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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?

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?

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 defining $\tilde{\phi}(p) = \phi(\frac{k}{\hbar})$.

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

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?