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For a particle in potential $U(x)$ in 1D which equation is correct

$$i\hbar\frac{\partial\psi}{\partial t}=(cp \sigma_x+mc^2\sigma_z+U(x))\psi$$ or $$i\hbar\frac{\partial\psi}{\partial t}=(cp \sigma_x+mc^2\sigma_z+U(x)\sigma_z)\psi$$

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    $\begingroup$ @Dimensio1n0, Life is too short to think about how much off-topic is this question! $\endgroup$ – richard Jan 6 '14 at 19:29
  • $\begingroup$ Actually, I'm retracting my close vote. It's fine, as long as the answers give a proper explanation. $\endgroup$ – Abhimanyu Pallavi Sudhir Jan 7 '14 at 11:54
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Hints :

a) Find a representation for the gamma matrices for a space-time with one spatial dimension, from their defining relation $\gamma^\mu \gamma^\nu + \gamma^\nu \gamma^\mu= 2 g^{\mu \nu}$

b) Remember the Dirac equation in presence of an electromagnetic field :

$[i \gamma^\mu(\partial_\mu + ie A_\mu) - m] \psi=0$

c) Think to the potential $U$ as a time component, that is, "corresponding" to $ e A_0$

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The second equation implies that particles with spin down see a different potential with respect to spin up particles, since $\sigma_z$(spin up)=(spin up), while $\sigma_z$(spin down)=-(spin down). Specifically, spin up particles see it as repulsive, while spin down particles see it as attractive. So, the add on $\sigma_z$ definitely changes the form of the potential. For a generic potential, equation 1 looks the correct one. You find lots of literature in google if you look for "one-dimensional Dirac equation". For example, look at here (http://arxiv.org/pdf/1309.0308v4.pdf), where the 1D Dirac equation is solved with some potential. You will not see any $\sigma_z$ beside the potential.

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  • $\begingroup$ Wait, though. In the link above (eq.1), now that I looked at it carefully, V(x) is the vector potential, while S(x) is the scalar potential. This latter does have a $\sigma_z$ beside. So, my conclusions are probably wrong. $\endgroup$ – Wizzerad May 14 '14 at 17:05

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