Questions concerning some parts of the section on one-particle states in Weinberg's first volume on QFT Below are the scan copies of some pages of Weinberg which are relevant to my doubts. My doubts basically concern the determination of normalization constant defined in (2.5.5). 


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*Isn't (2.5.12) true directly from the fact that $\Psi_{p,\sigma}$ are the eigenstates of a complete set of commuting hermitian operators? Or is it that because of the definition (2.5.5) the states $\Psi_{p,\sigma}$ are no longer orthonormal which doesn't seem to be the case to me as U's are unitary. Moreover, doesn't (2.5.5) imply that N(p) is just a phase factor by the unitarity of U?

*I can't see why (2.5.13) is an immediate consequence of (2.5.12) but isn't that just a restatement of the fact that U's are unitary? If these are all the usual things already clear, then why has Weinberg got to write it this way? And how do you conclude that (2.5.13) is an immediate consequence of (2.5.12)? Kindly clarify even if it is dumb.

*Then, he goes on to consider scalar product of states having arbitrary momenta p and p'. He defines $k'=L^{-1}(p)p'$. Now, if p and p' belong to the same class and are not equal, then we would have two standard momenta for the same class which shouldn't be the case. Also, as I vary p' (now p' need not be in same class), k' varies. So, there is no fixed standard momentum for the class in which p' belongs. Please clarify what's going on. You might say that k' has got nothing to do with standard momentum, but I have strong reasons to believe so as you get the second equality in the equation between (2.5.13) and (2.5.14) by setting N(k') (where k' is as defined above in this paragraph) as 1 which should be the case only if k' is the standard momentum for the class.





 A: An answer from a (slow) student:
1. Eqn (2.5.12) defines a choice of basis, where $k$ is the standard standard momentum and $k'$ is an arbitrary momentum state.  It's possible to choose such a basis but it's not automatic.
2. Again, the operator $U$ being unitary (length-preserving) does not guarantee that its rep's matrices will be unitary ($D^{-1}=D^\dagger $) for all choices of basis.  
As an elementary example, consider a 90 degree rotation $R$ in the x-y plane, which is clearly orthogonal.  With the usual orthonormal basis $\boldsymbol{x_1=\hat{x}, \, x_2=\hat{y}}$,
$$ R_1 = \left(\begin{array}{cc}  0 & -1 \\ 1 & 0  \end{array} \right) $$
and this matrix representation of $R$ is orthogonal:
$$ R_1^{-1} = R^T $$
However, one could instead have picked basis vectors $\boldsymbol{x_1=\hat{x}, \, x_2=(1/\sqrt{2})(\hat{x}+\hat{y})}$ which are independent and normalized but not orthogonal.  With respect to this basis (a similarity transformation from the first basis), the operator matrix is:
$$ R_2 = \left(\begin{array}{cc} -1 & -\sqrt{2} \\ \sqrt{2} & 1 \end{array} \right) $$
The operation is still orthogonal (it's the same 90 degree rotation), but $R_2$ is not an orthogonal matrix:
$$ R_2^{-1} =  \left( \begin{array}{cc} 1 & \sqrt{2} \\ -\sqrt{2} & -1 \end{array} \right) 
 \neq R_2^{T}$$
The upshot is that one needs to draw a distinction between an operator being unitary and its matrix rep being unitary (in the matrix sense of the term).
So, U being unitary is necessary, but in addition Professor Weinberg's orthonormal basis choice of (2.5.12) is needed for the resulting matrix rep $D$ to be unitary.  
3. Professor W's strategy here is to get the scalar product of two arbitrary momenta into the form of (2.5.12).  Hence he applies $L^{-1}(p)$ to both states, which 1) takes $p$ to the standard momentum $k$ and 2) maintains the "angle" between the two states, with $p'$ being transformed to $k'$.  Unless $p'=p, \, k'$ is not the standard momentum.  (There's only one such standard momentum for each value of $p^2$.)
You are correct that there is an $N(k')$ missing from the second equality between (2.5.13) and (2.5.14), but note the delta function $\delta^3(\boldsymbol{k'-k})$ in that expression:  by the properties of the delta function we can set $N(k')=N(k)=1$.  
A: ''Choosing'' the momentum states orthonormal works because momentum states are joint eigenstates of the (commuting, Hermitian) components of momentum, not because one can make arbitrary choices. 
For example, one cannot choose to have coherent states (eigenvectors of an annihilation operator) to be orthonormal. 
On the remaining queries:


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*The momentum delta function in (2.5.12) comes from having joint eigenstates of a set of commuting observables (momentum components). The label delta means that we choose in each eigenspace of fixed momentum an orthonormal basis. This can be done in may ways, hence involves an arbitrary choice. 
The factor $N(p)$ in (2.5.5) is arbitrary (not necessarily a phase), as it defines the left hand side. Note that all these states are in the continuum,
hence cannot be normalized. Thus there is an arbitrary scale factor. One could choose them all equal to 1 but this would produce a weird measure. By choosing $N(p)$ intelligently (which Weinberg promises in (2.5.5) but apparently does only on a later page that you didn't scan), one can make the measure look nice.

*The ''immediate'' consequence is explained in the footnote **. It is immediate only given enough practivce with handling representations, and since Weinberg had appaarently doubts whether it would be immediate for anyone, he added the footnote. That an arbitrary unitary irrep on states written in the form (2.5.5) should satisfy (2.5.13) is something that cannot be assumed without an argument. With a different choice in (2.5.12) this would not be the case. Weinberg makes the nice choice in order to get a nice condition on the $D$'s. 

*The standard momentum is a fixed vector independent of the momenta; just one representative of the possible orbits (classes in Weinbergs terminology) of the Lorentz group on momenta. It is one of the six cases in the table. The $k'$ introduced after (2.5.13) are usually nonstandard vectors. $L(p)$ transforms 
the standard vector $k$ into $p$ but some other, nonstandard vector $k'$ transforms under $L(p)$ into $p'$. (The standard vector is transformed into $p'$ by $L(p')$, not by $L(p)$!)
