This is an extremely naive question (based on a knowledge of chapter 2 of peskin and schroeder) so apologies for any things that seem obvious. The complex scalar field, when quantized, has a conserved charge $$Q \propto\int d^3x\left(\phi^*(x)\dot\phi(x)-\dot\phi^*(x)\phi(x)\right)=\int\frac{d^3p} {(2\pi)^3}\Big(a^\dagger(\vec p)a(\vec p)-b^\dagger(\vec p)b(\vec p)\Big)$$ (up to an infinite constant, etc). We interpret the $a$ particles as positively charged and the $b$ particles as equally negatively charged, and further we see that $$Q\propto N(a)-N(b)$$ where $N(a,b)$ are the number operators. According to this, charge conservation predicts that if we create an $a$ particle then a $b$ particle must be created also to cancel out charge (and similarly for destruction).

  1. In general, does this mean that no charged particles can be created in a QFT without a corresponding antiparticle appearing also (and is this the mathematical apparatus usually used to display this?)

  2. Also, if this is all true, how is it even possible that there are more electrons in the universe than positrons?

  • 1
    $\begingroup$ The problem you pose in 2. is called leptogenesis. $\endgroup$
    – ACuriousMind
    Apr 21, 2015 at 20:48

1 Answer 1


For the first question: there may be several species of particle in a theory, so charge conservation can be upheld without creating a particle and its antiparticle together. For example, in beta decay, an electron is `created', but no positron, and charge conservation is upheld by a neutron being exchanged for a proton.

This partially answers question two (for example, at high energies it is expected that lepton and baryon conservation should be violated, so it may be possible to create an electron and a proton starting with neutral stuff), though it still remains a problem to explain matter/antimatter asymmetry.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.