Questions concerning some parts of the section on one-particle states in Weinberg's first volume on QFT

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 $N(k')=N(k)=1$.