# Interaction picture (Is there a mistake in QFT books?)

1. Dec 27, 2008

### jdstokes

It is often stated that the transition amplitude between eigenstates of the free-field Hamiltonian $H_0$ is encoded by the S-matrix, defined by

$\langle \mathrm{f} | U{_\mathrm{I}} (\infty,-\infty)|\mathrm{i} \rangle$.

where $U_{\mathrm{I}}$ is the time-evolution operator in the interaction picture (e.g., [1]).

I would argue that this is not correct, and that the correct expression should be

$S_{\mathrm{fi}}=\langle \mathrm{f} | U_0(\infty,-\infty) U{_\mathrm{I}} (\infty,-\infty)|\mathrm{i} \rangle$

where $U_0(t,t') = \mathrm{e}^{-\mathrm{i} H_0 (t-t')}$ is the time-evolution operator for the free-field.

My argument is quite simple, the transition amplitude in the Schrodinger picture is

$\langle \mathrm{f} | U_{\mathrm{S}} (\infty,-\infty)|\mathrm{i} \rangle$.

It is easy to show, however, that the Schrodinger evolution operator is related to the interaction evolution operator by $U_{\mathrm{S}} = U_0 U_{\mathrm{I}}$. My argument seems to be supported by [2]. However, it concerns me that this is not addressed in mainstream QFT books.

Does anyone have any word on this apparent discrepancy?

[1] Mandl and Shaw, Quantum Field Theory.
[2] http://www.nyu.edu/classes/tuckerman/stat.mechII/lectures/lecture_21/node3.html

2. Dec 27, 2008

### olgranpappy

The amplitudes you have written differ only by an overall phase, which appears to be
$$e^{-iE_f(2\infty)}\;.$$

I think the difference may come from the fact that one should ask for the transition amplitude to the unperturbed final state propagated by the unperturbed hamiltonian from the initial time to the final time. This can be seen to cancel the factor in question, if the times are not taken to be infinite. Maybe a book on scattering theory like Taylor's or Goldberger and Watson would be good to consult in this matter. Cheers.

3. Dec 27, 2008

### jdstokes

If you have a look at Eq. (33) of my second reference, the phase is still there: no cancelation.

4. Dec 27, 2008

### tim_lou

i think the problem is that the f, and i states you refer to (in different instances) are different states.

In the first instance, the f and i are the states in the unperturbed Hamiltonian. This allow calculations using Feynman diagrams.

In the second instance, the f and i states are the one particle state in the complete Hamiltonian. Here I quote Peskin and Schroeder (p. 109):

"the problem is a deep one, and it is associated with one of the most fundamental difficulties of field theory, that interactions affect not only the scattering of distinct particles but also the form of the single particle states themselves"

So, indeed, your expression regarding Schrodinger picture is correct, but the f and i are not the simple one particle states you have in mind. They are related to the unperturbed ones in a nontrivial manner. In fact, the limit of the Us should be -∞(1-iϵ) to +∞(1+iϵ) in the first equation you gave, for some small positive ϵ. This cannot be obtained by the argument you gave above.

Last edited: Dec 27, 2008