Consider the Rutherford scattering $e^-p^+ \rightarrow e^-p^+$

If a proton is a treated as a heavier positron:

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If a proton is treated as a heavier electron with an opposite charge:

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These two amplitudes (I have omitted spin index) look different. At first, I think they should be the same when squared, but I failed to prove it, so the different treatments remain distinct to me. Could it be that treating a positron as the heavier electron with an opposite charge is the physical choice? Treating a positron as a heavier electron with opposite charge is what appears in the textbook. It makes certain sense because electron and protons are both particles rather than antiparticles. But why can't a proton be the antiparticle? What's the mathematical and physical argument that rules out this choice?

  • 1
    $\begingroup$ This reminds me. First: I'm a late-career solid-state / optics physicist. I'm supposed to be smart. I attended a public lecture by Ed Witten at the Institute of Advance Studies. During the Q and A at the end someone asked "Why does a proton have a different mass than the electron?" I was embarrassed for the poor guy. Quarks, etc. Witten began his reply, "That is a good question." I didn't understand a word he said after that. $\endgroup$
    – garyp
    Oct 6, 2023 at 13:01
  • $\begingroup$ @garyp I don't think you really understand my question. My question is more of a technical issue rather than the wildly open one you mentioned. $\endgroup$
    – Bababeluma
    Oct 6, 2023 at 13:33
  • $\begingroup$ You may accommodate the proton, ab initio, in a novel Dirac equation with a large mass, where it simulates a heavy "electron", or a heavy "positron": your choice. Its novel conserved baryon or anti baryon number is your option, just as choosing the electron or positron to be fundamental is... $\endgroup$ Oct 6, 2023 at 13:49
  • $\begingroup$ @CosmasZachosC But what about the amplitude I wrote? There is a distinct difference whether i take proton to be a heavy electron with positive charge or a heavy positron $\endgroup$
    – Bababeluma
    Oct 6, 2023 at 14:08
  • $\begingroup$ @CosmasZachos I think the title I gave was too misleading. I've corrected that. $\endgroup$
    – Bababeluma
    Oct 6, 2023 at 14:10

1 Answer 1


A proton "does not know" that it might be an anti-particle of another particle. We can describe it with a bispinor $u(p)e^{-ipx}$, a positive frequency 1-particle solution of the Dirac-equation (of course we do not consider here the inner structure of the proton).

And this is also true for a positron, it is also described by a bispinor based on a positive frequency 1-particle solution $u(p)e^{-ipx}$.

Because a positron is exactly the same as an electron apart from the charge. So if it is not interacting it behaves exactly the same as an electron. The solution of the Dirac equation $u(p)e^{-ipx}$ is a 1-particle solution without interaction, so it should work for an electron as well as for a positron.

Last statement might create some confusion, because one might wonder of the negative frequency solutions of the Dirac equation $v(p)e^{ipx}$. What about them? First of all let's us make clear: These solutions are electron-solutions. Because they are solutions of the same equation, why should an equation describe particles of different charge? This is even more clear if we consider the Dirac equation minimally coupled to the EM-field:

$$ (\gamma^\mu (i\partial_\mu -e A_\mu) -m)\psi =0$$

Positive and negative frequency solution of this equation couple to the same charge $+e$ and not to 2 different charges. One could object that $\psi$ is an interacting 1-particle solution. Then assume that the electromagnetic field is extremely weak. One would obtain in good approximation $u(p)e^{-ipx}$ and $v(p)e^{ipx}$ and both couple to the same charge with the same sign.

What then are the $v(p)e^{ipx}$ solutions? There are related to the "real" positron solutions via charge-conjugation:

$$C \overline{v(p)}^T = u(p)$$

But indeed one could consider them according to the Feynman-Stückelberg interpretation as electrons of negative energy traveling backwards in time.

In order to support this we will make the following exercise (Bjorken Drell I chapter 7.3).

Convert a vertex of 2 negative frequency solutions into a 2 positive frequency solutions:

$$\overline{v(p_2)}\gamma^\nu v(p_4)=u(p_2)^T C^{-1}\gamma^\nu C \overline{u(p_4)}^T=-u(p_2)^T (\gamma^\nu)^T \overline{u(p_4)}^T =-(\overline{u(p_4)}\gamma^\nu u(p_2))^T = - \overline{u(p_4)}\gamma^\nu u(p_2)$$

which shows that both expression given to the 2 Feynman diagrams in the post are identical. In order to achieve this equality we used the "defining" property of the charge conjugation matrix $C$:

$$C^{-1}\gamma^\mu C = -\gamma^{\mu T}$$

Furthermore note that $\overline{u(p)}\gamma^\mu u(q)$ is a number and the transpose of a number is the number itself.

So the answer to the post is that both expression yield the same amplitude.

  • $\begingroup$ Thanks for the detailed answer! $\endgroup$
    – Bababeluma
    Oct 7, 2023 at 14:18
  • $\begingroup$ I rechecked and found myself having trouble proving $(u_{\uparrow})^c=v_{\downarrow}$, this is obvious for zero momentum, how is that true for general momentum, I can't prove it $\endgroup$
    – Bababeluma
    Oct 10, 2023 at 13:29
  • $\begingroup$ @Bababeluma This is shown in all standard textbooks on QFT, Peskin & Schroeder, Srednicki, Landau & Lifshitz IV, Bjorken & Drell and many others. I also would have to look it up and when I will have got it the question will be probably closed as homework :-( $\endgroup$ Oct 10, 2023 at 14:02
  • $\begingroup$ Can you point to which section of Srednicki since you are familiar with it? I'm reading Schwartz and he skipped the proof. $\endgroup$
    – Bababeluma
    Oct 10, 2023 at 14:25
  • $\begingroup$ Somewhere around chapter 37. $\endgroup$ Oct 10, 2023 at 14:56

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