I'm working with the signature ##(+,-,-,-)## and with a Minkowski space-stime Lagrangian(adsbygoogle = window.adsbygoogle || []).push({});

##

\mathcal{L}_M = \Psi^\dagger\left(i\partial_0 + \frac{\nabla^2}{2m}\right)\Psi

##

The Minkowski action is

##

S_M = \int dt d^3x \mathcal{L}_M

##

I should obtain the Euclidean action by Wick rotation.

My question is about the way with that I should perform the Wick rotation.

Since the spacetime interval is defined by ##ds^2 = dt^2 - d\vec{x}^2##, If I perform a Wick rotation (just rotating the time axis) I get a negative Euclidean interval.

1. What's the sense of that? What's the connection between physical actions calculated in two different signature?

2. I can perform the rotation with different signs ##t =\pm i\tau##. I know that, if there exist any poles, I must choose the correct sign in order to not cross them. In this case, apparently I can choose both and I get always the same result.

If I choose ## t = i\tau ## I get

##

i\int_{+i\infty}^{-i\infty} d\tau d^3x \Psi^\dagger\left(i\frac{\partial}{\partial i\tau} + \frac{\nabla^2}{2m}\right)\Psi = -i\int_{-i\infty}^{+i\infty} d\tau d^3x \Psi^\dagger\left(\frac{\partial}{\partial \tau} + \frac{\nabla^2}{2m}\right)\Psi(x,i\tau)

##

If I choose ## t = -i\tau ## I get

##

-i\int_{-i\infty}^{+i\infty} d\tau d^3x \Psi^\dagger\left(-i\frac{\partial}{\partial i\tau} + \frac{\nabla^2}{2m}\right)\Psi = -i\int_{-i\infty}^{+i\infty} d\tau d^3x \Psi^\dagger\left(-\frac{\partial}{\partial \tau} + \frac{\nabla^2}{2m}\right)\Psi(x,-i\tau) = -i\int_{-i\infty}^{+i\infty} d\tau d^3x \Psi^\dagger\left(\frac{\partial}{\partial \tau} + \frac{\nabla^2}{2m}\right)\Psi(x,i\tau)

##

But, I should obtain

##

S_E = \int d\tau d^3x \Psi^\dagger(x,\tau)(\partial_\tau - \frac{\nabla^2}{2m})\Psi(x,\tau)

##

not?

3. Is the Euclidean action defined by ##S_M = i S_E## or by ##S_M = -iS_E##?

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# Performing Wick Rotation to get Euclidean action of scalar f

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