flyusx
- 71
- 12
- Homework Statement
- The Klein paradox is illustrated with particles traveling along the -axis incident on a potential step of strength where and . Determine the momentum in the classically forbidden region.
- Relevant Equations
- $$E^{2}=m^{2}c^{4}+p^{2}c^{2}$$ $$v_{g}=\frac{\partial E}{\partial p}$$
I'm currently reading over a little bit of relativistic quantum mechanics (with an emphasis on the need to transition to QFT) and have stumbled upon something I can't seem to figure out.
I was taught the Klein paradox as an example for why we need QFT. Pure relativistic quantum mechanics leads to (for instance) reflection coefficients exceeding unity and negative probability densities. My textbook follows very closely to the lecture notes published here (spinless case) and here (spin-1/2 case). They both come to the same conclusion.
My confusion amounts to the following. Consider the potential step $$V(x)=\begin{cases}V_{0}&z>0\\0&z\leq0\end{cases}$$ In the spinless (Klein–Gordon) case with ##E+mc^{2}>V_{0}##, we take momentum in the classically forbidden region to be $$-\sqrt{\frac{\left(E-V_{0}\right)^{2}}{c^{2}}-m^{2}c^{2}}$$ We specifically pick the momentum to be negative rather than positive. This is because the group velocity is $$\left(E-V_{0}\right)^{2}=p^{2}c^{2}+m^{2}c^{4}\to\frac{\partial E}{\partial p}=v_{g}=\frac{pc^{2}}{E-V_{0}}$$ Thus a negative momentum yields a positive group velocity and a rightwards-propagating wave. If we took a positive linear momentum instead, we would have a leftwards-propagating wave. Additionally, if we were considering an energy regime other than the Klein paradox, so ##E+mc^{2}>V_{0}##, a positive momentum would correspond to a positive group velocity.
The logic above makes sense to me. When I move on to the case for spin-1/2 particles, I see in the lecture notes and my textbook that the transmitted wave is given the positive momentum $$\sqrt{\frac{\left(V_{0}-E\right)^{2}}{c^{2}}-m^{2}c^{2}}$$ The subsequent calculations in both sources yield results consistent with the Klein paradox. For instance, the magnitude of the reflected probability current exceeds that of the incident probability current. However, nowhere did we enforce a positive group velocity. By picking the momentum in the classically forbidden region to be positive, doesn't this correspond to a negative group velocity and hence a leftwards-propagating wave? Why was the group velocity analysis method not used here?
Doing some further digging, I came across an article titled Motion of a wave packet in the Klein paradox by Nitta, Kudo, Minowa (Am J Phys, Vol 67, No 11, Nov 1999); I've linked a PDF here. On page 969, the authors present the possibility of taking the momentum in the classically forbidden region, what they call region II, to be negative instead of positive. They claim that doing so 'assumes that the negative-energy states in region II are empty', which is inconsistent with Dirac's hole model. Could this be the resolution to my confusion?
I was taught the Klein paradox as an example for why we need QFT. Pure relativistic quantum mechanics leads to (for instance) reflection coefficients exceeding unity and negative probability densities. My textbook follows very closely to the lecture notes published here (spinless case) and here (spin-1/2 case). They both come to the same conclusion.
My confusion amounts to the following. Consider the potential step $$V(x)=\begin{cases}V_{0}&z>0\\0&z\leq0\end{cases}$$ In the spinless (Klein–Gordon) case with ##E+mc^{2}>V_{0}##, we take momentum in the classically forbidden region to be $$-\sqrt{\frac{\left(E-V_{0}\right)^{2}}{c^{2}}-m^{2}c^{2}}$$ We specifically pick the momentum to be negative rather than positive. This is because the group velocity is $$\left(E-V_{0}\right)^{2}=p^{2}c^{2}+m^{2}c^{4}\to\frac{\partial E}{\partial p}=v_{g}=\frac{pc^{2}}{E-V_{0}}$$ Thus a negative momentum yields a positive group velocity and a rightwards-propagating wave. If we took a positive linear momentum instead, we would have a leftwards-propagating wave. Additionally, if we were considering an energy regime other than the Klein paradox, so ##E+mc^{2}>V_{0}##, a positive momentum would correspond to a positive group velocity.
The logic above makes sense to me. When I move on to the case for spin-1/2 particles, I see in the lecture notes and my textbook that the transmitted wave is given the positive momentum $$\sqrt{\frac{\left(V_{0}-E\right)^{2}}{c^{2}}-m^{2}c^{2}}$$ The subsequent calculations in both sources yield results consistent with the Klein paradox. For instance, the magnitude of the reflected probability current exceeds that of the incident probability current. However, nowhere did we enforce a positive group velocity. By picking the momentum in the classically forbidden region to be positive, doesn't this correspond to a negative group velocity and hence a leftwards-propagating wave? Why was the group velocity analysis method not used here?
Doing some further digging, I came across an article titled Motion of a wave packet in the Klein paradox by Nitta, Kudo, Minowa (Am J Phys, Vol 67, No 11, Nov 1999); I've linked a PDF here. On page 969, the authors present the possibility of taking the momentum in the classically forbidden region, what they call region II, to be negative instead of positive. They claim that doing so 'assumes that the negative-energy states in region II are empty', which is inconsistent with Dirac's hole model. Could this be the resolution to my confusion?