Is a Distribution Function a Ratio of Differentials?

[flyusx]

Homework Statement

Show that $\vert\tilde{\phi}\vert^{2}=\frac{1}{\sqrt{\hbar}}\vert\phi\vert^{2}$ where $\tilde{\phi}$ is a momentum-basis and $\phi$ is a k-space-basis wave packet.

Relevant Equations

$p=k\hbar$

I read on a post here titled 'Understanding Fourier Transform for Wavefunction Representation in K Space' that if one represents the squared-amplitude as a ratio of differentials, the solution is given. Letting the squared-amplitude be $\phi$.
$$\frac{d\phi}{dp}=\frac{d\phi}{dk}\frac{dk}{dp}$$
Where $$\frac{dk}{dp}=\frac{1}{\hbar}$$
And therefore
$$\vert\tilde{\phi}\vert^{2}=\frac{1}{\hbar}\vert\phi\vert^{2}$$

Additionally, when I represent the Planck distribution with respect to frequency as $\frac{du}{df}$ and multiply by $\left\vert\frac{df}{d\lambda}\right\vert$, I get the correct expression for the Planck distribution with respect to wavelength. Is this just distribution functions being represented as a ratio of derivatives?

 

[kuruman]

I would say, more simply, that the relation follows because $p=\hbar k$ and $\int \vert\tilde{\phi}\vert^{2}dp$ is a dimensionless probability and so is $\int \vert\phi\vert^{2} dk$.

 

[hutchphd]

For OP: In statement of problem, why is it $\sqrt(\hbar)$ and not just $\hbar$????

 

[flyusx]

My apologies. Yes, it is $\hbar$ that shouldn't be squared.

 

[hutchphd]

shouldn't be square-rooted.