I'm sorry for my mistake, I'm glad you noticed it! You're correct, helicity is not conserved but pseudospin is. To make this explicit, consider an electron hitting a symmetric np-junction perpendicularly. The state left of the barrier is
[tex] \psi_1 (x) = [1,+1] \exp(ikx) [/tex]
and on the right
[tex] \psi_2 (x) = [1,-\exp(i\pi)] \exp(-ikx) = [1,1] \exp(-ikx). [/tex]
So now helicities are
[tex] \Sigma \psi_1 = + \psi_1, \ \Sigma \psi_2 = - \psi_2 [/tex],
whereas the pseudospins are
[tex] \sigma_x \psi_1 = + \psi_1, \ \sigma_x \psi_2 = + \psi_2 [/tex].
So flipping the momentum requires flipping the helicity to conserve pseudospin. Thanks, I think we are on the right track now!
Any reflected wave would of the form
[tex] \psi_r (x) = [1,\exp(i\pi) ] \exp(-ikx) [/tex],
giving the helicity and pseudospin
[tex] \Sigma \psi_r = + \psi_r, \ \sigma_x \psi_r = - \psi_r [/tex].
For normal incidence the reflected part vanishes (Klein tunneling), but for other angles the transmission probability is [tex]T=\cos^2 \phi [/tex].