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Evolution of Field operators

by sumeetkd
Tags: peskin and schroeder, qft
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sumeetkd
#1
Feb19-12, 04:46 AM
P: 9
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 dont have the simple Eψ=Hψ structure so this shouldn't directly hold.

I'm sorry if this is too trivial
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tom.stoer
#2
Feb19-12, 06:06 AM
Sci Advisor
P: 5,366
Quote Quote by sumeetkd View Post
This is a doubt straight from Peskin, eq 2.43
∅(x,t) = eiHt∅(x)e-iHt.
What is ∅?

Quote Quote by sumeetkd View Post
We dont have the simple Eψ=Hψ structure so this shouldn't directly hold.
What's your problem with E|ψ>=H|ψ> in QFT?
rubbergnome
#3
Feb19-12, 06:22 AM
P: 12
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.

sumeetkd
#4
Feb19-12, 06:55 AM
P: 9
Evolution of Field operators

Quote Quote by tom.stoer View Post
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 [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
rubbergnome
#5
Feb19-12, 07:43 AM
P: 12
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.
tom.stoer
#6
Feb19-12, 08:23 AM
Sci Advisor
P: 5,366
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,t0) ψ(t0); in QFT ψ is the field operator and its time evolution ψ(t) = U(t) ψ0 U*(t); but this is not derived from the Schrödinger equation but from the Heisenberg equation of motion for operators.


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