Since Newton's law of gravitation can be gotten out of Einstein's field equations as an approximation, I was wondering whether the same applies for the electromagnetic force being the exchange of virtual photons. Is there an equation governing the force from the exchange of virtual photons? Are there any links which would show how the Coulomb force comes out of the equations for virtual photon exchange? I know that my question is somewhat similar to the one posted here The exchange of photons gives rise to the electromagnetic force, but it doesn't really have an answer to my question specifically.
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2$\begingroup$ For Rutherford scattering in QM & QED, see e.g. M.D. Schwartz, QFT & the standard model, 2014; section 13.4. $\endgroup$– Qmechanic ♦Commented Sep 14, 2021 at 7:37
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3 Answers
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The classical Coulomb potential can be recovered in the non-relativistic limit of the tree-level Feynman diagram between two charged particles.
Applying the Born approximation to QM scattering, we find that the scattering amplitude for a process with interaction potential $V(x)$ is
$$\mathcal{A}(\lvert p \rangle \to \lvert p'\rangle) - 1 = 2\pi \delta(E_p - E_{p'})(-\mathrm{i})\int V(\vec r)\mathrm{e}^{-\mathrm{i}(\vec p - \vec p')\vec r}\mathrm{d}^3r$$
This is to be compared to the amplitude obtained from the Feynman diagram:
$$ \int \mathrm{e}^{\mathrm{i}k r_0}\langle p',k \rvert S \lvert p,k \rangle \frac{\mathrm{d}^3k}{(2\pi)^3}$$
where we look at the (connected) S-matrix entry for two electrons scattering off each other, treating one with "fixed" momentum as the source of the potential, and the other scattering off that potential. Using the Feynman rules to compute the S-matrix element, we obtain in the non-relativistic limit with $m_0 \gg \lvert \vec p \rvert$
$$ \langle p',k \rvert S \lvert p,k \rangle \rvert_{conn} = -\mathrm{i}\frac{e^2}{\lvert \vec p -\vec p'\rvert^2 - \mathrm{i}\epsilon}(2m)^2\delta(E_{p,k} - E_{p',k})(2\pi)^4\delta(\vec p - \vec p')$$
Comparing with the QM scattering, we have to discard the $(2m)^2$ as they arise due to differing normalizations of momentum eigenstate in QFT compared to QM and obtain:
$$ \int V(\vec r)\mathrm{e}^{-\mathrm{i}(\vec p - \vec p')\vec r}\mathrm{d}^3r = \frac{e^2}{\lvert \vec p -\vec p'\rvert^2 - \mathrm{i}\epsilon}$$
where Fourier transforming both sides, solving the integral and taking $\epsilon \to 0$ at the end will yield
$$ V(r) = \frac{e^2}{4\pi r}$$
as the Coulomb potential.
answered Oct 19, 2014 at 21:12
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1$\begingroup$ Could you explain whether or not this method yields the overall sign, so that we can see that opposite charges attract whereas like charges repel? $\endgroup$ Commented Aug 11, 2022 at 21:58
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1$\begingroup$ @AndrewSteane It does, but in order to demonstrate how I'd have to explain in much more detail how the S-matrix element is computed from the Feynman diagram. $\endgroup$– ACuriousMind ♦Commented Aug 12, 2022 at 11:35
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$\begingroup$ Hmm. I find the answer rather circular: the photon propagator is the Green's function for the Laplacian (by definition, because the Laplacian is the wave eqn for E&M.). Written in momentum space, it is the familiar $1/{|p-p^{'}|}^{2}$ form and its fourier transform of the coulomb potential. (The massive Klein-Gordon eqn works exactly the same way) This is a mathematical statement, not a physical one. It is different from the physics question of how to understand the much vaguer idea of "a thermal bath of an infinite number of zero-energy photons" giving rise to electrostatics. $\endgroup$– LinasCommented Apr 10, 2023 at 3:49
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$\begingroup$ Are you sure about that $\delta(p-p')$, which seems to be enforcing that the cross section is zero unless $p=p'$? Or is that some relic of the way you enforced that one of the electrons (the one with momentum $k$) kept its momentum the same? $\endgroup$– AXensenCommented Apr 10, 2023 at 9:30
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1$\begingroup$ @AndrewChristensen Yes, that's an artifact of me setting $k' = k$. In "reality" that $\delta$-function just enforces momentum conservation. It was perhaps not the best choice of doing that. $\endgroup$– ACuriousMind ♦Commented Apr 10, 2023 at 10:25
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There is the (non-genetal) relation between the free energy of interacting of two currents $J^{a}, J^{b}$ and the propagator: $$ U = -\frac{1}{2} \int d^{4}xd^{4}y J^{a}(x) D_{ab}(x - y)J^{b}(y). $$ It's not general, but it realizes the simple example which can help you to understand how to get the expression for force.
The structure of field which causes structure of propagator helps us to get the expression for the force. For example, for intercation via scalar field (by setting $D_{ab}(x - y) = \frac{1}{p^{2} - m^{2}}$) after simple transformations for "point-like" currents $J(x) = \delta (\mathbf x - \mathbf x_{0})$ we can get $$ U = -\frac{1}{4 \pi |\mathbf r|}e^{-mr}. $$ For case $m = 0$ we get the Coulomb-like law of interaction.
Absolutely the same thing you may do with the case of EM field (in Feynman gauge $D_{\mu \nu} = -\frac{g_{\mu \nu}}{p^{2}}$).
If you need some explicit derivation I'll give it later.
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2$\begingroup$ Hi, I have a question – do you know how the first formula in this answer generalizes to non-abelian gauge theories? I presume the integral becomes a Wilson loop somehow, because as it stands it is not gauge-invariant. Is that true? $\endgroup$ Commented Jul 4, 2018 at 17:03
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1$\begingroup$ @SolenodonParadoxus that's pretty much right. In QED calculating the effective action for a static source is essentially the expectation value of a Wilson loop. If you run the same derivation through for a nonabelian gauge theory, you discover the equivalent of Coulombs law, which is basically to the old formula only with a factor of the quadratic casimir for the rep of the gauge group $\endgroup$– user213887Commented Nov 30, 2018 at 17:28
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$\begingroup$ The expression $e^{-mr}/r$ is called the Yukawa potential and has a long and storied history in nuclear physics. That wikipedia article provides the Fourier transform. $\endgroup$– LinasCommented Apr 12, 2023 at 17:49
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I've gone through the same thought process as you, so I suspect the question you're really trying to answer is:
"Can the formula for the electric Coulomb potential be generated as a natural consequence of quantum wave functions alone?"
If that's what you're wondering, then the answer is "no".
After Schrodinger had developed the wave function of the electron, he had hoped that it would somehow represent the electric charge and field. Alas, it does not. It is nothing more than a probability. Particle wave function doesn't give us the field potential.
Force potentials have to be added to the calculation of the interaction. The formulas that define the potential are experimentally derived. Physicists have come up with formulas that match observed behavior. Different forces have different formulas to define the potential. And that formula gets baked into the scattering calculation.
For a massless particle exchange, as in the photon, the potential simplifies to 1/r, as in classical physics. But this is really just because the potential was deliberately written to behave that way, in order to match experimental observation.
This leaves us no better or worse off than we were before with classical physics. The force potential is arbitrary, but this was also the case with the Coulomb potential. There's no particular reason the Coulomb potential should be 1/r in classical physics. It just is.
By contrast, the potential of the gravitational field potential naturally arises from solutions to the Einstein field equations. This is very aesthetically pleasing. Nothing comparable to this exists in quantum physics to derive the behavior of the other forces.
So if you were hoping that the potentials would naturally jump out of the particle wave functions, as Schrodinger had - sorry, no luck.
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1$\begingroup$ I don't understand why you say that the "1/r" potential for gravity is "aesthetically pleasing" because you can derive it from Einstein's equations, while you say that the "1/r" potential for electostatics is "arbitrary" despite the fact that it can be derived from Maxwell's equations. $\endgroup$– AndrewCommented Sep 28, 2023 at 17:16
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$\begingroup$ This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review $\endgroup$– MiyaseCommented Sep 28, 2023 at 17:19