I'm reading the Wikipedia page for the Dirac equation:

The Dirac equation is superficially similar to the Schrödinger equation for a free massive particle:

A) $-\frac{\hbar^2}{2m}\nabla^2\phi = i\hbar\frac{\partial}{\partial t}\phi.$

The left side represents the square of the momentum operator divided by twice the mass, which is the non-relativistic kinetic energy. Because relativity treats space and time as a whole, a relativistic generalization of this equation requires that space and time derivatives must enter symmetrically, as they do in the Maxwell equations that govern the behavior of light — the equations must be differentially of the same order in space and time. In relativity, the momentum and the energy are the space and time parts of a space-time vector, the 4-momentum, and they are related by the relativistically invariant relation

B) $\frac{E^2}{c^2} - p^2 = m^2c^2$

which says that the length of this vector is proportional to the rest mass m. Substituting the operator equivalents of the energy and momentum from the Schrödinger theory, we get an equation describing the propagation of waves, constructed from relativistically invariant objects,

C) $\left(\nabla^2 - \frac{1}{c^2}\frac{\partial^2}{\partial t^2}\right)\phi = \frac{m^2c^2}{\hbar^2}\phi$

I am not sure how the equation A and B lead to equation C. It seems that it is related to substituting special relativity value into quantum mechanics operators, but I just keep failing to get a result...

  • 1
    $\begingroup$ Dirac himself talks about how he derived most of his ideas in this great lecture video he did in 1973. A little shaky, but very informative on the background of his ideas. youtube.com/… $\endgroup$ Aug 31, 2013 at 14:40
  • $\begingroup$ Related: physics.stackexchange.com/q/10837/2451 and links therein. $\endgroup$
    – Qmechanic
    Apr 25, 2015 at 19:14

1 Answer 1


First, C) isn't the Dirac Equation, it's the Klein-Gordon equation

Now, to your main question. A) comes from the classical equation for a free massive particle:

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

by making the operator (operating on $\phi$) substitutions:

$p^2 \rightarrow - \hbar^2 \nabla^2$

$E \rightarrow i \hbar \dfrac{\partial}{\partial t}$

C) comes from B) by further recognizing that:

$E^2 \rightarrow -\hbar^2 \dfrac{\partial^2}{\partial t^2}$


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