How to derive the Unruh effect (or the thermofield double state) from the path integral?

Question

I have been reading about the path integral approach to deriving the thermofield double state for the Minkowski vacuum in terms of the Rindler states: \begin{equation} \left|0_{M}(t=0)\right\rangle=\sum_{n} \frac{e^{-\frac{\beta}{2} E_{n}}}{\sqrt{Z(\beta)}}\left|n_{R}\right\rangle \otimes\left[\Theta\left|n_{L}\right\rangle\right]. \end{equation} According to https://arxiv.org/abs/2001.09869, this result can be derived by considering \begin{equation} \begin{aligned} \left\langle\phi_{M} \mid 0_{M}(t=0)\right\rangle & \propto \int_{\phi(\theta=-\pi)=\phi_{I}}^{\phi(\theta=0)=\phi_{D}} D \phi e^{-I_{E}} \\ & \propto\left\langle\phi_{R}\left|e^{-\pi H^{R}}\right| \phi_{L}\right\rangle \end{aligned}. \end{equation} But my other reference is https://arxiv.org/abs/1409.1231, which claims we should be studying \begin{equation} \left\langle\phi_{L} \phi_{R} \mid \Omega\right\rangle \propto\left\langle\phi_{R}\left|e^{-\pi K_{R}} \Theta\right| \phi_{L}\right\rangle_{L} \end{equation} Which is different because of the CPT operator $\Theta$. (As far as I can tell $K_R$ and $H^R$ are the same thing).

Which of these is correct?

Answer 1

It seems I was confused by notation here. After all in the first expression the $| \phi_L \rangle $ can only be evolved by $H^R$ and projected onto $| \phi_R \rangle$ if it is in the right-Rindler wedge states. So $| \phi_L \rangle $ must live in the same space as $| \phi_R \rangle $ which can be achieved by applying $\Theta$ to a left-Rindler wedge state.