Method of Steepest Descent and Contour Deformation

Question

In the book An Introduction to Quantum Field Theory by Peskin and Schroeder, p. 14 in section 2.1, it is stated that, in looking at the asymptotic behavior for $x^{2} \gg t^{2}$ of the integral \begin{equation} \int_{-\infty}^{\infty} \mathrm{d}p \; p \, \mathrm{e}^{\phi(p)}, \qquad \phi(p) = i\left( p x - t \sqrt{p^{2} + m^{2}}\right),\tag{1} \end{equation} one can implement the method of stationary phase.

However, the stationary points of $\phi(p)$, $$p_{\pm} = \pm imx/\sqrt{x^{2} - t^{2}},\tag{2}$$ are imaginary, which would seem to indicate to me that the method of stationary phase is not applicable as is. The book then mentions that one can "freely push the contour upward", but exactly how and why is not clear to me.

After looking around, it seems that the asymptotic behavior can be obtained by using the method of the steepest descent. However, I don't understand the following:

1. How does one chooses a contour in the complex plane such that, through Cauchy's theorem I suppose, relates to the integral above, while at the same time passing through one of the stationary points.

2. Should one choose only one of the stationary points? If yes, why?

Any help is greatly appreciated.

Answer 1

1. At space-like distances $$x^2~>~t^2\tag{A}$$ let us consider the integral $$\begin{align} I(x,t)~:=~~~~&\int_{\mathbb{R}}\! \mathrm{d}p~e^{\phi(p)} ~=~2\int_{\mathbb{R}_+}\! \mathrm{d}p~\cos(px)e^{-it\sqrt{p^2 + m^2}}\cr ~\stackrel{(GZ3.914.1)}{=}&~2\frac{itm}{\sqrt{x^2-t^2}}K_1\left(m\sqrt{x^2-t^2}\right)\cr ~\stackrel{(AS9.7.2)}{\sim}~&2\frac{itm}{\sqrt{x^2-t^2}} \sqrt{\frac{\pi }{2m\sqrt{x^2-t^2}}} e^{-m\sqrt{x^2-t^2}}\cr ~=~~~~&it\frac{\sqrt{2\pi m}}{(x^2-t^2)^{3/4}} e^{-m\sqrt{x^2-t^2}} \quad\text{for}\quad \sqrt{x^2-t^2}~\to~ \infty.\end{align}\tag{B} $$ [This is the relevant integral for a 1+1D spacetime. OP's integral (1), which originates from a 3+1D spacetime, can then easily be obtained via differentiation of the integral (B) wrt. $x$.] Here we have followed the hint in a footnote of Ref. [PS] that the integral is just a Bessel function, cf. Ref. [GZ]. In the process we have regularized the integral $I(x,t)$ by assuming that the imaginary part of time $${\rm Im}(t)~<~0 \tag{C}$$ is slightly negative. This regularization (C) is consistent with the physics paradigm that the Minkowskian formulation is a Wick rotation of the Euclidean formulation.

2. OP's quote about "pushing the contour upwards" indicates that Ref. [PS] is using the method of steepest descent aka. the saddle point method with $$x,t~\propto~\lambda~\text{large}. \tag{D}$$ Let us now sketch how. We calculate $$\begin{align} \phi(p) ~:=~& i\left( p x - t \sqrt{p^2 + m^2}\right),\cr \phi^{\prime}(p) ~=~& i\left( x - \frac{tp}{\sqrt{p^2 + m^2}}\right),\cr \phi^{\prime\prime}(p) ~=~& -i\frac{tm^2}{(p^2 + m^2)^{3/2}}.\end{align}\tag{E}$$

3. For large $|p|\gg m$, the function ${\rm Re}\phi(p)$ is dominated by its first term $-{\rm Im}(p)x$, cf. eq. (A), so the sector of exponential decay is the upper (lower) half plane if $x>0$ ($x<0$), respectively$^1$.

4. The stationary points are $$\begin{align} p_{\pm}~:=~&\pm\frac{imx}{\sqrt{x^2-t^2}},\cr \sqrt{p_{\pm}^2 + m^2}~=~&\pm\frac{imt}{\sqrt{x^2-t^2}},\cr \phi(p_{\pm})~=~&\mp m\sqrt{x^2-t^2},\cr \phi^{\prime}(p_{\pm})~=~&0,\cr \phi^{\prime\prime}(p_{\pm})~=~&\pm\frac{(x^2-t^2)^{3/2}}{mt^2}. \end{align}\tag{F}$$

5. The direction of steepest descent from the stationary point $p_{\pm}$ is$^2$ vertical (horizontal) in the complex $p$-plane, respectively. One might naively think that the horizontal descent is relevant for the integral $I(x,t)$, but the horizontal descent contour crosses into the wrong Riemann sheet of the square root $\sqrt{p^2 + m^2}$.

6. In more detail, we put the branch cut along the imaginary $p$-axis with $|{\rm Im}(p)|\geq m$, cf. Fig. 1. Note that both stationary points $p_{\pm}$ lie next to the branch cut. [Without the regularization (C), $p_{\pm}$ would lie precisely on top of the chosen branch cut.] The stationary points are in quadrant I and III (II and IV) of the complex $p$-plane if $t>0$ ($t<0$), respectively.

   Im(p) ^    /
         |   / 
         |  /     
         | |        
         | x p+      
         | |        
         | |        
    \ im x |        
     \----/          
    --------------------------------> Re(p)
   -im x 
         |
         |
     --x-|
      p- |
         |
         |
   

   $\uparrow$ Fig. 1. The complex $p$-plane with the branch cut along the imaginary axis outside $\mp im$. Fig.1 depicts the case where $x>0$ and $t>0$. Then the stationary point $p_+$ with a vertical steepest decent direction lie in quadrant I, while the stationary point $p_-$ with a horizontal steepest decent direction lie in quadrant III. In this case the dominant contribution to the integral $I(x,t)$ comes from $p_+$.

   

   $\uparrow$ Fig. 2. In Fig. 2 we have chosen the parameters $x\!=\!1.3$, $t\!=\!1\!-\!0.1i$, and $m\!=\!3$. Four blue anti-Stokes curves ${\rm Re}\phi(p)={\rm Re}\phi(p_+)$ emanate from the stationary point $p_+\approx 0.63 + 4.53 i$ in the complex $p$-plane. The two yellow curves refer to extensions to the other Riemann sheet of the square root $\sqrt{p^2 + m^2}$. Note that the red spot is not an actual intersection point as the curves belong to different Riemann sheets. If we were to rotate $180^{\circ}$ the anti-Stokes curves for $p_+$ around the origin we would get the anti-Stokes curves ${\rm Re}\phi(p)={\rm Re}\phi(p_-)$ for $p_-$. The anti-Stokes curves divide the complex $p$-plane into sectors. It turns out that the steepest descent contour should only cross anti-Stokes curves at stationary points.

7. So the dominant contribution is instead actually given by the vertical descent contour around $p_+$: $$\begin{align} I(x,t)~\sim~&i{\rm sgn}(t)\sqrt{\frac{2\pi}{\phi^{\prime\prime}(p_+)}}e^{\phi(p_+)}\cr ~=~&\frac{it\sqrt{2\pi m}}{(x^2-t^2)^{3/4}}e^{-m\sqrt{x^2-t^2}} \quad\text{for}\quad \sqrt{x^2-t^2}~\to~ \infty.\end{align}\tag{G}$$ Here the direction of integration near $p_+$ is up (down) if $t>0$ ($t<0$), respectively.

   In conclusion: We should push the contour upwards (downwards) if $x>0$ ($x<0$), respectively.

References:

• [PS] M.E. Peskin & D.V. Schroeder, An Intro to QFT, 1995; p. 14.

• [GZ] I.S.Gradshteyn & I.M. Ryzhik, Table of Integrals, Series, and Products; eq. (3.914.1).

• [AS] Abramowitz & Stegun, Handbook of Mathematical Functions, eq. (9.7.2).

---

$^1$ In hindsight, an alternative method is to take the integration contour along the real $p$-axis and deform it to cling to both sides of the upper (lower) branch cut of the square root $\sqrt{p^2 + m^2}$ if $x>0$ ($x<0$), respectively. Then the integral simplifies to $$\begin{align}I(x,t)~=~& i\int_m^{\infty}\! \mathrm{d}p_y~e^{-p_y|x|}2\sinh\left(t\sqrt{p_y^2-m^2}\right)\cr ~=~&i\sum_{\pm}\pm\int_m^{\infty}\! \mathrm{d}p_y~e^{-p_y|x|\pm t\sqrt{p_y^2-m^2}}\cr ~=~&i\sum_{\pm}\pm\int_m^{\infty}\! \mathrm{d}p_y~e^{-S_{\mp}(p_y)}. \end{align}\tag{H}$$ [For $x<0$ we have made a substitution of the integration variable $p_y\to -p_y$ in eq. (H).] Here $$\begin{align} S_{\mp}(p_y)~=~& p_y|x|\mp t\sqrt{p_y^2-m^2},\cr S^{\prime}_{\mp}(p_y)~=~& |x|\mp \frac{tp_y}{\sqrt{p_y^2-m^2}},\cr S^{\prime\prime}_{\mp}(p_y)~=~& \pm \frac{tm^2}{(p_y^2-m^2)^{3/2}}. \end{align}\tag{I}$$ Next use Laplace's method. $S_{\mp}(p_y)$ has a stationary point if $t>0$ ($t<0$), respectively. In detail $$\begin{align} p_y~=~&\frac{m|tx|}{\sqrt{x^2-t^2}},\cr \sqrt{p_y^2-m^2}~=~&\frac{m|t|}{\sqrt{x^2-t^2}},\cr S_{\mp}(p_y)~=~&m\sqrt{x^2-t^2},\cr S^{\prime}_{\mp}(p_y)~=~&0,\cr S^{\prime\prime}_{\mp}(p_y)~=~&\frac{(x^2-t^2)^{3/2}}{mt^2}.\end{align}\tag{J}$$

This leads to the same answer as eq. (G) $$\begin{align} I(x,t)~\sim~&i{\rm sgn}(t)\sqrt{\frac{2\pi}{S^{\prime\prime}_{\mp}(p_y)}}e^{-S_{\mp}(p_y)}\cr ~=~&\frac{it\sqrt{2\pi m}}{(x^2-t^2)^{3/4}}e^{-m\sqrt{x^2-t^2}} \quad\text{for}\quad \sqrt{x^2-t^2}~\to~ \infty.\end{align}\tag{K}$$

$^2$ Strictly speaking, the direction is only truly vertical (horizontal), when we turn off the regularization (C).