Next we explore the various asymptotic limits of $\mathcal{G}_k(\eta, \eta_{\text{in}})$. First we note that in the limit $z_{\text{in}} \gg 1$ (so that $\eta_{\text{in}} \to -\infty$ in the distant past) we have

$$ \mathcal{G}_k(\eta, \eta_{\text{in}}) \approx 4\pi e^{2i\kappa z} \left[ \frac{16}{\kappa z^4} - \frac{24i}{z^3} - \frac{8(2\kappa^2 - 1)}{\kappa z^2} + \frac{i(2\kappa^2 - 1)^2}{\kappa^2 z} \right] \\ + 8\pi e^{i(2\kappa+1)z} \left[ -\frac{8}{\kappa z^4} + \frac{4i(3\kappa + 2)}{\kappa z^3} + \frac{8\kappa + 7}{z^2} - \frac{2i\kappa(\kappa + 1)}{z} \right] + \mathcal{O}\left(\frac{1}{z_{\text{in}}}\right), \quad (\text{C.35}) $$

and note that if we furthermore take the super-Hubble limit of this then we get

$$ \mathcal{G}_k(\eta, \eta_{\text{in}}) \approx -\frac{40\pi}{z^2} - \frac{4i\pi(24\kappa^3 + 6\kappa^2 + 16\kappa - 3)}{3\kappa^2 z} + \frac{92\pi}{3} + \mathcal{O}(z). \quad (\text{C.36}) $$

Next let us fix arbitrary $z_{\text{in}}$ and take the super-Hubble limit $z \ll 1$ so that

$$ \begin{aligned} \mathcal{G}_k(\eta, \eta_{\text{in}}) \approx & -\frac{40\pi}{z^2} + \frac{16}{3z} \left\{ e^{-2i\kappa z_{\text{in}}} \left[ -\frac{2i(\kappa^2 - 3)}{\kappa z_{\text{in}}} + \frac{4i}{\kappa z_{\text{in}}^3} - \frac{5}{z_{\text{in}}^2} \right] + e^{-i(2\kappa+1)z_{\text{in}}} \left[ -\frac{4i}{\kappa z_{\text{in}}^3} + \frac{5\kappa + 4}{\kappa z_{\text{in}}^2} \right. \right. \\ & \left. \left. + \frac{i(4\kappa^2 - 5\kappa - 8)}{2\kappa z_{\text{in}}} \right] + \left( \kappa^2 - 9 - \frac{3}{4\kappa^2} \right) \text{Ei}(-2i\kappa z_{\text{in}}) + \left( -\kappa^2 + 6\kappa + \frac{4}{\kappa} + \frac{21}{2} \right) \right. \\ & \left. \times \text{Ei}[-i(2\kappa + 1)z_{\text{in}}] \right\} + \frac{92\pi}{3} + \mathcal{O}(z). \end{aligned} \quad (\text{C.37}) $$

Let us now take the above expression and take $z_{\text{in}} \ll 1$ (assuming the hierarchy $z \ll z_{\text{in}} \ll 1$)

$$ \begin{aligned} \mathcal{G}_k(\eta, \eta_{\text{in}}) \approx & -\frac{40\pi}{z^2} + \frac{1}{z} \left[ -\frac{40i}{z_{\text{in}}} + 2 \left( -\frac{1}{\kappa^2} + 8\kappa + \frac{16}{3\kappa} + 2 \right) \{2\log[e^\gamma(2\kappa+1)z_{\text{in}}] - i\pi\} \right. \\ & \left. + 4 \left( -\frac{4\kappa^2}{3} + \frac{1}{\kappa^2} + 12 \right) \log \left( \frac{2\kappa+1}{2\kappa} \right) - \frac{16(39\kappa + 15 + 16/\kappa)}{9} + \mathcal{O}(z_{\text{in}}) \right] \\ & + \frac{92\pi}{3} + \mathcal{O}(z) \end{aligned} \quad (\text{C.38}) $$

where $\gamma$ is Euler-Mascheroni constant.

Finally let us take the coincident limit for $z \to z_{\text{in}}$ where we have

$$ \begin{aligned} \mathcal{G}_k(\eta, \eta_{\text{in}}) \approx & -\frac{6iz_{\text{in}}^2}{(z_{\text{in}} - z)^4} + \frac{4iz_{\text{in}}}{(z_{\text{in}} - z)^3} + \frac{2i(33 - 5z_{\text{in}}^2)}{3(z_{\text{in}} - z)^2} + i \left( \frac{40}{z_{\text{in}}^2} - \frac{92}{3} + \frac{43}{15}z_{\text{in}}^2 \right) \\ & \times \log[e^\gamma(2\kappa+1)(z_{\text{in}} - z)] + \pi \left\{ -zz_{\text{in}}\delta'''(z_{\text{in}} - z) + (5z - 4z_{\text{in}})\delta''(z_{\text{in}} - z) \right. \\ & \left. - \frac{2}{3} \left( 5zz_{\text{in}} + \frac{6z}{z_{\text{in}}} + \frac{6z_{\text{in}}}{z} - 30 \right) \delta'(z_{\text{in}} - z) + 2 \left[ \frac{10}{z} - \frac{8}{z_{\text{in}}} - 6z_{\text{in}} \right. \right. \\ & \left. \left. + \left( \frac{23}{3} - \frac{2}{z_{\text{in}}^2} \right) z \right] \delta(z_{\text{in}} - z) \right\} + \mathcal{O}\left[ (z_{\text{in}} - z)^0 \right]. \end{aligned} \quad (\text{C.39}) $$

Most important for us is the real part of the above, where we have

$$ \begin{aligned} \text{Re} [\mathcal{G}_k(\eta, \eta_{\text{in}})] \approx & \pi \left\{ -zz_{\text{in}}\delta'''(z_{\text{in}} - z) + (5z - 4z_{\text{in}})\delta''(z_{\text{in}} - z) - \frac{2}{3} \left( 5zz_{\text{in}} + \frac{6z}{z_{\text{in}}} + \frac{6z_{\text{in}}}{z} - 30 \right) \right. \\ & \left. \times \delta'(z_{\text{in}} - z) + 2 \left[ \frac{10}{z} - \frac{8}{z_{\text{in}}} - 6z_{\text{in}} + \left( \frac{23}{3} - \frac{2}{z_{\text{in}}^2} \right) z \right] \delta(z_{\text{in}} - z) \right\} + \mathcal{O}\left[ (z_{\text{in}} - z)^0 \right]. \end{aligned} \quad (\text{C.40}) $$

Notice that the singular part of the above is only coming from the $\delta$-functions.
## C.2 The function $\mathcal{I}$

Here we compute the integral (C.8), repeated here for convenience

$$ \mathcal{I}_k(\eta) := \int_0^1 da \int_{a+2\kappa}^\infty db \, z (a^2 + b^2 - 2)^2 \left[ 1 + \frac{2i}{(b-a)z} \right] \left[ 1 + \frac{2i}{(b+a)z} \right] \\ \times \left[ -\frac{1}{b^2} - \frac{iz}{b} - \frac{4}{b^2 - a^2} + \frac{4e^{ibz}\mathrm{Ei}(-ibz)}{b^2 - a^2} \right]. \qquad (C.41) $$

Luckily the computation of this is straightforward since we already know what the function $\mathcal{G}_k(\eta, \eta_{\text{in}})$ is — the reason for this is that the integrand of $\mathcal{I}_k(\eta)$ and $\mathcal{G}_k(\eta, \eta_{\text{in}})$ are identical in the limit $\eta_{\text{in}} \to \eta$ (one must only be careful about the divergences in making this replacement).

As before, we split apart the integral $\mathcal{I}_k(\eta)$ into a triangular and rectangular region,

$$ \mathcal{I}_k(\eta) = \mathcal{I}_k^{\Delta}(\eta) + \mathcal{I}_k^{\Box}(\eta) \qquad (C.42) $$

where we define

$$ \mathcal{I}_k^{\Delta}(\eta) := \int_{2\kappa}^{2\kappa+1} db \int_{0}^{b-2r} da \, f(a, b, z, z) \qquad (C.43) $$

$$ \mathcal{I}_k^{\Box}(\eta) := \int_{2\kappa+1}^{\infty} db \int_{0}^{1} da \, f(a, b, z, z), \qquad (C.44) $$

cf. eqs. (C.11) and (C.12) where we use

$$ f(a, b, z, z) = z(a^2 + b^2 - 2)^2 \left[ 1 + \frac{2i}{(b-a)z} \right] \left[ 1 + \frac{2i}{(b+a)z} \right] \\ \times \left[ -\frac{1}{b^2} - \frac{iz}{b} - \frac{4}{b^2 - a^2} + \frac{4e^{ibz}\mathrm{Ei}(-ibz)}{b^2 - a^2} \right], \qquad (C.45) $$

where $f(a, b, z, z_{\text{in}})$ is the integrand of $\mathcal{G}_k(\eta, \eta_{\text{in}})$, defined in eq. (C.13).

Note that

$$ \mathcal{I}_k^{\Delta}(\eta) = \lim_{z_{\text{in}} \to z} \left[ \mathcal{G}_k^{\Delta}(\eta, \eta_{\text{in}}) \right] \qquad (C.46) $$

where we have the explicit formula for $\mathcal{G}_k^{\Delta}(\eta, \eta_{\text{in}})$ given in eq. (C.17) and so we straightforwardly have an answer for $\mathcal{I}_k^{\Delta}(\eta)$ (for brevity we avoid writing it down here).

For computing $\mathcal{I}_k^{\Box}(\eta)$ in eq. (C.44) we use the formula (C.18) (after $a$ integration has been performed) and take the limit $z_{\text{in}} \to z$ giving

$$ \mathcal{I}_k^{\Box}(\eta) = \mathcal{I}_k^{\Box, \text{div}}(\eta) + \mathcal{I}_k^{\Box, \text{reg}}(\eta), \qquad (C.47) $$

where

$$ \mathcal{I}_k^{\Box, \text{div}}(\eta) := \int_{2\kappa+1}^{\infty} db \left\{ 4\pi e^{ibz} \left[ -izb^2 + 4b + i \left( 3z + \frac{4}{z} \right) \right] + \left[ -iz^2b^3 - zb^2 + \frac{2i(5z^2 - 18)}{3}b + \frac{10z}{3} + \frac{i}{b} \left( -\frac{40}{z^2} + \frac{92}{3} - \frac{43}{15}z^2 \right) \right] \right\} \qquad (C.48) $$
$$
\begin{equation}
\begin{aligned}
\mathcal{I}_k^{\square, \text{reg}}(\eta) := & \int_{2\kappa+1}^{\infty} \mathrm{d}b \left( -4\pi e^{+ibz} \left[-izb^2 + 4b + i\left(3z + \frac{4}{z}\right)\right] - 36ib + \frac{36}{z} + \frac{4i(z^2+30)}{3bz^2} \right. \\
& \left. \qquad + \frac{-43z^2 - 260}{15b^2z} + \frac{16(2ib^5z - b^4 - 2b^2 - 2ibz + 3)}{b^3z} \coth^{-1}(b) + 4 \left\{ -3b^2z + 12ib \right. \right. \\
& \left. \qquad \left. + \frac{11z^2 - 36}{3z} - \frac{8i}{b} + \frac{8}{b^2z} + \left[ \frac{4(b^2 - 1)^2z}{b} + \frac{8}{b^2}(b^4 - 1)\left(\frac{1}{bz} - i\right) \right] \coth^{-1}(b) \right\} \right. \\
& \left. \qquad \left. \times e^{ibz} \mathrm{Ei}(-ibz) \right)
\end{aligned}
\tag{C.49}
\end{equation}
$$

cf. the definitions (C.20) and (C.21). Note in particular that

$$
\mathcal{I}_{k}^{\square, \mathrm{reg}}(\eta) = \lim_{z_{\mathrm{in}} \to z} \left[ \mathcal{G}_{k}^{\square, \mathrm{reg}}(\eta, \eta_{\mathrm{in}}) \right], \qquad (\mathrm{C.50})
$$

where we can straightforwardly write down an expression for $\mathcal{I}_k^{\square,\text{reg}}(\eta)$ given we know what $\mathcal{G}_k^{\square,\text{reg}}(\eta, \eta_{\text{in}})$ is in eq. (C.32) (although once again we avoid writing this down here for brevity).

One can almost make a similar statement for $\mathcal{I}_k^{\square,1}(\eta)$, but not exactly because this integral is formally divergent — we need to use the distributions (C.26) to evaluate the first terms in eq. (C.48), as well as the following dimensional regularizations of various powers of $b$:

$$
\int_{2\kappa+1}^{\infty} db \, b^j \left(\frac{b}{M}\right)^{\epsilon} = - \frac{(2\kappa + 1)^{\epsilon + j + 1}}{M^{\epsilon}(\epsilon + j + 1)} \approx - \frac{(2\kappa + 1)^{j+1}}{j+1} + \mathcal{O}(\epsilon), \quad (C.51)
$$

$$
\int_{2\kappa+1}^{\infty} db \frac{1}{b} \left(\frac{b}{M}\right)^{\epsilon} = - \frac{(2\kappa + 1)^{\epsilon}}{M^{\epsilon}\epsilon} \approx -\frac{1}{\epsilon} - \log\left(\frac{2\kappa + 1}{M}\right) + \mathcal{O}(\epsilon) \quad (C.52)
$$

where the first relation applies for $j \in \{0, 1, 2, 3\}$ and $M = \mu/k > 0$ ($\epsilon \in \mathbb{C}$) for some mass scale $\mu > 0$ and where we have expanded the above integrals for $0 < |\epsilon| \ll 1$ (although the integrals are technically only convergent for $\text{Re}(\epsilon) < -j-1$ and $\text{Re}(\epsilon) < 0$ respectively). With this, we find that

$$
\begin{equation}
\begin{aligned}
\mathcal{I}_k^{\square, \text{div}}(\eta) ={}& i \left(\frac{40}{z^2} - \frac{92}{3} + \frac{43}{15}z^2\right) \left[\log\left(\frac{2\kappa + 1}{M}\right) + \frac{1}{\epsilon}\right] + 8\pi e^{i(2\kappa + 1)z} \left[-\frac{5}{z^2} + \frac{3i(2\kappa + 1)}{z}\right. \\
& \left. \qquad + 2\kappa^2 + 2\kappa - 1\right] + 6i(2\kappa + 1)^2 + \frac{(2\kappa + 1)(4\kappa^2 + 4\kappa - 9)}{3}z \\
& \qquad \left. + \frac{i(2\kappa + 1)^2(12\kappa^2 + 12\kappa - 17)}{12}z^2,
\end{aligned}
\tag{C.53}
\end{equation}
$$

where we have ignored the Dirac functions. With the above, we can at last sum together all the pieces to get our result

$$
\begin{align*}
\mathcal{I}_k(\eta) &= \mathcal{I}_k^{\Delta}(\eta) + \mathcal{I}_k^{\square, \text{div}}(\eta) + \mathcal{I}_k^{\square, \text{reg}}(\eta) \\
&= i \left( \frac{40}{z^2} - \frac{92}{3} + \frac{43}{15} z^2 \right) \left[ \frac{1}{\epsilon} + \log \left( \frac{2k_{\text{UV}} + k}{\mu} \right) \right] + \frac{64}{\kappa z^3} - \frac{4i(\kappa + 4)}{\kappa z^2} + \frac{2i}{9\kappa} (84\kappa^3 + 60\kappa^2 \\
&\quad + 79\kappa + 27) + \frac{(32\kappa^3 - 40\kappa + 1)z}{3} + \frac{i}{900} (-3611 - 3360\kappa + 360\kappa^2 + 4320\kappa^3 + 720\kappa^4)
\end{align*}
$$
$$ 
\begin{aligned}
& + \left[ \frac{64i}{\kappa z^4} - \frac{32}{z^3} + \frac{32i}{\kappa z^2} + \frac{8i(4\kappa^4 - 12\kappa^2 - 3)}{3\kappa} - 4(2\kappa^2 - 1)^2 z + \frac{8i\kappa(12\kappa^4 - 20\kappa^2 + 15)}{15} z^2 \right] \\
& \times \log \left( \frac{2\kappa + 1}{2\kappa} \right) + 8e^{i(2\kappa+1)z} \operatorname{Ei} [-i(2\kappa+1)z] \left[ -\frac{8i}{\kappa z^4} - \frac{4(3\kappa+2)}{\kappa z^3} + \frac{i(8\kappa+7)}{z^2} + \frac{2\kappa(\kappa+1)}{z} \right] \\
& + 8e^{2i\kappa z} \operatorname{Ei} (-2i\kappa z) \left[ \frac{8i}{\kappa z^4} + \frac{12}{z^3} - \frac{4i(2\kappa^2-1)}{\kappa z^2} - \frac{(2\kappa^2-1)^2}{2\kappa^2 z} \right].
\end{aligned} 
\quad (C.54) $$

Writing out the real and imaginary parts of the above function gives

$$ 
\begin{aligned}
\operatorname{Re}[\mathcal{I}_k(\eta)] = & \frac{64}{\kappa z^3} + \frac{(32\kappa^3 - 40\kappa + 1)z}{3} + \left[-\frac{32}{z^3} - 4(2\kappa^2 - 1)^2 z\right] \log\left(\frac{2\kappa + 1}{2\kappa}\right) \\
& + \operatorname{Re}\left\{8e^{i(2\kappa+1)z}\operatorname{Ei}[-i(2\kappa+1)z]\left[-\frac{8i}{\kappa z^4} - \frac{4(3\kappa+2)}{\kappa z^3} + \frac{i(8\kappa+7)}{z^2} + \frac{2\kappa(\kappa+1)}{z}\right]\right\} \\
& + \operatorname{Re}\left\{8e^{2i\kappa z}\operatorname{Ei}(-2i\kappa z)\left[\frac{8i}{\kappa z^4} + \frac{12}{z^3} - \frac{4i(2\kappa^2-1)}{\kappa z^2} - \frac{(2\kappa^2-1)^2}{2\kappa^2 z}\right]\right\}
\end{aligned} 
\quad (C.55) $$

while the imaginary part is

$$ 
\begin{aligned}
\operatorname{Im}[\mathcal{I}_k(\eta)] = & \left(\frac{40}{z^2} - \frac{92}{3} + \frac{43}{15}z^2\right) \left[\frac{1}{\epsilon} + \log\left(\frac{2k_{UV} + k}{\mu}\right)\right] - \frac{4(\kappa + 4)}{\kappa z^2} + \frac{2}{9\kappa} \left(84\kappa^3 + 60\kappa^2\right. \\
& \left. + 79\kappa + 27\right) + \frac{-3611 - 3360\kappa + 360\kappa^2 + 4320\kappa^3 + 720\kappa^4}{900} z^2 + \left[\frac{64}{\kappa z^4} + \frac{32}{\kappa z^2}\right. \\
& \left. + \frac{8(4\kappa^4 - 12\kappa^2 - 3)}{3\kappa} + \frac{8\kappa(12\kappa^4 - 20\kappa^2 + 15)}{15} z^2\right] \log\left(\frac{2\kappa + 1}{2\kappa}\right) \\
& + \operatorname{Im}\left\{8e^{i(2\kappa+1)z}\operatorname{Ei}[-i(2\kappa+1)z]\left[-\frac{8i}{\kappa z^4} - \frac{4(3\kappa+2)}{\kappa z^3} + \frac{i(8\kappa+7)}{z^2} + \frac{2\kappa(\kappa+1)}{z}\right]\right\} \\
& + \operatorname{Im}\left\{8e^{2i\kappa z}\operatorname{Ei}(-2i\kappa z)\left[\frac{8i}{\kappa z^4} + \frac{12}{z^3} - \frac{4i(2\kappa^2-1)}{\kappa z^2} - \frac{(2\kappa^2-1)^2}{2\kappa^2 z}\right]\right\}.
\end{aligned} 
\quad (C.56) $$

Next we explore the various asymptotic limits of $\mathcal{I}_k(\eta)$. First we consider the super-Hubble limit $z \ll 1$ (which also assumes $\kappa z = -k_{UV}\eta \ll 1$)

$$ 
\begin{aligned}
\mathcal{I}_k(\eta) \approx & i \left(\frac{40}{z^2} - \frac{92}{3}\right) \left[\frac{1}{\epsilon} + \log\left(\frac{2k_{UV} + k}{\mu}\right)\right] + \frac{-20\pi + 28i - 40i \log[e^\gamma(2\kappa + 1)z]}{z^2} \\
& + \frac{1}{z} \left(\frac{4}{9} \left\{-4 \left(39\kappa + 15 + \frac{16}{\kappa}\right) + 3 \left(-4\kappa^2 + 36 + \frac{3}{\kappa^2}\right) \log\left(\frac{2\kappa + 1}{2\kappa}\right) + \left(72\kappa + 18\right.\right. \\
& \left. \left. + \frac{48}{\kappa} - \frac{9}{\kappa^2}\right) \log[e^\gamma(2\kappa + 1)z]\right\} - 2i\pi \left(8\kappa + 2 + \frac{16}{3\kappa} - \frac{1}{\kappa^2}\right)\right) + \frac{1}{3} \left\{46\pi\right. \\
& \left. + 92i \log[e^\gamma(2\kappa + 1)z] - 128i\right\} + \mathcal{O}(z)
\end{aligned} 
\quad (C.57) $$

Next we consider the early-time expansion $z \gg 1$ (which implicitly assumes that $z_{in} \gg z \gg$
1), where

$$
\begin{equation}
\begin{aligned}
\mathcal{I}_k(\eta) \approx {}& i \left(-\frac{92}{3} + \frac{43}{15} z^2\right) \left[\frac{1}{\epsilon} + \log\left(\frac{2k_{\text{UV}} + k}{\mu}\right)\right] + i \left[\frac{720\kappa^4 + 4320\kappa^3 + 360\kappa^2 - 3360\kappa - 3611}{900}\right. \\
& \left. \qquad + \frac{8\kappa(12\kappa^4 - 20\kappa^2 + 15)}{15} \log\left(\frac{2\kappa + 1}{2\kappa}\right) \right] z^2 \\
& \qquad + \left[\frac{32\kappa^3 - 40\kappa + 1}{3} - 4(1 - 2\kappa^2)^2\right. \\
& \qquad \left. \qquad \times \log\left(\frac{2\kappa + 1}{2\kappa}\right) \right] z + \frac{2i}{9\kappa} \left[84\kappa^3 + 60\kappa^2 + 79\kappa + 27 + 12(4\kappa^4 - 12\kappa^2 - 3)\log\left(\frac{2\kappa + 1}{2\kappa}\right)\right] \\
& \qquad \left. + \mathcal{O}\left(\frac{1}{z}\right)\right].
\end{aligned}
\tag{C.58}
\end{equation}
$$

### C.3 Asymptotics of $\mathfrak{F}$

We can now take the above asymptotic expressions for $\mathcal{G}_k(\eta, \eta_{\text{in}})$ and $\mathcal{I}_k(\eta)$ and use formula (C.7) to obtain expressions for $\mathfrak{F}_k(\eta, \eta_{\text{in}})$ in various limits used in the main text. Most importantly we consider the limit $z \ll z_{\text{in}} \ll 1$ in which, using eqs. (C.37) and (C.57), we have

$$
\begin{equation}
\begin{aligned}
\mathfrak{F}_k(\eta, \eta_{\text{in}}) ={}& \frac{\epsilon_1 H^2 k^2}{1024 \pi^2 M_p^2} \left( \frac{40i}{z^2} \left[ \frac{1}{\epsilon} + \log \left( \frac{2k_{\text{UV}} + k}{\mu} \right) \right] - \frac{20\pi}{z^2} + \frac{i}{z^2} \left\{ 28 - 40 \log [e^{\gamma}(2\kappa + 1)z] \right\} \right. \\
& \left. \qquad + \mathcal{O}\left(\frac{1}{z}\right) - \left[ -\frac{40\pi}{z^2} + \mathcal{O}\left(\frac{1}{z}\right) \right] \right).
\end{aligned}
\tag{C.59}
\end{equation}
$$

This expression (in fact an even more accurate version of this equation with higher order terms written explicitly) is given in eq. (3.40). Eq. (3.38), which is also used in the main text, also corresponds to the limit $z \ll 1$ but for arbitrary $z_{\text{in}}$.

We now turn to address the question raised in footnote 11. This footnote asks whether the singularity of Re $[\mathfrak{F}_k(\eta, \eta_{\text{in}})]$ in the coincident limit might mean that integrating through the coincident limit might compete in the expression for $\Xi_k$ even if $k\eta$ is not small. We show here that these corrections are in fact subdominant in $k\eta_{\text{in}}$. In the coincidence limit, the integral in the denominator of eq. (4.13) can be written as

$$
\begin{equation}
\begin{split}
& - \int_{z_0}^{z_{\text{in}}} dz' \operatorname{Re} \left[ \mathcal{G}_k \left( -\frac{z'}{k}, -\frac{z_{\text{in}}}{k} \right) \right] \left[ 1 + \frac{1}{(z')^2} \right] \approx \\
& \quad -\pi \int_{z_0}^{z_{\text{in}}} dz' \left\{ -z' z_{\text{in}} \delta'''(z_{\text{in}} - z') \right. \\
& \quad \left. + (5z' - 4z_{\text{in}}) \delta''(z_{\text{in}} - z') - \frac{2}{3} \left( 5z' z_{\text{in}} + \frac{6z'}{z_{\text{in}}} + \frac{6z_{\text{in}}}{z'} - 30 \right) \delta'(z_{\text{in}} - z') + 2 \right. \\
& \quad \left. \left[ \frac{10}{z'} - \frac{8}{z_{\text{in}}} - 6z_{\text{in}} + \left( \frac{23}{3} - \frac{2}{z_{\text{in}}^2} \right) z' \right] \delta(z_{\text{in}} - z') \right\} \left[ 1 + \frac{1}{(z')^2} \right]
\end{split}
\tag{C.60}
\end{equation}
$$

where we have used $\mathfrak{F}_k \propto \mathcal{I}_k - \mathcal{G}_k$ and where $z \simeq z_0$ is close to the coincident limit so that $|z_0 - z_{\text{in}}| \ll 1$ and where we keep the $\delta$-functions because all other terms are regular in the coincident limit. The above integral can be calculated exactly by using the regularization $\delta_{\epsilon}(x) = \pi^{-1} \lim_{\epsilon \to 0} \epsilon/(x^2 + \epsilon^2)$. Then, one obtains

$$
\begin{equation}
\begin{split}
& - \int_{z_0}^{z_{\text{in}}} dz' \operatorname{Re} \left[ \mathcal{G}_k \left( -\frac{z'}{k}, -\frac{z_{\text{in}}}{k} \right) \right] \left[ 1 + \frac{1}{(z')^2} \right] \approx \\
& \qquad \frac{2(z_{\text{in}}^2 + 1)}{\epsilon^3} - \frac{10z_{\text{in}}^4 - 41z_{\text{in}}^2 - 39}{3z_{\text{in}}^2 \epsilon} \\
& \qquad - \frac{2\pi(5z_{\text{in}}^2 - 24)}{3z_{\text{in}}^3} + \mathcal{O}(\epsilon).
\end{split}
\tag{C.61}
\end{equation}
$$

We see that the $\epsilon$-dependence is power-law, and any $z_{\text{in}}$-dependence is subdominant to the result quoted in the main text.
### C.4 The validity coefficient

Here we compute the integral $\mathfrak{M}_k$ defined in eq. (3.42)

$$ \mathfrak{M}_k(\eta, \eta_{\text{in}}) := (2\pi)^{3/2} \int_{\eta_{\text{in}}}^{\eta} d\eta' G(\eta)G(\eta')\mathcal{C}_k(\eta, \eta') (\eta' - \eta). \quad (C.62) $$

Noting the earlier definition (3.37) of $\mathfrak{F}_k(\eta, \eta_{\text{in}})$, it turns out that we can decompose $\mathfrak{M}_k$ as follows

$$ \mathfrak{M}_k(\eta, \eta_{\text{in}}) = \mathfrak{N}_k(\eta, \eta_{\text{in}}) - \eta \mathfrak{F}_k(\eta, \eta_{\text{in}}) \quad (C.63) $$

with $\mathfrak{N}_k$ is defined by

$$ \mathfrak{N}_k(\eta, \eta_{\text{in}}) := (2\pi)^{3/2} \int_{\eta_{\text{in}}}^{\eta} d\eta' G(\eta) G(\eta') \mathcal{C}_k(\eta, \eta') \eta'. \quad (C.64) $$

Following similar steps for the computation as for $\mathfrak{F}_k(\eta, \eta_{\text{in}})$ as in appendix C, we find that $\mathfrak{N}_k(\eta, \eta_{\text{in}})$ is given by the following expression

$$
\begin{aligned}
\mathfrak{N}_k(\eta, \eta_{\text{in}}) ={}& \frac{\varepsilon_1 H^2 k}{1024 \pi^2 M_{\text{P}}^2} \left( -32\kappa - 8 - \frac{64}{3\kappa} + \frac{4}{\kappa^2} - \frac{2z^2}{3} \right) \\
& \left[ 32\kappa^3 - 40\kappa + 1 - 12(2\kappa^2 - 1)^2 \right. \\
& \left. \times \log\left(\frac{2\kappa + 1}{2r}\right) \right] - iz \left(\frac{40}{z^2} - 24 + \frac{43z^2}{15}\right) \left[\frac{1}{\epsilon} + \log\left(\frac{2\kappa + 1}{M}\right)\right] + \frac{i}{z} \left[ 2\left(\frac{1}{\kappa^3} + \frac{2}{2\kappa + 1} + \frac{4}{\kappa}\right) \right. \\
& \left. - \frac{32}{\kappa} \log\left(\frac{2\kappa + 1}{2\kappa}\right) \right] + \frac{iz}{\kappa} \left[ 8(4\kappa^4 + 1) \log\left(\frac{2\kappa + 1}{2\kappa}\right) - \frac{2(36\kappa^4 + 24\kappa^3 + 6\kappa^2 + 9\kappa + 4)}{2\kappa + 1} \right] \\
& + \frac{iz^3}{15} \left[ -12\kappa^4 - 72\kappa^3 - 6\kappa^2 + 56\kappa + \frac{3611}{60} - 8(12\kappa^4 - 20\kappa^2 + 15)\kappa \log\left(\frac{2\kappa + 1}{2r}\right) \right] \\
& - 8 \left[ \frac{4i}{\kappa z} - i\left(4\kappa^3 + \frac{1}{\kappa}\right)z - \left(1 - 2\kappa^2\right)^2 z^2 + \frac{i\kappa(12\kappa^4 - 20\kappa^2 + 15)}{15}z^3 \right] \text{Ei}[-2i\kappa(z_{\text{in}} - z)] \\
& - 8 \left[ -\frac{i}{z}\left(\frac{4}{\kappa} + 5\right) + \frac{iz}{\kappa}(4\kappa^4 + 3\kappa + 1) + (1 - 2\kappa^2)^2 z^2 - \frac{iz^3}{120}(96\kappa^5 - 160\kappa^3 + 120\kappa + 43) \right] \text{Ei}[i(2\kappa + 1)(z - z_{\text{in}})] \\
& - \pi \left( \frac{43z^3}{15} - 24z + \frac{40}{z} \right) \\
& + e^{-2i\kappa(z_{\text{in}} - z)} \left\{ -\frac{224z_{\text{in}}^3}{5(z_{\text{in}} - z)^5} + \frac{16z_{\text{in}}}{5(z_{\text{in}} - z)^4} \left(-13i\kappa z_{\text{in}}^2 - \frac{15i}{\kappa} + 7z_{\text{in}}\right) + \frac{16}{15(z_{\text{in}} - z)^3} \right. \\
& \left. \times \left[ 2(12\kappa^2 - 5)z_{\text{in}}^3 + 27i\kappa z_{\text{in}}^2 - \frac{15i}{\kappa} + 84z_{\text{in}} \right] + \frac{16}{15(z_{\text{in}} - z)^2} \left[ i\kappa(6\kappa^2 - 5)z_{\text{in}}^3 - 12\kappa^2 z_{\text{in}}^2 \right. \right. \\
& \left. \left. + \frac{3i(21\kappa^2 - 5)z_{\text{in}}}{\kappa} + \frac{30i}{\kappa z_{\text{in}}} + 27 \right] + \frac{4}{15(z_{\text{in}} - z)} \left[ -12i\kappa(6\kappa^2 - 5)z_{\text{in}}^2 - 24(3\kappa^2 - 5)z_{\text{in}} \right. \\
& \left. + \frac{12i(3\kappa^2 - 5)}{\kappa} - (12\kappa^4 - 20\kappa^2 + 15)z_{\text{in}}^3 + \frac{120i}{\kappa z_{\text{in}}^2} - \frac{120}{z_{\text{in}}} \right] + \frac{2}{z} \left[ -\frac{i(2\kappa^2 - 1)^2}{\kappa^3} + \frac{16i}{\kappa z_{\text{in}}^2} \right] \right\}
\end{aligned}
$$
$$
\begin{equation}
\begin{aligned}
& -\frac{16}{z_{\text{in}}} \Biggl[ \frac{4}{15} \Biggl[ 8i\kappa (6\kappa^2 - 5) z_{\text{in}} + (12\kappa^4 - 20\kappa^2 + 15) z_{\text{in}}^2 - \frac{36\kappa^4 + 20\kappa^2 + 15}{\kappa^2} \Biggr] \\
& \qquad + \frac{4z^2}{15} (12\kappa^4 - 20\kappa^2 + 15) + \frac{2z}{15} \Biggl[ 2(12\kappa^4 - 20\kappa^2 + 15) z_{\text{in}} + \frac{i(108\kappa^4 - 100\kappa^2 + 15)}{\kappa} \Biggr] \Biggr\} \\
& \qquad + e^{-i(2\kappa+1)(z_{\text{in}}-z)} \left\{ \frac{224z_{\text{in}}^3}{5(z_{\text{in}}-z)^5} + \frac{2z_{\text{in}}}{5(z_{\text{in}}-z)^4} \Biggl[ i(104\kappa + 97)z_{\text{in}}^2 + \frac{120i}{\kappa} - 56z_{\text{in}} \Biggr] \right. \\
& \qquad \left. + \frac{1}{15(z_{\text{in}}-z)^3} \Biggl[ -2(192\kappa^2 + 222\kappa + 43) z_{\text{in}}^3 - 18i(24\kappa + 17)z_{\text{in}}^2 - \frac{48(28\kappa + 15)z_{\text{in}}}{\kappa} \right. \\
& \qquad \left. + \frac{240i}{\kappa} \Biggr] + \frac{1}{15(z_{\text{in}}-z)^2} \Biggl[ -\frac{6i(168\kappa^2 + 169\kappa + 20)}{\kappa} z_{\text{in}} - i(96\kappa^3 + 204\kappa^2 + 52\kappa - 43) z_{\text{in}}^3 \right. \\
& \qquad \left. + 6(2\kappa + 1)(16\kappa + 23)z_{\text{in}}^2 - \frac{480i}{\kappa z_{\text{in}}} - \frac{48(9\kappa + 5)}{\kappa} \Biggr] + \frac{1}{15(z_{\text{in}}-z)} \Biggl[ 3i(96\kappa^3 + 44\kappa^2 \\
& \qquad - 28\kappa - 3) z_{\text{in}}^2 + (48\kappa^4 - 24\kappa^3 - 68\kappa^2 + 34\kappa + 43) z_{\text{in}}^3 - \frac{480i}{\kappa z_{\text{in}}^2} \right. \\
& \qquad \left. + \frac{6(2\kappa + 1)(3\kappa + 4)(8\kappa - 5)z_{\text{in}}}{\kappa} + \frac{480(\kappa + 1)}{\kappa z_{\text{in}}} - \frac{6i(3\kappa + 4)(8\kappa - 5)}{\kappa} \right] \\
& \qquad + \frac{16}{z} \Biggl[ -\frac{2i}{\kappa z_{\text{in}}^2} + \frac{2(\kappa + 1)}{\kappa z_{\text{in}}} + \frac{i\kappa(\kappa + 1)}{2\kappa + 1} \Biggr] + \frac{1}{15} \Biggl[ -2i(96\kappa^3 - 36\kappa^2 - 68\kappa + 17) z_{\text{in}} \\
& \qquad + \frac{2(72\kappa^3 - 18\kappa^2 + 43\kappa + 60)}{\kappa} - (48\kappa^4 - 24\kappa^3 - 68\kappa^2 + 34\kappa + 43) z_{\text{in}}^2 \Biggr] - \frac{z}{15} \Biggl[ (48\kappa^4 \\
& \qquad - 24\kappa^3 - 68\kappa^2 + 34\kappa + 43) z_{\text{in}} + \frac{i(432\kappa^4 + 24\kappa^3 - 412\kappa^2 - 34\kappa + 77)}{2\kappa + 1} \Biggr] \\
& \qquad - \frac{z^2}{15} (48\kappa^4 - 24\kappa^3 - 68\kappa^2 + 34\kappa + 43) \Biggr\} + \pi \left\{ -zz_{\text{in}}^2 \delta'''(z_{\text{in}} - z) \right. \\
& \qquad \left. + 2z_{\text{in}}(3z - 2z_{\text{in}})\delta''(z_{\text{in}} - z) + \frac{2}{3} \Biggl[ -\frac{6z_{\text{in}}^2}{z} - 5z(z_{\text{in}}^2 + 3) + 36z_{\text{in}} \Biggr] \delta'(z_{\text{in}} - z) \right. \\
& \qquad \left. + \frac{4}{3} \left( 14zz_{\text{in}} + \frac{18z_{\text{in}}}{z} - 9z_{\text{in}}^2 - 30 \right) \delta(z_{\text{in}} - z) \right\}.
\end{aligned}
\tag{C.65}
\end{equation}
$$

The limit $z \ll z_{\text{in}} \ll 1$ of the above is

$$
\begin{equation}
\begin{split}
\mathfrak{N}_k(\eta, \eta_{\text{in}}) \approx {}& \frac{\varepsilon_1 H^2 k}{1024 \pi^2 M_p^2} \left( -\frac{20\pi}{z} + \frac{4i}{z} \left\{ 10 \log[e^\gamma(2\kappa+1)z_{\text{in}}] - 7 \right\} + \left( -32\kappa - 8 - \frac{64}{3\kappa} + \frac{4}{\kappa^2} \right) \right. \\
& \left. -i \frac{40}{z} \left[ \frac{1}{\epsilon} + \log\left(\frac{2\kappa+1}{M}\right) \right] + \mathcal{O}(z) \right).
\end{split}
\tag{C.66}
\end{equation}
$$
With this formula, along with the earlier formula (C.59) in the same limit, we get

$$
\begin{align}
\mathfrak{M}_k(\eta, \eta_{\text{in}}) &= \mathfrak{N}_k(\eta, \eta_{\text{in}}) - \eta \mathfrak{F}_k(\eta, \eta_{\text{in}}) = \mathfrak{N}_k(\eta, \eta_{\text{in}}) + \frac{z}{k} \mathfrak{F}_k(\eta, \eta_{\text{in}}) \nonumber \\
&= \frac{\varepsilon_1 H^2 k}{1024 \pi^2 M_{\text{P}}^2} \left\{ -\frac{40i}{z} \log \left( \frac{z}{z_{\text{in}}} \right) + \frac{40i}{z_{\text{in}}} - \frac{20i}{3} z \left[ \frac{1}{\epsilon} + \log \left( \frac{2\kappa + 1}{M} \right) \right] \right. \nonumber \\
&\qquad \left. + 4 \left( -\frac{1}{\kappa^2} + 8\kappa + \frac{16}{3\kappa} + 2 \right) \log \left( \frac{z}{ez_{\text{in}}} \right) + \mathcal{O}(z, z_{\text{in}}) \right\}, \tag{C.67}
\end{align}
$$

where (as before) $M = \mu/k$.

D Infrared volume factors

After splitting apart the Nakajima-Zwanzig equation into different (continuous) momenta $k \in \mathbb{R}^{3+}$ in eq. (3.20) there appear volume factors $\mathcal{V}$ on the l.h.s. so that the equations make sense dimensionally. This appendix derives the necessity of these volume factors, in the simpler setting where there are no interactions at all where

$$
\frac{\mathcal{V}}{(2\pi)^3} \frac{\partial \varrho_{sk}}{\partial \eta} = -i [\mathcal{H}_{sk}(\eta), \varrho_{sk}(\eta)] \quad (\text{D.1})
$$

which is derived from the free Liouville equation. For clarity of notation, we omit the label
$\alpha = \text{R, I}$ from eq. (2.21) in the main text in this appendix. This is done by assuming an
ansatz of the form (2.42) for the reduced density matrix, repeated here,

$$
\varrho(\eta) = \bigotimes_{q<k_{UV}} \varrho_q(\eta), \qquad (\text{D.2})
$$

which assumes the momentum label is continuous. We justify the presence of the volume
factors in eq. (D.1) here by passing to the limit where discrete momenta are considered, so
that the system is placed inside a box of volume $\mathcal{V}$. In this case the conversion between
discrete and continuum normalization is given by

$$
\sum_k = \frac{\mathcal{V}}{(2\pi)^3} \int d^3\mathbf{k}, \quad C_k(\eta_{\text{in}}) = \left[\frac{(2\pi)^3}{\mathcal{V}}\right]^{1/2} c_k(\eta_{\text{in}}), \quad V_k(\eta) = \left[\frac{(2\pi)^3}{\mathcal{V}}\right]^{1/2} v_k(\eta), \quad (\text{D.3})
$$

where $C_k(\eta_{\text{in}})$ are the discretely normalized annihilation operators, and $V_k(\eta)$ are the discretely normalized Fourier transforms of the Mukhanov-Sasaki field, and so on. With this, continuum momentum field expansions like

$$
v(\eta, \boldsymbol{x}) = \int \frac{d^3\boldsymbol{k}}{(2\pi)^{3/2}} v_k(\eta) e^{i\boldsymbol{k}\cdot\boldsymbol{x}} = \int \frac{d^3\boldsymbol{k}}{(2\pi)^{3/2}} \left[ u_k(\eta) c_k(\eta_{\text{in}}) + u_k^*(\eta) c_{-\boldsymbol{k}}^\dagger(\eta_{\text{in}}) \right] e^{i\boldsymbol{k}\cdot\boldsymbol{x}} \quad (\text{D.4})
$$

become instead

$$
v(\eta, \boldsymbol{x}) = \frac{1}{\sqrt{\mathcal{V}}} \sum_k V_k(\eta) e^{i\boldsymbol{k}\cdot\boldsymbol{x}} = \frac{1}{\sqrt{\mathcal{V}}} \sum_k \left[ u_k(\eta) C_k(\eta_{\text{in}}) + u_k^*(\eta) C_{-k}^\dagger(\eta_{\text{in}}) \right] e^{i\boldsymbol{k}\cdot\boldsymbol{x}} . \quad (\text{D.5})
$$
The tensor product structure of the ansatz (D.2) is now the same but over a discrete label — this allows us to compute the time-derivative of $\rho_S$ as

$$ 
\begin{aligned}
\frac{\partial \rho_S}{\partial \eta} &= \frac{\partial}{\partial \eta} \left[ \bigotimes_{k<k_{UV}} \rho_{Sk}(\eta) \right] = \sum_{k<k_{UV}} \frac{\partial \rho_{Ask}}{\partial \eta} \bigotimes_{q<k_{UV}, q \neq k} \rho_{Sq}(\eta) \\
&= \frac{V}{(2\pi)^3} \int_{k<k_{UV}} d^3 k \frac{\partial \rho_{Ask}}{\partial \eta} \bigotimes_{q<k_{UV}, q \neq k} \rho_{Sq}(\eta).
\end{aligned}
\quad (D.6) $$

On the other hand, the r.h.s. of the free Liouville equation does not end up with a volume factor by the above logic. To see why, recall the form of the free Hamiltonian (2.18)

$$ 
\begin{aligned}
\mathcal{H}_{AS}(\eta) &= \frac{1}{2} \int_{k<k_{UV}} d^3 k \left[ p_{Sk}(\eta) p_{Sk}^{\dagger}(\eta) + \omega^2(\mathbf{k}, \eta) v_{Sk}(\eta) v_{Sk}^{\dagger}(\eta) \right] = \frac{1}{2} \int_{k<k_{UV}} d^3 k \mathcal{H}_{Sk}(\eta) \\
&= \frac{1}{2} \sum_{k<k_{UV}} \left[ P_{Sk}(\eta) P_{Sk}^{\dagger}(\eta) + \omega^2(\mathbf{k}, \eta) V_{Sk}(\eta) V_{Sk}^{\dagger}(\eta) \right] = \sum_{k<k_{UV}} \mathfrak{h}_{Sk}(\eta)
\end{aligned}
\quad (D.7) $$

where we define the shorthand $\mathfrak{h}_{Sk}(\eta) := \frac{1}{2}[P_{Sk}(\eta)P_{Sk}^{\dagger}(\eta) + \omega^2(\mathbf{k}, \eta)V_{Sk}(\eta)V_{Sk}^{\dagger}(\eta)]$ in the last line, which is the discrete normalized version of the (free) Hamiltonian density $\mathcal{H}_{Sk}(\eta)$ — note that there is no volume factor since this operator is built from two fields. The above implies that:

$$ -i[\mathcal{H}_{AS}(\eta), \rho_{AS}(\eta)] = -i \left[ \sum_{k<k_{UV}} \mathfrak{h}_{Sk}(\eta), \bigotimes_{q<k_{UV}} \rho_{Sq}(\eta) \right] \quad (D.8) $$

$$ = -i \sum_{k<k_{UV}} [\mathfrak{h}_{Sk}(\eta), \rho_{Sk}(\eta)] \bigotimes_{q<k_{UV}, q \neq k} \rho_{Sq}(\eta) \quad (D.9) $$

$$ = -i \int_{k<k_{UV}} d^3 k [\mathcal{H}_{Sk}(\eta), \rho_{Sk}(\eta)] \bigotimes_{q<k_{UV}, q \neq k} \rho_{Sq}(\eta). \quad (D.10) $$

Using the above in the free Liouville equation yields eq. (D.1) with the desired volume factors.

# E Scalar decoherence from a tensor environment

## E.1 Correlation function in real space

Here we consider the $\zeta\gamma\gamma$ interaction,

$$ S_{\text{int}} = \frac{M_{\text{P}}^2}{8} \int dt \, d^3 \mathbf{x} \, a \, \varepsilon_1 \, \zeta \partial_{\ell} \gamma_{ij} \partial_{\ell} \gamma_{ij} \quad (\text{E.1}) $$

which, after using $a \, d\eta = dt$ corresponds to the interaction Hamiltonian

$$ \mathcal{H}_{\text{int}}(\eta) = -\frac{M_{\text{P}}^2 \varepsilon_1}{8} a^2 \int d^3 \mathbf{x} \, \zeta(\eta, \mathbf{x}) \otimes \partial_{\ell} \gamma_{ij}(\eta, \mathbf{x}) \partial_{\ell} \gamma_{ij}(\eta, \mathbf{x}). \quad (\text{E.2}) $$

Our interest is in how the environment of short-wavelength tensors decohering long-wavelength scalar fluctuations.
In terms of the canonical fields $v = aM_p\sqrt{2\epsilon_1}\zeta$ and $v_{ij} = \frac{1}{2}aM_p\gamma_{ij}$ — see eq. (A.17) and using $a^{-1} = -H\eta$ — we have

$$ \begin{aligned} \mathcal{H}_{\text{int}}(\eta) &= -\frac{\sqrt{\epsilon_1}}{2\sqrt{2}a(\eta)M_p} \int d^3\mathbf{x} v(\eta, \mathbf{x}) \otimes \partial_\ell v_{ij}(\eta, \mathbf{x}) \partial_\ell v_{ij}(\eta, \mathbf{x}) \\ &= G(\eta) \int d^3\mathbf{x} v(\eta, \mathbf{x}) \otimes B_{\text{T}}(\eta, \mathbf{x}) \end{aligned} \quad (E.3) $$

where $G(\eta)$ is the same as that defined in eq. (3.3) and

$$ B_{\text{T}}(\eta, \mathbf{x}) := \partial_{\ell}v_{ij}(\eta, \mathbf{x})\partial_{\ell}v_{ij}(\eta, \mathbf{x}). \quad (E.4) $$

Next we concern ourselves with the mode expansion for the graviton, noting that the free part of the graviton action is

$$ \begin{aligned} {}^{(2)}S[\gamma] &= \frac{M_p^2}{8} \int d\eta \, d^3\mathbf{x} \, a^2 \left( \gamma'_{ij} \gamma'_{ij} - \partial_\ell \gamma_{ij} \partial_\ell \gamma_{ij} \right) = \frac{1}{2} \int d\eta \, d^3\mathbf{x} \left( v'_{ij} v'_{ij} - \partial_\ell v_{ij} \partial_\ell v_{ij} + \frac{2}{\eta^2} v_{ij} v_{ij} \right) \\ &= \frac{1}{2} \sum_{P=+,\times} \int d\eta \, d^3\mathbf{k} \left[ |(v'_{\mathbf{k}})'|^2 - \left( k^2 - \frac{2}{\eta^2} \right) |v'_{\mathbf{k}}|^2 \right] \end{aligned} \quad (E.5) $$

which uses $\dot{\gamma}_{ij} = a^{-1}\gamma'_{ij}$ as well as $a\gamma'_{ij} = \frac{2}{M_p}(v'_{ij} + v_{ij}/\eta)$, and the expansion in terms of graviton modes, already introduced in eq. (A.18), and repeated here for convenience

$$ v_{ij}(\eta, \mathbf{x}) = \int \frac{d^3\mathbf{k}}{(2\pi)^{3/2}} \sum_{P=+,-} \epsilon_{ij}^{P}(\mathbf{k}) v_{\mathbf{k}}^{P}(\eta) e^{i\mathbf{k}\cdot\mathbf{x}} \quad (E.6) $$

where we recall that $\epsilon_{ij}^P(\mathbf{k})$ is the polarization tensor. Let us explain how they are defined concretely. We here follow the conventions of ref. [75], up to numerical factors. For a given momentum vector $\hat{\mathbf{k}} = \mathbf{k}/k$ pointing along the direction $(\theta, \varphi)$ in polar coordinates we define the vectors

$$ \mathbf{e}^x(\hat{\mathbf{k}}) = \begin{pmatrix} \cos \theta \cos \varphi \\ \cos \theta \sin \varphi \\ -\sin \theta \end{pmatrix}, \quad \mathbf{e}^y(\hat{\mathbf{k}}) = \begin{pmatrix} -\sin \varphi \\ \cos \varphi \\ 0 \end{pmatrix}. \quad (E.7) $$

These vectors are perpendicular to $\mathbf{k}$ as well as to each other where

$$ \mathbf{e}^L(\hat{\mathbf{k}}) \cdot \mathbf{e}^{L'}(\hat{\mathbf{k}}) = \delta_{LL'}, \quad \mathbf{k} \cdot \mathbf{e}^L(\hat{\mathbf{k}}) = 0, \quad L, L' \in (x, y), \quad (E.8) $$

and under reflection $\mathbf{k} \to -\mathbf{k}$ these satisfy (this means taking $\theta \to \pi - \theta$ and $\varphi \to \varphi + \pi$)

$$ \mathbf{e}^x(-\hat{\mathbf{k}}) = \mathbf{e}^x(\hat{\mathbf{k}}), \quad \mathbf{e}^y(-\hat{\mathbf{k}}) = -\mathbf{e}^y(\hat{\mathbf{k}}). \quad (E.9) $$

Note that when $\hat{\mathbf{k}} = \hat{\mathbf{z}}$ with $\varphi = \theta = 0$ then the above reduce to $\mathbf{e}^x(\hat{\mathbf{z}}) = (1, 0, 0)$ and $\mathbf{e}^y(\hat{\mathbf{z}}) = (0, 1, 0)$. Next construct $\epsilon^+ := \frac{1}{\sqrt{2}}(\mathbf{e}^x \otimes \mathbf{e}^x - \mathbf{e}^y \otimes \mathbf{e}^y)$ and $\epsilon^\times := \frac{1}{\sqrt{2}}(\mathbf{e}^x \otimes \mathbf{e}^y + \mathbf{e}^y \otimes \mathbf{e}^x)$, or more simply in terms of components as:

$$ \epsilon_{ij}^{+}(\mathbf{k}) := \frac{1}{\sqrt{2}} \left[ e_i^x(\hat{\mathbf{k}}) e_j^x(\hat{\mathbf{k}}) - e_i^y(\hat{\mathbf{k}}) e_j^y(\hat{\mathbf{k}}) \right], \quad (E.10) $$

$$ \epsilon_{ij}^{\times}(\mathbf{k}) := \frac{1}{\sqrt{2}} \left[ e_i^x(\hat{\mathbf{k}}) e_j^y(\hat{\mathbf{k}}) + e_i^y(\hat{\mathbf{k}}) e_j^x(\hat{\mathbf{k}}) \right]. \quad (E.11) $$
These are the standard linear polarizations, which for the familiar case of $k \propto \hat{z}$ simplify to

$$ \epsilon^{+}(\hat{z}) = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{pmatrix}, \qquad \epsilon^{\times}(\hat{z}) = \frac{1}{\sqrt{2}} \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}. \qquad (E.12) $$

Note that these are normalized as

$$ \epsilon_{ij}^{P}(\mathbf{k}) \epsilon_{ij}^{P'}(\mathbf{k}) = \delta_{PP'}. \qquad (E.13) $$

Importantly the symmetry under $k \to -k$ means that $\epsilon_{ij}^{+}(-k) = \epsilon_{ij}^{+}(k)$ as well as $\epsilon_{ij}^{\times}(-k) = -\epsilon_{ij}^{\times}(k)$. This implies that

$$ \epsilon_{ij}^{+}(-\mathbf{k})\epsilon_{ij}^{+}(\mathbf{k}) = 1, \qquad \epsilon_{ij}^{\times}(-\mathbf{k})\epsilon_{ij}^{\times}(\mathbf{k}) = -1. \qquad (E.14) $$

Furthermore we have the identify, see eq. (2.21) of ref. [76]

$$ \sum_{P} \epsilon_{ij}^{P}(\mathbf{k}) \epsilon_{nm}^{P}(\mathbf{k}) = \frac{1}{2} [\perp_{in}(\mathbf{k}) \perp_{jm}(\mathbf{k}) + \perp_{im}(\mathbf{k}) \perp_{jn}(\mathbf{k}) - \perp_{ij}(\mathbf{k}) \perp_{nm}(\mathbf{k})] \qquad (E.15) $$

where $\perp_{ij}(\mathbf{k}) = \delta_{ij} - k_i k_j / k^2$ is a (symmetric) projection tensor such that $k_i \perp_{ij} = 0$.

As for scalars, the reality of $v_{ij}(\eta, \mathbf{x})$ implies that $v_{\mathbf{k}}^{+*}(\eta) = v_{-\mathbf{k}}^{+*}(\eta)$ and $v_{\mathbf{k}}^{\times*}(\eta) = -v_{-\mathbf{k}}^{\times*}(\eta)$. This has the same mode expansion as the Mukhanov-Sasaki field with an extra polarization label summing over $P = +, \times$. We then expand

$$ v_{\mathbf{k}}^{P}(\eta) = u_{\mathbf{k}}(\eta) c_{\mathbf{k}}^{P} + s_{P} u_{\mathbf{k}}^{*}(\eta) c_{-\mathbf{k}}^{P\dagger} \qquad (E.16) $$

with $s_P = 1$ for $P = +$ and $s_P = -1$ for $P = \times$, where $u_k$ are the Bunch-Davies mode functions (same as for the scalar) and the ladder operators satisfy $[c_k^P, c_{-k}^{P'\dagger}] = \delta(\mathbf{k} - \mathbf{q})\delta_{PP'}$ and so on. The above then implies that the operator $B_T(\eta, \mathbf{x})$ has the expansion

$$ B_T(\eta, \mathbf{x}) := - \sum_{P, P' = +, \times} \int_{k, q > k_{UV}} \frac{d^3 \mathbf{k} \, d^3 \mathbf{q}}{(2\pi)^3} (\mathbf{k} \cdot \mathbf{q}) \epsilon_{ij}^{P}(\mathbf{k}) \epsilon_{ij}^{P'}(\mathbf{q}) v_{\mathbf{k}}^{P}(\eta) v_{\mathbf{q}}^{P'}(\eta) e^{i(\mathbf{k}+\mathbf{q})\cdot\mathbf{x}}, \qquad (E.17) $$

which can be used to simplify the one-point function,

$$ \mathcal{B}_T(\eta) := \langle 0_B | B_T(\eta, \mathbf{x}) | 0_B \rangle \qquad (E.18) $$

$$ = - \sum_{P, P' = +, \times} \int_{k, q > k_{UV}} \frac{d^3 \mathbf{k} \, d^3 \mathbf{q}}{(2\pi)^3} (\mathbf{k} \cdot \mathbf{q}) \epsilon_{ij}^{P}(\mathbf{k}) \epsilon_{ij}^{P'}(\mathbf{q}) s_{P'} u_{\mathbf{k}}(\eta) u_{\mathbf{q}}^{*}(\eta) \delta(\mathbf{k} + \mathbf{q}) \delta_{PP'} e^{i(\mathbf{k} + \mathbf{q}) \cdot \mathbf{x}} \qquad (E.19) $$

$$ = 2 \int_{k > k_{UV}} \frac{d^3 \mathbf{k}}{(2\pi)^3} k^2 |u_{\mathbf{k}}(\eta)|^2, \qquad (E.20) $$

where we have used $\epsilon_{ij}^{P}(-\mathbf{k})s_P = \epsilon_{ij}^{P}(\mathbf{k})$ and then $\epsilon_{ij}^{P}(\mathbf{k})\epsilon_{ij}^{P'}(\mathbf{k}) = 1$ for each $P = +, \times$. Using the one-point function $\mathcal{B}(\eta)$ defined in eq. (B.2), we find that

$$ \mathcal{B}_T(\eta) = 2 \mathcal{B}(\eta). \qquad (E.21) $$
Let us now turn to the calculation of the two-point correlation function. Using the expectation values

$$ \langle 0_B | c_k^P c_{-q}^{P'\dagger} c_p^{P''} c_{-\ell}^{P'''\dagger} | 0_B \rangle = \delta(\mathbf{k} + \mathbf{q})\delta(\mathbf{p} + \mathbf{l})\delta_{PP'}\delta_{P''P''}, \quad (E.22) $$

$$ \langle 0_B | c_k^P c_q^{P'} c_{-p}^{P''\dagger} c_{-\ell}^{P'''\dagger} | 0_B \rangle = \delta(\mathbf{k} + \mathbf{l})\delta(\mathbf{p} + \mathbf{q})\delta_{PP''}\delta_{P'P''} + \delta(\mathbf{k} + \mathbf{p})\delta(\mathbf{l} + \mathbf{q})\delta_{PP''}\delta_{P'P''} \quad (E.23) $$

this quantity simplifies to

$$ \begin{aligned} \langle 0_B | B_T(\eta, \mathbf{x}) B_T(\eta', \mathbf{x}') | 0_B \rangle = & \mathcal{B}_T(\eta)\mathcal{B}_T(\eta') + 2 \sum_{P,P',P'',P'''} \int_{k,q,p,\ell>k_{UV}} \frac{d^3\mathbf{k} \, d^3\mathbf{q} \, d^3\mathbf{p} \, d^3\mathbf{l}}{(2\pi)^6} \\ & \times \epsilon_{ij}^P(\mathbf{k}) \epsilon_{ij}^{P'}(\mathbf{q}) \epsilon_{nm}^{P''}(\mathbf{p}) \epsilon_{nm}^{P'''}(\mathbf{l}) (\mathbf{k} \cdot \mathbf{q})(\mathbf{p} \cdot \mathbf{l}) e^{i(\mathbf{k}+\mathbf{q})\cdot\mathbf{x}+i(\mathbf{p}+\mathbf{l})\cdot\mathbf{x}'} \\ & \times u_k(\eta)u_q(\eta)u_p^*(\eta')u_l^*(\eta')s_{P''}s_{P'''}\delta(\mathbf{k}+\mathbf{l})\delta(\mathbf{p}+\mathbf{q})\delta_{PP''}\delta_{P'P''}. \end{aligned} \quad (E.24) $$

Our interest is in the centred two-point function, defined by

$$ \begin{aligned} C_T(\eta, \eta'; \mathbf{x} - \mathbf{x}') &:= \langle 0_B | [B_T(\eta, \mathbf{x}) - \mathcal{B}_T(\eta)] [B_T(\eta', \mathbf{x}') - \mathcal{B}_T(\eta')] | 0_B \rangle \\ &= \langle 0_B | B_T(\eta, \mathbf{x}) B_T(\eta', \mathbf{x}') | 0_B \rangle - \mathcal{B}_T(\eta) \mathcal{B}_T(\eta'), \end{aligned} \quad (E.25) $$

which after using eq. (E.24), integrating over the $\delta$-functions and using $s_P \epsilon_{nm}^P(-\mathbf{k}) = \epsilon_{nm}^P(\mathbf{k})$ gives rise to

$$ C_T(\eta, \eta'; \mathbf{y}) = \sum_{P,P'} \int_{k,q>k_{UV}} \frac{d^3\mathbf{k} \, d^3\mathbf{q}}{(2\pi)^6} \epsilon_{ij}^P(\mathbf{k}) \epsilon_{ij}^{P'}(\mathbf{q}) \epsilon_{nm}^{P''}(\mathbf{q}) \epsilon_{nm}^{P'''}(\mathbf{k}) 2(\mathbf{k} \cdot \mathbf{q})^2 \\ \times u_k(\eta) u_q(\eta) u_q^*(\eta') u_k^*(\eta') e^{i(\mathbf{k}+\mathbf{q})\cdot\mathbf{y}}. \quad (E.26) $$

Using the identity (E.15) involving the projection $\perp_{ij}(\mathbf{k}) := \delta_{ij} - k_i k_j / k^2$ the summations over polarizations can be simplified to

$$ \begin{aligned} C_T(\eta, \eta'; \mathbf{y}) = & \int_{k,q>k_{UV}} \frac{d^3\mathbf{k} \, d^3\mathbf{q}}{(2\pi)^6} \frac{1}{2} [\perp_{\text{in}}(\mathbf{k}) \perp_{jm}(\mathbf{k}) + \perp_{im}(\mathbf{k}) \perp_{jn}(\mathbf{k}) - \perp_{ij}(\mathbf{k}) \perp_{nm}(\mathbf{k})] \\ & \times \frac{1}{2} [\perp_{\text{in}}(\mathbf{q}) \perp_{jm}(\mathbf{q}) + \perp_{im}(\mathbf{q}) \perp_{jn}(\mathbf{q}) - \perp_{ij}(\mathbf{q}) \perp_{nm}(\mathbf{q})] 2(\mathbf{k} \cdot \mathbf{q})^2 \\ & \times u_k(\eta) u_q(\eta) u_q^*(\eta') u_k^*(\eta') e^{i(\mathbf{k}+\mathbf{q})\cdot\mathbf{y}}. \end{aligned} \quad (E.27) $$

There is a sum over the indices $i, j, n, m$, and so to this end we note that

$$ \perp_{\text{in}}(\mathbf{k}) \perp_{\text{in}}(\mathbf{q}) = 1 + \frac{(\mathbf{k} \cdot \mathbf{q})^2}{k^2 q^2}, \quad (E.28) $$

$$ \perp_{\text{in}}(\mathbf{k}) \perp_{\text{in}}(\mathbf{q}) \perp_{jm}(\mathbf{k}) \perp_{jm}(\mathbf{q}) = \left[ 1 + \frac{(\mathbf{k} \cdot \mathbf{q})^2}{k^2 q^2} \right]^2, \quad (E.29) $$

$$ \perp_{\text{in}}(\mathbf{k}) \perp_{im}(\mathbf{q}) \perp_{jm}(\mathbf{k}) \perp_{jn}(\mathbf{q}) = 1 + \frac{(\mathbf{k} \cdot \mathbf{q})^4}{k^4 q^4}, \quad (E.30) $$
which are the only two types of contractions that occur in the above (after re-labeling i, j, n, m in various ways). After some manipulation the above implies that

$$C_T(\eta, \eta'; \mathbf{y}) = \int_{k,q>k_{UV}} \frac{d^3\mathbf{k} d^3\mathbf{q}}{(2\pi)^6} \frac{1}{4} \left[ 1 + 6 \frac{(\mathbf{k} \cdot \mathbf{q})^2}{k^2 q^2} + \frac{(\mathbf{k} \cdot \mathbf{q})^4}{k^4 q^4} \right] 2(\mathbf{k} \cdot \mathbf{q})^2 \\ \times u_k(\eta) u_q(\eta) u_q^*(\eta') u_k^*(\eta') e^{i(\mathbf{k}+\mathbf{q}) \cdot \mathbf{y}}. \qquad (E.31)$$

## E.2 Fourier transform of $C_T$

What appears in the Lindblad equation for this interaction is of course the Fourier transform of the above correlator which we define as

$$\mathcal{T}_k(\eta, \eta') := \int \frac{d^3\mathbf{y}}{(2\pi)^{3/2}} C_T(\eta, \eta'; \mathbf{y}) e^{-i\mathbf{k} \cdot \mathbf{y}}. \qquad (E.32)$$

We again are restricted to modes with $0 < k < k_{UV}$ and using eq. (E.31) we find

$$\mathcal{T}_k(\eta, \eta') = \frac{2}{(2\pi)^{9/2}} \int_{q,p>k_{UV}} d^3\mathbf{q} d^3\mathbf{p} (\mathbf{q} \cdot \mathbf{p})^2 \left[ \frac{1}{4} + \frac{3}{2} \frac{(\mathbf{q} \cdot \mathbf{p})^2}{q^2 p^2} + \frac{1}{4} \frac{(\mathbf{q} \cdot \mathbf{p})^4}{q^4 p^4} \right] \\ \times u_q(\eta) u_p(\eta) u_p^*(\eta') u_q^*(\eta') \delta(\mathbf{q} + \mathbf{p} - \mathbf{k}). \qquad (E.33)$$

From here the integration over the angles goes over exactly in the same manner as in appendix B.2, eventually giving rise to the expression

$$\mathcal{T}_k(\eta, \eta') = \frac{1}{32(2\pi)^{7/2}k} \int_0^k dQ \int_{Q+2k_{UV}}^{\infty} dP \left[ \frac{1}{4} + \frac{3}{2} \frac{(P^2+Q^2-2k^2)^2}{(P^2-Q^2)^2} + \frac{1}{4} \frac{(P^2+Q^2-2k^2)^4}{(P^2-Q^2)^4} \right] \\ \times (P^2+Q^2-2k^2)^2 \left[ 1 - \frac{2i}{(P-Q)\eta} \right] \left[ 1 + \frac{2i}{(P-Q)\eta'} \right] \left[ 1 - \frac{2i}{(P+Q)\eta} \right] \\ \times \left[ 1 + \frac{2i}{(P+Q)\eta'} \right] e^{-i(\eta-\eta')P}. \qquad (E.34)$$

## E.3 Super-Hubble limit of Lindblad coefficient

Next we must compute the super-Hubble limit of the Lindblad coefficient

$$\mathfrak{F}_k(\eta, \eta_{\text{in}}) := (2\pi)^{3/2} \int_{\eta_{\text{in}}}^{\eta} d\eta' G(\eta)G(\eta') [\mathcal{C}_k(\eta, \eta') + \mathcal{T}_k(\eta, \eta')] . \qquad (E.35)$$

We use the earlier representations (B.31) and (E.34) of $C_k$ and $\mathcal{T}_k$ above, as well as $G(\eta)G(\eta') = \varepsilon_1 H^2 \eta \eta' / (8M_p^2)$, which expresses $\mathfrak{F}_k$ as the triple integral

$$\mathfrak{F}_k(\eta, \eta_{\text{in}}) = \frac{\varepsilon_1 H^2}{1024 \pi^2 M_p^2 k} \int_{\eta_{\text{in}}}^{\eta} d\eta' \int_0^k dQ \int_{Q+2k_{\text{UV}}}^{\infty} dP \eta \eta' \left[ \frac{5}{4} + \frac{3}{2} \frac{(P^2+Q^2-2k^2)^2}{(P^2-Q^2)^2} \right. \\ \left. + \frac{1}{4} \frac{(P^2+Q^2-2k^2)^4}{(P^2-Q^2)^4} \right] (P^2+Q^2-2k^2)^2 \left[ 1 - \frac{2i}{(P-Q)\eta} \right] \left[ 1 + \frac{2i}{(P-Q)\eta'} \right] \\ \times \left[ 1 - \frac{2i}{(P+Q)\eta} \right] \left[ 1 + \frac{2i}{(P+Q)\eta'} \right] e^{-iP(\eta-\eta')}. \qquad (E.36)$$
We first evaluate the $\eta'$-integral using the formula (C.3),

$$ \int_{\eta_{\text{in}}}^{\eta} d\eta' \eta' \left[ 1 + \frac{2i}{(P-Q)\eta'} \right] \left[ 1 + \frac{2i}{(P+Q)\eta'} \right] e^{-i(\eta-\eta')P} = \left[ e^{-iP(\eta-\eta')} \left( \frac{1}{P^2} - \frac{i\eta'}{P} + \frac{4}{P^2-Q^2} \right) - \frac{4e^{-iP\eta}\text{Ei}(iP\eta')}{P^2-Q^2} \right] \Bigg|_{\eta' \to \eta_{\text{in}}}^{\eta' \to \eta} \quad (\text{E.37}) $$

$$ = \frac{1}{P^2} - \frac{i\eta}{P} + \frac{4}{P^2 - Q^2} - \frac{4e^{-iP\eta}[\text{Ei}(iP\eta) + i\pi]}{P^2 - Q^2} - e^{-iP(\eta-\eta_{\text{in}})} \left( \frac{1}{P^2} - \frac{i\eta_{\text{in}}}{P} + \frac{4}{P^2 - Q^2} \right) + \frac{4e^{-iP\eta}[\text{Ei}(iP\eta_{\text{in}}) + i\pi]}{P^2 - Q^2}, \quad (\text{E.38}) $$

where we have added and subtracted an extra factor of $4i\pi e^{-iP\eta}/(P^2 - Q^2)$ for convenience later on. With this we find that

$$ \mathfrak{F}_k(\eta, \eta_{\text{in}}) = \frac{\varepsilon_1 H^2}{1024 \pi^2 M_{\text{P}}^2} [h(\eta) + g(\eta, \eta_{\text{in}})] \quad (\text{E.39}) $$

with the definitions

$$ h(\eta) := \int_{0}^{k} dQ \int_{Q+2k_{\text{UV}}}^{\infty} dP \eta \left[ \frac{5}{4} + \frac{3}{2} \frac{(P^2+Q^2-2k^2)^2}{(P^2-Q^2)^2} + \frac{1}{4} \frac{(P^2+Q^2-2k^2)^4}{(P^2-Q^2)^4} \right] \\ \times \frac{(P^2+Q^2-2k^2)^2}{k} \left[ 1 - \frac{2i}{(P-Q)\eta} \right] \left[ 1 - \frac{2i}{(P+Q)\eta} \right] \left\{ \frac{1}{P^2} - \frac{i\eta}{P} + \frac{4}{P^2-Q^2} - \frac{4e^{-iP\eta}[\text{Ei}(iP\eta) + i\pi]}{P^2-Q^2} \right\}, \quad (\text{E.40}) $$

$$ g(\eta, \eta_{\text{in}}) := \int_{0}^{k} dQ \int_{Q+2k_{\text{UV}}}^{\infty} dP \eta \left[ \frac{5}{4} + \frac{3}{2} \frac{(P^2+Q^2-2k^2)^2}{(P^2-Q^2)^2} + \frac{1}{4} \frac{(P^2+Q^2-2k^2)^4}{(P^2-Q^2)^4} \right] \\ \times \frac{(P^2+Q^2-2k^2)^2}{k} \left[ 1 - \frac{2i}{(P-Q)\eta} \right] \left[ 1 - \frac{2i}{(P+Q)\eta} \right] \left\{ -e^{-iP(\eta-\eta_{\text{in}})} \left( \frac{1}{P^2} - \frac{i\eta_{\text{in}}}{P} + \frac{4}{P^2-Q^2} \right) + \frac{4e^{-iP\eta}[\text{Ei}(iP\eta_{\text{in}}) + i\pi]}{P^2-Q^2} \right\}. \quad (\text{E.41}) $$

Our goal will be written down the $0 \ll -k\eta \ll -k\eta_{\text{in}} \ll 1$ limit of $\mathfrak{F}_k(\eta, \eta_{\text{in}})$.

Let us now study the super-Hubble limit of $h$. As before, we make use of the variables $z = -k\eta$, $z_{\text{in}} = -k\eta_{\text{in}}$ and $\kappa := k_{\text{UV}}/k$ and change the (positive) integration variables to $x = -Q\eta$ and $y = -P\eta$ which turns eq. (E.40) into

$$ h(\eta) := \int_{0}^{z} dx \frac{f(x, z)}{z^3} \quad (\text{E.42}) $$

with $f$ defined by

$$ f(x, z) := \int_{x+2\kappa z}^{\infty} dy \left[ \frac{5}{4} + \frac{3}{2} \frac{(x^2+y^2-2z^2)^2}{(x^2-y^2)^2} + \frac{1}{4} \frac{(x^2+y^2-2z^2)^4}{(x^2-y^2)^4} \right] (x^2+y^2-2z^2)^2 \\ \times \left( 1 - \frac{2i}{x-y} \right) \left( 1 + \frac{2i}{x+y} \right) \left\{ -\frac{1}{y^2} - \frac{i}{y} + \frac{4}{x^2-y^2} - \frac{4e^{iy}[\text{Ei}(-iy) + i\pi]}{x^2-y^2} \right\}. \quad (\text{E.43}) $$
We next notice the Taylor series about $z = 0$,

$$ \int_0^z dx f(x, z) \approx f(0, 0)z + \mathcal{O}(z^{-1}), \qquad (E.44) $$

for $0 < z \ll 1$, which implies that $f$ contributes to $h$ in eq. (E.42) at order $z^{-2}$ for small $z$, such that

$$ h(\eta) \approx \frac{f(0, 0)}{z^2} + \mathcal{O}(z^{-1}) \qquad (E.45) $$

with coefficient through eq. (E.43) given as

$$ f(0, 0) := 3 \int_0^\infty dy \, y^4 \left(1 + \frac{2i}{y}\right)^2 \left\{ -\frac{5 + iy}{y^2} + \frac{4e^{iy}}{y^2} [\mathrm{Ei}(-iy) + i\pi] \right\}. \qquad (E.46) $$

In order to compare to the earlier section, let us focus on the real part of the above where

$$ \mathrm{Re}[f(0, 0)] = 3 \int_0^\infty dy (20 - y^2 + 4(y^2 - 4)) \mathrm{Re}\left\{e^{iy}[\mathrm{Ei}(-iy) + i\pi]\right\} \\ -16y \mathrm{Im}\left\{e^{iy}[\mathrm{Ei}(-iy) + i\pi]\right\}. \qquad (E.47) $$

We next use (for $y > 0$)

$$ e^{iy} [\mathrm{Ei}(-iy) + i\pi] = \mathrm{Ci}(y) \cos y + \left[\mathrm{Si}(y) - \frac{\pi}{2}\right] \sin(y) + i \left\{\mathrm{Ci}(y) \sin y - \left[\mathrm{Si}(y) - \frac{\pi}{2}\right] \cos y\right\}, \qquad (E.48) $$

where the functions $\mathrm{Ci}(y)$ and $\mathrm{Si}(y)$ are defined by

$$ \mathrm{Ci}(y) = - \int_y^\infty dt \frac{\cos t}{t}, \qquad \mathrm{Si}(y) = \int_0^y dt \frac{\sin t}{t}, \qquad (E.49) $$

to write

$$ \mathrm{Re}[f(0, 0)] = 3 \int_0^\infty dy \left( -y^2 + 20 + 4(y^2 - 4) \left\{ \mathrm{Ci}(y) \cos y + \left[ \mathrm{Si}(y) - \frac{\pi}{2} \right] \sin y \right\} \right. \\ \left. -16y \left\{ \mathrm{Ci}(y) \sin y - \left[ \mathrm{Si}(y) - \frac{\pi}{2} \right] \cos y \right\} \right). \qquad (E.50) $$

For $y \gg 1$ the integrand in this expression behaves as

$$ \begin{aligned} & -y^2 + 20 + 4(y^2 - 4) \left\{ \mathrm{Ci}(y) \cos y + \left[ \mathrm{Si}(y) - \frac{\pi}{2} \right] \sin y \right\} - 16y \left\{ \mathrm{Ci}(y) \sin y - \left[ \mathrm{Si}(y) - \frac{\pi}{2} \right] \cos y \right\} \\ & \approx -y^2 + \frac{72}{y^2} + \mathcal{O}(y^{-4}) \end{aligned} \qquad (E.51) $$

and so the integral diverges in the UV. A similar divergence also arose when summing over scalar fluctuations, and corresponded to a distributional singularity in the correlation function near $\eta = \eta'$, see the discussion surrounding eq. (C.22). The distributional singularities do not contribute to the integration over $\eta'$ performed here, and for the present purposes it is convenient to have an alternative approach that provides a short-cut to the small-$z$ behaviour without fully evaluating the position-space correlation function for all values of its
arguments. We adopt here a regularization of this divergence that properly reproduces the small-z behaviour found in the more complete treatment of appendix C.

To this end we regulate the divergence by first isolating the UV-divergent part $\propto y^2$ of the function, leading to

$$ \mathrm{Re}[f(0,0)] = -3 \int_0^\infty dy \, y^2 + 3 \int_0^\infty dy \left( 20 - 4(y^2 - 4) \left\{ \mathrm{Ci}(y) \cos y + \left[ \mathrm{Si}(y) - \frac{\pi}{2} \right] \sin y \right\} \right. \\ \left. -16y \left\{ \mathrm{Ci}(y) \sin(y) - \left[ \mathrm{Si}(y) - \frac{\pi}{2} \right] \cos y \right\} \right). \qquad (\mathrm{E.52}) $$

We regulate the divergent integral $\int_0^\infty dy \, y^2$ in the spirit of dimensional regularization by writing

$$ \int_0^\infty dy \, y^2 \rightarrow \lim_{q \to 0^+} \lim_{n \to 0} \int_0^\infty dy \, y^n \left( \frac{y^4}{y^2 + q^2} \right) = \lim_{q \to 0^+} \lim_{n \to 0} \left[ \frac{\pi q^{n+3}}{2 \cos(n\pi/2)} \right] = 0, \qquad (\mathrm{E.53}) $$

where the initial integral only converges when $\mathrm{Re}(n) < -5$ and we introduce a parameter $q > 0$ to regulate the associated divergence in the IR (whose presence ultimately cancels the UV divergence). This leaves the convergent integral

$$ \mathrm{Re}[f^{\mathrm{reg}}(0,0)] = 3 \int_0^\infty dy \left( 20 + 4(y^2 - 4) \left\{ \mathrm{Ci}(y) \cos y + \left[ \mathrm{Si}(y) - \frac{\pi}{2} \right] \sin y \right\} \right. \\ \left. -16y \left\{ \mathrm{Ci}(y) \sin y - \left[ \mathrm{Si}(y) - \frac{\pi}{2} \right] \cos y \right\} \right) \qquad (\mathrm{E.54}) $$

$$ = \left\{ -4y + 4\mathrm{Ci}(y) \left[ 6y \cos y + (y^2 - 10) \sin y \right] - 4 \left[ (y^2 - 10) \cos y - 6y \sin y \right] \right. \\ \left. \times \left[ \mathrm{Si}(y) - \frac{\pi}{2} \right] \right|_{y \to 0}^{y \to \infty} = 60\pi \qquad (\mathrm{E.55}) $$

which is the result used in eq. (E.45).

Let us finally compute the super-Hubble behaviour of $g$ defined in eq. (E.41). We again use the variables $z$ and $z_{\mathrm{in}}$ and we here argue that

$$ g(\eta, \eta_{\mathrm{in}}) \sim \mathcal{O}(z^{-1}) \qquad (\mathrm{E.56}) $$

in the $z \ll 1$ limit and so is subdominant to $h$. To see why it is easier to use the integration $a := Q/k$ and $b := P/k$ which turns eq. (E.41) into

$$ g(\eta, \eta_{\mathrm{in}}) := \int_0^1 da \int_{a+2\kappa}^\infty db \left[ \frac{5}{4} + \frac{3(a^2+b^2-2)^2}{2(a^2-b^2)^2} + \frac{1(a^2+b^2-2)^4}{4(a^2-b^2)^4} \right] (a^2+b^2-2)^2 z \\ \times \left[ 1 - \frac{2i}{(a-b)z} \right] \left[ 1 + \frac{2i}{(a+b)z} \right] \left\{ e^{ib(z-z_{\mathrm{in}})} \left( \frac{1}{b^2} + \frac{iz_{\mathrm{in}}}{b} - \frac{4}{a^2-b^2} \right) \right. \\ \left. + \frac{4e^{izb}[\mathrm{Ei}(-iz_{\mathrm{in}}b) + i\pi]}{a^2-b^2} \right\}. \qquad (\mathrm{E.57}) $$

We then write the above as

$$ g(z, z_{\mathrm{in}}) := \int_0^1 da \, \mathcal{G}(a, z, z_{\mathrm{in}}) \qquad (\mathrm{E.58}) $$
where we have

$$ \mathcal{G}(a, z, z_{\text{in}}) := \int_{a+2\kappa}^{\infty} \mathrm{d}b \left[ \frac{\chi_{-1}(a, b, z_{\text{in}})}{z} + \chi_{0}(a, b, z_{\text{in}}) + \chi_{+1}(a, b, z_{\text{in}})z \right] e^{ibz} \quad (\text{E.59}) $$

with the definitions:

$$ \chi_{-1}(a, b, z_{\text{in}}) := \frac{4}{a^2 - b^2} \left[ \frac{5}{4} + \frac{3}{2} \frac{(a^2 + b^2 - 2)^2}{(a^2 - b^2)^2} + \frac{1}{4} \frac{(a^2 + b^2 - 2)^4}{(a^2 - b^2)^4} \right] (a^2 + b^2 - 2)^2 \\ \times \left\{ e^{-ibz_{\text{in}}} \left( \frac{1}{b^2} + \frac{iz_{\text{in}}}{b} - \frac{4}{a^2 - b^2} \right) + \frac{4[\mathrm{Ei}(-iz_{\text{in}}b) + i\pi]}{a^2 - b^2} \right\} \quad (\text{E.60}) $$

$$ \chi_0(a, b, z_{\text{in}}) := -\frac{4ib}{a^2 - b^2} \left[ \frac{5}{4} + \frac{3}{2} \frac{(a^2 + b^2 - 2)^2}{(a^2 - b^2)^2} + \frac{1}{4} \frac{(a^2 + b^2 - 2)^4}{(a^2 - b^2)^4} \right] (a^2 + b^2 - 2)^2 \\ \times \left\{ e^{-ibz_{\text{in}}} \left( \frac{1}{b^2} + \frac{iz_{\text{in}}}{b} - \frac{4}{a^2 - b^2} \right) + \frac{4[\mathrm{Ei}(-iz_{\text{in}}b) + i\pi]}{a^2 - b^2} \right\} \quad (\text{E.61}) $$

$$ \chi_{+1}(a, b, z_{\text{in}}) := \left[ \frac{5}{4} + \frac{3}{2} \frac{(a^2 + b^2 - 2)^2}{(a^2 - b^2)^2} + \frac{1}{4} \frac{(a^2 + b^2 - 2)^4}{(a^2 - b^2)^4} \right] (a^2 + b^2 - 2)^2 \\ \times \left\{ e^{-ibz_{\text{in}}} \left( \frac{1}{b^2} + \frac{iz_{\text{in}}}{b} - \frac{4}{a^2 - b^2} \right) + \frac{4[\mathrm{Ei}(-iz_{\text{in}}b) + i\pi]}{a^2 - b^2} \right\}. \quad (\text{E.62}) $$

Since eq. (E.59) reveals that $\mathcal{G}$ is a Fourier transform, we note that its $0 < z \ll 1$ limit (for fixed $z_{\text{in}}$ and $\kappa$) is governed by the $b \gg 1$ behaviour of the functions $\chi_j$, in which limit we have:

$$ \chi_{-1}(a, b, z_{\text{in}}) \approx e^{-ibz_{\text{in}}} \{-12ib - 60z_{\text{in}} + \mathcal{O}(b^{-1})\}, \quad (\text{E.63}) $$

$$ \chi_0(a, b, z_{\text{in}}) \approx e^{-ibz_{\text{in}}} \left\{ -12z_{\text{in}}b^2 + 60ib + \left[ (80 - 68a^2)z_{\text{in}} + \frac{48}{z_{\text{in}}} \right] + \mathcal{O}(b^{-1}) \right\}, \quad (\text{E.64}) $$

$$ \chi_{+1}(a, b, z_{\text{in}}) \approx e^{-ibz_{\text{in}}} \left\{ 3iz_{\text{in}}b^3 + 15b^2 + \frac{2i[(7a^2 - 10)z_{\text{in}}^2 - 6]}{z_{\text{in}}}b + \frac{12}{z_{\text{in}}^2} + 82a^2 - 100 + \mathcal{O}(b^{-1}) \right\}. \quad (\text{E.65}) $$

To derive the required asymptotics of each of the above Fourier transforms, we write

$$ \int_{a+2\kappa}^{\infty} \mathrm{d}b \chi_{-1}(a, b, z_{\text{in}}) e^{ibz} = \int_{a+2\kappa}^{\infty} \mathrm{d}b e^{-ibz_{\text{in}}} (-12ib - 60z_{\text{in}}) e^{ibz} + \Xi_{-1}(z, a, z_{\text{in}}) \quad (\text{E.66}) $$

with the definition

$$ \Xi_{-1}(z, a, z_{\text{in}}) := \int_{a+2\kappa}^{\infty} \mathrm{d}b \left[ \chi_{-1}(a, b, z_{\text{in}}) - e^{-ibz_{\text{in}}}(-12ib - 60z_{\text{in}}) \right] e^{ibz}. \quad (\text{E.67}) $$

First note that $\Xi_{-1}(z, a, z_{\text{in}}) \sim \mathcal{O}(z^0)$ in the $z \ll 1$ limit, which follows from the fact that its integrand converges to a $z$-independent constant when $z \to 0^+$. What remains then is to evaluate the distribution

$$ \int_{a+2\kappa}^{\infty} \mathrm{d}b e^{-ibz_{\text{in}}} (-12ib - 60z_{\text{in}}) e^{ibz} = e^{-i(z_{\text{in}}-z)(a+2\kappa)} \left[ \frac{12i}{(z_{\text{in}}-z)^2} - \frac{12(a+2\kappa-5i)}{z_{\text{in}}-z} \right] \\ + 12\pi\delta'(z_{\text{in}}-z) - 60\pi z_{\text{in}}\delta(z_{\text{in}}-z), \quad (\text{E.68}) $$
which follows from eqs. (B.49) and (B.50). Fixing $a$ and $z_{\text{in}}$ in the above, and taking the $0 < z \ll 1$ limits then implies that (also assuming that $\kappa z \ll 1$)

$$ \int_{a+2\kappa}^{\infty} db \, e^{-ibz_{\text{in}}} (-12ib - 60z_{\text{in}}) e^{+ibz} \approx e^{-iz_{\text{in}}(a+2\kappa)} \left[ \frac{12i}{z_{\text{in}}^2} - \frac{12(a+2\kappa-5i)}{z_{\text{in}}} \right] + \mathcal{O}(z), \qquad (\text{E.69}) $$

which implies that

$$ \int_{a+2\kappa}^{\infty} db \, \chi_{-1}(a, b, z_{\text{in}}) e^{ibz} \sim \mathcal{O}(z^0). \qquad (\text{E.70}) $$

Similar computations show that in the super-Hubble limit $0 < z \ll 1$ with $\kappa z \ll 1$, $\int_{a+2\kappa}^{\infty} db \, \chi_{0,+1}(a, b, z_{\text{in}}) e^{ibz} \sim \mathcal{O}(z^0)$, which when combined with eq. (E.59) show that $\mathcal{G}(a, z, z_{\text{in}}) \sim \mathcal{O}(z^{-1})$ in the same limit. Integrating from $a=0$ to $a=1$ in eq. (E.58) leaves the same $z$-dependence and so we conclude that $g(z, z_{\text{in}}) \sim \mathcal{O}(z^{-1})$.

Combining the above dependences for $h$ and $g$ in eq. (E.39) implies

$$ \tilde{\mathcal{F}}_k(\eta, \eta_{\text{in}}) \approx \frac{\varepsilon_1 H^2 k^2}{1024 \pi^2 M_{\text{P}}^2} \frac{60\pi}{(-k\eta)^2} + \mathcal{O}((-k\eta)^{-1}) \qquad (\text{E.71}) $$

in the super-Hubble limit, as claimed in the main text, see eq. (3.61).

# F Tensor decoherence from a scalar environment

## F.1 General considerations

Here we consider the interaction Hamiltonian (5.2)

$$ \mathcal{H}_{\text{int}}(\eta) = \tilde{G}(\eta) \int d^3\mathbf{x} \, v_{ij}(\eta, \mathbf{x}) \otimes B^{ij}(\eta, \mathbf{x}), \qquad (\text{F.1}) $$

with $\tilde{G} = -(2M_{\text{P}}a)^{-1}$ defined in eq. (5.4), $v_{ij}$ the canonical tensor mode for the system modes (related to $\gamma_{ij}$ by eq. (A.17)) and the environmental operator $B^{ij}(\mathbf{x}) = \delta^{ik}\delta^{jl}\partial_k v(\eta, \mathbf{x})\partial_l v(\eta, \mathbf{x})$ defined in eq. (5.3).

Using an analogous setup as that used for the scalar, the Nakajima-Zwanzig equation for the reduced density matrix (now for the graviton) at second-order in $\tilde{G}$ is given by

$$ \begin{aligned} \frac{\partial \varrho}{\partial \eta} \approx & -i \tilde{G}(\eta) \mathcal{B}^{ia}(\eta) \int d^3 \mathbf{x} \, [v_{ia}(\eta, \mathbf{x}), \varrho(\eta)] \\ & - \int d^3 \mathbf{x} \int d^3 \mathbf{x}' \int_{\eta_{\text{in}}}^{\eta} d\eta' \tilde{G}(\eta) \tilde{G}(\eta') \{[v_{ia}(\eta, \mathbf{x}), v_{jb}(\eta', \mathbf{x}')\varrho(\eta')] C^{iajb}(\eta, \eta'; \mathbf{x} - \mathbf{x}') \\ & + [\varrho(\eta')v_{jb}(\eta', \mathbf{x}'), v_{ia}(\eta, \mathbf{x})] C^{iajb*}(\eta, \eta'; \mathbf{x} - \mathbf{x}') \} \end{aligned} \qquad (\text{F.2}) $$

cf. eq. (3.16), with the definitions

$$ \mathcal{B}^{ia}(\eta) := \langle 0_B | B^{ia}(\eta, \mathbf{x}) | 0_B \rangle \qquad (\text{F.3}) $$

and

$$ C^{iajb}(\eta, \eta'; \mathbf{x} - \mathbf{x}') = \langle 0_B | [B^{ia}(\eta, \mathbf{x}) - \mathcal{B}^{ia}(\eta)] [B^{jb}(\eta', \mathbf{x}') - \mathcal{B}^{jb}(\eta')] | 0_B \rangle. \qquad (\text{F.4}) $$
Using the mode expansion of the scalar as usual one finds that

$$ \mathcal{B}^{ia}(\eta) = \int_{k>k_{UV}} \frac{d^3\mathbf{k}}{(2\pi)^3} k^i k^a |u_{\mathbf{k}}(\eta)|^2 = \frac{1}{3} \delta^{ia} \mathcal{B}(\eta) \quad (F.5) $$

with $\mathcal{B}$ given in eq. (3.17)<fnref n="14" /> and

$$ C^{iajb}(\eta, \eta'; \mathbf{x} - \mathbf{x}') = 2 \int_{q,p>k_{UV}} \frac{d^3\mathbf{q} \, d^3\mathbf{p}}{(2\pi)^6} p^i q^a p^j q^b u_q(\eta) u_p(\eta) u_q^*(\eta') u_p^*(\eta') e^{i(\mathbf{q}+\mathbf{p})\cdot(\mathbf{x}-\mathbf{x}')} \quad (F.6) $$

Using the mode expansion for $v_{ij}$ given in eq. (A.18), repeated here for convenience

$$ v_{ij}(\eta, \mathbf{x}) = \int \frac{d^3\mathbf{k}}{(2\pi)^{3/2}} \sum_{P=+,-} \epsilon_{ij}^{P}(\mathbf{k}) v_{\mathbf{k}}^{P}(\eta) e^{i\mathbf{k}\cdot\mathbf{x}}, \quad (F.7) $$

one can re-write the above equation as

$$ \frac{\partial \rho}{\partial \eta} = -(2\pi)^{3/2} \sum_{P,P'} \int_{k<k_{UV}} d^3\mathbf{k} \int_{\eta_{in}}^{\eta} d\eta' \tilde{G}(\eta) \tilde{G}(\eta') \epsilon_{ia}^{P}(\mathbf{k}) \epsilon_{jb}^{P'}(-\mathbf{k}) \\ \times \left\{ \left[ v_{\mathbf{k}}^{P}(\eta), v_{-\mathbf{k}}^{P'}(\eta') \rho(\eta') \right] \mathcal{C}_{-\mathbf{k}}^{iajb}(\eta, \eta') + \left[ \rho(\eta') v_{-\mathbf{k}}^{P'}(\eta'), v_{\mathbf{k}}^{P}(\eta) \right] \mathcal{C}_{-\mathbf{k}}^{iajb*}(\eta, \eta') \right\}, \quad (F.8) $$

where we define the Fourier transform $\mathcal{C}_{\mathbf{k}}^{iajb}$ of the correlator $C^{iajb}(\eta, \eta'; \mathbf{y})$ in the variable $\mathbf{y}$ where

$$ \begin{aligned} \mathcal{C}_{\mathbf{k}}^{iajb}(\eta, \eta') &= \int \frac{d^3\mathbf{y}}{(2\pi)^{3/2}} C^{iajb}(\eta, \eta'; \mathbf{y}) e^{-i\mathbf{k}\cdot\mathbf{y}} \\ &= \frac{2}{(2\pi)^{9/2}} \int_{q,p>k_{UV}} d^3\mathbf{q} \, d^3\mathbf{p} \, p^i q^a p^j q^b u_q(\eta) u_p(\eta) u_q^*(\eta') u_p^*(\eta') \delta(\mathbf{p} + \mathbf{q} - \mathbf{k}) . \end{aligned} \quad (F.9) $$

From here we note that eq. (F.9) implies $\mathcal{C}_{-\mathbf{k}}^{iajb}(\eta, \eta') = \mathcal{C}_{\mathbf{k}}^{iajb}(\eta, \eta')$, and we also use the symmetry $\epsilon_{jb}^{P'}(-\mathbf{k})v_{-\mathbf{k}}^{P'}(\eta') = \epsilon_{jb}^{P'}(\mathbf{k})v_{\mathbf{k}}^{P'}(\eta')$ for each polarization $P'$ giving

$$ \frac{\partial \rho}{\partial \eta} = -(2\pi)^{3/2} \sum_{P,P'} \int_{k<k_{UV}} d^3\mathbf{k} \int_{\eta_{in}}^{\eta} d\eta' \tilde{G}(\eta) \tilde{G}(\eta') \epsilon_{ia}^{P}(\mathbf{k}) \epsilon_{jb}^{P'}(\mathbf{k}) \\ \times \left\{ \left[ v_{\mathbf{k}}^{P}(\eta), v_{\mathbf{k}}^{P'\dagger}(\eta') \rho(\eta') \right] \mathcal{C}_{\mathbf{k}}^{iajb}(\eta, \eta') + \left[ \rho(\eta') v_{\mathbf{k}}^{P'\dagger}(\eta'), v_{\mathbf{k}}^{P}(\eta) \right] \mathcal{C}_{\mathbf{k}}^{iajb*}(\eta, \eta') \right\}. \quad (F.10) $$

We next split the density matrix into a product over modes as before, with mode labels $\mathbf{k}$ and polarizations $P = +, \times$

$$ \rho = \bigotimes_{\mathbf{k},P} \rho_{\mathbf{k}}^{P} \quad (F.11) $$

and repeating the arguments used in the scalar case leads to the separate evolution equation for each label:

$$ \frac{\mathcal{V}}{(2\pi)^{3/2}} \frac{\partial \rho_{\mathbf{k}}^{P}}{\partial \eta} = -(2\pi)^{3/2} \int_{\eta_{in}}^{\eta} d\eta' \tilde{G}(\eta) \tilde{G}(\eta') \{ \mathcal{S}_{\mathbf{k}}(\eta, \eta') [\tilde{v}_{\mathbf{k}}^{P}(\eta), \tilde{v}_{\mathbf{k}}^{P}(\eta')] \rho_{\mathbf{k}}^{P}(\eta') \} + \text{h.c.} \quad (F.12) $$

with $\tilde{v}$ a proxy for the real and imaginary parts of the field (as in the text below (2.21)) and where we define

$$ \epsilon_{ia}^{P}(\mathbf{k}) \epsilon_{jb}^{P'}(\mathbf{k}) \mathcal{C}_{\mathbf{k}}^{iajb}(\eta, \eta') = \delta^{PP'} \mathcal{S}_{\mathbf{k}}(\eta, \eta') . \quad (F.13) $$

<footnote n="14">¹⁴Notice that since $\mathcal{B}^{ia} \propto \delta^{ia}$ then the contraction in the first term of eq. (F.2) vanishes since we assume the graviton is traceless, where $\delta^{ia}\gamma_{ia} = 0$.</footnote>
## F.2 Computing the correlation function

We first note that eq. (F.9) implies $\mathcal{C}_k^{iajb}(\eta, \eta')$ has the symmetries $\mathcal{C}_k^{iajb} = \mathcal{C}_k^{jaib} = \mathcal{C}_k^{ibja} = \mathcal{C}_k^{aibj}$. The most general form possible consistent with these symmetries is

$$ \mathcal{C}_k^{iajb} = \frac{A}{2} (\delta^{ia}\delta^{jb} + \delta^{ib}\delta^{ja}) + B \delta^{ij}\delta^{ab} + \frac{C}{4} (\delta^{ia}k^j k^b + \delta^{ja}k^i k^b + \delta^{ib}k^j k^a + \delta^{jb}k^i k^a) \\ + \frac{D}{2} (\delta^{ij}k^a k^b + \delta^{ab}k^i k^j) + E k^i k^j k^a k^b, \qquad (\text{F.14}) $$

where the coefficients A through E are functions of $\eta, \eta'$ and $k^2$. Contracting the above form with two polarization tensors as in eq. (F.13) yields

$$ \epsilon_{ia}^{P}(\mathbf{k}) \epsilon_{jb}^{P'}(\mathbf{k}) \mathcal{C}_{\mathbf{k}}^{iajb}(\eta, \eta') = \delta^{PP'} \left(\frac{A}{2} + B\right), \qquad (\text{F.15}) $$

which uses $\epsilon_{ij}^P(\mathbf{k}) = \epsilon_{ji}^P(\mathbf{k})$, $\epsilon_{ij}^P(\mathbf{k})\epsilon_{ij}^{P'}(\mathbf{k}) = \delta^{PP'}$ and $\epsilon_{ii}^P(\mathbf{k}) = k^i\epsilon_{ij}(\mathbf{k}) = 0$, showing that

$$ \mathcal{S}_k(\eta, \eta') = \frac{A}{2} + B \qquad (\text{F.16}) $$

is the function appearing in eq. (F.13). The required coefficients A and B can be computed by inverting the following five expressions for rotation-invariant integrals:

$$ \delta_{ij}\delta_{ab}\mathcal{C}_{\mathbf{k}}^{iajb} = 3A + 9B + k^2C + 3k^2D + k^4E, \qquad (\text{F.17}) $$

$$ \delta_{ia}\delta_{jb}\mathcal{C}_{\mathbf{k}}^{iajb} = \mathcal{C}_{\mathbf{k}} = 6A + 3B + 2k^2C + k^2D + k^4E, \qquad (\text{F.18}) $$

$$ k_i k_j \delta_{ab} \mathcal{C}_{\mathbf{k}}^{iajb} = k^2 A + 3k^2 B + k^4 C + 2k^4 D + k^6 E, \qquad (\text{F.19}) $$

$$ \delta_{ia}k_j k_b \mathcal{C}_{\mathbf{k}}^{iajb} = 2k^2 A + k^2 B + \frac{3}{2}k^4 C + k^4 D + k^6 E, \qquad (\text{F.20}) $$

and

$$ k_i k_a k_j k_b \mathcal{C}_{\mathbf{k}}^{iajb} = k^4 A + k^4 B + k^6 C + k^6 D + k^8 E. \qquad (\text{F.21}) $$

Inverting these formulas gives the coefficients A through E in terms of integrals, and using the result in eq. (F.16) yields

$$ \mathcal{S}_k(\eta, \eta') = \frac{\delta_{ij}\delta_{ab}\mathcal{C}_k^{iajb}}{4} - \frac{k_i k_j \delta_{ab}\mathcal{C}_k^{iajb}}{2k^2} + \frac{k_i k_a k_j k_b \mathcal{C}_k^{iajb}}{4k^4} \qquad (\text{F.22}) $$

$$ = \frac{2}{(2\pi)^{9/2}} \int_{q,p>k_{UV}} d^3\mathbf{q} d^3\mathbf{p} u_q(\eta) u_p(\eta) u_q^*(\eta') u_p^*(\eta') \left[ \frac{q^2 p^2}{4} - \frac{(\mathbf{k} \cdot \mathbf{p})^2 q^2}{2k^2} + \frac{(\mathbf{k} \cdot \mathbf{q})^2 (\mathbf{k} \cdot \mathbf{p})^2}{4k^4} \right] \delta(\mathbf{q} + \mathbf{p} - \mathbf{k}) \qquad (\text{F.23}) $$

$$ = \frac{2}{(2\pi)^{9/2}} \int_{q,p>k_{UV}} dq dp p^4 q^4 u_q(\eta) u_p(\eta) u_q^*(\eta') u_p^*(\eta') \int_0^{4\pi} d^2\Omega_q d^2\Omega_p \left[ \frac{1}{4} - \frac{(\mathbf{k} \cdot \mathbf{p})^2}{2k^2 p^2} + \frac{(\mathbf{k} \cdot \mathbf{q})^2 (\mathbf{k} \cdot \mathbf{p})^2}{4k^4 p^2 q^2} \right] \delta(\mathbf{q} + \mathbf{p} - \mathbf{k}), \qquad (\text{F.24}) $$
which writes the integrals in polar coordinates. The three terms mainly differ in their angular integrations. Rotation invariance allows us to choose **k** (as in appendix B) to point along the z axis: **k** = (0, 0, k). Using this choice, the angular integrals become

$$
\begin{align}
\mathcal{J}(p, q, k) &:= \int_0^{4\pi} \mathrm{d}^2\Omega_q \mathrm{d}^2\Omega_p \left[ \frac{1}{4} - \frac{(\mathbf{k} \cdot \mathbf{p})^2}{2k^2p^2} + \frac{(\mathbf{k} \cdot \mathbf{q})^2(\mathbf{k} \cdot \mathbf{p})^2}{4k^4p^2q^2} \right] \delta(\mathbf{q} + \mathbf{p} - \mathbf{k}) \tag{F.25} \\
&= \frac{1}{q^2} \int_{-1}^{1} \mathrm{d}\cos\theta_q \int_0^{2\pi} \mathrm{d}\varphi_q \int_{-1}^{1} \mathrm{d}\cos\theta_p \int_0^{2\pi} \mathrm{d}\varphi_p \left[ \frac{1}{4} - \frac{\cos^2\theta_p}{2} + \frac{\cos^2\theta_q\cos^2\theta_p}{4} \right] \nonumber \\
&\quad \times \delta\left( q - \sqrt{p^2 + k^2 - 2kp\cos\theta_p} \right) \delta\left( \cos\theta_q - \frac{k - p\cos\theta_p}{\sqrt{p^2 + k^2 - 2kp\cos\theta_p}} \right) \nonumber \\
&\quad \times \delta[\varphi_q - (\varphi_p + \pi)]. \tag{F.26}
\end{align}
$$

Integrating over $\varphi_q$, $\varphi_p$ and then $\theta_q$, and writing $\mu = \cos \theta_p$ yields

$$
\mathcal{J}(p, q, k) = \frac{2\pi}{pqk} \int_{-1}^{1} \mathrm{d}\mu \left[ \frac{1}{4} - \frac{\mu^2}{2} + \frac{(k-p\mu)^2\mu^2}{4(p^2+k^2-2kp\mu)} \right] \delta\left(\mu - \frac{p^2+k^2-q^2}{2pk}\right), \quad (\text{F.27})
$$

which uses the identity $\delta(q - \sqrt{p^2 + k^2 - 2kp\mu}) = \frac{q}{kp} \delta\left(\mu - \frac{p^2+k^2-q^2}{2pk}\right)$. The result vanishes unless $-2pk \le p^2 + k^2 - q^2 \le 2pk$, which requires the p and q integrals to run over the region U depicted in figure 5. For p, q in this region the final angular integral gives

$$
\mathcal{J}(p, q, k) = \frac{\pi}{2pqk} \left\{ 1 - \frac{(p^2 + k^2 - q^2)^2}{2p^2k^2} + \frac{[k^4 - (p^2 - q^2)^2]^2}{16p^2q^2k^4} \right\}, \quad (\text{F.28})
$$

which when used in eq. (F.22) together with the Bunch-Davies mode functions (2.40) gives

$$
\begin{equation}
\begin{split}
\mathcal{S}_k(\eta, \eta') ={}& \frac{1}{8(2\pi)^{7/2}k} \iint_U \mathrm{d}q \mathrm{d}p \left\{ q^2 p^2 - \frac{(k^2 + p^2 - q^2)^2 q^2}{2k^2} + \frac{[k^4 - (p^2 - q^2)^2]^2}{16k^4} \right\} \\
& \times \left(1 - \frac{i}{q\eta}\right) \left(1 + \frac{i}{q\eta'}\right) \left(1 - \frac{i}{p\eta}\right) \left(1 + \frac{i}{p\eta'}\right) e^{-i(q+p)(\eta-\eta')}
\end{split}
\tag{F.29}
\end{equation}
$$

The integral over the region $U$ is performed using the coordinate change $p = \frac{1}{2}(P+Q)$ and $q = \frac{1}{2}(P-Q)$, leading to

$$
\begin{equation}
\begin{split}
\mathcal{S}_k(\eta, \eta') ={}& \frac{1}{32(2\pi)^{7/2}k} \int_{-k}^{k} \mathrm{d}Q \int_{|Q|+2k_{UV}}^{\infty} \mathrm{d}P \left[ \frac{(P^2-Q^2)^2}{8} - \frac{(k^2+PQ)^2(P-Q)^2}{4k^2} \right. \\
& \qquad \left. + \frac{(k^4-P^2Q^2)^2}{8k^4} \right] \left[ 1 - \frac{2i}{(P-Q)\eta} \right] \left[ 1 + \frac{2i}{(P-Q)\eta'} \right] \\
& \qquad \times \left[ 1 - \frac{2i}{(P+Q)\eta} \right] \left[ 1 + \frac{2i}{(P+Q)\eta'} \right] e^{-i(\eta-\eta')P}.
\end{split}
\tag{F.30}
\end{equation}
$$

Note that this integrand is *not* symmetric under $Q \to -Q$. By splitting up the $Q$-integral to be over $[-k, 0]$ and $[0, k]$ and then taking $Q \to -Q$ in the first piece the above simplifies to

$$
\begin{equation}
\begin{split}
\mathcal{S}_k(\eta, \eta') ={}& \frac{1}{128(2\pi)^{7/2}k^5} \int_0^k \mathrm{d}Q (k^2 - Q^2)^2 \int_{Q+2k_{UV}}^{\infty} \mathrm{d}P (k^2 - P^2)^2 \left[1 - \frac{2i}{(P-Q)\eta}\right] \\
& \times \left[1 + \frac{2i}{(P-Q)\eta}\right] \left[1 - \frac{2i}{(P+Q)\eta}\right] \left[1 + \frac{2i}{(P+Q)\eta'}\right] e^{-i(\eta-\eta')P}.
\end{split}
\tag{F.31}
\end{equation}
$$
Our interest is in the leading behaviour as for small $-k\eta$ and so we focus on the $k \to 0$ (and so also $Q \to 0$) limit of eq. (F.31), which is

$$ \mathcal{S}_k(\eta, \eta') \approx \frac{1}{128(2\pi)^{7/2}k^5} \int_0^k dQ (k^2 - Q^2)^2 \int_{2k_{UV}}^{\infty} dP P^4 \left(1 - \frac{2i}{P\eta}\right)^2 \left(1 + \frac{2i}{P\eta'}\right)^2 e^{-i(\eta-\eta')P} \quad (F.32) $$

$$ \approx \frac{1}{240(2\pi)^{7/2}} \int_{2k_{UV}}^{\infty} dP P^4 \left(1 - \frac{2i}{P\eta}\right)^2 \left(1 + \frac{2i}{P\eta'}\right)^2 e^{-i(\eta-\eta')P}, \quad (F.33) $$

where the last expression is valid in the limit $k \to 0$. This is to be compared with the same limit for the scalar contribution (B.31), which is

$$ \mathcal{C}_k(\eta, \eta') = \frac{1}{32(2\pi)^{7/2}} \int_0^k dQ \int_{Q+2k_{UV}}^{\infty} dP \frac{(P^2+Q^2-2k^2)^2}{k} \left[1-\frac{2i}{(P-Q)\eta}\right] \left[1+\frac{2i}{(P-Q)\eta'}\right] \times \left[1-\frac{2i}{(P+Q)\eta}\right] \left[1+\frac{2i}{(P+Q)\eta'}\right] e^{-i(\eta-\eta')P} \quad (F.34) $$

$$ \approx \frac{1}{32(2\pi)^{7/2}} \int_{2k_{UV}}^{\infty} dP P^4 \left(1 - \frac{2i}{P\eta}\right)^2 \left(1 + \frac{2i}{P\eta'}\right)^2 e^{-i(\eta-\eta')P}, \quad (F.35) $$

where, again, the last expression is valid for $k \to 0$. Therefore, we see that for small $k$

$$ \mathcal{S}_k(\eta, \eta') \approx \frac{2}{15} \mathcal{C}_k(\eta, \eta'). \quad (F.36) $$

This can also be derived directly by taking the $k \to 0$ limit of eq. (F.25), which becomes

$$ \mathcal{J}(p, q, k \to 0) := \frac{1}{q^2} \delta(p-q) \int_0^{4\pi} d^2\Omega_q d^2\Omega_p \left[ \frac{1}{4} - \frac{(\mathbf{k} \cdot \mathbf{p})^2}{2k^2p^2} + \frac{(\mathbf{k} \cdot \mathbf{q})^2(\mathbf{k} \cdot \mathbf{p})^2}{4k^4p^2q^2} \right] \delta\left(\frac{\mathbf{q}}{q} + \frac{\mathbf{p}}{p}\right) \quad (F.37) $$

$$ = \frac{2\pi}{q^2} \delta(p-q) \int_{-1}^{1} d\mu \left( \frac{1}{4} - \frac{\mu^2}{2} + \frac{\mu^4}{4} \right) = \frac{4}{15} \left(\frac{2\pi}{q^2}\right) \delta(p-q), \quad (F.38) $$

which is $\frac{2}{15}$ times the result obtained when the square bracket is 1.

## F.3 Lindblad coefficient

Beginning with the Nakajima-Zwanzig equation (F.39), and applying the Markovian approximation in the same spirit as earlier gives

$$ \frac{\mathcal{V}}{(2\pi)^{3/2}} \frac{\partial \varrho_k^P}{\partial \eta} = -\mathcal{T}_k(\eta, \eta_{\text{in}}) [\tilde{v}_k^P(\eta), \tilde{v}_k^P(\eta') \varrho_k^P(\eta')] - \mathcal{T}_k^*(\eta, \eta_{\text{in}}) [\varrho_k^P(\eta'), \tilde{v}_k^P(\eta'), \tilde{v}_k^P(\eta)] , \quad (F.39) $$

where we define

$$ \mathcal{T}_k(\eta, \eta_{\text{in}}) = (2\pi)^{3/2} \int_{\eta_{\text{in}}}^{\eta} d\eta' \tilde{G}(\eta) \tilde{G}(\eta') \mathcal{S}_k(\eta, \eta') \quad (F.40) $$

and we note that again Re $[\mathcal{T}_k(\eta, \eta_{\text{in}})]$ drives the decoherence. Using $\tilde{G}(\eta) = H\eta/M_p$ as well as eq. (F.31) means that eq. (F.40) can be written as the triple integral

$$ \mathcal{T}_k(\eta, \eta_{\text{in}}) = \frac{H^2}{512\pi^2 M_p^2 k^5} \int_{\eta_{\text{in}}}^{\eta} d\eta' \int_0^k dQ \int_{Q+2k_{\text{UV}}}^{\infty} dP \eta \eta' (k^2 - P^2)^2 (k^2 - Q^2)^2 \left[1 - \frac{2i}{(P-Q)\eta}\right] \times \left[1 + \frac{2i}{(P-Q)\eta'}\right] \left[1 - \frac{2i}{(P+Q)\eta}\right] \left[1 + \frac{2i}{(P+Q)\eta'}\right] e^{-i(\eta-\eta')P} . \quad (F.41) $$
We first evaluate the $\eta'$-integral using the formula (C.3), giving

$$ \begin{aligned} \mathcal{T}_k(\eta, \eta_{\text{in}}) = & \frac{H^2}{512\pi^2 M_p^2 k^5} \int_0^k dQ \int_{Q+2k_{\text{UV}}}^{\infty} dP \eta (k^2 - P^2)^2 (k^2 - Q^2)^2 \left[1 - \frac{2i}{(P-Q)\eta}\right] \\ & \times \left[1 - \frac{2i}{(P+Q)\eta}\right] \left\{ \frac{1}{P^2} - \frac{i\eta}{P} + \frac{4}{P^2 - Q^2} - \frac{4e^{-iP\eta}[\text{Ei}(iP\eta) + i\pi]}{P^2 - Q^2} \right. \\ & \left. - e^{-iP(\eta-\eta_{\text{in}})} \left(\frac{1}{P^2} - \frac{i\eta_{\text{in}}}{P} + \frac{4}{P^2 - Q^2}\right) + \frac{4e^{-iP\eta}[\text{Ei}(iP\eta_{\text{in}}) + i\pi]}{P^2 - Q^2} \right\}. \end{aligned} \quad (F.42) $$

The terms involving $\eta_{\text{in}}$ can be handled in precisely the same way as in the discussion around eq. (E.56), showing that they contribute only subdominantly for small ($-k\eta$), and so for our purposes can be neglected. This leaves

$$ \begin{aligned} \mathcal{T}_k(\eta, \eta_{\text{in}}) \approx & \frac{H^2}{512\pi^2 M_p^2 k^5} \int_0^k dQ \int_{Q+2k_{\text{UV}}}^{\infty} dP \eta (k^2 - P^2)^2 (k^2 - Q^2)^2 \left[1 - \frac{2i}{(P-Q)\eta}\right] \\ & \times \left[1 - \frac{2i}{(P+Q)\eta}\right] \left\{ \frac{1}{P^2} - \frac{i\eta}{P} + \frac{4}{P^2 - Q^2} - \frac{4e^{-iP\eta}[\text{Ei}(iP\eta) + i\pi]}{P^2 - Q^2} \right\}, \end{aligned} \quad (F.43) $$

which can be rewritten using the change of variables already used before, reproduced here for convenience, $z = -k\eta$, $\kappa = k_{\text{UV}}/k$, $x = -Q\eta$ and $y = -P\eta$ as

$$ \mathcal{T}_k \approx \frac{H^2 k^2}{512 \pi^2 M_p^2 z^3} \int_0^z dx \left(\frac{x^2}{z^2} - 1\right)^2 h(x, z) \quad (F.44) $$

with

$$ h(x, z) := \int_{x+2\kappa z}^{\infty} dy (y^2 - z^2)^2 \left(1 - \frac{2i}{x-y}\right) \left(1 + \frac{2i}{x+y}\right) \left\{-\frac{1}{y^2} - \frac{i}{y} + \frac{4-4e^{iy}[\text{Ei}(-iy) + i\pi]}{x^2 - y^2}\right\}. \quad (F.45) $$

This integral diverges (as did the previous examples), since the integrand approaches $-y^2$ for large $y$. We handle this divergence the same way as before by subtracting the divergent large-$y$ form, leaving a convergent result (see the discussion around eq. (E.53)).

Taking the $z \to 0$ limit (which also implies $x \to 0$) then gives the leading contribution

$$ \mathcal{T}_k \approx \frac{H^2 k^2}{512 \pi^2 M_p^2 z^3} h^{\text{reg}}(0, 0) \int_0^z dx \left(\frac{x^2}{z^2} - 1\right)^2 = \frac{H^2 k^2}{512 \pi^2 M_p^2 z^2} \left[\frac{8h^{\text{reg}}(0, 0)}{15}\right] \quad (F.46) $$

where (compare to eqs. (E.46) and (E.54))

$$ h^{\text{reg}}(0, 0) = \int_0^\infty dy \left( y^4 \left( 1 + \frac{2i}{y} \right)^2 \left\{ -\frac{1}{y^2} - \frac{i}{y} - \frac{4 - 4e^{iy} [\text{Ei}(-iy) + i\pi]}{y^2} \right\} + y^2 \right) \quad (F.47) $$

$$ = \frac{f^{\text{reg}}(0, 0)}{3} = 20\pi. \quad (F.48) $$

This leads to the following $\eta_{\text{in}}$-independent leading behaviour

$$ \mathrm{Re}[\mathcal{T}_k(\eta, \eta_{\text{in}})] \approx \frac{H^2 k^2}{48 \pi M_p^2 z^2} = \frac{H^2 k^2}{1024 \pi^2 M_p^2} \left[ \frac{64\pi}{3z^2} + \mathcal{O}(z^{-1}) \right] \quad (F.49) $$
and so comparing with eq. (3.38) we see that

$$ \mathrm{Re}[\mathfrak{F}_k(\eta, \eta_{\mathrm{in}})] \approx \frac{16}{15\varepsilon_1} \mathrm{Re}[\mathfrak{F}_k(\eta, \eta_{\mathrm{in}})] \quad (\mathrm{F.50}) $$

in the super-Hubble limit, as claimed in the main text. The factor 16/(15ε₁) has two sources: the factor of 2/15 seen in eq. (F.36) together with the factor of 8/ε₁ coming from the change in effective coupling noted in eq. (5.9).

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