Quantum Mechanics – How to Obtain the Relativistic Dispersion Relation from Wave Function

Question

If we have a plane wave, $$\psi(t,\textbf{r})=Ae^{i(\textbf{k}\cdot\textbf{r}-\omega t)}$$ we can apply the QM energy operator ($\hat{\mathcal{E}}=i\hbar\partial_{t}$) to the wave function, we get:$$\hat{\mathcal{E}}\psi(t,\textbf{r})=\hbar \omega\psi(t,\textbf{r})$$Similarly, applying the momentum operator ($\hat{\mathcal{P}}=-i\hbar\nabla$) to the plane wave, we have $$\hat{\mathcal{P}}\psi(t,\textbf{r})=\hbar\textbf{k}\psi(t,\textbf{r})$$ The phase velocity of the plane wave should be $$v=\frac{\omega}{k}$$ So if $E=\hbar \omega$ and $p=\hbar k$, $E=\hbar vk=v p$, then if the particle has velocity that is the speed of light, $E^{2}=c^{2}p^{2}$, as you would expect. However i can't see the equivalence between:
$$E^{2}=v^{2}p^{2}$$ and
$$E^{2}=m^2c^{4}+p^{2}c^{2}$$ Is there a way to demonstrate this equivalence? or is there no equivalence? Even more confusingly, by boosting a massive particle at rest in 1+1D, 4-momentum which is initially $$P= \begin{bmatrix}
mc^{2} \\
0 \\
\end{bmatrix}$$
Transforms under a Lorentz boost as
$$\Lambda^{\nu}_{\mu} P^{\mu}=\begin{bmatrix}
\gamma && \beta\gamma\\
\beta\gamma && \gamma \\
\end{bmatrix}
\begin{bmatrix}
mc^{2} \\
0 \\
\end{bmatrix}=
\begin{bmatrix}
\gamma mc^{2} \\
\beta\gamma mc^{2} \\
\end{bmatrix}$$
Thus, $$p=\beta E$$ which is seemingly the oppsite of what I derived previously.

Answer 1

You are using the Schrödinger equation for the free non-relativistic particle, where $$ H= \frac{p^2}{2 m} $$ therefore, the dispersion relation that you obtain can only be consistent with $$ E= \hbar \omega = \frac{\hbar^2 k^2}{2 m} $$ This is the usual kinetic energy of a non-relativistic particle, nothing more to say: it is the non-relativistic limit of the usual relativistic dispersion relation, in the sense that $$ E=\sqrt{m^2c^{4}+p^{2}c^{2}} \approx \frac{p^2}{2 m} + m c^2 \, \qquad (\text{for } \quad p\ll c\, m) $$