strangerep said:
So what? We all knew the Dirac eqn was correct. Indeed, Dirac originally formulated it such that it would satisfy relativistic invariance.
Indeed, I consider it obvious that the Dirac equation describes the 4-vector ##\textbf{k}## but I was asked to prove it.
Now I will demonstrate how the Schrodinger equation is a component of the Dirac equation just as ##k_0## is a component of ##\textbf{k}##.
I note:
\begin{equation}
\begin{split}
i \hbar \gamma^{\mu} \partial_{\mu}\psi &= \left[ i \hbar \gamma^0 \partial_0 \psi, i \hbar \gamma^j \partial_j \psi \right]
\\
&= \left[ \hbar(\pm k_0) \psi, i \hbar (\pm \vec{k}) \psi \right]
\end{split}
\end{equation}
where ##j=1,2,3##
The notation ##\pm## is merely a statement that an observable can take positive or negative values. In most of mathematics, this is assumed to be true unless specified otherwise. Energy and momentum can take negative values according to the Dirac equation. Of course, negative energy can be considered as positive energy with a negative charge, and negative momentum can represented as positive momentum with a negative spin state. In this context, the spin state and charge are represented by the sign of ##\vec{k}## and ##k_0##, respectively. Thus, the ##\pm## notation can be dropped such that:
\begin{equation}
\begin{split}
i \hbar \gamma^{\mu} \partial_{\mu}\psi &= \left[ i \hbar \gamma^0 \partial_0 \psi, i \hbar \gamma^j \partial_j \psi \right]
\\
&= \left[ \hbar k_0 \psi, i \hbar \vec{k} \psi \right]
\end{split}
\end{equation}
where ##j=1,2,3##
The recognition that an observable can be positive or negative (or instead of negative, at least positive with a negative spin state or charge) also means that the Dirac matrix is unnecessary in the Dirac equation when writing the Dirac equation utilizing ##k_0## and ##\vec{k}## form. Without the Dirac matrix, the charge and spin state become part of the observable ##k_0## and ##\vec{k}## as their sign, such that:
\begin{equation}
\begin{split}
\gamma^0 \partial_0 &= \partial_t
\end{split}
\end{equation}
and
\begin{equation}
\begin{split}
\gamma^j \partial_j &= \partial_{\vec{x}}
\end{split}
\end{equation}
Thus,
\begin{equation}
\begin{split}
i \hbar \gamma^{\mu} \partial_{\mu}\psi &= \left[ i \hbar \partial_t \psi, i \hbar \partial_{\vec{x}} \psi \right]
\\
&= \left[ \hbar k_0 \psi, i \hbar \vec{k} \psi \right]
\end{split}
\end{equation}
Given
\begin{equation}
\begin{split}
\hat{\mathcal{H}} &= i \hbar \partial_t
\end{split}
\end{equation}
and
\begin{equation}
\begin{split}
E &= \hbar k_0
\end{split}
\end{equation}
I note:
\begin{equation}
\begin{split}
i \hbar \gamma^{\mu} \partial_{\mu}\psi &= \left[ \hat{\mathcal{H}} \psi, i \hbar \partial_{\vec{x}} \psi \right]
\\
&= \left[ E \psi, i \hbar \vec{k} \psi \right]
\end{split}
\end{equation}
Thus, the Schrodinger equation is a component of the Dirac equation, where the Dirac equation demonstrates that energy in the Schrodinger equation can take negative values (or positive values with negative charge).
