de Broglie Relation
In classical mechanics, the momentum of a particle \(p\) can be found by finding the product of a particle’s mass \(m\) and the particle’s velocity \(v\):
\[p = m \cdot v\]The quantum definition of particle momentum takes Planck’s constant \(h\) and divides it by the wavelength \(\lambda\):
\[p = \frac{h}{\lambda}\]Let’s take a look at how we find this definition…
Consider this sine wave:
Can we express the speed of this light in terms of frequency?
We know that speed is the ratio of distance travelled over time. Thus if we let wavelength \(\lambda\) be distance travelled and time at \(2\pi \ t_{2\pi}\) be the time interval of interest, we get
\[c = \frac{\lambda}{t_{2\pi}}\]If frequency \(\nu\) is the inverse of time \(t\)
\[\nu=\frac{1}{t_{2\pi}} \Rightarrow \frac{1}{\nu}=t_{2\pi}\]Then we can write the speed of light \(c\) in terms of wavelength \(\lambda\) and frequency \(\nu\)
\[c=\frac{\lambda}{1/\nu} \Rightarrow c=\lambda \cdot \nu\]
Recall Functions for Kinetic Energy \(E\)
\[c=\lambda \cdot \nu \Rightarrow \nu = \frac{c}{\lambda}\]\(\begin{align} E & = \frac{1}{2}mv^{2} \\ E & = mc^{2} \\ E & = h\nu \end{align}\) From the above equation for speed of light, we can find a formula for frequency \(\nu\)
We can now use this in our eqations for energy
\[E=h \left( \frac{c}{\lambda} \right) = mc^{2} \Rightarrow mc = \frac{h}{\lambda} = p\]Thus the de Broglie Relation is writted as
\[p=\frac{h}{\lambda}\]This relation only becomes meaningful for very small objects (such as particles) as it breaks classical mechanics. Wavelengths of tangible objects are so small that they aren’t relevant to consider when determining momentum.
Page Author: Reagan McNeill Womack