Quantum momentum (De Broglie)

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

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.

• I need a clarification regarding this, but my comment is way too long , [thus I moved it to chat][1] [1]: chat.stackexchange.com/rooms/26241/… Jul 26 '15 at 15:24

The assumption that Plank made from his Planck's law of blackbody radiation was that the electromagnetic radiation is quantized, it takes only integer multiples of the quantity $hf$, where $h$ is the Plank constant (with units of energy $\times$ time) and $f$ is the frequency of radiation (with units 1/time). This is the energy of the quanta of the electromagnetic radiation - a photon.

So, firstly the Plank constant $h$ was defined by Plank as a proportionality constant between the energy $E$ of a photon and the frequency of its associated electromagnetic wave.

$E=hf$

This can be rewritten in terms of the wavelength, instead of the frequency, using the relation $f=c/\lambda$. We get:

$E=\frac{hc}{\lambda}$ (1)

Also, from special relativity energy-momentum relation $E^2=m^2 c^4 + p^2 c^2$ we know that for a photon with zero rest mass, $m=0$, its energy is related to its momentum through the relation:

$E=pc$ (2)

Combining the equations (1) and (2) we get:

$p=\frac{h}{\lambda}=\hbar k$, where $\hbar=\frac{h}{2\pi}$ and $k=\frac{2\pi}{\lambda}$.

De Broglie made the assumption that the particles have wave properties too, i.e an frequency and wave length and this has been experimentally proved. They are particles and waves at the same time. The wave properties of particles are given by the de Broglie equations:

$f=\frac{E}{h}$ and $\lambda=\frac{h}{p}$

So, for a given particle's energy $E$ and momentum $p$ we can find its frequency and wave length.

• I don't think the OP was asking for a derivation. He was asking how (or whether) it applies to macroscopic objects.
– user4552
Jul 21 '11 at 18:20