# What effect does multiplying $\mathscr{L}$ by $-1$ have on the propagator?

I am following along Ashok Das' development of Thermofield dynamics in his book Finite Temperature Field Theory. Here you have two real scalar fields $\phi_1$ and $\phi_2$ with Lagrangian density $$\mathscr{L}_0[\phi_1,\phi_2] = -\tfrac{1}{2} ( \partial_\mu \phi_1 )( \partial^\mu \phi_1 ) - \tfrac{1}{2} m^2 \phi_1^2 + \tfrac{1}{2} ( \partial_\mu \phi_2 )( \partial^\mu \phi_2 ) + \tfrac{1}{2} m^2 \phi_2^2$$ Note the relative minus sign between the two pieces.

In this treatment, $\phi_1$ is the physical field and $\phi_2$ is an artificial field that will represent the presence of a heat bath.

He forms a doublet $\Phi = \left[ \begin{matrix} \phi_1 \\ \phi_2^{\ast} \end{matrix} \right]$ of the fields. But the fields are real, so this is really just $\Phi = \left[ \begin{matrix} \phi_1 \\ \phi_2 \end{matrix} \right]$ and then the overall time-ordered propagator of the system is the following $2 \times 2$ matrix: $$- i G(x;y) = \langle 0_1 0_2 | \mathcal{T}\big( \Phi(x) \Phi(y) \big) | 0_1 0_2\rangle$$

He then writes the answer in momentum-space with $\int \frac{d^{4}p}{(2\pi)^4} \Delta(p) e^{ip\cdot (x-y)}$, it is: $$- i \Delta(p) = \left[ \begin{matrix} \frac{-i}{p^2 + m^2 - i \epsilon} & 0 \\ 0 & \frac{i}{p^2 + m^2 + i \epsilon} \end{matrix} \right]$$

I am confused about the (22)-propagator $- i \Delta_{22}(p)=\frac{i}{p^2 + m^2 + i \epsilon}$ $\to$ this is the anti-time ordered propagator. How does this come about?

I know that the source of this is that he defines the doublet $\Phi$ first in terms of $\phi_2^{\ast}$. He in fact does say that ''the Feynman boundary condition is equivalent to adding an infinitesimal imaginary term to the quadratic part of the Lagrangian density'' $\to$ so complex conjugating the field will supposedly swap the sign of this infinitesimal $-i\epsilon \to + i \epsilon$ as we see.

But I don't really understand this explanation...is there a way of understanding the appearance of the $\frac{i}{p^2 + m^2 + i \epsilon}$ without resorting to this argument? Does the relative minus sign in the Lagrangian play a role in this?