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I am following Zee's QFT book in Section II.6. I have found the amplitude for an electron to scatter from a static Coulomb potential as \begin{align*} \mathcal{M}&=ie\!\int\!d^4x\,\big\langle P_2\big| \bar\psi\gamma^\mu\psi \big|P_1\big\rangle A_\mu(x)\\ &=ie\,C_1(P_1)\,C_2(P_2) \, \bar u (P_2)\gamma^\mu u(P_1) \int\!d^4x\, e^{ix(P_2-P_1)} A_\mu(x)~~. \end{align*}

The potential is static so $A_\mu(x)=A_\mu(\vec x)$ and since it is the Coulomb potential, I also have $A_i=0$. I separate the time part of the integral as \begin{align*} \mathcal{M}&=ieC_1C_2 \!\int\!d^3x\,\bar u (P_2)\gamma^0 u(P_1)e^{i\vec x\cdot(\vec P_2-\vec P_1)}A_0\int\!dx^0\, e^{ix^0[(P_2)_0-(P_1)_0]} \\[4pt] &=2\pi ieC_1C_2 \,\bar u (P_2)\gamma^0 u(P_1)\,\delta\big[(P_2)_0-(P_1)_0\big]\int\!d^3x\,e^{i\vec x\cdot(\vec P_2-\vec P_1)}A_0(\vec x) ~~. \end{align*}

The Coulomb potential (of a stationary proton) sets $A_0=-e/|\vec x|~~,$ so \begin{align*} \mathcal{M}&=-2\pi ie^2C_1C_2 \,\bar u (P_2)\gamma^0 u(P_1)\,\delta\big[(P_2)_0-(P_1)_0\big]\int\!d^3x\,\frac{e^{i\vec x\cdot(\vec P_2-\vec P_1)}}{|\vec x|} ~~. \end{align*}

I can simplify the integral a little as $$ \int\!d^3x\,\frac{e^{i\vec x\cdot(\vec P_2-\vec P_1)}}{|\vec x|} = \int\!d^3x\,\frac{\cos( \vec x\cdot\Delta\vec P)}{|\vec x|}+\underbrace{i\!\int\!d^3x\,\frac{\sin(\vec x\cdot\Delta\vec P)}{|\vec x|}}_{=0}~~,$$

but I do not see how to solve it and it is a little too complicated for the free online integrators. How do I proceed?

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    $\begingroup$ I believe you can change coordinates to polar where, defining $|\vec{x}| := r$, the integrand becomes $e^{i r \Delta P \cos{\theta}}/r$. Of course there will be an additional factor from the change of coordinates. $\endgroup$ – QuantumFieldMedalist Dec 3 '20 at 17:27
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You have to use the scalar product in your integral : \begin{equation} \int d^3 x \frac{e^{i\vec{x} \cdot (\vec{P}_2-\vec{P}_1)}}{|\vec{x}|}=\int_{\mathbb{R}^{+}} \iint_{\Omega} e^{i|\vec{x}||\vec{P}_2-\vec{P}_1|\cos \theta} |\vec{x}| \sin \theta \,d|\vec{x}|d\theta d\varphi \end{equation} Know it should be easier.

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  • $\begingroup$ Thanks! Where did the $i$ go in the exponential on the RHS? $\endgroup$ – hodop smith Dec 3 '20 at 17:30
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    $\begingroup$ I forgot it, sorry, I made an edit. $\endgroup$ – Jeanbaptiste Roux Dec 3 '20 at 17:32

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