# What is the form of the wave packet in terms of momentum?

The wave packet in terms of the wave number $$k$$ is:

$$\begin{equation} \Psi(x, t) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} \mathrm{d}k \ A(k) \ e^{-i(kx-\omega t)} \tag{1} \end{equation}$$

Knowing that $$p = \hbar k$$ and $$E = \hbar \omega$$ we can replace $$k$$ with $$p$$, and Eq. (1) becomes:

$$\begin{equation} \Psi(x, t) = \frac{1}{\hbar \sqrt{2\pi}} \int_{-\infty}^{+\infty} \mathrm{d}p \ A\left(\frac{p}{\hbar}\right) \ e^{-i(px- Et)/\hbar} = \frac{1}{\hbar \sqrt{2\pi}} \int_{-\infty}^{+\infty} \mathrm{d}p \ \phi(p) \ e^{-i(px- Et)/\hbar} \tag{2} \end{equation}$$

However, this appears to be wrong, and the equation is found in the literature as:

$$\begin{equation} \Psi(x, t) = \frac{1}{\sqrt{2 \pi \hbar}} \int_{-\infty}^{+\infty} \mathrm{d}p \ \phi(p) \ e^{-i(px- Et)/\hbar} \tag{3} \end{equation}$$

with the $$\hbar$$ under the square root. How does this happen? Shouldn't $$\mathrm{d}p = \hbar \ \mathrm{d}k$$?

In your equation 2 you implicitly make the following definition of $$\phi(p)$$ $$\begin{equation} A\left(\frac{p}{\hbar}\right) = \phi(p) \end{equation}$$ Let's think a bit more about how $$\phi(p)$$ should be defined.
We want $$\phi(p)$$ to be a properly normalized momentum space wavefunction, meaning that $$|\phi(p)|^2 dp$$ should be a dimensionless number, corresponding to the probability of finding the particle's momentum in an interval from $$p$$ to $$p+dp$$. Therefore, $$\phi(p)$$ should have dimensions of $$p^{-1/2}$$.
Now look at $$A(k)$$. From the same argument, we know that $$A$$ has dimensions of $$k^{-1/2}$$. However, then your implicit definition equating $$A$$ and $$\phi$$ above cannot be correct by dimensional analysis, because it equates two quantities with different dimensions.
Therefore, in order to relate $$A$$ and $$\phi$$, we need a factor of $$\sqrt{\hbar}$$, purely for dimensional reasons, leading to the correct transformation $$\begin{equation} A\left(\frac{p}{\hbar}\right) = \sqrt{\hbar} \ \phi(p) \end{equation}$$ Carrying this through leads to the usual expression.
To state this somewhat differently, if you impose standard normalization conditions on $$A$$ and $$\phi$$ as wavefunctions in $$k$$ and $$p$$ space, respectively: $$\begin{eqnarray} \int_{-\infty}^\infty dk |A(k)|^2 &=& 1 \\ \int_{-\infty}^\infty dp |\phi(p)|^2 &=& 1 \end{eqnarray}$$ you will find that $$A$$ and $$\phi$$ are related by a factor of $$\sqrt{\hbar}$$. By defining $$\phi=A$$, without the $$\sqrt{\hbar}$$ factor, you implicitly fixed an unconventional normalization of $$\phi(p)$$, which then explains why your final expression has a different overall normalization than the standard one.