Demystifier
Science Advisor
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By a cutoff. Now you will say that cutoff ruins Lorentz invariance, after which I will reply that cutoff should also be used without trajectories, so my theory is not less Lorentz invariant than standard QFT.A. Neumaier said:But the quantum equations for QED are mathematically meaningless because they use ill-defined bare coefficients. How do your particle trajectories survive renormalization when not even the number operator survives it?
