4
$\begingroup$

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...

$\endgroup$
  • 2
    $\begingroup$ $mc^2$ 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 $\endgroup$ – Phoenix87 Jan 6 '15 at 16:12
  • $\begingroup$ Related: physics.stackexchange.com/q/34214/2451 $\endgroup$ – Qmechanic Jan 6 '15 at 16:24
  • $\begingroup$ @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? $\endgroup$ – Calmarius Jan 6 '15 at 16:43
2
$\begingroup$

I believe this is simpler than you make it to be. If you want to substitute in the non-relativistic energy relation, then the correct energy term is just the kinetic energy:

$$ E = \frac{p^2}{2m}$$

Everything else follows from there:

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

$\endgroup$
  • $\begingroup$ Then the question is: why only the kinetic energy? It's the smaller portion of the total energy of the particle. $\endgroup$ – Calmarius Jan 6 '15 at 17:04
  • $\begingroup$ @Calmarius, Phoenix87 above makes a good point. A constant energy term will affect nothing since only energy differences are physically meaningful. Hence why he said you can redefine the 0 point energy and just ignore the constant term. $\endgroup$ – Constandinos Damalas Jan 6 '15 at 17:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.