# Weinberg QFT I: Lorentz Transformation with interaction

#### SeySchW

Hi,

a few lines below equation (3.1.5) Weinberg writes:
"The transformation rule (3.1.1) is only possible for particles that for one reason or another are not interacting."

I thought a lot about it, but dont see any possible reason. Can you help please?

After a few lines he defines the states $$\Phi_{\alpha}$$
then these one should transform like (3.1.1). But why?

Thanks a lot

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#### Bill_K

I believe he gives a reason for this in the very next sentence: "Eq (3.1.1) requires among other things that the state has an energy equal to the sum of the one-particle energies with no interaction terms that would involve more than one particle at a time."

#### SeySchW

"Eq (3.1.1) requires among other things that the state has an energy equal to the sum of the one-particle energies ... and with no interaction terms, terms that would involve more than one particle at a time."
Thank you for the answer, I'm not quite sure, if I understand this argument. So I will rephrase it. In a theory with no interactions we expect that the energy of a multiparticle state is the sum of the single energies. Whereas in an interacting theory this should be different. But I dont understand why this should be the case?

Thanks

#### meopemuk

In a theory with no interactions we expect that the energy of a multiparticle state is the sum of the single energies. Whereas in an interacting theory this should be different. But I dont understand why this should be the case?
But this is the *definition* of interaction! In an interacting theory the Hamiltonian has the form (3.1.8), where H_0 is the sum of single particle energies and V is the interaction potential energy.

Eugene.

#### SeySchW

Ok this makes sense. But then $\Psi_{\alpha}^{\pm}$ is not a direct product state of one particle states otherwise it would transform like (3.1.1). How can we then define something like the index $\alpha$ for $\Psi_{\alpha}^{\pm}$?

Thanks

#### meopemuk

SeySchW,

In interacting theory the total boost operator (3.3.20) is different from the non-interacting boost operator K_0. Therefore, boost transformation laws of n-particle states are different from (3.1.1) in the interacting case. For example, a 1-particle state may transform into a n-particle state under interacting boost.

Eugene.

#### weejee

Ok this makes sense. But then $\Psi_{\alpha}^{\pm}$ is not a direct product state of one particle states otherwise it would transform like (3.1.1). How can we then define something like the index $\alpha$ for $\Psi_{\alpha}^{\pm}$?