# Mental picture of interactions in QFT

Apologies in advance for the possible vagueness of my question. If it is not a good question, an explanation of why it isn't would be a very useful answer to me.

I am trying to find a useful mental picture of interactions in QFT. I think these kinds of mental pictures are an indispensable aid to grasp abstract concepts in some intuitive way that can guide you when doing hard calculations, but when the picture is inaccurate it can also lead you astray.

I have seen as the mental picture of interactions by particle exchange in QFT two people throwing a ball back and forth, resulting in a repulsive force. For attractive forces, they either throw a boomerang, or a ball with negative momentum.

I don't really see how this picture actually can guide us in any meaningful way, though maybe that is because I don't understand QFT well enough.

I was thinking that maybe a useful picture would be to still think in terms of point particles, but their interaction is through a field, as required by locality. In fact, both particles interact with the field, rather than directly with each other. Since this is a quantum field, changes in the field are quantized: the change is interpreted as the addition of a particle to the field, or the absorption of a particle from the field. If this picture is more or less correct, interacting particles don't really exchange particles, but rather they both interact with the boson field, thereby altering it, which will in turn do something to the other particle.

However, I think this cannot be really accurate: the particles added to the field or absorbed from it are virtual, hence they are not observable: they cannot originate from only one particle interacting with the field: unless a disturbance created by one particle is absorbed by the other, it was never really there. If there is any validity to the previous description, is there any way it can be made more accurate to account for the non-observability of the exchanged bosons?

Another refinement would be not to imagine the interacting particles as some classical kind of point particles, but rather as the quanta of disturbance of their own fields. In the straightforward picture it is not hard to imagine any kind of interaction, but it is not so clear what would be the role of a force-mediating field.

As a particular application I would be interested how asymptotic freedom in QCD could be imagined in terms of such a picture. Should be first translate high energy to short distance (only in the first picture in which the particles are points)? If so, can we see what it would mean that the coupling is low at very short separations? Or should be see the high energy as the matter field disturbance to consist of very high frequency fluctuations, and can we see what it means that at high frequencies the fields interact little?

Is there any validity to these mental pictures? If so, which one would be most accurate, and how could it be corrected or refined?

• "I was thinking that maybe a useful picture would be to still think in terms of point particles, but their interaction is through a field" - as far as I understand it, there is no particle interpretation for interacting quantum fields except effectively in the asymptotic limit of 'infinite' past or 'infinite' future. I recently asked the professor of the QFT class I'm currently taking what is effectively infinity since our detectors in, e.g., the LHC, are clearly not at temporal or spatial infinity. Sigh... At least he responded that it was a good question. Nov 9, 2016 at 1:20
• This sounds silly, but as I self study I need to do it when I read QFT books. At the top of every page I write " what are my assumptions" very slowly. Mental pictures, you can't stop them coming into your head, but if you question the assumptions behind them, they are a help, in my opinion.
– user108787
Nov 9, 2016 at 1:25
• See this excellent article about virtual particles. I think the moral is that your should pay attention to the field viewpoint too. Nov 9, 2016 at 1:28
• @AlfredCentauri: I would say that infinity here means much bigger than the spatial and temporal scales of interaction, which are extremely small for the LHC. So you could take "infinite" to mean "at least $1\ \mathrm{cm}$ or $1\ \mathrm{s}$". Nov 9, 2016 at 1:30
• @Javier, I'm thinking that's the correct answer too but where have you seen that stated? Nov 9, 2016 at 1:32

I also like to have "mental pictures" (as you call them) for abstract concept :)

I will try to share with you a picture which I have in mind when I do QFT calculations. Please ignore this answer if it doesn't help you.

The one for the interacting QFT that I have does not include particles at all. I imagine quantum fields as fluctuating fields over spacetime which we integrate over in the path integral. The question that we would like to answer in this approach is - given a particular functional of fields, say a product of fields at different spacetime points - what is the expectation value of the product? In another words, what is the value of the path integral

$$\left< \phi_1 (x_1) \dots \phi_k (x_k) \right> = \int {\mathcal D}\phi_1 \dots {\mathcal D}\phi_n \; e^{i S[\phi_1,\dots,\phi_n] \, / \, \hbar} \phi_1(x_1) \dots \phi_k(x_k).$$

The measures can be chosen such that $$\left< 1 \right> = 1,$$

which makes the infinity in the normalization factor disappear.

I know that you probably looking for something with less math. But I would like to try to convince you that this picture is incredibly useful and pretty simple to work with.

In this picture, the field kind of "disappears", meaning the following. The actual predictions of the theory, the things which we would like to compute, don't depend on $\phi(x)$ at all! Instead, they depend on $k$ spacetime points. The fluctuating field $\phi(x)$ is useful for deriving the results, but it does not enter them explicitly.

1. The free theory propagator and Wick's theorem follow from the path integral almost instantly. As an example of the formal proof, consider this. I would like to consider the expectation $$\left< \phi(y) \right> = \int {\mathcal D}\phi e^{i S[\phi]} \phi(y).$$ What if I change the dummy integration variable by a constant shift of $\delta \phi(x)$? The measure is formally invariant, and the value of the integral can't change: $$0 = \delta \left< \phi(y) \right> = \int {\mathcal D}\phi \left( e^{i S[\phi]} i \delta S[\phi] \cdot \phi(y) + e^{i S[\phi]} \cdot \delta \phi(y) \right) = \left< i \delta S[\phi] \cdot \phi(y) + \delta \phi(y) \right>$$ $$= \int d^4 x \; \delta \phi(x) \; \left< i \frac{\delta S[\phi]}{\delta \phi(x)} \phi(y) + \delta^{(4)} (x - y) \right>.$$ For the free theory, $\delta S / \delta \phi(x) = \hat{\Theta} \phi(x)$ where $\hat{\Theta}$ is some linear differential operator generating the equations of motion, so we have $$\hat{\Theta}_x \left< \phi(x) \phi(y) \right> = i \delta^{(4)} (x - y),$$ which is exactly the expected result: the free theory propagator is a Green's function of the differential operator which generates the equations of motion.

2. Feynman rules follow from the path integral almost instantly. You simply Tailor-expand the exponential of the interaction term in the action. From now on we can think of Feynman diagrams as of terms in the series that approximate the path integral.

3. Regularization can be interpreted as a modification of the path integral measure ${\mathcal D}\phi$. It is pretty convenient, because you still have the equation above for the finite regularized theory.

4. You can easily derive covariant Feynman rules even for gauge theories via the Faddeev-Popov trick.

5. The Ward identities are straightforward to derive. The derivation basically resembles the one from point 1, but it uses a dummy integration variable relabeling through symmetry transformations instead of a constant shift by $\delta \phi(x)$.

6. Anomalies can be interpreted as the non-invariance of the path integral measure under symmetry transformations. The anomalous Ward identity can be derived via the regularization of the measure.

7. The approach is explicitly Lorentz invariant, and in fact, explicitly invariant under all symmetry transformations. Regularizations can break this property, but I would argue that broken upon regularization and expected to reappear after taking the regularization off Lorentz invariance is still much easier to deal with in the path integral approach.

There are drawbacks to this picture, ofcourse. Despite the fact that is extremely simple and intuitive to think in terms of the fluctuating field which we integrate over, it doesn't offer the complete picture. For example, the space of asymptotic states (Fock space) has to be derived independently, and the reduction formula should be used in order to express the transition amplitudes between asymptotic states in terms of the path integral expectation of some functional.

• Thanks for your reply, +1. I'll have to read it carefully still, but I think it could be a useful point of view. However, to me it raises a new problem: that of a mental picture of the path integral! I'll think about that and maybe post another question at some later time. Nov 9, 2016 at 0:16
• @doetoe You are welcome! Path integrals are just like ordinary integrals, but over a huge domain of all possible fluctuating fields :) Nov 9, 2016 at 0:19
• +1 You have written an answer that I would be proud of. Although I am nearly there at trying at to get rid of physical pictures, when I use them I just always go wrong, possibly because I self study. But if "cartoons" worked for Feynman, but not for Schwinger, then it just depends on the individual. Great answer.
– user108787
Nov 9, 2016 at 1:14
• as an experimentalist, I am curious what QFT calculations you are making? Do they have as an output numbers to be compared with data as evaluating Feynman diagrams. Nov 9, 2016 at 5:30
• @annav I am not sure that I understand your question. I've been evaluating regularized multi-loop corrections to the terms in effective actions for various models of gauge fields and modified perturbative gravity on the generic background, for example. Feynman diagrams are a computational tool, while the picture which I described in my answer is just a useful interpretation to keep in mind. I don't claim it to have great physical meaning, it is only meant to help you to develop intuition. Nov 9, 2016 at 6:40