Performing Wick Rotation to get Euclidean action of scalar f

[FrancescoS]

I'm working with the signature $(+,-,-,-)$ and with a Minkowski space-stime Lagrangian

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

 

[vanhees71]

Despite the fact that you problem is non-relativistic, and there's no Minkowski symmetry your calculation shows that you get the same Euclidean action for both choices for the imaginary-time formlism. The reason is that the time-ordered propagator in the non-relativistic case is the same as the retarded propagator.

 

[FrancescoS]

Despite the fact that you problem is non-relativistic, and there's no Minkowski symmetry your calculation shows that you get the same Euclidean action for both choices for the imaginary-time formlism. The reason is that the time-ordered propagator in the non-relativistic case is the same as the retarded propagator.

And then how do you derive the euclidean action?
I'm reading a paper which is following this method...

 

[vanhees71]

Which paper? But as far as I can see, you have already calculated the effective action. Where is the problem?

 

[FrancescoS]

Which paper? But as far as I can see, you have already calculated the effective action. Where is the problem?

Because it should be
$S_E = \int d\tau d^3 x \Psi^\dagger(x,\tau)(\partial_t - \frac{\nabla^2}{2m})\Psi(x,t)$
which is different from what I obtained.

This is the paper (pag.7)
http://arxiv.org/abs/nucl-th/0510023

 

[vanhees71]

Ok, let's see. The correct sign for the Wick rotation in the time domain is obtained from the analogy between QFT and (grand-)canonical statistical mechanics. The grand-canonical statistical operator is $\propto \exp(-\beta \hat{H})$ which is a time-evolution operator with imaginary time $t=-\mathrm{i} \tau$. Then
$$\partial_t=\frac{\partial \tau}{\partial t} \partial_{\tau}=\mathrm{i} \partial_{\tau}$$
and thus
$$-S_{\text{E}}=\psi^* \left(\mathrm{i} \partial_t +\frac{\Delta}{2m} \right ) \psi=-\psi^* \left (\partial_{\tau}-\frac{\Delta}{2m} \right) \psi.$$
In momentum space the equation of motion reads
$$\partial_{\tau} \psi=-\frac{p^2}{2m}\psi \; \Rightarrow \psi(\tau,p)=A \exp\left (-\frac{p^2}{2m} \tau \right),$$
i.e., a falling exponential as it should be.

 

[FrancescoS]

Ok, let's see. The correct sign for the Wick rotation in the time domain is obtained from the analogy between QFT and (grand-)canonical statistical mechanics. The grand-canonical statistical operator is $\propto \exp(-\beta \hat{H})$ which is a time-evolution operator with imaginary time $t=-\mathrm{i} \tau$. Then
$$\partial_t=\frac{\partial \tau}{\partial t} \partial_{\tau}=\mathrm{i} \partial_{\tau}$$
and thus
$$-S_{\text{E}}=\psi^* \left(\mathrm{i} \partial_t +\frac{\Delta}{2m} \right ) \psi=-\psi^* \left (\partial_{\tau}-\frac{\Delta}{2m} \right) \psi.$$
In momentum space the equation of motion reads
$$\partial_{\tau} \psi=-\frac{p^2}{2m}\psi \; \Rightarrow \psi(t,p)=A \exp\left (-\frac{p^2}{2m} t \right),$$
i.e., a falling exponential as it should be.

Ok, this is the same of the first calculation I wrote. But, can you explicit the time-dependence of the field $\Psi$. We start with a field $\Psi(x,t)$ and then we get... ?

 

[vanhees71]

Argh. There was a typo in the last equation. In the solution, of course, I have to write $\tau$ instead of $t$ (it's corrected in my posting). Then you get
$$\psi(\tau,p):=\psi(\mathrm{i} t,p)=A \exp \left (-\frac{p^2}{2m} \tau \right )=A \exp \left (-\frac{p^2}{2m} \mathrm{i} t \right ),$$
i.e., you come from the Euclidean to the "Minkowskian" (or better the imaginary-time to the real-time) wave function by simply substituting $\tau=\mathrm{i} t$. Only if there are non-trivial analytic structures in some function like branching points, you have to be careful in specifying the analytic continuation from imaginary to real time (i.e., the Riemann sheet of the mulitvalued function).

 

[FrancescoS]

Thank you. So the keyword is "analytic continuation" If think. I will go to review, thank you again