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  How to understand the equivalence between Andreev reflection and Cooper pair injection?

+ 5 like - 0 dislike

  It is well know that Andreev reflection dominates the subgap transport at the normal metal-superconductor interface. An incident electron can be reflected as a hole in the Nambu space, which effectively describes two electrons transmit into the superconductor and form a Cooper pair. It seems reasonable and intuitive that the two pictures are equivalent to each other at first glance. However, it is not so easy for me to understand this fact.

  In the electron-hole picture, an electron in \(|+k\rangle\) state is incident from the normal side. Then it will transmit into the superconductor by pulling another electron in \(|-k\rangle\) state to form a Cooper pair. The empty state in \(|-k\rangle\) is called a hole, and the moving direction of the hole is opposite to the incident electron, so that this phenomenon is called Andreev "reflection". Based on this picture, the incident electron and reflected hole possess nearly the same momentum \(+k\) (note that the empty state in \(|-k\rangle\) has a momentum of \(+k\)). Therefore, the whole Andreev process is momentum conserved.

  Now let's turn to the two-electron picture. In this picture, two electron incident from the normal side both transmit into the superconductor, and they possess nonzero total momentum. So when they form a Cooper pair in the superconductor, their total momentum must be changed, for the Cooper pair has zero momentum.

  Based on the argument above, we see that the Andreev reflection is a process with the momentum conservation, while the two-electron injection process breaks the momentum conservation. I believe that, if the two pictures are equivalent, then the momentum of the two physical processes——Andreev reflection and two-electron injection, must be equal to each other. In my opinion, using the Andreev reflection picture to study the transport problems is always right. However, the Andreev picture can not deal with all relevant problems. For example, when we are interested in the two electron entanglement, then the single particle process described by Andreev reflection is powerless. Therefore, to find good equivalence between the two pictures are highly nontrivial.

asked Jun 2, 2014 in Theoretical Physics by pchenweis (40 points) [ no revision ]
retagged Jun 2, 2014

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