Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

up vote 1 down vote accepted

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

$$ \begin{align} |N(E)| &= \left(\frac m{2E}\right)^{1/4}\frac1{\sqrt{2\pi\hbar}} \\ &= \left(\frac m{2Eh^2}\right)^{1/4}\;. \end{align}$$

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.