Scalar field propagator in euclidean field theory

Question

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?

Answer 1

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.