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.
Best Answer
The delta distribution has the property $f(x)\delta(x)=f(0)\delta(x)$. Thus it doesn't matter whether you use $E$ or $E'$, since in either case you have $2\sqrt E\delta(E-E')=2\sqrt{E'}\delta(E-E')$.
[Edit in response to the comment:]
You seem to be missing a number of factors here.
First, since $E=p^2/2m$ and $p=\hbar k$, it should be $k=\sqrt{2mE}/\hbar$, with $\hbar$ where you had $h$.
Then you dropped all the factors in converting the delta functions. The relationship you need is
$$\delta\left(\frac{\sqrt{2mE}}\hbar-\frac{\sqrt{2mE'}}\hbar\right)=\sqrt{\frac{2E}m}\hbar\delta(E-E')\;.$$
Finally, you're missing a factor of $2\pi$ from the normalization of the momentum eigenfunctions themselves,
$$ \int_{-\infty}^\infty\mathrm e^{\mathrm i(k-k')x/\hbar}\,\mathrm dx=2\pi\delta(k-k')\;. $$
Putting this all together, we have
$$ \begin{align} \langle E'|E\rangle &= \int_{-\infty}^\infty \psi_{E'}^*(x)\psi_E(x)\,\mathrm dx \\ &= N(E')^*N(E)\int_{-\infty}^\infty \mathrm e^{\mathrm i(k-k')x}\,\mathrm dx \\ &= N(E')^*N(E)2\pi\delta(k-k') \\ &= |N(E)|^22\pi\sqrt{\frac{2E}m}\hbar\,\delta(E-E')\;. \end{align} $$
If you want this to be $\delta(E-E')$, you need
$$ \begin{align} |N(E)| &= \left(\frac m{2E}\right)^{1/4}\frac1{\sqrt{2\pi\hbar}} \\ &= \left(\frac m{2Eh^2}\right)^{1/4}\;. \end{align}$$