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The goal is to see what is the asymptotic behaviour of

$$U(t)=\frac{1}{2\pi^2|\vec{x}-\vec{x}_0|}\int_0^{\infty}dp~p \sin(p|\vec{x}-\vec{x}_0|)e^{-it\sqrt{p^2+m^2}}\tag{p.14}$$ for $x^2\gg t^2$.

In my book ('Introduction to Quantum Field Theory' of Peskin and Schröder page 14) they do steps I have great problems to understand: They use the method of stationary phase. They get the stationary point $$p=imx/\sqrt{x^2-t^2}$$ of the phase function $$f(p)=px-t\sqrt{p^2+m^2}.$$ Now they write

We may freely push the contour upward so that it goes through this point. Plugging in this value for $p$, we find that, up to a rational function of $x$ and $t$, $$U(t)\sim e^{-m\sqrt{x^2-t^2}}.\tag{p.14} $$

I don't understand this method, I looked at Wikipedia and in the searching machine, but I don't get it for this example and I don't quite understand what a phase function is. I changed the function $U(t)$ with the relation $\sin(x)=\frac{1}{2i}(e^{ix}-e^{-ix})$. In this way one of the the exponents is $$i(p|\vec{x}-\vec{x}_0|-t\sqrt{p^2+m^2}).$$ I calculated the first and second derivative, which have to be zero for a saddle point and so I hoped to get the stationary point written above. But if I plug $$p=imx/\sqrt{x^2-t^2}$$ in $$f'(p)=x-pt/\sqrt{p^2+m^2}$$ this derivative is not zero, but it is $f'(imx/\sqrt{x^2-t^2})=x-t$.

Do I understand all this totally wrong? I also don't understand what they mean in the cited sentences.

Qmechanic
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Kathi
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1 Answers1

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I think you've plugged it wrong. For $p=\frac{imx}{\sqrt{x^2-t^2}}$ we have $$ p^2+m^2 = \frac{-m^2x^2}{x^2-t^2}+m^2= \frac{-m^2t^2}{x^2-t^2}$$ so one of the roots that we have is $$ \sqrt{p^2+m^2} = \frac{imt}{\sqrt{x^2-t^2}}$$ and then $$ \frac{p}{\sqrt{p^2+m^2}} = \frac{x}{t}$$ $$ f'(p) = x - \frac{pt}{\sqrt{p^2+m^2}} = 0$$ $$ if(p) = ixp-it\sqrt{p^2+m^2} = -\frac{mx^2}{\sqrt{x^2-t^2}} + \frac{mt^2}{\sqrt{x^2-t^2}} = -m\sqrt{x^2-t^2}$$

Adam Latosiński
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