How Does Field Operator Evolution Hold in QFT?

sumeetkd
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This is a doubt straight from Peskin, eq 2.43
∅(x,t) = eiHt∅(x)e-iHt.

This had been derived in Quantum Mechanics.
How does this hold in the QFT framework?
We don't have the simple Eψ=Hψ structure so this shouldn't directly hold.

I'm sorry if this is too trivial
 
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sumeetkd said:
This is a doubt straight from Peskin, eq 2.43
∅(x,t) = eiHt∅(x)e-iHt.
What is ∅?

sumeetkd said:
We don't have the simple Eψ=Hψ structure so this shouldn't directly hold.
What's your problem with E|ψ>=H|ψ> in QFT?
 
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.
 
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tom.stoer said:
What is ∅?


What's your problem with E|ψ>=H|ψ> in QFT?

I'm sorry ∅ is the field operator.

The problem is that the Schrodinger equation goes as i\hbar \frac{∂}{∂ t} \Psi = H\Psi
With which we can just write \Psi(x,t) = e-iHt\Psi and hence the Heinsenberg picture.
but this doesn't 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.
 
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