Exploring Quantum Physics - Week 1 Question 2

Question:

Recall how the Schrödinger equation was motivated by the non-relativistic dispersion relation $E = \frac{p^2}{2m}$. If we follow the same procedure for the case of a relativistic dispersion relation $E^2 = p^2 c^2 + m^2 c^4$), what equation do we arrive at? (For simplicity consider the one-dimensional case)

Solution:

Here comes the math. This problem scare me at first, but once I go through the lecture on deriving the Schrödinger equation it wasn't so hard to copy cat the approach here.

First of all, the operator $i \hbar \frac{\partial}{\partial t} = E$, and $-i \hbar \frac{\partial}{\partial x} = p$, we have:

$\begin{eqnarray*} E^2 &=& p^2 c^2 + m^2 c^4 \\ (i \hbar \frac{\partial}{\partial t})^2 &=& (-i \hbar \frac{\partial}{\partial x})^2 c^2 + m^2 c^4 \\ - \hbar^2 \frac{\partial^2}{\partial t^2} &=& - \hbar^2 c^2 \frac{\partial^2}{\partial x^2} + m^2 c^4 \\ \hbar^2 \frac{\partial^2}{\partial t^2} - \hbar^2 c^2 \frac{\partial^2}{\partial x^2} + m^2 c^4 &=& 0 \\ \frac{1}{c^2} \frac{\partial^2}{\partial t^2} - \frac{\partial^2}{\partial x^2} + \frac{m^2 c^2}{\hbar^2} &=& 0 \\ (\frac{1}{c^2} \frac{\partial^2}{\partial t^2} - \frac{\partial^2}{\partial x^2} + \frac{m^2 c^2}{\hbar^2}) \phi &=& 0 \end{eqnarray*}$

To be honest, I don't really sure what I am doing, I am just copying what the lecture did. I will explain what these means in the next post.