Question 2. For nonrelativistic free particle $E=\frac{p^{2}}{2m}$ and using the de Broglie relations the corresponding dispersive relation becomes $$\omega(k)=\frac{\hbar k^{2}}{2m}$$ Then following the relativistic free particle energy relation $E=\sqrt{p^{2}c^{2}+m^{2}c^{4}}$ and then using the de Broglie relations can we say that the corresponding dispersive relation is given by $$\omega(k)=\sqrt{k^{2}c^{2}+\frac{m^{2}c^{4}}{\hbar^{2}}}~?$$

  • $\begingroup$ In short, yes we can. $\endgroup$ – Nihar Karve Nov 23 '20 at 5:27

Yes, You can!

The de Broglie equations relate the wavelength $λ$ to the momentum $p$, and frequency f to the total energy $E$ of a free particle:

$$\mathbf{p}=\hbar\mathbf{k}$$ $$E=\hbar \omega $$

Using two formulas from special relativity, one for the relativistic mass-energy and one for the relativistic momentum

$$E=\gamma m_0c^2$$ $$\mathbf{p}=\gamma m_0\mathbf{v}$$ also $$E^2=p^2c^2+(m_0c^2)^2$$ which lead to dispersion relation $$\hbar^2\omega^2=\hbar^2k^2+(m_0c^2)^2\Rightarrow \omega(k)=\sqrt{k^2+\frac{(m_0c^2)^2}{\hbar^2}}$$

which is sometimes written as $$\left(\frac{\omega}{c}\right)^2=k^2+\left(\frac{m_0c^2}{\hbar}\right)^2$$ This can also be derived from the magnitude of the four-wavevector


Since the reduced Planck constant $ħ$ and the speed of light $c$ both appear and clutter this equation, this is where natural units are especially helpful. Normalizing them so that $ħ = c = 1$, we have: $$\omega^2=k^2+m_0^2$$


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