# Confusion (5) from Weinberg's QFT.(invariance of S-matrix)

1. Aug 10, 2011

### kof9595995

At the beginning of section 3.3, he says lorentz invariance of S-matrix means the same unitary operator acts on both in and out states. I feel a bit blur about this since he doesn't give any concrete example. Say the eletron-positron annihilation process, we have 1 electron, 1 positron, 0 photon as in state, and 0 electron, 0 positron, 2 photons as out state, we know lorentz group has different representations on electrons and photons, then how shall we have the same unitary operator of both in and out states in such process?

2. Aug 10, 2011

### meopemuk

I prefer a different definition of the Lorentz invariance of scattering. This definition is purely algebraical, does not involve states, and is applied directly to the S-operator:

$$U_0(\Lambda, a) S U_0^{-1}(\Lambda, a) = S$$

where $U_0$ is the non-interacting representation of the Poincare group.

Eugene.

3. Aug 11, 2011

### dextercioby

The in/out states are multi-particle states, so the unitary operator acting on them is actually an operator acting on a product of different (sub)spaces. For example, the unitary operator corresponding to a photon acts trivially (no change) on a subspace of an electron representation.

4. Aug 11, 2011

### IRobot

I don't have the book with me, but I think several times Weinberg gives explicit examples of $U(\lambda,a)$ acting on states with $\sum_{\sigma'}D_{\sigma,\sigma'} \psi_\sigma$ where the D matrix correspond to the specific representation of the Lorentz Group.

5. Aug 15, 2011

### kof9595995

Thank you all, but somehow I still get a strange feeling somewhere which I can't express clearly. Perhaps I'll come back to this later.