1
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ The dispersion relation is different. The Propagator in Euclidean spacetime is the green’s function of a different operator. $\endgroup$ – adithya May 13 '20 at 6:27
  • $\begingroup$ Can you please write dispersion relation in Euclidean spacetime@adithya $\endgroup$ – ROBIN RAJ May 14 '20 at 18:48
  • $\begingroup$ It's actually given in Wein Eld's answer below. I suggest you go through that. $\endgroup$ – adithya May 14 '20 at 20:45
5
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ How you got $S_E [\phi]$ $\endgroup$ – ROBIN RAJ May 14 '20 at 17:03
  • $\begingroup$ Wick rotation just changes the metric that we used , metric change is balance by the substitution of complex time @Wein Eld $\endgroup$ – ROBIN RAJ May 14 '20 at 17:27
  • $\begingroup$ @ROBIN RAJ It is just the definition for the Euclidean action. You can have more motivation for that definition if you derive the path integral representation of the Euclidean transition amplitude. $\endgroup$ – Wein Eld May 14 '20 at 23:13
  • $\begingroup$ Can you give the dispersion relation and klein jordan equation in eucledian space.@Wein Eld $\endgroup$ – ROBIN RAJ May 15 '20 at 13:35
  • $\begingroup$ @ROBINRAJ I updated the answer. $\endgroup$ – Wein Eld May 16 '20 at 9:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.