# Why did Dirac want a first-order equation?

• I
If I understand it correctly Dirac developed his equation because he was looking for a relativistic first order wave equation for the electron, rather than a second-order one like the Klein-Gordon equation.

Why did he wanted a first-order equation? Is it because the probability current is not positive-definite for higher than second order equations?

## Answers and Replies

jedishrfu
Mentor
The wikipedia article gets into why Dirac chose first order instead of second order here:
Dirac's coup

Dirac thus thought to try an equation that was first order in both space and time. One could, for example, formally take the relativistic expression for the energy replace p by its operator equivalent, expand the square root in an infinite series of derivative operators, set up an eigenvalue problem, then solve the equation formally by iterations. Most physicists had little faith in such a process, even if it were technically possible.

As the story goes, Dirac was staring into the fireplace at Cambridge, pondering this problem, when he hit upon the idea of taking the square root of the wave operator...

https://en.wikipedia.org/wiki/Dirac_equation
There's also some discussion on physics stackexchange about it here having to do with the 1/2 spin of fermions which implies a first order equation:

http://physics.stackexchange.com/qu...-dirac-equation-and-bosons-a-second-order-one

bhobba and carllacan
Ok, so to sum up in case someone googles his way to this post:

If the equation for the wavefunction is of second order the boundary conditions must include the initial value of its first time derivative, which can be negative. The expression for the current density involves the time derivative of the wavefunction, so if the value of that derivative was negative at some point (which as we have said cold happen) then you would have a non positive definite probability current, which is nonsensical.

vanhees71