Suppose $\psi_E (x)=N(E)\exp (ikx)$
where $\psi_E (x)$ is a momentum eigenfunction, $N(E)$ is the normalization constant on the energy scale such that $\langle E'|E\rangle=\int_{-\infty}^\infty \psi_{E'}^*(x)\psi_E(x) dx=\delta (E-E')$, $k$ is the wave number corresponding to energy $E$ so that $k={\sqrt{2mE}\over h}$.
I wish to know how one can find $N(E)$ explicitly.
$\delta(\sqrt{E}-\sqrt{E'})=2\sqrt{E} \delta(E-E')$ right? This is obtained using the property $\delta (f(x))=\sum_i{\delta(x-x_i)\over |f'(x_i)|}$ where $x_i$ is a zero of $f(x)$. Here I have taken $x$ to be $E$.
But then we could equally have taken $E'$ instead, giving $\delta(\sqrt{E}-\sqrt{E'})=2\sqrt{E'} \delta(E-E')$. Is there a way to resolve the breaking of symmetry in the expression? -- since I need $N^*(E')N(E)={1\over2\sqrt{E}}={1\over2\sqrt{E'}}$.
Thank you.