which one can check, for example, by using a specific representation of the
Dirac matrices. We then obtain

$\displaystyle S_{1}(\vec{x}_{1}-\vec{x}_{2};\tau_{1},\tau_{2})=$
$\displaystyle~{}a_{1}^{-\frac{d-1}{2}}a_{2}^{-\frac{d-1}{2}}\frac{\pi\sqrt{\tau_{1}\tau_{2}}}{4}\int\frac{d^{d-1}k}{(2\pi)^{d-1}}e^{i\vec{k}\cdot(\vec{x}_{1}-\vec{x}_{2})}$
$\displaystyle\times$
$\displaystyle\left[k\left(e^{i\pi\frac{\nu_{+}-\nu_{-}}{2}}H_{\nu_{+}}^{(1)}(z_{1})H_{\nu_{-}}^{(2)}(z_{2})\frac{1+\gamma^{0}}{2}-e^{-i\pi\frac{\nu_{+}-\nu_{-}}{2}}H_{\nu_{-}}^{(1)}(z_{1})H_{\nu_{+}}^{(2)}(z_{2})\frac{1-\gamma^{0}}{2}\right)\right.$
$\displaystyle~{}\left.-\vec{k}\cdot\vec{\gamma}\left(H_{\nu_{-}}^{(1)}(z_{1})H_{\nu_{-}}^{(2)}(z_{2})\frac{1+\gamma^{0}}{2}+H_{\nu_{+}}^{(1)}(z_{1})H_{\nu_{+}}^{(2)}(z_{2})\frac{1-\gamma^{0}}{2}\right)\right]\,,$
(C.50)

and

$\displaystyle S_{2}(\vec{x}_{1}-\vec{x}_{2};\tau_{1},\tau_{2})=$
$\displaystyle~{}a_{1}^{-\frac{d-1}{2}}a_{2}^{-\frac{d-1}{2}}\frac{\pi\sqrt{\tau_{1}\tau_{2}}}{4}\int\frac{d^{d-1}k}{(2\pi)^{d-1}}e^{i\vec{k}\cdot(\vec{x}_{1}-\vec{x}_{2})}$
$\displaystyle\times$
$\displaystyle\left[k\left(-e^{-i\pi\frac{\nu_{+}-\nu_{-}}{2}}H_{\nu_{+}}^{(2)}(z_{1})H_{\nu_{-}}^{(1)}(z_{2})\frac{1+\gamma^{0}}{2}+e^{i\pi\frac{\nu_{+}-\nu_{-}}{2}}H_{\nu_{-}}^{(2)}(z_{1})H_{\nu_{+}}^{(1)}(z_{2})\frac{1-\gamma^{0}}{2}\right)\right.$
$\displaystyle~{}\left.-\vec{k}\cdot\vec{\gamma}\left(H_{\nu_{-}}^{(2)}(z_{1})H_{\nu_{-}}^{(1)}(z_{2})\frac{1+\gamma^{0}}{2}+H_{\nu_{+}}^{(2)}(z_{1})H_{\nu_{+}}^{(1)}(z_{2})\frac{1-\gamma^{0}}{2}\right)\right]\,,$
(C.51)

where $a_{i}=a(\tau_{i})$ and $z_{i}=-k\tau_{i}$. Eq. (C.5) tells us that

$\displaystyle
ke^{i\pi\frac{\nu_{\lambda}-\nu_{-\lambda}}{2}}H_{\nu_{\lambda}}^{(1)}(z)=i\left(\frac{d}{d\tau}+\frac{\nu_{-\lambda}}{\tau}\right)H_{\nu_{-\lambda}}^{(1)}(z),\quad
ke^{-i\pi\frac{\nu_{\lambda}-\nu_{-\lambda}}{2}}H_{\nu_{\lambda}}^{(2)}(z)=-i\left(\frac{d}{d\tau}+\frac{\nu_{-\lambda}}{\tau}\right)H_{\nu_{-\lambda}}^{(2)}(z),$
(C.52)

and this allows us to extract the derivatives as

$\displaystyle S_{1}(\vec{x}_{1}-\vec{x}_{2};\tau_{1},\tau_{2})$
$\displaystyle=a_{1}^{-\frac{d-1}{2}}a_{2}^{-\frac{d-1}{2}}\left[i\gamma^{b}\partial_{b}+am\right]_{1}$
$\displaystyle\times\int\frac{d^{d-1}k}{(2\pi)^{d-1}}e^{i\vec{k}\cdot(\vec{x}_{1}-\vec{x}_{2})}\frac{\pi\sqrt{\tau_{1}\tau_{2}}}{4}\left[H_{\nu_{-}}^{(1)}(z_{1})H_{\nu_{-}}^{(2)}(z_{2})\frac{1+\gamma^{0}}{2}+H_{\nu_{+}}^{(1)}(z_{1})H_{\nu_{+}}^{(2)}(z_{2})\frac{1-\gamma^{0}}{2}\right],$
(C.53) $\displaystyle S_{2}(\vec{x}_{1}-\vec{x}_{2};\tau_{1},\tau_{2})$
$\displaystyle=a_{1}^{-\frac{d-1}{2}}a_{2}^{-\frac{d-1}{2}}\left[i\gamma^{b}\partial_{b}+am\right]_{1}$
$\displaystyle\times\int\frac{d^{d-1}k}{(2\pi)^{d-1}}e^{i\vec{k}\cdot(\vec{x}_{1}-\vec{x}_{2})}\frac{\pi\sqrt{\tau_{1}\tau_{2}}}{4}\left[H_{\nu_{-}}^{(2)}(z_{1})H_{\nu_{-}}^{(1)}(z_{2})\frac{1+\gamma^{0}}{2}+H_{\nu_{+}}^{(2)}(z_{1})H_{\nu_{+}}^{(1)}(z_{2})\frac{1-\gamma^{0}}{2}\right],$
(C.54)

where the subscript “1” indicates that $a$ is evaluated at $\tau=\tau_{1}$ and
the derivatives are acting on $\tau_{1},\vec{x}_{1}$. Now the momentum
integral is of the form (C.8), and by rewriting the ordinary derivative to the
covariant derivative, we obtain

$\displaystyle S_{1}(\vec{x}_{1}-\vec{x}_{2};\tau_{1},\tau_{2})$
$\displaystyle=\frac{H^{d-2}}{(4\pi)^{d/2}}\left[a\left(i\not{\nabla}+m\right)\right]_{1}\frac{1}{\sqrt{a_{1}a_{2}}}\left[\frac{1+\gamma^{0}}{2}I_{\nu_{-}}(Z_{12})+\frac{1-\gamma^{0}}{2}I_{\nu_{+}}(Z_{12})\right],$
(C.55) $\displaystyle S_{2}(\vec{x}_{1}-\vec{x}_{2};\tau_{1},\tau_{2})$
$\displaystyle=\frac{H^{d-2}}{(4\pi)^{d/2}}\left[a\left(i\not{\nabla}+m\right)\right]_{1}\frac{1}{\sqrt{a_{1}a_{2}}}\left[\frac{1+\gamma^{0}}{2}I_{\nu_{-}}(Z_{21})+\frac{1-\gamma^{0}}{2}I_{\nu_{+}}(Z_{21})\right].$
(C.56)

From this, we obtain the four propagators in the Schwinger-Keldysh formalism
as

$\displaystyle
S_{\lambda\lambda^{\prime}}(\vec{x}_{1}-\vec{x}_{2};\tau_{1},\tau_{2})$
$\displaystyle=\frac{H^{d-2}}{(4\pi)^{d/2}}\left[a\left(i\not{\nabla}+m\right)\right]_{1}\frac{1}{\sqrt{a_{1}a_{2}}}\left[\frac{1+\gamma^{0}}{2}I_{\nu_{-}}(Z_{\lambda\lambda^{\prime}})+\frac{1-\gamma^{0}}{2}I_{\nu_{+}}(Z_{\lambda\lambda^{\prime}})\right]\,,$
(C.57)

where $\lambda,\lambda^{\prime}=\pm$ correspond to the different contours in
the Schwinger-Keldysh formalism. The time-ordered propagator correctly
reproduces the results in [98, 54]. Note that the propagators differ only in
the sign of the $i\epsilon$ prescription in the embedding distance, as in the
flat spacetime case. This indicates that, in the late-time expansion, i.e.
$\tau_{1},\tau_{2}\to 0$ while keeping $x_{12}$ finite, these differences
disappear and all the propagators give the same result.

###  C.4 Massive gauge boson

Finally, we review the derivation of the massive gauge boson propagator in de
Sitter spacetime [99, 100]. We consider the action in the $R_{\xi}$-gauge,
given by

$\displaystyle S_{A}\;=\;\int
d^{d}x\sqrt{-g}\left[-\frac{1}{4}F^{\mu\nu}F_{\mu\nu}+\frac{m^{2}}{2}A_{\mu}A^{\mu}-\frac{1}{2\xi_{g}}\left(\nabla^{\mu}A_{\mu}\right)^{2}\right]\,,$
(C.58)

where $\xi_{g}$ is the gauge fixing parameter. Here we do not write down the
kinetic mixing part and the pure Goldstone part in the gauge fixing. The
former cancels with the mixing from the kinetic term of the Higgs, while the
latter (together with the Goldstone modes themselves) can be ignored in the
unitary gauge $\xi_{g}\to\infty$, which we take in this paper. Below we derive
the propagator in the standard canonical quantization method, following Ref.
[100].

#### Mode equation

In de Sitter spacetime, the action is given by

$\displaystyle S_{A}\;=\;\int d\tau d^{d-1}x$
$\displaystyle\left[\frac{a^{d-4}}{2}\left(\partial_{\tau}A_{i}-\partial_{i}A_{0}\right)^{2}-\frac{a^{d-4}}{2}\left((\partial_{i}A_{j})^{2}-(\partial_{i}A_{i})^{2}\right)\right.$
$\displaystyle\left.+\frac{m^{2}}{2}a^{d-2}\left(A_{0}^{2}-A_{i}^{2}\right)-\frac{a^{d-4}}{2\xi_{g}}\left(\partial_{\tau}A_{0}+(d-2)\frac{da/d\tau}{a}A_{0}-\partial_{i}A_{i}\right)^{2}\right]\,.$
(C.59)

Note that $A_{0}$ has the kinetic term since we keep $\xi_{g}$ finite at this
moment. It is convenient to define the conjugate momenta to solve the mode
equations. They are given by

$\displaystyle\pi^{0}\;=\;-\frac{a^{d-4}}{\xi_{g}}\left[\partial_{\tau}A_{0}+(d-2)\frac{da/d\tau}{a}A_{0}-\partial_{i}A_{i}\right]\,,\quad\pi^{i}\;=\;a^{d-4}\left(\partial_{\tau}A_{i}-\partial_{i}A_{0}\right)\,,$
(C.60)

and with them the equation of motion is given by

$\displaystyle 0$
$\displaystyle=\partial_{\tau}\pi^{i}-a^{d-4}(\nabla^{2}A_{i}-\partial_{i}\partial_{j}A_{j})+m^{2}a^{d-2}A_{i}-\partial_{i}\pi^{0}\,,$
(C.61) $\displaystyle 0$
$\displaystyle=\partial_{\tau}\pi^{0}-(d-2)\frac{da/d\tau}{a}\pi^{0}-m^{2}a^{d-2}A_{0}-\partial_{i}\pi^{i}\,,$
(C.62)

where $\nabla^{2}=\partial_{i}^{2}$. To disentangle the equations, we may
define the transverse and longitudinal modes as

$\displaystyle
A_{i}\;=\;A_{Ti}+\partial_{i}A_{L},\quad\pi^{i}\;=\;\pi_{T}^{i}+\partial^{i}\pi_{L}=\pi_{T}^{i}-\partial_{i}\pi_{L}\,.$
(C.63)

The transverse part is conveniently written in terms of $A_{Ti}$ as

$\displaystyle 0$
$\displaystyle=\left[\partial_{\tau}^{2}+(d-4)\frac{da/d\tau}{a}\partial_{\tau}-\nabla^{2}+m^{2}a^{2}\right]A_{Ti},\quad\pi^{i}_{T}=a^{d-4}\partial_{\tau}A_{Ti}.$
(C.64)

The longitudinal and temporal modes are mixed in terms of $A_{L}$ and $A_{0}$,
but are decoupled in terms of $\pi_{L}$ and $\pi^{0}$ as

$\displaystyle 0$
$\displaystyle=\left[\partial_{\tau}^{2}-(d-2)\frac{da/d\tau}{a}\partial_{\tau}-\nabla^{2}+m^{2}a^{2}\right]\pi_{L},$
(C.65) $\displaystyle 0$
$\displaystyle=\left[\partial_{\tau}^{2}-(d-2)\frac{da/d\tau}{a}\partial_{\tau}-\nabla^{2}+\xi_{g}m^{2}a^{2}-(d-2)\frac{d}{d\tau}\left(\frac{da/d\tau}{a}\right)\right]\pi^{0}.$
(C.66)

After solving these equations, $A_{L}$ and $A_{0}$ are given by

$\displaystyle
A_{L}\;=\;\frac{a^{2-d}}{m^{2}}\left(\partial_{\tau}\pi_{L}+\pi^{0}\right)\,,\quad
A_{0}\;=\;\frac{a^{2-d}}{m^{2}}\left[\partial_{\tau}\pi^{0}-(d-2)\frac{da/d\tau}{a}\pi^{0}+\nabla^{2}\pi_{L}\right]\,.$
(C.67)

We quantize the fields as

$\displaystyle\hat{A}_{Ti}(\vec{x},t)=a^{-\frac{d-4}{2}}\int\frac{d^{d-1}k}{(2\pi)^{d-1}}e^{i\vec{k}\cdot\vec{x}}\sum_{\lambda=\pm}\left[\epsilon_{i}^{(\lambda)}(\vec{k})A_{T}(\vec{k},\tau)\hat{a}_{\vec{k},\lambda}+{\epsilon_{i}^{(\lambda)}}^{*}(-\vec{k})A_{T}^{*}(-\vec{k},\tau)\hat{a}^{\dagger}_{-\vec{k},\lambda}\right]\,,$
(C.68)
$\displaystyle\hat{\pi}_{L}(\vec{x},t)=a^{\frac{d-2}{2}}\int\frac{d^{d-1}k}{(2\pi)^{d-1}}e^{i\vec{k}\cdot\vec{x}}\left[\pi_{L}(\vec{k},\tau)\hat{a}_{\vec{k},L}+{\pi_{L}^{*}(-\vec{k},\tau)}\hat{a}_{-\vec{k},L}^{\dagger}\right]\,,$
(C.69)
$\displaystyle\hat{\pi}^{0}(\vec{x},t)=a^{\frac{d-2}{2}}\int\frac{d^{d-1}k}{(2\pi)^{d-1}}e^{i\vec{k}\cdot\vec{x}}\left[\pi^{0}(\vec{k},\tau)\hat{a}_{\vec{k},0}+{\pi^{0}}^{*}(-\vec{k},\tau)\hat{a}_{-\vec{k},0}^{\dagger}\right]\,,$
(C.70)

where the polarization vector and creation-annihilation operators satisfy

$\displaystyle\sum_{\lambda=\pm}\epsilon_{i}^{(\lambda)}{\epsilon_{j}^{(\lambda)}}^{*}\;=\;\delta_{ij}-\frac{k_{i}k_{j}}{k^{2}}\,,\quad\left[\hat{a}_{\vec{k},\lambda},\hat{a}_{\vec{k}^{\prime},\lambda^{\prime}}^{\dagger}\right]$
$\displaystyle=-(2\pi)^{d-1}\eta_{\lambda\lambda^{\prime}}\delta^{(d-1)}(\vec{k}-\vec{k}^{\prime})\,,$
(C.71)

where we take $\eta_{++}=\eta_{--}=\eta_{LL}=-\eta_{00}=-1$ with all the other
components vanishing. The temporal component has the negative metric, but it
does not cause an issue since this degree of freedom is killed by the gauge
redundancy (or more precisely the BRST symmetry) and is unphysical. The mode
equations are given by

$\displaystyle 0$
$\displaystyle=\left[\frac{d^{2}}{d\tau^{2}}+k^{2}-\frac{\nu^{2}-1/4}{\tau^{2}}\right]A_{T},\quad
0=\left[\frac{d^{2}}{d\tau^{2}}+k^{2}-\frac{\nu^{2}-1/4}{\tau^{2}}\right]\pi_{L},\quad
0=\left[\frac{d^{2}}{d\tau^{2}}+k^{2}-\frac{\nu_{\xi}^{2}-1/4}{\tau^{2}}\right]\pi^{0},$
(C.72)

where

$\displaystyle\nu\;=\;\sqrt{\frac{(d-3)^{2}}{4}-\frac{m^{2}}{H^{2}}}\,,\quad\nu_{\xi}\;=\;\sqrt{\frac{(d-1)^{2}}{4}-\frac{\xi_{g}m^{2}}{H^{2}}}\,.$
(C.73)

Note that $\pi^{0}$ becomes infinitely heavy in the unitarity gauge
$\xi_{g}\to\infty$. These are again Bessel differential equations (C.1) and
the solutions can be written in terms of the Hankel functions. The Bunch-
Davies vacuum condition together with the canonical commutation relation

$\displaystyle\left[\hat{A}_{\mu}(\vec{x},\tau),\hat{\pi}^{\nu}(\vec{y},\tau)\right]\;=\;i\delta_{\mu}^{\nu}\delta^{(d-1)}(\vec{x}-\vec{y})\,,$
(C.74)

fixes the mode functions as

$\displaystyle
A_{T}\;=\;\sqrt{\frac{-\pi\tau}{4}}e^{i\pi\nu/2}H_{\nu}^{(1)}(-k\tau)\,,\quad\pi_{L}\;=\;\frac{m}{k}\sqrt{\frac{-\pi\tau}{4}}e^{i\pi\nu/2}H_{\nu}^{(1)}(-k\tau)\,,\quad\pi^{0}\;=\;m\sqrt{\frac{-\pi\tau}{4}}e^{i\pi\nu_{\xi}/2}H_{\nu_{\xi}}^{(1)}(-k\tau)\,.$
(C.75)

#### Propagators

We are now ready to compute the propagators. The transverse part is easy to
compute as

$\displaystyle\langle
A_{Ti}(\vec{x}_{1},\tau_{1})A_{Tj}(\vec{x}_{2},\tau_{2})\rangle$
$\displaystyle=\frac{H^{d-2}a_{1}a_{2}}{(4\pi)^{d/2}}\left(\delta_{ij}-\frac{\partial_{1i}\partial_{1j}}{\nabla_{1}^{2}}\right){I}_{\nu}(Z_{12})\,,$
(C.76)

where we used Eqs. (C.6) and (C.8), and the subscripts “1” of the derivatives
indicate that they act on $\vec{x}_{1}$. To derive the longitudinal and
temporal parts, we note that

$\displaystyle\langle\pi_{L}(\vec{x}_{1},\tau_{1})\pi_{L}(\vec{x}_{2},\tau_{2})\rangle$
$\displaystyle=-\frac{(Ha_{1}a_{2})^{d-2}}{(4\pi)^{d/2}}\frac{m^{2}}{\nabla_{1}^{2}}I_{\nu}(Z_{12}),~{}~{}\langle\pi^{0}(\vec{x}_{1},\tau_{1})\pi^{0}(\vec{x}_{2},\tau_{2})\rangle=-\frac{m^{2}(Ha_{1}a_{2})^{d-2}}{(4\pi)^{d/2}}I_{\nu_{\xi}}(Z_{12}),$
(C.77)

where we again used Eqs. (C.6) and (C.8). From this we obtain

$\displaystyle\langle\partial_{1i}A_{L}(\vec{x}_{1},\tau_{1})\partial_{2j}A_{L}(\vec{x}_{2},\tau_{2})\rangle=-\frac{H^{d-2}}{m^{2}(4\pi)^{d/2}}\left[\frac{1}{(a_{1}a_{2})^{d-2}}\frac{\partial_{1i}\partial_{2j}}{\nabla_{1}^{2}}\partial_{\tau_{1}}\partial_{\tau_{2}}\left((a_{1}a_{2})^{d-2}I_{\nu}(Z_{12})\right)+\partial_{1i}\partial_{2j}I_{\nu_{\xi}}(Z_{12})\right],$
(C.78) $\displaystyle\langle
A_{0}(\vec{x}_{1},\tau_{1})A_{0}(\vec{x}_{2},\tau_{2})\rangle=-\frac{H^{d-2}}{m^{2}(4\pi)^{d/2}}\left[\nabla_{1}^{2}I_{\nu}(Z_{12})+\partial_{\tau_{1}}\partial_{\tau_{2}}I_{\nu_{\xi}}(Z_{12})\right],$
(C.79)

and

$\displaystyle\langle
A_{0}(\vec{x}_{1},\tau_{1})\partial_{2i}A_{L}(\vec{x}_{2},\tau_{2})\rangle=-\frac{H^{d-2}}{m^{2}(4\pi)^{d/2}}\left[\frac{1}{a_{2}^{d-2}}\partial_{\tau_{2}}\partial_{2i}\left(a^{d-2}_{2}I_{\nu}(Z_{12})\right)+\partial_{\tau_{1}}\partial_{2i}I_{\nu_{\xi}}(Z_{12})\right]\,,$
(C.80)
$\displaystyle\langle\partial_{1i}A_{L}(\vec{x}_{1},\tau_{1})A_{0}(\vec{x}_{2},\tau_{2})\rangle=-\frac{H^{d-2}}{m^{2}(4\pi)^{d/2}}\left[\frac{1}{a_{1}^{d-2}}\partial_{\tau_{1}}\partial_{1i}\left(a^{d-2}_{1}I_{\nu}(Z_{12})\right)+\partial_{\tau_{2}}\partial_{1i}I_{\nu_{\xi}}(Z_{12})\right]\,.$
(C.81)

Next, we make these expressions more concise. With the relations (C.14) and
(C.15), it is relatively easy to show that

$\displaystyle\langle
A_{\alpha}(\vec{x}_{1},\tau_{1})A_{0}(\vec{x}_{2},\tau_{2})\rangle$
$\displaystyle=\frac{H^{d-2}}{m^{2}(4\pi)^{d/2}}\left[K_{\alpha
0}(Z_{12})-\partial_{1\alpha}\partial_{\tau_{2}}I_{\nu_{\xi}}(Z_{12})\right]\,,$
(C.82) $\displaystyle\langle
A_{0}(\vec{x}_{1},\tau_{1})A_{\alpha}(\vec{x}_{2},\tau_{2})\rangle$
$\displaystyle=\frac{H^{d-2}}{m^{2}(4\pi)^{d/2}}\left[K_{0\alpha}(Z_{12})-\partial_{\tau_{1}}\partial_{2\alpha}I_{\nu_{\xi}}(Z_{12})\right]\,,$
(C.83)

where the bitensor function $K_{\alpha\beta}$ is defined as

$\displaystyle K_{\alpha\beta}(Z_{12})$
$\displaystyle=\left[Z_{12}I_{\nu}^{\prime\prime}+(d-1)I_{\nu}^{\prime}\right](\partial_{1\alpha}Z_{12})(\partial_{2\beta}Z_{12})+\left[(1-Z_{12}^{2})I_{\nu}^{\prime\prime}-(d-1)Z_{12}I_{\nu}^{\prime}\right]\partial_{1\alpha}\partial_{2\beta}Z_{12}\,,$
(C.84)

with the prime denoting the derivatives with respect to $Z_{12}$. The pure
spatial component is more complicated. The part that includes $\nu_{\xi}$ is
trivial, and thus we focus on the other part. Our goal is to show that

$\displaystyle\tilde{K}_{ij}(Z_{12})=\left[\frac{m^{2}}{H^{2}\tau_{1}\tau_{2}}\left(\delta_{ij}+\frac{\partial_{1i}\partial_{2j}}{\nabla_{1}^{2}}\right)-\frac{1}{(a_{1}a_{2})^{d-2}}\frac{\partial_{1i}\partial_{2j}}{\nabla_{1}^{2}}\partial_{\tau_{1}}\partial_{\tau_{2}}(a_{1}a_{2})^{d-2}\right]I_{\nu}\,,$
(C.85)

is equivalent to

$\displaystyle
K_{ij}(Z_{12})\;=\;\frac{\delta_{ij}}{\tau_{1}\tau_{2}}\left[(1-Z_{12}^{2})I_{\nu}^{\prime\prime}-(d-1)Z_{12}I_{\nu}^{\prime}\right]-\frac{(x_{12})_{i}(x_{12})_{j}}{\tau_{1}^{2}\tau_{2}^{2}}\left[Z_{12}I_{\nu}^{\prime\prime}+(d-1)I_{\nu}^{\prime}\right]\,,$
(C.86)

where $\vec{x}_{12}=\vec{x}_{1}-\vec{x}_{2}$. For this purpose, we act
$\nabla_{1}^{2}$ on both expressions and check if they
agree.\natural18\natural18$\natural$18 This is fine up to the functions that
have vanishing Laplacian, but these functions do not allow the Fourier
transform and are not important.  By acting $\nabla_{1}^{2}$, we obtain

$\displaystyle\nabla_{1}^{2}\left(K_{ij}-\tilde{K}_{ij}\right)$
$\displaystyle=\frac{\delta_{ij}}{\tau_{1}\tau_{2}}\nabla_{1}^{2}\left[(1-Z_{12}^{2})I_{\nu}^{\prime\prime}-(d-1)Z_{12}I_{\nu}^{\prime}-\frac{m^{2}}{H^{2}}I_{\nu}\right]-\nabla_{1}^{2}\left[\frac{(x_{12})_{i}(x_{12})_{j}}{\tau_{1}^{2}\tau_{2}^{2}}\left(Z_{12}I_{\nu}^{\prime\prime}+(d-1)I_{\nu}^{\prime}\right)\right]$
$\displaystyle+\left[-\frac{m^{2}}{H^{2}\tau_{1}\tau_{2}}+(\tau_{1}\tau_{2})^{d-2}\partial_{\tau_{1}}\partial_{\tau_{2}}\frac{1}{(\tau_{1}\tau_{2})^{d-2}}\right]\left[-\frac{(x_{12})_{i}(x_{12})_{j}}{\tau_{1}^{2}\tau_{2}^{2}}I_{\nu}^{\prime\prime}+\frac{\delta_{ij}}{\tau_{1}\tau_{2}}I_{\nu}^{\prime}\right]\,.$
(C.87)

There are two structures, $\delta_{ij}$ and $(x_{12})_{i}(x_{12})_{j}$. The
former is given by

$\displaystyle\left.\nabla_{1}^{2}\left(K_{ij}-\tilde{K}_{ij}\right)\right|_{\delta_{ij}}$
$\displaystyle=\frac{\nabla_{1}^{2}}{\tau_{1}\tau_{2}}\left[(1-Z_{12}^{2})I_{\nu}^{\prime\prime}-(d-1)Z_{12}I_{\nu}^{\prime}-\frac{m^{2}}{H^{2}}I_{\nu}\right]-\frac{2}{\tau_{1}^{2}\tau_{2}^{2}}\left(Z_{12}I_{\nu}^{\prime\prime}+(d-1)I_{\nu}^{\prime}\right)$
$\displaystyle+\frac{1}{\tau_{1}\tau_{2}}\left[-\frac{m^{2}}{H^{2}\tau_{1}\tau_{2}}+(\tau_{1}\tau_{2})^{d-1}\partial_{\tau_{1}}\partial_{\tau_{2}}\frac{1}{(\tau_{1}\tau_{2})^{d-1}}\right]I_{\nu}^{\prime}\,.$
(C.88)

After some computation, we obtain

$\displaystyle\left.\nabla_{1}^{2}\left(K_{ij}-\tilde{K}_{ij}\right)\right|_{\delta_{ij}}$
$\displaystyle=\frac{1}{\tau_{1}\tau_{2}}\left[\frac{1}{\tau_{1}\tau_{2}}\frac{d}{dZ_{12}}+\nabla_{1}^{2}\right]\left[(1-Z_{12}^{2})I_{\nu}^{\prime\prime}-dI_{\nu}^{\prime}-\left(d-2+\frac{m^{2}}{H^{2}}\right)I_{\nu}\right]=0\,,$
(C.89)

where we used Eq. (C.11) in the last equality. The latter is given by

$\displaystyle\left.\nabla_{1}^{2}\left(K_{ij}-\tilde{K}_{ij}\right)\right|_{(x_{12})_{i}(x_{12})_{j}}=-\frac{1}{\tau_{1}^{2}\tau_{2}^{2}}$
$\displaystyle\left[\left(\nabla_{1}^{2}-\frac{4}{\tau_{1}\tau_{2}}\frac{d}{dZ_{12}}\right)\left(Z_{12}I_{\nu}^{\prime\prime}+(d-1)I_{\nu}^{\prime}\right)\right.$
$\displaystyle\left.+\left(-\frac{m^{2}}{H^{2}\tau_{1}\tau_{2}}+(\tau_{1}\tau_{2})^{d}\partial_{\tau_{1}}\partial_{\tau_{2}}\frac{1}{(\tau_{1}\tau_{2})^{d}}\right)I_{\nu}^{\prime\prime}\right]\,.$
(C.90)

After several steps, we obtain

$\displaystyle\left.\nabla_{1}^{2}\left(K_{ij}-\tilde{K}_{ij}\right)\right|_{(x_{12})_{i}(x_{12})_{j}}$
$\displaystyle=-\frac{1}{\tau_{1}^{3}\tau_{2}^{3}}\frac{d^{2}}{d^{2}Z_{12}}\left[(1-Z_{12}^{2})I_{\nu}^{\prime\prime}-dI_{\nu}^{\prime}-\left(d-2+\frac{m^{2}}{H^{2}}\right)I_{\nu}\right]=0\,,$
(C.91)

where we again used Eq. (C.11). Therefore we conclude that
$\tilde{K}_{ij}=K_{ij}$, and we find

$\displaystyle\langle
A_{\alpha}(\vec{x}_{1},\tau_{1})A_{\beta}(\vec{x}_{2},\tau_{2})\rangle$
$\displaystyle=\frac{H^{d-2}}{m^{2}(4\pi)^{d/2}}\left[K_{\alpha\beta}(Z_{12})-\partial_{1\alpha}\partial_{2\beta}I_{\nu_{\xi}}(Z_{12})\right]\,.$
(C.92)

In the unitarity gauge $\xi_{g}\to\infty$, $\pi^{0}$ becomes infinitely
massive, and we can simply ignore $I_{\nu_{\xi}}$. Furthermore, different
propagators correspond to merely different choices of $Z$ in the argument of
$K_{\alpha\beta}$. Therefore, we obtain the Schwinger-Keldysh propagator of
the gauge bosons

$\displaystyle
G_{\lambda\lambda^{\prime};\alpha\beta}(\vec{x}_{1}-\vec{x}_{2};\tau_{1},\tau_{2})$
$\displaystyle=\frac{H^{d-2}}{m^{2}(4\pi)^{d/2}}K_{\alpha\beta}(Z_{\lambda\lambda^{\prime}})\,,$
(C.93)

with $\lambda,\lambda^{\prime}=\pm$. One can show its equivalence to the
expressions in [99, 54], as demonstrated in [100].

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