UFIFT-QG-21-05

Large Logarithms from Quantum Gravitational Corrections to a Massless,
Minimally Coupled Scalar on de Sitter

D. Glavan1∗, S. P. Miao2⋆, T. Prokopec3† and R. P. Woodard4‡

1 CEICO, Institute of Physics of the Czech Academy of Sciences (FZU),
Na Slovance 1999/2, 182 21 Prague 8, CZECH REPUBLIC

2 Department of Physics, National Cheng Kung University,
No. 1 University Road, Tainan City 70101, TAIWAN

3 Institute for Theoretical Physics, Spinoza Institute & EMME$\Phi$,
Utrecht University, Postbus 80.195, 3508 TD Utrecht, THE NETHERLANDS

4 Department of Physics, University of Florida,
Gainesville, FL 32611, UNITED STATES

ABSTRACT

We consider single graviton loop corrections to the effective field equation
of a massless, minimally coupled scalar on de Sitter background in the
simplest gauge. We find a large temporal logarithm in the approach to freeze-
in at late times, but no correction to the feeze-in amplitude. We also find a
large spatial logarithm (at large distances) in the scalar potential generated
by a point source, which can be explained using the renormalization group with
one of the higher derivative counterterms regarded as a curvature-dependent
field strength renormalization. We discuss how these results set the stage for
a project to purge gauge dependence by including quantum gravitational
corrections to the source which disturbs the effective field and to the
observer who measures it.

PACS numbers: 04.50.Kd, 95.35.+d, 98.62.-g

∗<EMAIL_ADDRESS>
⋆ e-mail<EMAIL_ADDRESS>
† e-mail<EMAIL_ADDRESS>
‡ e-mail<EMAIL_ADDRESS>
##  1 Introduction

One of the most profound predictions of primordial inflation is that the
accelerated expansion literally rips long wavelength quanta out of the vacuum
[1]. This is what produced the tensor power spectrum $\Delta_{h}^{2}(k)$ [2].
The occupation number $N(\eta,k)$ of a single polarization of wave number
$\vec{k}$ at conformal time $\eta$ is simply staggering,

$N(\eta,k)=\frac{\pi\Delta_{h}^{2}(k)}{64Gk^{2}}\\!\times\\!a^{2}(\eta)\;,$
(1)

where $G$ is Newton’s constant and $a(\eta)$ is the scale factor, which we
remind the reader grows exponentially rapidly in co-moving time.

The tensor power spectrum is the primary, tree order signal of the production
of inflationary gravitons. However, there must be secondary, loop effects from
the interactions of these gravitons with each other and with other particles.
Among these effects are:

  1. 1.

The self-gravitation between inflationary gravitons may slow the expansion
rate [3, 4];

  2. 2.

Inflationary gravitons correct the linearized Einstein equation [5, 6], which
enhances the field strength of gravitational radiation [7, 8] and has the
potential to change the force of gravity [9];

  3. 3.

Inflationary gravitons correct the linearized Dirac equation [10, 11], which
enhances the field strength of fermions [12];

  4. 4.

Inflationary gravitons correct the field equations for a massless, minimally
coupled scalar [13], but makes no significant change in the field strength of
scalar radiation [14];

  5. 5.

Inflationary gravitons correct Maxwell’s equation [15], which enhances the
field strength of electromagnetic radiation [16] and makes significant changes
to the response to charges and currents at large distances and late times
[17]; and

  6. 6.

Inflationary gravitons correct the field equation for a massless, conformally
coupled scalar [18, 19, 20], but do not make significant changes, either in
the propagation of dynamical scalars or in the scalar exchange potential [21].

No one doubts that a classical ensemble of gravitational radiation would
change kinematics and forces; this is the basis for the proposal to detect
gravitational radiation using the timing of pulsars [22, 23]. However,
graviton propagators do require gauge fixing, and it has been argued that the
apparent effects of inflationary gravitons are artifacts of the gauge [24, 25,
26, 27, 28, 29, 30, 31]. These doubts persist in spite of the fact that
similar effects derive from loops of massless, minimally coupled scalars [32,
9], which experience the same growth (1) as inflationary gravitons and require
no gauge fixing.

A technique has been developed for removing gauge dependence from effective
field equations by including quantum gravitational correlations with the
source which disturbs the effective field and the observer who detects it
[33]. The procedure is to build the same diagrams (Figure 1 gives two
examples) that would go into an S-matrix element, and then simplify them with
a series of relations derived by Donoghue [34, 35, 36] to capture the infrared
physics. In the end only vestigial traces of the source and observer remain,
and each of the simplified diagrams can be regarded as a correction to the 1PI
2-point function in the linearized, effective field equation.

Figure 1: The left diagram shows how the self-mass (3) contributes to massive
scalar scattering. The diagram on the right gives the contribution from
graviton correlations between the vertices. Solid lines represent the massless
scalar, wavy lines represent the graviton, and dashed lines are massive
scalars.

It is worthwhile describing this for a massless, minimally coupled scalar on
flat space background ($g_{\mu\nu}\equiv\eta_{\mu\nu}+\sqrt{16\pi
G}\,h_{\mu\nu}$) in the 2-parameter family of covariant gauge fixing
functions,

$\mathcal{L}_{GF}=-\frac{1}{2\alpha}\eta^{\mu\nu}F_{\mu}F_{\nu}\qquad,\qquad
F_{\mu}=\eta^{\rho\sigma}\Bigl{(}h_{\mu\rho,\sigma}-\frac{\beta}{2}h_{\rho\sigma,\mu}\Bigr{)}\;.$
(2)

The renormalized self-mass is [33],

$\displaystyle-iM^{2}(x;x^{\prime})$ $\displaystyle=$
$\displaystyle\mathcal{C}_{0}(\alpha,\beta)\times\frac{G\partial^{6}}{4\pi^{3}}\Biggl{[}\frac{\ln(\mu^{2}\Delta
x^{2})}{\Delta x^{2}}\Biggr{]}\quad,\quad\Delta
x^{2}\equiv(x\\!-\\!x^{\prime})^{2}\;,\qquad$ (3)
$\displaystyle\mathcal{C}_{0}(\alpha,\beta)$ $\displaystyle=$
$\displaystyle+\frac{3}{4}-\frac{3}{4}\times\alpha-\frac{3}{2}\times\frac{1}{\beta\\!-\\!2}+\frac{3}{4}\times\frac{(\alpha\\!-\\!3)}{(\beta\\!-\\!2)^{2}}\;.$
(4)

Now imagine quantum gravitational corrections to the scattering of two massive
scalars by the exchange of such a massless scalar. Figure 1 shows two of the
many diagrams which contribute. After applying the Donoghue Identities, each
of these contributions can be regarded as a correction to the self-mass,
having the same spacetime dependence as (3) but multiplied by different gauge
dependent coefficients. Table 1 lists each contribution, and one can see that
the sum is indeed independent of $\alpha$ and $\beta$.

$i$ | $1$ | $\alpha$ | $\frac{1}{\beta-2}$ | $\frac{(\alpha-3)}{(\beta-2)^{2}}$ | Description
---|---|---|---|---|---
0 | $+\frac{3}{4}$ | $-\frac{3}{4}$ | $-\frac{3}{2}$ | $+\frac{3}{4}$ | scalar exchange
1 | $0$ | $0$ | $0$ | $+1$ | vertex-vertex
2 | $0$ | $0$ | $0$ | $0$ | vertex-source,observer
3 | $0$ | $0$ | $+3$ | $-2$ | vertex-scalar
4 | $+\frac{17}{4}$ | $-\frac{3}{4}$ | $0$ | $-\frac{1}{4}$ | source-observer
5 | $-2$ | $+\frac{3}{2}$ | $-\frac{3}{2}$ | $+\frac{1}{2}$ | scalar-source,observer
Total | $+3$ | $0$ | $0$ | $0$ |
Table 1: The gauge dependent factors $C_{i}(\alpha,\beta)$ for each
contribution to the invariant scalar self-mass-squared, and their gauge-
independent sum. Figure 1 shows the $i=0$ and $i=1$ diagrams.

In position space all diagrams consist of products of (possibly
differentiated) massive and massless propagators (from the scalar and the
graviton), $i\Delta_{m}(x;x^{\prime})$ and $i\Delta(x;x^{\prime})$,
respectively. All the 3-point and 4-point diagrams can be reduced to 2-point
form by applying the Donoghue Identities [33],

$\displaystyle
i\Delta_{m}(x;y)i\Delta(x;x^{\prime})i\Delta(y;x^{\prime})\longrightarrow\frac{i\delta^{D}(x\\!-\\!y)}{2m^{2}}\Bigl{[}i\Delta(x;x^{\prime})\Bigr{]}^{2}\;,$
(5)
$\displaystyle\partial^{x}_{\mu}i\Delta(x;x^{\prime})\partial_{y}^{\mu}i\Delta(y;y^{\prime})i\Delta_{m}(x;y)i\Delta_{m}(x^{\prime};y^{\prime})$
(6) $\displaystyle\hskip
128.0374pt\longrightarrow\frac{i\delta^{D}(x\\!-\\!y)i\delta^{D}(x^{\prime}\\!-\\!y^{\prime})}{2m^{2}}\Bigl{[}i\Delta(x;x^{\prime})\Bigr{]}^{2}\;,\qquad$
$\displaystyle\partial^{x}_{\mu}i\Delta(x;y^{\prime})\partial^{\mu}_{y}i\Delta(y;x^{\prime})i\Delta_{m}(x;y)i\Delta_{m}(x^{\prime};y^{\prime})$
(7) $\displaystyle\hskip
128.0374pt\longrightarrow-\frac{i\delta^{D}(x\\!-\\!y)i\delta^{D}(x^{\prime}\\!-\\!y^{\prime})}{2m^{2}}\Bigl{[}i\Delta(x;x^{\prime})\Bigr{]}^{2}\;.\qquad$

Any 2-point contribution so obtained can be regarded as a correction to the
self-mass through a trivial identity based the massless propagator equation
$\partial^{2}i\Delta(x;x^{\prime})=i\delta^{D}(x-x^{\prime})$,

$f(x;x^{\prime})=-\int d^{D}z\,i\Delta(x;z)\int
d^{D}z^{\prime}\,i\Delta(x^{\prime};z^{\prime})\times\partial_{z}^{2}\partial_{z^{\prime}}^{2}f(z;z^{\prime})\;.$
(8)

The procedure just described has been implemented on flat space background for
scalars [33] and for electromagnetism [37]. Generalizing it to de Sitter will
be challenging because the Hubble parameter $H$ permits more varied spacetime
dependence than (3) on the dimensionless product $H^{2}\Delta x^{2}$. Our
program is therefore to find the simplest venue for implementing the gauge
purge on de Sitter background, and then check gauge independence using the
family of de Sitter breaking gauges analogous to (2) [38]. In addition to
simplicity, we require a system for which the potentially gauge dependent
computation shows big effects, because there is little point to removing gauge
dependence from a small or null effect. Previous studies have revealed that
graviton corrections to massless, conformally coupled scalars are simple but
do not engender significant effects [20, 21]. In this paper we show that the
massless, minimally coupled scalar provides the system we seek.

In section 2 of this paper we compute the single graviton loop contribution to
the self-mass $-iM^{2}(x;x^{\prime})$ of a massless, minimally coupled scalar
on de Sitter background. Section 3 uses this result to quantum-correct the
linearized effective scalar field equation. Solving this equation reveals no
significant 1-loop correction to the field strength of scalar radiation, but a
large logarithmic correction to the scalar exchange potential. We also explain
the large logarithm using a version of the renormalization group. Our
conclusions comprise section 4.

##  2 Graviton Loop Contribution to $-iM^{2}(x;x^{\prime})$

The purpose of this section is to compute the 1-graviton loop contribution to
the 1PI (one-particle-irreducible) 2-point function of a massless, minimally
coupled scalar on de Sitter background. We begin by precisely defining
$-iM^{2}(x;x^{\prime})$, analytically and diagrammatically, and by giving the
required Feynman rules. We next employ dimensional regularization to evaluate
first the simplest diagram and then the more complicated one. The section
closes with a discussion of renormalization.

###  2.1 Feynman Rules

The bare Lagrangian in $D$ spacetime dimensions is,

$\mathcal{L}=\frac{[R\\!-\\!(D\\!-\\!2)\Lambda]\sqrt{-g}}{16\pi
G}-\frac{1}{2}\partial_{\mu}\phi\partial_{\nu}\phi
g^{\mu\nu}\sqrt{-g}\qquad,\qquad\Lambda\equiv(D\\!-\\!1)H^{2}\;,$ (9)

where $G$ is Newton’s constant and $\Lambda$ is the cosmological constant. We
define the graviton field $h_{\mu\nu}(x)$ as a perturbation of the conformally
rescaled metric,

$g_{\mu\nu}(x)\equiv a^{2}(\eta)\widetilde{g}_{\mu\nu}(x)\equiv
a^{2}(\eta)\Bigl{[}\eta_{\mu\nu}+\kappa h_{\mu\nu}\Bigr{]}\qquad,\qquad
a(\eta)\equiv-\frac{1}{H\eta}\;,$ (10)

where $\kappa^{2}\equiv 16\pi G$ is the loop-counting parameter and $\eta<0$.
Our signature is spacelike, and we employ an overlined tensor to denote the
suppression of its temporal components,

$\overline{\eta}_{\mu\nu}\equiv\eta_{\mu\nu}+\delta^{0}_{~{}\mu}\delta^{0}_{~{}\nu}\qquad,\qquad\overline{\partial}_{\mu}\equiv\partial_{\mu}-\delta^{0}_{~{}\mu}\partial_{0}\;.$
(11)

The 1-graviton loop contribution to the scalar self-mass can be represented as
the in-out matrix element of variations of the action $S[g,\phi]$ and the
counterterm action $\Delta S[g,\phi]$ (discussed in subsection 2.4),

$\displaystyle-iM^{2}(x;x^{\prime})=\Bigl{\langle}\Omega^{\rm
out}\Bigl{|}T^{*}\Biggl{\\{}\Bigl{[}\frac{i\delta
S[g,\phi]}{\delta\phi(x)}\Bigr{]}_{h\phi}\\!\times\\!\Bigl{[}\frac{i\delta
S[g,\phi]}{\delta\phi(x^{\prime})}\Bigr{]}_{h\phi}$ (12) $\displaystyle\hskip
119.50148pt+\Bigl{[}\frac{i\delta^{2}S[g,\phi]}{\delta\phi(x)\delta\phi(x^{\prime})}\Bigr{]}_{hh}+\Bigl{[}\frac{i\delta^{2}\Delta
S[g,\phi]}{\delta\phi(x)\delta\phi(x^{\prime})}\Bigr{]}_{1}\Biggr{\\}}\Bigr{|}\Omega^{\rm
in}\Bigr{\rangle}.\qquad$

The $T^{*}$-ordering symbol in expression (12) indicates that derivatives are
taken outside time-ordering; the subscripts of square-bracketed variations
indicate how many perturbative fields contribute to the 1-loop result. The
associated diagrams are shown in Figure 2.

Figure 2: 1-graviton loop contributions to $-iM^{2}(x;x^{\prime})$
corresponding to the three terms of expression (12). Graviton lines are wavy,
scalar lines are straight and counterterms are denoted by a cross.

The propagators mostly depend on the de Sitter length function
$y(x;x^{\prime})$,

$y(x;x^{\prime})\equiv aa^{\prime}H^{2}\Delta x^{2}(x;x^{\prime})\equiv
aa^{\prime}H^{2}\Bigl{[}\Bigl{\|}\vec{x}\\!-\\!\vec{x}^{\prime}\Bigr{\|}^{2}-\Bigl{(}|\eta\\!-\\!\eta^{\prime}|\\!-\\!i\epsilon\Bigr{)}^{2}\Bigr{]}\;.$
(13)

The scalar propagator is [41, 42],

$i\Delta_{A}(x;x^{\prime})=A(y)+k\ln(aa^{\prime})\qquad,\qquad
k\equiv\frac{H^{D-2}}{(4\pi)^{\frac{D}{2}}}\frac{\Gamma(D\\!-\\!1)}{\Gamma(\frac{D}{2})}\;,$
(14)

where the derivative of $A(y)$ is,

$\displaystyle
A^{\prime}(y)=-\frac{H^{D-2}}{4(4\pi)^{\frac{D}{2}}}\Biggl{\\{}\Gamma\Bigl{(}\frac{D}{2}\Bigr{)}\Bigl{(}\frac{4}{y}\Bigr{)}^{\frac{D}{2}}+\Gamma\Bigl{(}\frac{D}{2}\\!+\\!1\Bigr{)}\Bigl{(}\frac{4}{y}\Bigr{)}^{\frac{D}{2}-1}$
(15) $\displaystyle\hskip
56.9055pt+\sum_{n=0}^{\infty}\Biggl{[}\frac{\Gamma(n\\!+\\!\frac{D}{2}\\!+\\!2)}{\Gamma(n\\!+\\!3)}\Bigl{(}\frac{y}{4}\Bigr{)}^{n-\frac{D}{2}+2}-\frac{\Gamma(n\\!+\\!D)}{\Gamma(n\\!+\\!\frac{D}{2}\\!+\\!1)}\Bigl{(}\frac{y}{4}\Bigr{)}^{n}\Biggr{]}\Biggr{\\}}.\qquad$

The undifferentiated result can be inferred from the coincidence limit,

$i\Delta_{A}(x;x)=k\Bigl{[}-\pi{\rm
cot}\Bigl{(}\frac{D\pi}{2}\Bigr{)}+2\ln(a)\Bigr{]}\;.$ (16)

Our gauge fixing term is a de Sitter breaking analog of (2) for
$\alpha=\beta=1$ [39, 40],

$\mathcal{L}_{GF}=-\frac{1}{2}a^{D-2}\eta^{\mu\nu}F_{\mu}F_{\nu}\;\;,\;\;F_{\mu}=\eta^{\rho\sigma}\Bigl{[}h_{\mu\rho,\sigma}\\!-\\!\frac{1}{2}h_{\rho\sigma,\mu}\\!+\\!(D\\!-\\!2)aHh_{\mu\rho}\delta^{0}_{~{}\sigma}\Bigr{]}.$
(17)

In this gauge the graviton propagator is the sum of three constant tensor
factors times scalar propagators,

$i\Bigl{[}\mbox{}_{\mu\nu}\Delta_{\rho\sigma}\Bigr{]}(x;x^{\prime})=\sum_{I=A,B,C}\Bigl{[}\mbox{}_{\mu\nu}T^{I}_{\rho\sigma}\Bigr{]}\\!\times\\!i\Delta_{I}(x;x^{\prime})\;.$
(18)

The $A$-type propagator is the same as the scalar propagator (14). The
$B$-type and $C$-type propagators are for minimally coupled scalars with
masses $M^{2}_{B}=(D-2)H^{2}$ and $M^{2}_{C}=2(D-3)H^{2}$, which can be
expressed as,

$i\Delta_{B}=-\frac{[(4y\\!-\\!y^{2})A^{\prime}(y)\\!+\\!(2\\!-\\!y)k]}{2(D\\!-\\!2)}\quad,\quad
i\Delta_{C}=\frac{(2\\!-\\!y)i\Delta_{B}}{2}+\frac{k}{D\\!-\\!3}\;.$ (19)

And the constant tensor factors $[\mbox{}_{\mu\nu}T^{I}_{\rho\sigma}]$ are,

$\displaystyle\Bigl{[}\mbox{}_{\mu\nu}T^{A}_{\rho\sigma}\Bigr{]}=2\overline{\eta}_{\mu(\rho}\overline{\eta}_{\sigma)\nu}-\frac{2}{D\\!-\\!3}\overline{\eta}_{\mu\nu}\overline{\eta}_{\rho\sigma}\quad,\quad\Bigl{[}\mbox{}_{\mu\nu}T^{B}_{\rho\sigma}\Bigr{]}=-4\delta^{0}_{~{}(\mu}\overline{\eta}_{\nu)(\rho}\delta^{0}_{~{}\sigma)}\;,\qquad$
(20)
$\displaystyle\Bigl{[}\mbox{}_{\mu\nu}T^{C}_{\rho\sigma}\Bigr{]}=\frac{2E_{\mu\nu}E_{\rho\sigma}}{(D\\!-\\!2)(D\\!-\\!3)}\quad,\quad
E_{\mu\nu}\equiv(D\\!-\\!3)\delta^{0}_{~{}\mu}\delta^{0}_{~{}\nu}+\overline{\eta}_{\mu\nu}\;.\qquad$
(21)

Here and henceforth, parenthesized indices are symmetrized.

A hatted 2nd rank tensor denotes the trace-reversal,

$\widehat{h}_{\mu\nu}\equiv h_{\mu\nu}-\frac{1}{2}\eta_{\mu\nu}h\;.$ (22)

Trace-reversing a single index of the graviton propagator gives,

$i\Bigl{[}\mbox{}_{\widehat{\mu\nu}}\Delta_{\rho\sigma}\\!\Bigr{]}\\!=2\eta_{\mu(\rho}\eta_{\sigma)\nu}i\Delta_{A}+4\delta^{0}_{~{}(\mu}\overline{\eta}_{\nu)(\rho}\delta^{0}_{~{}\sigma)}(i\Delta_{A}\\!-\\!i\Delta_{B})-\frac{2\delta^{0}_{~{}\mu}\delta^{0}_{~{}\nu}E_{\rho\sigma}}{D\\!-\\!3}(i\Delta_{A}\\!-\\!i\Delta_{C})\;.$
(23)

This form is desirable because the noncovariant tensor factors multiply
differences, $(i\Delta_{A}-i\Delta_{B})$ and $(i\Delta_{A}-i\Delta_{C})$,
which are only logarithmically singular at coincidence. Trace-reversing on
both indices gives,

$\displaystyle
i\Bigl{[}\mbox{}_{\widehat{\mu\nu}}\Delta_{\widehat{\rho\sigma}}\Bigr{]}=\Bigl{[}2\eta_{\mu(\rho}\eta_{\sigma)\nu}\\!-\\!\eta_{\mu\nu}\eta_{\rho\sigma}\Bigr{]}i\Delta_{A}$
(24) $\displaystyle\hskip
19.91684pt+4\delta^{0}_{~{}(\mu}\overline{\eta}_{\nu)(\rho}\delta^{0}_{~{}\sigma)}(i\Delta_{A}\\!-\\!i\Delta_{B})-2\Bigl{(}\frac{D\\!-\\!2}{D\\!-\\!3}\Bigr{)}\delta^{0}_{~{}\mu}\delta^{0}_{~{}\nu}\delta^{0}_{~{}\rho}\delta^{0}_{~{}\sigma}(i\Delta_{A}\\!-\\!i\Delta_{C})\;.\qquad$

###  2.2 The Primitive 4-Point Contribution

We might call the middle diagram of Figure 2 $-iM^{2}_{4}(x;x^{\prime})$. From
expression (12) we see that it involves the in-out matrix element of,

$\displaystyle\Bigl{[}\frac{i\delta^{2}S[g,\phi]}{\delta\phi(x)\delta\phi(x^{\prime})}\Bigr{]}_{hh}=i\partial_{\mu}\Bigl{[}\sqrt{-g(x)}\,g^{\mu\nu}(x)\partial_{\nu}\delta^{D}(x\\!-\\!x^{\prime})\Bigr{]}_{hh}\;,$
(26) $\displaystyle\hskip
22.76228pt=i\kappa^{2}\partial_{\mu}\Bigl{[}a^{D-2}\Bigl{(}\widehat{h}^{\mu\rho}(x)h^{\nu}_{~{}\rho}(x)\\!-\\!\frac{1}{4}\eta^{\mu\nu}\widehat{h}^{\rho\sigma}(x)h_{\rho\sigma}(x)\Bigr{)}\partial_{\nu}\delta^{D}(x\\!-\\!x^{\prime})\Bigr{]}\;.\qquad$

The matrix element involves the single trace-reversed propagator (23),

$-iM^{2}_{4}=i\kappa^{2}\partial_{\mu}\Biggl{\\{}a^{D-2}\Biggl{(}i\Bigl{[}\mbox{}^{\widehat{\mu\rho}}\Delta^{\nu}_{~{}\rho}\Bigr{]}(x;x)-\frac{1}{4}\eta^{\mu\nu}i\Bigl{[}\mbox{}^{\widehat{\rho\sigma}}\Delta_{\rho\sigma}\Bigr{]}(x;x)\Biggr{)}\partial_{\nu}\delta^{D}(x\\!-\\!x^{\prime})\Biggr{\\}}.$
(27)

It can be reduced to give,

$\displaystyle-
iM^{2}_{4}(x;x^{\prime})=i\kappa^{2}\partial_{\mu}\Biggl{\\{}a^{D-2}\Biggl{(}\eta^{\mu\nu}\Bigl{[}-\frac{(D\\!-\\!5)D}{4}\,i\Delta_{A}(x;x)+\frac{1}{2}\Bigl{(}\frac{D\\!-\\!4}{D\\!-\\!3}\Bigr{)}k\Bigr{]}$
(28) $\displaystyle\hskip
56.9055pt+\delta^{\mu}_{~{}0}\delta^{\nu}_{~{}0}\Bigl{[}Di\Delta_{A}(x;x)+\frac{(D\\!-\\!1)(D\\!-\\!4)}{(D\\!-\\!2)(D\\!-\\!3)}k\Bigr{]}\Biggr{)}\partial_{\nu}\delta^{D}(x\\!-\\!x^{\prime})\Biggr{\\}},\qquad$

where we used the coincidence limits of $i\Delta_{B}$ and $i\Delta_{C}$
inferred from (19).

###  2.3 The Primitive 3-Point Contribution

The left hand diagram of Figure 2 might be called $-iM^{2}_{3}(x;x^{\prime})$.
From expression (12) we see that it involves the product of two first
variations,

$\Bigl{[}\frac{i\delta
S[g,\phi]}{\delta\phi(x)}\Bigr{]}_{h\phi}=i\partial_{\mu}\Bigl{[}\sqrt{-g}\,g^{\mu\nu}\partial_{\nu}\phi\Bigr{]}_{h\phi}=-i\kappa\partial_{\mu}\Bigl{[}a^{D-2}\widehat{h}^{\mu\nu}\partial_{\nu}\phi\Bigr{]}\;.$
(29)

The 3-point contribution involves the twice trace-reversed propagator (24),

$-iM^{2}_{3}(x;x^{\prime})=-\kappa^{2}\partial_{\mu}\partial^{\prime}_{\rho}\Bigl{\\{}(aa^{\prime})^{D-2}i\Bigl{[}\mbox{}^{\widehat{\mu\nu}}\Delta^{\widehat{\rho\sigma}}\Bigr{]}(x;x^{\prime})\,\partial_{\nu}\partial^{\prime}_{\sigma}i\Delta_{A}(x;x^{\prime})\Bigr{\\}}.$
(30)

Expression (24) suggests a natural decomposition into four parts,

$\displaystyle-iM^{2}_{3A}$ $\displaystyle\\!\\!\\!\equiv\\!\\!\\!$
$\displaystyle-\kappa^{2}\partial\\!\cdot\\!\partial^{\prime}\Bigl{\\{}(aa^{\prime})^{D-2}i\Delta_{A}\\!\times\\!\partial\\!\cdot\\!\partial^{\prime}i\Delta_{A}\Bigr{\\}}\;,\qquad$
(31) $\displaystyle-iM^{2}_{3B}$ $\displaystyle\\!\\!\\!\equiv\\!\\!\\!$
$\displaystyle-\kappa^{2}\partial^{\mu}{\partial^{\prime}}^{\rho}\Bigl{\\{}(aa^{\prime})^{D-2}i\Delta_{A}\\!\times\\!(\partial_{\rho}\partial^{\prime}_{\mu}\\!-\\!\partial_{\mu}\partial^{\prime}_{\rho})i\Delta_{A}\Bigr{\\}}\;,\qquad$
(32) $\displaystyle-iM^{2}_{3C}$ $\displaystyle\\!\\!\\!\equiv\\!\\!\\!$
$\displaystyle-\kappa^{2}\partial^{\mu}{\partial^{\prime}}^{\rho}\Bigl{\\{}(aa^{\prime})^{D-2}(i\Delta_{A}\\!-\\!i\Delta_{B})$
(33) $\displaystyle\hskip
56.9055pt\times\Bigl{[}\overline{\eta}_{\mu\rho}\partial_{0}\partial^{\prime}_{0}\\!-\\!\delta^{0}_{~{}\mu}\overline{\partial}_{\rho}\partial^{\prime}_{0}\\!+\\!\delta^{0}_{~{}\rho}\overline{\partial}_{\mu}\partial_{0}\\!-\\!\delta^{0}_{~{}\mu}\delta^{0}_{~{}\rho}\nabla^{2}\Bigr{]}i\Delta_{A}\Bigr{\\}}\;,\qquad$
$\displaystyle-iM^{2}_{3D}$ $\displaystyle\\!\\!\\!\equiv\\!\\!\\!$
$\displaystyle
2\Bigl{(}\frac{D\\!-\\!2}{D\\!-\\!3}\Bigl{)}\kappa^{2}\partial_{0}\partial^{\prime}_{0}\Bigl{\\{}(aa^{\prime})^{D-2}(i\Delta_{A}\\!-\\!i\Delta_{C})\partial_{0}\partial_{0}^{\prime}i\Delta_{A}\Bigr{\\}}\;.\qquad$
(34)

The terms inside the curly brackets of expression (31) are quadratically
divergent, whereas the curly brackets of (32-34) are only logarithmically
divergent. This means we only need a few terms of $i\Delta_{A}(x;x^{\prime})$,

$\displaystyle
i\Delta_{A}(x;x^{\prime})=\frac{\Gamma(\frac{D}{2}\\!-\\!1)}{4\pi^{\frac{D}{2}}(aa^{\prime})^{\frac{D}{2}-1}}\Biggl{\\{}\frac{1}{\Delta
x^{D-2}}+\frac{D(D\\!-\\!2)}{8(D\\!-\\!4)}\frac{aa^{\prime}H^{2}}{\Delta
x^{D-4}}+\dots\Biggr{\\}}$ (35) $\displaystyle\hskip
156.49014pt+k\Biggl{\\{}-\pi{\rm
cot}\Bigl{(}\frac{D\pi}{2}\Bigr{)}+\ln(aa^{\prime})+\dots\Biggr{\\}}.\qquad$

The 4-step procedure for reducing expressions (31-34) is,

  1. 1.

Act the two inner derivatives on $i\Delta_{A}$;111Note that this can produce a
delta function when acting on the most singular term,
$\displaystyle\partial_{\alpha}\partial^{\prime}_{\beta}\Bigl{[}\frac{1}{\Delta
x^{D-2}}\Bigr{]}=\frac{\delta^{0}_{~{}\alpha}\delta^{0}_{~{}\beta}4\pi^{\frac{D}{2}}i\delta^{D}(x\\!-\\!x^{\prime})}{\Gamma(\frac{D}{2}\\!-\\!1)}+(D\\!-\\!2)\Bigl{[}\frac{\eta_{\alpha\beta}}{\Delta
x^{D}}-\frac{D\Delta x_{\alpha}\Delta x_{\beta}}{\Delta x^{D+2}}\Bigr{]}\;.$

  2. 2.

Multiply by the scalar propagators from the graviton propagator, and retain
only those terms which are nonzero in the unregulated limit;

  3. 3.

Extract derivatives from the quadratically divergent terms to reach a
logarithmically divergent form,

$\frac{1}{\Delta
x^{2D-2}}=\frac{\partial^{2}}{2(D\\!-\\!2)^{2}}\Bigl{[}\frac{1}{\Delta
x^{2D-4}}\Bigr{]}\qquad{\rm;and}$ (36)

  4. 4.

Localize the divergence and take the unregulated limit on the remainder using
the flat space propagator equation [41],

$\displaystyle\frac{1}{\Delta
x^{2D-4}}=\frac{\partial^{2}}{2(D\\!-\\!3)(D\\!-\\!4)}\Bigl{[}\frac{1}{\Delta
x^{2D-6}}\Bigr{]}\;,$ (39)
$\displaystyle=\frac{\mu^{D-4}4\pi^{\frac{D}{2}}i\delta^{D}(x\\!-\\!x^{\prime})}{2(D\\!-\\!3)(D\\!-\\!4)\Gamma(\frac{D}{2}\\!-\\!1)}+\frac{\partial^{2}}{2(D\\!-\\!3)(D\\!-\\!4)}\Bigl{[}\frac{1}{\Delta
x^{2D-6}}\\!-\\!\frac{\mu^{D-4}}{\Delta x^{D-2}}\Bigr{]},\qquad$
$\displaystyle=\frac{\mu^{D-4}4\pi^{\frac{D}{2}}i\delta^{D}(x\\!-\\!x^{\prime})}{2(D\\!-\\!3)(D\\!-\\!4)\Gamma(\frac{D}{2}\\!-\\!1)}-\frac{1}{4}\partial^{2}\Bigl{[}\frac{\ln(\mu^{2}\Delta
x^{2})}{\Delta x^{2}}\Bigr{]}+O(D\\!-\\!4)\;.$

Before applying the 4-step procedure on $-iM^{2}_{3B}(x;x^{\prime})$ it is
useful to expand the traces over $\mu$ and $\rho$,

$-iM^{2}_{3B}(x;x^{\prime})=-\kappa^{2}(\partial_{0}\\!+\\!\partial^{\prime}_{0})\partial_{i}\Bigl{\\{}(aa^{\prime})^{D-2}i\Delta_{A}(x;x^{\prime})(\partial_{0}\\!+\\!\partial^{\prime}_{0})\partial_{i}i\Delta_{A}(x;x^{\prime})\Bigr{\\}}\;.$
(40)

A similar expansion of $-iM^{2}_{3C}(x;x^{\prime})$ gives,

$\displaystyle-
iM^{2}_{C}=\kappa^{2}\nabla^{2}\Bigl{\\{}(aa^{\prime})^{D-2}(i\Delta_{A}\\!-\\!i\Delta_{B})\partial_{0}\partial^{\prime}_{0}i\Delta_{A}\Bigr{\\}}+\kappa^{2}\partial_{0}\partial_{i}\Bigl{\\{}(aa^{\prime})^{D-2}$
(41)
$\displaystyle\times(i\Delta_{A}\\!-\\!i\Delta_{B})\partial^{\prime}_{0}\partial_{i}i\Delta_{A}\Bigr{\\}}+\kappa^{2}\partial^{\prime}_{0}\partial_{i}\Bigl{\\{}(aa^{\prime})^{D-2}(i\Delta_{A}\\!-\\!i\Delta_{B})\partial_{0}\partial_{i}i\Delta_{A}\Bigr{\\}}\qquad$
$\displaystyle\hskip
142.26378pt+\kappa^{2}\partial_{0}\partial^{\prime}_{0}\Bigl{\\{}(aa^{\prime})^{D-2}(i\Delta_{A}\\!-\\!i\Delta_{B})\nabla^{2}i\Delta_{A}\Bigr{\\}}.\qquad$

Also note that each divergence is proportional to one of two constants,

$A_{0}\equiv\kappa^{2}k\pi{\rm cot}\Bigl{(}\frac{D\pi}{2}\Bigr{)}\qquad,\qquad
A_{1}\equiv\frac{\kappa^{2}H^{2}}{4\pi^{\frac{D}{2}}}\frac{\mu^{D-4}\Gamma(\frac{D}{2})}{(D\\!-\\!3)(D\\!-\\!4)}\;.$
(42)

Now apply the 4-step procedure to find,

$\displaystyle-
iM^{2}_{3A}(x;x^{\prime})=\Bigl{[}A_{0}+\frac{{}_{3}}{{}^{4}}A_{1}\Bigr{]}\partial^{\mu}\Bigl{[}a^{2}\partial_{\mu}i\delta^{D}(x\\!-\\!x^{\prime})\Bigr{]}+\frac{\kappa^{2}H^{2}\partial\\!\cdot\\!\partial^{\prime}}{64\pi^{4}}\Bigl{\\{}aa^{\prime}\partial_{0}^{2}\Bigl{[}\frac{1}{\Delta
x^{2}}\Bigr{]}$ (43)
$\displaystyle+\frac{{}_{3}}{{}^{2}}aa^{\prime}\partial^{2}\Bigl{[}\frac{\ln(\mu^{2}\Delta
x^{2})}{\Delta
x^{2}}\Bigr{]}+4a^{2}{a^{\prime}}^{2}H^{2}\Bigl{[}\frac{\ln(\frac{1}{4}H^{2}\Delta
x^{2})\\!+\\!1}{\Delta x^{2}}\Bigr{]}$ $\displaystyle\hskip
99.58464pt+\frac{{}_{1}}{{}^{2}}a^{2}{a^{\prime}}^{2}H^{2}\partial_{0}^{2}\Bigl{[}\ln^{2}(\frac{{}_{1}}{{}^{4}}H^{2}\Delta
x^{2})\\!+\\!3\ln(\frac{{}_{1}}{{}^{4}}H^{2}\Delta
x^{2})\Bigr{]}\Bigr{\\}},\qquad$ $\displaystyle-
iM^{2}_{3B}(x;x^{\prime})=\frac{{}_{1}}{{}^{2}}A_{1}\partial_{i}\Bigl{[}a^{2}\partial_{i}i\delta^{D}(x\\!-\\!x^{\prime})\Bigr{]}-\frac{\kappa^{2}H^{2}\nabla^{2}}{64\pi^{4}}\Bigl{\\{}aa^{\prime}\partial^{2}\Bigl{[}\frac{\ln(\mu^{2}\Delta
x^{2})}{\Delta x^{2}}\Bigr{]}$ (44)
$\displaystyle+12a^{2}{a^{\prime}}^{2}H^{2}\Bigl{[}\frac{\ln(\frac{1}{4}H^{2}\Delta
x^{2})\\!+\\!\frac{3}{2}}{\Delta
x^{2}}\Bigr{]}+\frac{{}_{1}}{{}^{2}}a^{2}{a^{\prime}}^{2}H^{2}\Delta\eta^{2}\partial^{2}\Bigl{[}\frac{\ln(\mu^{2}\Delta
x^{2})}{\Delta x^{2}}\Bigr{]}$ $\displaystyle\hskip
156.49014pt+4a^{3}{a^{\prime}}^{3}H^{4}\Delta\eta^{2}\Bigl{[}\frac{\ln(\frac{1}{4}H^{2}\Delta
x^{2})\\!+\\!\frac{3}{2}}{\Delta x^{2}}\Bigr{]}\Bigr{\\}},\qquad$
$\displaystyle-
iM^{2}_{3C}(x;x^{\prime})=A_{1}\Bigl{\\{}-\frac{{}_{1}}{{}^{4}}\partial_{i}\Bigl{[}a^{2}\partial_{i}i\delta^{D}(x\\!-\\!x^{\prime})\Bigr{]}+\frac{{}_{1}}{{}^{4}}(D\\!-\\!1)\partial_{0}\Bigl{[}a^{2}\partial_{0}i\delta^{D}(x\\!-\\!x^{\prime})\Bigr{]}\Bigr{\\}}$
(45)
$\displaystyle+\frac{\kappa^{2}H^{2}\nabla^{2}}{64\pi^{4}}\Bigl{\\{}\frac{{}_{1}}{{}^{2}}aa^{\prime}\partial^{2}\Bigl{[}\frac{\ln(\mu^{2}\Delta
x^{2})}{\Delta
x^{2}}\Bigr{]}+6aa^{\prime}\partial_{0}^{2}\Bigl{[}\frac{\ln(\frac{1}{4}H^{2}\Delta
x^{2})\\!+\\!2}{\Delta x^{2}}\Bigr{]}$
$\displaystyle+\frac{{}_{1}}{{}^{2}}a^{2}{a^{\prime}}^{2}H^{2}\partial^{2}\Bigl{[}\ln^{2}(\frac{{}_{1}}{{}^{4}}H^{2}\Delta
x^{2})\\!+\\!\frac{{}_{1}}{{}^{2}}\ln(\frac{{}_{1}}{{}^{4}}H^{2}\Delta
x^{2})\Bigr{]}\\!-\\!a^{2}{a^{\prime}}^{2}H^{2}\partial_{0}^{2}\Bigl{[}\ln^{2}(\frac{{}_{1}}{{}^{4}}H^{2}\Delta
x^{2})$ $\displaystyle\hskip
14.22636pt+\frac{{}_{7}}{{}^{2}}\ln(\frac{{}_{1}}{{}^{4}}H^{2}\Delta
x^{2})\Bigr{]}+a^{3}{a^{\prime}}^{3}H^{4}\Bigl{[}\ln^{2}(\frac{{}_{1}}{{}^{4}}H^{2}\Delta
x^{2})\\!+\\!3\ln(\frac{{}_{1}}{{}^{4}}H^{2}\Delta x^{2})\Bigr{]}$
$\displaystyle\hskip
71.13188pt+\frac{{}_{1}}{{}^{2}}a^{3}{a^{\prime}}^{3}H^{4}\partial_{0}^{2}\Bigl{(}\Delta
x^{2}\Bigl{[}\ln^{2}(\frac{{}_{1}}{{}^{4}}H^{2}\Delta
x^{2})\\!+\\!\ln(\frac{{}_{1}}{{}^{4}}H^{2}\Delta
x^{2})\\!-\\!1\Bigr{]}\Bigr{)}\Bigr{\\}}$
$\displaystyle+\frac{\kappa^{2}H^{2}\partial_{0}\partial_{0}^{\prime}}{64\pi^{4}}\Bigl{\\{}\frac{{}_{3}}{{}^{2}}aa^{\prime}\partial^{2}\Bigl{[}\frac{\ln(\mu^{2}\Delta
x^{2})}{\Delta
x^{2}}\Bigr{]}\\!-\\!2aa^{\prime}\nabla^{2}\Bigl{[}\frac{\ln(\frac{1}{4}H^{2}\Delta
x^{2})\\!+\\!2}{\Delta
x^{2}}\Bigr{]}\\!-\\!\frac{{}_{3}}{{}^{2}}a^{2}{a^{\prime}}^{2}H^{2}$
$\displaystyle\hskip
8.5359pt\times\partial^{2}\ln(\frac{{}_{1}}{{}^{4}}H^{2}\Delta
x^{2})\\!+\\!\frac{{}_{1}}{{}^{2}}a^{2}{a^{\prime}}^{2}H^{2}\nabla^{2}\Bigl{[}\ln^{2}(\frac{{}_{1}}{{}^{4}}H^{2}\Delta
x^{2})\\!+\\!3\ln(\frac{{}_{1}}{{}^{4}}H^{2}\Delta
x^{2})\Bigr{]}\Bigr{\\}},\qquad$ $\displaystyle-
iM^{2}_{3D}(x;x^{\prime})=A_{1}\partial_{0}\Bigl{[}a^{2}\partial_{0}i\delta^{D}(x\\!-\\!x^{\prime})\Bigr{]}+\frac{\kappa^{2}H^{2}\partial_{0}\partial_{0}^{\prime}}{64\pi^{4}}\Bigl{\\{}2aa^{\prime}\partial^{2}\Bigl{[}\frac{\ln(\mu^{2}\Delta
x^{2})}{\Delta x^{2}}\Bigr{]}$ (46) $\displaystyle\hskip
199.16928pt+8aa^{\prime}\partial_{0}^{2}\Bigl{[}\frac{\ln(\frac{1}{4}H^{2}\Delta
x^{2})\\!+\\!2}{\Delta x^{2}}\Bigr{]}\Bigr{\\}}.\qquad$

###  2.4 Renormalization

Reducing the 4-point contribution (28) to the 3-point form (43-46) gives,

$-iM^{2}_{4}(x;x^{\prime})=\frac{{}_{1}}{{}^{4}}D(D\\!-\\!5)A_{0}\partial^{\mu}\Bigl{[}a^{2}\partial_{\mu}i\delta^{D}(x\\!-\\!x^{\prime})\Bigr{]}-DA_{0}\partial_{0}\Bigl{[}a^{2}\partial_{0}i\delta^{D}(x\\!-\\!x^{\prime})\Bigr{]}\;.$
(47)

We can now sum the divergent parts from expressions (43-47),

$\displaystyle-iM^{2}_{\rm
div}(x;x^{\prime})=\Bigl{[}\frac{{}_{1}}{{}^{4}}(D\\!-\\!1)(D\\!-\\!4)A_{0}\\!+\\!A_{1}\Bigr{]}\partial^{\mu}\Bigl{[}a^{2}\partial_{\mu}i\delta^{D}(x\\!-\\!x^{\prime})\Bigr{]}$
(48) $\displaystyle\hskip
113.81102pt+\Bigl{[}-DA_{0}\\!+\\!\frac{{}_{1}}{{}^{4}}(D\\!+\\!4)A_{1}\Bigr{]}\partial_{0}\Bigl{[}a^{2}\partial_{0}i\delta^{D}(x\\!-\\!x^{\prime})\Bigr{]}\;.\qquad$

Recall that $A_{0}$ and $A_{1}$ were defined in (42).

The 1-loop divergences (48) are canceled by three counterterms,

$\Delta\mathcal{L}=-\frac{1}{2}\alpha_{1}\kern 1.0pt\vbox{\hrule
height=1.2pt\hbox{\vrule width=1.2pt\hskip 3.0pt\vbox{\vskip 6.0pt}\hskip
3.0pt\vrule width=0.6pt}\hrule height=0.6pt}\kern 1.0pt\phi\kern
1.0pt\vbox{\hrule height=1.2pt\hbox{\vrule width=1.2pt\hskip 3.0pt\vbox{\vskip
6.0pt}\hskip 3.0pt\vrule width=0.6pt}\hrule height=0.6pt}\kern
1.0pt\phi\sqrt{-g}-\frac{1}{2}\alpha_{2}R\partial_{\mu}\phi\partial_{\nu}\phi
g^{\mu\nu}\sqrt{-g}-\frac{1}{2}\alpha_{3}R\partial_{0}\phi\partial_{0}\phi
g^{00}\sqrt{-g}\;.$ (49)

Hence the final diagram of Figure 2 is,

$\displaystyle-iM^{2}_{\rm
ctm}(x;x^{\prime})\equiv\Bigl{[}\frac{i\delta^{2}\Delta
S}{\delta\phi(x)\delta\phi(x^{\prime})}\Bigr{]}_{1}=-\alpha_{1}\partial^{\mu}{\partial^{\prime}}^{\rho}\Biggl{[}(aa^{\prime})^{D-2}\partial_{\mu}\partial^{\prime}_{\rho}\Bigl{(}\frac{i\delta^{D}(x\\!-\\!x^{\prime})}{a^{D}}\Bigr{)}\Biggr{]}$
(50) $\displaystyle\hskip
42.67912pt+\alpha_{2}\partial^{\mu}\Bigl{[}Ra^{D-2}\partial_{\mu}i\delta^{D}(x\\!-\\!x^{\prime})\Bigr{]}-\alpha_{3}\partial_{0}\Bigl{[}Ra^{D-2}\partial_{0}i\delta^{D}(x\\!-\\!x^{\prime})\Bigr{]}\;,\qquad$

where the Ricci scalar is $R=D(D-1)H^{2}$. Comparison between expressions (48)
and (50) implies,

$\displaystyle\alpha_{1}$ $\displaystyle\\!\\!\\!=\\!\\!\\!$ $\displaystyle
0\;,$ (51) $\displaystyle\alpha_{2}R$ $\displaystyle\\!\\!\\!=\\!\\!\\!$
$\displaystyle-\frac{\kappa^{2}H^{2}\mu^{D-4}}{4\pi^{\frac{D}{2}}}\Bigl{\\{}\frac{\Gamma(D)}{\Gamma(\frac{D}{2})}\frac{(D\\!-\\!4)}{16}\pi{\rm
cot}\Bigl{(}\frac{D\pi}{2}\Bigr{)}+\frac{\Gamma(\frac{D}{2})}{(D\\!-\\!3)(D\\!-\\!4)}\Bigr{\\}},\qquad$
(52) $\displaystyle\alpha_{3}R$ $\displaystyle\\!\\!\\!=\\!\\!\\!$
$\displaystyle-\frac{\kappa^{2}H^{2}\mu^{D-4}}{4\pi^{\frac{D}{2}}}\Bigl{\\{}\frac{\Gamma(D\\!-\\!1)}{\Gamma(\frac{D}{2})}\frac{D}{4}\pi{\rm
cot}\Bigl{(}\frac{D\pi}{2}\Bigr{)}-\frac{(D\\!+\\!4)\Gamma(\frac{D}{2})}{4(D\\!-\\!3)(D\\!-\\!4)}\Bigr{\\}}.\qquad$
(53)

The vanishing of $\alpha_{1}$ is an artifact of $\alpha=\beta=1$ gauge in the
flat space limit (3). The counterterms proportional to $\alpha_{2}$ and
$\alpha_{3}$ vanish in the flat space limit and their potential gauge
dependence is not known.

Combining the primitive divergence with the counterterms and taking the
unregulated limit gives,

$\displaystyle-iM^{2}_{\rm div}(x;x^{\prime})-iM^{2}_{\rm
ctm}(x;x^{\prime})=-\frac{\kappa^{2}H^{2}}{4\pi^{2}}\partial^{\mu}\Bigl{[}a^{2}\ln(a)\partial_{\mu}i\delta^{4}(x\\!-\\!x^{\prime})\Bigr{]}$
(54) $\displaystyle\hskip
85.35826pt+\frac{\kappa^{2}H^{2}}{2\pi^{2}}\partial_{0}\Bigl{[}a^{2}\ln\Bigl{(}\frac{4\mu^{2}a}{H^{2}}\Bigr{)}\partial_{0}i\delta^{4}(x\\!-\\!x^{\prime})\Bigr{]}+O(D\\!-\\!4)\;.\qquad$

The renormalized self-mass comes from adding these local terms to the nonlocal
parts of (43-46), and then simplifying the sum,

$\displaystyle-iM^{2}_{\rm
ren}(x;x^{\prime})=-\frac{\kappa^{2}H^{2}}{4\pi^{2}}\partial^{\mu}\Bigl{[}a^{2}\ln(a)\partial_{\mu}i\delta^{4}(x\\!-\\!x^{\prime})\Bigr{]}$
(55)
$\displaystyle+\frac{\kappa^{2}H^{2}}{2\pi^{2}}\partial_{0}\Bigl{[}a^{2}\ln\Bigl{(}\frac{4\mu^{2}a}{H^{2}}\Bigr{)}\partial_{0}i\delta^{4}(x\\!-\\!x^{\prime})\Bigr{]}+\frac{\kappa^{2}H^{2}\partial_{0}\partial_{0}^{\prime}}{64\pi^{4}}\Biggr{\\{}aa^{\prime}\partial_{0}\partial_{0}^{\prime}\Bigl{[}\frac{1}{\Delta
x^{2}}\Bigr{]}$
$\displaystyle+2aa^{\prime}(\partial_{0}\partial_{0}^{\prime}\\!+\\!\nabla^{2})\Bigl{[}\frac{\ln(\mu^{2}\Delta
x^{2})}{\Delta
x^{2}}\Bigr{]}-2aa^{\prime}(4\partial_{0}\partial_{0}^{\prime}\\!+\\!\nabla^{2})\Bigl{[}\frac{\ln(\frac{1}{4}H^{2}\Delta
x^{2})\\!+\\!2}{\Delta x^{2}}\Bigr{]}\Biggr{\\}}$
$\displaystyle+\frac{\kappa^{2}H^{2}\nabla^{2}}{64\pi^{4}}\Biggl{\\{}-2aa^{\prime}(\partial_{0}\partial_{0}^{\prime}\\!+\\!\nabla^{2})\Bigl{[}\frac{\ln(\mu^{2}\Delta
x^{2})}{\Delta
x^{2}}\Bigr{]}-6aa^{\prime}\partial_{0}\partial_{0}^{\prime}\Bigl{[}\frac{\ln(\frac{1}{4}H^{2}\Delta
x^{2})\\!+\\!\frac{11}{6}}{\Delta x^{2}}\Bigr{]}$
$\displaystyle+\frac{{}_{1}}{{}^{4}}a^{2}{a^{\prime}}^{2}H^{2}(\partial_{0}\partial_{0}^{\prime}\\!-\\!\nabla^{2})\ln(\mu^{2}\Delta
x^{2})+\frac{{}_{15}}{{}^{4}}a^{2}{a^{\prime}}^{2}H^{2}\partial_{0}\partial_{0}^{\prime}\ln(\frac{{}_{1}}{{}^{4}}H^{2}\Delta
x^{2})$ $\displaystyle\hskip
85.35826pt-a^{2}{a^{\prime}}^{2}H^{2}\nabla^{2}\Bigl{[}\frac{{}_{3}}{{}^{2}}\ln^{2}(\frac{{}_{1}}{{}^{4}}H^{2}\Delta
x^{2})+\frac{{}_{5}}{{}^{4}}\ln(\frac{{}_{1}}{{}^{4}}H^{2}\Delta
x^{2})\Bigr{]}\Biggr{\\}}.\qquad$

##  3 The Linearized Effective Field Equation

The purpose of this section is to use the renormalized self-mass (55) to
quantum-correct the linearized effective field equation and then solve this
equation for scalar radiation and for the scalar exchange potential. We begin
by explaining the Schwinger-Keldysh formalism that is used to produce a causal
and real effective field equation. The equation is then solved perturbatively,
first for scalar radiation and then for the exchange potential. The section
closes by using the renormalization group to explain the latter.

###  3.1 Schwinger-Keldysh Formalism

The linearized effective field equation is,

$\sqrt{-g}\kern 1.0pt\vbox{\hrule height=1.2pt\hbox{\vrule width=1.2pt\hskip
3.0pt\vbox{\vskip 6.0pt}\hskip 3.0pt\vrule width=0.6pt}\hrule
height=0.6pt}\kern
1.0pt\phi(x)=\partial^{\mu}\Bigl{[}a^{2}\partial_{\mu}\phi(x)\Bigr{]}\equiv\mathcal{D}\phi(x)=J(x)+\int\\!\\!d^{4}x^{\prime}M^{2}(x;x^{\prime})\phi(x^{\prime})\;,$
(56)

where $J(x)$ is the source. Substituting expression (55) for the self-mass
results in an equation with three peculiar properties:

  * •

It isn’t local because $M^{2}_{\rm ren}(x;x^{\prime})$ fails to vanish for
${x^{\prime}}^{\mu}\neq x^{\mu}$;

  * •

It isn’t causal because $M^{2}_{\rm ren}(x;x^{\prime})$ fails to vanish for
${x^{\prime}}^{\mu}$ outside the past light-cone of $x^{\mu}$; and

  * •

It isn’t real because $M^{2}_{\rm ren}(x;x^{\prime})$ has a nonzero imaginary
part.

Effective field equations are unavoidably nonlocal but the other two
properties derive from $-iM^{2}_{\rm ren}(x;x^{\prime})$ representing an in-
out amplitude rather than a true expectation value. Of course that is what the
Feynman rules produce, and it is exactly the right thing for scattering
amplitudes. However, the “in” and “out” vacua disagree due to the very
cosmological particle production (1) whose effect we seek to study, and
causality precludes the $S$-matrix from being an observable. It is therefore
more sensible to study the evolution of the expectation value of $\phi(x)$ in
the presence of a state which was empty in the distant past. The Schwinger-
Keldysh formalism provides a diagrammatic procedure for computing this which
is almost as simple to use as the Feynman rules [43, 44, 45, 46, 47]. This
expectation value obeys the Schwinger-Keldysh effective field equations, which
are both causal and real [48, 49, 50].

It is straightforward to convert the in-out effective field equations to the
in-in equations of the Schwinger-Keldysh formalism. The rules are [51]:

  * •

End points of lines in the diagrammatic formalism have $\pm$ polarizations,
resulting in four propagators and $2^{N}$ 1PI $N$-point functions;

  * •

The $++$ propagator is the same as the Feynman propagator, and the $--$
propagator is its complex conjugate;

  * •

The $+-$ and $-+$ propagators are homogeneous solutions of the propagator
equation, which are obtained from the Feynman propagator by changing the
$i\epsilon$ in the conformal coordinate interval from (13) to,

$\displaystyle\Delta x^{2}_{+-}$ $\displaystyle\equiv$
$\displaystyle\|\vec{x}\\!-\\!\vec{x}^{\prime}\|^{2}-(\eta\\!-\\!\eta^{\prime}\\!+\\!i\epsilon)^{2}\;,$
(57) $\displaystyle\Delta x^{2}_{-+}$ $\displaystyle\equiv$
$\displaystyle\|\vec{x}\\!-\\!\vec{x}^{\prime}\|^{2}-(\eta\\!-\\!\eta^{\prime}\\!-\\!i\epsilon)^{2}\qquad{\rm;and}$
(58)

  * •

Vertices carry only a single polarity, so all their lines are either $+$ or
$-$, with the $+$ vertices being the same as those of the Feynman rules and
the $-$ vertices being their complex conjugates.

The term “$M^{2}(x;x^{\prime})$” in the Schwinger-Keldysh effective field
equation (56) is $M^{2}_{++}(x;x^{\prime})+M^{2}_{+-}(x;x^{\prime})$. It is
real because the the $\pm$ vertex at ${x^{\prime}}^{\mu}$ results in a
relative minus sign, and because $\Delta x^{2}_{+-}=\Delta x^{2}_{++}$ for
$\eta<\eta^{\prime}$ whereas $\Delta x^{2}_{+-}=(\Delta x^{2}_{++})^{*}$ for
$\eta>\eta^{\prime}$. To see causality one first eliminates inverse powers of
$\Delta x^{2}$,

$\displaystyle\frac{1}{\Delta x^{2}}$ $\displaystyle=$
$\displaystyle\frac{{}_{1}}{{}^{4}}\partial^{2}\ln(\mu^{2}\Delta x^{2})\;,$
(59) $\displaystyle\frac{\ln(\mu^{2}\Delta x^{2})}{\Delta x^{2}}$
$\displaystyle=$
$\displaystyle\frac{{}_{1}}{{}^{8}}\partial^{2}\Bigl{[}\ln^{2}(\mu^{2}\Delta
x^{2})-2\ln(\mu^{2}\Delta x^{2})\Bigr{]}\;.$ (60)

Now note that differences of powers of $++$ and $+-$ logarithms are
proportional to $\theta(\Delta\eta-\Delta r)$, where
$\Delta\eta\equiv\eta-\eta^{\prime}$ and $\Delta
r\equiv\|\vec{x}-\vec{x}^{\prime}\|$,

$\displaystyle\ln(\mu^{2}\Delta x^{2}_{++})-\ln(\mu^{2}\Delta x^{2}_{+-})$
$\displaystyle=$ $\displaystyle 2\pi i\theta(\Delta\eta\\!-\\!\Delta r)\;,$
(61) $\displaystyle\ln^{2}(\mu^{2}\Delta x^{2}_{++})-\ln^{2}(\mu^{2}\Delta
x^{2}_{+-})$ $\displaystyle=$ $\displaystyle 4\pi
i\theta(\Delta\eta\\!-\\!\Delta r)\ln[\mu^{2}(\Delta\eta^{2}\\!-\\!\Delta
r^{2})]\;.$ (62)

In converting expression (55) to Schwinger-Keldysh form we will employ the
notation $\Theta\equiv\theta(\Delta\eta-\Delta r)$ to achieve a more compact
form,

$\displaystyle M^{2}_{\rm
SK}(x;x^{\prime})=\frac{\kappa^{2}H^{2}}{4\pi^{2}}\Biggl{\\{}\partial^{\mu}\Bigl{[}a^{2}\ln(a)\partial_{\mu}\Bigr{]}\\!-\\!2\partial_{0}\Bigl{[}a^{2}\ln\Bigl{(}\frac{4\mu^{2}a}{H^{2}}\Bigr{)}\partial_{0}\Bigr{]}\Biggr{\\}}\delta^{4}(x\\!-\\!x^{\prime})$
(63)
$\displaystyle+\frac{\kappa^{2}H^{2}\partial_{0}\partial^{\prime}_{0}}{128\pi^{3}}\Biggl{\\{}-2aa^{\prime}(\partial_{0}\partial^{\prime}_{0}\\!+\\!\nabla^{2})\partial^{2}\Bigl{[}\Theta\ln[\mu^{2}(\Delta\eta^{2}\\!-\\!\Delta
r^{2})]\Bigr{]}+9aa^{\prime}\partial_{0}\partial^{\prime}_{0}\partial^{2}\Theta$
$\displaystyle\hskip
56.9055pt+4aa^{\prime}\nabla^{2}\partial^{2}\Theta+2aa^{\prime}(4\partial_{0}\partial^{\prime}_{0}\\!+\\!\nabla^{2})\partial^{2}\Bigl{[}\Theta\ln[\frac{{}_{1}}{{}^{4}}H^{2}(\Delta\eta^{2}\\!-\\!\Delta
r^{2})]\Bigr{]}\Biggr{\\}}$
$\displaystyle+\frac{\kappa^{2}H^{2}\nabla^{2}}{128\pi^{3}}\Biggl{\\{}2aa^{\prime}(\partial_{0}\partial^{\prime}_{0}\\!+\\!\nabla^{2})\partial^{2}\Bigl{[}\Theta\ln[\mu^{2}(\Delta\eta^{2}\\!-\\!\Delta
r^{2})]\Bigr{]}+3aa^{\prime}\partial_{0}\partial^{\prime}_{0}\partial^{2}\Theta$
$\displaystyle-2aa^{\prime}\nabla^{2}\partial^{2}\Theta+6aa^{\prime}\partial_{0}\partial^{\prime}_{0}\partial^{2}\Bigl{[}\Theta\ln[\frac{{}_{1}}{{}^{4}}H^{2}(\Delta\eta^{2}\\!-\\!\Delta
r^{2})]\Bigr{]}-16a^{2}{a^{\prime}}^{2}H^{2}\partial_{0}\partial^{\prime}_{0}\Theta$
$\displaystyle\hskip
51.21504pt+6a^{2}{a^{\prime}}^{2}H^{2}\nabla^{2}\Theta+12a^{2}{a^{\prime}}^{2}H^{2}\nabla^{2}\Bigl{[}\Theta\ln[\frac{{}_{1}}{{}^{4}}H^{2}(\Delta\eta^{2}\\!-\\!\Delta
r^{2})]\Bigr{]}\Biggr{\\}}.\qquad$

###  3.2 The Scalar Mode Function

Scalar radiation corresponds to $J(x)=0$ and solutions take the form,

$\phi(x)=u(\eta,k)e^{i\vec{k}\cdot\vec{x}}\qquad,\qquad k\equiv\|\vec{k}\|\;.$
(64)

The spatial exponential can be factored out using translation invariance,

$\mathcal{D}u(\eta,k)\equiv-a^{2}\Bigl{[}\partial_{0}^{2}+2aH\partial_{0}+k^{2}\Bigr{]}u(\eta,k)=\int\\!\\!d^{4}x^{\prime}M^{2}_{\rm
SK}(x;x^{\prime})u(\eta^{\prime},k)e^{-i\vec{k}\cdot\Delta\vec{x}}.$ (65)

Here any spatial derivatives in the self-mass are replaced by
$\partial_{i}\longrightarrow ik_{i}$.

Because we only have the 1-loop self-mass, equation (65) must be solved
perturbatively ($u=u_{0}+u_{1}+\dots$) in powers of the loop-counting
parameter $\kappa^{2}\equiv 16\pi G$. The tree order solution is,

$u_{0}(\eta,k)=\frac{H}{\sqrt{2k^{3}}}\Bigl{[}1-\frac{ik}{aH}\Bigr{]}\exp\Bigl{[}\frac{ik}{aH}\Bigr{]}\xrightarrow{a\to\infty}\frac{H}{\sqrt{2k^{3}}}\Bigl{[}1+\frac{k^{2}}{2a^{2}H^{2}}+\dots\Bigr{]}\;,$
(66)

and it is useful to note,

$\partial_{0}u_{0}(\eta,k)=\frac{H}{\sqrt{2k^{3}}}\Bigl{[}-\frac{k^{2}}{aH}\Bigr{]}\exp\Bigl{[}\frac{ik}{aH}\Bigr{]}\quad\Longrightarrow\quad(\partial_{0}^{2}+k^{2})\Bigl{[}a\partial_{0}u_{0}(\eta,k)\Bigr{]}=0\;.$
(67)

Relation (67) means that the 2nd and 3rd lines of expression (63) for
$M^{2}_{\rm SK}$ make no contribution to the 1-loop correction,

$\displaystyle\mathcal{D}u_{1}(\eta,k)=\frac{\kappa^{2}H^{2}}{4\pi^{2}}\Bigl{\\{}-3a^{3}H\partial_{0}u_{0}(\eta,k)+2\ln\Bigl{(}\frac{4\mu^{2}a}{H^{2}}\Bigr{)}a^{2}k^{2}u_{0}(\eta,k)\Bigr{\\}}$
(68)
$\displaystyle-\frac{\kappa^{2}H^{4}k^{2}a}{64\pi^{3}}\Biggl{\\{}2(\partial_{0}^{2}\\!+\\!k^{2})\\!\int\\!\\!d^{4}x^{\prime}\Theta\ln[\mu^{2}(\Delta\eta^{2}\\!-\\!\Delta
r^{2})]{a^{\prime}}^{3}u_{0}(\eta^{\prime},k)e^{-i\vec{k}\cdot\Delta\vec{x}}$
$\displaystyle+(3\partial_{0}^{2}\\!-\\!2k^{2})\\!\\!\int\\!\\!d^{4}x^{\prime}\Theta{a^{\prime}}^{3}u_{0}(\eta^{\prime},k)e^{-i\vec{k}\cdot\Delta\vec{x}}\\!+\\!6\partial_{0}^{2}\\!\\!\int\\!\\!d^{4}x^{\prime}\Theta\ln[\frac{{}_{1}}{{}^{4}}H^{2}(\Delta\eta^{2}\\!-\\!\Delta
r^{2})]$ $\displaystyle\hskip
42.67912pt\times{a^{\prime}}^{3}u_{0}(\eta^{\prime},k)e^{-i\vec{k}\cdot\Delta\vec{x}}+a(8\partial_{0}^{2}\\!-\\!3k^{2})\\!\\!\int\\!\\!d^{4}x^{\prime}\Theta{a^{\prime}}^{2}u_{0}(\eta^{\prime},k)e^{-i\vec{k}\cdot\Delta\vec{x}}$
$\displaystyle\hskip
71.13188pt-6ak^{2}\\!\\!\int\\!\\!d^{4}x^{\prime}\Theta\ln[\frac{{}_{1}}{{}^{4}}H^{2}(\Delta\eta^{2}\\!-\\!\Delta
r^{2})]{a^{\prime}}^{2}u_{0}(\eta^{\prime},k)e^{-i\vec{k}\cdot\Delta\vec{x}}\Biggr{\\}}.\qquad$

In reaching this form we have also used,

$(\partial_{0}^{2}\\!+\\!k^{2})\Bigl{[}au_{0}(\eta,k)\Bigr{]}=2a^{3}H^{2}u_{0}(\eta,k)\;.$
(69)

Equation (68) has the general form,

$-a^{2}\Bigl{[}\partial_{0}^{2}+2aH\partial_{0}+k^{2}\Bigr{]}u_{1}(\eta,k)=S(\eta)\;.$
(70)

It is important to understand the relation between the asymptotic late time
form of the source $S(\eta)$ and the late time form it induces in
$u_{1}(\eta,k)$,

$\displaystyle S=a^{4}H^{2}\ln(a)\Longrightarrow
u_{1}\rightarrow-\frac{{}_{1}}{{}^{6}}\ln^{2}(a)$ , $\displaystyle
S=a^{4}H^{2}\Longrightarrow
u_{1}\rightarrow-\frac{{}_{1}}{{}^{3}}\ln(a)\;,\qquad$ (71) $\displaystyle
S=a^{3}H^{2}\ln(a)\Longrightarrow u_{1}\rightarrow+\frac{\ln(a)}{2a}$ ,
$\displaystyle S=a^{3}H^{2}\Longrightarrow
u_{1}\rightarrow+\frac{1}{2a}\;,\qquad$ (72) $\displaystyle
S=a^{2}H^{2}\ln(a)\Longrightarrow u_{1}\rightarrow+\frac{\ln(a)}{2a^{2}}$ ,
$\displaystyle S=a^{2}H^{2}\Longrightarrow
u_{1}\rightarrow+\frac{1}{2a^{2}}\;.\qquad$ (73)

One can see that the leading asymptotic form of the local source terms on the
first line of (68) is,

$\frac{\kappa^{2}H^{2}}{4\pi^{2}}\Bigl{\\{}-3a^{3}H\partial_{0}u_{0}+2\ln\Bigl{(}\frac{4\mu^{2}a}{H^{2}}\Bigr{)}a^{2}k^{2}u_{0}\Bigr{\\}}\longrightarrow\frac{\kappa^{2}H^{2}}{4\pi^{2}}\times
2a^{2}\ln(a)k^{2}u_{0}(0,k)\;.$ (74)

Because all the nonlocal terms contain at least one factor of $k^{2}$, we can
suppress higher powers of $k^{2}$, which carry factors of $1/a^{2}$. This
simplifies the integrations,

$\displaystyle-\frac{\kappa^{2}H^{4}k^{2}u_{0}(0,k)a}{64\pi^{3}}\Biggl{\\{}2\partial_{0}^{2}\\!\\!\int\\!\\!d^{4}x^{\prime}\Theta\ln[\mu^{2}(\Delta\eta^{2}\\!-\\!\Delta
r^{2})]{a^{\prime}}^{3}+3\partial_{0}^{2}\\!\\!\int\\!\\!d^{4}x^{\prime}\Theta{a^{\prime}}^{3}$
(75) $\displaystyle\hskip
85.35826pt+6\partial_{0}^{2}\\!\\!\int\\!\\!d^{4}x^{\prime}\Theta\ln[\frac{{}_{1}}{{}^{4}}H^{2}(\Delta\eta^{2}\\!-\\!\Delta
r^{2})]{a^{\prime}}^{3}+8a\partial_{0}^{2}\\!\\!\int\\!\\!d^{4}x^{\prime}\Theta{a^{\prime}}^{2}\Biggr{\\}}$
$\displaystyle\longrightarrow\frac{\kappa^{2}H^{2}}{4\pi^{2}}\times\Bigl{[}\ln\Bigl{(}\frac{H}{2\mu}\Bigr{)}\Bigr{]}\\!+\\!\frac{5}{4}\Bigr{]}a^{2}k^{2}u_{0}(0,k)\;.\qquad$

In view of (73), expressions (74) and (75) imply,

$u_{1}(\eta,k)\longrightarrow\frac{\kappa^{2}H^{2}}{4\pi^{2}}\\!\times\\!\frac{k^{2}\ln(a)}{a^{2}H^{2}}\\!\times\\!u_{0}(0,k)\;.$
(76)

Hence 1-loop graviton corrections do not change the constant freeze-in value
of the mode function (66), but they do slow down the rate of approach to this
constant.

###  3.3 The Response to a Point Source

The scalar exchange potential corresponds to $J(x)=Ka\delta^{3}(\vec{x})$. The
solution has the form $\Phi(\eta,r)$, so this system is fundamentally
2-dimensional, unlike the 1-dimensional problem of the mode function
$u(\eta,k)$. The tree order response is [52],

$\Phi_{0}(t,r)=\frac{KH}{4\pi}\Bigl{\\{}-\frac{1}{aHr}+\ln\Bigl{(}Hr\\!+\\!\frac{1}{a}\Bigr{)}\Bigr{\\}}\xrightarrow{a\to\infty}\frac{HK}{4\pi}\Bigl{\\{}\ln(Hr)-\frac{1}{2a^{2}H^{2}r^{2}}+\dots\Bigr{\\}}\;.$
(77)

Derivatives of $\Phi_{0}(\eta,r)$ are,

$\partial_{0}\Phi_{0}(\eta,r)=\frac{KH^{2}}{4\pi}\Bigl{\\{}\frac{1}{Hr}-\frac{1}{Hr\\!+\\!\frac{1}{a}}\Bigr{\\}}\quad,\quad\partial_{0}^{2}\Phi_{0}(\eta,r)=-\frac{KH^{3}}{4\pi}\frac{1}{(Hr\\!+\\!\frac{1}{a})^{2}}\;,$
(78)
$\nabla^{2}\Phi_{0}(\eta,r)=\frac{K\delta^{3}(\vec{x})}{a}+\frac{1}{a^{2}}\partial_{0}\Bigl{[}a^{2}\partial_{0}\Phi_{0}\Bigr{]}.$
(79)

In addition to $\mathcal{D}\Phi_{0}=Ka\delta^{3}(\vec{x})$, two useful
consequences are,

$\partial^{2}\Bigl{[}a\Phi_{0}\Bigr{]}=Ka\delta^{3}(\vec{x})+2a^{3}H^{2}\Phi_{0}\qquad,\qquad\partial^{2}\Bigl{[}a\partial_{0}\Phi_{0}\Bigr{]}=-HKa\delta^{3}(\vec{x})\;.$
(80)

The second identity of (80) can be used to perform the spatial integrations in
the 1-loop sources induced by the 2nd and 3rd lines of (63),

$\displaystyle\mathcal{D}\Phi_{1}(\eta,r)=\frac{\kappa^{2}H^{2}}{4\pi^{2}}\Bigl{\\{}Ka\ln(a)\delta^{3}(\vec{x})+a^{3}H\partial_{0}\Phi_{0}-2\ln\Bigl{(}\frac{4\mu^{2}a}{H^{2}}\Bigr{)}\partial_{0}\Bigl{[}a^{2}\partial_{0}\phi_{0}\Bigr{]}\Bigr{\\}}$
(81)
$\displaystyle+\frac{\kappa^{2}H^{3}K\partial_{0}}{128\pi^{3}}\Biggl{\\{}-2a\partial^{2}\\!\\!\int_{\eta_{i}}^{\eta-r}\\!\\!\\!\\!\\!\\!\\!\\!d\eta^{\prime}a^{\prime}\ln[\mu^{2}(\Delta\eta^{2}\\!-\\!r^{2})]+a(4\nabla^{2}\\!-\\!9\partial_{0}^{2})\\!\\!\int_{\eta_{i}}^{\eta-r}\\!\\!\\!\\!\\!\\!\\!\\!d\eta^{\prime}a^{\prime}$
$\displaystyle+2a(\nabla^{2}\\!-\\!4\partial_{0}^{2})\\!\\!\int_{\eta_{i}}^{\eta-r}\\!\\!\\!\\!\\!\\!\\!\\!d\eta^{\prime}a^{\prime}\ln[\frac{{}_{1}}{{}^{4}}H^{2}(\Delta\eta^{2}\\!-\\!r^{2})]\Biggr{\\}}+\frac{\kappa^{2}H^{2}\nabla^{2}}{128\pi^{3}}\Biggl{\\{}2a\partial^{2}\\!\\!\int\\!\\!d^{4}x^{\prime}\Theta$
$\displaystyle\hskip 25.6073pt\times\ln[\mu^{2}(\Delta\eta^{2}\\!-\\!\Delta
r^{2})]{\partial^{\prime}}^{2}[a^{\prime}\Phi_{0}(x^{\prime})]-a(2\nabla^{2}\\!+\\!3\partial_{0}^{2})\\!\\!\int\\!\\!d^{4}x^{\prime}\Theta{\partial^{\prime}}^{2}[a^{\prime}\Phi_{0}(x^{\prime})]$
$\displaystyle-6a\partial_{0}^{2}\\!\\!\int\\!\\!d^{4}x^{\prime}\Theta\ln[\frac{{}_{1}}{{}^{4}}H^{2}(\Delta\eta^{2}\\!-\\!\Delta
r^{2})]{\partial^{\prime}}^{2}[a^{\prime}\Phi_{0}(x^{\prime})]\\!+\\!a^{2}H^{2}(16\partial_{0}^{2}\\!+\\!6\nabla^{2})\\!\\!\int\\!\\!d^{4}x^{\prime}\Theta$
$\displaystyle\hskip
25.6073pt\times{a^{\prime}}^{2}\Phi_{0}(x^{\prime})+12a^{2}H^{2}\nabla^{2}\\!\\!\int\\!\\!d^{4}x^{\prime}\Theta\ln[\frac{{}_{1}}{{}^{4}}H^{2}(\Delta\eta^{2}\\!-\\!\Delta
r^{2})]{a^{\prime}}^{2}\Phi_{0}(x^{\prime})\Biggr{\\}}.\qquad$

Each of the source terms on the right hand side of (81) can be evaluated, at
least for late times, and then its contribution to $\Phi_{1}(\eta,r)$ can be
derived by integrating against the retarded Green’s function associated with
the differential operator
$\mathcal{D}\equiv\partial^{\mu}a^{2}\partial_{\mu}$,

$G_{\rm
ret}(x;x^{\prime})=-\frac{1}{4\pi}\Bigl{\\{}\frac{\delta(\Delta\eta\\!-\\!\Delta
r)}{aa^{\prime}\Delta r}+H^{2}\theta(\Delta\eta\\!-\\!\Delta r)\Bigr{\\}}\;.$
(82)

However, it turns out that only the first source term in equation (81) makes a
significant contribution at late times,

$\displaystyle\int\\!\\!d^{4}x^{\prime}G_{\rm
ret}(x;x^{\prime})\\!\times\\!\frac{\kappa^{2}H^{2}}{4\pi^{2}}Ka^{\prime}\ln(a^{\prime})\delta^{3}(\vec{x}^{\prime})$
(83) $\displaystyle\hskip
113.81102pt=-\frac{\kappa^{2}H^{3}K}{32\pi^{3}}\Bigl{\\{}\ln^{2}\Bigl{(}Hr\\!+\\!\frac{1}{a}\Bigr{)}-\frac{2\ln(Hr\\!+\\!\frac{1}{a})}{aHr}\Bigr{\\}}\;.\qquad$

The easiest way to see that the other source terms in equation (81) do not
contribute at late times is by changing the time variable to $a$ and the space
variable to $aHr$, and then extracting a factor of $a^{4}H^{2}$ from both
sides of equation (81). The left hand side becomes,

$\mathcal{D}=a^{4}H^{2}\Bigl{[}-a^{2}\frac{\partial^{2}}{\partial
a^{2}}-4a\frac{\partial}{\partial
a}+\frac{1}{a^{2}H^{2}}\frac{\partial^{2}}{\partial
r^{2}}+\frac{2}{a^{2}H^{2}r}\frac{\partial}{\partial r}\Bigr{]}\;.$ (84)

For an example of the right hand side, consider the second of the nonlocal
source terms,

$\displaystyle\frac{\kappa^{2}H^{3}K\partial_{0}}{128\pi^{3}}\Bigl{\\{}\\!a(4\nabla^{2}\\!-\\!9\partial_{0}^{2})\\!\\!\int_{\eta_{i}}^{\eta-r}\\!\\!\\!\\!\\!\\!\\!\\!d\eta^{\prime}a^{\prime}\\!\Bigr{\\}}=\frac{\kappa^{2}H^{2}K\partial_{0}}{128\pi^{3}}\Bigl{\\{}\\!a(4\nabla^{2}\\!-\\!9\partial_{0}^{2})\ln\Bigl{(}\\!\frac{1}{Hr\\!+\\!\frac{1}{a}}\\!\Bigr{)}\\!\Bigr{\\}},$
(86) $\displaystyle\hskip
19.91684pt=a^{4}H^{2}\\!\times\\!\frac{\kappa^{2}H^{3}K}{128\pi^{3}}\Bigl{\\{}-\frac{16}{aHr}\\!+\\!\frac{16}{aHr\\!+\\!1}\\!+\\!\frac{3}{(aHr\\!+\\!1)^{2}}\\!-\\!\frac{10}{(aHr\\!+\\!1)^{3}}\Bigr{\\}}.\qquad$

The part of expression (86) inside the curly brackets goes like $1/(aHr)^{2}$
at late times, which corresponds to a late time contribution to
$\Phi_{1}(\eta,r)$ of the same strength according to (84). The strongest
nonlocal source goes like $\ln(a)/aHr$. Hence the leading ($a\gg 1$, $aHr\gg
1$) form comes from (83),

$\Phi_{1}(t,r)\longrightarrow-\frac{\kappa^{2}H^{2}}{8\pi^{2}}\ln(Hr)\\!\times\\!\frac{HK}{4\pi}\ln(Hr)\;.$
(87)

###  3.4 Renormalization Group Explanation

The $h\partial\phi\partial\phi$ interaction of gravity with our scalar is very
similar to the $A\partial B\partial B$ interaction of a nonlinear sigma model
which was recently studied [53]. It was shown that the leading inflationary
logarithms of that model could all be explained by combining a variant of
Starobinsky’s stochastic formalism [54, 55], based on curvature-dependent
effective potentials, with a variant of the renormalization group, based on
curvature-dependent field strength renormalizations. The
$\phi\rightarrow\phi+{\rm constant}$ shift symmetry of our Lagrangian (9)
precludes there being any effective potential for $\phi$ but it does seem
possible to identify a curvature-dependent field strength renormalization
among the three 1-loop counterterms of expression (49),

$\Delta\mathcal{L}=-\frac{1}{2}\alpha_{1}\kern 1.0pt\vbox{\hrule
height=1.2pt\hbox{\vrule width=1.2pt\hskip 3.0pt\vbox{\vskip 6.0pt}\hskip
3.0pt\vrule width=0.6pt}\hrule height=0.6pt}\kern 1.0pt\phi\kern
1.0pt\vbox{\hrule height=1.2pt\hbox{\vrule width=1.2pt\hskip 3.0pt\vbox{\vskip
6.0pt}\hskip 3.0pt\vrule width=0.6pt}\hrule height=0.6pt}\kern
1.0pt\phi\sqrt{-g}-\frac{1}{2}\alpha_{2}R\partial_{\mu}\phi\partial_{\nu}\phi
g^{\mu\nu}\sqrt{-g}-\frac{1}{2}\alpha_{3}R\partial_{0}\phi\partial_{0}\phi
g^{00}\sqrt{-g}\;.$ (88)

This is the $\alpha_{2}$ counterterm, which can be viewed as a field strength
renormalization of the original Lagrangian (9) with $\delta Z=\alpha_{2}R$.
Just as for $A$-$B$ nonlinear sigma model of Ref. [53], the higher derivative
counterterm proportional to $\alpha_{1}$ plays no role because it cannot be
viewed as a renormalization of the bare Lagrangian. Nor can the noncovariant
counterterm proportional to $\alpha_{3}$, whose existence is partially due to
our use of the simple, de Sitter breaking gauge [39, 40], and partly to the
time-ordering of interactions which seems unavoidable in the Schwinger-Keldysh
formalism [56].

If we accept the $\alpha_{2}$ counterterm as a field strength renormalization,
and employ our result (52) for $\alpha_{2}R$, the associated $\gamma$ function
is,

$Z=1+\alpha_{2}R+O(\kappa^{4})\qquad\Longrightarrow\qquad\gamma\equiv\frac{\partial\ln(Z)}{\partial\ln(\mu^{2})}=-\frac{\kappa^{2}H^{2}}{8\pi^{2}}+O(\kappa^{4})\;.$
(89)

The exchange potential $\Phi(\eta,r)$ represents an integral of the 1PI
2-point function, so the Callan-Symanzik equation for it should read,

$\Bigl{[}\frac{\partial}{\partial\ln(\mu)}-2\gamma\Bigr{]}\Phi(t,r)=0\;.$ (90)

If we replace the scale parameter $\mu$ by $r$, it will be seen that equation
(90) exactly explains the leading 1-loop logarithm from the known tree order
result (77),

$\frac{\partial\Phi_{1}}{\partial\ln(r)}=-2\\!\times\\!\frac{\kappa^{2}H^{2}}{8\pi^{2}}\\!\times\\!\Phi_{0}\quad\Longrightarrow\quad\Phi_{1}(t,r)\longrightarrow-\frac{\kappa^{2}H^{3}K}{32\pi^{3}}\ln^{2}(Hr)\;.$
(91)

No similar explanation can be given for the late time correction (76) to the
mode function. This seems to be because $u_{1}(\eta,k)$ is not a leading
logarithm effect; indeed, it vanishes at late times. Replacing $\mu$ by $a$,
or any other time variable, is also problematic because most of the time
dependence of the mode function comes in the form of $k/aH$.

##  4 Conclusions

Our long term goal is to establish the reality of large loop corrections from
inflationary gravitons (1) by purging the effective field equations of gauge
dependence. The Introduction described a procedure which has already been
implemented on flat space background for massless, minimally coupled scalars
[33], and for electromagnetism [37]. We plan to generalize this procedure to
de Sitter background and, for that purpose, we have sought the simplest system
which exhibits large graviton loop corrections before the gauge purge.
Although conformally coupled scalars are very simple, neither their mode
function nor their exchange potential shows a large correction at 1-loop order
[21]. The next simplest system is the massless, minimally coupled scalar which
we analyzed in this paper. We found that there is no large correction to the
1-loop mode function (76), but the 1-loop exchange potential (87) does
experience such a correction. Our main conclusion is therefore that the
massless, minimally coupled scalar (9) is the system we have been seeking, and
its exchange potential is the proper thing to study.

Section 2 employed dimensional regularization to compute the fully
renormalized 1-loop graviton contribution to the scalar self-mass (55).
Although we discovered some small mistakes in a previous computation [14],
which do not alter the conclusion of that work that there are no growing
secular corrections to the mode function, our biggest improvement is the use
of a simple representation which is not burdened with cumbersome de Sitter
invariant inverse differential operators. In section 3 we used the result to
quantum-correct the linearized, Schwinger-Keldysh effective field equation
(56). Specializing this equation to scalar radiation gave relation (68) for
the 1-loop mode function, whose asymptotic late time solution is (76). We find
no correction to the freeze-in amplitude of the mode function, but we do find
a large temporal logarithm correction to the approach to freeze-in.
Specializing (56) to the response to a point source gave relation (81) for the
1-loop exchange potential, whose asymptotic late time solution is (87). We
find a large spatial logarithm correction to the exchange potential. It is
significant that this correction derives entirely from the first of the source
terms on the right hand side of equation (81). Of course that means we can
focus on just this term when carrying out the gauge purge, which is a huge
simplification and justifies the effort put into this study.

In section 4.4 we used the renormalization group to explain the large
logarithm in the 1-loop exchange potential (87). This is significant for two
reasons:

  * •

It is the first time a large logarithm from inflationary gravitons has been
explained using the renormalization group; and

  * •

It ties the appearance of a large inflationary logarithm to the existence of
the $\alpha_{2}$ counterterm in (49).

The second point is relevant to the continuing controversy over the reality of
graviton-induced logarithms because it means that the absence of inflationary
logarithms at $\ell$-loop order requires that divergences proportional to
$(GR)^{\ell}\partial_{\mu}\phi\partial_{\nu}\phi g^{\mu\nu}\sqrt{-g}$ must
vanish, for all $\ell$, and in the absence of any symmetry argument. That
seems to strain credulity.

Acknowledgements

This work was partially supported by the Czech Science Foundation (GAČR) grant
20-28525S; by Taiwan MOST grant 108-2112-M-006-004; by the D-ITP consortium, a
program of the Netherlands Organization for Scientific Research (NWO) that is
funded by the Dutch Ministry of Education, Culture and Science (OCW); by NSF
grant PHY-1912484; and by the Institute for Fundamental Theory at the
University of Florida.

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#  $SO(4)$ Symmetry in Hydrogen Atom with Spin

Xing-Yan Fan  Theoretical Physics Division, Chern Institute of Mathematics and
LPMC, Nankai University, Tianjin 300071, People’s Republic of China  Xiang-Ru
Xie  School of Physics, Nankai University, Tianjin 300071, People’s Republic
of China  Sheng-Ming Li  School of Physics, Nankai University, Tianjin 300071,
People’s Republic of China  Jing-Ling Chen<EMAIL_ADDRESS>Theoretical
Physics Division, Chern Institute of Mathematics and LPMC, Nankai University,
Tianjin 300071, People’s Republic of China

(August 28, 2024)

###### Abstract

As the simplest atom in nature, the hydrogen atom has been explored thoroughly
from the perspective of non-relativistic quantum mechanics to relativistic
quantum mechanics. Among the research on hydrogen atom, its energy level is
the most basic, which can be obtained more conveniently predicated on the
$SO(4)$ symmetry than the wave-equation resolution. Moreover, “spin” is
another indispensable topic in quantum mechanics, appearing as an intrinsic
degree of freedom. In this work, we generalize the quantum Runge-Lenz vector
to a spin-dependent one, and then extract a novel Hamiltonian of hydrogen atom
with spin based on the requirement of $SO(4)$ symmetry. Furthermore, the
energy spectrum of hydrogen atom with spin potentials is also determined by
the remarkable approach of $SO(4)$ symmetry. Our findings extend the ground of
hydrogen atom, and may contribute to other complicated models based on
hydrogen atom.

##  I Introduction

After the birth of quantum mechanics (QM), the Schrödinger equation was used
to address the problem of hydrogen atom (HA), which underlies the solution to
other problem involving Coulomb interaction in QM, such as the problem of
hydrogen-like atom GrinerQMHAtom . Furthermore, the triumph of computing its
energy spectrum prompt physicists to research on other questions covering HA
using quantum theory, for example, the low-energy scattering through HAs
1962RMPBurke , the perturbations of HA 2010RMPEfstathiou , or the long-range
interaction of HAs at finite temperatures 2024PRAZalialiutdinov .

As is well known that symmetry is instrumental importantly in solving the HA
problem 2022A1CourSym . Explicitly, the spherical symmetry helps us decompose
the Schrödinger equation into the radial and angular parts, thus simplifying
the complicated partial differential equation to its ordinary partner;
certainly if one merely focuses on the energy spectrum of HA, a succinct
method can be adopted concerning $SO(4)$ symmetry 1988PauliSO4 ; GrinerSymSO4
; 1991Lange . Thereafter, this hidden symmetry drew more attention
1966RMPBANDER1 ; 1966RMPBANDER2 . For instance, it was used to explicate the
Stark states of HA as an irreducible representation of the $O(4)$ group
1967PPSHughes ; more its subgroup reductions were considered in distinct bases
1976SIAMJAM ; it was investigated even in virtue of the computer algebra
system 2021CPCSzriftgiser .

Another essential conception “spin” is also connected with the symmetry in QM
intimately. Under the circumstance of non-relativistic QM (NQM), the $SU(2)$
description for spin-1/2 system was revealed by Pauli 1927Pauli . Soon then
Dirac finished the relativistic quantum-mechanical treatment for electron
1928Dirac ; emphatically, the “spinors” as well as the Dirac matrices
(satisfying Clifford algebra) appeared for the first time, which were then
developed as a branch of geometry, dubbed spin geometry 1989SpinGeo . It is
worth mentioning that the spin-based symmetry is not just a theoretical
object, but an experimentally-observed property in condensed matter physics,
e.g., the spin-group symmetry in magnetic materials 2022PRXLiu .

For a more prevalent taste of physicists, combining different symmetry will
refresh their understanding of diverse models in QM. In this work, we advance
the study of HA by pondering $SO(4)$ symmetry and “spin” simultaneously, in
turn, we successfully attain a spin-dependent Hamiltonian for HA that still
possesses the $SO(4)$ symmetry.

The article is organized as follows: In Sec. II, we make a brief review of
$SO(4)$ symmetry by starting from the Hamiltonian of HA; In Sec. III, we
consider an inverse problem, i.e., how to determine the Hamiltonian of the
usual HA if one starts from the quantum Runge-Lenz (QRL) vector; In Sec. IV,
we generalize the quantum Runge-Lenz vector with spin, and extract a
Hamiltonian of HA with spin based on the requirement of $SO(4)$ symmetry; In
Sec. V, the energy spectrum of HA with spin potentials is determined by the
approach of $SO(4)$ symmetry; Conclusion and discussion are made in the Sec.
VI.

##  II Brief Review of $SO(4)$ Symmetry in Hydrogen Atom

The Hamiltonian of HA is given by GrinerQMHAtom

$H=\dfrac{\vec{p}^{\,2}}{2M}-\dfrac{\kappa}{r},$ (1)

where $\vec{p}=(p_{x},p_{y},p_{z})$ is the linear momentum operator;
$M=[m_{\rm e}m_{\rm p}/(m_{\rm e}+m_{\rm p})]$ is the reduced mass with
$m_{\rm e}$ and $m_{\rm p}$ representing the mass of electron and proton,
respectively; $\kappa=Ze^{2}$, and for HA, $Z=1$; $e$ depicts the absolute
value of the charge involving an electron; $r=|\vec{r}|$ indicates the norm of
coordinate operator $\vec{r}=(r_{x},r_{y},r_{z})$ regarding the relative
motion between electron and proton.

One may define the orbital angular momentum operator as

$\vec{\ell}=\vec{r}\times\vec{p},$ (2)

which satisfies the definition of angular momentum

$\vec{\ell}\times\vec{\ell}={\rm i}\hbar\,\vec{\ell},$ (3)

or in a commutator form as

$\left[{\ell}_{u},{\ell}_{v}\right]={\rm i}\hbar\,\epsilon_{uvw}{\ell}_{w},$
(4)

with $\epsilon_{uvw}$ being the Levi-Civita symbol, and
$\epsilon_{xyz}=\epsilon_{yzx}=\epsilon_{zxy}=+1$,
$\epsilon_{zyx}=\epsilon_{yxz}=\epsilon_{xzy}=-1$.

Due to

$\displaystyle\left[{\ell}_{u},r_{v}\right]={\rm
i}\hbar\,\epsilon_{uvw}r_{w},\;\;\;\;\left[{\ell}_{u},p_{v}\right]={\rm
i}\hbar\,\epsilon_{uvw}p_{w},$ (5)

it is easy to have

$\left[\vec{\ell},\,r\right]=0,\;\;\;\left[\vec{\ell},\,\vec{p}^{\,2}\right]=0,$
(6)

and then one proves that

$\left[H,\,\vec{\ell}\,\right]=0.$ (7)

Notably, Eq. (4) implies that a HA system possesses $SO(3)$ symmetry.

However, the HA can have a hidden symmetry of $SO(4)$ owing to the existence
of the quantum Runge-Lenz (QRL) vector, defined by 1988PauliSO4 ; GrinerSymSO4

$\displaystyle\vec{R}=\dfrac{1}{2M}\left(\vec{p}\times\vec{\ell}-\vec{\ell}\times\vec{p}\right)-\kappa\,\dfrac{\vec{r}}{r},$
(8)

which satisfies

$\displaystyle\vec{R}\times\vec{R}=\dfrac{-2H}{M}{\rm i}\hbar\,\vec{\ell},$
(9)

and

$\displaystyle\left[H,\vec{R}\right]=0,$ (10a)
$\displaystyle\vec{\ell}\cdot\vec{R}=\vec{R}\cdot\vec{\ell}=0,$ (10b)
$\displaystyle\vec{R}^{\,2}=\dfrac{2H}{M}\left(\vec{\ell}^{\,2}+\hbar^{2}\right)+\kappa^{2},$
(10c) $\displaystyle\left[{\ell}_{u},\,{R}_{v}\right]={\rm
i}\hbar\,\epsilon_{uvw}R_{w}.$ (10d)

Since $\\{\vec{\ell},\vec{R}\\}$ satisfy the following $SO(4)$ algebraic
relations (up to a re-scaling factor $\sqrt{M/(-2H)}$ multiplied together
$\vec{R}$)

$\displaystyle\left[{\ell}_{u},{\ell}_{v}\right]={\rm
i}\hbar\,\epsilon_{uvw}{\ell}_{w},$ (11a)
$\displaystyle\left[{\ell}_{u},{R}_{v}\right]={\rm
i}\hbar\,\epsilon_{uvw}{R}_{w},$ (11b)
$\displaystyle\left[{R}_{u},{R}_{v}\right]=\frac{-2H}{M}{\rm
i}\hbar\,\epsilon_{uvw}{\ell}_{w},$ (11c)

and thus the HA system possesses $SO(4)$ symmetry.

##  III An Inverse Problem

A HA is a real physical system existing in nature. The Hamiltonian (1) of a HA
can be recast to

$H=\dfrac{\vec{p}^{\,2}}{2M}+V_{\rm cp}(r),$ (12)

where

$V_{\rm cp}(r)=-\dfrac{\kappa}{r}$ (13)

represents the well-known Coulomb potential (CP). In the previous section, we
have considered a _forward problem_ : Given the Hamiltonian of a HA, can one
determine its underlying symmetry? The answer is positive. The underlying
symmetry is just $SO(4)$ symmetry, as shown in Eqs. (11a)-(11c).

In this section, let us consider an _inverse problem_ : Given the underlying
$SO(4)$ symmetry, can one construct the Hamiltonian of a HA (or derive the
CP)? The answer is still positive. The key point is the QRL vector; if one
focuses on Eq. (9), he will find that the Hamiltonian $H$ can be extracted
from $\vec{R}\times\vec{R}$ by requiring the result proportional to
$H\,\vec{\ell}$. The following is the procedure of construction.

Without lose of generality, let us define the QRL vector in the following
form:

$\displaystyle\vec{R}=\dfrac{1}{2M}\left(\vec{p}\times\vec{\ell}-\vec{\ell}\times\vec{p}\right)+f(r)\,\vec{r},$
(14)

where $f(r)$ is a certain function depending on $r$, and we only require

$\displaystyle[\vec{\ell},f(r)]=0.$ (15)

If one compares Eq. (8) and Eq. (14), one may notice that

$\displaystyle f(r)=-\dfrac{\kappa}{r}.$ (16)

However, suppose we do not know it in the beginning, thus temporarily treating
$f(r)$ as an unknown function.

By direct calculation, we easily have

$\displaystyle[{\ell}_{u},\ {\ell}_{v}]={\rm
i}\hbar\,\epsilon_{uvw}\,{\ell}_{w},$ (17) $\displaystyle[{\ell}_{u},\
{R}_{v}]={\rm i}\hbar\,\epsilon_{uvw}\,{R}_{w}.$ (18)

The underlying $SO(4)$ symmetry requires the following commutative relation

$\displaystyle[{R}_{u},\ {R}_{v}]\propto{\rm
i}\hbar\,\epsilon_{uvw}\,{\ell}_{w},$ (19)

i.e., $[{R}_{u},\ {R}_{v}]$ is required to be proportional to $\ell_{w}$, and
the proportional coefficient is related to the Hamiltonian $H$ of a HA. To
reach this purpose, let us compute the commutator $[{R}_{x},\ {R}_{y}]$, or
$\vec{R}\times\vec{R}$.

The result gives

$\displaystyle\vec{R}\times\vec{R}$ $\displaystyle=$ $\displaystyle-{\rm
i}\hbar\dfrac{2}{M}\biggl{(}\dfrac{\vec{p}^{\,2}}{2\,M}+\dfrac{1}{2}\Bigl{\\{}\big{[}\vec{\nabla}f(r)\big{]}\cdot\vec{r}+3\,f(r)\Bigr{\\}}\biggr{)}\vec{\ell}$
(20) $\displaystyle=$ $\displaystyle\dfrac{-2H}{M}{\rm i}\hbar\,\vec{\ell},$

based on which we can immediately extract a Hamiltonian as

$\displaystyle H=\dfrac{\vec{p}^{\,2}}{2M}+{\cal V}(r),$ (21)

with the potential function

$\displaystyle{\cal V}(r)$ $\displaystyle=$
$\displaystyle\dfrac{1}{2}\Bigl{\\{}\big{[}\vec{\nabla}f(r)\big{]}\cdot\vec{r}+3\,f(r)\Bigr{\\}}$
(22) $\displaystyle=$
$\displaystyle\dfrac{1}{2}\Bigl{\\{}rf^{\prime}(r)+3\,f(r)\Bigr{\\}}.$

It is easy to prove that

$\displaystyle\left[\vec{\ell},H\right]=0.$ (23)

Furthermore, we use the following condition

$\displaystyle\left[\vec{R},H\right]=0$ (24)

to determine the function $f(r)$, for $\vec{R}$ is the conservation quantity
of the Hamiltonian system. One can have

$\displaystyle[\vec{R},H]$ $\displaystyle=$ $\displaystyle\dfrac{-{\rm
i}\hbar}{M}\left[\vec{\nabla}\left[{\cal
V}(r)-f(r)\right]\cdot\vec{p}\right]\vec{r}$ (25) $\displaystyle+\dfrac{{\rm
i}\hbar}{M}\left[\left(\vec{\nabla}{\cal
V}(r)\right)\cdot\vec{r}+f(r)\right]\vec{p}$
$\displaystyle+\dfrac{\hbar^{2}}{2M}\left(\vec{\nabla}^{2}f(r)\right)\vec{r},$

and thus

$\displaystyle\vec{\nabla}\left[{\cal V}(r)-f(r)\right]=0,$ (26a)
$\displaystyle\left(\vec{\nabla}{\cal V}(r)\right)\cdot\vec{r}+f(r)=0,$ (26b)
$\displaystyle\vec{\nabla}^{2}f(r)=0,$ (26c)

which lead to

$\displaystyle{\cal V}(r)=f(r)=\dfrac{C}{r}.$ (27)

Consequently, by choosing $C=-\kappa$ we have successfully constructed the
Hamiltonian of a HA by starting from the QRL vector.

This significant finding of the inverse problem is very helpful for studying a
more complicated Hamiltonian under the constraint of $SO(4)$ symmetry. In the
next section, based on $SO(4)$ symmetry we will construct the Hamiltonian of
HA with spin.

##  IV Constructing the Hamiltonian of Hydrogen Atom with Spin

To construct the Hamiltonian of HA with spin, we need to generalize the
orbital angular momentum $\vec{\ell}$ and the spin-independent QPL vector
$\vec{R}$ to the spin-dependent ones, namely, $\vec{\ell}\Rightarrow\vec{J}$
and $\vec{R}\Rightarrow\vec{\mathcal{R}}$, where $\vec{J}$ and
$\vec{\mathcal{R}}$ depend on the spin angular momentum operator $\vec{S}$,
which satisfies the definition of angular momentum

$\vec{S}\times\vec{S}={\rm i}\hbar\,\vec{S},$ (28)

or in a commutator form as

$\left[{S}_{u},{S}_{v}\right]={\rm i}\hbar\,\epsilon_{uvw}{S}_{w}.$ (29)

Our main result is the following theorem:

######  Theorem 1.

Let

$\displaystyle\vec{J}=\vec{\ell}+\mu\vec{S},$ (30a) $\displaystyle\vec{\cal
R}=\dfrac{1}{2\,M}\big{(}\vec{\Pi}\times\vec{J}-\vec{J}\times\vec{\Pi}\big{)}+h(r)\,\vec{r},$
(30b)

and then they satisfy the $SO(4)$ commutation relations (up to a re-scaling
factor multiplied together $\vec{\cal R}$)

$\displaystyle[J_{u},\ J_{v}]={\rm i}\hbar\,\epsilon_{uvw}J_{w},$ (31a)
$\displaystyle[J_{u},\ {\cal R}_{v}]={\rm i}\hbar\,\epsilon_{uvw}{\cal
R}_{w},$ (31b) $\displaystyle[{\cal R}_{u},\ {\cal R}_{v}]\propto{\rm
i}\hbar\,\epsilon_{uvw}{\cal H}\,J_{w},$ (31c)

and

$\displaystyle[\vec{J},\,{\cal H}]=0,$ (32a) $\displaystyle[\vec{\cal
R},\,{\cal H}]=0.$ (32b)

Here the Hamiltonian of HA with spin reads

$\displaystyle{\cal H}=\dfrac{\vec{\Pi}^{2}}{2\,M}+{\cal V}(r),$ (33)

with

$\displaystyle\vec{\Pi}=\vec{p}-\vec{A},$ (34a)
$\displaystyle\vec{A}=\mu\dfrac{(\vec{r}\times\vec{S})}{r^{2}},$ (34b)
$\displaystyle
h(r)=\dfrac{k_{1}}{r}+k_{2}\,\mu\,\dfrac{\vec{r}\cdot\vec{S}}{r^{2}},$ (34c)
$\displaystyle{\cal
V}(r)=h(r)+\mu\dfrac{1}{2\,M}\dfrac{(\vec{r}\cdot\vec{S})^{2}}{r^{4}},$ (34d)

$k_{1}$ and $k_{2}$ are two constants, and $\mu=0,1$.

###### Proof.

We divide five steps to complete the proof.

_Step 1: Construction of $\vec{J}$ and Proving Eq. (31a)._—Since the orbital
angular momentum $\vec{\ell}$ and the spin operator $\vec{S}$ are defined in
the external and internal spaces respectively, they are mutually commutative.
If we define the “total” angular momentum operator as

$\displaystyle\vec{J}=\vec{\ell}+\vec{S},$ (35)

then one immediately proves Eq. (31a). Here we have designed $\vec{J}$ in the
form of Eq. (30a) with the parameter $\mu$, to treat the HAs with and without
spin in a unified way; namely, $\vec{J}$ reduces to $\vec{\ell}$ if $\mu=0$,
and becomes Eq. (35) if $\mu=1$.

_Step 2: Construction of $\vec{\cal R}$ and Proving Eq. (31b)._—Similar to
Ref. 2023FRSpinV , one can recast $\vec{J}$ to the following form

$\displaystyle\vec{J}$ $\displaystyle=$ $\displaystyle\vec{\ell}+\mu\vec{S}$
(36) $\displaystyle=$
$\displaystyle\vec{r}\times(\vec{p}-\vec{A})+\mu\,\vec{e}_{r}S_{r}$
$\displaystyle=$ $\displaystyle\vec{r}\times\vec{\Pi}+\mu\,\vec{e}_{r}S_{r};$

then one extracts “canonical” momentum $\vec{\Pi}$ of the electron in HA as in
Eq. (34a), and the spin vector potential $\vec{A}$ induced by spin operator
$\vec{S}$ of the proton in HA as Eq. (34b). Here the operator

$\displaystyle\hat{S}_{r}=\hat{r}\cdot\vec{S},$ (37)

and $\hat{r}=\vec{e}_{r}=\vec{r}/r$.

After performing the following replacement

$\displaystyle\vec{p}\Rightarrow\vec{\Pi},\quad\vec{\ell}\Rightarrow\vec{J},\quad
f(r)\Rightarrow h(r),$ (38)

the QPL vector $\vec{R}$ is turned into the generalized QPL vector $\vec{\cal
R}$ as shown in Eq. (30b), with $h(r)$ being a function of $r$ and
$(\vec{r}\cdot\vec{S})$ satisfying the commutation relation

$\displaystyle[\vec{J},\,h(r)]=0.$ (39)

Due to the relations

$\displaystyle[J_{u},\ J_{v}]={\rm i}\hbar\,\epsilon_{uvw}J_{w},$ (40a)
$\displaystyle[J_{u},\ \Pi_{v}]={\rm i}\hbar\,\epsilon_{uvw}\Pi_{w},$ (40b)
$\displaystyle[J_{u},\ (\vec{\Pi}\times\vec{J})_{v}]={\rm
i}\hbar\,\epsilon_{uvw}(\vec{\Pi}\times\vec{J})_{w},$ (40c)
$\displaystyle[J_{u},\ (\vec{J}\times\vec{\Pi})_{v}]={\rm
i}\hbar\,\epsilon_{uvw}(\vec{J}\times\vec{\Pi})_{w},$ (40d)
$\displaystyle[J_{u},\ r_{v}]={\rm i}\hbar\,\epsilon_{uvw}r_{w},$ (40e)

one directly proves Eq. (31b).

_Step 3: Proving Eq. ( 31c)._—After careful deductions, one has

$\displaystyle\vec{\cal R}\times\vec{\cal R}$ $\displaystyle=$
$\displaystyle{\rm
i}\hbar\,\dfrac{-2}{M}\Bigg{\\{}\dfrac{\vec{\Pi}^{2}}{2\,M}+\dfrac{1}{2}\bigg{[}-\dfrac{1}{{\rm
i}\hbar}\big{[}\vec{\Pi},\
h(r)\big{]}\cdot\vec{r}+3\,h(r)+\dfrac{\mu}{M}\dfrac{(\vec{r}\cdot\vec{S})^{2}}{r^{4}}\bigg{]}\Bigg{\\}}\vec{J}-\dfrac{\mu}{M}\biggl{\\{}\big{[}\vec{\Pi},\
h(r)\big{]}-{\rm
i}\hbar\dfrac{\vec{r}}{r^{2}}h(r)\biggr{\\}}(\vec{r}\cdot\vec{S}).$ (41)

To fulfill the requirement of Eq. (31c), one must have

$\displaystyle{\cal H}=\dfrac{\vec{\Pi}^{2}}{2\,M}+{\cal V}(r),$ (42a)
$\displaystyle{\cal V}(r)=\dfrac{1}{2}\bigg{[}-\dfrac{1}{{\rm
i}\hbar}\big{[}\vec{\Pi},\
h(r)\big{]}\cdot\vec{r}+3\,h(r)+\dfrac{\mu}{M}\dfrac{(\vec{r}\cdot\vec{S})^{2}}{r^{4}}\bigg{]},$
(42b) $\displaystyle\dfrac{\mu}{M}\biggl{\\{}\big{[}\vec{\Pi},\
h(r)\big{]}-{\rm
i}\hbar\dfrac{\vec{r}}{r^{2}}h(r)\biggr{\\}}(\vec{r}\cdot\vec{S})=0.$ (42c)

For $\mu=0$, Eq. (41) reduces to Eq. (20), and thus Eq. (31c) is proved. For
$\mu=1$, from Eq. (42c) we obtain

$\displaystyle\big{[}\vec{\Pi},\ h(r)\big{]}-{\rm
i}\hbar\dfrac{\vec{r}}{r^{2}}h(r)=0.$ (43)

By setting

$\displaystyle h(r)=\xi_{1}(r)+\xi_{2}(r)\left(\vec{r}\cdot\vec{S}\right),$
(44)

one obtains the solution of $h(r)$ as shown in Eq. (34c) with
$\xi_{1}(r)=k_{1}/r$ and $\xi_{2}(r)=\mu k_{2}/r^{2}$, and hence Eq. (31c) is
proved. After substituting Eq. (42c) into Eq. (42b), one has the potential
function ${\cal V}$ as in Eq. (34d).

_Step 4: Proving Eq. ( 32a)._— Due to the relations

$\displaystyle[\vec{J},\,\vec{\Pi}^{2}]=0,$ (45a)
$\displaystyle[\vec{J},\,r]=0,$ (45b)
$\displaystyle[\vec{J},\,\vec{r}\cdot\vec{S}]=0,$ (45c)

one directly proves Eq. (32a).

_Step 5: Proving Eq. ( 32b)._— Due to Eq. (40a)-Eq. (40e), Eq. (45a)-Eq.
(45c), and the following relations

$\displaystyle[\Pi_{u},\ r_{v}]=-{\rm i}\hbar\,\delta_{uv},$ (46a)
$\displaystyle\bigg{[}\vec{\Pi},\ \dfrac{1}{r}\bigg{]}={\rm
i}\hbar\dfrac{\vec{r}}{r^{3}},$ (46b) $\displaystyle\bigg{[}\vec{\Pi},\
\dfrac{1}{r^{2}}\bigg{]}={\rm i}\hbar\dfrac{2\,\vec{r}}{r^{4}},$ (46c)
$\displaystyle[\Pi_{u},\,\Pi_{v}]={\rm i}\hbar\,\epsilon_{uvw}{\cal B}_{w},$
(46d) $\displaystyle\vec{\cal B}={\rm
i}\hbar\,\mu(\mu-2)\dfrac{\vec{r}\cdot\vec{S}}{r^{4}}\vec{r},$ (46e)
$\displaystyle[\vec{\Pi},\,\vec{\Pi}^{2}]={\rm
i}\hbar\big{[}(\vec{\Pi}\times\vec{\cal B})-(\vec{\cal
B}\times\vec{\Pi})\big{]},$ (46f)
$\displaystyle\vec{\Pi}\times\vec{J}+\vec{J}\times\vec{\Pi}=2{\rm
i}\hbar\,\vec{\Pi},$ (46g) $\displaystyle\bigg{[}\vec{\Pi},\
\dfrac{\vec{r}\cdot\vec{S}}{r^{2}}\bigg{]}={\rm
i}\hbar(\mu-1)\dfrac{\vec{S}}{r^{2}}+{\rm
i}\hbar(2-\mu)\frac{\vec{r}\cdot\vec{S}}{r^{4}}\vec{r},$ (46h)

with $\delta_{uv}$ denoting the Kronecker delta symbol, one can directly prove
Eq. (32b). ∎

##  V Energy Spectrum of Hydrogen Atom with Spin

In this section, we come to solve the energy spectrum of HA with spin based on
the approach of $SO(4)$ symmetry. We restrict our problem to the Hilbert space
of Hamiltonian ${\cal H}$’s eigenstates, where we can replace ${\cal H}$ by
its eigenvalue $E$, and then Eq. (31c) becomes

$\displaystyle[{\cal R}_{u},{\cal R}_{v}]=\dfrac{-2E}{M}{\rm
i}\hbar\,\epsilon_{uvw}\,J_{w}.$ (47)

If we define a re-scaling operator as

$\displaystyle\vec{\cal R}^{\prime}=\sqrt{\dfrac{M}{-2E}}\,\vec{\cal R},$ (48)

then $\vec{J}$ and $\vec{\cal R}^{\prime}$ fit the standard SO(4) algebra. Let
us further define two new quantities as

$\displaystyle\vec{W}=\dfrac{1}{2}(\vec{J}+\vec{\cal R}^{\prime}),$ (49a)
$\displaystyle\vec{K}=\dfrac{1}{2}(\vec{J}-\vec{\cal R}^{\prime}).$ (49b)

Then they satisfy the following commutators

$\displaystyle[W_{u},\ W_{v}]={\rm i}\hbar\,\epsilon_{uvw}W_{w},$ (50a)
$\displaystyle[K_{u},\ K_{v}]={\rm i}\hbar\,\epsilon_{uvw}K_{w},$ (50b)
$\displaystyle[W_{u},\ K_{v}]=0,$ (50c)

which implies that $\vec{W}$ and $\vec{K}$ construct two independent $SU(2)$
algebras GrinerSymSO4 ; hence the eigenvalues are given as
$\vec{W}^{2}=w(w+1)\hbar^{2}$, and $\vec{K}^{2}=k(k+1)\hbar^{2}$, with
$k,w=0,1/2,1,3/2,\dots$.

We can have

$\displaystyle\vec{W}^{2}$ $\displaystyle=$
$\displaystyle\dfrac{1}{4}(\vec{J}^{2}+\vec{{\cal
R}^{\prime}}^{2}+2\vec{J}\cdot\vec{{\cal R}^{\prime}}),$ (51a)
$\displaystyle\vec{K}^{2}$ $\displaystyle=$
$\displaystyle\dfrac{1}{4}(\vec{J}^{2}+\vec{{\cal
R}^{\prime}}^{2}-2\vec{J}\cdot\vec{{\cal R}^{\prime}}).$ (51b)

Afterwards, by using

$\displaystyle\vec{J}\cdot\vec{{\cal R}^{\prime}}=\vec{{\cal
R}^{\prime}}\cdot\vec{J}=\mu\sqrt{\dfrac{M}{-2E}}\,h(r)(\vec{r}\cdot\vec{S}),$
(52a) $\displaystyle\vec{{\cal
R}^{\prime}}^{2}=\left[\mu\dfrac{(\vec{r}\cdot\vec{S})^{2}}{r^{2}}-\vec{J}^{\,2}-\hbar^{2}\right]-\dfrac{M}{2E}\big{[}h(r)\big{]}^{2}r^{2},$
(52b)

we get

$\displaystyle\vec{W}^{2}+\vec{K}^{2}$ $\displaystyle=$
$\displaystyle\dfrac{1}{2}\bigg{[}{\mu}\dfrac{(\vec{r}\cdot\vec{S})^{2}}{r^{2}}-\hbar^{2}-\dfrac{M}{2E}\big{[}h(r)\big{]}^{2}r^{2}\bigg{]},$
(53a) $\displaystyle\vec{W}^{2}-\vec{K}^{2}$ $\displaystyle=$
$\displaystyle\mu\sqrt{\dfrac{M}{-2E}}\,h(r)\,(\vec{r}\cdot\vec{S}),$ (53b)

Furthermore, one can verify that

$\displaystyle\left[\hat{S}_{r},\ {\cal H}\right]=0,$ (54)

and therefore $\hat{S}_{r}$ is also a conserved quantity for the Hamiltonian
system, and we denote its eigenvalue as $s_{r}\hbar$. In the case of

$\displaystyle\vec{S}=\frac{\hbar}{2}\vec{\sigma}$ (55)

i.e., the spin-$1/2$ operator, one has $S_{r}^{2}=(\hbar^{2}/4)\openone$,
where $\vec{\sigma}$ and $\openone$ express the vector of Pauli matrices and
identity matrix respectively, which means that $s_{r}=\pm 1/2$.

After replacing the operators by their corresponding eigenvalues, Eq. (53a)
and Eq. (53b) become

$\displaystyle
w(w+1)\hbar^{2}+k(k+1)\hbar^{2}=\dfrac{1}{2}\bigg{[}{\mu}(s_{r}\hbar)^{2}-\hbar^{2}-\dfrac{M}{2E}(k_{1}+{\mu}\,k_{2}s_{r}\hbar)^{2}\bigg{]},$
(56a) $\displaystyle
w(w+1)\hbar^{2}-k(k+1)\hbar^{2}=\mu\sqrt{\dfrac{M}{-2E}}\,\big{(}k_{1}+{\mu}\,k_{2}s_{r}\hbar\big{)}s_{r}\hbar.$
(56b)

There is no solution for the energy $E$ if $k_{1}+{\mu}\,k_{2}s_{r}\hbar=0$.
For $k_{1}+k_{2}s_{r}\hbar\neq 0$, one has the energy spectrum of HA with spin
as

$\displaystyle E\equiv E_{n}$ $\displaystyle=$
$\displaystyle-\dfrac{M}{2\hbar^{2}}\dfrac{(k_{1}+\mu\,k_{2}s_{r}\hbar)^{2}}{\big{(}n\pm\mu\,s_{r}\big{)}^{2}},$
(57)

with $n=2\,w+1=1,2,3,....$ For $\mu=0$,

$\displaystyle E_{n}$ $\displaystyle=$
$\displaystyle-\dfrac{Mk_{1}^{2}}{2\hbar^{2}}\frac{1}{n^{2}},$ (58)

which reduces to the energy spectrum of the usual HA.

##  VI Conclusion and Discussion

To sum up, we have written down the spin-dependent QRL vector, constructed the
Hamiltonian of HA with spin under the requirement of $SO(4)$ symmetry. The
energy spectrum of the novel hamiltonian system is also determined by the
approach of $SO(4)$ symmetry, which reduces to the usual HA without
considering spin. Our computations extend the investigation of HA from the
perspective of spin potential, which may be applied to HA-based models in
atomic physics and condensed matter physics.

Considering the relativistic treatments for HA with $SO(4)$ approach have been
fully studied, such as 2008PRAChen ; 2008PRAZhang ; 2023APEremko , it is our
next object to explore the energy spectrum of HA with spin potential under the
framework of relativistic quantum mechanics. Certainly, the connection of HA
and harmonic oscillator is intimate (see e.g., 2020JPABars ), so our future
work also includes the quantum harmonic oscillator with spin potential.
Finally, we hope our consideration of HA with spin potential can be realized
experimentally, or can be achieved through quantum simulation 2014QSim , in
that a more complex model (i.e., Rydberg manifold 2022RydbQSim ) under $SO(4)$
symmetry has been simulated successfully.

###### Acknowledgements.

J.L.C. was supported by the National Natural Science Foundations of China
(Grant Nos. 12275136 and 12075001) and the 111 Project of B23045. X.Y.F. was
supported by the Nankai Zhide Foundations.

X.Y.F. and X.R.X. contributed equally to this work.

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