# Inner product of standard-momentum one-particle states in Weinberg

My question has essentially already been addressed in Questions concerning some parts of the section on one-particle states in Weinberg's first volume on QFT (third question), but unfortunately not been answered satisfactorily (at least for me).

Weinberg writes that we can choose states with standard momentum ($k = (M, 0, 0, 0)$ for massive, $k = (1, 0, 0, 1)$ for massless particles) to be orthonormal in the sense that

$$(\Psi_{k',\sigma'}, \Psi_{k,\sigma}) = \delta^3(\vec k' - \vec k) \delta_{\sigma'\sigma}$$

What I don't understand: If we allow the standard momenta to be from different "classes" (value of $k^2$), then the inner product of two states with different mass would be non-zero by this formula. So I conclude that the whole discussion takes place only within one class, which means that there is only one standard momentum. But why appear then two different standard momenta $k$ and $k'$ in the equation?

Both states $\Psi_{k,\sigma}$ and $\Psi_{k',\sigma'}$ are meant to be states of the same particle species i.e. they have the same values of the squared mass $k^2$. The inner product of one-particle states from different species $s$ is zero which one might indicate by additional $s,s'$ labels and a Kronecker symbol $\delta_{s,s'}$.
Weinberg claims about the orthogonality for the standard momenta is only meant to explain why it's safe to include the factor $\delta_{\sigma' \sigma}$ for $k=k'$. For $k\neq k'$, this discussion is unnecessary or vacuous and no problem can arise because the inner product is zero because of the $\delta^{(3)}(k-k')$ factor, anyway.
His discussion says how to treat the polarizations $\sigma'$ in the case when $k=k'$ and moreover both $k=k'$ are the "standard momenta". The standard momenta are just a "without a loss of generality" trick. Once one sees that this is how it's possible to choose the basis for the standard momenta, one can choose the polarizations in the same way for any allowed momenta.