# 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 $$K_x$$ on $$x$$. 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.

What you are asking is depicted in figure 8 in https://arxiv.org/pdf/1409.1231.pdf. The basic principle of QFT says that the value of the correlator should not depend on the foliation you choose. In Euclidean space, the only restriction on the slice is that the slice should sweep out all space when acted on by a symmetry operator of your theory. A Hilbert space will be defined on each slice and the symmetry operator will move you from one slice to another or, said differently, from one Hilbert to the next. In the usual foliation we choose the constant $$t$$ plane as the slice and the Hamiltonian $$H$$ as the symmetry operator. In the calculations that are done in these papers instead of foliating our space by constant t slices, we choose a constant-angular-variable slice. Now we treat the left part of the x-axis as the starting slice and the right part as the finishing slice and the boost plays the role of the symmetry operator.
Of course, the decomposition $$|\phi \rangle = |\phi_L \rangle |\phi_R\rangle$$ taken at face value is incorrect as it forgets about a boundary condition at the origin where they meet. So, the actual justification is more involved, but morally this story should make sense.