1
$\begingroup$

I'm a layperson currently reading through Matt Strassler and on his website, he states, in relation to using positronium as an example of vacuum fluctuation effects:

The stronger force between the electron and positron makes it more common for virtual photons to be present

My background is based primarily on Griffith's Quantum Mechanics and Particle Physics texts, I also think I understand the basic application of simple harmonic oscillators as creation and annihilation operators.

I don't want to appear lazy, but of the several reasons I can think of to answer my question myself, I just don't know enough to say which of them is correct. For example, do the charges add energy to the field, or is there an interference effect between the electron and positron, if we treat them as wave packets?

I know that the number of virtual particles increases as we approach the source, but how this is related to force in the way described is not immediately clear to me.

Obviously, I am as confused as I am ignorant on this topic, my apologies.

Is it possible to answer my title question, at the level of basic QFT, or to refer me to a source that does so?

$\endgroup$
0
$\begingroup$

Quantum field theory is formulated in terms of fields, not particles. Particles are phenomena that the theory predicts when the conditions are right, but "virtual particles" refers to something else.

Another page by Matt Strassler (https://profmattstrassler.com/articles-and-posts/particle-physics-basics/virtual-particles-what-are-they) hints at the fact that "virtual particles" are really just terms in a calculation, not discrete physical entities. The type of calculation that involves "virtual particles" is one that starts by solving the equations in the trivial case where nothing interacts with anything else, and then successively adding in more and more "corrections" to account for the interaction. The stronger the interaction, the more "corrections" are needed. These "corrections" are mathematical expressions that are often represented graphically as Feynman diagrams. Superficially, these diagrams depict processes in which the interaction between physical objects is mediated by one or more "virtual particles." However, what these diagrams really represent are mathematical expressions that arise in this particular method of calculation.

Here's an analogy: The words "carrying the $1$" refer to a step in the usual manual procedure for adding two integers: \begin{align*} \begin{matrix} \phantom{+}\scriptsize{1}\phantom{0} \\ \phantom{+}25 \\ +17 \\ \hline \phantom{+}42 \end{matrix} \end{align*} "Carrying the $1$" is important for getting the right answer when the calculation is done this way, but this isn't the only way to do the calculation, and it's not a good way to think about the meaning of addition. Similarly, the type of calculation to which the words "virtual particle" refer is a type of manual calculation that is mostly useful for describing scattering experiments, where the interacting objects are close to each other only for a brief instant. The longer the objects are close to each other, the more of these "virtual particle" terms we need to add up in order to get a good approximation. This is roughly analogous to the fact that when we're adding two large integers by hand, we need to "carry more $1$s" than when we're adding two small integers. The excerpt from Matt Strassler's site, shown in the OP's question, is presumably referring to something like this.

There are situations in which a small number of these "correction" terms give an excellent approximation to the exact answer, and in those situations, there can be short-lived physically-observable particle-like phenomena that correspond relatively directly to the computational devices called "virtual particles". In general, though, there is no such correspondence.

$\endgroup$
  • $\begingroup$ Thanks very much for taking the time to answer Dan, our answers crossed. I am at the stage of reading the text, but not doing the exercises, and it shows......the wording force made me assume things I should have already known $\endgroup$ – user213900 Nov 23 '18 at 22:31
1
$\begingroup$

Thinking more deeply (for me:) about the word force, I guess without a constant exchange of force carriers, self evidently there would be no force. Each force carrier (photons in this case) has an amplitude to act as a loop in a Feynmann diagram and produce virtual particles.

Oh boy, I'm new to this.

$\endgroup$
  • 1
    $\begingroup$ We were all new to this at some point! The whole reason I got involved with Physics SE is because I struggled to learn this stuff, too, and I'm still struggling to learn more. Welcome to Physics SE! And I'm sorry our answers got crossed -- I didn't know you were already posting an answer. (+1) $\endgroup$ – Chiral Anomaly Nov 23 '18 at 22:46
  • 1
    $\begingroup$ I think it's also useful to remember that this "exchange of virtual particles causing a force" should not be taken too literally. If you try to explain the force through exchanging particles, where the "force" arises from recoil because of conservation of momentum, you can "explain" repulsion, but not attractive forces, between opposite charges, for instance. I believe a less ambiguous (maybe not as intuitive) way to explain the origin of the force is by saying that the photon field $A_\mu$ couples to the fermion field, such that 2 fermionic sources are coupled through $A_\mu$ $\endgroup$ – Avantgarde Nov 24 '18 at 10:24
  • $\begingroup$ @avantgarde thanks for your comment, I will always go wrong trying to apply a classical concept to a quantum mechanical process. More and more, I am thinking in math terms. I did ask in my local university about explaining attraction, and was told to think in terms of "inverse momentum".......:) but I'm happy to follow the diagrams and the math relating to propagators from now on. $\endgroup$ – user213900 Nov 24 '18 at 12:25
  • $\begingroup$ @Xander Yeah, following the math is a good idea. By the way, what is 'inverse' momentum? $\endgroup$ – Avantgarde Nov 24 '18 at 12:36
  • 1
    $\begingroup$ @Xander Heh. But I guess you figured it out yourself that it cannot explain attraction. Sometimes, you have to abandon classical intuition for explaining quantum processes. $\endgroup$ – Avantgarde Nov 24 '18 at 14:22
0
$\begingroup$

Strassler says this in a section called “What would happen if $\alpha$ were closer to 1?” The fine structure constant is, roughly, the probability that an electron or positron “emits a virtual photon”. (In the mathematical expressions for the Feynman diagrams, each vertex contributes a factor of $\alpha$ to the probability of whatever process one is calculating.) So all he is really saying is that when you increase $\alpha$ you increase the probability of virtual photons. He isn’t comparing positronium to other bound systems like H atoms; he is comparing higher-$\alpha$ positronium to lower-$\alpha$ positronium.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy