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


1 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.