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
<br /> S_{\beta\,\alpha}=\langle\Psi_{\beta}^-\mid\Psi_{\alpha}^+\rangle<br /> =\langle\Psi_{\beta}^- \mid U^{\dagger}U\mid \Psi_{\alpha}^+\rangle<br />
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?