Is the S-matrix in Weinberg's book always unitary?

[eoghan]

Hi,
I've read a lot of posts about how Weinberg describes the S-matrix invariance in his book, but none of theme answered my questions.
At page 116, sec 3.3 - "Lorentz Invariance" of Quantum theory of fields vol.1 Weinberg says:
"Since the operator $U(\Lambda, a)$ is unitary we may write
$$S_{\beta\,\alpha}=\langle\Psi_{\beta}^-\mid\Psi_{\alpha}^+\rangle =\langle\Psi_{\beta}^- \mid U^{\dagger}U\mid \Psi_{\alpha}^+\rangle$$
From this equation he gets some conditions that the S-matrix has to fulfill.
But if the operator $U(\Lambda, a)$ is unitary, then shouldn't be
$U^{\dagger}U=1$?
And so the equation above is always satisfied no matter the form of the S matrix!

 

[Bill_K]

Yes, this equation is an obvious identity. Lorentz invariance is the next assertion - that when you apply U to Ψ⁺ and Ψ^- they transform as representations of the inhomogeneous Lorentz group (Eq. 3.1.1), leading to Eq 3.3.1.

 

[eoghan]

Uhm... ok, but now another question arises...
$U^{\dagger}U=1$, but $\Psi_{\alpha}$ and $\Psi_{\beta}$
are two different states, so they can transform with two different irreducible representations.
In that case $U^{\dagger}$ and $U$ are two matrices of different kind, so how can i say that $U^{\dagger}U=1$?