Sign of Wick rotation

Question

Suppose you have the integral

$$i \int^\infty_{-\infty} L_M(t) dt$$

and that $L_M$ contains two poles: when $t>0$ the pole lies above the t-axis and when $t<0$ the poles lies below the t-axis. Therefore you can rotate the contour from the real axis to the contour going from $i\infty$ to $-i\infty$. This path can be parameterized as $z=i\tau$ where $\tau$ is from $\infty$ to $-\infty$:

$$i \int^\infty_{-\infty} L_M(t) dt=i \int L_M(z) dz= i\int^{-\infty}_{\infty} L_M(i\tau) id\tau= \int^{\infty}_{-\infty} L_M(i\tau) d\tau \\\equiv-\int^{\infty}_{-\infty} L_E(\tau) d\tau $$

However, textbooks write instead:

$$\int^{\infty}_{-\infty} L_M(-i\tau) d\tau \equiv-\int^{\infty}_{-\infty} L_E(\tau) d\tau $$

so they get the sign wrong (or I got the sign wrong).

Why did you switch the integral limits in your second step? Also, if the pole lies above the x-axis for positive $t$, shouldn't you be doing $t\mapsto -\mathrm{i}\tau$ (i.e. rotate clockwise) to not hit it? In any case, hunting sign conventions and errors in your (or the textbook's) calculation seems to me to be off-topic as homework-like. — ACuriousMind, Feb 06 '16 at 01:18
related (or duplicate?): Performing Wick Rotation to get Euclidean action of scalar field — AccidentalFourierTransform, Feb 06 '16 at 18:01

score 1 · Answer 1 · answered Feb 06 '16 at 15:54

1

Thanks for your help. I agree this was homework-like - should've posted it somewhere else.

This is an equality

$$\int^\infty_{-\infty}f(x)dx=\int^\infty_{-\infty}f(-x)dx$$

so I guess mathematically it doesn't matter the sign of your substitution.

answered Feb 06 '16 at 15:54

Jim Burns

329

Sign of Wick rotation

1 Answers1