathbf{k}}$
$\displaystyle=\frac{\sqrt{\sqrt{e_{\mathbf{k}}^{2}+B_{\mathbf{k}}^{2}}+e_{\mathbf{k}}}}{\sqrt{2e_{\mathbf{k}}}},\quad\gamma_{\mathbf{k}}=\frac{\sqrt{\sqrt{e_{\mathbf{k}}^{2}+B_{\mathbf{k}}^{2}}-e_{\mathbf{k}}}}{\sqrt{2e_{\mathbf{k}}}},$
$\displaystyle e_{\mathbf{k}}$
$\displaystyle:=\sqrt{\frac{1}{4}|{\mathbf{k}}|^{4}+B_{\mathbf{k}}|{\mathbf{k}}|^{2}},\
\ \ B_{\mathbf{k}}:=\frac{\hat{v}({\mathbf{k}})}{\hat{v}(0)}\mu.$

The inverse relation is

$\displaystyle a_{\mathbf{k}}$ $\displaystyle=$
$\displaystyle\sigma_{\mathbf{k}}b_{\mathbf{k}}-\gamma_{\mathbf{k}}b_{-{\mathbf{k}}}^{*},$
$\displaystyle a_{\mathbf{k}}^{*}$ $\displaystyle=$
$\displaystyle\sigma_{\mathbf{k}}b_{\mathbf{k}}^{*}-\gamma_{\mathbf{k}}b_{-{\mathbf{k}}}.$

It is well known that (50) diagonalizes $H_{2}^{\rm exc}$ in terms of the
quasi-particle operators:

$\displaystyle H_{2}^{{\rm exc}}$ $\displaystyle=$ $\displaystyle E_{\rm
Bog}+H_{\mathrm{Bog}},$ (51)

where

$\displaystyle E_{\rm Bog}$ $\displaystyle:=-\frac{1}{2}\sum_{{\mathbf{k}}\neq
0}\left(\frac{1}{2}|{\mathbf{k}}|^{2}+\frac{\hat{v}({\mathbf{k}})}{\hat{v}(0)}\mu-
e_{\mathbf{k}}\right),$ (52) $\displaystyle\qquad H_{\mathrm{Bog}}$
$\displaystyle:=\sum_{{\mathbf{k}}\neq
0}e_{\mathbf{k}}b_{\mathbf{k}}^{*}b_{\mathbf{k}}.$ (53)

We also express $H_{3}^{\rm exc}$ in terms of quasiparticles:

$\displaystyle H_{3}^{\rm exc}$
$\displaystyle=\frac{1}{L^{3/2}}\sum_{{\mathbf{k}}_{1},{\mathbf{k}}_{2},{\mathbf{k}}_{1}+{\mathbf{k}}_{2}\neq
0}\frac{\sqrt{\mu}\hat{v}({\mathbf{k}}_{1})}{\sqrt{\hat{v}(0)}}(A({\mathbf{k}}_{1},{\mathbf{k}}_{2})+A^{*}({\mathbf{k}}_{1},{\mathbf{k}}_{2}),$
$\displaystyle A({\mathbf{k}}_{1},{\mathbf{k}}_{2})=$
$\displaystyle\quad\sigma_{{\mathbf{k}}_{1}}\sigma_{{\mathbf{k}}_{2}}\sigma_{{\mathbf{k}}_{1}+{\mathbf{k}}_{2}}b_{{\mathbf{k}}_{1}}^{*}b_{{\mathbf{k}}_{2}}^{*}b_{{\mathbf{k}}_{1}+{\mathbf{k}}_{2}}-\gamma_{{\mathbf{k}}_{1}}\sigma_{{\mathbf{k}}_{2}}\sigma_{{\mathbf{k}}_{1}+{\mathbf{k}}_{2}}b_{{\mathbf{k}}_{2}}^{*}b_{-{\mathbf{k}}_{1}}b_{{\mathbf{k}}_{1}+{\mathbf{k}}_{2}}$
$\displaystyle-\gamma_{{\mathbf{k}}_{1}}\gamma_{{\mathbf{k}}_{2}}\sigma_{{\mathbf{k}}_{1}+{\mathbf{k}}_{2}}b_{{\mathbf{k}}_{1}}^{*}b_{-{\mathbf{k}}_{2}}b_{{\mathbf{k}}_{1}+{\mathbf{k}}_{2}}+\gamma_{{\mathbf{k}}_{1}}\gamma_{{\mathbf{k}}_{2}}\sigma_{{\mathbf{k}}_{1}+{\mathbf{k}}_{2}}b_{-{\mathbf{k}}_{1}}b_{-{\mathbf{k}}_{2}}b_{{\mathbf{k}}_{1}+{\mathbf{k}}_{2}}$
$\displaystyle-\sigma_{{\mathbf{k}}_{1}}\sigma_{{\mathbf{k}}_{2}}\gamma_{{\mathbf{k}}_{1}+{\mathbf{k}}_{2}}b_{{\mathbf{k}}_{1}}^{*}b_{{\mathbf{k}}_{2}}^{*}b_{-{\mathbf{k}}_{1}-{\mathbf{k}}_{2}}^{*}-\gamma_{{\mathbf{k}}_{1}}\sigma_{{\mathbf{k}}_{2}}\gamma_{{\mathbf{k}}_{1}+{\mathbf{k}}_{2}}b_{{\mathbf{k}}_{2}}^{*}b_{-{\mathbf{k}}_{1}-{\mathbf{k}}_{2}}^{*}b_{-{\mathbf{k}}_{1}}$
$\displaystyle+\sigma_{{\mathbf{k}}_{1}}\gamma_{{\mathbf{k}}_{2}}\gamma_{{\mathbf{k}}_{1}+{\mathbf{k}}_{2}}b_{{\mathbf{k}}_{1}}^{*}b_{-{\mathbf{k}}_{1}-{\mathbf{k}}_{2}}^{*}b_{-{\mathbf{k}}_{2}}-\gamma_{{\mathbf{k}}_{1}}\gamma_{{\mathbf{k}}_{2}}\gamma_{{\mathbf{k}}_{1}+{\mathbf{k}}_{2}}b_{-{\mathbf{k}}_{1}-{\mathbf{k}}_{2}}^{*}b_{-{\mathbf{k}}_{1}}b_{-{\mathbf{k}}_{2}}.$

Thus

$\displaystyle H_{3}^{\rm exc}$ $\displaystyle=H_{3,1}^{\rm exc}+H_{3,2}^{\rm
exc},$ (54) $\displaystyle H_{3,1}^{\rm
exc}=\sum_{{\mathbf{k}}_{1},{\mathbf{k}}_{2},{\mathbf{k}}_{1}+{\mathbf{k}}_{2}\neq
0}\frac{\sqrt{\mu}\hat{v}({\mathbf{k}}_{1})}{L^{3/2}\sqrt{\hat{v}(0)}}$
$\displaystyle\Big{(}\sigma_{{\mathbf{k}}_{1}}\sigma_{{\mathbf{k}}_{2}}\sigma_{{\mathbf{k}}_{1}+{\mathbf{k}}_{2}}b_{{\mathbf{k}}_{1}}^{*}b_{{\mathbf{k}}_{2}}^{*}b_{{\mathbf{k}}_{1}+{\mathbf{k}}_{2}}-\gamma_{{\mathbf{k}}_{1}}\sigma_{{\mathbf{k}}_{2}}\sigma_{{\mathbf{k}}_{1}+{\mathbf{k}}_{2}}b_{{\mathbf{k}}_{2}}^{*}b_{-{\mathbf{k}}_{1}}b_{{\mathbf{k}}_{1}+{\mathbf{k}}_{2}}$
$\displaystyle-$
$\displaystyle\gamma_{{\mathbf{k}}_{1}}\gamma_{{\mathbf{k}}_{2}}\sigma_{{\mathbf{k}}_{1}+{\mathbf{k}}_{2}}b_{{\mathbf{k}}_{1}}^{*}b_{-{\mathbf{k}}_{2}}b_{{\mathbf{k}}_{1}+{\mathbf{k}}_{2}}-\gamma_{{\mathbf{k}}_{1}}\sigma_{{\mathbf{k}}_{2}}\gamma_{{\mathbf{k}}_{1}+{\mathbf{k}}_{2}}b_{{\mathbf{k}}_{2}}^{*}b_{-{\mathbf{k}}_{1}-{\mathbf{k}}_{2}}^{*}b_{-{\mathbf{k}}_{1}}$
$\displaystyle+$
$\displaystyle\sigma_{{\mathbf{k}}_{1}}\gamma_{{\mathbf{k}}_{2}}\gamma_{{\mathbf{k}}_{1}+{\mathbf{k}}_{2}}b_{{\mathbf{k}}_{1}}^{*}b_{-{\mathbf{k}}_{1}-{\mathbf{k}}_{2}}^{*}b_{-{\mathbf{k}}_{2}}-\gamma_{{\mathbf{k}}_{1}}\gamma_{{\mathbf{k}}_{2}}\gamma_{{\mathbf{k}}_{1}+{\mathbf{k}}_{2}}b_{-{\mathbf{k}}_{1}-{\mathbf{k}}_{2}}^{*}b_{-{\mathbf{k}}_{1}}b_{-{\mathbf{k}}_{2}}$
$\displaystyle+$
$\displaystyle\sigma_{{\mathbf{k}}_{1}}\sigma_{{\mathbf{k}}_{2}}\sigma_{{\mathbf{k}}_{1}+{\mathbf{k}}_{2}}b_{{\mathbf{k}}_{1}+{\mathbf{k}}_{2}}^{*}b_{{\mathbf{k}}_{1}}b_{{\mathbf{k}}_{2}}-\gamma_{{\mathbf{k}}_{1}}\sigma_{{\mathbf{k}}_{2}}\sigma_{{\mathbf{k}}_{1}+{\mathbf{k}}_{2}}b_{-{\mathbf{k}}_{1}}^{*}b_{{\mathbf{k}}_{1}+{\mathbf{k}}_{2}}^{*}b_{{\mathbf{k}}_{2}}$
$\displaystyle-$
$\displaystyle\gamma_{{\mathbf{k}}_{1}}\gamma_{{\mathbf{k}}_{2}}\sigma_{{\mathbf{k}}_{1}+{\mathbf{k}}_{2}}b_{-{\mathbf{k}}_{2}}^{*}b_{{\mathbf{k}}_{1}+{\mathbf{k}}_{2}}^{*}b_{{\mathbf{k}}_{1}}-\gamma_{{\mathbf{k}}_{1}}\sigma_{{\mathbf{k}}_{2}}\gamma_{{\mathbf{k}}_{1}+{\mathbf{k}}_{2}}b_{-{\mathbf{k}}_{1}}^{*}b_{{\mathbf{k}}_{2}}b_{-{\mathbf{k}}_{1}-{\mathbf{k}}_{2}}$
$\displaystyle+$
$\displaystyle\sigma_{{\mathbf{k}}_{1}}\gamma_{{\mathbf{k}}_{2}}\gamma_{{\mathbf{k}}_{1}+{\mathbf{k}}_{2}}b_{-{\mathbf{k}}_{2}}^{*}b_{{\mathbf{k}}_{1}}b_{-{\mathbf{k}}_{1}-{\mathbf{k}}_{2}}-\gamma_{{\mathbf{k}}_{1}}\gamma_{{\mathbf{k}}_{2}}\gamma_{{\mathbf{k}}_{1}+{\mathbf{k}}_{2}}b_{-{\mathbf{k}}_{1}}^{*}b_{-{\mathbf{k}}_{2}}^{*}b_{-{\mathbf{k}}_{1}-{\mathbf{k}}_{2}}\Big{)},$
$\displaystyle H_{3,2}^{\rm
exc}=\sum_{{\mathbf{k}}_{1},{\mathbf{k}}_{2},{\mathbf{k}}_{1}+{\mathbf{k}}_{2}\neq
0}\frac{\sqrt{\mu}\hat{v}({\mathbf{k}}_{1})}{L^{3/2}\sqrt{\hat{v}(0)}}$
$\displaystyle\Big{(}\gamma_{{\mathbf{k}}_{1}}\gamma_{{\mathbf{k}}_{2}}\sigma_{{\mathbf{k}}_{1}+{\mathbf{k}}_{2}}b_{-{\mathbf{k}}_{1}}b_{-{\mathbf{k}}_{2}}b_{{\mathbf{k}}_{1}+{\mathbf{k}}_{2}}-\sigma_{{\mathbf{k}}_{1}}\sigma_{{\mathbf{k}}_{2}}\gamma_{{\mathbf{k}}_{1}+{\mathbf{k}}_{2}}b_{{\mathbf{k}}_{1}}^{*}b_{{\mathbf{k}}_{2}}^{*}b_{-{\mathbf{k}}_{1}-{\mathbf{k}}_{2}}^{*}$
$\displaystyle+$
$\displaystyle\gamma_{{\mathbf{k}}_{1}}\gamma_{{\mathbf{k}}_{2}}\sigma_{{\mathbf{k}}_{1}+{\mathbf{k}}_{2}}b_{-{\mathbf{k}}_{1}}^{*}b_{-{\mathbf{k}}_{2}}^{*}b_{{\mathbf{k}}_{1}+{\mathbf{k}}_{2}}^{*}-\sigma_{{\mathbf{k}}_{1}}\sigma_{{\mathbf{k}}_{2}}\gamma_{{\mathbf{k}}_{1}+{\mathbf{k}}_{2}}b_{{\mathbf{k}}_{1}}b_{{\mathbf{k}}_{2}}b_{-{\mathbf{k}}_{1}-{\mathbf{k}}_{2}}\Big{)}.$

We could also compute $H_{4}$, but we will not need it.

##  5\. Effective Friedrichs Hamiltonian

Let $\Omega_{\mathrm{Bog}}:=U_{\mathrm{Bog}}^{*}\Omega_{\alpha}$ be the
quasiparticle vacuum. Introduce the space $\mathcal{F}^{\mathrm{exc}}$
consisting of the Bogoliubov vacuum and quasiparticle excitations, and its
$n$-quasiparticle sector:

$\displaystyle\mathcal{F}^{\mathrm{exc}}:=$
$\displaystyle\operatorname{Span}^{\mathrm{cl}}\\{b_{{\mathbf{k}}_{1}}^{*}\cdots
b_{{\mathbf{k}}_{n}}^{*}\Omega_{\mathrm{Bog}}\ |\
{\mathbf{k}}_{1},\dots,{\mathbf{k}}_{n}\neq 0,\quad n=0,1,\dots\\},$
$\displaystyle\mathcal{F}_{n}^{\mathrm{exc}}:=$
$\displaystyle\operatorname{Span}^{\mathrm{cl}}\\{b_{{\mathbf{k}}_{1}}^{*}\cdots
b_{{\mathbf{k}}_{n}}^{*}\Omega_{\mathrm{Bog}}\ |\
{\mathbf{k}}_{1},\dots,{\mathbf{k}}_{n}\neq 0\\}.$

The most “violent” approximation that we are going to make is compressing the
Hamiltonian $H$ into the space $\mathcal{F}^{\mathrm{exc}}$. We also drop the
uninteresting constant $\kappa^{-1}H_{0}$ and the (somewhat more interesting)
constant $E_{\mathrm{Bog}}$. Thus we introduce the excitation Hamiltonian

$H^{\mathrm{exc}}:=I^{\mathrm{exc}*}\big{(}H-\kappa^{-1}H_{0}-E_{\mathrm{Bog}}\big{)}I^{\mathrm{exc}},$

where $I^{\mathrm{exc}}$ denotes the embedding of $\mathcal{F}^{\mathrm{exc}}$
in $\mathcal{F}$. Thus $H^{\mathrm{exc}}$ is an operator on
$\mathcal{F}^{\mathrm{exc}}$ and

$H^{\mathrm{exc}}=H_{\mathrm{Bog}}+\sqrt{\kappa}H_{3}^{\mathrm{exc}}+\kappa
H_{4}^{\mathrm{exc}},$ (55)

where $H_{3}^{\mathrm{exc}}$ and $H_{4}^{\mathrm{exc}}$ are defined in (47)
and 48.

We make two more approximations. We drop $\kappa H_{4}$, which is of higher
order in $\kappa$ than $\sqrt{\kappa}H_{3}$. We also drop $H_{3,2}$, which
involves $3$-quasiparticle creation/annihilation operators, and does not
contribute to the damping rate (see Section 6 for a justification). Thus
$H^{\mathrm{exc}}$ is replaced with

$H^{\mathrm{eff}}:=H_{\mathrm{Bog}}+\sqrt{\kappa}H_{3,1}^{\mathrm{exc}}.$ (56)

To make our following discussion consistent with Sect. 3 about the Fermi
Golden Rule, we introduce a new coupling constant

$\lambda:=\sqrt{\kappa}.$ (57)

Let ${\mathbf{k}}\neq 0$. Clearly, $b_{\mathbf{k}}^{*}\Omega_{\mathrm{Bog}}$
is an eigenstate of $H^{\mathrm{eff}}$ for $\lambda=0$. We would like to
compute the self-energy for the vector
$b_{\mathbf{k}}^{*}\Omega_{\mathrm{Bog}}$ and the Hamiltonian
$H^{\mathrm{eff}}$:

$\lambda^{2}\Sigma^{\mathrm{eff}}_{\mathbf{k}}(z):=\frac{-1}{(b_{\mathbf{k}}^{*}\Omega_{\mathrm{Bog}}|(z-H^{\rm
eff})^{-1}b_{\mathbf{k}}^{*}\Omega_{\mathrm{Bog}})}+z-e_{\mathbf{k}}.$ (58)

Introduce the subspaces of $\mathcal{F}^{\mathrm{exc}}$ and
$\mathcal{F}_{n}^{\mathrm{exc}}$ with the total momentum ${\mathbf{k}}$:

$\displaystyle\mathcal{F}^{\mathrm{exc}}({\mathbf{k}})$
$\displaystyle:=\operatorname{Span}^{\mathrm{cl}}\\{b_{{\mathbf{k}}_{1}}^{*}\cdots
b_{{\mathbf{k}}_{n}}^{*}\Omega_{\mathrm{Bog}},\qquad{\mathbf{k}}_{1}+\cdots{\mathbf{k}}_{n}={\mathbf{k}},\
{\mathbf{k}}_{1},\dots,{\mathbf{k}}_{n}\neq 0,\quad n=0,1,\dots\\},$
$\displaystyle\mathcal{F}_{n}^{\mathrm{exc}}({\mathbf{k}})$
$\displaystyle:=\operatorname{Span}^{\mathrm{cl}}\\{b_{{\mathbf{k}}_{1}}^{*}\cdots
b_{{\mathbf{k}}_{n}}^{*}\Omega_{\mathrm{Bog}},\qquad{\mathbf{k}}_{1}+\cdots{\mathbf{k}}_{n}={\mathbf{k}},\
{\mathbf{k}}_{1},\dots,{\mathbf{k}}_{n}\neq 0\\}.$

$b_{\mathbf{k}}^{*}\Omega_{\mathrm{Bog}}$ is contained in the space
$\mathcal{F}^{\mathrm{exc}}({\mathbf{k}})$, which is preserved by
$H^{\mathrm{eff}}$. Let $H^{\mathrm{eff}}({\mathbf{k}})$ denote the operator
$H^{\mathrm{eff}}$ restricted to $\mathcal{F}^{\mathrm{exc}}({\mathbf{k}})$.
Thus we can restrict ourselves to the fiber space
$\mathcal{F}^{\mathrm{exc}}({\mathbf{k}})$ and the fiber Hamiltonian
$H^{\mathrm{eff}}({\mathbf{k}})$. In particular, in (58) we can replace
$H^{\mathrm{eff}}$ with $H^{\mathrm{eff}}({\mathbf{k}})$.

For our analysis it is enough to know only $H^{\mathrm{eff}}$ (or
$H^{\mathrm{eff}}({\mathbf{k}})$) compressed to
$\mathcal{F}_{1}^{\mathrm{exc}}({\mathbf{k}})\oplus\mathcal{F}_{2}^{\rm
exc}({\mathbf{k}})$. Note that the one-quasiparticle state
$b_{\mathbf{k}}^{*}|\Omega_{\mathrm{Bog}}\rangle$ spans
$\mathcal{F}_{1}^{\mathrm{exc}}({\mathbf{k}})$, and $\mathcal{F}_{2}^{\rm
exc}({\mathbf{k}})$ is spanned by
$b_{\mathbf{p}}^{*}b_{{\mathbf{k}}-{\mathbf{p}}}^{*}\Omega_{\mathrm{Bog}}$
with ${\mathbf{p}},{\mathbf{k}}-{\mathbf{p}}\neq 0.$ We compute

$\displaystyle(b_{\mathbf{k}}^{*}\Omega_{\mathrm{Bog}}|H^{\rm
eff}b_{\mathbf{k}}^{*}\Omega_{\mathrm{Bog}})$ $\displaystyle=e_{\mathbf{k}},$
$\displaystyle(b_{\mathbf{p}}^{*}b_{{\mathbf{k}}-{\mathbf{p}}}^{*}\Omega_{\mathrm{Bog}}|H^{\rm
eff}b_{\mathbf{p}}^{*}b_{{\mathbf{k}}-{\mathbf{p}}}^{*}\Omega_{\mathrm{Bog}})$
$\displaystyle=e_{\mathbf{p}}+e_{{\mathbf{k}}-{\mathbf{p}}},$
$\displaystyle(b_{\mathbf{p}}^{*}b_{{\mathbf{k}}-{\mathbf{p}}}^{*}\Omega_{\mathrm{Bog}}|H^{\rm
eff}b_{\mathbf{k}}^{*}\Omega_{\mathrm{Bog}})$
$\displaystyle=\frac{\lambda}{L^{3/2}}h_{\mathbf{k}}({\mathbf{p}}),$ (59)
$\displaystyle(b_{\mathbf{k}}^{*}\Omega_{\mathrm{Bog}}|H^{\rm
eff}b_{\mathbf{p}}^{*}b_{{\mathbf{k}}-{\mathbf{p}}}^{*}\Omega_{\mathrm{Bog}})$
$\displaystyle=\frac{\lambda}{L^{3/2}}h_{\mathbf{k}}({\mathbf{p}})$ (60)

with

$\displaystyle h_{\mathbf{k}}({\mathbf{p}})$
$\displaystyle=2\sqrt{\frac{\mu\hat{v}^{2}({\mathbf{k}})}{\hat{v}(0)}}\Big{(}\sigma_{\mathbf{p}}\gamma_{-{\mathbf{k}}}\gamma_{{\mathbf{p}}-{\mathbf{k}}}+\sigma_{{\mathbf{k}}-{\mathbf{p}}}\gamma_{-{\mathbf{k}}}\gamma_{\mathbf{p}}+\sigma_{\mathbf{p}}\sigma_{{\mathbf{k}}-{\mathbf{p}}}\sigma_{\mathbf{k}}$
(61)
$\displaystyle\qquad-\gamma_{\mathbf{p}}\sigma_{-{\mathbf{k}}}\sigma_{{\mathbf{p}}-{\mathbf{k}}}-\gamma_{{\mathbf{k}}-{\mathbf{p}}}\sigma_{-{\mathbf{k}}}\sigma_{\mathbf{p}}-\gamma_{\mathbf{p}}\gamma_{{\mathbf{k}}-{\mathbf{p}}}\gamma_{\mathbf{k}}\Big{)}.$

The Hamiltonian $H^{\mathrm{eff}}$ compressed to
$\mathcal{F}_{1}^{\mathrm{exc}}({\mathbf{k}})\oplus\mathcal{F}_{2}^{\rm
exc}({\mathbf{k}})$ will be called the effective Friedrichs Hamiltonian (for
volume $L^{3}$ and momentum ${\mathbf{k}}$). It is denoted
$H_{\rm{Fried}}^{L}({\mathbf{k}})$ and given by

$\displaystyle H_{\rm{Fried}}^{L}({\mathbf{k}})$
$\displaystyle:=\begin{bmatrix}e_{\mathbf{k}}&\frac{\lambda}{L^{3/2}}(h_{\bf
k}|\\\ \frac{\lambda}{L^{3/2}}|h_{\bf
k})&e_{\mathbf{p}}+e_{{\mathbf{k}}-{\mathbf{p}}}\end{bmatrix},$ (62)
$\displaystyle\text{on
}\quad\mathcal{F}_{1}^{\mathrm{exc}}({\mathbf{k}})\oplus\mathcal{F}_{2}^{\rm
exc}({\mathbf{k}})$ $\displaystyle\simeq\mathbb{C}\oplus
l^{2}\Big{(}\frac{2\pi}{L}\mathbb{Z}^{3}\setminus\\{0,{\mathbf{k}}\\}\Big{)},$
(63)

where we explicitly introduced a reference to the volume $L^{3}$ in the
notation. Thus we end up in a situation described in Section 3. According to
the Fermi Golden Rule (37) we want to compute

$\displaystyle\Sigma_{\mathbf{k}}^{L}(z)$
$\displaystyle=\frac{1}{L^{3}}\sum_{{\mathbf{p}},{\mathbf{k}}-{\mathbf{p}}\neq
0}\frac{h_{\bf
k}^{2}({\mathbf{p}})}{(z-e_{\mathbf{p}}-e_{{\mathbf{k}}-{\mathbf{p}}})},$ (64)

Unfortunately, the sum (64) is divergent. To cure the divergence we can
introduce a cut-off. The cut-off is to a large extent arbitrary. It is
convenient to use $|{\mathbf{p}}|+|{\mathbf{k}}-{\mathbf{p}}|<\Lambda$. Thus
we replace (62), (61) and (64) with

$\displaystyle H_{\rm{Fried}}^{L,\Lambda}({\mathbf{k}})$
$\displaystyle:=\begin{bmatrix}e_{\mathbf{k}}&\frac{\lambda}{L^{3/2}}(h_{\bf
k}^{\Lambda}|\\\ \frac{\lambda}{L^{3/2}}|h_{\bf
k}^{\Lambda})&e_{\mathbf{p}}+e_{{\mathbf{k}}-{\mathbf{p}}}\end{bmatrix},$ (65)
$\displaystyle h_{\mathbf{k}}^{\Lambda}({\mathbf{p}})$
$\displaystyle:=h({\mathbf{p}}){\mathds{1}}_{\\{|{\mathbf{p}}|+|{\mathbf{k}}-{\mathbf{p}}|<\Lambda\\}}({\mathbf{p}}),$
(66) $\displaystyle\Sigma_{\mathbf{k}}^{L,\Lambda}(z)$
$\displaystyle:=\frac{1}{L^{3}}\sum_{{\mathbf{p}},{\mathbf{k}}-{\mathbf{p}}\neq
0}\frac{h_{\mathbf{k}}^{\Lambda}({\mathbf{p}})^{2}}{(z-e_{\mathbf{p}}-e_{{\mathbf{k}}-{\mathbf{p}}})}.$
(67)

The functions ${\mathbf{p}}\mapsto
e_{\mathbf{p}},h_{\mathbf{k}}({\mathbf{p}}),h_{\mathbf{k}}^{\Lambda}({\mathbf{p}})$
are well defined for all ${\mathbf{p}}\in\mathbb{R}^{3}\setminus\\{0\\}$, and
not only for $\frac{2\pi}{L}\mathbb{Z}^{3}\setminus\\{0,{\mathbf{k}}\\}$. The
expression (67) can be interpreted as the Riemann sum converging as
$L\to\infty$ to the integral

$\displaystyle\Sigma_{\mathbf{k}}^{\Lambda}(z)$
$\displaystyle=\frac{1}{(2\pi)^{3}}\int\frac{h_{\mathbf{k}}^{\Lambda}({\mathbf{p}})^{2}\,{\rm
d}{\mathbf{p}}}{(z-e_{\mathbf{p}}-e_{{\mathbf{k}}-{\mathbf{p}}})}.$ (68)

We can also introduce the infinite volume effective Friedrichs Hamiltonian

$\displaystyle H_{\rm{Fried}}^{\Lambda}({\mathbf{k}})$
$\displaystyle:=\begin{bmatrix}e_{\mathbf{k}}&\lambda(h_{\mathbf{k}}^{\Lambda}|\\\
\lambda|h_{\mathbf{k}}^{\Lambda})&e_{\mathbf{p}}+e_{{\mathbf{k}}-{\mathbf{p}}}\end{bmatrix},$
(69) on $\displaystyle\mathbb{C}\oplus L^{2}(\mathbb{R}^{3}),$

The Fermi Golden Rule predicts that
$\Sigma_{\mathbf{k}}^{\Lambda}(e_{\mathbf{k}}+i0)$ describes the energy shift
of the eigenvalue of the infinite volume cut-off Hamiltonian
$H_{\rm{Fried}}^{\Lambda}({\mathbf{k}})$. Unfortunately, in our case
$\lim\limits_{\Lambda\to\infty}\mathrm{Re}\Sigma_{\mathbf{k}}^{\Lambda}(e_{\mathbf{k}}+\mathrm{i}0)$
is infinite. However, we will see that
$\mathrm{Im}\Sigma_{\mathbf{k}}^{\Lambda}(e_{\mathbf{k}}+\mathrm{i}0)$ is
finite and for large $\Lambda$ is independent of $\Lambda$. Physically it
describes the decay of the quasiparticle at momentum ${\mathbf{k}}$.

##  6\. The shape of the quasiparticle spectrum

If ${\mathbf{k}}\mapsto e_{\mathbf{k}}$ is a dispersion relation of
quasiparticles, then the infimum of the $n$-quasiparticle spectrum is

$\inf\\{e_{{\mathbf{p}}_{1}}+\cdots e_{{\mathbf{p}}_{n}}\ |\
{\mathbf{p}}_{1}+\cdots+{\mathbf{p}}_{n}={\mathbf{k}}\\}.$ (70)

Sometimes, it is possible to compute (70) exactly, as shown in the following
lemma.

######  Lemma 1.

Let ${\mathbf{k}}\mapsto e_{\mathbf{k}}$ be a convex function. Then

$\inf_{{\mathbf{p}}}\\{e_{\mathbf{p}}+e_{{\mathbf{k}}-{\mathbf{p}}}\\}=2e_{{\mathbf{k}}/2}.$
(71)

In particular,

$\inf_{{\mathbf{p}}}\\{e_{\mathbf{p}}+e_{{\mathbf{k}}-{\mathbf{p}}}\\}\leq
e_{{\mathbf{k}}}.$ (72)

If in addition ${\mathbf{k}}\mapsto e_{\mathbf{k}}$ is a strictly convex
function, then

$\inf_{{\mathbf{p}}}\\{e_{\mathbf{p}}+e_{{\mathbf{k}}-{\mathbf{p}}}\\}<e_{{\mathbf{k}}},\quad{\mathbf{k}}\neq
0.$ (73)

###### Proof.

The left hand side of (71) is called infimal involution and is often denoted
as

$e\square
e({\mathbf{k}}):=\inf_{{\mathbf{p}}}\\{e_{\mathbf{p}}+e_{{\mathbf{k}}-{\mathbf{p}}}\\}.$
(74)

Since $e_{\mathbf{k}}$ is a convex function so is $e\square e({\mathbf{k}})$
[1, Chapter 12] and it satisfies

$(e\square e)^{*}=e^{*}+e^{*}=2e^{*}$ (75)

where $e^{*}$ denotes the Legendre–Fenchel transform of $e$. Hence

$\displaystyle\inf_{{\mathbf{p}}}\\{e_{\mathbf{p}}+e_{{\mathbf{k}}-{\mathbf{p}}}\\}=e\square
e({\mathbf{k}})=(e\square
e)^{**}({\mathbf{k}})=(2e^{*})^{*}({\mathbf{k}})=2e_{{\mathbf{k}}/2}$

which proves (71). Now (72) follows from convexity. Indeed,

$2e_{{\mathbf{p}}/2}=2e_{{\mathbf{p}}/2+0/2}\leq e_{{\mathbf{p}}}.$

∎

Now $e_{{\mathbf{k}}}$ in (11) is strictly convex. Therefore, (73) is true,
and so the dispersion relation is embedded inside the 2-quasiparticle
spectrum.

If $e_{\mathbf{k}}$ is given by (53), then it is strictly convex for small
${\mathbf{k}}$. Therefore, the dispersion relation is embedded inside the
2-quasiparticle spectrum at least for small momenta. The same is true for the
cutoff effective Friedrichs Hamiltonian $H_{\mathrm{Fried}}^{\Lambda}$ for
large enough $\Lambda$.

The Hamiltonian $H^{\mathrm{exc}}$ couples
$b_{\mathbf{k}}^{*}\Omega_{\mathrm{Bog}}$ with 4-quasiparticle states through
$H_{3,2}^{\mathrm{exc}}$. The bottom of 4-quasiparticle spectrum lies below
the dispersion relation (in fact, if it is given by (11), it is equal to
$4e_{{\mathbf{k}}/4}<e_{\mathbf{k}}$). However, $H_{3,2}^{\mathrm{exc}}$ does
not couple $b_{\mathbf{k}}^{*}\Omega_{\mathrm{Bog}}$ to all possible
4-quasiparticle states with the total momentum ${\mathbf{k}}$, but only to
states of the form
$b_{{\mathbf{p}}_{1}}b_{{\mathbf{p}}_{1}}b_{{\mathbf{p}}_{1}}b_{\mathbf{k}}\Omega_{\mathrm{Bog}}$
with ${\mathbf{p}}_{1}+{\mathbf{p}}_{2}+{\mathbf{p}}_{3}=0$. Their energy is

$e_{\mathbf{k}}+e_{{\mathbf{p}}_{1}}+e_{{\mathbf{p}}_{2}}+e_{{\mathbf{p}}_{3}}\geq
e_{\mathbf{k}}.$ (76)

Thus the state $b_{\mathbf{k}}^{*}\Omega_{\mathrm{Bog}}$ is situated at the
boundary of the energy-momentum spectrum and the only coupling is through
${\mathbf{p}}_{1}={\mathbf{p}}_{2}={\mathbf{p}}_{3}=0$. Before going to the
thermodynamic limit this is excluded, because on the excited space all momenta
are different from zero. Assuming that this effect survives the thermodynamic
limit, we expect that the term $H^{\mathrm{exc}}_{3,2}$ does not lead to
damping and we therefore drop it from $H_{\mathrm{Fried}}^{\Lambda}$, even
though in terms of the coupling parameter $\kappa$ this term is of the same
order as $H^{\mathrm{exc}}_{3,1}$, which we keep in our analysis.

##  7\. Computing the self-energy

In the remaining part of our paper, the main goal will be to compute
approximately the 3-dimensional integral (68). To do this efficiently it is
important to choose a convenient coordinate system.

Let us introduce the notation $k=|{\mathbf{k}}|$, $p=|{\mathbf{p}}|$,
$l=|{\mathbf{l}}|$, where ${\mathbf{l}}={\mathbf{k}}-{\mathbf{p}}$. One could
try to compute (68) using the spherical coordinates for ${\mathbf{p}}$ with
respect to the axis determined by ${\mathbf{k}}$. This means using
$p=|{\mathbf{p}}|,w=\cos\theta,\phi$, so that
${\mathbf{p}}=(p\sqrt{1-w^{2}}\cos\phi,p\sqrt{1-w^{2}}\sin\phi,pw)$. The self-
energy in these coordinates is

$\displaystyle\Sigma_{\mathbf{k}}^{\Lambda}(z)$
$\displaystyle=\frac{1}{(2\pi)^{3}}\int_{0}^{\infty}\int_{-1}^{1}\int_{0}^{2\pi}\frac{h_{\mathbf{k}}^{\Lambda}(p,w)^{2}p^{2}\,{\rm
d}p\,{\rm d}w\,{\rm d}\phi}{(z-e_{p}-e_{l(p,w)})}$ (77)

where, with abuse of notation, $h_{\mathbf{k}}^{\Lambda}(p,w)$ is the function
$h_{\mathbf{k}}^{\Lambda}({\mathbf{p}})$ in the variables $p,w,\phi$. The
variable $\phi$ can be easily integrated out.
$h_{\mathbf{k}}^{\Lambda}({\mathbf{p}})$ depends only on $k,p,l$ and (77) can
be rewritten as

$\displaystyle\Sigma_{{\mathbf{k}}}^{\Lambda}(z)$
$\displaystyle=\frac{1}{(2\pi)^{2}}\int_{0}^{\infty}\int_{-1}^{1}\frac{(h_{k}^{\Lambda}(p,l(p,w)))^{2}p^{2}\,{\rm
d}p\,{\rm d}w}{(z-e_{p}-e_{l(p,w)})},$

The coordinates $p,w$ are not convenient because they break the natural
symmetry ${\mathbf{p}}\to{\mathbf{k}}-{\mathbf{p}}$ of the system. Instead of
$p,w$ it is much better to use the variables $p,l$. Note the constraints

$\displaystyle|p-l|$ $\displaystyle\leq k,$ (78) $\displaystyle k$
$\displaystyle\leq p+l,$ (79)

that follow from the triangle inequality. We have
$w=\frac{k^{2}+p^{2}-l^{2}}{2kp}$. The Jacobian is easily computed:

$p^{2}\,{\rm d}p\,{\rm d}w=\frac{pl}{k}\,{\rm d}p\,{\rm d}l=\frac{1}{4k}\,{\rm
d}p^{2}\,{\rm d}l^{2}.$ (80)

Let us make another change of variables:

$t=p+l,\quad s=p-l;\qquad p=\frac{t+s}{2},\quad l=\frac{t-s}{2};$ (81) $\,{\rm
d}p^{2}\,{\rm d}l^{2}=\frac{t^{2}-s^{2}}{2}\,{\rm d}t\,{\rm d}s.$ (82)

The limits of integration following from the constraints (78) and (79) are
very easy to impose:

$\displaystyle\Sigma_{{\mathbf{k}}}^{\Lambda}(z)$
$\displaystyle=\frac{1}{(2\pi)^{2}}\int_{k}^{\Lambda}\,{\rm
d}t\int_{-k}^{k}\,{\rm
d}s\frac{(h_{k}^{\Lambda}(t,s))^{2}(t^{2}-s^{2})}{8k(z-e_{\frac{t+s}{2}}-e_{\frac{t-s}{2}})},$
(83)

Another choice of variables can also be useful. If $k\mapsto e_{k}$ is an
increasing function, which is always the case for small $k$, but also for the
important case of constant $\frac{\hat{v}({\mathbf{k}})}{\hat{v}(0)}$, we can
use the variables $u:=e_{p}$ and $w:=e_{l}$. Set

$f(e_{k}):=\frac{\,{\rm d}k^{2}}{\,{\rm d}e_{k}^{2}}.$ (84)

Thus we change the variables

$\frac{1}{4k}\,{\rm d}p^{2}\,{\rm d}l^{2}=\frac{1}{2k}f(u)f(w)\,{\rm
d}u^{2}\,{\rm d}w^{2}.$ (85) $\displaystyle\Sigma_{{\mathbf{k}}}^{\Lambda}(z)$
$\displaystyle=\frac{1}{(2\pi)^{2}}\int\frac{(h_{k}^{\Lambda}(u,w))^{2}f(u)f(w)\,{\rm
d}u^{2}\,{\rm d}w^{2}}{4k(z-u-w)},$

We then perform a further change of variable

$x=u+w,\quad y=u-w;\qquad u=\frac{x+y}{2},\quad w=\frac{x-y}{2};$ (86) $\,{\rm
d}u^{2}\,{\rm d}w^{2}=\frac{x^{2}-y^{2}}{2}\,{\rm d}x\,{\rm d}y.$ (87)

Now we can write

$\displaystyle\Sigma_{{\mathbf{k}}}^{\Lambda}(z)$
$\displaystyle=\frac{1}{8\pi^{2}k}\iint\frac{(h_{k}^{\Lambda}(x,y))^{2}f(\frac{x+y}{2})f(\frac{x-y}{2})(x^{2}-y^{2})\,{\rm
d}y\,{\rm d}x}{4(z-x)},$

where the limits of integration are somewhat more difficult to describe.

When $\frac{\hat{v}({\mathbf{k}})}{\hat{v}(0)}$ is a constant, so that

$e_{k}=k\sqrt{\mu+\frac{k^{2}}{4}},\qquad
k^{2}=2\big{(}\sqrt{e_{k}^{2}+\mu^{2}}-\mu\big{)},$ (88)

we can compute the function $f$:

$f(u)=\frac{1}{\sqrt{u^{2}+\mu^{2}}}.$ (89)

We also have

$\displaystyle\sigma_{k}$
$\displaystyle=\sqrt{\frac{\frac{{k}^{2}}{2}+\mu+\sqrt{\frac{{k}^{4}}{4}+\mu{k}^{2}}}{2\sqrt{\frac{{k}^{4}}{4}+\mu{k}^{2}}}},\quad\gamma_{k}=\sqrt{\frac{\frac{{k}^{2}}{2}+\mu-\sqrt{\frac{{k}^{4}}{4}+\mu{k}^{2}}}{2\sqrt{\frac{{k}^{4}}{4}+\mu{k}^{2}}}}.$
(90)

##  8\. Damping rate

The following theorem is the main result of this paper.

######  Theorem 2.

Suppose that the dispersion relation is given by (11). Then
$\mathrm{Im}\Sigma_{{\mathbf{k}}}^{\Lambda}$ does not depend on $\Lambda$ for
large $\Lambda$ and we have

$\lim_{\Lambda\to\infty}\mathrm{Im}\Sigma_{{\mathbf{k}}}^{\Lambda}(e_{k}+\mathrm{i}0)=-c_{\mathrm{Bel}}k^{5}+O(k^{6})\qquad\text{as}\qquad
k\to 0,\qquad c_{\mathrm{Bel}}=\frac{3\hat{v}(0)}{640\pi^{2}\mu}k^{5}.$ (91)

###### Proof of Theorem 2.

To prove Theorem 2 we will use the variables $x,y$:

$\displaystyle\Sigma_{{\mathbf{k}}}^{\Lambda}(e_{k}+\mathrm{i}0)$
$\displaystyle=\frac{1}{8\pi^{2}k}\iint\frac{(h_{{\mathbf{k}}}^{\Lambda}(x,y))^{2}(x^{2}-y^{2})\,{\rm
d}y\,{\rm
d}x}{(e_{k}-x+\mathrm{i}0)\sqrt{(x+y)^{2}+4\mu^{2}}\sqrt{(x-y)^{2}+4\mu^{2}}}.$
(92)

It follows from (92) and the Sochocki-Plemelj formula that

$\displaystyle\Sigma_{{\mathbf{k}}}^{\Lambda}(e_{k}+\mathrm{i}0)$
$\displaystyle=\mathrm{Re}\Sigma_{{\mathbf{k}}}^{\Lambda}(e_{k}+\mathrm{i}0)+\mathrm{i}\mathrm{Im}\Sigma_{{\mathbf{k}}}^{\Lambda}(e_{k}+\mathrm{i}0),$
$\displaystyle\mathrm{Re}\Sigma_{{\mathbf{k}}}k^{\Lambda}(e_{k}+\mathrm{i}0)$
$\displaystyle=\frac{1}{8\pi^{2}k}\iint\frac{(h_{{\mathbf{k}}}^{\Lambda}(x,y))^{2}(x^{2}-y^{2})\,{\rm
d}y\,{\rm d}x}{(e_{k}-x)\sqrt{(x+y)^{2}+4\mu^{2}}\sqrt{(x-y)^{2}+4\mu^{2}}}$
(93)
$\displaystyle\mathrm{Im}\Sigma_{{\mathbf{k}}}^{\Lambda}(e_{k}+\mathrm{i}0)$
$\displaystyle=-\frac{\pi}{8\pi^{2}k}\iint\frac{(h_{{\mathbf{k}}}^{\Lambda}(x,y))^{2}(x^{2}-y^{2})\delta(e_{k}-x)\,{\rm
d}y\,{\rm d}x}{\sqrt{(x+y)^{2}+4\mu^{2}}\sqrt{(x-y)^{2}+4\mu^{2}}}$ (94)
$\displaystyle=-\frac{\pi}{8\pi^{2}k}\int\frac{(h_{{\mathbf{k}}}^{\Lambda}(e_{k},y))^{2}(e_{k}^{2}-y^{2})\,{\rm
d}y}{\sqrt{(e_{k}+y)^{2}+4\mu^{2}}\sqrt{(e_{k}-y)^{2}+4\mu^{2}}}.$ (95)

Our starting point is the expression (95). Obviously, we first need to
establish the integration limits in $y$. Recall that $y=e_{p}-e_{l}$ but under
the additional constraint that $e_{k}=e_{p}+e_{l}$ which comes from the
constraint $\delta(x-e_{k})$ in (94). It follows immediately that $-e_{k}\leq
y\leq e_{k}$. Thus, for $\Lambda$ large enough,
$\mathrm{Im}\Sigma_{{\mathbf{k}}}^{\Lambda}(e_{k}+\mathrm{i}0)$ will not
depend on $\Lambda$.

Let us first compute $(h_{{\mathbf{k}}}(x,y))^{2}$. For further reference we
will keep $x$ as a variable. Recall we assume
$\hat{v}({\mathbf{k}})=\hat{v}(0)$. From the definition of
$h_{{\mathbf{k}}}({\mathbf{p}})$ we get

$\displaystyle\frac{h_{{\mathbf{k}}}({\mathbf{p}})}{2\sqrt{\mu\hat{v}(0)}}$
$\displaystyle=\sigma_{k}(\sigma_{p}\sigma_{l}-\sigma_{l}\gamma_{p}-\sigma_{p}\gamma_{l})+\gamma_{k}(\sigma_{p}\gamma_{l}+\sigma_{l}\gamma_{p}-\gamma_{p}\gamma_{l}).$
$\displaystyle=\frac{\sigma_{k}}{2\sqrt{uw}}\bigg{(}\sqrt{\sqrt{u^{2}+\mu^{2}}+u}\sqrt{\sqrt{w^{2}+\mu^{2}}+w}-\sqrt{\sqrt{w^{2}+\mu^{2}}+w}\sqrt{\sqrt{u^{2}+\mu^{2}}-u}$
$\displaystyle-\sqrt{\sqrt{u^{2}+\mu^{2}}+u}\sqrt{\sqrt{w^{2}+\mu^{2}}-w}\bigg{)}$
$\displaystyle+\frac{\gamma_{k}}{2\sqrt{uw}}\bigg{(}\sqrt{\sqrt{u^{2}+\mu^{2}}+u}\sqrt{\sqrt{w^{2}+\mu^{2}}-w}+\sqrt{\sqrt{w^{2}+\mu^{2}}+w}\sqrt{\sqrt{u^{2}+\mu^{2}}-u}$
$\displaystyle-\sqrt{\sqrt{u^{2}+\mu^{2}}-u}\sqrt{\sqrt{w^{2}+\mu^{2}}-w}\bigg{)}$
(96)
$\displaystyle=\frac{1}{2\sqrt{x^{2}-y^{2}}}\bigg{(}\sigma_{k}\sqrt{(A_{1}+x+y)(A_{2}+x-y))}-\gamma_{k}\sqrt{(A_{1}-x-y)(A_{2}-x+y)}$
$\displaystyle+(\gamma_{k}-\sigma_{k})\sqrt{(A_{1}-x-y)(A_{2}+x-y))}+(\gamma_{k}-\sigma_{k})\sqrt{(A_{1}+x+y)(A_{2}-x+y))}\bigg{)},$
(97)

where

$A_{1}:=A_{1}(x,y)=\sqrt{(x+y)^{2}+4\mu^{2}},\qquad
A_{2}:=A_{2}(x,y)=\sqrt{(x-y)^{2}+4\mu^{2}}.$ (98)

Therefore the integrand in (92) becomes

$\displaystyle\frac{(h_{{\mathbf{k}}}(x,y))^{2}(x^{2}-y^{2})}{\sqrt{(x+y)^{2}+4\mu^{2}}\sqrt{(x-y)^{2}+4\mu^{2}}}$
(99)
$\displaystyle=\frac{\mu\hat{v}(0)}{A_{1}A_{2}}\bigg{(}\sigma_{k}\sqrt{(A_{1}+x+y)(A_{2}+x-y))}-\gamma_{k}\sqrt{(A_{1}-x-y)(A_{2}-x+y)}$
$\displaystyle+(\gamma_{k}-\sigma_{k})\sqrt{(A_{1}-x-y)(A_{2}+x-y))}+(\gamma_{k}-\sigma_{k})\sqrt{(A_{1}+x+y)(A_{2}-x+y))}\bigg{)}^{2}.$
$\displaystyle=\frac{\mu\hat{v}(0)}{A_{1}A_{2}}\bigg{(}\sigma_{k}^{2}\left(3A_{1}A_{2}+(x+y)A_{2}+(x-y)A_{1}-(x^{2}-y^{2})-4\mu(A_{1}+A_{2}+2x)+8\mu^{2}\right)$
$\displaystyle+\gamma_{k}^{2}\left(3A_{1}A_{2}-(x+y)A_{2}-(x-y)A_{1}-(x^{2}-y^{2})-4\mu(A_{1}+A_{2}-2x)+8\mu^{2}\right)$
$\displaystyle+2\sigma_{k}\gamma_{k}\left(4\mu A_{1}+4\mu
A_{2}-2A_{1}A_{2}+2(x^{2}-y^{2})-12\mu^{2}\right)\bigg{)}.$ (100)

Thus

$\displaystyle\int_{-e_{k}}^{e_{k}}\,{\rm
d}y\frac{h_{{\mathbf{k}}}^{2}(x,y)(x^{2}-y^{2})}{\sqrt{(x+y)^{2}+4\mu^{2}}\sqrt{(x-y)^{2}+4\mu^{2}}}$
(101) $\displaystyle=\mu\hat{v}(0)\int_{-e_{k}}^{e_{k}}\,{\rm
d}y\bigg{(}\left(3\sigma_{k}^{2}+3\gamma_{k}^{2}-4\sigma_{k}\gamma_{k}\right)+(\sigma_{k}^{2}-\gamma_{k}^{2})\left(\frac{x-y}{A_{2}}+\frac{x+y}{A_{1}}-\frac{8\mu
x}{A_{1}A_{2}}\right)$
$\displaystyle+(-\sigma_{k}^{2}-\gamma_{k}^{2}+4\sigma_{k}\gamma_{k})\frac{x^{2}-y^{2}}{A_{1}A_{2}}-4\mu(\sigma_{k}-\gamma_{k})^{2}\frac{A_{1}+A_{2}}{A_{1}A_{2}}+8\mu^{2}(\sigma_{k}^{2}+\gamma_{k}^{2}-3\sigma_{k}\gamma_{k})\frac{1}{A_{1}A_{2}}\bigg{)}.$
(102)

The integrals involving $\frac{x\pm y}{A_{\pm}}$ and $\frac{1}{A_{\pm}}$
(where $A_{+}=A_{1}$ and $A_{-}=A_{2}$) can be computed explicitly. Setting
$x=e_{k}$ this implies

$\displaystyle\int_{-e_{k}}^{e_{k}}\,{\rm d}y\frac{e_{k}\pm
y}{A_{\pm}(e_{k},y)}=\int_{-e_{k}}^{e_{k}}\,{\rm d}y\left(\frac{e_{k}\pm
y}{\sqrt{(e_{k}\pm y)^{2}+4\mu^{2}}}\right)=2\sqrt{\mu^{2}+e_{k}^{2}}-2\mu,$
(103) $\displaystyle\int_{-e_{k}}^{e_{k}}\,{\rm
d}y\frac{1}{A_{\pm}(e_{k},y)}=\int_{-e_{k}}^{e_{k}}\,{\rm
d}y\left(\frac{1}{\sqrt{(e_{k}\pm
y)^{2}+4\mu^{2}}}\right)=\log\left(\frac{e_{k}}{\mu}+\sqrt{1+\frac{e_{k}^{2}}{\mu^{2}}}\right).$
(104)

This yields

$\displaystyle\int_{-e_{k}}^{e_{k}}\,{\rm
d}y\left(\frac{(h^{\Lambda}_{{\mathbf{k}}}(x,y))^{2}(x^{2}-y^{2})}{\sqrt{(x+y)^{2}+4\mu^{2}}\sqrt{(x-y)^{2}+4\mu^{2}}}\right)$
(105)
$\displaystyle=\mu\hat{v}(0)\left(2\left(3\sigma_{k}^{2}+3\gamma_{k}^{2}-4\sigma_{k}\gamma_{k}\right)e_{k}+4\sqrt{\mu^{2}+e_{k}^{2}}-4\mu-8\mu(\sigma_{k}-\gamma_{k})^{2}\log\left(\frac{e_{k}}{\mu}+\sqrt{1+\frac{e_{k}^{2}}{\mu^{2}}}\right)\right)$
$\displaystyle+\mu\hat{v}(0)\int_{-e_{k}}^{e_{k}}\,{\rm
d}y\bigg{(}\frac{-(\sigma_{k}^{2}-4\sigma_{k}\gamma_{k}+\gamma_{k}^{2})(e_{k}^{2}-y^{2})-8\mu
e_{k}+8\mu^{2}(\sigma_{k}^{2}+\gamma_{k}^{2}-3\sigma_{k}\gamma_{k})}{A_{1}A_{2}}\bigg{)}.$
(106)

where two types of integrals, namely

$\int\left(\frac{-y^{2}}{A_{1}A_{2}}\right)\,{\rm
d}y\,\,\,\text{and}\,\,\,\qquad\int\left(\frac{1}{A_{1}A_{2}}\right)\,{\rm
d}y,$ (107)

still appear as they cannot be computed explicitly.

Since we are interested in the expansion in $e_{k}$ (which is small, as $k$ is
small) we write

$\sigma_{k}=\sqrt{\frac{\sqrt{e_{k}^{2}+\mu^{2}}+e_{k}}{2e_{k}}},\qquad\gamma_{k}=\sqrt{\frac{\sqrt{e_{k}^{2}+\mu^{2}}-e_{k}}{2e_{k}}},$
(108)

which gives

$\sigma_{k}^{2}+\gamma_{k}^{2}=\frac{\sqrt{e_{k}^{2}+\mu^{2}}}{e_{k}},\qquad\sigma_{k}\gamma_{k}=\frac{\mu}{2e_{k}}.$
(109)

Then (106) equals to

$\displaystyle\mu\hat{v}(0)\left(2\left(3\sigma_{k}^{2}+3\gamma_{k}^{2}-4\sigma_{k}\gamma_{k}\right)e_{k}+4\sqrt{\mu^{2}+e_{k}^{2}}-4\mu-8\mu(\sigma_{k}-\gamma_{k})^{2}\log\left(\frac{e_{k}}{\mu}+\sqrt{1+\frac{e_{k}^{2}}{\mu^{2}}}\right)\right)$
$\displaystyle+\mu\hat{v}(0)\int_{-e_{k}}^{e_{k}}\,{\rm
d}y\bigg{(}\frac{-(\sigma_{k}^{2}-4\sigma_{k}\gamma_{k}+\gamma_{k}^{2})(e_{k}^{2}-y^{2})-8\mu
e_{k}+8\mu^{2}(\sigma_{k}^{2}+\gamma_{k}^{2}-3\sigma_{k}\gamma_{k})}{A_{1}A_{2}}\bigg{)}$
$\displaystyle=\mu\hat{v}(0)\left(2(3\sqrt{e_{k}^{2}+\mu^{2}}-2\mu)+2(2\sqrt{\mu^{2}+e_{k}^{2}}-2\mu)-8\mu\frac{\sqrt{e_{k}^{2}+\mu^{2}}-\mu}{e_{k}}\log\left(\frac{e_{k}}{\mu}+\sqrt{1+\frac{e_{k}^{2}}{\mu^{2}}}\right)\right)$
$\displaystyle-\mu\hat{v}(0)\frac{\sqrt{e_{k}^{2}+\mu^{2}}-2\mu}{e_{k}}\int_{-e_{k}}^{e_{k}}\,{\rm
d}y\left(\frac{e_{k}^{2}-y^{2}}{A_{1}A_{2}}\right)$
$\displaystyle-\left(8\mu^{2}\hat{v}(0)e_{k}-8\mu^{3}\hat{v}(0)\frac{2\sqrt{e_{k}^{2}+\mu^{2}}-3\mu}{2e_{k}}\right)\int_{-e_{k}}^{e_{k}}\,{\rm
d}y\left(\frac{1}{A_{1}A_{2}}\right)$
$\displaystyle=\mu\hat{v}(0)\left(10\mu\sqrt{(e_{k}/\mu)^{2}+1}-8\mu-8\mu\frac{\sqrt{(e_{k}/\mu)^{2}+1}-1}{e_{k}/\mu}\log\left(\frac{e_{k}}{\mu}+\sqrt{1+\frac{e_{k}^{2}}{\mu^{2}}}\right)\right)$
(110)
$\displaystyle-\mu\hat{v}(0)\frac{\sqrt{(e_{k}/\mu)^{2}+1}-2}{e_{k}/\mu}\int_{-e_{k}}^{e_{k}}\,{\rm
d}y\left(\frac{e_{k}^{2}-y^{2}}{A_{1}A_{2}}\right)$ (111)
$\displaystyle-\mu\hat{v}(0)\left(\frac{8\mu
e_{k}^{2}-4\mu^{3}(2\sqrt{(e_{k}/\mu)^{2}+1}-3)}{e_{k}}\right)\int_{-e_{k}}^{e_{k}}\,{\rm
d}y\left(\frac{1}{A_{1}A_{2}}\right).$ (112)

We expand (110) up to order $O(e_{k}^{8})$. A tedious computation yields

$\eqref{intable-
part}=\mu\hat{v}(0)\left(2\mu+\frac{e_{k}^{2}}{\mu}+\frac{5e_{k}^{4}}{12\mu^{3}}-\frac{41e_{k}^{6}}{120\mu^{5}}+O(e_{k}^{8})\right).$
(113)

We shall now deal with the terms (111) and (112). To this end we write

$\displaystyle A_{1}A_{2}$
$\displaystyle=\sqrt{4\mu^{2}+(e_{k}+y)^{2}}\sqrt{4\mu^{2}+(e_{k}-y)^{2}}$
(114)
$\displaystyle=4\mu^{2}\sqrt{1+\left(\frac{e_{k}+y}{2\mu}\right)^{2}}\sqrt{1+\left(\frac{e_{k}-y}{2\mu}\right)^{2}}$
(115)
$\displaystyle=4\mu^{2}\sqrt{1+\frac{e_{k}^{2}+y^{2}}{2\mu^{2}}+\left(\frac{e_{k}^{2}-y^{2}}{4\mu^{2}}\right)^{2}}$
(116) $\displaystyle=4\mu^{2}\sqrt{1+Q_{1}}$ (117)
$\displaystyle=4\mu^{2}\left(1+\frac{1}{2}Q_{1}-\frac{1}{8}Q_{1}^{2}+\frac{1}{16}Q_{1}^{3}\right)+O(Q_{1}^{4}).$
(118)

where

$Q_{1}:=\frac{e_{k}^{2}+y^{2}}{2\mu^{2}}+\left(\frac{e_{k}^{2}-y^{2}}{4\mu^{2}}\right)^{2}$
(119)

Then

$\frac{1}{A_{1}A_{2}}=\frac{1}{4\mu^{2}(1+Q_{2})}=\frac{1}{4\mu^{2}}(1-Q_{2}+Q_{2}^{2}-Q_{2}^{3})+O(Q_{2}^{4})$
(120)

where

$Q_{2}:=\frac{1}{2}Q_{1}-\frac{1}{8}Q_{1}^{2}+\frac{1}{16}Q_{1}^{3}.$ (121)

This leads to

$\frac{1}{A_{1}A_{2}}=\frac{1}{4\mu^{2}}-\frac{e_{k}^{2}}{16\mu^{4}}+\frac{e_{k}^{4}}{64\mu^{6}}-\frac{e_{k}^{6}}{256\mu^{8}}-\frac{y^{2}}{16\mu^{4}}+\frac{e_{k}^{2}y^{2}}{16\mu^{6}}-\frac{9e_{k}^{4}y^{2}}{256\mu^{8}}+\frac{y^{4}}{64\mu^{6}}-\frac{9e_{k}^{2}y^{4}}{256\mu^{8}}-\frac{y^{6}}{256\mu^{8}}+O(e_{k}^{\iota_{1}}y^{\iota_{2}})$
(122)

where $\iota_{1}+\iota_{2}=7$. In turn

$\int_{-e_{k}}^{e_{k}}\frac{1}{A_{1}A_{2}}\,{\rm
d}y=\frac{e_{k}}{2\mu^{2}}-\frac{e_{k}^{3}}{6\mu^{4}}+\frac{19e_{k}^{5}}{240\mu^{6}}-\frac{13e_{k}^{7}}{280\mu^{8}}+O(e_{k}^{8})$
(123)

and

$\int_{-e_{k}}^{e_{k}}\frac{e_{k}^{2}-y^{2}}{A_{1}A_{2}}\,{\rm
d}y=\frac{e_{k}^{3}}{3\mu^{2}}-\frac{e_{k}^{5}}{10\mu^{4}}+\frac{11e_{k}^{7}}{280\mu^{6}}+O(e_{k}^{8}).$
(124)

This implies

$\eqref{nonintable-
part-1}=-\mu\hat{v}(0)\left(-\frac{e_{k}^{2}}{3\mu}+\frac{4e_{k}^{4}}{15\mu^{3}}-\frac{11e_{k}^{6}}{84\mu^{5}}\right)+O(e_{k}^{8}),$
(125)

and

$\eqref{nonintable-
part-2}=-\mu\hat{v}(0)\left(2\mu+\frac{4e_{k}^{2}}{3\mu}+\frac{3e_{k}^{4}}{20\mu^{3}}-\frac{2e_{k}^{6}}{7\mu^{5}}\right)+O(e_{k}^{8}).$
(126)

Combining (125), (126) and (113) we obtain

$\displaystyle-\frac{1}{8\pi
k}\int_{-e^{k}}^{e_{k}}\frac{(h^{\Lambda}_{{\mathbf{k}}}(e_{k},y))^{2}(e_{k}^{2}-y^{2})}{\sqrt{(e_{k}+y)^{2}+4\mu^{2}}\sqrt{(e_{k}-y)^{2}+4\mu^{2}}}\,{\rm
d}y$ $\displaystyle=-\frac{\mu\hat{v}(0)}{16\pi
k}\left(\frac{5}{12}-\frac{41}{120}\right)\frac{e_{k}^{6}}{\mu^{5}}=-\frac{3\hat{v}(0)}{640\pi^{2}\mu^{4}}\frac{e_{k}^{6}}{k}.$
(127)

This yields (91). ∎

##  9\. Renormalization of the full self-energy

In this section we will try to make sense of the real part of the energy
shift. We will see that it is much more problematic. Actually, our result will
be negative: The Fermi Golden Rule starting from the Bogoliubov approximation
does not allow us to compute the energy shift of the dispersion relation.

We start with a seemingly positive result, which may suggest that one can hope
for a removal of the ultraviolet cutoff in the self-energy:

######  Theorem 3.

For ${\mathbf{k}}\neq 0$, the cutoff self-energy at $z=0$, that is
$\Sigma_{{\mathbf{k}}}^{\Lambda}(0)$, is finite. Moreover, for
$\mathrm{Im}z>0$ there exists the limit

$\tilde{\Sigma}_{{\mathbf{k}}}(z):=\lim_{\Lambda\to\infty}\big{(}\Sigma_{{\mathbf{k}}}^{\Lambda}(z)-\Sigma_{{\mathbf{k}}}^{\Lambda}(0)\big{)}.$
(128)

One can also take the limit of (128) on the real line:

$\tilde{\Sigma}_{{\mathbf{k}}}(e_{\mathbf{k}}+\mathrm{i}0):=\lim_{\Lambda\to\infty}\big{(}\Sigma_{{\mathbf{k}}}^{\Lambda}(e_{\mathbf{k}}+\mathrm{i}0)-\Sigma_{{\mathbf{k}}}^{\Lambda}(0)\big{)}=\lim_{\varepsilon\searrow
0}\tilde{\Sigma}_{{\mathbf{k}}}(e_{\mathbf{k}}+\mathrm{i}\varepsilon).$ (129)

What is the physical meaning of $\tilde{\Sigma}_{\mathbf{k}}(z)$ and
$\tilde{\Sigma}_{{\mathbf{k}}}(e_{\mathbf{k}}+\mathrm{i}0)$? Probably none.
The counterterm $\Sigma_{\mathbf{k}}^{\Lambda}(0)$ depends on $k$. We conclude
that the quantity
$\mathrm{Re}\Sigma_{\mathbf{k}}^{\mathrm{ren}}(e_{\mathbf{k}}+\mathrm{i}0)$
probably has little to with the real energy shift as we do not see how one can
justify that we are using the “right” counterterm. Indeed, in principle, one
could add to this counterterm an arbitrary function of $k$.

If one could find a $k$-independent counterterm $c^{\Lambda}$ such that

$\Sigma_{\mathbf{k}}^{\mathrm{ren}}(z):=\lim_{\Lambda\to\infty}\big{(}\Sigma_{{\mathbf{k}}}^{\Lambda}(z)-c^{\Lambda}\big{)}$
(130)

exists, then imposing $\Sigma_{0}^{\mathrm{ren}}(0+\mathrm{i}0)=0$ one could
hope that $\Sigma_{\mathbf{k}}^{\mathrm{ren}}(e_{k}+\mathrm{i}0)$ yields the
real part of the energy shift. Unfortunately, the next theorem excludes this
possibility.

######  Theorem 4.

We have

$\lim\limits_{k\to 0}\Sigma_{{\mathbf{k}}}^{\Lambda}(0)=-\infty.$ (131)

Proof of Theorem 3. In this section we will use the variables $t:=p+l$ and
$s:=p-l$ for integration. Recall from (83) that in these variables

$\displaystyle\Sigma_{{\mathbf{k}}}^{\Lambda}(z)$
$\displaystyle=\frac{1}{(2\pi)^{2}}\int_{k}^{\Lambda}\,{\rm
d}t\int_{-k}^{k}\,{\rm
d}s\frac{(h_{k}^{\Lambda}(p,l))^{2}pl}{8k(z-e_{p}-e_{l})},$ (132)

Hence,

$\displaystyle\Sigma_{k}^{\Lambda}(0)$
$\displaystyle=-\frac{1}{(2\pi)^{2}}\int_{k}^{\Lambda}\,{\rm
d}t\int_{-k}^{k}\,{\rm d}s\frac{h_{k}^{2}(p,l)pl}{8k(e_{p}+e_{l})}.$ (133)

Note that for some $c>0$, we have

$e_{p}+e_{l}\geq c(p+l)=ct.$ (134)

Let $k\neq 0$. Using (134) we see that (133) is an integral of a continuous
function over a compact region, hence finite.

Subtracting (133) from (132) we obtain

$\displaystyle\Sigma_{k}^{\Lambda}(z)-\Sigma_{k}^{\Lambda}(0)$
$\displaystyle=\frac{1}{(2\pi)^{2}}\int_{k}^{\Lambda}\,{\rm
d}t\int_{-k}^{k}\,{\rm
d}s\frac{zh_{k}^{2}(p,l)pl}{8k(z-e_{p}-e_{l})(e_{p}+e_{l})},$ (135)

For small $t$ the integrand is bounded, using again (134). For large $t$ we
have $e_{p}\simeq\frac{t^{2}}{2},$ $e_{l}\simeq\frac{t^{2}}{2}$. Moreover,
$h_{k}(p,l)$ is bounded. Therefore, the integrand of (135) behaves as
$t^{-2}$. Hence it is integrable for large $t$ and we can take the limit
$\Lambda\to\infty$ obtaining

$\displaystyle\Sigma_{k}^{\mathrm{ren}}(z):$
$\displaystyle=\lim_{\Lambda\to\infty}\big{(}\Sigma_{k}^{\Lambda}(z)-\Sigma_{k}^{\Lambda}(0)\big{)}$
(136) $\displaystyle=\frac{1}{(2\pi)^{2}}\int_{k}^{\infty}\,{\rm
d}t\int_{-k}^{k}\,{\rm
d}s\frac{zh_{k}^{2}(p,l)pl}{8k(z-e_{p}-e_{l})(e_{p}+e_{l})}.$ (137)

This ends the proof of the theorem. ∎

Before we show Theorem 4 we prove some lemmas.

######  Lemma 5.

For small $p,l$, we have

$\displaystyle\frac{e_{\frac{t}{2}}}{e_{p}+e_{l}}-\frac{1}{2}$
$\displaystyle=O(s^{2}),$ (138)
$\displaystyle\frac{pl}{e_{p}e_{l}}-\frac{t^{2}}{4e_{\frac{t}{2}}^{2}}$
$\displaystyle=O(s^{2}),$ (139)
$\displaystyle\sigma_{p}\sigma_{l}\sqrt{e_{p}e_{l}}-\sigma_{\frac{t}{2}}^{2}e_{\frac{t}{2}}$
$\displaystyle=O(s^{2}),$ (140)
$\displaystyle\gamma_{p}\gamma_{l}\sqrt{e_{p}e_{l}}-\gamma_{\frac{t}{2}}^{2}e_{\frac{t}{2}}$
$\displaystyle=O(s^{2}).$ (141)

###### Proof.

We can assume that $s\geq 0$.

$e_{p}^{\prime}=\big{(}\tfrac{p^{2}}{2}+\mu\big{)}\big{(}\tfrac{p^{2}}{4}+\mu\big{)}^{-\frac{1}{2}},\quad
e_{p}^{\prime\prime}=p\big{(}\tfrac{p^{2}}{8}+\tfrac{3\mu}{4}\big{)}\big{(}\tfrac{p^{2}}{4}+\mu\big{)}^{-\frac{3}{2}}=O(p).$
(142)

Therefore,

$2e_{\frac{t}{2}}-e_{p}-e_{l}=-\int_{-\frac{s}{2}}^{\frac{s}{2}}\big{(}\tfrac{s}{2}-|v|\big{)}e_{\frac{t}{2}+v}^{\prime\prime}\,{\rm
d}v=O(ts^{2}),$

and hence

$\frac{e_{\frac{t}{2}}}{e_{p}+e_{l}}-\frac{1}{2}=\frac{2e_{\frac{t}{2}}-e_{p}-e_{l}}{2(e_{p}+e_{l})}$

is $O(s^{2})$, which proves (138).

Next, set $f(p):=\frac{p}{e_{p}}$. We have

$\frac{\,{\rm d}}{\,{\rm
d}p}f(p)=\frac{-2p}{(p^{2}+4\mu)^{\frac{3}{2}}}=O(p),\qquad\frac{\,{\rm
d}^{2}}{\,{\rm
d}p^{2}}f(p)=\frac{4(p^{2}-2\mu)}{(p^{2}+4\mu)^{\frac{5}{2}}}=O(1).$ (143)

Hence

$\displaystyle\frac{pl}{e_{p}e_{l}}-\frac{t^{2}}{4e_{\frac{t}{2}}^{2}}=f(p)f(l)-f\big{(}\tfrac{t}{2}\big{)}^{2}$
(144) $\displaystyle=$
$\displaystyle\int_{0}^{\frac{s}{2}}\big{(}\tfrac{s}{2}-v\big{)}\Big{(}f^{\prime\prime}\big{(}\tfrac{t}{2}+v\big{)}f\big{(}\tfrac{t}{2}-v\big{)}-2f^{\prime}\big{(}\tfrac{t}{2}+v\big{)}f^{\prime}\big{(}\tfrac{t}{2}-v\big{)}+f\big{(}\tfrac{t}{2}+v\big{)}f^{\prime\prime}\big{(}\tfrac{t}{2}-v\big{)}\Big{)}\,{\rm
d}v,$

which is $O(s^{2})$, which proves (139).

We check that the 0th, 1st and 2nd derivatives of

$\displaystyle\sigma_{p}\sqrt{e_{p}}=\frac{1}{\sqrt{2}}\sqrt{\tfrac{p^{2}}{2}+\mu+\sqrt{\tfrac{p^{4}}{4}+\mu
p^{2}}},$ (145)
$\displaystyle\gamma_{p}\sqrt{e_{p}}=\frac{1}{\sqrt{2}}\sqrt{\tfrac{p^{2}}{2}+\mu-\sqrt{\tfrac{p^{4}}{4}+\mu
p^{2}}}$ (146)

are bounded. Then we argue as in (144), proving (140) and (141). ∎

######  Lemma 6.

$\displaystyle\lim_{k\to 0}\int_{k}^{\Lambda}\,{\rm d}t\int_{-k}^{k}\,{\rm
d}s\frac{(\sigma_{p}\sigma_{l}-\gamma_{p}\gamma_{l})^{2}pl}{8k(e_{p}+e_{l})}=\int_{0}^{\Lambda}\,{\rm
d}t\frac{t^{2}}{64e_{\frac{t}{2}}},$ (147)

where the right hand side is a finite positive number.

###### Proof.

We have

$\displaystyle\frac{(\sigma_{p}\sigma_{l}-\gamma_{p}\gamma_{l})^{2}pl}{8k(e_{p}+e_{l})}-\frac{t^{2}}{8\cdot
8ke_{\frac{t}{2}}}$ (148) $\displaystyle=$
$\displaystyle\frac{\big{(}(\sigma_{p}\sigma_{l}-\gamma_{p}\gamma_{l})\sqrt{e_{p}e_{l}}+e_{\frac{t}{2}}\big{)}pl}{8k(e_{p}+e_{l})e_{p}e_{l}}\Big{(}(\sigma_{p}\sigma_{l}-\gamma_{p}\gamma_{l})\sqrt{e_{p}e_{l}}-e_{\frac{t}{2}}\big{)}\Big{)}$
(149) $\displaystyle+$
$\displaystyle\frac{e_{\frac{t}{2}}^{2}}{8k(e_{p}+e_{l})}\Big{(}\frac{pl}{e_{p}e_{l}}-\frac{t^{2}}{4e_{\frac{t}{2}}^{2}}\Big{)}$
(150) $\displaystyle+$
$\displaystyle\frac{t^{2}}{32ke_{\frac{t}{2}}}\Big{(}\frac{e_{\frac{t}{2}}}{e_{p}+e_{l}}-\frac{1}{2}\Big{)}.$
(151)

By Lemma 5 the terms in the big brackets on the right of (149), (150) and
(151) are $O(s^{2})$. The terms in (150), (151) on the left are all
$\frac{1}{k}O(t)$. The most singular in $t$ term is the one on the left of
(149) and it is of order $\frac{1}{k}O(t^{-1})$. Therefore,

$\displaystyle\int_{k}^{\Lambda}\,{\rm d}t\int_{-k}^{k}\,{\rm
d}s\Bigg{(}\frac{(\sigma_{p}\sigma_{l}-\gamma_{p}\gamma_{l})^{2}pl}{8k(e_{p}+e_{l})}-\frac{t^{2}}{64e_{\frac{t}{2}}}\Bigg{)}$
(152) $\displaystyle=$ $\displaystyle\int_{k}^{\Lambda}\,{\rm
d}t\int_{-k}^{k}\,{\rm d}sO(t^{-1})\frac{O(s^{2})}{k}=\int_{k}^{\Lambda}\,{\rm
d}tO(t^{-1}k^{2})=O(k^{2}\ln k)\to 0.$ (153)

∎

Proof of Theorem 4. Recall (61). We have

$\displaystyle\frac{h_{{\mathbf{k}}}({\mathbf{p}})}{2\sqrt{\mu\hat{v}(0)}}$
$\displaystyle=\frac{1}{2}(\sigma_{k}+\gamma_{k})(\sigma_{p}\sigma_{l}-\gamma_{p}\gamma_{l})+\frac{1}{2}(\sigma_{k}-\gamma_{k})(\sigma_{p}\sigma_{l}+\gamma_{p}\gamma_{l}-2\sigma_{p}\gamma_{l}-2\gamma_{p}\sigma_{l}).$
(154)

Thus, using (83), we obtain

$\displaystyle-\frac{(2\pi)^{2}}{\mu\hat{v}(0)}\Sigma_{k}^{\Lambda}(0)$ (155)
$\displaystyle=$
$\displaystyle(\sigma_{k}+\gamma_{k})^{2}\int_{k}^{\Lambda}\,{\rm
d}t\int_{-k}^{k}\,{\rm
d}s\frac{(\sigma_{p}\sigma_{l}-\gamma_{p}\gamma_{l})^{2}pl}{2k(e_{p}+e_{l})}$
(156) $\displaystyle+$ $\displaystyle 2\int_{k}^{\Lambda}\,{\rm
d}t\int_{-k}^{k}\,{\rm
d}s\frac{(\sigma_{p}\sigma_{l}-\gamma_{p}\gamma_{l})(\sigma_{p}\sigma_{l}+\gamma_{p}\gamma_{l}-2\sigma_{p}\gamma_{l}-2\gamma_{p}\sigma_{l})pl}{2k(e_{p}+e_{l})}$
(157) $\displaystyle+$
$\displaystyle(\sigma_{k}-\gamma_{k})^{2}\int_{k}^{\Lambda}\,{\rm
d}t\int_{-k}^{k}\,{\rm
d}s\frac{(\sigma_{p}\sigma_{l}+\gamma_{p}\gamma_{l}-2\sigma_{p}\gamma_{l}-2\gamma_{p}\sigma_{l})^{2}pl}{2k(e_{p}+e_{l})}$
(158)

where we used that $\sigma_{k}^{2}-\gamma_{k}^{2}=1$. Since $\Lambda$ is fixed
we are only interested in the small $t$ region. Since $k$ is small too, this
implies also $p$ and $l$ are small. For such we have

$\displaystyle(\sigma_{k}+\gamma_{k})^{2}$ $\displaystyle\geq Ck^{-1},\quad
C>0$ (159) $\displaystyle(\sigma_{k}-\gamma_{k})^{2}$ $\displaystyle=O(k),$
(160) $\displaystyle(\sigma_{p}\sigma_{l}-\gamma_{p}\gamma_{l})\sqrt{pl}$
$\displaystyle=O(p)+O(l)=O(t),$ (161)
$\displaystyle(\sigma_{p}\sigma_{l}+\gamma_{p}\gamma_{l}-2\sigma_{p}\gamma_{l}-2\gamma_{p}\sigma_{l})\sqrt{pl}$
$\displaystyle=O(1),$ (162) $\displaystyle\frac{1}{e_{p}+e_{l}}$
$\displaystyle=O(t^{-1}).$ (163)

By Lemma 6 and (159),

$\displaystyle|\eqref{quw1}|$ $\displaystyle\geq C_{1}k^{-1}\to+\infty.$ (164)

By (161), (162) and (163),

$\displaystyle|\eqref{quw2}|$ $\displaystyle\leq C\int_{k}^{\Lambda}\,{\rm
d}t\int_{-k}^{k}\,{\rm d}s\frac{1}{k}\rightarrow
C_{\Lambda}\qquad\text{as}\,\,\,k\to 0.$ (165)

Here $C_{\Lambda}$ is a constant depending on $\Lambda$ (which is fixed). By
(160), (162) and (163),

$\displaystyle|\eqref{quw}|$ $\displaystyle\leq Ck\int_{k}^{\Lambda}\,{\rm
d}t\int_{-k}^{k}\,{\rm d}s\frac{1}{kt}\leq Ck|\ln(k)|\to 0,$ (166)

Hence (155) converges to $+\infty$. ∎

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