# Question about path integral step of the rindler decomposition

In most papers where I've read about Rindler decomposition and the Unruh effect ( see for example [1] or [2]) they start by saying that they want to find the wavefunction of the vacuum state in the basis of states $$\phi_L$$ and $$\phi_R$$ living on the left and right sides of Minkowski space. They write the wavefunction in basis $$\phi$$ as

$$$$\Psi\big(\phi \big)=\langle \phi |\Omega\rangle \propto\lim_{T\rightarrow \infty} \langle \phi_L \phi_R | e^{-HT}|\chi\rangle \\ \propto \int_{\phi(t_E=-\infty)=0}^{\phi(t_E=0)=\phi} \mathcal{D}\phi e^{-S_E}$$$$ So far so good. Then, every review I found claims that you can make a change of variables such that instead of integrating from $$\phi(t_E=-\infty)=0$$ to $$\phi(t_E=0)=\phi$$ we integrate from $$\phi(0)=\phi_L^*$$ to $$\phi(\pi)=\phi_R$$ rotating with the boost operator $$K_x$$ in euclidean space. How does this work? I would think that the state $$\phi$$ that you want at the end is already $$|\phi\rangle= |\phi_L \rangle |\phi_R\rangle$$ so I don't see how you end with $$\phi_L$$ on one side of the boundary conditions and $$\phi_R$$ on the other one. Even more, it's not clear at all which coordinate change they are doing or why the euclidean action suddenly becomes the euclidean boost on x $$K_x$$. Once they asume that they write

$$$$\Psi\big(\phi_L \phi_R \big)=\langle \phi_R | e^{-\pi K_x} \Theta | \phi_L\rangle$$$$

where $$\Theta$$ is the CPT operator. Once I trust the previous step where they "rotate", this result looks reasonable... but the rotation step seems very unclear to me.