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During the calculation to compute the shift of the energy of the hydrogen atom due to the photons by perturbation theory Schwartz states that

At second order in perturbation theory, only one photon can be created and destroyed, but we have to integrate over this photon's momentum. We are interested in the integration region where the photon has a very large momentum. By momentum conservation in OFPT (Old fashioned perturbation theory) since the ground state only has support for small momentum, the excited state of the atom must have large momentum roughly backward to that of the photon.

So I have 3 doubts in the above-mentioned paragraph.

$1$. Why only one photon? Since this is my first time handling calculation of QFT I don't know what does second-order perturbation convey to field theorist. Because in QM we just expand $H$ in terms of $H_O$ and $H_{int}$ collect the respective terms of $E^n$ and $\psi^n$ to get first, second ... nth order correction. There is as such no meaning given to nth order perturbation. They are there to just give more accurate calculation.

$2$. Why are we interested in only very large momentum state of photon?

$3$. What does the term support mean in the passage?

Qmechanic
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aitfel
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1 Answers1

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1) The order of perturbation is essentially the number of vertices. To create a photon requires a vertex. To re-absorb it requires another. We must re-absorb it to keep the system at the same energy.

Some place in your text there should be some Feynman diagrams. Second order is one loop. If you start with an electron then one loop is the electron giving off and re-absorbing a photon.

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2) Interested in large momentum because the integrals diverge at large momentum. If you are doing loops you need to do renormalization. Eventually you will be subtracting two terms, each of which diverges in the large momentum limit. You need to set up those limits so the result is meaningful and finite. The infinity is then pushed off to the next order of perturbation. If you can do that without introducing new parameters then your theory is renormalizable.

3) Renormalized quantum field theory still conserves energy and momentum, indeed does so at each vertex. So the energy round the loop has to come from some place. It comes from the electron moving off the mass shell. While it has emitted a photon it is no longer true that the mass of the electron is the mass we measure. But note that the photon and the electron are each virtual at this point, meaning they are not on the mass shell.

https://en.wikipedia.org/wiki/On_shell_and_off_shell

Before the electron emits the photon it satisfies the mass shell equation. But after it emits it and before it re-absorbs it, it does not. However, each component of 4-momentum is conserved at each vertex. So the photon and electron have 4-momentum components that sum back to what the electron had before it emitted the photon. Since the electron started close to at rest, then the momentum of each has to be nearly opposite in the loop.