2 added 393 characters in body edited Aug 22 '17 at 17:04 SergioHC 933 bronze badges If you forget about the $$U(1)$$ gauge group of QED you can certainly take a theory of fermions (e.g. electrons) and add all sorts of coupling terms (so long as these are sensible terms respecting Poincaré and global $$U(1)$$) that will give you vacuum graphs and pair creation-annihilation. For example, you could have a quartic coupling of the form $$(\bar{\psi}\psi)^2$$ in your theory of electrons and positrons. In a free theory, there would be no interactions (by definition), and no pair-creation or annihilation. EDIT: By the way, I think you are misunderstanding what "free" means. A QFT is free if it is at most quadratic in its fields (such that equations of motion are linear), with no two different fields coupled together. Such theory only has kinetic and mass terms. You can have a theory without photons which is still not a free theory (e.g. fermions with a quartic vertex like the one above). If you forget about the $$U(1)$$ gauge group of QED you can certainly take a theory of fermions (e.g. electrons) and add all sorts of coupling terms (so long as these are sensible terms respecting Poincaré and global $$U(1)$$) that will give you vacuum graphs and pair creation-annihilation. For example, you could have a quartic coupling of the form $$(\bar{\psi}\psi)^2$$ in your theory of electrons and positrons. In a free theory, there would be no interactions (by definition), and no pair-creation or annihilation. If you forget about the $$U(1)$$ gauge group of QED you can certainly take a theory of fermions (e.g. electrons) and add all sorts of coupling terms (so long as these are sensible terms respecting Poincaré and global $$U(1)$$) that will give you vacuum graphs and pair creation-annihilation. For example, you could have a quartic coupling of the form $$(\bar{\psi}\psi)^2$$ in your theory of electrons and positrons. In a free theory, there would be no interactions (by definition), and no pair-creation or annihilation. EDIT: By the way, I think you are misunderstanding what "free" means. A QFT is free if it is at most quadratic in its fields (such that equations of motion are linear), with no two different fields coupled together. Such theory only has kinetic and mass terms. You can have a theory without photons which is still not a free theory (e.g. fermions with a quartic vertex like the one above). 1 answered Aug 22 '17 at 16:55 SergioHC 933 bronze badges If you forget about the $$U(1)$$ gauge group of QED you can certainly take a theory of fermions (e.g. electrons) and add all sorts of coupling terms (so long as these are sensible terms respecting Poincaré and global $$U(1)$$) that will give you vacuum graphs and pair creation-annihilation. For example, you could have a quartic coupling of the form $$(\bar{\psi}\psi)^2$$ in your theory of electrons and positrons. In a free theory, there would be no interactions (by definition), and no pair-creation or annihilation.