Question

QFT: What does “finiteness” mean?

Answer 1

hey ❕ I think this is your answer
As above: what is the definition of a QFT to be "finite"?

That all UV corrections are finite and there are no divergences at all?

That there are divergences, but these divergences can be absorbed in the (suppose renormalisable) Lagrangian with counterterms, leaving just a finite quantum correction?

may it helps you

Answer 2

In relativistic quantum field theories, the traditional renormalization approach leads to Hamiltonians with infinite counterterms. The infinities cancel when this Hamiltonian is used to calculate the renormalized finite S-matrix and related observable properties in a good agreement with experiment. However, time evolution of state vectors and observables cannot be studied without a well-defined finite Hamiltonian. Based on the “clothed particle” approach (O. W. Greenberg and S. S. Schweber, 1958, Nuovo Cimento8, 378), we reformulate the theory in such a way that ultraviolet infinities appear neither in the S-matrix nor in the Hamiltonian. In this formulation the Hamiltonian is finite and allows us to calculate the time evolution of wave functions, the S-matrix, and other properties by a straightforward application of quantum mechanical rules without renormalization. A rigorous approach to the bound states in quantum field theory is also discussed using the hydrogen atom as an example.