# Why don't the De Broglie dispersion relation contain a constant term?

Wikipedia says that the dispersion relation for a non-relativistic particle is:

$$\omega = \frac{\hbar k^2}{2m}.$$

But when I tried to calculate it myself, I seem to get a constant term in that formula. My derivation is the following:

Reordering the De Broglie relations I have a generic dispersion relation:

$$\omega = \frac{E k}{p}$$

Substituting the non-relativistic energy limit:

$$\omega = \frac{\left( m c^2 + \frac{p^2}{2m} \right)k}{p}$$

Substituting the momentum:

$$\omega = \frac{\left( m c^2 + \frac{\hbar^2 k^2}{2m} \right)}{\hbar }$$

Performing the division, I get:

$$\omega = \frac{m c^2}{\hbar} + \frac{\hbar k^2}{2m}$$

Maybe I miss something obvious. The relation in the Wikipedia doesn't contain that constant term why? Maybe in the non-relativistic case the mass energy is not considered as energy at all? That would be interesting...

• $mc^2$ is the energy enclosed in the mass of the particle, so it's like "frozen" in it. So you can just redefine energy by removing this value and take energy levels by referring to the constant term as the 0-energy level – Phoenix87 Jan 6 '15 at 16:12
• – Qmechanic Jan 6 '15 at 16:24
• @Phoenix87 Why am I allowed to bias energy levels that way? If I bias frequencies with some arbitrary constant, why does the model remains correct? – Calmarius Jan 6 '15 at 16:43

$$E = \frac{p^2}{2m}$$
$$\omega = \frac{\hbar^2 k^3}{2m} \times \frac{1}{\hbar k} = \frac{\hbar k^2}{2m}$$