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I am learning QFT, and for electron-positron annihilation, the differential cross section is infinite when $\theta=0$: enter image description here The usual explanation is that it means the two particles pass through each other. My confusion is that if two particles pass through each other, shouldn't the differential cross section be 0? I cannot understand an infinite cross section.

ZHANG Juenjie
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    Always explain your notation - I could guess it, but what is $\theta$ here? – ACuriousMind Mar 18 '17 at 13:53
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    "The usual explanation is that it means the two particles pass through each other without any interaction." Really? Think how the scattering angle varies as a function of the impact parameter. – dmckee --- ex-moderator kitten Mar 18 '17 at 15:15
  • related question http://physics.stackexchange.com/questions/111072/regarding-the-infinite-cross-section-in-rutherford-scattering – anna v Mar 28 '17 at 06:31

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The answer is NO, the differential cross section for the electron-position annihilation is not approaching infinity as θ->0. The general expression of the differential cross section for this type of QED process (e+e- -> ff(anti), which is a pair production of charged fermions) is

enter image description here

which is measured in the center of mass frame (but it is Lorentz invariant) with natural units, where s is the center of mass energy, Qf is the charge of the final state particle, a is the fine structure constant which is equal to 1/137, and B is the relative velocity v/c. So as you can see, cosθ can vary from -1 to 1, but it doesn't blow up at θ=0. In the ultra-relativistic limit, B~1, so you will have (1+cos^2θ) left in the parenthesis.

Back to your question. It looks like that you are referring to a scattering process (given that you provided a plot with Rutherford differential cross section vs. angle), which is a low energy non-relativistic process. Thus you will get your usual Rutherford differential cross section expression (for a electron-nucleus scattering) with a (1-cosθ)^-2 term, which indeed will blow up at θ=0. Why? Pause for a second and think about it yourself, as it requires some physical insights (BTW, your passing-through-each-other argument is totally wrong). And you know, that's what you need as a soon-to-be physicist.

So here's the reason. Imagine the positron is fixed in space, and you shoot it with an electron. Since the E field extends to all space, the electron will always feel a force from the positron. So imagine position is staying on an axis, and you shoot your electron in the direction parallel to the positron axis, no matter how far the two axes are from each other, the electron will always be deflected by a little bit. Thus, you will never quite get a θ=0 for the electron trajectory as measured from the electron axis. Now imagine again that the two axes are infinitely apart from each other, only then you will get your θ=0, but then what's your differential cross section at θ=0? It's pi*inf^2, which is infinite.

I hope this helps, and I think you are really dealing with a classical picture here instead of a relativitic quantum one.

Cheers.

Noitall
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    well the classical Rutherford scattering formula is reproduced in the quantum mechanical formulation http://www.fisica.net/quantica/ebooks/MERZBACHER%20-%20Quantum%20Mechanics%202ed.pdf – anna v Mar 28 '17 at 06:38
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    the formula is 11.105 and the page 262 . So one has to deal with it by the fact that since the coulomb potential covers the whole space the zero angle impinging and scattering sees the whole space , as you state – anna v Mar 28 '17 at 06:49