# Wick rotation is consistent with caculus?

1. Aug 23, 2009

### 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 $$+i\infty$$

2. Aug 23, 2009

### humanino

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.

3. Aug 23, 2009

### PRB147

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

4. Aug 23, 2009

### RedX

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.

5. Aug 23, 2009

### Parlyne

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.

6. Aug 23, 2009

### humanino

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