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This is a line from Chapter 2 of Peskin:

"We have no right to assume that any relativistic process can explained in terms of a single particle, since the Einstein relation $E = mc^2$ allows for the creation of particle-antiparticle pairs."

Does this simply mean that two particles can be produced if the energy threshold for creation because of their masses is met by the preceding interaction? Or is there something deeper inherent to the particle-antiparticle pair specifically?

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  • $\begingroup$ I don’t understand the statement in Peskin. $E=mc^2$ also allows for the creation of particle-particle pairs etc etc. $\endgroup$
    – my2cts
    Commented Nov 8 at 17:07
  • $\begingroup$ A good answer is already there, but my two cents: it's probably just unnecessary words distracting readers from the main idea. From the context, the point might only be "$E=mc^2$ allowing creation and single-particle assumption does not hold". However, the particle-antiparticle pair production is fun and common, so they chose to write it down. Knowing the multi-particle possibility in relativistic cases should be enough. $\endgroup$
    – Yang Xiaosheng
    Commented Nov 9 at 15:19

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Let's suppose we start with an energy eigenstate of the free Hamiltonian of a relativistic system (which is what we would consider an electron). This energy eigenstate could be in a degenerate subspace. Generically, evolution of the Schrodinger equation -- including the interaction Hamiltonian -- will cause the quantum state to evolve and become some kind of superposition of all the energy eigenstates within this subspace.

Then what can happen if you start with a quantum particle of mass $m$ with a given energy $E > 2mc^2$ (which is generically true for relativistic processes where energies are much larger than the mass), is that the degenerate subspace of energy eigestates will include states where both the particle and its antiparticle are present. If no conservation law or other obstruction prevents the original "particle-only" state from evolving to include a superposition of this new "particle+antiparticle" state, then the Schrodinger equation will generically cause the state to evolve to be a superposition with a non-zero coefficient with the "particle+antiparticle" state.

In more flowery language, "anything not forbidden is mandatory" in quantum mechanics. The quantum state will generically evolve into a superposition over every basis state it can consistent with the constraints of unitary Schrodinger evolution.

It's worth pointing out that a single electron will never spontaneously produce an electron-positron pair as a final state because you could not satisfy all the constraints of momentum and energy conservation in this process, but you need to consider those kinds of multi-particle virtual states in Feynman diagrams and they have the effect of renormalizing the electron mass and charge.

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  • $\begingroup$ If the initial state is already an "energy eigenstate", how can it include other eigenstate components by evolution? $\endgroup$
    – Yang Xiaosheng
    Commented Nov 7 at 1:07
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    $\begingroup$ @XiaoshengYang If you have a degenerate subspace of states that all have the same energy, Schrodinger evolution will generically lead the state to explore that subspace. $\endgroup$
    – Andrew
    Commented Nov 7 at 1:25
  • $\begingroup$ I don't think this is true. If you're talking about an energy eigenstate $\psi$, i.e. $\hat{H}\psi = E\psi$, then it's evolution will be $\mathrm{e}^{-\mathrm{i} E t} \psi$, which means it always stays in its own 1-dimensional subspace, rather than spreading to the degenerate eigen-subspace. However, I have heard of a situation similar to your description. An eigenstate of an un-perturbed Hamiltonian $H_0$ will spread to the degenerate subspace of a perturbed Hamiltonian $H_0+H_{\mathrm{int}}$. Is this what you're talking about or probably I misunderstood something you say. $\endgroup$
    – Yang Xiaosheng
    Commented Nov 8 at 1:54
  • $\begingroup$ @YangXiaosheng Sorry, you are completely right. Let me think about whether I can rephrase what I wrote or if I should delete this answer. But yes you are right, for a free theory if you started with an electron with energy $>2 mc^2$, you would never get a state with a particle-anti-particle pair, there needs to be some interaction. So the interaction/perturbation allows the state to evolve within this degenerate subspace. $\endgroup$
    – Andrew
    Commented Nov 8 at 14:18
  • $\begingroup$ I added some language to include the interaction Hamiltonian. I'm still not sure I said it in the best way but I'll think a bit more on it. What I'm basically getting at is that if a final state is energetically allowed (and consistent with other conservation laws), then generically there will be some non-zero transition amplitude that will take the initial state to that final state. Normally in particle physics that's phrased in terms of an $S$ matrix but that should be equivalent to Schrodinger evolution. But, you're right, in a free theory the $S$ matrix is trivial. $\endgroup$
    – Andrew
    Commented Nov 8 at 14:21

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