# How to describe electron-electron repulsion using virtual photon exchange? [duplicate]

Electron-electron repulsion can be described deterministically using Coulomb's law $$F = k\frac{e^2}{r^2}$$

Given two initially stationary electrons, the complete time evolution (distances apart, velocities, acceleration) of each electron at any later time is specified by this model.

Suppose we would now like to describe this exact same repulsion scenario by an exchange of virtual photons between the two electrons.

The first electron recoil by emitting the virtual photon. A little later, the virtual photon is absorbed by the second electron, again causing it to recoil. The total result is an apparent repulsion between the two electrons.

I am unsure about some aspects of this model:

1. Photons are localized wavepackets and travel as a very specific local disturbance. How it is that the virtual photon emitted by the first electron can be aimed so precisely as to hit the second electron every time?

2. The Coulomb model describes much more than simple repulsion between the electrons, it prescribes the precise strength of the repulsive force (acceleration) at every instant. It also prescribes the exact trajectory in the time evolution of the two electrons. By what mechanism can the electron trajectories be perfectly reproduced using the virtual photon exchange model of electron repulsion?

• A Feynman diagram like you drew is actually a shorthand for an integral over infinitely many possibilities. All that matters is the connected-ness of the vertices -- you can rotate it and deform the lines, and it still represents the same thing, so you shouldn't get too caught up on "time" being the y axis, nor should you interpret the x axis as a physical direction in space. See this article for the actual form of the integral (over all possible photon directions and momenta) it represents. Commented Dec 13, 2022 at 18:59
• "...how does the virtual photon find the 'direction' in 3D space to reach the second electron?" It doesn't. Similarly, the potential $1/r$ doesn't have to "find" either of the electrons.
– hft
Commented Dec 13, 2022 at 19:26
• @James Quantum mechanically, you are not guaranteed to take the shortest path. In this metaphorical example, the path integral tells you that the probability of your friend observing that you took a particular path will be smaller if the path is longer. Commented Dec 13, 2022 at 19:32
• I would recommend Feynman and Hibbs' Quantum Mechanics and Path Integrals for a rather elegant (and less math-heavy) introduction to this way of thinking. Commented Dec 13, 2022 at 19:39
• Commented Dec 13, 2022 at 22:04

Virtual particle exchange isn't a model, per se - it's essentially a way to associate intuitive pictures to complicated calculations.

To give you a very loose cartoon picture of the kinds of calculations which occur in QFT, we can consider a QFT in which we have two fields $$\phi$$ and $$\psi$$, which do not interact with themselves but do interact with each other. The fundamental quantities which we'd like to calculate generally take the form

$$\mathcal M \sim \langle q_1,q_2 ;\emptyset|\mathbb S| k_1,k_2;\emptyset\rangle$$

This expression can be broken down as follows:

• $$|k_1,k_2;\emptyset\rangle$$ describes an initial state of the system which consists of two $$\phi$$ particles with momenta $$k_1$$ and $$k_2$$, respectively, and no $$\psi$$ particles.
• $$\langle q_1,q_2;\emptyset|$$ describes a final state of the system which consists of two $$\phi$$ particles with momenta $$q_1$$ and $$q_2$$, respectively, and no $$\psi$$ particles, where in general $$(q_1,q_2)\neq (k_1,k_2)$$.
• $$\mathbb S$$ is the scattering operator (or scattering matrix), which depends on the details of the fields and the ways in which they interact. This operator serves to "evolve" the initial state into the final state.

In total, the square of this probability amplitude $$|\mathcal M|^2$$ gives us the probability that the initial state will evolve into the final state. We interpret this as the initial $$\phi$$ particles interacting with each other indirectly via the $$\psi$$ field.

In the absence of an interaction between the $$\phi$$ and $$\psi$$ fields, the $$\phi$$ particles will continue on with precisely the same momenta forever. However, if the fields are allowed to interact then "ripples" in the $$\phi$$ field create distortions in the $$\psi$$ field, which can interact with the original ripples and change their momenta. As a very rough classical analogy, you might imagine two ships on the ocean, each interacting with the wake produced by the other. This is essentially what scattering is.

The problem is that $$\mathbb S$$ is generally much too complicated to calculate exactly; as a result, we approximate it as $$\mathbb S = 1 + \lambda \hat T_1 + \lambda^2\hat T_2 + \ldots$$ where $$\lambda$$ (called the coupling constant) describes the strength of the interaction between the fields. If $$\lambda$$ is small, each successive term in this expansion is smaller than the last and so terminating this expansion after the second (or third, or fourth, or ... ) term yields a reasonable approximation for $$\mathbb S$$.

In turn, we find that each $$T_i$$ can be broken down into a sum of terms, e.g. $$\hat T_1 = \hat A + \hat B + \hat C + \ldots$$. Therefore, if we are interested only in the "first order" approximation to $$\mathcal M$$, we need to compute all of the terms $$\mathcal M \approx \langle q_1,q_2;\emptyset|1|k_1,k_2;\emptyset\rangle + \lambda\bigg(\langle q_1,q_2;\emptyset|\hat A|k_1,k_2;\emptyset\rangle+\langle q_1,q_2;\emptyset|\hat B|k_1,k_2;\emptyset\rangle+\ldots \bigg)$$

Up to this point, there has been no talk whatsoever of virtual particles. We only have two fields which interact in a complicated way. However, Feynman realized that each term (e.g. $$\langle q_1,q_2;\emptyset|\hat A|k_1,k_2;\emptyset\rangle$$) can be associated with a pretty diagram, and that once you understand the rules of the theory, you could simply read off the value of each term just by looking at the corresponding diagram.

The diagrams also have a nice, physical(ish) interpretation - they can be thought of as the $$\phi$$ particles "exchanging" some number of $$\psi$$ particles. However, one finds during the calculation that the particles being exchanged don't act like freely-propagating particles do (more specifically, they don't obey the energy-momentum relationship $$E^2=p^2 c^2 + m^2 c^4$$). As a result, they are called virtual particles.

In this context, it is hopefully clear that the virtual particles are essentially computational tools which correspond to terms in a complicated approximation. The initial and final states $$|k_1,k_2;\emptyset\rangle$$ and $$\langle q_1,q_2;\emptyset|$$ are genuine states of the fields, and therefore must obey the proper energy-momentum relationships; the "internal" terms of the expansion don't actually correspond to real states and therefore are not constrained in the same way.

So in summary - you should not take Feynman diagrams literally. They do not represent actual, dynamical exchanges of real particles. They are nothing more than suggestive and intuitive pictures which can be used to simplify what is otherwise a very complex calculation. What's actually happening when e.g. two electrons scatter by "exchanging a photon" is that the electron field interacts with the electromagnetic field to produce a localized distortion, and that the electrons (i.e. "ripples in the electron field") interact with this distortion and change their momentum.

Finally, I would reiterate that this is a very loose, hand-wavy sketch of perturbative QFT. There are many textbooks worth of technical (and not-so-technical) details which I either simplified enormously or completely ignored. My purpose here was merely to paint a picture of how the virtual particle concept arises in QFT, and to demonstrate that virtual particles and Feynman diagrams are just flowery language and pictures through which we may choose to organize terms in an otherwise mundane approximation scheme.

• thank you vm. The answer is expansive and and clear to me. One question if you will, since the two electrons are already in the same electron field, why can't they interact directly using ripples in this electron field, but have to make use of an intermediary ripple in the electromagnetic field in order to interact with each other? Commented Dec 14, 2022 at 2:31
• @James That's a good question. You're asking why electrons interact electromagnetically but not directly. The answer is that (a) that seems to be the way the universe operates, and (b) that one of the details I left out is that if we follow the naive approach I have outlined, we find that all of the terms from $\lambda^2$ onward are formally infinite. This seemingly enormous problem is resolved through the techniques of renormalization, but it turns out that any QFT which features direct interaction between fermions is non-renormalizable. Commented Dec 14, 2022 at 3:04
• thank you. Lastly, suppose the two electrons' trajectories are about to cross each other in the electron field... will they pass through each other without taking any notice in this electron field, supposing we can suppress their electromagnetic field interaction somehow? Commented Dec 14, 2022 at 4:01
• @James If you turn off the electron/photon interaction (e.g. by letting the coupling constant go to zero) then the electron field is what is known as a free field. In that case, yes electrons simply pass through each other. Commented Dec 14, 2022 at 5:37