Klein Paradox - Dirac equation with step potentional

[Drew Carey]

Hi all!

I was reading up on the Klein paradox in Itzykson & Zuber's Quantum Field Theory (but I think this is a pretty standard part that's probably present in most QFT textbooks) and on page 62 they have a pretty straight forward solution to the Dirac equation with a step potential.

I've noticed something odd though - they're solution is based upon solving for the two regions (V=0 and V>0) and requiring continuity at the origin. They do not however require continuity of the first derivative, and in fact their solution's first derivative isn't continuous.

When solving Schrödinger's equation with a step potential it is always customary to enforce continuity of the first derivative as the equation contains a second space derivative.

I've tried to solve the Dirac problem myself with this added constraint, but is seems unsolvable (over constrained. Though I may have some mistake in the algebra). I've searched online for other solutions to this problem, and they all seem to ignore continuity of the first derivative, without even mentioning it.

What am I missing? How can we give up on continuity of the first derivative? Especially in light of the fact that the Dirac equation is supposed to be equal to Schrödinger's equation in a non-relativistic limit these two constraints need to be satisfied by the Dirac equation solution as well, no?

Thanks in advance!

 

[ChrisVer]

Why do you take the continuity of the first spatial derivative for the Shrodinger's equation?
Because it's a 2nd order differential equation (wrt spatial components).
This is of course not the case for Dirac's equation, where you only have one spatial derivative.
(\slash{p} \pm m) \psi=0

The fact that Dirac's equation in the NRL yields the classical results, should not be confused of that fact that it gives even more extra information over physics. For example the non relativistic limit will drop your spinors from 4 to 2 component, while in fact they exist in one single 4component spinor.