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

Question

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?

Answer 1

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.

Answer 2

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.