This is a doubt straight from Peskin, eq 2.43 ∅(x,t) = e^{iHt}∅(x)e^{-iHt}. This had been derived in Quantum Mechanics. How does this hold in the QFT framework? We dont have the simple Eψ=Hψ structure so this shouldn't directly hold. I'm sorry if this is too trivial
I think this is just an analogy taken from QM to QFT. I assume ∅ here is an operator. This is just the definition of the Heisenberg picture, which one derives from the way the time evolution operator acts. And the way the time evolution operator acts is pretty much the Schrodinger equation... The time-indipendent version you wrote there is a perfectly valid way to calculate energy eigenvalues in QFT. I hope I didn't write any big mistakes.
I'm sorry ∅ is the field operator. The problem is that the Schrodinger equation goes as [itex]i\hbar \frac{∂}{∂ t} \Psi = H\Psi[/itex] With which we can just write [itex]\Psi(x,t) = e[/itex]^{-iHt}[itex]\Psi[/itex] and hence the Heinsenberg picture. but this doesnt directly hold for QFT
Why not? I mean, the Schrodinger equation expresses nothing more than the fact that the Hamiltonian is the infinitesimal generator of time translations. In Dyson's formula, one uses the time integral of the Hamiltonian density from initial to final time, rather than Ht where H is time-indipendent. I think that's the only real difference, but the evolution operator is used in the same way to compute S-matrix elements.
you should be careful; in QM ψ is the wave function and its time evolution (derived from the Schrödinger equation) is ψ(t) = U(t,t_{0}) ψ(t_{0}); in QFT ψ is the field operator and its time evolution ψ(t) = U(t) ψ_{0} U*(t); but this is not derived from the Schrödinger equation but from the Heisenberg equation of motion for operators.