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saaskis
#6
Jun21-09, 06:11 AM
P: 66
Quote Quote by tirrel View Post
the world I underlined in red is it a typo? I would have written "pseudospin" instead of helicity... in fact (as far as I've understood), if we fix the K valley, there electrons have helicity positive whereas holes have negative helicity, so that changing the sign of the momentum of a hole (actually pseudo-momentum, since we talk about brillouin zones) doesn't affect its helicity but only the direction of the pseudospin... is it right all that?
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].