Wick rotation and contour integral

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The discussion centers on the conditions necessary for performing a Wick rotation in contour integrals within the complex q plane. It emphasizes that the integrand must vanish rapidly as q approaches infinity to allow for a counterclockwise rotation of the contour without crossing any poles. The participants clarify that while fast vanishing is essential, it does not inherently prevent the contour from passing over poles; this is a separate requirement. The fast vanishing condition ensures that contributions from the semicircular contours at infinity are negligible, enabling the integral to be analytically continued from the real to the imaginary axis. Overall, understanding these conditions is crucial for correctly applying Wick rotation in complex analysis.
koolmodee
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We have an integral over q from -\infty to +\infty as a contour integral in the complex q plane. If the integrand vanishes fast enough as the absolute value of q goes to infinity, we can rotate this contour counterclockwise by 90 degrees, so that it runs from -i\infty to +i\infty.
In making this, the contour does not pass over any poles. Why?

When I make a contour integral, the big semi-circle that I use in the upper half has the -x+iy pole in it. How does the fast vanishing of the integral and the rotation make the integral not pass over the pole? I don't see it.

thank you
 
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koolmodee said:
We have an integral over q from -\infty to +\infty as a contour integral in the complex q plane. If the integrand vanishes fast enough as the absolute value of q goes to infinity, we can rotate this contour counterclockwise by 90 degrees, so that it runs from -i\infty to +i\infty.
In making this, the contour does not pass over any poles. Why?

When I make a contour integral, the big semi-circle that I use in the upper half has the -x+iy pole in it. How does the fast vanishing of the integral and the rotation make the integral not pass over the pole? I don't see it.

thank you
I think the fast vanishing of the integral is a required condition for performing Wick rotation. Only when the fast vanishing boundary condition is satisfied, one can perform the Wick rotation.
The condition that the rotation of the contour does not sweep through any poles is not a consequence of fast vanishing condition. Fast vanishing does not imply that the rotation sweeps no poles. Correctly speaking, the rotation passing through no poles is another condition one needs in order to be able to perform Wick rotation.
 
thanks ismaili!

I misread something in my book. It says the rotation does not pass over no pole ( in the specific integral given). I misunderstood that as after the rotation there are no poles in the contour.

But why is the fast vanishing of the integral necessary for Wick rotation?
 
koolmodee said:
thanks ismaili!

I misread something in my book. It says the rotation does not pass over no pole ( in the specific integral given). I misunderstood that as after the rotation there are no poles in the contour.

But why is the fast vanishing of the integral necessary for Wick rotation?
If the integral vanishes fast toward the infinity, the big semicircles contribute nothing so that the contour integral from -\infty to \infty in the real axis can be analytic continuated to the imaginary axis.
(This is my understanding, everyone is welcome to correct me or supplement.)
 
Makes sense.

thanks again
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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