Scalar field propagator in euclidean field theory We have a scalar field propagator in minkowski space with signature $(+,-,-,-)$ as
$$ G (k)={1\over 
k^2-m^2 }.$$
But in Euclidean space the scalar field propagator is
$$G (k)={1\over k^2+m^2 }.$$
Can anyone please explain how this conversion  happen?
 A: The action for the free scalar field theory in Minkowski space is 
$$S_M[\phi]=\int d^4x \left[\frac{1}{2}\eta^{\mu\nu}(\partial_\mu\phi)\partial_\nu\phi -\frac{1}{2}m^2\phi^2\right]=\int d^4x \left[\frac{1}{2}\phi(-\partial^2-m^2)\phi\right],$$
where $\partial^2=\eta^{\mu\nu}\partial_\mu\partial_\nu$.
The Green's function is defined as $(-\partial^2-m^2) G(x,y)=\delta^{(4)}(x-y)$. In momentum space, the Green's function takes $1/(k^2-m^2)$ where $k^2=(k^0)^2-\vec{k}^2$ (using the (+,-,-,-) signature for the Minkowski metric).
The Euclidean theory can be obtained by a Wick rotation $t\rightarrow -i\tau$ together with $iS_M\rightarrow -S_E$. Then we have
$$S_E[\phi]=\int d^4x \left[\frac{1}{2}\delta^{\mu\nu}(\partial_\mu\phi)\partial_\nu\phi +\frac{1}{2}m^2\phi^2\right].$$
Now you have the operator $-\delta^{\mu\nu}\partial_\mu\partial_\nu+m^2$. The Green's function in momentum space is $1/(k^2+m^2)$ where $k^2=(k^0)^2+\vec{k}^2$.
The Klein-Gordon equation is simply
$$\left(-\delta^{\mu\nu}\partial_\mu\partial_\nu+m^2\right)\phi=0.$$
Taking $\phi\sim e^{-i\omega t+i\vec{k}\cdot\vec{x}}$, one has the dispersion relation $\omega^2+\vec{k}^2+m^2=0$, i.e. $\omega=\pm i\sqrt{\vec{k}^2+m^2}$. This means that in Euclidean space, one has decaying modes instead of oscillating modes. 
