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In QFT we work with Lagrangians which contain terms $m$ such that the relativistic relation $E^2 = p^2 + m^2$ is satisfied. By classical analogy $m$ is called the 'mass'. We note that due to the existence of the Higgs field, spontaneous symmetry breaking can explain why some fields have zero and others nonzero values of $m$.

Presumably this quantity happens also to be the same $m$ that one would measure by, say, whacking a field quantum with a baseball bat and then studying its trajectory. My question is: why should this be? I.e. how does the rest mass of special relativity emerge from the mass terms of quantum fields?

Even more lovely would be an interpretation of how the Higgs mechanism might be understood from this perspective, or an explanation of for what reason no such interpretation is possible/useful. But this is a secondary goal.

Related: The interpretation of mass in quantum field theories.

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  • $\begingroup$ I'm not sure what you're asking for. You correctly observe that the QFT mass of a state fulfills $E^2 = p^2 + m^2$. Since there's no $\hbar$ in there, it still does so in the classical limit, meaning the $m^2$ still fulfills that relation in the classical theory we might associate, meaing it's the special relativistic rest mass. You can't talk about "whacking a quantum with a baseball bat and then studying its trajectory" because quantum objects don't have a well-defined trajectory, and it's unclear what "whacking" would mean - they are not classical objects upon which a force could act. $\endgroup$ – ACuriousMind Nov 16 '15 at 20:42
  • $\begingroup$ Fair enough. What I mean is how does the quantum mechanical mass influence the classical behaviours of systems of very large particle number. If someone who had heard of QFT but not of Newtonian physics were asked to predict the trajectory of, say, a thrown Coke can, what would their explanation be? $\endgroup$ – AGML Nov 16 '15 at 20:46
  • $\begingroup$ Well, that sounds as if your question is really more about how to take the classical limit of a QFT than about mass. $\endgroup$ – ACuriousMind Nov 16 '15 at 21:12
  • $\begingroup$ do not forget that the binding energy also contributes to the mass $\endgroup$ – user172184 Dec 11 '17 at 2:43

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