# Normalization parameter, properties of Dirac delta functions

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.

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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}

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Thank you, joriki. So would $N(E)={1\over \sqrt{2\sqrt{E}}}$? – ma mere l'Oye Aug 2 '12 at 15:12
@mamerel'Oye: Not quite; I've edited the post to calculate the factor. Note that you can check such things by checking dimensions. The normalization factor needs to have dimensions of reciprocal energy (for the delta) times reciprocal length (for the integral). – joriki Aug 6 '12 at 8:22