## Physics Friday 44

Quantum Mechanics and Momentum
Part 1: The De Broglie Relations

Planck’s relation tells us that the energy of a photon is proportional to the frequency: E=, where h is Planck’s constant. In terms of angular frequency ω, we use ω=2πν to find that , where  is called the reduced Planck constant, or the Dirac constant.
Classical electrodynamics of an electromagnetic wave and the relativistic energy-momentum relation for a particle with zero rest mass both give the same result for the momentum of a photon: that , and so the momentum is
, where λ is the wavelength. Solving for wavelength, we have . This wavelength can be found for any particle, not just photons, and is called the de Broglie wavelength (see here for details on the historical context and experimental support for this).

Now, let us consider the corresponding angular wavenumber . Then we have

or .

The two equations,  and , are known as the de Broglie relations, and the latter one will be important in the later parts of this series.

### 3 Responses to “Physics Friday 44”

1. Physics Friday 46 « Twisted One 151’s Weblog Says:

[…] is also normalized. The reason we want to use such a Fourier transformed function goes back to our de Broglie relations, particularly . Using this, we see that our “wavenumber domain,” via this equation, […]

2. Physics Friday 48 « Twisted One 151’s Weblog Says:

[…] constant of integration) and is the wavenumber corresponding to the wavefunction momentum p0 via our de Broglie relation . Note that for all x, so fixed momentum requires equal probability for all positions (when […]

3. Physics Friday 107 « Twisted One 151's Weblog Says:

[…] and scattering is in the xy plane, with θ in the usual direction in that plane. As discussed here, the energy and momentum of a photon are given by , and , giving us an energy-momentum 4-vector for […]