# Expressing a particle's matter wave in terms of its momentum

I'm following Zettili's QM book and on p. 39 the following manipulation is done,

Given a localized wave function (called a wave packet), it can be expressed as $$\psi(x,t) = \frac{1}{\sqrt{ 2 \pi}} \int_{-\infty}^{\infty} \phi(k) e^{i(kx-\omega t)} dk$$ Now use the de broglie relations: $p = \hbar k$ and $E = \hbar \omega$ and define $\tilde{\phi}(p) = \phi(\frac{k}{\hbar})$.

This should yield $$\psi(x,t) = \frac{1}{\sqrt{ 2 \pi \hbar}}\int_{-\infty}^{\infty} \tilde{\phi}(p) e^{i(px-E t)/ \hbar} dp$$ but I get $$\psi(x,t) = \frac{1}{\sqrt{ 2 \pi} \hbar}\int_{-\infty}^{\infty} \phi\biggl(\frac{p}{h}\biggr) e^{i(px-E t)/ \hbar} dp$$ when I make the change-of-variable $k=\frac{p}{\hbar}$. What am I missing?

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Are you just talking about the factor of $\sqrt{\hbar}$? – Colin K Jan 24 '12 at 22:01
It's correct as written. So no, $\tilde{\phi}$ does not align either. – user19192 Jan 24 '12 at 23:54

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Sorry, this is a really weak question. It's a straightforward and trivial algebra to get the right power of $\hbar$. Moreover, mature physicists - probably including the author of the textbook you mention - use units with $\hbar=1$, so missing powers of $\hbar$ are "not really mistakes". On the other hand, you have real mistakes in your text, e.g. the claim $\tilde \phi(p)=\phi(k/\hbar)$ at the top. You meant $p/\hbar$ on the right hand side, right? When you make this fix, you do see that the "obtained" and "desired" expressions only differ by $\sqrt{\hbar}$, don't you? – Luboš Motl Jan 25 '12 at 6:30

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I'm sorry the author of an introductory textbook doesn't use your convention for mature physicists. And no, the text states $k/\hbar$, not $p/\hbar$. This is not in the errata either. – user19192 Jan 25 '12 at 15:58

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– Maksim Zholudev Jan 25 '12 at 16:56

## 2 Answers

In bra-ket notation your formulas should look as follows $$\left<x\right.\left|\psi\right> = \int_{-\infty}^\infty \left<x\right.\left|k\right> \left<k\right.\left|\psi\right> dk = \int_{-\infty}^\infty \left<x\right.\left|p\right> \left<p\right.\left|\psi\right> dp$$ where
$\left<k\right.\left|\psi\right> = \phi(k)\exp(-i\omega t)$ and $\left<p\right.\left|\psi\right> = \tilde\phi(p)\exp(-i\omega t)$
are the wavefunctions in terms of $k$ and $p$;
$\left<x\right.\left|k\right>$ and $\left<x\right.\left|p\right>$ are the eigenfunctions of the operators $\hat{k}$ and $\hat{p}$ respectively.

These eigenfunctions should be normalized and the usual normalization for continuous quantum numbers ($k$ and $p$) is the one with delta function: $$\left<k'\right.\left|k\right> = \int_{-\infty}^\infty \left<k'\right.\left|x\right> \left<x\right.\left|k\right> dx = \delta(k'-k)$$ The eigenfunctions normalized like this are $$\left<x\right.\left|k\right> = \frac{1}{\sqrt{2\pi}} e^{ikx}$$ and $$\left<x\right.\left|p\right> = \frac{1}{\sqrt{2\pi\hbar}} e^{ipx/\hbar}$$ So here is the lost $\sqrt{\hbar}$, in the normalization coefficient.

## Edit: version without bra-kets

$$\psi(x, t) = \int_{-\infty}^\infty \xi_k(x)\varphi(k,t) dk = \int_{-\infty}^\infty \tilde\xi_p(x)\tilde\varphi(p,t) dp$$ where
$$\varphi(k,t) = \phi(k)\exp(-i\omega t) = \int_{-\infty}^\infty \xi_k^*(x)\psi(x, t) dx$$ and $$\tilde\varphi(p,t) = \tilde\phi(p)\exp(-i\omega t) = \int_{-\infty}^\infty \tilde\xi_p^*(x)\psi(x, t) dx$$
are the wavefunctions in terms of $k$ and $p$;
$\xi_k(x)$ and $\tilde\xi_p(x)$ are the eigenfunctions of the operators $\hat{k}$ and $\hat{p}$ respectively.

These eigenfunctions should be normalized and the usual normalization for continuous quantum numbers ($k$ and $p$) is the one with delta function: $$\int_{-\infty}^\infty \xi_{k'}^*(x)\xi_k(x) dx = \delta(k'-k)$$ The eigenfunctions normalized like this are $$\xi_k(x) = \frac{1}{\sqrt{2\pi}} e^{ikx}$$ and $$\tilde\xi_p(x) = \frac{1}{\sqrt{2\pi\hbar}} e^{ipx/\hbar}$$ So here is the lost $\sqrt{\hbar}$, in the normalization coefficient.

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I haven't studied bra-kets yet so I will have to save this to read later. But you are basically saying to just re-normalize the wave function? That makes notation tricky because then the $\psi(x,t)$ are no longer equal to themselves. – user19192 Jan 25 '12 at 16:06
@user19192, nothing happens to $\psi(x,t)$ here. You have wrong assumption that functions $\tilde\phi(p)$ and $\phi(p/\hbar)$ are equal. They differ by $\sqrt{\hbar}$ because the basis functions are also different. See the 2nd and 3rd formulas in the updated part of my answer. – Maksim Zholudev Jan 25 '12 at 16:45

The culprit seems to be a typo

$$\widetilde{\phi}(p) ~=~ \phi(\frac{k}{\hbar}) \qquad\qquad ({\rm Wrong!})$$

right above eq. (1.98) in the 1st edition of N. Zettili, Quantum Mechanics: Concepts and Applications, which is corrected to

$$\widetilde{\phi}(p) ~=~ \frac{\phi(k)}{\sqrt{\hbar}}$$

in the 2nd edition, see also a comment from Maksim Zholudev.

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