The de broglie hypothesis suggests a particle can be associated with a wave of
momentum $p = \hbar k$
my question is the following: how does one arrive at this concept of the momentum of a wave?
I understand the classical definition of momentum, however I do not see an obvious extension to an object such as a vibrating guitar string.
De Broglie is saying that if I have a vibrating guitar string and the wavelenth of vibration decreases then the momentum of the guitar string increases? Is it the case that a guitar string at a higher frequency(time) has higher energy?
is the case that a guitar string vibrating at a higher spatial frequency will also have higher energy? etc.
I'm sorry if this guitar string example is bad… i'm just trying to understand this properly
Best Answer
In the de Broglie relation $p=\hbar k$, the left-hand side represents the momentum of a particle, while the $k$ on the right is the wavenumber of a wave. The key here is that if the relation is going to be valid, the wave has to have the right amplitude to represent exactly one particle.
For example, suppose we have a light wave that represents exactly one photon's worth of light. It has a certain energy and frequency, related by $E=\hbar \omega$, and a certain momentum and wavenumber, related by $p=\hbar k$.
Now suppose we take that light wave and scale up its amplitude (i.e., the strength of its electric and magnetic fields) by a factor of 2. Since the energy of a wave is proportional to the square of its amplitude, this quadruples the energy. It also quadruples the momentum. (You can see this either by using the relativistic relation $E=pc$ for m=0, or by using the Poynting vector.) But we haven't changed $\omega$ or $k$, so this wave violates the de Broglie relations. The key here is that this wave is no longer the right strength to represent one photon. It represents four photons.
Let's play with some numbers in the case of the guitar string. Say $\omega$ is 1000 Hz, and $E$ is 1 J. When you calculate $\hbar\omega$, you get something many orders of magnitude smaller than this $E$. That's because there isn't just one particle in the guitar string, there are many. The same idea applies to momentum.