- #1
diraq
- 14
- 0
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:
[itex]S_{\beta\alpha}=(\Psi_\beta^-,\Psi_\alpha^+)[/itex]
When talking about the Lorentz invariance of S-matrix, the Lorentz transformation induced unitary operator [itex]U(\Lambda,a)[/itex] 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 [itex]\Psi_\alpha^\pm[/itex] are the eigenstates of the full Hamiltonian with a non-zero interaction term, how can [itex]\Psi_\alpha^\pm[/itex] 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.
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:
[itex]S_{\beta\alpha}=(\Psi_\beta^-,\Psi_\alpha^+)[/itex]
When talking about the Lorentz invariance of S-matrix, the Lorentz transformation induced unitary operator [itex]U(\Lambda,a)[/itex] 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 [itex]\Psi_\alpha^\pm[/itex] are the eigenstates of the full Hamiltonian with a non-zero interaction term, how can [itex]\Psi_\alpha^\pm[/itex] 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.