Re: Finite time calculation in QED



Igor Khavkine wrote: 
On 2005-04-15, Eugene Stefanovich <eugenev@xxxxxxxxxxxx> wrote:

Igor Khavkine wrote: 


The Gell-Mann-Low theorem says that if interaction is adiabatically
turned on, and |Phi> is eigenvector of H_0, then

|Psi> = exp[iH(0 - -oo)] |Phi>

is eigenvector of H with the same
eigenvalue (let us assume for simplicity that the spectra of H_0 and
H are the same). This works for vacuum and one-particle states.
So, it seems that you achieved the desired dressing. Not so fast.
It appears that two-, three-, etc. particle states transformed by
exp[iH(0 - -oo)] also become eigenstates of H. E.g.,

exp[iH(0 - -oo)] a^*(p) a^*(q) |0>

is an eigenstate of H.
So,  you found a representation that diagonalizes H.
This means that there is no scattering anymore.
The scattering operator is identically unity S=1.



You misunderstand the theorem. First, you cannot use exp(iHt) as the
time evolution operator since H is time dependent (it contains the
adiabatic switching function). Second, |Phi> and |Psi> will not have the
same energy, they'll be different because of the interaction. Third, you
can't make any conclusions about what kind of state |Psi> is just from
knowing what kind of state |Phi> is. The bare ground state need not
evolve into the true ground state (superconductivity or electroweak
theory). An n-particle state need not evolve into an n-particle state
(QED, only charge is conserved, not particle number in fact, the
particle number will be indeterminate). You have no way of controlling
localization properties of particle states at finite time.

Some states can still be identified by their quantum numbers. For
instance, in my proposed model, if at t=-oo a particle of charge +1 is
place in each well, then at t=0 we'll have an eigenstate of the
interacting Hamiltonian with charge +1 in each well, but we can't say
much more. However, this is sufficient for the purposes of the proposed
experiment.


I basically agree with everything you are saying. However, you haven't
addressed my point. I said that if you form the state of two physical
particles at t=0 by formula (adiabatic switching of interaction is
assumed)

exp[iH(0 - -oo)] a^*(p) a^*(q) |0>

then according to the Gell-Mann-Low theorem this state is an eigenstate
of the full Hamiltonian H. Then the time evolution of this state is
just a multiplication by a trivial phase factor, which is clearly wrong.

Can different observers still agree on what's going to
happen to the ball? How?


OK, we agreed (I hope) to consider the "spaceship + ball" as
our physical system. [...]



Leave preparation to the preparers and observation to the observers.
There is only one spaceship, but some observers see it move and others
don't. They can all still compare notes and agree on what trajectory the
ball took if the relative motion of the ship is taken into account as
well as how they measured time intervals and distances. In short, for
simplicity and without loss of generality, we can choose to do
calculations only in a frame where the ship is stationary.

Do you need any more justification for the use of an external potential?
If not, my offer is still open and you have the following steps to do:

1. Find the stationary states of the free classical system.
2. Promote the mode coefficients (and their conjugate momenta)
   to operators and obtain a diagonal expression for the Hamiltonian.
3. Write down the equations of motion for states and operators in the
   interaction picture. Write down the formal solution. (This does not
   require an explicit expression for the Hamiltonian).

More later, but at least some of the above has to be done first. 

I am sorry, I am starting to lose the thread of this discussion.
I don't see what's the connection between balls, spaceships, external
potentials, etc. and the simple question I posed: describe the time
evolution of a single particle in QED.

Let me try to pose the question in another way. Do you agree or not
agree that the Hamiltonian (8.4.1) derived in Weinberg's section 8.4
(plus counterterms that can be obtained from formula (11.1.9)) is the
correct Hamiltonian of QED?
Your choices are:
1) You agree, and we start discussion right away based on this
   Hamiltonian
2) You disagree, then please explain which of the following you think is
   true:
   a) The above Hamiltonian should be modified (please explain how)
   b) The above Hamiltonian is useless for time evolution calculations,
      such calculations should be done not by formula exp(iHt), but in
      entirely different way.

My feeling is that your choice is 2)b). Am I right?

Eugene Stefanovich.





.

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