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

Question

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 $\mathrm{d}p = \hbar \ \mathrm{d}k$?

Answer 1

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.