Wick rotation is consistent with caculus?

[PRB147]

knowing the action
$$S[x(t)]=\int_0^\infty dt'\left[\frac{m}{2}\left(\frac{dx}{dt'}\right)^2-V(x)\right]$$
After the so-called wick's rotation$$t'=-i\tau$$ with $$\tau$$ being real,
the action becomes
$$S[x(t)]=i\int_0^\infty d\tau\left[\frac{m}{2}\left(\frac{dx}{d\tau}\right)^2+V(x)\right]$$
my question is why the upper limit of the integral in the secong equation is still $$\infty$$?
I think ist should be $$+i\infty$$

 

[humanino]

my question is why the upper limit of the integral in the secong equation is still $$\infty$$?

Because it is not merely a change of variable : it is really a rotation in the complex plane. A Wick rotation is usually equivalent to a so-called "i$\epsilon$" prescription : you want to define a contour in the complex plane avoiding the singularities of your integrand on a precise side (up or down, right or left). Usually you do that because your integrand is well behaved at infinity only in certain directions.

If I understand correctly, "calculus" refers to "real analysis". A Wick rotation involves complex analysis. If that can cheer you up, complex analysis is very different from real analysis, and in many regards more powerful, so I would say more fun to use.

 

[PRB147]

avoiding the singularities of your integrand on a precise side (up or down, right or left). Usually you do that because your integrand is well behaved at infinity only in certain directions.

Do you mean the origin, the positive real axis and positive imaginary axis constitute a
closed contour?

 

[RedX]

Because it is not merely a change of variable : it is really a rotation in the complex plane. A Wick rotation is usually equivalent to a so-called "i$\epsilon$" prescription

So by "usually" do you mean that a Wick rotation in position space is almost the same as doing the Wick rotation in momentum space, but not always?

Because when you have the $$i\epsilon$$ prescription in momentum space, you know where your singularities are, so you can perform a Wick rotation. But in position space you just perform a Wick rotation without knowing where the singularities are.

 

[Parlyne]

Do you mean the origin, the positive real axis and positive imaginary axis constitute a
closed contour?

No. Wick rotation generally proceeds through a quadrant of the complex plane where the integrand is strictly zero at infinity. So, the contour actually involves (in the case you cite) the positive real axis, a 90 degree arc at infinity and the positive complex axis.

 

[humanino]

So by "usually" do you mean that a Wick rotation in position space is almost the same as doing the Wick rotation in momentum space, but not always?

Because when you have the $$i\epsilon$$ prescription in momentum space, you know where your singularities are, so you can perform a Wick rotation. But in position space you just perform a Wick rotation without knowing where the singularities are.

Well I hope that the potential V can be treated pertubatively around a minimum (steepest descent) as usual in momentum space.