Understanding graviton propagators in Minkowski space In Minkowski space, the graviton propagator is given by:
$$\Pi_{\mu \nu\rho \sigma} = \frac{\mathcal{P}^2_{\mu\nu\rho\sigma}}{k^2} - \frac{\mathcal{P}^0_{s,\mu\nu\rho\sigma}}{2k^2} \tag{1}$$
where (see page 6 of this paper)
$$\mathcal{P}^2_{\mu\nu\rho\sigma} = \frac{1}{2}\left( \theta_{\mu\rho}\theta_{\nu\sigma} + \theta_{\mu\sigma}\theta_{\nu\rho}\right) - \frac{1}{3}\theta_{\mu\nu}\theta_{\rho\sigma} \tag{2}$$
$$
\mathcal{P}^0_{s,\mu\nu\rho\sigma} = \frac{1}{3}\theta_{\mu\nu}\theta_{\rho\sigma} \tag{3}
$$
and
$$\theta_{\mu \nu} = \eta_{\mu \nu} - \frac{k_{\mu}k_{\nu}}{k^2} \tag{4}$$
I am interested in calculating the $0000$ element to compute a scattering amplitude which is given as:
$$ -\kappa^2\int \frac{d^3\textbf{k}}{(2\pi)^3}T^{00}(\textbf{k})\text{ }\Pi_{0000}(\textbf{k})\text{ }T^{00}(-\textbf{k})e^{i\textbf{k}\cdot \textbf{x}}  \tag{5}$$
Using (2),(3) in (1) we have:
$$\Pi_{\mu\nu\rho\sigma} = \frac{1}{k^2}\left( \frac{1}{2}\left( \theta_{\mu\rho}\theta_{\nu\sigma} + \theta_{\mu\sigma}\theta_{\nu\rho}\right) - \frac{1}{3}\theta_{\mu\nu}\theta_{\rho\sigma} - \frac{1}{6}\theta_{\mu\nu}\theta_{\rho\sigma}\right) = \frac{1}{2k^2}\left(\theta_{\mu\rho}\theta_{\nu\sigma} + \theta_{\mu\sigma}\theta_{\nu\rho} - \theta_{\mu\nu}\theta_{\rho\sigma}\right) \tag{6}$$
Therefore,
$$\Pi_{0000} = \frac{1}{2k^2}\left(\theta_{00}^2\right) \tag{7}$$ But from (4) we can calculate this as:
$$ \theta_{00} = \eta_{00} - \frac{k_0k_0}{k^2} = -1 - \frac{\textbf{k}^2}{k^2} \tag{8}$$
Thus,
$$\Pi_{0000} = \frac{1}{2k^2}\left(1 + \frac{\textbf{k}^2}{k^2}\right)^2 \tag{9}$$
However, the graviton propagator can also be written as:
$$\Pi_{\mu\nu\rho\sigma} = \frac{1}{2\textbf{k}^2}\left(\eta_{\mu\rho}\eta_{\nu\sigma} + \eta_{\mu\sigma}\eta_{\nu\rho} - \eta_{\mu\nu}\eta_{\rho\sigma}\right) \implies \Pi_{0000} = \frac{1}{2\textbf{k}^2} \tag{10}$$
How do I reconcile the two forms for the propagator? In particular, to do the integration in (5), I will need an integrand in the 3-momentum $\textbf{k}$ so I don't know how to do the integration using (9) but with (10) it becomes a simple Fourier transform.
 A: Taking your expression (6) for the propagator
$$\Pi_{\mu\nu\rho \sigma} = \frac{1}{2 k^2} \left( \theta_{\mu\rho} \theta_{\nu \sigma} + \theta_{\mu\sigma} \theta_{\nu \rho} - \theta_{\mu\nu} \theta_{\rho \sigma} \right)$$
I substitute the transversal projector $\theta_{\alpha \beta} = \eta_{\alpha \beta} - k_\alpha k_\beta/k^2$ and obtain an expression which disagrees with (10) by a number of terms. I assume (10) comes from a different source?
The propagator (10) will give the same results as (6) only if

*

*The incoming state for the gravitons already has only physical polarizations such that $\bar{h}_{\mu\nu} k^\nu = 0$ in momentum space. As a consequence, $\theta^{\kappa\mu} \bar{h}_{\mu\nu} = \eta^{\kappa \mu} \bar{h}_{\mu\nu}$.

*The sources $T^{\mu\nu}$ fulfill the equations of motion (conservation of stress-energy) $\partial_\mu T^{\mu\nu} = 0$. In momentum space, one then also has $k_\mu T^{\mu\nu} = 0$ and $\theta_{\lambda \mu} T^{\mu\nu} = \eta_{\lambda \mu} T^{\mu\nu}$.

I don't know what is the exact approach you are intending to use, but beware that one generally has to be more precise about the construction of the perturbative expansion once going beyond tree level.
