Spectrum of the Hamiltonian in QFT

unchained1978
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I know in ordinary QM, the spectrum of the Hamiltonian \{ E_{n}\} gives you just about everything you need for the system in question (roughly speaking). So what happens to this spectrum in QFT where |\psi\rangle is now a multiparticle wavefunction in some Fock space? I've been trying to understand this, but I don't yet have a clear grasp. Essentially, what's wrong with writing \hat H |\psi_{n}\rangle=E_{n}|\psi_{n}\rangle in QFT where the psi's are now multiparticle states?
 
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It's nothing wrong, but it's not really useful. The QFT has two types of states: free / asymptotic ones for which the spectral equation for the Hamiltonian has solutions - free particles on their mass sheet (according to the representations of SL(2,C) semidirect product with R^4), while for the interacting states there's no use for the spectral equation, since the states are no longer stationary -> S-matrix formalism.
 
What's a mass sheet? (Or did you mean mass shell?) Also, I read elsewhere that determining the spectrum corresponds to finding the spectrum of m^{2} or something, but I don't quite understand what that means or why it's important.
 
unchained1978 said:
Essentially, what's wrong with writing \hat H |\psi_{n}\rangle=E_{n}|\psi_{n}\rangle in QFT where the psi's are now multiparticle states?

There is nothing wrong. In some favourable cases you can diagonalize the multiparticle hamiltonian and you get everything you want from it, like e.g. for the strong coupling hamiltonian in superconductors.
The problem with relativistic QFT's like QED is that in 3+1 dimensions no one even has shown that the QFT exists at all as a well defined theory and the hamiltonian is unknown.
 
DrDu said:
The problem with relativistic QFT's like QED is that in 3+1 dimensions no one even has shown that the QFT exists at all as a well defined theory and the hamiltonian is unknown.

Can you elaborate a bit please? Or provide some links? I don't quite understand what you mean here.
 
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