How Can the S-Matrix Transform Like Free Particle States in Weinberg's QFT?

[diraq]

Hi All,

The S-matrix is defined as the inner product of the in- and out-states, as in Eq. (3.2.1) in Weinberg's QFT vol 1:
S_{\beta\alpha}=(\Psi_\beta^-,\Psi_\alpha^+)

When talking about the Lorentz invariance of S-matrix, the Lorentz transformation induced unitary operator U(\Lambda,a) is applied both on the in- and out-states, and the transformation rule is the same as that for free particle states, i.e., Eq. (3.1.1).

However, since \Psi_\alpha^\pm are the eigenstates of the full Hamiltonian with a non-zero interaction term, how can \Psi_\alpha^\pm be transformed according to the same rule for the states composed of free particles? In the paragraph under Eq. (3.1.5), Weinberg explicitly indicates that the rule of Eq. (3.1.1) can only be applied to non-interacting particle states.

I appreciate any help from you to eliminate my miss/non-understanding. Thanks.

 

[bobloblaw]

I think actually that 3.1.1 doesn't apply to the in and out states although the notation seems to indicate this. If you read the paragraph under 3.1.7 Weinberg says:

"On the other hand, the transformation rule (3.1.1) does apply in scattering processes at t\rightarrow \pm \infty."​

In other words in general it does not apply to In and Out states. In and Out states appear non-interacting only to observers at t=\pm\infty

 

[diraq]

Thank you very much.

I am still confused. In- and out-states are non-interacting but according to Weinberg they are still the eigenstates of the full Hamiltonian.

It looks like that Weinberg took a different perspective. The asymptotic hypothesis that will turn off the interaction at t\rightarrow\pm\infty is not included here. The Hamiltonian is is time dependent in asymptotic hypothesis, but Weinberg treats it as time-independent--only the state vector evolves with time.

I think actually that 3.1.1 doesn't apply to the in and out states although the notation seems to indicate this. If you read the paragraph under 3.1.7 Weinberg says:

"On the other hand, the transformation rule (3.1.1) does apply in scattering processes at t\rightarrow \pm \infty."​

In other words in general it does not apply to In and Out states. In and Out states appear non-interacting only to observers at t=\pm\infty

 

[bobloblaw]

I am still confused. In- and out-states are non-interacting but according to Weinberg they are still the eigenstates of the full Hamiltonian

In and Out states only appear non-interecting to observers sitting at temporal infinity. To all other observers they appear to be "interacting" in the sense that they no longer have definite particle content. That is how they are able to be eigenstates of the full Hamiltonian. Equation (3.1.12) makes this more precise. The point is that we can find states that satisfy (3.1.12) despite the Hamiltonian being time independent, ie we can find states that look like they are non-interacting to observers at infinity.

If you look at eq. (3.1.7) you can see that at any finite t the In and Out states have components along every multiparticle state but at t\rightarrow\pm\infty \Psi^\mp \rightarrow\Phi. Also as V\rightarrow 0 we also get \Psi^\mp \rightarrow\Phi. So in this sense t\rightarrow\pm\infty is equivalent to V\rightarrow 0.

This method of doing things appears also in scattering in non-relativistic QM. For example it is discussed in Sakurai's book in Chapter 7: Scattering theory.

 

[diraq]

Thank you very much. I think I have got the idea from your help.

 

[bobloblaw]

sweet! glad I could help.