The Wick rotation in position space

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1. Apr 30, 2015

Quantioner

The Feynman propagator is
$$G_{F}(x) = \int d^4p \, \frac{e^{-ip x}}{p^2 - m^2 + i\epsilon}.$$
I want to understand why the directions of Wick rotation in position space and momentum space are contrary. Every book I find says something like "we should keep $xp$ unchanged", but why?

As we know the poles $p_0 = \pm (\omega - i\epsilon)$ of the propagator in the momentum space decide the direction of Wick rotation in the momentum space. Thus, what's the poles of the propagator in the position space? If we know them, the direction of Wick rotation is decided directly. But no book talks about them and Wick rotation in this way, why?

2. May 6, 2015

Greg Bernhardt

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

3. May 7, 2015

Quantioner

Because
$$\Delta(p) = \frac{1}{p^2-m^2+i\epsilon} = -i \int_0^\infty d\alpha ~e^{i(p^2 - m^2 +i\epsilon)\alpha}$$
Thus
$$\Delta(x) = \int \frac{d^4 p}{(2\pi)^4} e^{-ipx} \Delta(p) \\ = -i \int_0^\infty d\alpha \int \frac{d^4 p}{(2\pi)^4} ~e^{-ipx+i(p^2 - m^2 +i\epsilon)\alpha} \\ = -i \int_0^\infty d\alpha \frac{1}{(2\pi)^4} [-i\pi^2\alpha^{-2} e^{\frac{-ix^2}{4\alpha}-i(m^2-i\epsilon)\alpha}]$$
Let $\beta = \frac{1}{\alpha}$, then we get
$$\frac{-1}{16\pi^2} \int_0^\infty d\beta~ e^{-\frac{i\beta x^2}{4}-\frac{i(m^2-i\epsilon)}{\beta}}$$
There is an integration formula (see "Table of integrals, series and products" 7ed, p337 section3.324 1st integral)
$$\int_0^\infty d\beta \exp\left[-\frac{A}{4\beta}-B\beta\right]=\sqrt{\frac{A}{B}}K_1\left(\sqrt{AB}\right)\qquad [\mathrm{Re}A\ge0, \mathrm{Re}B>0].$$
If $\mathrm{Re}A\ge0, \mathrm{Re}B>0$ is violated, the integral will be divergence.

In my case, $A=4(im^2+\epsilon)$ and $B=ix^2/4$, so $\mathrm{Re}A=4\epsilon>0$ and $\mathrm{Re}B=0$ which does not satisfy the convergent condition. Therefore, to guarantee the convergence of the integral, we should treat $B=ix^2/4$ as the limit $B=\lim_{\epsilon'\rightarrow0+}i(x^2-i\epsilon')/4$. Thus we have
$$\Delta(x)=\lim_{\epsilon,\epsilon'\rightarrow0+}\frac{-1}{16\pi^2}\int_0^\infty d\beta \exp\left[-\frac{i\beta (x^2-i\epsilon')}{4}-\frac{i(m^2-i\epsilon)}{\beta}\right]\\ =\lim_{\epsilon,\epsilon'\rightarrow0+}\frac{-1}{4\pi^2}\sqrt{\frac{m^2-i\epsilon}{x^2-i\epsilon'}}K_1\left(\sqrt{-(m^2-i\epsilon)(x^2-i\epsilon')}\right)\\ =\lim_{\epsilon'\rightarrow0+}\frac{-m}{4\pi^2\sqrt{x^2-i\epsilon'}}K_1\left(m\sqrt{-(x^2-i\epsilon')}\right)$$
As a result, the singularity of the propagator is at $x^2-i\epsilon=t^2-\mathbf{x}^2-i\epsilon=0$, i.e. $t=\pm(|\mathbf{x}|+i\epsilon)$.

Actually, the convergent condition of the integral restricts the analytic regime of $\Delta(x)$:
$$0<\mathrm{Re}(ix^2)=\mathrm{Re}(it^2)=-\mathrm{Im}(t^2)$$
i.e.
$$(2n-1)\pi\le\arg(t^2)=2\arg(t)\le 2n\pi\\ (n-\frac{1}{2})\pi\le\arg(t)\le n\pi$$
Therefore, $\Delta(x)$ only can be analytically continued to the second and the forth quadrants in the complex plane of $t$. In conclusion, the wick rotation in $t$ plane should be clockwise.