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

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- #1

- 274

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

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jedishrfu

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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:Dirac's coup

Dirac thus thought to try an equation that wasfirst orderin both space and time. One could, for example, formally take the relativistic expression for the energy replacepby 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

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

- #3

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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.

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