# Why did Dirac choose a linear equation in momentum for formulating a relativistic wavefunction?

• The Dirac equation arose out of the need for a quantum mechanical wave equation that treats position and time on an equal basis, just as relativity expects them to be. Now the Schrodinger equation is second order in space and first order in time rendering it non-relativistic.

• One possibility for converting the non-relativistic Schrödinger equation for a free particle to a relativistic equation is to replace the classical energy $$H=\frac{p^2}{2m}$$by the relativistic energy $$H=\sqrt{\mathbf p^2c^2+m^2c^4}$$ and replacing $$\mathbf p \rightarrow -i \hbar\nabla.$$

• Dirac’s solution was essentially to turn the problem on its head. Rather than trying to find the square root of an existing quantity, he postulated that the quantity inside the square root is the perfect square of an expression that is linear in the momentum, so that both space and time are involved as first order. $$H = -i \vec \alpha.\mathbf p + \beta m$$

• My question is could we have also chosen a scenario where we had both space and time derivatives of second order instead of linear equations? Apart from possibly making the life complex, would there be any physical considerations not allowing this formulation?

• – anna v Apr 26 '20 at 7:45
• @annav, thanks for the link. It was an interesting read to know that he pondered all this while staring at fire! But my question is why not second order just like how position is in Schrodinger equation? – Abhay Hegde Apr 26 '20 at 7:51
• Related/possible duplicate: physics.stackexchange.com/q/111401/50583 – ACuriousMind Apr 26 '20 at 8:08

The scenario in which both time and space derivatives where second order was firstly studied by Klein-Gordon. The Klein-Gordon equation was in fact a first try at a relativistic quantum theory, but it had some problems. One of those problems, which really bothered Dirac, was the fact that the probability given by the wavefunction wasn't positive definite. The current density associated to the Klein-Gordon field is in fact given by $$\phi^\dagger \overset{\leftrightarrow}{\partial_\mu}\phi = \phi^\dagger\partial_\mu\phi-\phi\partial_\mu\phi^\dagger$$ which is easy to see how it could become negative.